Added July 7,2019
Solve for \(u(x,t)\) for \(t>0\) and \(0<x<L\) \[ u_{tt} = c^2 u_{xx} \] With boundary condition both ends fixed \begin {align*} u(0,t) &= 0 \\ u(L,t) &=0 \end {align*}
And initial conditions \begin {align*} u(x,0) &= f(x) \\ u_t(x,0) &= g(x) \\ \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], {t, 2}] == c^2*D[u[x, t], {x, 2}]; ic = {u[x, 0] == f[x], Derivative[0, 1][u][x, 0] == g[x]}; bc = {u[0, t] == 0, u[L, t] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t}, Assumptions->L>0], 60*10]];
\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}\sqrt {2} \sqrt {\frac {1}{L}} \sin \left (\frac {\pi x K[1]}{L}\right ) \left (\cos \left (\pi t \sqrt {\frac {c^2 K[1]^2}{L^2}}\right ) \int _0^L \frac {\sqrt {2} f(x) \sin \left (\frac {\pi x K[1]}{L}\right )}{\sqrt {L}} \, dx+\frac {L \left (\int _0^L \frac {\sqrt {2} g(x) \sin \left (\frac {\pi x K[1]}{L}\right )}{\sqrt {L}} \, dx\right ) \sin \left (\pi t \sqrt {\frac {c^2 K[1]^2}{L^2}}\right )}{\pi | c| | K[1]| }\right ) & K[1]\in \mathbb {Z}\land ((L<0\land ((c<0\land K[1]\geq 1)\lor (c>0\land K[1]\geq 1)))\lor (L>0\land ((c<0\land K[1]\geq 1)\lor (c>0\land K[1]\geq 1)))) \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
Maple ✓
restart; pde := diff(u(x,t),t$2)=c^2*diff(u(x,t),x$2); bc := u(0,t)=0,u(L,t)=0; ic := u(x,0)=f(x),eval(diff(u(x,t),t),t=0)=g(x); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic, bc],u(x,t)) assuming L>0),output='realtime'));
\[u \left ( x,t \right ) =\sum _{n=1}^{\infty } \left ( 2\,{\frac {1}{n\pi \,cL}\sin \left ( {\frac {n\pi \,x}{L}} \right ) \left ( \pi \,\cos \left ( {\frac {n\pi \,ct}{L}} \right ) \int _{0}^{L}\!\sin \left ( {\frac {n\pi \,x}{L}} \right ) f \left ( x \right ) \,{\rm d}xnc+L\int _{0}^{L}\!\sin \left ( {\frac {n\pi \,x}{L}} \right ) g \left ( x \right ) \,{\rm d}x\sin \left ( {\frac {n\pi \,ct}{L}} \right ) \right ) } \right ) \]
Hand solution
Solving for \(t>0,0<x<L\)\[ u_{tt}=c^{2}u_{xx}\] With BC \begin {align*} u\left ( 0,t\right ) & =0\\ u\left ( L,t\right ) & =0 \end {align*}
And initial conditions\begin {align*} u\left ( x,0\right ) & =f\left ( x\right ) \\ u_{t}\left ( x,0\right ) & =g\left ( x\right ) \end {align*}
Let \(u=X\left ( x\right ) T\left ( t\right ) \). The PDE becomes\begin {align*} \frac {T^{\prime \prime }X}{c^{2}} & =X^{\prime \prime }T\\ \frac {1}{c^{2}}\frac {T^{\prime \prime }}{T} & =\frac {X^{\prime \prime }}{X}=-\lambda \end {align*}
Where \(\lambda \) is separation constant. Hence the eigenvalue ODE is \begin {align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X\left ( L\right ) & =0 \end {align*}
From the boundary conditions, we see that \(\lambda >0\) is the only possible value. Therefore the solution to the above ODE is\[ X\left ( x\right ) =A\cos \left ( \sqrt {\lambda }x\right ) +B\sin \left ( \sqrt {\lambda }x\right ) \] Since \(X\left ( 0\right ) =0\) then \(A=0\) and the solution becomes \(X\left ( x\right ) =B\sin \left ( \sqrt {\lambda }x\right ) \). Since \(X\left ( L\right ) =0\) then for non trivial solution we want \(\sqrt {\lambda }L=n\pi \) or \[ \lambda _{n}=\left ( \frac {n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \] Hence the eigenfunctions are \[ \Phi _{n}\left ( x\right ) =\sin \left ( \frac {n\pi }{L}x\right ) \qquad n=1,2,3,\cdots \] The time ODE now becomes\[ T^{\prime \prime }+c^{2}\left ( \frac {n\pi }{L}\right ) ^{2}T=0 \] Which has the solution\[ T\left ( t\right ) =D_{n}\cos \left ( c\frac {n\pi }{L}t\right ) +E_{n}\sin \left ( c\frac {n\pi }{L}t\right ) \] Therefore the complete solution becomes\begin {equation} u\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( D_{n}\cos \left ( c\frac {n\pi }{L}t\right ) +E_{n}\sin \left ( c\frac {n\pi }{L}t\right ) \right ) \Phi _{n}\left ( x\right ) \tag {1} \end {equation} At \(t=0\) the above becomes\[ f\left ( x\right ) =\sum _{n=1}^{\infty }D_{n}\Phi _{n}\left ( x\right ) \] Applying orthogonality gives\begin {align*} \int _{0}^{L}f\left ( x\right ) \Phi _{n}\left ( x\right ) dx & =D_{n}\int _{0}^{L}\Phi _{n}^{2}\left ( x\right ) dx\\ & =\frac {L}{2}D_{n} \end {align*}
Hence\begin {equation} D_{n}=\frac {2}{L}\int _{0}^{L}f\left ( x\right ) \Phi _{n}\left ( x\right ) dx\tag {2} \end {equation} Taking time derivative of (1) gives\[ u_{t}\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( -c\frac {n\pi }{L}D_{n}\sin \left ( c\frac {n\pi }{L}t\right ) +E_{n}c\frac {n\pi }{L}\cos \left ( c\frac {n\pi }{L}t\right ) \right ) \Phi _{n}\left ( x\right ) \] At \(t=0\) the above becomes\[ g\left ( x\right ) =\sum _{n=1}^{\infty }E_{n}c\frac {n\pi }{L}\Phi _{n}\left ( x\right ) \] Applying orthogonality gives\begin {align*} \int _{0}^{L}g\left ( x\right ) \Phi _{n}\left ( x\right ) dx & =E_{n}c\frac {n\pi }{L}\int _{0}^{L}\Phi _{n}^{2}\left ( x\right ) dx\\ & =\frac {L}{2}E_{n}c\frac {n\pi }{L}\\ & =\frac {1}{2}E_{n}cn\pi \end {align*}
Hence\begin {equation} E_{n}=\frac {2}{cn\pi }\int _{0}^{L}g\left ( x\right ) \Phi _{n}\left ( x\right ) dx\tag {3} \end {equation} Using (2,3) in (1) gives the final solution as\begin {align*} u\left ( x,t\right ) & =\frac {2}{L}\sum _{n=1}^{\infty }\left ( \int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) \cos \left ( c\frac {n\pi }{L}t\right ) \sin \left ( \frac {n\pi }{L}x\right ) \\ & +\frac {2}{c\pi }\sum _{n=1}^{\infty }\frac {1}{n}\left ( \int _{0}^{L}g\left ( s\right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) \sin \left ( c\frac {n\pi }{L}t\right ) \sin \left ( \frac {n\pi }{L}x\right ) \end {align*}
____________________________________________________________________________________
Added July 8,2019
Solve for \(u(x,t)\) for \(t>0\) and \(0<x<L\) \[ u_{tt} = c^2 u_{xx} \] With boundary condition both ends fixed \begin {align*} u(0,t) &= 0 \\ u(L,t) &=0 \end {align*}
And initial conditions \begin {align*} u(x,0) &= f(x) \\ u_t(x,0) &= g(x) \\ \end {align*}
Using the following values \begin {align*} L&=10\\ c&=2\\ f(x)&=0\\ g(x)&= \frac {8 x (L-x)^2}{L^3} \end {align*}
Mathematica ✓
ClearAll["Global`*"]; L=10; c=2; f=0; g=(8*x*(L-x)^2)/L^3; pde = D[u[x, t], {t, 2}] == c^2*D[u[x, t], {x, 2}]; ic = {u[x, 0] == f, Derivative[0, 1][u][x, 0] == g}; bc = {u[0, t] == 0, u[L, t] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t}], 60*10]]; sol = sol/.K[1]->n;
\[\left \{\left \{u(x,t)\to \underset {n=1}{\overset {\infty }{\sum }}\frac {160 \left (2+(-1)^n\right ) \sin \left (\frac {n \pi t}{5}\right ) \sin \left (\frac {n \pi x}{10}\right )}{n^4 \pi ^4}\right \}\right \}\]
Maple ✓
restart; L:=10; c:=2; f:=0; g:=(8*x*(L-x)^2)/L^3; pde := diff(u(x,t),t$2)=c^2*diff(u(x,t),x$2); bc := u(0,t)=0,u(L,t)=0; ic := u(x,0)=f,eval(diff(u(x,t),t),t=0)=g; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic, bc],u(x,t))),output='realtime'));
\[u \left ( x,t \right ) =\sum _{n=1}^{\infty }160\,{\frac { \left ( \left ( -1 \right ) ^{n}+2 \right ) \sin \left ( 1/10\,n\pi \,x \right ) \sin \left ( 1/5\,n\pi \,t \right ) }{{\pi }^{4}{n}^{4}}}\]
Hand solution
Solving the wave PDE on string with both ends fixed \[ u_{tt}=c^{2}u_{xx}\qquad t>0,x>0 \] With BC \begin {align*} u\left ( 0,t\right ) & =0\\ u\left ( L,t\right ) & =0 \end {align*}
And initial conditions\begin {align*} u\left ( x,0\right ) & =f\left ( x\right ) =0\\ u_{t}\left ( x,0\right ) & =g\left ( x\right ) =\frac {8x\left ( L-x\right ) ^{2}}{L^{3}} \end {align*}
Using \(c=2,L=10\).
The general problem PDE was solved in 5.1.1.1 on page 1217 and the solution is\begin {align*} u\left ( x,t\right ) & =\frac {2}{L}\sum _{n=1}^{\infty }\left ( \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {n\pi }{L}x\right ) dx\right ) \cos \left ( c\frac {n\pi }{L}t\right ) \sin \left ( \frac {n\pi }{L}x\right ) \\ & +\frac {2}{c\pi }\sum _{n=1}^{\infty }\frac {1}{n}\left ( \int _{0}^{L}g\left ( x\right ) \sin \left ( \frac {n\pi }{L}x\right ) dx\right ) \sin \left ( c\frac {n\pi }{L}t\right ) \sin \left ( \frac {n\pi }{L}x\right ) \end {align*}
Substituting the specific values given above into this solution gives\[ u\left ( x,t\right ) =\frac {1}{\pi }\sum _{n=1}^{\infty }\frac {1}{n}\left ( \int _{0}^{10}\frac {8x\left ( 10-x\right ) ^{2}}{10^{3}}\sin \left ( \frac {n\pi }{10}x\right ) dx\right ) \sin \left ( 2\frac {n\pi }{10}t\right ) \sin \left ( \frac {n\pi }{10}x\right ) \] But \(\int _{0}^{10}\frac {8x\left ( 10-x\right ) ^{2}}{10^{3}}\sin \left ( \frac {n\pi }{10}x\right ) dx=\frac {160\left ( 2+\left ( -1\right ) ^{n}\right ) }{n^{3}\pi ^{3}}\), hence the solution becomes\[ u\left ( x,t\right ) =\frac {1}{\pi ^{4}}\sum _{n=1}^{\infty }\frac {160\left ( 2+\left ( -1\right ) ^{n}\right ) }{n^{4}}\sin \left ( \frac {n\pi }{5}t\right ) \sin \left ( \frac {n\pi }{10}x\right ) \] Animation is below
Source code used for the above
____________________________________________________________________________________
Added July 8,2019
Solve for \(u(x,t)\) for \(t>0\) and \(0<x<L\) \[ u_{tt} = c^2 u_{xx} \] With boundary condition both ends fixed \begin {align*} u(0,t) &= 0 \\ u(L,t) &=0 \end {align*}
And initial conditions \begin {align*} u(x,0) &= f(x) \\ u_t(x,0) &= g(x) \\ \end {align*}
Using the following values \begin {align*} L&=10\\ c&=2\\ f(x)&=\frac {8 x (L-x)^2}{L^3}\\ g(x)&= 0 \end {align*}
Mathematica ✓
ClearAll["Global`*"]; L=10; c=2; g=0; f=(8*x*(L-x)^2)/L^3; pde = D[u[x, t], {t, 2}] == c^2*D[u[x, t], {x, 2}]; ic = {u[x, 0] == f, Derivative[0, 1][u][x, 0] == g}; bc = {u[0, t] == 0, u[L, t] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t}], 60*10]]; sol = sol/.K[1]->n;
\[\left \{\left \{u(x,t)\to \underset {n=1}{\overset {\infty }{\sum }}\frac {32 \left (2+(-1)^n\right ) \cos \left (\frac {n \pi t}{5}\right ) \sin \left (\frac {n \pi x}{10}\right )}{n^3 \pi ^3}\right \}\right \}\]
Maple ✓
restart; L:=10; c:=2; g:=0; f:=(8*x*(L-x)^2)/L^3; pde := diff(u(x,t),t$2)=c^2*diff(u(x,t),x$2); bc := u(0,t)=0,u(L,t)=0; ic := u(x,0)=f,eval(diff(u(x,t),t),t=0)=g; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic, bc],u(x,t))),output='realtime'));
\[u \left ( x,t \right ) =\sum _{n=1}^{\infty }32\,{\frac { \left ( \left ( -1 \right ) ^{n}+2 \right ) \sin \left ( 1/10\,n\pi \,x \right ) \cos \left ( 1/5\,n\pi \,t \right ) }{{n}^{3}{\pi }^{3}}}\]
Hand solution
Solving the wave PDE on string with both ends fixed \[ u_{tt}=c^{2}u_{xx}\qquad t>0,x>0 \] With BC \begin {align*} u\left ( 0,t\right ) & =0\\ u\left ( L,t\right ) & =0 \end {align*}
And initial conditions\begin {align*} u\left ( x,0\right ) & =f\left ( x\right ) =\frac {8x\left ( L-x\right ) ^{2}}{L^{3}}\\ u_{t}\left ( x,0\right ) & =g\left ( x\right ) =0 \end {align*}
Using \(c=2,L=10\).
The general problem PDE was solved in 5.1.1.1 on page 1217 and the solution is\begin {align*} u\left ( x,t\right ) & =\frac {2}{L}\sum _{n=1}^{\infty }\left ( \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {n\pi }{L}x\right ) dx\right ) \cos \left ( c\frac {n\pi }{L}t\right ) \sin \left ( \frac {n\pi }{L}x\right ) \\ & +\frac {2}{c\pi }\sum _{n=1}^{\infty }\frac {1}{n}\left ( \int _{0}^{L}g\left ( x\right ) \sin \left ( \frac {n\pi }{L}x\right ) dx\right ) \sin \left ( c\frac {n\pi }{L}t\right ) \sin \left ( \frac {n\pi }{L}x\right ) \end {align*}
Substituting the specific values given above into this solution gives\[ u\left ( x,t\right ) =\frac {1}{5}\sum _{n=1}^{\infty }\left ( \int _{0}^{10}\frac {8x\left ( 10-x\right ) ^{2}}{10^{3}}\sin \left ( \frac {n\pi }{10}x\right ) dx\right ) \cos \left ( \frac {n\pi }{5}t\right ) \sin \left ( \frac {n\pi }{10}x\right ) \] But \(\int _{0}^{10}\frac {8x\left ( 10-x\right ) ^{2}}{10^{3}}\sin \left ( \frac {n\pi }{10}x\right ) dx=\frac {160\left ( 2+\left ( -1\right ) ^{n}\right ) }{n^{3}\pi ^{3}}\), hence the solution becomes\[ u\left ( x,t\right ) =32\sum _{n=1}^{\infty }\frac {\left ( 2+\left ( -1\right ) ^{n}\right ) }{n^{3}\pi ^{3}}\cos \left ( \frac {n\pi }{5}t\right ) \sin \left ( \frac {n\pi }{10}x\right ) \] Animation is below
Source code used for the above
____________________________________________________________________________________
Added sept 23 ,2019
Solve for \(u(x,t)\) for \(t>0\) and \(-\pi <x<\pi \) \[ u_{tt} = c^2 u_{xx} \] With boundary condition \begin {align*} u(-\pi ,t) &= 0 \\ u(\pi ,t) &=0 \end {align*}
And initial conditions \begin {align*} u(x,0) &= 0 \\ u_t(x,0) &= \sin (x)^2 \\ \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], {t, 2}] == c^2*D[u[x, t], {x, 2}]; ic = {u[x, 0] == 0, Derivative[0, 1][u][x, 0] == Sin[x]^2}; bc = {u[-Pi, t] == 0, u[Pi, t] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t}], 60*10]];
\[\left \{\left \{u(x,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {32 \left (-1+(-1)^{K[1]}\right ) \sin \left (\frac {1}{2} \sqrt {c^2} t K[1]\right ) \sin \left (\frac {1}{2} (x+\pi ) K[1]\right )}{\sqrt {c^2} \pi K[1]^2 \left (K[1]^2-16\right )}\right \}\right \}\]
Maple ✓
restart; pde := diff(u(x,t),t$2)=c^2*diff(u(x,t),x$2); bc := u(-Pi,t)=0,u(Pi,t)=0; ic := u(x,0)=0,D[2](u)(x,0)=sin(x)^2; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic, bc],u(x,t))),output='realtime'));
\[u \left ( x,t \right ) ={\frac {1}{315\,\pi \,c} \left ( 315\,\sum _{n=5}^{\infty }32\,{\frac {\sin \left ( 1/2\,n \left ( x+\pi \right ) \right ) \sin \left ( 1/2\,cnt \right ) \left ( \left ( -1 \right ) ^{n}-1 \right ) }{\pi \,{n}^{2}c \left ( {n}^{2}-16 \right ) }}\pi \,c-320\,\cos \left ( 3/2\,x \right ) \sin \left ( 3/2\,ct \right ) +1344\,\cos \left ( x/2 \right ) \sin \left ( 1/2\,ct \right ) \right ) }\]
Hand solution
Solve
\[ u_{tt}=c^{2}u_{xx}\] With BC \begin {align*} u\left ( -\pi ,t\right ) & =0\\ u\left ( \pi ,t\right ) & =0 \end {align*}
And initial conditions\begin {align*} u\left ( x,0\right ) & =0\\ u_{t}\left ( x,0\right ) & =\sin ^{2}\left ( x\right ) \end {align*}
Let \(\xi =x+\pi \). When \(x=-\pi ,\xi =0\) and when \(x=\pi ,\xi =2\pi \). In terms of \(\xi \), the new pde in \(U\left ( \xi ,t\right ) \) becomes
\[ U_{tt}=c^{2}U_{\xi \xi }\] With BC \begin {align*} U\left ( 0,t\right ) & =0\\ U\left ( 2\pi ,t\right ) & =0 \end {align*}
And initial conditions\begin {align*} U\left ( \xi ,0\right ) & =0\\ U_{t}\left ( \xi ,0\right ) & =\sin ^{2}\left ( \xi \right ) \end {align*}
Let \(U=X\left ( \xi \right ) T\left ( t\right ) \). The PDE becomes\begin {align*} \frac {T^{\prime \prime }X}{c^{2}} & =X^{\prime \prime }T\\ \frac {1}{c^{2}}\frac {T^{\prime \prime }}{T} & =\frac {X^{\prime \prime }}{X}=-\lambda \end {align*}
Where \(\lambda \) is separation constant. Hence the eigenvalue ODE is \begin {align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X\left ( 2\pi \right ) & =0 \end {align*}
From the boundary conditions, we see that \(\lambda >0\) is the only possible value. Therefore the solution to the above ODE is\[ X\left ( x\right ) =A\cos \left ( \sqrt {\lambda }\xi \right ) +B\sin \left ( \sqrt {\lambda }\xi \right ) \] Since \(X\left ( 0\right ) =0\) then \(A=0\) and the solution becomes \(X\left ( \xi \right ) =B\sin \left ( \sqrt {\lambda }\xi \right ) \). Since \(X\left ( 2\pi \right ) =0\) then for non trivial solution we want \(\sqrt {\lambda }2\pi =n\pi \) or \[ \lambda _{n}=\left ( \frac {n}{2}\right ) ^{2}\qquad n=1,2,3,\cdots \] Hence the eigenfunctions are \[ X_{n}\left ( \xi \right ) =\sin \left ( \frac {n}{2}\xi \right ) \qquad n=1,2,3,\cdots \] The time ODE now becomes\begin {align*} T_{n}^{\prime \prime }+c^{2}\lambda _{n}T_{n} & =0\\ T_{n}^{\prime \prime }+c^{2}\left ( \frac {n}{2}\right ) ^{2}T_{n} & =0\\ T_{n}^{\prime \prime }+\frac {c^{2}n^{2}}{4}T_{n} & =0 \end {align*}
Which has the solution\[ T_{n}\left ( t\right ) =D_{n}\cos \left ( \frac {cn}{2}t\right ) +E_{n}\sin \left ( \frac {cn}{2}t\right ) \] Therefore the complete solution becomes\begin {equation} U\left ( \xi ,t\right ) =\sum _{n=1}^{\infty }\left ( D_{n}\cos \left ( \frac {cn}{2}t\right ) +E_{n}\sin \left ( \frac {cn}{2}t\right ) \right ) \sin \left ( \frac {n}{2}\xi \right ) \tag {1} \end {equation}
Switching back to \(x\) the above becomes
\begin {equation} u\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( D_{n}\cos \left ( \frac {cn}{2}t\right ) +E_{n}\sin \left ( \frac {cn}{2}t\right ) \right ) \sin \left ( \frac {n}{2}\left ( x+\pi \right ) \right ) \tag {1A} \end {equation}
At \(t=0\) the above becomes\begin {align*} 0 & =\sum _{n=1}^{\infty }D_{n}\sin \left ( \frac {n}{2}\left ( x+\pi \right ) \right ) \\ D_{n} & =0 \end {align*}
The solution (1A) simplifies to
\begin {equation} u\left ( x,t\right ) =\sum _{n=1}^{\infty }E_{n}\sin \left ( \frac {cn}{2}t\right ) \sin \left ( \frac {n}{2}\left ( x+\pi \right ) \right ) \tag {2} \end {equation}
Taking time derivative of (2) gives\[ u_{t}\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( E_{n}\frac {cn}{2}\cos \left ( \frac {cn}{2}t\right ) \right ) \sin \left ( \frac {n}{2}\left ( x+\pi \right ) \right ) \] At \(t=0\) the above becomes\[ \sin ^{2}\left ( x\right ) =\sum _{n=1}^{\infty }E_{n}\frac {cn}{2}\sin \left ( \frac {n}{2}\left ( x+\pi \right ) \right ) \] Applying orthogonality gives\begin {align*} \int _{-\pi }^{\pi }\sin ^{2}\left ( x\right ) \sin \left ( \frac {n}{2}\left ( x+\pi \right ) \right ) dx & =E_{n}\frac {cn}{2}\int _{-\pi }^{\pi }\sin ^{2}\left ( \frac {n}{2}\left ( x+\pi \right ) \right ) dx\\ & =E_{n}\frac {\pi cn}{2} \end {align*}
For the LHS, for \(n\) even \(\int _{-\pi }^{\pi }\sin ^{2}\left ( x\right ) \sin \left ( \frac {n}{2}\left ( x+\pi \right ) \right ) dx=0\). Hence \(E_{n}=0\) for all \(n\) even. For \(n\) odd \begin {align*} \int _{-\pi }^{\pi }\sin ^{2}\left ( x\right ) \sin \left ( \frac {n}{2}\left ( x+\pi \right ) \right ) dx & =\frac {16\left ( \cos \left ( n\pi \right ) -1\right ) }{\left ( n^{2}-16\right ) n}\\ & =\frac {16\left ( \left ( -1\right ) ^{n}-1\right ) }{\left ( n^{2}-16\right ) n} \end {align*}
But \(n\) is odd, hence the above simplifies more to
\[ \int _{-\pi }^{\pi }\sin ^{2}\left ( x\right ) \sin \left ( \frac {n}{2}\left ( x+\pi \right ) \right ) dx=\frac {-32}{\left ( n^{2}-16\right ) n}\]
Therefore\begin {align} E_{n} & =\frac {2}{n\pi c}\frac {-32}{\left ( n^{2}-16\right ) n}\nonumber \\ & =\frac {-64}{n^{2}\pi c\left ( n^{2}-16\right ) }\nonumber \end {align}
Therefore the final solution (2) now becomes\[ u\left ( x,t\right ) =\sum _{n=1,3,5,\cdots }^{\infty }\frac {-64}{n^{2}\pi c\left ( n^{2}-16\right ) }\sin \left ( \frac {cn}{2}t\right ) \sin \left ( \frac {n}{2}\left ( x+\pi \right ) \right ) \]
The following is an animation of the solution for \(c=2\). Using \(c=2\) then the solution above becomes
\[ u\left ( x,t\right ) =\sum _{n=1,3,5,\cdots }^{\infty }\frac {-32}{n^{2}\pi \left ( n^{2}-16\right ) }\sin \left ( nt\right ) \sin \left ( \frac {n}{2}\left ( x+\pi \right ) \right ) \]
Source code used for the above
____________________________________________________________________________________
Added January 8 ,2020
Problem 6.3.31 Introduction to Partial Dfferential Equations by Peter Olver, ISBN 9783319020983.
Solve
\begin {align*} u_{tt} & =u_{xx}\\ u\left ( -1,t\right ) & =0\\ u\left ( 1,t\right ) & =0\\ u\left ( x,0\right ) & =\delta \left ( x\right ) \\ \frac {\partial u\left ( x,0\right ) }{\partial t} & =0 \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], {t, 2}] == D[u[x, t], {x, 2}]; bc = {u[-1, t] == 0, u[1, t] == 0}; ic = {u[x, 0] == DiracDelta[x], Derivative[0, 1][u][x, 0] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t}], 60*10]];
\[\left \{\left \{u(x,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\cos \left (\frac {1}{2} \pi t K[1]\right ) \sin \left (\frac {1}{2} \pi K[1]\right ) \sin \left (\frac {1}{2} \pi (x+1) K[1]\right )\right \}\right \}\]
Maple ✓
restart; pde := diff(u(x,t),t$2)=diff(u(x,t),x$2); bc:=u(-1,t)=0,u(1,t)=0; ic := u(x,0)=Dirac(x), D[2](u)(x,0)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic, bc],u(x,t))),output='realtime'));
\[u \left ( x,t \right ) =\sum _{n=1}^{\infty }\sin \left ( {\frac {n\pi }{2}} \right ) \sin \left ( {\frac {n\pi \, \left ( x+1 \right ) }{2}} \right ) \cos \left ( {\frac {n\pi \,t}{2}} \right ) \]
Hand solution
Since the boundary conditions are at \(x=-1\) and at \(x=1\), it is a little easier to solve this by first shifting the boundaries to \(x=0\) and \(x=2\). This is done by transformation. Let\[ z=x+1 \] When \(x=-1\) then \(z=0\) and when \(x=1\) then \(z=2\). The PDE in terms of \(z\) remains the same but the B.C. are shifted. Hence we want to solve for \(v\left ( z,t\right ) \) in\begin {align*} v_{tt} & =v_{zz}\\ v\left ( 0,t\right ) & =0\\ v\left ( 2,t\right ) & =0 \end {align*}
No need to worry about initial conditions now, since we will transform back to \(x\) before applying initial conditions and therefore will use the original initial conditions. This PDE is now solved by separation. Let \(v=Z\left ( z\right ) T\left ( t\right ) \). Substituting into the PDE gives\begin {align*} T^{\prime \prime }Z & =Z^{\prime \prime }T\\ \frac {T^{\prime \prime }}{T} & =\frac {Z^{\prime \prime }}{Z}=-\lambda \end {align*}
This gives the boundary value ODE \begin {align} Z^{\prime \prime }+\lambda Z & =0\tag {1}\\ Z\left ( 0\right ) & =0\nonumber \\ Z\left ( 2\right ) & =0\nonumber \end {align}
And the time ODE\begin {equation} T^{\prime \prime }+\lambda T=0\tag {2} \end {equation} Solving (1). From the boundary conditions we know only \(\lambda >0\) is an eigenvalue. Hence for \(\lambda >0\) the solution is \[ Z\left ( z\right ) =A\cos \left ( \sqrt {\lambda }z\right ) +B\sin \left ( \sqrt {\lambda }z\right ) \] At \(z=0\) this gives \(A=0\). Hence the solution now becomes \(Z\left ( z\right ) =B\sin \left ( \sqrt {\lambda }z\right ) \). At \(z=2\) the above gives \(0=B\sin \left ( 2\sqrt {\lambda }\right ) \). For non-trivial solution we want \(\sin \left ( 2\sqrt {\lambda }\right ) =0\) which implies \(2\sqrt {\lambda }=n\pi \) or\[ \lambda _{n}=\left ( \frac {n\pi }{2}\right ) ^{2}\qquad n=1,2,3,\cdots \] And the corresponding eigenfunctions\[ Z_{n}\left ( z\right ) =\sin \left ( \frac {n\pi }{2}z\right ) \qquad n=1,2,3,\cdots \] The time ODE (2) now becomes\[ T^{\prime \prime }+\left ( \frac {n\pi }{2}\right ) ^{2}T=0 \] Which has solution\[ T_{n}\left ( t\right ) =A_{n}\cos \left ( \frac {n\pi }{2}t\right ) +B_{n}\sin \left ( \frac {n\pi }{2}t\right ) \] Hence the complete solution is\[ v\left ( z,t\right ) =\sum _{n=1}^{\infty }\left ( A_{n}\cos \left ( \frac {n\pi }{2}t\right ) +B_{n}\sin \left ( \frac {n\pi }{2}t\right ) \right ) \sin \left ( \frac {n\pi }{2}z\right ) \] We are now ready to switch back from \(z\) to \(x\). Since \(z=x+1\) then the above becomes\begin {equation} u\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( A_{n}\cos \left ( \frac {n\pi }{2}t\right ) +B_{n}\sin \left ( \frac {n\pi }{2}t\right ) \right ) \sin \left ( \frac {n\pi }{2}\left ( x+1\right ) \right ) \tag {3} \end {equation} Now we apply initial conditions to find \(A_{n},B_{n}\). At \(t=0,u\left ( x,0\right ) =\delta \left ( x\right ) \). Hence the above gives\[ \delta \left ( x\right ) =\sum _{n=1}^{\infty }A_{n}\sin \left ( \frac {n\pi }{2}\left ( x+1\right ) \right ) \] Multiplying both sides by \(\sin \left ( \frac {m\pi }{2}\left ( x+1\right ) \right ) \) and Integrating gives\[ \int _{-1}^{1}\delta \left ( x\right ) \sin \left ( \frac {m\pi }{2}\left ( x+1\right ) \right ) dx=\sum _{n=1}^{\infty }A_{n}\int _{-1}^{1}\sin \left ( \frac {n\pi }{2}\left ( x+1\right ) \right ) \sin \left ( \frac {m\pi }{2}\left ( x+1\right ) \right ) dx \] By orthogonality of \(\sin \) functions only term survives and the above simplifies to\begin {align*} \int _{-1}^{1}\delta \left ( x\right ) \sin \left ( \frac {m\pi }{2}\left ( x+1\right ) \right ) dx & =A_{m}\overset {1}{\overbrace {\int _{-1}^{1}\sin ^{2}\left ( \frac {m\pi }{2}\left ( x+1\right ) \right ) dx}}\\ & =A_{m} \end {align*}
But \(\int _{-1}^{1}\delta \left ( x\right ) \sin \left ( \frac {m\pi }{2}\left ( x+1\right ) \right ) dx=\sin \left ( \frac {m\pi }{2}\right ) \) since that is where \(x=0\). The above reduces to\[ A_{n}=\sin \left ( \frac {n\pi }{2}\right ) \qquad n=1,2,3,\cdots \] The solution (1) becomes\begin {equation} u\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( \sin \left ( \frac {n\pi }{2}\right ) \cos \left ( \frac {n\pi }{2}t\right ) +B_{n}\sin \left ( \frac {n\pi }{2}t\right ) \right ) \sin \left ( \frac {n\pi }{2}\left ( x+1\right ) \right ) \tag {4} \end {equation} Taking time derivatives\[ \frac {\partial }{\partial t}u\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( -\frac {n\pi }{2}\sin \left ( \frac {n\pi }{2}\right ) \sin \left ( \frac {n\pi }{2}t\right ) +\frac {n\pi }{2}B_{n}\cos \left ( \frac {n\pi }{2}t\right ) \right ) \sin \left ( \frac {n\pi }{2}\left ( x+1\right ) \right ) \] At \(t=0\) the above becomes\[ 0=\sum _{n=1}^{\infty }\frac {n\pi }{2}B_{n}\sin \left ( \frac {n\pi }{2}\left ( x+1\right ) \right ) \] Therefore \(B_{n}=0\). Hence the solution (4) becomes\begin {equation} u\left ( x,t\right ) =\sum _{n=1}^{\infty }\sin \left ( \frac {n\pi }{2}\right ) \cos \left ( \frac {n\pi }{2}t\right ) \sin \left ( \frac {n\pi }{2}\left ( x+1\right ) \right ) \tag {5} \end {equation} Notice that \(\sin \left ( \frac {n\pi }{2}\right ) \) is zero when \(n\) is even.
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This is problem at page 28, David J Logan textbook, applied PDE textbook. No initial conditions given \[ u_{tt} = c^2 u_{xx} \] With boundary condition \begin {align*} u(0,t) &= 0 \\ u(L,t) &= 0 \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], {t, 2}] == c^2*D[u[x, t], {x, 2}]; bc = {u[0, t] == 0, u[L, t] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, t], {x, t}, Assumptions -> {L > 0}], 60*10]];
\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}\sqrt {2} \sqrt {\frac {1}{L}} \sin \left (\frac {\pi x K[1]}{L}\right ) \left (\cos \left (\frac {\pi t | c| K[1]}{L}\right ) \int _0^L \sqrt {2} \sqrt {\frac {1}{L}} \sin \left (\frac {\pi x K[1]}{L}\right ) u(x,0) \, dx+\frac {L \left (\int _0^L \frac {\sqrt {2} \sin \left (\frac {\pi x K[1]}{L}\right ) u^{(0,1)}(x,0)}{\sqrt {L}} \, dx\right ) \sin \left (\frac {\pi t | c| K[1]}{L}\right )}{\pi | c| K[1]}\right ) & K[1]\in \mathbb {Z}\land K[1]\geq 1\land c^2 K[1]^2>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
Maple ✓
restart; interface(showassumed=0); pde := diff(u(x,t),t$2)=c^2*diff(u(x,t),x$2); bc := u(0,t)=0,u(L,t)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(x,t)) assuming L>0),output='realtime'));
\[u \left ( x,t \right ) =\sum _{n=1}^{\infty }\sin \left ( {\frac {n\pi \,x}{L}} \right ) \left ( \sin \left ( {\frac {cn\pi \,t}{L}} \right ) {\it \_C1} \left ( n \right ) +\cos \left ( {\frac {cn\pi \,t}{L}} \right ) {\it \_C5} \left ( n \right ) \right ) \]
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Added Feb 25, 2019. Exam 1 problem, MATH 4567 Applied Fourier Analysis, University of Minnesota, Twin Cities.
Solve for \(u(x,t)\) \[ u_{tt} = u_{xx} -u \] With boundary condition \begin {align*} u(0,t) &= 0 \\ u(\pi ,t) &=0 \end {align*}
And initial conditions \begin {align*} u(x,0) &= 0 \\ u_t(x,0) &= 1 \\ \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], {t, 2}] == D[u[x, t], {x, 2}] - u[x, t]; ic = {u[x, 0] == 0, Derivative[0, 1][u][x, 0] == 1}; bc = {u[0, t] == 0, u[Pi, t] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t}], 60*10]];
\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {2 \left (1+(-1)^{K[1]+1}\right ) \sin (x K[1]) \sin \left (t \sqrt {K[1]^2+1}\right )}{\sqrt {\pi } K[1] \sqrt {\pi K[1]^2+\pi }} & K[1]\in \mathbb {Z}\land K[1]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
Maple ✓
restart; pde := diff(u(x,t),t$2)=diff(u(x,t),x$2)-u(x,t); bc := u(0,t)=0,u(Pi,t)=0; ic := u(x,0)=0,eval(diff(u(x,t),t),t=0)=1; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic, bc],u(x,t))),output='realtime'));
\[u \left ( x,t \right ) =\sum _{n=1}^{\infty }-2\,{\frac { \left ( \left ( -1 \right ) ^{n}-1 \right ) \sin \left ( nx \right ) \sin \left ( \sqrt {{n}^{2}+1}t \right ) }{\pi \,\sqrt {{n}^{2}+1}n}}\]
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This is problem at page 149, David J Logan textbook, applied PDE textbook.
\[ \frac {\partial ^2 u}{\partial t^2} = c^2 \frac {\partial ^2 u}{\partial x^2} + p(x,t) \] With boundary conditions \begin {align*} u(\pi ,0) &=0 \\ u(0,t) &= 0 \end {align*}
With initial conditions \begin {align*} \frac {\partial u}{\partial t}(x,0) &=0 \\ u(x,0) &= 0 \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], {t, 2}] == c^2*D[u[x, t], {x, 2}] + p[x, t]; bc = {u[0, t] == 0, u[Pi, t] == 0}; ic = {u[x, 0] == 0, Derivative[0, 1][u][x, 0] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}], 60*10]];
\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}\sqrt {\frac {2}{\pi }} \left (\int _0^t \frac {\left (\int _0^{\pi } \sqrt {\frac {2}{\pi }} p(x,K[2]) \sin (x K[1]) \, dx\right ) \sin \left (\sqrt {c^2 K[1]^2} (t-K[2])\right )}{\sqrt {c^2 K[1]^2}} \, dK[2]\right ) \sin (x K[1]) & K[1]\in \mathbb {Z}\land ((c<0\land K[1]\geq 1)\lor (c>0\land K[1]\geq 1)) \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
Maple ✓
restart; interface(showassumed=0); pde := diff(u(x,t),t$2)=c^2*diff(u(x,t),x$2)+p(x,t); bc := u(0,t)=0,u(Pi,t)=0; ic := u(x,0)=0,D[2](u)(x,0)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc,ic],u(x,t))),output='realtime'));
\[u \left ( x,t \right ) =\int _{0}^{t}\!\sum _{n=1}^{\infty } \left ( 2\,{\frac {\int _{0}^{\pi }\!\sin \left ( nx \right ) p \left ( x,\tau \right ) \,{\rm d}x\sin \left ( nx \right ) \sin \left ( cn \left ( t-\tau \right ) \right ) }{n\pi \,c}} \right ) \,{\rm d}\tau \]
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Added Nov 25, 2018.
This is problem 8.5.2 (a), Richard Haberman applied partial differential equations book, 5th edition
Both ends fixed end, initial position given, zero initial velocity, with source that depends on time and space.
Consider a vibrating string with time-dependent forcing: \[ \frac {\partial ^2 u}{\partial t^2} = c^2 \frac {\partial ^2 u}{\partial x^2} + Q(x,t) \] With boundary conditions \begin {align*} u(0,t) &=0 \\ u(L,t) &= 0 \end {align*}
With initial conditions \begin {align*} u_t(x,0) &=0 \\ u(x,0) &= f(x) \end {align*}
my hand solution in in the file feb_24_2019_4_24_pm.tex, but I need to go over my solution
again to make sure it is correct.
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], {t, 2}] == c^2*D[u[x, t], {x, 2}] + Q[x, t]; bc = {u[0, t] == 0, u[L, t] == 0}; ic = {u[x, 0] == f[x], Derivative[0, 1][u][x, 0] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}], 60*10]];
\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}\sqrt {2} \sqrt {\frac {1}{L}} \left (\cos \left (\pi t \sqrt {\frac {c^2 K[1]^2}{L^2}}\right ) \int _0^L \sqrt {2} \sqrt {\frac {1}{L}} f(x) \sin \left (\frac {\pi x K[1]}{L}\right ) \, dx+\int _0^t \frac {\left (\int _0^L \sqrt {2} \sqrt {\frac {1}{L}} Q(x,K[2]) \sin \left (\frac {\pi x K[1]}{L}\right ) \, dx\right ) \sin \left (\pi \sqrt {\frac {c^2 K[1]^2}{L^2}} (t-K[2])\right )}{\pi \sqrt {\frac {c^2 K[1]^2}{L^2}}} \, dK[2]\right ) \sin \left (\frac {\pi x K[1]}{L}\right ) & K[1]\in \mathbb {Z}\land ((L<0\land ((c<0\land K[1]\geq 1)\lor (c>0\land K[1]\geq 1)))\lor (L>0\land ((c<0\land K[1]\geq 1)\lor (c>0\land K[1]\geq 1)))) \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
Maple ✓
restart; interface(showassumed=0); pde := diff(u(x,t),t$2)=c^2*diff(u(x,t),x$2)+Q(x,t); bc := u(0,t)=0,u(L,t)=0; ic := u(x,0)=f(x), eval( diff(u(x,t),t),t=0)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc,ic],u(x,t)) assuming L>0),output='realtime'));
\[u \left ( x,t \right ) =\sum _{n=1}^{\infty } \left ( 2\,{\frac {1}{L}\int _{0}^{L}\!\sin \left ( {\frac {n\pi \,x}{L}} \right ) f \left ( x \right ) \,{\rm d}x\sin \left ( {\frac {n\pi \,x}{L}} \right ) \cos \left ( {\frac {cn\pi \,t}{L}} \right ) } \right ) +\int _{0}^{t}\!\sum _{n=1}^{\infty } \left ( 2\,{\frac {1}{cn\pi }\int _{0}^{L}\!\sin \left ( {\frac {n\pi \,x}{L}} \right ) Q \left ( x,\tau \right ) \,{\rm d}x\sin \left ( {\frac {n\pi \,x}{L}} \right ) \sin \left ( {\frac {cn\pi \, \left ( t-\tau \right ) }{L}} \right ) } \right ) \,{\rm d}\tau \]
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Added Nov 25, 2018.
This is problem 8.5.2 (b), Richard Haberman applied partial differential equations book, 5th edition.
Both ends fixed end, initial position given, zero initial velocity, with source that depends on time and space.
Consider a vibrating string with time-dependent forcing: \[ u_{tt} = c^2 u_{xx} + g(x) \cos (\omega t) \] With boundary conditions \begin {align*} u(0,t) &=0 \\ u(L,t) &= 0 \end {align*}
With initial conditions \begin {align*} u_t(x,0) &=0 \\ u(x,0) &= f(x) \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], {t, 2}] == c^2*D[u[x, t], {x, 2}] + g[x]*Cos[w*t]; bc = {u[0, t] == 0, u[L, t] == 0}; ic = {u[x, 0] == f[x], Derivative[0, 1][u][x, 0] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}], 60*10]];
\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}\sqrt {2} \sqrt {\frac {1}{L}} \left (\cos \left (\pi t \sqrt {\frac {c^2 K[1]^2}{L^2}}\right ) \int _0^L \sqrt {2} \sqrt {\frac {1}{L}} f(x) \sin \left (\frac {\pi x K[1]}{L}\right ) \, dx+\int _0^t \frac {\left (\int _0^L \sqrt {2} \sqrt {\frac {1}{L}} \cos (w K[2]) g(x) \sin \left (\frac {\pi x K[1]}{L}\right ) \, dx\right ) \sin \left (\pi \sqrt {\frac {c^2 K[1]^2}{L^2}} (t-K[2])\right )}{\pi \sqrt {\frac {c^2 K[1]^2}{L^2}}} \, dK[2]\right ) \sin \left (\frac {\pi x K[1]}{L}\right ) & K[1]\in \mathbb {Z}\land ((L<0\land ((c<0\land K[1]\geq 1)\lor (c>0\land K[1]\geq 1)))\lor (L>0\land ((c<0\land K[1]\geq 1)\lor (c>0\land K[1]\geq 1)))) \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
Maple ✓
restart; interface(showassumed=0); pde := diff(u(x,t),t$2)=c^2*diff(u(x,t),x$2)+ g(x)*cos(w*t); bc := u(0,t)=0,u(L,t)=0; ic := u(x,0)=0, eval( diff(u(x,t),t),t=0)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc,ic],u(x,t)) assuming L>0),output='realtime'));
\[u \left ( x,t \right ) ={\frac {1}{\pi \,c} \left ( -2\,\sum _{n=1}^{\infty } \left ( {\frac {L\cos \left ( wt \right ) \pi \,c}{-{\pi }^{2}{c}^{2}{n}^{2}+{L}^{2}{w}^{2}}\int _{0}^{L}\!\sin \left ( {\frac {n\pi \,x}{L}} \right ) g \left ( x \right ) \,{\rm d}x\sin \left ( {\frac {n\pi \,x}{L}} \right ) } \right ) +2\,\sum _{n=1}^{\infty } \left ( {\frac {L\pi \,c}{-{\pi }^{2}{c}^{2}{n}^{2}+{L}^{2}{w}^{2}}\int _{0}^{L}\!\sin \left ( {\frac {n\pi \,x}{L}} \right ) g \left ( x \right ) \,{\rm d}x\sin \left ( {\frac {n\pi \,x}{L}} \right ) \cos \left ( {\frac {cn\pi \,t}{L}} \right ) } \right ) \right ) }\]
Hand solution
Let \[ u\left ( x,t\right ) =\sum _{n=1}^{\infty }A_{n}\left ( t\right ) \phi _{n}\left ( x\right ) \] Where we used \(=\) instead of \(\sim \) above, since the PDE given has homogeneous B.C. We know that \(\phi _{n}\left ( x\right ) =\sin \left ( \sqrt {\lambda _{n}}x\right ) \) for \(n=1,2,3,\cdots \) where \(\lambda _{n}=\left ( \frac {n\pi }{L}\right ) ^{2}\). Substituting the above in the given PDE gives\[ \sum _{n=1}^{\infty }A_{n}^{\prime \prime }\left ( t\right ) \phi _{n}\left ( x\right ) =c^{2}\sum _{n=1}^{\infty }A_{n}\left ( t\right ) \frac {d^{2}\phi _{n}\left ( x\right ) }{dx^{2}}+Q\left ( x,t\right ) \] But \(Q\left ( x,t\right ) =\sum _{n=1}^{\infty }q_{n}\left ( t\right ) \phi _{n}\left ( x\right ) \), hence the above becomes\[ \sum _{n=1}^{\infty }A_{n}^{\prime \prime }\left ( t\right ) \phi _{n}\left ( x\right ) =c^{2}\sum _{n=1}^{\infty }A_{n}\left ( t\right ) \frac {d^{2}\phi _{n}\left ( x\right ) }{dx^{2}}+\sum _{n=1}^{\infty }g_{n}\left ( t\right ) \phi _{n}\left ( x\right ) \] But \(\frac {d^{2}\phi _{n}\left ( x\right ) }{dx^{2}}=-\lambda _{n}\phi _{n}\left ( x\right ) \), hence\[ \sum _{n=1}^{\infty }A_{n}^{\prime \prime }\left ( t\right ) \phi _{n}\left ( x\right ) =-c^{2}\sum _{n=1}^{\infty }\lambda _{n}A_{n}\left ( t\right ) \phi _{n}\left ( x\right ) +\sum _{n=1}^{\infty }g_{n}\left ( t\right ) \phi _{n}\left ( x\right ) \] Multiplying both sides by \(\phi _{m}\left ( x\right ) \) and integrating gives\begin {align*} \int _{0}^{L}\sum _{n=1}^{\infty }A_{n}^{\prime \prime }\left ( t\right ) \phi _{m}\left ( x\right ) \phi _{n}\left ( x\right ) dx & =-c^{2}\int _{0}^{L}\sum _{n=1}^{\infty }\lambda _{n}A_{n}\left ( t\right ) \phi _{m}\left ( x\right ) \phi _{n}\left ( x\right ) dx+\int _{0}^{L}\sum _{n=1}^{\infty }g_{n}\left ( t\right ) \phi _{m}\left ( x\right ) \phi _{n}\left ( x\right ) dx\\ A_{n}^{\prime \prime }\left ( t\right ) \int _{0}^{L}\phi _{n}^{2}\left ( x\right ) dx & =-c^{2}\lambda _{n}A_{n}\left ( t\right ) \int _{0}^{L}\phi _{n}^{2}\left ( x\right ) dx+g_{n}\left ( t\right ) \int _{0}^{L}\phi _{n}^{2}\left ( x\right ) dx \end {align*}
Hence\[ A_{n}^{\prime \prime }\left ( t\right ) +c^{2}\lambda _{n}A_{n}\left ( t\right ) =g_{n}\left ( t\right ) \] Now we solve the above ODE. Let solution be \[ A_{n}\left ( t\right ) =A_{n}^{h}\left ( t\right ) +A_{n}^{p}\left ( t\right ) \] Which is the sum of the homogenous and particular solutions. The homogenous solution is \[ A_{n}^{h}\left ( t\right ) =c_{1_{n}}\cos \left ( c\sqrt {\lambda _{n}}t\right ) +c_{2_{n}}\sin \left ( c\sqrt {\lambda _{n}}t\right ) \] And the particular solution depends on \(q_{n}\left ( t\right ) \). Once we find \(q_{n}\left ( t\right ) \), we plug-in everything back into \(u\left ( x,t\right ) =\sum _{n=1}^{\infty }A_{n}\left ( t\right ) \phi _{n}\left ( x\right ) \) and then use initial conditions to find \(c_{1_{n}},c_{2_{n}}\), the two constant of integrations. Now we are given that \(Q\left ( x,t\right ) =g\left ( x\right ) \cos \left ( \omega t\right ) \). Hence\[ g_{n}\left ( t\right ) =\frac {\int _{0}^{L}Q\left ( x,t\right ) \phi _{n}\left ( x\right ) dx}{\int _{0}^{L}\phi _{n}^{2}\left ( x\right ) dx}=\frac {\cos \left ( \omega t\right ) \int _{0}^{L}g\left ( x\right ) \phi _{n}\left ( x\right ) dx}{\int _{0}^{L}\phi _{n}^{2}\left ( x\right ) dx}=\cos \left ( \omega t\right ) \gamma _{n}\] Where\[ \gamma _{n}=\frac {\int _{0}^{L}g\left ( x\right ) \phi _{n}\left ( x\right ) dx}{\int _{0}^{L}\phi _{n}^{2}\left ( x\right ) dx}\] is constant that depends on \(n\). Now we use the above in result found in part (a)\begin {equation} A_{n}^{\prime \prime }\left ( t\right ) +c^{2}\lambda _{n}A_{n}\left ( t\right ) =\gamma _{n}\cos \left ( \omega t\right ) \tag {1} \end {equation} We know the homogenous solution from part (a). \[ A_{n}^{h}\left ( t\right ) =c_{1_{n}}\cos \left ( c\sqrt {\lambda _{n}}t\right ) +c_{2_{n}}\sin \left ( c\sqrt {\lambda _{n}}t\right ) \] We now need to find the particular solution. Will solve using method of undetermined coefficients.
Case 1 \(\omega \neq c\sqrt {\lambda _{n}}\) (no resonance)
We can now guess \[ A_{n}^{p}\left ( t\right ) =z_{1}\cos \left ( \omega t\right ) +z_{2}\sin \left ( \omega t\right ) \] Plugging this back into (1) gives\begin {align*} \left ( z_{1}\cos \left ( \omega t\right ) +z_{2}\sin \left ( \omega t\right ) \right ) ^{\prime \prime }+c^{2}\lambda _{n}\left ( z_{1}\cos \left ( \omega t\right ) +z_{2}\sin \left ( \omega t\right ) \right ) & =\gamma _{n}\cos \left ( \omega t\right ) \\ \left ( -\omega z_{1}\sin \left ( \omega t\right ) +\omega z_{2}\cos \left ( \omega t\right ) \right ) ^{\prime }+c^{2}\lambda _{n}\left ( z_{1}\cos \left ( \omega t\right ) +z_{2}\sin \left ( \omega t\right ) \right ) & =\gamma _{n}\cos \left ( \omega t\right ) \\ -\omega ^{2}z_{1}\cos \left ( \omega t\right ) -\omega ^{2}z_{2}\sin \left ( \omega t\right ) +c^{2}\lambda _{n}\left ( z_{1}\cos \left ( \omega t\right ) +z_{2}\sin \left ( \omega t\right ) \right ) & =\gamma _{n}\cos \left ( \omega t\right ) \end {align*}
Collecting terms\[ \cos \left ( \omega t\right ) \left ( -\omega ^{2}z_{1}+c^{2}\lambda _{n}z_{1}\right ) +\sin \left ( \omega t\right ) \left ( -\omega ^{2}z_{2}+c^{2}\lambda _{n}z_{2}\right ) =\gamma _{n}\cos \left ( \omega t\right ) \] Therefore we obtain two equations in two unknowns\begin {align*} -\omega ^{2}z_{1}+c^{2}\lambda _{n}z_{1} & =\gamma _{n}\\ -\omega ^{2}z_{2}+c^{2}\lambda _{n}z_{2} & =0 \end {align*}
From the second equation, \(z_{2}=0\) and from the first equation\begin {align*} z_{1}\left ( c^{2}\lambda _{n}-\omega ^{2}\right ) & =\gamma _{n}\\ z_{1} & =\frac {\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}} \end {align*}
Hence \begin {align*} A_{n}^{p}\left ( t\right ) & =z_{1}\cos \left ( \omega t\right ) +z_{2}\sin \left ( \omega t\right ) \\ & =\frac {\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\cos \left ( \omega t\right ) \end {align*}
Therefore\begin {align*} A_{n}\left ( t\right ) & =A_{n}^{h}\left ( t\right ) +A_{n}^{p}\left ( t\right ) \\ & =c_{1_{n}}\cos \left ( c\sqrt {\lambda _{n}}t\right ) +c_{2_{n}}\sin \left ( c\sqrt {\lambda _{n}}t\right ) +\frac {\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\cos \left ( \omega t\right ) \end {align*}
Now we need to find \(c_{1_{n}},c_{2_{n}}\). Since\begin {align*} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }A_{n}\left ( t\right ) \phi _{n}\left ( x\right ) \\ & =\sum _{n=1}^{\infty }\left ( c_{1_{n}}\cos \left ( c\sqrt {\lambda _{n}}t\right ) +c_{2_{n}}\sin \left ( c\sqrt {\lambda _{n}}t\right ) +\frac {\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\cos \left ( \omega t\right ) \right ) \sin \left ( \frac {n\pi }{L}x\right ) \end {align*}
At \(t=0\) the above becomes\begin {align*} f\left ( x\right ) & =\sum _{n=1}^{\infty }\left ( c_{1_{n}}+\frac {\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\right ) \sin \left ( \frac {n\pi }{L}x\right ) \\ & =\sum _{n=1}^{\infty }c_{1_{n}}\sin \left ( \frac {n\pi }{L}x\right ) +\sum _{n=1}^{\infty }\frac {\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\sin \left ( \frac {n\pi }{L}x\right ) \end {align*}
Applying orthogonality\begin {align*} \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {m\pi }{L}x\right ) dx & =\int _{0}^{L}\sum _{n=1}^{\infty }c_{1_{n}}\sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{L}x\right ) dx+\int _{0}^{L}\sum _{n=1}^{\infty }\frac {\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{L}x\right ) dx\\ \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {m\pi }{L}x\right ) dx & =c_{1_{n}}\int _{0}^{L}\sin ^{2}\left ( \frac {n\pi }{L}x\right ) dx+\frac {\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\int _{0}^{L}\sin ^{2}\left ( \frac {n\pi }{L}x\right ) dx \end {align*}
Rearranging\begin {align*} \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {m\pi }{L}x\right ) dx-\frac {\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\int _{0}^{L}\sin ^{2}\left ( \frac {n\pi }{L}x\right ) dx & =c_{1_{n}}\int _{0}^{L}\sin ^{2}\left ( \frac {n\pi }{L}x\right ) dx\\ c_{1_{n}} & =\frac {\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {m\pi }{L}x\right ) dx}{\int _{0}^{L}\sin ^{2}\left ( \frac {n\pi }{L}x\right ) dx}-\frac {\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\\ & =\frac {2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {m\pi }{L}x\right ) dx-\frac {\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}} \end {align*}
We now need to find \(c_{2_{n}}\). For this we need to differentiate the solution once.\[ \frac {\partial u\left ( x,t\right ) }{\partial t}=\sum _{n=1}^{\infty }\left ( -c\sqrt {\lambda _{n}}c_{1_{n}}\sin \left ( c\sqrt {\lambda _{n}}t\right ) +c\sqrt {\lambda _{n}}c_{2_{n}}\cos \left ( c\sqrt {\lambda _{n}}t\right ) -\frac {\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\omega \sin \left ( \omega t\right ) \right ) \sin \left ( \frac {n\pi }{L}x\right ) \] Applying initial conditions \(\frac {\partial u\left ( x,0\right ) }{\partial t}=0\) gives\[ 0=\sum _{n=1}^{\infty }c\sqrt {\lambda _{n}}c_{2_{n}}\sin \left ( \frac {n\pi }{L}x\right ) \] Hence \[ c_{2_{n}}=0 \] Therefore the final solution is\[ A_{n}\left ( t\right ) =c_{1_{n}}\cos \left ( c\sqrt {\lambda _{n}}t\right ) +\frac {\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\cos \left ( \omega t\right ) \] And\[ u\left ( x,t\right ) =\sum _{n=1}^{\infty }A_{n}\left ( t\right ) \sin \left ( \frac {n\pi }{L}x\right ) \] Where \[ c_{1_{n}}=\frac {2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {m\pi }{L}x\right ) dx-\frac {\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\] Case 2 \(\omega =c\sqrt {\lambda _{n}}\) Resonance case. Now we can’t guess \(A_{n}^{p}\left ( t\right ) =z_{1}\cos \left ( \omega t\right ) +z_{2}\sin \left ( \omega t\right ) \) so we have to use \[ A_{n}^{p}\left ( t\right ) =z_{1}t\cos \left ( \omega t\right ) +z_{2}t\sin \left ( \omega t\right ) \] Substituting this in \(A_{n}^{\prime \prime }\left ( t\right ) +c^{2}\lambda _{n}A_{n}\left ( t\right ) =\gamma _{n}\cos \left ( \omega t\right ) \) gives\begin {equation} \left ( z_{1}t\cos \left ( \omega t\right ) +z_{2}t\sin \left ( \omega t\right ) \right ) ^{\prime \prime }+c^{2}\lambda _{n}\left ( z_{1}t\cos \left ( \omega t\right ) +z_{2}t\sin \left ( \omega t\right ) \right ) =\gamma _{n}\cos \left ( \omega t\right ) \tag {2} \end {equation} But \begin {align*} \left ( z_{1}t\cos \left ( \omega t\right ) +z_{2}t\sin \left ( \omega t\right ) \right ) ^{\prime \prime } & =\left ( z_{1}\cos \left ( \omega t\right ) -z_{1}\omega t\sin \left ( \omega t\right ) +z_{2}\sin \left ( \omega t\right ) +z_{2}\omega t\cos \left ( \omega t\right ) \right ) ^{\prime }\\ & =-z_{1}\omega \sin \left ( \omega t\right ) -\left ( z_{1}\omega \sin \left ( \omega t\right ) +z_{1}\omega ^{2}t\cos \left ( \omega t\right ) \right ) \\ & +z_{2}\omega \cos \left ( \omega t\right ) +\left ( z_{2}\omega \cos \left ( \omega t\right ) -z_{2}\omega ^{2}t\sin \left ( \omega t\right ) \right ) \\ & =-2z_{1}\omega \sin \left ( \omega t\right ) -z_{1}\omega ^{2}t\cos \left ( \omega t\right ) +2z_{2}\omega \cos \left ( \omega t\right ) -z_{2}\omega ^{2}t\sin \left ( \omega t\right ) \end {align*}
Hence (2) becomes\[ -2z_{1}\omega \sin \left ( \omega t\right ) -z_{1}\omega ^{2}t\cos \left ( \omega t\right ) +2z_{2}\omega \cos \left ( \omega t\right ) -z_{2}\omega ^{2}t\sin \left ( \omega t\right ) +c^{2}\lambda _{n}\left ( z_{1}t\cos \left ( \omega t\right ) +z_{2}t\sin \left ( \omega t\right ) \right ) =\gamma _{n}\cos \left ( \omega t\right ) \] Comparing coefficients we see that \(2z_{2}\omega =\gamma _{n}\) or \[ z_{2}=\frac {\gamma _{n}}{2\omega }\] And \(z_{1}=0\). Therefore \[ A_{n}^{p}\left ( t\right ) =\frac {\gamma _{n}}{2\omega }t\sin \left ( \omega t\right ) \] Therefore\begin {align*} A_{n}\left ( t\right ) & =A_{n}^{h}\left ( t\right ) +A_{n}^{p}\left ( t\right ) \\ & =c_{1_{n}}\cos \left ( c\sqrt {\lambda _{n}}t\right ) +c_{2_{n}}\sin \left ( c\sqrt {\lambda _{n}}t\right ) +\frac {\gamma _{n}}{2c\sqrt {\lambda _{n}}}t\sin \left ( \omega t\right ) \end {align*}
We now can find \(c_{1_{n}},c_{2_{n}}\) from initial conditions.\begin {align} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }A_{n}\left ( t\right ) \phi _{n}\left ( x\right ) \nonumber \\ & =\sum _{n=1}^{\infty }\left ( c_{1_{n}}\cos \left ( c\sqrt {\lambda _{n}}t\right ) +c_{2_{n}}\sin \left ( c\sqrt {\lambda _{n}}t\right ) +\frac {\gamma _{n}}{2c\sqrt {\lambda _{n}}}t\sin \left ( \omega t\right ) \right ) \sin \left ( \frac {n\pi }{L}x\right ) \tag {4} \end {align}
At \(t=0\)\begin {align*} f\left ( x\right ) & =\sum _{n=1}^{\infty }c_{1_{n}}\sin \left ( \frac {n\pi }{L}x\right ) \\ c_{1n} & =\frac {2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {n\pi }{L}x\right ) dx \end {align*}
Taking time derivative of (4) and setting it to zero will give \(c_{2n}\). Since initial speed is zero then \(c_{2_{n}}=0\). Hence\[ A_{n}\left ( t\right ) =c_{1_{n}}\cos \left ( c\sqrt {\lambda _{n}}t\right ) +\frac {\gamma _{n}}{2c\sqrt {\lambda _{n}}}t\sin \left ( \omega t\right ) \] This completes the solution.
Summary of solution
The solution is given by\[ u\left ( x,t\right ) =\sum _{n=1}^{\infty }A_{n}\left ( t\right ) \phi _{n}\left ( x\right ) \] Case \(\omega \neq c\sqrt {\lambda _{n}}\)\[ A_{n}\left ( t\right ) =c_{1_{n}}\cos \left ( c\sqrt {\lambda _{n}}t\right ) +\frac {\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\cos \left ( \omega t\right ) \] And\[ c_{1_{n}}=\frac {2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {m\pi }{L}x\right ) dx-\frac {\gamma _{n}}{c^{2}\lambda _{n}-\omega ^{2}}\] And\[ \gamma _{n}=\frac {\int _{0}^{L}g\left ( x\right ) \phi _{n}\left ( x\right ) dx}{\int _{0}^{L}\phi _{n}^{2}\left ( x\right ) dx}\] And \(\lambda _{n}=\left ( \frac {n\pi }{L}\right ) ^{2},n=1,2,3,\)
Case \(\omega =c\sqrt {\lambda _{n}}\) (resonance)\[ A_{n}\left ( t\right ) =c_{1_{n}}\cos \left ( c\sqrt {\lambda _{n}}t\right ) +\frac {\gamma _{n}}{2c\sqrt {\lambda _{n}}}t\sin \left ( \omega t\right ) \] And\[ c_{1_{n}}=\frac {2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {n\pi }{L}x\right ) dx \]
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Added July 2, 2018.
Taken from Maple 2018.1 improvements to PDE’s document. Solve \[ v_{tt} = v_{xx} \] For \(t>0\) and \(0<x<1\). With boundary conditions \begin {align*} v(0,t)&=0\\ v(1,0)&=0 \end {align*}
With initial conditions \begin {align*} v( x,0) & =f(x) \\ \frac {\partial v}{\partial t}(x,0) &=g(x) \\ \end {align*}
Where \(f(x)=-{\frac {{{\rm e}^{2}}x-{{\rm e}^{x+1}}-x+{{\rm e}^{1-x}}}{{{\rm e}^{2}}-1}}\) and \(g(x)=1+{\frac {{{\rm e}^{2}}x-{{\rm e}^{x+1}}-x+{{\rm e}^{1-x}}}{{{\rm e}^{2}}-1}}\)
Mathematica ✓
ClearAll["Global`*"]; pde = D[v[x, t], {t, 2}] == D[v[x, t], {x, 2}]; bc = {v[0, t] == 0, v[1, t] == 0}; ic = {v[x, 0] == -((Exp[2]*x - Exp[x + 1] - x + Exp[1 - x])/(Exp[2] - 1)), Derivative[0, 1][v][x, 0] == 1 + (Exp[2]*x - Exp[x + 1] - x + Exp[1 - x])/(Exp[2] - 1)}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, v[x, t], {x, t}], 60*10]]; sol = sol /. K[1] -> n;
\[\left \{\left \{v(x,t)\to \underset {n=1}{\overset {\infty }{\sum }}-\frac {2 \left ((-1)^{n+1} n \pi \cos (n \pi t)+\left (\left (-1+(-1)^n\right ) \pi ^2 n^2+2 (-1)^n-1\right ) \sin (n \pi t)\right ) \sin (n \pi x)}{\pi ^4 n^4+\pi ^2 n^2}\right \}\right \}\]
Maple ✓
restart; pde := diff(v(x, t), t, t)=(diff(v(x, t), x, x)); bc := v(0, t) = 0, v(1, t) = 0; ic := v(x, 0) = -(exp(2)*x-exp(x+1)-x+exp(1-x))/(exp(2)-1), (D[2](v))(x, 0) = 1+(exp(2)*x-exp(x+1)-x+exp(1-x))/(exp(2)-1); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde,ic,bc],v(x,t))),output='realtime'));
\[v \left ( x,t \right ) =\sum _{n=1}^{\infty }-2\,{\frac {\sin \left ( n\pi \,x \right ) \left ( \left ( \left ( -1 \right ) ^{n}{\pi }^{2}{n}^{2}-{\pi }^{2}{n}^{2}+2\, \left ( -1 \right ) ^{n}-1 \right ) \sin \left ( n\pi \,t \right ) - \left ( -1 \right ) ^{n}\cos \left ( n\pi \,t \right ) \pi \,n \right ) }{{\pi }^{2}{n}^{2} \left ( {\pi }^{2}{n}^{2}+1 \right ) }}\]
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Added July 2, 2018.
Third example, from Maple 2018.1 improvements to PDE’s document. What_is_New_after_Maple_2018.pdf
Solve \[ \frac {\partial ^2 u}{\partial t^2} = c^2 \frac {\partial ^2 u}{\partial x^2} + 1 \] For \(t>0\) and \(0<x<L\). With boundary conditions \begin {align*} u(0,t)&=0\\ u(L,0)&=0 \end {align*}
With initial conditions \begin {align*} u ( x,0) & =f(x) \\ \frac {\partial u}{\partial t}(x,0) &=g(x) \\ \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], {t, 2}] == c^2*D[u[x, t], {x, 2}] + 1; bc = {u[0, t] == 0, u[L, t] == 0}; ic = {u[x, 0] == f[x], Derivative[0, 1][u][x, 0] == g[x]}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}, Assumptions -> L > 0], 60*10]];
\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}\fbox {$\sqrt {2} \sqrt {\frac {1}{L}} \sin \left (\frac {\pi x K[1]}{L}\right ) \left (\frac {\sqrt {2} \left (1+(-1)^{K[1]+1}\right ) \left (L-L \cos \left (\frac {c \pi t K[1]}{L}\right )\right ) L^{3/2}}{c^2 \pi ^3 K[1]^3}+\frac {\left (\int _0^L \frac {\sqrt {2} g(x) \sin \left (\frac {\pi x K[1]}{L}\right )}{\sqrt {L}} \, dx\right ) \sin \left (\pi t \sqrt {\frac {c^2 K[1]^2}{L^2}}\right ) L}{\pi | c| | K[1]| }+\cos \left (\pi t \sqrt {\frac {c^2 K[1]^2}{L^2}}\right ) \int _0^L \frac {\sqrt {2} f(x) \sin \left (\frac {\pi x K[1]}{L}\right )}{\sqrt {L}} \, dx\right )\text { if }\left (t\left |\frac {1}{\sqrt {c^2 K[1]^2}}\right .\right )\in \mathbb {R}$} & K[1]\in \mathbb {Z}\land ((L<0\land ((c<0\land K[1]\geq 1)\lor (c>0\land K[1]\geq 1)))\lor (L>0\land ((c<0\land K[1]\geq 1)\lor (c>0\land K[1]\geq 1)))) \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
Maple ✓
restart; interface(showassumed=0); f='f'; pde :=diff(u(x, t), t, t) = c^2* diff(u(x, t), x, x) + 1; bc := u(0, t) = 0, u(L, t) = 0; ic := u(x, 0) = f(x), (D[2](u))(x, 0) = g(x); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde, ic, bc],u(x,t)) assuming L>0),output='realtime'));
\[u \left ( x,t \right ) =\sum _{n=1}^{\infty } \left ( {\frac {1}{\pi \,n{c}^{2}L}\sin \left ( {\frac {n\pi \,x}{L}} \right ) \left ( 2\,L\int _{0}^{L}\!\sin \left ( {\frac {n\pi \,x}{L}} \right ) g \left ( x \right ) \,{\rm d}x\sin \left ( {\frac {cn\pi \,t}{L}} \right ) c-\pi \,\int _{0}^{L}\!\sin \left ( {\frac {n\pi \,x}{L}} \right ) \left ( -2\,f \left ( x \right ) {c}^{2}+Lx-{x}^{2} \right ) \,{\rm d}x\cos \left ( {\frac {cn\pi \,t}{L}} \right ) n \right ) } \right ) -{\frac {{x}^{2}}{2\,{c}^{2}}}+{\frac {Lx}{2\,{c}^{2}}}\]
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This is problem at page 213, David J Logan textbook, applied PDE textbook. Both ends fixed end, with source.
\[ u_{tt} = c^2 u_{xx}+ A x \] With boundary conditions \begin {align*} u(L,0) &=0 \\ u(0,t) &= 0 \end {align*}
With initial conditions \begin {align*} \frac {\partial u}{\partial t}(x,0) &=0 \\ u(x,0) &= 0 \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], {t, 2}] == c^2*D[u[x, t], {x, 2}] + A*x; bc = {u[0, t] == 0, u[L, t] == 0}; ic = {u[x, 0] == 0, Derivative[0, 1][u][x, 0] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}], 60*10]];
\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} -\frac {i A L^3 \left (\text {PolyLog}\left (3,-e^{-\frac {i \pi (c t-x)}{L}}\right )-\text {PolyLog}\left (3,-e^{\frac {i \pi (c t-x)}{L}}\right )+2 \text {PolyLog}\left (3,-e^{-\frac {i \pi x}{L}}\right )-2 \text {PolyLog}\left (3,-e^{\frac {i \pi x}{L}}\right )-\text {PolyLog}\left (3,-e^{-\frac {i \pi (c t+x)}{L}}\right )+\text {PolyLog}\left (3,-e^{\frac {i \pi (c t+x)}{L}}\right )\right )}{2 c^2 \pi ^3} & K[1]\in \mathbb {Z}\land c\neq 0\land L\neq 0\land K[1]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
Maple ✓
restart; interface(showassumed=0); pde := diff(u(x,t),t$2)=c^2*diff(u(x,t),x$2)+A* x; bc := u(0,t)=0,u(L,t)=0; ic := u(x,0)=0,D[2](u)(x,0)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde,bc,ic],u(x,t)) assuming L>0),output='realtime'));
\[u \left ( x,t \right ) =\sum _{n=1}^{\infty }2\,{\frac {{L}^{3} \left ( -1 \right ) ^{n}A}{{n}^{3}{\pi }^{3}{c}^{2}}\sin \left ( {\frac {n\pi \,x}{L}} \right ) \cos \left ( {\frac {cn\pi \,t}{L}} \right ) }-{\frac {A{x}^{3}}{6\,{c}^{2}}}+{\frac {A{L}^{2}x}{6\,{c}^{2}}}\]
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Both ends fixed with damping Solve \[ u_{tt} + 2 u_t = c^2 u_{xx} \] With boundary conditions \begin {align*} u(0,t) &= 0\\ u(\pi ,0) &=0 \end {align*}
With initial conditions \begin {align*} \frac {\partial u}{\partial t}(x,0) &=0 \\ u(x,0) &= f(x) \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], {t, 2}] + 2*D[u[x, t], t] == D[u[x, t], {x, 2}]; bc = {u[0, t] == 0, u[Pi, t] == 0}; ic = {Derivative[0, 1][u][x, 0] == 0, u[x, 0] == f[x]}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], x, t], 60*10]];
\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}e^{-t} \sqrt {\frac {2}{\pi }} \sin (x K[1]) \left (\cos \left (\frac {1}{2} t \sqrt {4 K[1]^2-4}\right ) \int _0^{\pi } \sqrt {\frac {2}{\pi }} f(x) \sin (x K[1]) \, dx+\frac {\sin \left (\frac {1}{2} t \sqrt {4 K[1]^2-4}\right ) \int _0^{\pi } \sqrt {\frac {2}{\pi }} f(x) \sin (x K[1]) \, dx}{\sqrt {K[1]^2-1}}\right ) & K[1]\in \mathbb {Z}\land K[1]\geq 2 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
Maple ✓
restart; interface(showassumed=0); pde := diff(u(x,t),t$2)+2*diff(u(x,t),t)=diff(u(x,t),x$2); ic :=D[2](u)(x,0)=0,u(0,t)=0,u(x,0)=f(x); bc := u(0,t)=0,u(Pi,t)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde,ic,bc],u(x,t)) assuming t>0),output='realtime'));
\[u \left ( x,t \right ) =\sum _{n=1}^{\infty } \left ( \begin {cases}2\,{\frac {\sin \left ( x \right ) \int _{0}^{\pi }\!\sin \left ( x \right ) f \left ( x \right ) \,{\rm d}x{{\rm e}^{-t}} \left ( t+1 \right ) }{\pi }} & n=1\\{\frac {\sin \left ( nx \right ) \int _{0}^{\pi }\!\sin \left ( nx \right ) f \left ( x \right ) \,{\rm d}x}{\pi } \left ( {{\rm e}^{- \left ( \sqrt {-{n}^{2}+1}+1 \right ) t}}\sqrt {-{n}^{2}+1}+{{\rm e}^{ \left ( -1+\sqrt {-{n}^{2}+1} \right ) t}}\sqrt {-{n}^{2}+1}-{{\rm e}^{- \left ( \sqrt {-{n}^{2}+1}+1 \right ) t}}+{{\rm e}^{ \left ( -1+\sqrt {-{n}^{2}+1} \right ) t}} \right ) {\frac {1}{\sqrt {-{n}^{2}+1}}}} & \text {otherwise}\end {cases} \right ) \]
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Added July 12, 2019.
Solve \[ u_{tt}+ \gamma ^2 u(x,t) = c^2 u_{xx} \] Dispersion term \(\gamma ^2 u(x,t)\) causes the shape of the original wave to distort with time. With \(0<x<L\) and \(t>0\) and with boundary conditions \begin {align*} u(0,t) &= 0\\ u(L,0) &=0 \end {align*}
With initial conditions \begin {align*} \frac {\partial u}{\partial t}(x,0) &=0 \\ u(x,0) &= f(x) \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], {t, 2}] + gamma^2*u[x, t] == c^2 D[u[x, t], {x, 2}]; bc = {u[0, t] == 0, u[L, t] == 0}; ic = {Derivative[0, 1][u][x, 0] == 0, u[x, 0] == f[x]}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t},Assumptions->L>0], 60*10]];
\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}\sqrt {2} \sqrt {\frac {1}{L}} \cos \left (t \sqrt {c^2 \left (\frac {\gamma ^2}{c^2}+\frac {\pi ^2 K[1]^2}{L^2}\right )}\right ) \left (\int _0^L \frac {\sqrt {2} f(x) \sin \left (\frac {\pi x K[1]}{L}\right )}{\sqrt {L}} \, dx\right ) \sin \left (\frac {\pi x K[1]}{L}\right ) & K[1]\in \mathbb {Z}\land \gamma \in \mathbb {R}\land ((L<0\land ((c<0\land K[1]\geq 1)\lor (c>0\land K[1]\geq 1)))\lor (L>0\land ((c<0\land K[1]\geq 1)\lor (c>0\land K[1]\geq 1)))) \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\] Due to adding dispersion term
Maple ✓
restart; interface(showassumed=0); pde :=diff(u(x,t),t$2)+gamma^2*u(x,t)=c^2*diff(u(x,t),x$2); bc := u(0,t)=0,u(L,t)=0; ic := u(x,0)=f(x),D[2](u)(x,0)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(x,t)) assuming L>0),output='realtime'));
\[u \left ( x,t \right ) =\sum _{n=1}^{\infty } \left ( 2\,{\frac {1}{L}\sin \left ( {\frac {n\pi \,x}{L}} \right ) \cos \left ( {\frac {\sqrt {{\pi }^{2}{c}^{2}{n}^{2}+{L}^{2}{\gamma }^{2}}t}{L}} \right ) \int _{0}^{L}\!\sin \left ( {\frac {n\pi \,x}{L}} \right ) f \left ( x \right ) \,{\rm d}x} \right ) \]
Hand solution
Solving for \(t>0,0<x<L\)\[ \frac {\partial ^{2}u}{\partial t^{2}}+\gamma ^{2}u=c^{2}\frac {\partial ^{2}u}{\partial x^{2}}\qquad 0<x<L,t>0 \] With BC \begin {align*} u\left ( 0,t\right ) & =0\\ u\left ( L,t\right ) & =0 \end {align*}
And initial conditions\begin {align*} u\left ( x,0\right ) & =f\left ( x\right ) \\ u_{t}\left ( x,0\right ) & =0 \end {align*}
Dispersion term \(\gamma ^{2}u\,\) causes the shape of the original wave to distort with time. Using separation of variables, Let \(u=X\left ( x\right ) T\left ( t\right ) .\) Substituting this back in the PDE gives\begin {align*} T^{\prime \prime }X+\gamma ^{2}XT & =c^{2}X^{\prime \prime }T\\ \frac {1}{c^{2}}\left ( \frac {T^{\prime \prime }}{T}+\gamma ^{2}\right ) & =\frac {X^{\prime \prime }}{X}=-\lambda \end {align*}
The eigenvalue ODE is\begin {align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X\left ( L\right ) & =0 \end {align*}
The eigenvalues are \(\lambda _{n}=\left ( \frac {n\pi }{L}\right ) ^{2},n=1,2,3,\cdots \) and the eigenfunctions are \(X_{n}\left ( x\right ) =c_{n}\sin \left ( \frac {n\pi }{L}x\right ) \). The time ODE becomes\[ T^{\prime \prime }+\left ( \gamma ^{2}+c^{2}\lambda _{n}\right ) T=0 \] The solution is\[ T_{n}\left ( t\right ) =A_{n}\cos \left ( \sqrt {\gamma ^{2}+c^{2}\lambda _{n}}t\right ) +B_{n}\sin \left ( \sqrt {\gamma ^{2}+c^{2}\lambda _{n}}t\right ) \] Taking time derivatives gives\[ T_{n}^{\prime }\left ( t\right ) =-\sqrt {\gamma ^{2}+c^{2}\lambda _{n}}A_{n}\sin \left ( \sqrt {\gamma ^{2}+c^{2}\lambda _{n}}t\right ) +B_{n}\sqrt {\gamma ^{2}+c^{2}\lambda _{n}}\cos \left ( \sqrt {\gamma ^{2}+c^{2}\lambda _{n}}t\right ) \] At time \(t=0\), the above is zero (initial velocity is zero), which gives\[ 0=B_{n}\sqrt {\gamma ^{2}+c^{2}\lambda _{n}}\] Hence \(B_{n}=0\) and the time ODE solution becomes\[ T_{n}\left ( t\right ) =A_{n}\cos \left ( \sqrt {\gamma ^{2}+c^{2}\lambda _{n}}t\right ) \] Hence the fundamental solution is\begin {align*} u_{n}\left ( x,t\right ) & =T_{n}X_{n}\\ & =c_{n}\cos \left ( \sqrt {\gamma ^{2}+c^{2}\left ( \frac {n\pi }{L}\right ) ^{2}}t\right ) \sin \left ( \frac {n\pi }{L}x\right ) \end {align*}
Therefore the solution is\[ u\left ( x,t\right ) =\sum _{n=1}^{\infty }c_{n}\cos \left ( \frac {1}{L}\sqrt {\left ( L^{2}\gamma ^{2}+c^{2}n^{2}\pi ^{2}\right ) }t\right ) \sin \left ( \frac {n\pi }{L}x\right ) \] \(c_{n}\) is found from initial position. At \(t=0\) the above becomes\[ f\left ( x\right ) =\sum _{n=1}^{\infty }c_{n}\sin \left ( \frac {n\pi }{L}x\right ) \] Applying orthogonality gives\begin {align*} \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {n\pi }{L}x\right ) dx & =c_{n}\frac {L}{2}\\ c_{n} & =\frac {2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {n\pi }{L}x\right ) dx \end {align*}
Hence solution is\[ u\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) \cos \left ( \frac {1}{L}\sqrt {\left ( L^{2}\gamma ^{2}+c^{2}n^{2}\pi ^{2}\right ) }t\right ) \sin \left ( \frac {n\pi }{L}x\right ) \]
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Solve \[ u_{tt}+ \gamma ^2 u(x,t) = c^2 u_{xx} \] Dispersion term \(\gamma ^2 u(x,t)\) causes the shape of the original wave to distort with time. With \(0<x<\pi \) and \(t>0\) and with boundary conditions \begin {align*} u(0,t) &= 0\\ u(\pi ,0) &=0 \end {align*}
With initial conditions \begin {align*} \frac {\partial u}{\partial t}(x,0) &=0 \\ u(x,0) &= \sin ^2(x) \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], {t, 2}] + gamma^2*u[x, t] == c^2*D[u[x, t], {x, 2}]; bc = {u[0, t] == 0, u[Pi, t] == 0}; ic = {Derivative[0, 1][u][x, 0] == 0, u[x, 0] == Sin[x]^2}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}], 60*10]];
\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}\sqrt {\frac {2}{\pi }} \cos \left (t \sqrt {c^2 \left (\frac {\gamma ^2}{c^2}+K[1]^2\right )}\right ) \left (\begin {array}{cc} \{ & \begin {array}{cc} 0 & K[1]=2 \\ \frac {2 \left (-1+(-1)^{K[1]}\right ) \sqrt {\frac {2}{\pi }}}{K[1]^3-4 K[1]} & \text {True} \\\end {array} \\\end {array}\right ) \sin (x K[1]) & K[1]\in \mathbb {Z}\land \gamma \in \mathbb {R}\land ((c<0\land K[1]\geq 1)\lor (c>0\land K[1]\geq 1)) \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\] Due to adding dispersion term
Maple ✓
restart; interface(showassumed=0); pde :=diff(u(x,t),t$2)+gamma^2*u(x,t)=c^2*diff(u(x,t),x$2); bc := u(0,t)=0,u(Pi,t)=0; ic := u(x,0)=sin(x)^2,(D[2](u))(x,0)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(x,t))),output='realtime'));
\[u \left ( x,t \right ) ={\frac {8\,\sin \left ( x \right ) }{3\,\pi }\cos \left ( \sqrt {{c}^{2}+{\gamma }^{2}}t \right ) }+\sum _{n=3}^{\infty }4\,{\frac {\sin \left ( nx \right ) \cos \left ( \sqrt {{c}^{2}{n}^{2}+{\gamma }^{2}}t \right ) \left ( \left ( -1 \right ) ^{n}-1 \right ) }{\pi \,n \left ( {n}^{2}-4 \right ) }}\]
Hand solution
Solving for \(t>0,0<x<L\)\begin {equation} \frac {\partial ^{2}u}{\partial t^{2}}+\gamma ^{2}u=\frac {\partial ^{2}u}{\partial x^{2}}\qquad 0<x<L,t>0\tag {1} \end {equation} With BC \begin {align*} u\left ( 0,t\right ) & =0\\ u\left ( L,t\right ) & =0 \end {align*}
And initial conditions\begin {align*} u\left ( x,0\right ) & =f\left ( x\right ) \\ u_{t}\left ( x,0\right ) & =0 \end {align*}
Where now \(L=\pi ,f\left ( x\right ) =\sin ^{2}\left ( x\right ) \).
The general solution for (1) was found in problem 5.1.1.15 on page 1278 as\[ u\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) \cos \left ( \frac {t}{L}\sqrt {L^{2}\gamma ^{2}+c^{2}n^{2}\pi ^{2}}\right ) \sin \left ( \frac {n\pi }{L}x\right ) \] Using the above specific values for this problem, the solution becomes\begin {align*} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }\left ( \frac {2}{\pi }\int _{0}^{\pi }f\left ( s\right ) \sin \left ( \frac {n\pi }{\pi }s\right ) ds\right ) \cos \left ( \frac {t}{\pi }\sqrt {\pi ^{2}\gamma ^{2}+c^{2}n^{2}\pi ^{2}}\right ) \sin \left ( \frac {n\pi }{\pi }x\right ) \\ & =\sum _{n=1}^{\infty }\left ( \frac {2}{\pi }\int _{0}^{\pi }f\left ( s\right ) \sin \left ( ns\right ) ds\right ) \cos \left ( \sqrt {\gamma ^{2}+c^{2}n^{2}}t\right ) \sin \left ( nx\right ) \end {align*}
But \begin {align*} \int _{0}^{\pi }\sin ^{2}\left ( s\right ) \sin \left ( ns\right ) ds & =\int _{0}^{\pi }\left ( \frac {1}{2}-\frac {1}{2}\cos \left ( 2s\right ) \right ) \sin \left ( ns\right ) ds\\ & =\int _{0}^{\pi }\frac {1}{2}\sin \left ( ns\right ) ds-\frac {1}{2}\int _{0}^{\pi }\cos \left ( 2s\right ) \sin \left ( ns\right ) ds\\ & =\frac {1-\left ( -1\right ) ^{n}}{2n}-\frac {1}{2}\int _{0}^{\pi }\cos \left ( 2s\right ) \sin \left ( ns\right ) ds \end {align*}
\(\int _{0}^{\pi }\cos \left ( 2s\right ) \sin \left ( ns\right ) ds=-\frac {2}{3}\)for \(n=1\) and \(\int _{0}^{\pi }\cos \left ( 2s\right ) \sin \left ( ns\right ) ds=0\) For \(n=2\) and \(\int _{0}^{\pi }\cos \left ( 2s\right ) \sin \left ( ns\right ) ds=\frac {\left ( 1-\left ( -1\right ) ^{n}\right ) n}{n^{2}-4}\ \)for \(n>2\). Hence \[ \int _{0}^{\pi }\sin ^{2}\left ( s\right ) \sin \left ( ns\right ) ds=\left \{ \begin {array} [c]{ccc}\frac {1-\left ( -1\right ) }{2}+\frac {1}{3}=\frac {4}{3} & & n=1\\ \frac {1-\left ( -1\right ) ^{n}}{2n}=0 & & n=2\\ \frac {1-\left ( -1\right ) ^{n}}{2n}-\frac {1}{2}\frac {\left ( 1-\left ( -1\right ) ^{n}\right ) n}{n^{2}-4}=\frac {2}{n}\frac {\left ( -1\right ) ^{n}-1}{n^{2}-4} & & n>2 \end {array} \right . \]
The solution becomes\begin {align*} u\left ( x,t\right ) & =\left ( \frac {2}{\pi }\right ) \frac {4}{3}\cos \left ( \sqrt {\gamma ^{2}+c^{2}}t\right ) \sin \left ( x\right ) +\sum _{n=3}^{\infty }\left ( \frac {2}{\pi }\frac {2}{n}\frac {\left ( -1\right ) ^{n}-1}{n^{2}-4}\right ) \cos \left ( \sqrt {\gamma ^{2}+c^{2}n^{2}}t\right ) \sin \left ( nx\right ) \\ & =\frac {8}{3\pi }\cos \left ( \sqrt {\gamma ^{2}+c^{2}}t\right ) \sin \left ( x\right ) +\frac {4}{\pi }\sum _{n=3}^{\infty }\frac {\left ( -1\right ) ^{n}-1}{n\left ( n^{2}-4\right ) }\cos \left ( \sqrt {\gamma ^{2}+c^{2}n^{2}}t\right ) \sin \left ( nx\right ) \end {align*}
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Added July 12, 2019 Solve \[ u_{tt}+ \gamma ^2 u(x,t) = c^2 u_{xx} \] Dispersion term \(\gamma ^2 u(x,t)\) causes the shape of the original wave to distort with time. With \(0<x<L\) and \(t>0\) and with boundary conditions \begin {align*} u(0,t) &= 0\\ u(\pi ,0) &=0 \end {align*}
With initial conditions \begin {align*} \frac {\partial u}{\partial t}(x,0) &=0 \\ u(x,0) &= f(x) \end {align*}
Using the following values \begin {align*} L &= 10 \\ \gamma &= \frac {1}{8}\\ f(x) &=\left \{ \begin {array} [c]{ccc}x-4 & & 4\leq x\leq 5\\ 6-x & & 5\leq x\leq 6\\ 0 & & \text {otherwise}\end {array} \right .\\ c &= 1 \end {align*}
Mathematica ✓
ClearAll["Global`*"]; gamma=1/8; c=1; L=10; f=Piecewise[{{x-4,4<=x<=5},{6-x,5<x<=6},{0,True}}]; pde = D[u[x, t], {t, 2}] + gamma^2*u[x, t] == c^2*D[u[x, t], {x, 2}]; bc = {u[0, t] == 0, u[L, t] == 0}; ic = {Derivative[0, 1][u][x, 0] == 0, u[x, 0] == f}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}], 60*10]];
\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}-\frac {20 \cos \left (t \sqrt {\frac {1}{100} \pi ^2 K[1]^2+\frac {1}{64}}\right ) \left (\sin \left (\frac {2}{5} \pi K[1]\right )-2 \sin \left (\frac {1}{2} \pi K[1]\right )+\sin \left (\frac {3}{5} \pi K[1]\right )\right ) \sin \left (\frac {1}{10} \pi x K[1]\right )}{\pi ^2 K[1]^2} & K[1]\in \mathbb {Z}\land K[1]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\] Due to adding dispersion term
Maple ✓
restart; local gamma; gamma:=1/8; c:=1; L:=10; f:=piecewise(4<=x and x<=5, x-4,5<x and x<=6,6-x,true,0); pde :=diff(u(x,t),t$2)+gamma^2*u(x,t)=c^2*diff(u(x,t),x$2); bc := u(0,t)=0,u(L,t)=0; ic := u(x,0)=f,(D[2](u))(x,0)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(x,t))),output='realtime'));
\[u \left ( x,t \right ) =\sum _{n=1}^{\infty }-40\,{\frac { \left ( -1+\cos \left ( 1/2\,n\pi \right ) +2\,\cos \left ( 3/10\,n\pi \right ) +2\,\cos \left ( 1/10\,n\pi \right ) -2\,\cos \left ( 2/5\,n\pi \right ) -2\,\cos \left ( 1/5\,n\pi \right ) \right ) \sin \left ( 1/10\,n\pi \right ) \sin \left ( 1/10\,n\pi \,x \right ) \cos \left ( 1/40\,\sqrt {16\,{\pi }^{2}{n}^{2}+25}t \right ) }{{\pi }^{2}{n}^{2}}}\]
Hand solution
Solve for \(0<x<L,t>0\)\[ \frac {\partial ^{2}u}{\partial t^{2}}+\gamma ^{2}u=c^{2}\frac {\partial ^{2}u}{\partial x^{2}}\] Boundary conditions, \(t>0\) (both ends fixed)\begin {align*} u\left ( 0,t\right ) & =0\\ u\left ( L,t\right ) & =0 \end {align*}
Initial conditions, \(t=0\)\begin {align*} \frac {\partial u\left ( x,0\right ) }{\partial t} & =0\\ u\left ( x,0\right ) & =f\left ( x\right ) \end {align*}
Using \(L=10,\gamma =1/8,c=1\) and initial position \[ f\left ( x\right ) =\left \{ \begin {array} [c]{ccc}x-4 & & 4\leq x\leq 5\\ 6-x & & 5\leq x\leq 6\\ 0 & & \text {otherwise}\end {array} \right . \]
The general solution to the above PDE was given in problem 5.1.1.15 on page 1278 as
\[ u\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {n\pi }{L}s\right ) ds\right ) \cos \left ( \frac {1}{L}\sqrt {\left ( L^{2}\gamma ^{2}+c^{2}n^{2}\pi ^{2}\right ) }t\right ) \sin \left ( \frac {n\pi }{L}x\right ) \]
Replacing given values in the above solution results in
\begin {align} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }\left ( \frac {2}{10}\int _{0}^{10}f\left ( s\right ) \sin \left ( \frac {n\pi }{10}s\right ) ds\right ) \cos \left ( \frac {1}{10}\sqrt {\left ( 100\left ( \frac {1}{8}\right ) ^{2}+n^{2}\pi ^{2}\right ) }t\right ) \sin \left ( \frac {n\pi }{10}x\right ) \nonumber \\ & =\frac {2}{10}\sum _{n=1}^{\infty }\left ( \int _{0}^{10}f\left ( s\right ) \sin \left ( \frac {n\pi }{10}s\right ) ds\right ) \cos \left ( \frac {1}{10}\sqrt {\frac {1}{16}\left ( \frac {100}{4}+16n^{2}\pi ^{2}\right ) }t\right ) \sin \left ( \frac {n\pi }{10}x\right ) \tag {1}\\ & =\frac {2}{10}\sum _{n=1}^{\infty }\left ( \int _{0}^{10}f\left ( s\right ) \sin \left ( \frac {n\pi }{10}s\right ) ds\right ) \cos \left ( \frac {1}{40}\sqrt {25+16n^{2}\pi ^{2}}t\right ) \sin \left ( \frac {n\pi }{10}x\right ) \end {align}
But
\[ \int _{0}^{10}f\left ( x\right ) \sin \left ( \frac {n\pi }{10}x\right ) ds=\frac {100\left ( 2\sin \left ( \frac {n\pi }{2}\right ) -\sin \left ( \frac {2n\pi }{5}\right ) -\sin \left ( \frac {3n\pi }{5}\right ) \right ) }{n^{2}\pi ^{2}}\]
Hence the solution (1) becomes
\begin {align*} u\left ( x,t\right ) & =\frac {2}{10}\sum _{n=1}^{\infty }\frac {100\left ( 2\sin \left ( \frac {n\pi }{2}\right ) -\sin \left ( \frac {2n\pi }{5}\right ) -\sin \left ( \frac {3n\pi }{5}\right ) \right ) }{n^{2}\pi ^{2}}\cos \left ( \frac {1}{40}\sqrt {25+16n^{2}\pi ^{2}}t\right ) \sin \left ( \frac {n\pi }{10}x\right ) \\ & =\frac {20}{\pi ^{2}}\sum _{n=1}^{\infty }\frac {1}{n^{2}}\left ( 2\sin \left ( \frac {n\pi }{2}\right ) -\sin \left ( \frac {2n\pi }{5}\right ) -\sin \left ( \frac {3n\pi }{5}\right ) \right ) \cos \left ( \frac {1}{40}\sqrt {25+16n^{2}\pi ^{2}}t\right ) \sin \left ( \frac {n\pi }{10}x\right ) \end {align*}
Animation is below. The left one uses \(\gamma =\frac {1}{8}\) and the right one uses larger value of \(\gamma =\frac {5}{8}\) in order to show the effect of larger dispersion.
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Source code used for the above
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Added March 9, 2018. Solve \[ u_{tt} = 4 u_{xx} \] With boundary conditions \begin {align*} u(0,t) &= 0\\ u(\pi ,0) &=0 \end {align*}
With initial conditions \begin {align*} \frac {\partial u}{\partial t}(x,0) &=0 \\ u(x,0) &= \sin ^2(x) \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], {t, 2}] == 4*D[u[x, t], {x, 2}]; ic = {Derivative[0, 1][u][x, 0] == 0, u[x, 0] == Sin[x]^2}; bc = {u[0, t] == 0, u[Pi, t] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}], 60*10]]; sol = sol /. K[1] -> n;
\[\left \{\left \{u(x,t)\to \underset {n=1}{\overset {\infty }{\sum }}\frac {4 (\cos (n \pi )-1) \cos (2 n t) \sin (n x)}{\left (n^3-4 n\right ) \pi }\right \}\right \}\] But sum should not include \(n=2\)
Maple ✓
restart; interface(showassumed=0); pde := diff(u(x,t),t$2)= 4*diff(u(x,t),x$2); bc := u(0,t)=0,u(Pi,t)=0; ic := u(x,0)=sin(x)^2,D[2](u)(x,0)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc,ic],u(x,t))),output='realtime'));
\[u \left ( x,t \right ) ={\frac {8\,\sin \left ( x \right ) \cos \left ( 2\,t \right ) }{3\,\pi }}+\sum _{n=3}^{\infty }4\,{\frac {\sin \left ( nx \right ) \cos \left ( 2\,nt \right ) \left ( \left ( -1 \right ) ^{n}-1 \right ) }{n\pi \, \left ( {n}^{2}-4 \right ) }}\] Handled \(n=2\) case correctly
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Added December 20, 2018.
Example 18, Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018
Solve for \(u(x,t)\) with \(0<x<1\) and \(t>0\) \[ \frac {\partial ^2 u}{\partial t^2} = \frac {\partial ^2 u}{\partial x^2} + x e^{-t} \] With boundary conditions \begin {align*} u(0,t) &= 0\\ u(1,0) &=0 \end {align*}
With initial conditions \begin {align*} u(x,0) &= 0\\ \frac {\partial u}{\partial t}(x,0) &=1 \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], {t, 2}] == D[u[x, t], {x, 2}] + x*Exp[-t]; ic = {u[x, 0] == 0, Derivative[0, 1][u][x, 0] == 0}; bc = {u[0, t] == 0, u[1, t] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}], 60*10]];
\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {e^{-i ((-i+\pi ) t+\pi x)} \left (i e^{2 i \pi t+t} \pi ^2 (i+\pi ) \, _2F_1\left (1,\frac {-i+\pi }{\pi };2-\frac {i}{\pi };-e^{i \pi (t-x)}\right )+e^{i \pi t} (1-i \pi ) \pi ^2 \, _2F_1\left (1,\frac {-i+\pi }{\pi };2-\frac {i}{\pi };-e^{-i \pi x}\right )+i e^{i \pi (t+2 x)} \pi ^3 \, _2F_1\left (1,\frac {-i+\pi }{\pi };2-\frac {i}{\pi };-e^{i \pi x}\right )-e^{i \pi (t+2 x)} \pi ^2 \, _2F_1\left (1,\frac {-i+\pi }{\pi };2-\frac {i}{\pi };-e^{i \pi x}\right )-i e^{2 i \pi t+t+2 i \pi x} \pi ^3 \, _2F_1\left (1,\frac {-i+\pi }{\pi };2-\frac {i}{\pi };-e^{i \pi (t+x)}\right )+e^{2 i \pi t+t+2 i \pi x} \pi ^2 \, _2F_1\left (1,\frac {-i+\pi }{\pi };2-\frac {i}{\pi };-e^{i \pi (t+x)}\right )-i e^{t+2 i \pi x} \pi ^3 \, _2F_1\left (1,\frac {i+\pi }{\pi };2+\frac {i}{\pi };-e^{-i \pi (t-x)}\right )-e^{t+2 i \pi x} \pi ^2 \, _2F_1\left (1,\frac {i+\pi }{\pi };2+\frac {i}{\pi };-e^{-i \pi (t-x)}\right )-i e^{i \pi t} \pi ^3 \, _2F_1\left (1,\frac {i+\pi }{\pi };2+\frac {i}{\pi };-e^{-i \pi x}\right )-e^{i \pi t} \pi ^2 \, _2F_1\left (1,\frac {i+\pi }{\pi };2+\frac {i}{\pi };-e^{-i \pi x}\right )+i e^{i \pi (t+2 x)} \pi ^3 \, _2F_1\left (1,\frac {i+\pi }{\pi };2+\frac {i}{\pi };-e^{i \pi x}\right )+e^{i \pi (t+2 x)} \pi ^2 \, _2F_1\left (1,\frac {i+\pi }{\pi };2+\frac {i}{\pi };-e^{i \pi x}\right )+i e^t \pi ^3 \, _2F_1\left (1,\frac {i+\pi }{\pi };2+\frac {i}{\pi };-e^{-i \pi (t+x)}\right )+e^t \pi ^2 \, _2F_1\left (1,\frac {i+\pi }{\pi };2+\frac {i}{\pi };-e^{-i \pi (t+x)}\right )+i e^{i \pi t+t+i \pi x} \pi ^3 \log \left (1+e^{-i \pi (t-x)}\right )+i e^{i \pi t+t+i \pi x} \pi \log \left (1+e^{-i \pi (t-x)}\right )-i e^{i \pi t+t+i \pi x} \pi ^3 \log \left (1+e^{i \pi (t-x)}\right )-i e^{i \pi t+t+i \pi x} \pi \log \left (1+e^{i \pi (t-x)}\right )+2 i e^{i \pi (t+x)} \pi ^3 \log \left (1+e^{-i \pi x}\right )+2 i e^{i \pi (t+x)} \pi \log \left (1+e^{-i \pi x}\right )-2 i e^{i \pi (t+x)} \pi ^3 \log \left (1+e^{i \pi x}\right )-2 i e^{i \pi (t+x)} \pi \log \left (1+e^{i \pi x}\right )-i e^{i \pi t+t+i \pi x} \pi ^3 \log \left (1+e^{-i \pi (t+x)}\right )-i e^{i \pi t+t+i \pi x} \pi \log \left (1+e^{-i \pi (t+x)}\right )+i e^{i \pi t+t+i \pi x} \pi ^3 \log \left (1+e^{i \pi (t+x)}\right )+i e^{i \pi t+t+i \pi x} \pi \log \left (1+e^{i \pi (t+x)}\right )-e^{i \pi t+t+i \pi x} \text {PolyLog}\left (2,-e^{-i \pi (t-x)}\right )-e^{i \pi t+t+i \pi x} \pi ^2 \text {PolyLog}\left (2,-e^{-i \pi (t-x)}\right )-e^{i \pi t+t+i \pi x} \text {PolyLog}\left (2,-e^{i \pi (t-x)}\right )-e^{i \pi t+t+i \pi x} \pi ^2 \text {PolyLog}\left (2,-e^{i \pi (t-x)}\right )+e^{i \pi t+t+i \pi x} \text {PolyLog}\left (2,-e^{-i \pi (t+x)}\right )+e^{i \pi t+t+i \pi x} \pi ^2 \text {PolyLog}\left (2,-e^{-i \pi (t+x)}\right )+e^{i \pi t+t+i \pi x} \text {PolyLog}\left (2,-e^{i \pi (t+x)}\right )+e^{i \pi t+t+i \pi x} \pi ^2 \text {PolyLog}\left (2,-e^{i \pi (t+x)}\right )\right )}{2 \pi ^2 \left (1+\pi ^2\right )} & K[1]\in \mathbb {Z}\land K[1]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
Maple ✓
restart; pde := diff(u(x, t), t$2) = diff(u(x, t), x$2)+x*exp(-t); bc := u(0,t)=0,u(1,t)=0; ic := u(x,0)=0,eval(diff(u(x,t),t),t=0)=1; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc,ic],u(x,t))),output='realtime'));
\[u \left ( x,t \right ) =\sum _{n=1}^{\infty }-2\,{\frac {\sin \left ( n\pi \,x \right ) \left ( \left ( \left ( -1 \right ) ^{n}{\pi }^{2}{n}^{2}-{\pi }^{2}{n}^{2}+2\, \left ( -1 \right ) ^{n}-1 \right ) \sin \left ( n\pi \,t \right ) + \left ( -1 \right ) ^{n}\pi \,n \left ( {{\rm e}^{-t}}-\cos \left ( n\pi \,t \right ) \right ) \right ) }{{\pi }^{2}{n}^{2} \left ( {\pi }^{2}{n}^{2}+1 \right ) }}\]
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Added July 8, 2019 \[ u_{tt} = c^2 u_{xx} \] With boundary conditions \begin {align*} u(0,t) &= 0\\ u_x(L,t) &=0 \end {align*}
With initial conditions \begin {align*} u(x,0) &= f(x)\\ u_t(x,0) &= g(x) \\ \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], {t, 2}] == c^2*D[u[x, t], {x, 2}]; bc = {u[0, t] == 0, Derivative[1, 0][u][L, t] == 0}; ic = {Derivative[0, 1][u][x, 0] == g[x], u[x, 0] == f[x]}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}, Assumptions -> L>0], 60*10]];
\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {\sqrt {2} \sin \left (\frac {\pi x (2 K[1]-1)}{2 L}\right ) \left (\cos \left (\frac {1}{2} \pi t \sqrt {\frac {c^2 (2 K[1]-1)^2}{L^2}}\right ) \int _0^L \frac {\sqrt {2} f(x) \sin \left (\frac {\pi x (2 K[1]-1)}{2 L}\right )}{\sqrt {L}} \, dx+\frac {2 L \left (\int _0^L \frac {\sqrt {2} g(x) \sin \left (\frac {\pi x (2 K[1]-1)}{2 L}\right )}{\sqrt {L}} \, dx\right ) \sin \left (\frac {1}{2} \pi t \sqrt {\frac {c^2 (2 K[1]-1)^2}{L^2}}\right )}{\pi | c| | 1-2 K[1]| }\right )}{\sqrt {L}} & K[1]\in \mathbb {Z}\land ((L<0\land ((c<0\land K[1]\geq 1)\lor (c>0\land K[1]\geq 1)))\lor (L>0\land ((c<0\land K[1]\geq 1)\lor (c>0\land K[1]\geq 1)))) \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
Maple ✓
restart; interface(showassumed=0); pde := diff(u(x,t),t$2)=c^2*diff(u(x,t),x$2); bc := u(0,t)=0,D[1](u)(L,t)=0; ic := D[2](u)(x,0)=g(x),u(x,0)=f(x); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc,ic],u(x,t)) assuming L>0),output='realtime'));
\[u \left ( x,t \right ) =\sum _{n=0}^{\infty } \left ( 4\,{\frac {1}{\pi \,c \left ( 1+2\,n \right ) L} \left ( \cos \left ( 1/2\,{\frac {\pi \,c \left ( 1+2\,n \right ) t}{L}} \right ) \pi \,c \left ( n+1/2 \right ) \int _{0}^{L}\!\sin \left ( 1/2\,{\frac { \left ( 1+2\,n \right ) \pi \,x}{L}} \right ) f \left ( x \right ) \,{\rm d}x+\int _{0}^{L}\!\sin \left ( 1/2\,{\frac { \left ( 1+2\,n \right ) \pi \,x}{L}} \right ) g \left ( x \right ) \,{\rm d}x\sin \left ( 1/2\,{\frac {\pi \,c \left ( 1+2\,n \right ) t}{L}} \right ) L \right ) \sin \left ( 1/2\,{\frac { \left ( 1+2\,n \right ) \pi \,x}{L}} \right ) } \right ) \]
Hand solution
Solving for \(0<x<L\)\[ \frac {\partial ^{2}u}{\partial t^{2}}=c^{2}\frac {\partial ^{2}u}{\partial x^{2}}\qquad 0<x<L,t>0 \] Boundary conditions, \(t>0\)\begin {align*} u\left ( 0,t\right ) & =0\\ \left . \frac {\partial u}{\partial x}\right \vert _{x=L} & =0 \end {align*}
Initial conditions, \(t=0\)\begin {align*} u_{t}\left ( x,0\right ) & =g\left ( x\right ) \\ u\left ( x,0\right ) & =f\left ( x\right ) \end {align*}
Separation of variables gives the eigenvalue ODE\begin {align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X^{\prime }\left ( L\right ) & =0 \end {align*}
Only \(\lambda >0\) gives non-trivial solution from the nature of the boundary conditions. Hence solution is\[ X\left ( x\right ) =A\cos \left ( \sqrt {\lambda }x\right ) +B\sin \left ( \sqrt {\lambda }x\right ) \] Since \(X\left ( 0\right ) =0\) then the above gives \(0=A\) and the solution becomes\[ X\left ( x\right ) =B\sin \left ( \sqrt {\lambda }x\right ) \] Taking derivatives\[ X^{\prime }\left ( x\right ) =\sqrt {\lambda }B\cos \left ( \sqrt {\lambda }x\right ) \] Since \(X^{\prime }\left ( L\right ) =0\) the above becomes\[ 0=\sqrt {\lambda }B\cos \left ( \sqrt {\lambda }L\right ) \] Which implies \(\sqrt {\lambda }L=\frac {n\pi }{2}\) for \(n=1,3,5,\cdots \) or \[ \lambda _{n}=\left ( \frac {n\pi }{2L}\right ) ^{2}\qquad n=1,3,5,\cdots \] Hence the eigenfunctions are\[ \Phi _{n}\left ( x\right ) =\sin \left ( \sqrt {\lambda _{n}}x\right ) \qquad n=1,3,5,\cdots \]
The time ODE now becomes\[ T^{\prime \prime }+c^{2}\left ( \frac {n\pi }{2L}\right ) ^{2}T=0 \] Which has the solution\[ T\left ( t\right ) =D_{n}\cos \left ( c\frac {n\pi }{2L}t\right ) +E_{n}\sin \left ( c\frac {n\pi }{2L}t\right ) \] Therefore the complete solution becomes\begin {equation} u\left ( x,t\right ) =\sum _{n=1,3,5,\cdots }^{\infty }\left ( D_{n}\cos \left ( c\frac {n\pi }{2L}t\right ) +E_{n}\sin \left ( c\frac {n\pi }{2L}t\right ) \right ) \Phi _{n}\left ( x\right ) \tag {1} \end {equation} At \(t=0\) the above becomes\[ f\left ( x\right ) =\sum _{n=1,3,5,\cdots }^{\infty }D_{n}\Phi _{n}\left ( x\right ) \] Applying orthogonality gives\begin {align*} \int _{0}^{L}f\left ( x\right ) \Phi _{n}\left ( x\right ) dx & =D_{n}\int _{0}^{L}\Phi _{n}^{2}\left ( x\right ) dx\\ & =\frac {L}{2}D_{n} \end {align*}
Hence\begin {equation} D_{n}=\frac {2}{L}\int _{0}^{L}f\left ( x\right ) \Phi _{n}\left ( x\right ) dx\tag {2} \end {equation} Taking time derivative of (1) gives\[ u_{t}\left ( x,t\right ) =\sum _{n=1,3,5,\cdots }^{\infty }\left ( -c\frac {n\pi }{2L}D_{n}\sin \left ( c\frac {n\pi }{2L}t\right ) +E_{n}c\frac {n\pi }{L}\cos \left ( c\frac {n\pi }{2L}t\right ) \right ) \Phi _{n}\left ( x\right ) \] At \(t=0\) the above becomes\[ g\left ( x\right ) =\sum _{n=1,3,5,\cdots }^{\infty }E_{n}c\frac {n\pi }{2L}\Phi _{n}\left ( x\right ) \] Applying orthogonality gives\begin {align*} \int _{0}^{L}g\left ( x\right ) \Phi _{n}\left ( x\right ) dx & =E_{n}c\frac {n\pi }{L}\int _{0}^{L}\Phi _{n}^{2}\left ( x\right ) dx\\ & =\frac {L}{2}E_{n}c\frac {n\pi }{2L}\\ & =\frac {1}{4}E_{n}cn\pi \end {align*}
Hence\begin {equation} E_{n}=\frac {4}{cn\pi }\int _{0}^{L}g\left ( x\right ) \Phi _{n}\left ( x\right ) dx\tag {3} \end {equation} Using (2,3) in (1) gives the final solution as\begin {align*} u\left ( x,t\right ) & =\frac {2}{L}\sum _{n=1,3,5,\cdots }^{\infty }\left ( \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {n\pi }{2L}x\right ) dx\right ) \cos \left ( c\frac {n\pi }{2L}t\right ) \sin \left ( \frac {n\pi }{2L}x\right ) \\ & +\frac {4}{c\pi }\sum _{n=1,3,5,\cdots }^{\infty }\frac {1}{n}\left ( \int _{0}^{L}g\left ( x\right ) \sin \left ( \frac {n\pi }{2L}x\right ) dx\right ) \sin \left ( c\frac {n\pi }{2L}t\right ) \sin \left ( \frac {n\pi }{2L}x\right ) \end {align*}
Or
\begin {align*} u\left ( x,t\right ) & =\frac {2}{L}\sum _{n=0}^{\infty }\left ( \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) dx\right ) \cos \left ( c\frac {\left ( 2n+1\right ) \pi }{2L}t\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) \\ & +\frac {4}{c\pi }\sum _{n=0}^{\infty }\frac {1}{\left ( 2n+1\right ) }\left ( \int _{0}^{L}g\left ( x\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) dx\right ) \sin \left ( c\frac {\left ( 2n+1\right ) \pi }{2L}t\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) \end {align*}
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Added July 8, 2019 \[ u_{tt} = c^2 u_{xx} \] With boundary conditions \begin {align*} u(0,t) &= 0\\ u_x(L,t) &=0 \end {align*}
With initial conditions \begin {align*} u(x,0) &= f(x)\\ u_t(x,0) &=0 \\ \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], {t, 2}] == c^2*D[u[x, t], {x, 2}]; bc = {u[0, t] == 0, Derivative[1, 0][u][L, t] == 0}; ic = {Derivative[0, 1][u][x, 0] == 0, u[x, 0] == f[x]}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}, Assumptions -> L>0], 60*10]];
\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {\sqrt {2} \cos \left (\frac {1}{2} \pi t \sqrt {\frac {c^2 (2 K[1]-1)^2}{L^2}}\right ) \left (\int _0^L \frac {\sqrt {2} f(x) \sin \left (\frac {\pi x (2 K[1]-1)}{2 L}\right )}{\sqrt {L}} \, dx\right ) \sin \left (\frac {\pi x (2 K[1]-1)}{2 L}\right )}{\sqrt {L}} & K[1]\in \mathbb {Z}\land ((L<0\land ((c<0\land K[1]\geq 1)\lor (c>0\land K[1]\geq 1)))\lor (L>0\land ((c<0\land K[1]\geq 1)\lor (c>0\land K[1]\geq 1)))) \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
Maple ✓
restart; interface(showassumed=0); pde := diff(u(x,t),t$2)=c^2*diff(u(x,t),x$2); bc := u(0,t)=0,D[1](u)(L,t)=0; ic := D[2](u)(x,0)=0,u(x,0)=f(x); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc,ic],u(x,t)) assuming L>0),output='realtime'));
\[u \left ( x,t \right ) =\sum _{n=0}^{\infty } \left ( 2\,{\frac {1}{L}\int _{0}^{L}\!\sin \left ( 1/2\,{\frac { \left ( 1+2\,n \right ) \pi \,x}{L}} \right ) f \left ( x \right ) \,{\rm d}x\sin \left ( 1/2\,{\frac { \left ( 1+2\,n \right ) \pi \,x}{L}} \right ) \cos \left ( 1/2\,{\frac {c \left ( 1+2\,n \right ) \pi \,t}{L}} \right ) } \right ) \]
Hand solution
Solving for \(0<x<L\)\[ u_{tt}=c^{2}u_{xx}\qquad 0<x<L,t>0 \] Boundary conditions, \(t>0\)\begin {align*} u\left ( 0,t\right ) & =0\\ \left . \frac {\partial u}{\partial x}\right \vert _{x=L} & =0 \end {align*}
Initial conditions, \(t=0\)\begin {align*} u_{t}\left ( x,0\right ) & =g\left ( x\right ) =0\\ u\left ( x,0\right ) & =f\left ( x\right ) \end {align*}
The general PDE was solved in 5.1.1.20 on page 1302 and the solution is
\begin {align*} u\left ( x,t\right ) & =\frac {2}{L}\sum _{n=0}^{\infty }\left ( \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) dx\right ) \cos \left ( c\frac {\left ( 2n+1\right ) \pi }{2L}t\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) \\ & +\frac {4}{c\pi }\sum _{n=0}^{\infty }\frac {1}{\left ( 2n+1\right ) }\left ( \int _{0}^{L}g\left ( x\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) dx\right ) \sin \left ( c\frac {\left ( 2n+1\right ) \pi }{2L}t\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) \end {align*}
But here \(g\left ( x\right ) =0\), hence the above reduces to
\[ u\left ( x,t\right ) =\frac {2}{L}\sum _{n=0}^{\infty }\left ( \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) dx\right ) \cos \left ( c\frac {\left ( 2n+1\right ) \pi }{2L}t\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) \]
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Added July 8, 2019 \[ u_{tt} = c^2 u_{xx} \] With boundary conditions \begin {align*} u(0,t) &= 0\\ u_x(L,t) &=0 \end {align*}
With initial conditions \begin {align*} u(x,0) &= f(x)\\ u_t(x,0) &=0 \\ \end {align*}
Using the following values \begin {align*} c &= 4\\ L &=3 \\ h &= \frac {1}{10}\\ f(x) &= \left \{ \begin {array} [c]{ccc}\frac {3h}{L}x & & 0<x<\frac {L}{3}\\ h & & \frac {L}{3}<x<L \end {array} \right . \end {align*}
Mathematica ✓
ClearAll["Global`*"]; L=3; c=4; h=1/10; f=Piecewise[{{3*h/L*x,0<x<L/3},{h,L/3<x<L}}]; pde = D[u[x, t], {t, 2}] == c^2*D[u[x, t], {x, 2}]; bc = {u[0, t] == 0, Derivative[1, 0][u][L, t] == 0}; ic = {Derivative[0, 1][u][x, 0] == 0, u[x, 0] == f}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}], 60*10]];
\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}-\frac {12 \cos \left (\frac {1}{3} \pi (K[1]+1)\right ) \cos \left (\frac {2}{3} \pi t \sqrt {(2 K[1]-1)^2}\right ) \sin \left (\frac {1}{6} \pi x (2 K[1]-1)\right )}{5 \pi ^2 (1-2 K[1])^2} & K[1]\in \mathbb {Z}\land K[1]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
Maple ✓
restart; L:=3; c:=4; h:=1/10; f:=piecewise(0<x and x<L/3,3*h/L*x,L/3<x and x<L,h); pde := diff(u(x,t),t$2)=c^2*diff(u(x,t),x$2); bc := u(0,t)=0,D[1](u)(L,t)=0; ic := D[2](u)(x,0)=0,u(x,0)=f; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc,ic],u(x,t))),output='realtime'));
\[u \left ( x,t \right ) =\sum _{n=0}^{\infty }{\frac {6}{5\, \left ( 1+2\,n \right ) ^{2}{\pi }^{2}}\sin \left ( {\frac { \left ( 1+2\,n \right ) \pi \,x}{6}} \right ) \cos \left ( {\frac { \left ( 2+4\,n \right ) t\pi }{3}} \right ) \left ( \sqrt {3}\sin \left ( {\frac {n\pi }{3}} \right ) +\cos \left ( {\frac {n\pi }{3}} \right ) \right ) }\]
Hand solution
Solving the wave PDE on string \[ u_{tt}=c^{2}u_{xx}\qquad t>0,x>0 \] Boundary conditions, \(t>0\)\begin {align*} u\left ( 0,t\right ) & =0\\ \left . \frac {\partial u}{\partial x}\right \vert _{x=L} & =0 \end {align*}
Initial conditions, \(t=0\)\begin {align*} u\left ( x,0\right ) & =f\left ( x\right ) =\left \{ \begin {array} [c]{ccc}\frac {3h}{L}x & & 0<x<\frac {L}{3}\\ h & & \frac {L}{3}<x<L \end {array} \right . \\ u_{t}\left ( x,0\right ) & =0 \end {align*}
Using \(c=4,L=3,h=\frac {1}{10}\). Hence \(f\left ( x\right ) =\left \{ \begin {array} [c]{ccc}\frac {1}{10}x & & 0<x<1\\ \frac {1}{10} & & 1<x<3 \end {array} \right . \)
The general problem PDE was solved in 5.1.1.21 on page 1306 and the solution is\[ u\left ( x,t\right ) =\frac {2}{L}\sum _{n=0}^{\infty }\left ( \int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) \cos \left ( \frac {\left ( 2n+1\right ) \pi }{2L}ct\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) \] Substituting the specific values given above into this solution gives\[ u\left ( x,t\right ) =\frac {2}{3}\sum _{n=0}^{\infty }\left ( \int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}s\right ) ds\right ) \cos \left ( 4\frac {\left ( 2n+1\right ) \pi }{6}t\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}x\right ) \] But \begin {align*} \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}x\right ) dx & =\int _{0}^{\frac {L}{3}}f\left ( x\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}x\right ) dx+\int _{\frac {L}{3}}^{L}f\left ( x\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}x\right ) dx\\ & =\frac {1}{10}\int _{0}^{1}x\sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}x\right ) dx+\frac {1}{10}\int _{1}^{3}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}x\right ) dx\\ & =\frac {18}{5\pi ^{2}\left ( 2n+1\right ) ^{2}}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) \end {align*}
Hence the solution becomes\begin {align*} u\left ( x,t\right ) & =\frac {2}{3}\sum _{n=0}^{\infty }\frac {18}{5\pi ^{2}\left ( 2n+1\right ) ^{2}}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) \cos \left ( \frac {2}{3}\left ( 2n+1\right ) \pi t\right ) \sin \left ( \frac {1}{6}\left ( 2n+1\right ) \pi x\right ) \\ & =\frac {36}{15\pi ^{2}}\sum _{n=0}^{\infty }\frac {1}{\left ( 2n+1\right ) ^{2}}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) \cos \left ( \frac {2}{3}\left ( 2n+1\right ) \pi t\right ) \sin \left ( \frac {1}{6}\left ( 2n+1\right ) \pi x\right ) \end {align*}
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Source code used for the above
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Added July 9, 2019 \[ u_{tt} + b u_t = c^2 u_{xx} \] For \(t>0\) and \(0<x<L\) and boundary conditions \begin {align*} u(0,t) &= 0\\ u_x(L,t) &=0 \end {align*}
With initial conditions \begin {align*} u(x,0) &= f(x)\\ u_t(x,0) &= 0 \\ \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], {t, 2}] + b*D[u[x,t],t] == c^2*D[u[x, t], {x, 2}]; bc = {u[0, t] == 0, Derivative[1, 0][u][L, t] == 0}; ic = {Derivative[0, 1][u][x, 0] == 0, u[x, 0] == f[x]}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}, Assumptions->{b>0,L>0}], 60*10]];
\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {\sqrt {2} e^{-\frac {b t}{2}} \sin \left (\frac {\pi x (2 K[1]-1)}{2 L}\right ) \left (\cos \left (\frac {1}{2} t \sqrt {\frac {c^2 \pi ^2 (2 K[1]-1)^2}{L^2}-b^2}\right ) \int _0^L \frac {\sqrt {2} f(x) \sin \left (\frac {\pi x (2 K[1]-1)}{2 L}\right )}{\sqrt {L}} \, dx+\frac {2 \left (\int _0^L \frac {b f(x) \sin \left (\frac {\pi x (2 K[1]-1)}{2 L}\right )}{\sqrt {2} \sqrt {L}} \, dx\right ) \sin \left (\frac {1}{2} t \sqrt {\frac {c^2 \pi ^2 (2 K[1]-1)^2}{L^2}-b^2}\right )}{\sqrt {\frac {c^2 \pi ^2 (1-2 K[1])^2}{L^2}-b^2}}\right )}{\sqrt {L}} & K[1]\in \mathbb {Z}\land \left (\left (L<0\land \left (\left (c<0\land \left (\left (b<-\sqrt {\frac {c^2}{L^2}} \pi \land K[1]>\frac {\sqrt {\frac {b^2 L^2}{c^2}}}{2 \pi }+\frac {1}{2}\right )\lor \left (b=-\sqrt {\frac {c^2}{L^2}} \pi \land K[1]\geq 2\right )\lor \left (-\sqrt {\frac {c^2}{L^2}} \pi <b<\sqrt {\frac {c^2}{L^2}} \pi \land K[1]\geq 1\right )\lor \left (b=\sqrt {\frac {c^2}{L^2}} \pi \land K[1]\geq 2\right )\lor \left (b>\sqrt {\frac {c^2}{L^2}} \pi \land K[1]>\frac {\sqrt {\frac {b^2 L^2}{c^2}}}{2 \pi }+\frac {1}{2}\right )\right )\right )\lor \left (c>0\land \left (\left (b<-\sqrt {\frac {c^2}{L^2}} \pi \land K[1]>\frac {\sqrt {\frac {b^2 L^2}{c^2}}}{2 \pi }+\frac {1}{2}\right )\lor \left (b=-\sqrt {\frac {c^2}{L^2}} \pi \land K[1]\geq 2\right )\lor \left (-\sqrt {\frac {c^2}{L^2}} \pi <b<\sqrt {\frac {c^2}{L^2}} \pi \land K[1]\geq 1\right )\lor \left (b=\sqrt {\frac {c^2}{L^2}} \pi \land K[1]\geq 2\right )\lor \left (b>\sqrt {\frac {c^2}{L^2}} \pi \land K[1]>\frac {\sqrt {\frac {b^2 L^2}{c^2}}}{2 \pi }+\frac {1}{2}\right )\right )\right )\right )\right )\lor \left (L>0\land \left (\left (c<0\land \left (\left (b<-\sqrt {\frac {c^2}{L^2}} \pi \land K[1]>\frac {\sqrt {\frac {b^2 L^2}{c^2}}}{2 \pi }+\frac {1}{2}\right )\lor \left (b=-\sqrt {\frac {c^2}{L^2}} \pi \land K[1]\geq 2\right )\lor \left (-\sqrt {\frac {c^2}{L^2}} \pi <b<\sqrt {\frac {c^2}{L^2}} \pi \land K[1]\geq 1\right )\lor \left (b=\sqrt {\frac {c^2}{L^2}} \pi \land K[1]\geq 2\right )\lor \left (b>\sqrt {\frac {c^2}{L^2}} \pi \land K[1]>\frac {\sqrt {\frac {b^2 L^2}{c^2}}}{2 \pi }+\frac {1}{2}\right )\right )\right )\lor \left (c>0\land \left (\left (b<-\sqrt {\frac {c^2}{L^2}} \pi \land K[1]>\frac {\sqrt {\frac {b^2 L^2}{c^2}}}{2 \pi }+\frac {1}{2}\right )\lor \left (b=-\sqrt {\frac {c^2}{L^2}} \pi \land K[1]\geq 2\right )\lor \left (-\sqrt {\frac {c^2}{L^2}} \pi <b<\sqrt {\frac {c^2}{L^2}} \pi \land K[1]\geq 1\right )\lor \left (b=\sqrt {\frac {c^2}{L^2}} \pi \land K[1]\geq 2\right )\lor \left (b>\sqrt {\frac {c^2}{L^2}} \pi \land K[1]>\frac {\sqrt {\frac {b^2 L^2}{c^2}}}{2 \pi }+\frac {1}{2}\right )\right )\right )\right )\right )\right ) \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
Maple ✓
restart; interface(showassumed=0); pde := diff(u(x,t),t$2) + b*diff(u(x,t),t) = c^2*diff(u(x,t),x$2); bc := u(0,t)=0,D[1](u)(L,t)=0; ic := D[2](u)(x,0)=0,u(x,0)=f(x); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc,ic],u(x,t)) assuming L>0,b>0),output='realtime'));
\[u \left ( x,t \right ) =\sum _{n=0}^{\infty } \left ( {\frac {1}{L} \left ( \left ( Lb+\sqrt {-4\, \left ( 1/2+n \right ) ^{2}{c}^{2}{\pi }^{2}+{b}^{2}{L}^{2}} \right ) {{\rm e}^{-{\frac {t}{2\,L} \left ( Lb-\sqrt {-4\, \left ( 1/2+n \right ) ^{2}{c}^{2}{\pi }^{2}+{b}^{2}{L}^{2}} \right ) }}}-{{\rm e}^{-{\frac {t}{2\,L} \left ( Lb+\sqrt {-4\, \left ( 1/2+n \right ) ^{2}{c}^{2}{\pi }^{2}+{b}^{2}{L}^{2}} \right ) }}} \left ( Lb-\sqrt {-4\, \left ( 1/2+n \right ) ^{2}{c}^{2}{\pi }^{2}+{b}^{2}{L}^{2}} \right ) \right ) \sin \left ( {\frac { \left ( 1+2\,n \right ) \pi \,x}{2\,L}} \right ) \int _{0}^{L}\!\sin \left ( {\frac { \left ( 1+2\,n \right ) \pi \,x}{2\,L}} \right ) f \left ( x \right ) \,{\rm d}x{\frac {1}{\sqrt {-4\, \left ( 1/2+n \right ) ^{2}{c}^{2}{\pi }^{2}+{b}^{2}{L}^{2}}}}} \right ) \]
Hand solution
Solving for \(t>0,0<x<L\)\[ u_{tt}+bu_{t}=c^{2}u_{xx}\qquad 0<x<L,t>0 \] Boundary conditions\begin {align*} u\left ( 0,t\right ) & =0\\ \left . \frac {\partial u}{\partial x}\right \vert _{x=L} & =0 \end {align*}
Initial conditions, \(t=0\)\begin {align*} u_{t}\left ( x,0\right ) & =0\\ u\left ( x,0\right ) & =f\left ( x\right ) \end {align*}
Separation of variables gives\begin {align*} T^{\prime \prime }X+bT^{\prime }X & =c^{2}X^{\prime \prime }T\\ \frac {1}{c^{2}}\left ( \frac {T^{\prime \prime }}{T}+b\frac {T^{\prime }}{T}\right ) & =\frac {X^{\prime \prime }}{X}=-\lambda \end {align*}
The eigenvalue ODE is\begin {align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X^{\prime }\left ( L\right ) & =0 \end {align*}
Only \(\lambda >0\) gives non-trivial solution from the nature of the boundary conditions. Hence solution is\[ X\left ( x\right ) =A\cos \left ( \sqrt {\lambda }x\right ) +B\sin \left ( \sqrt {\lambda }x\right ) \] Since \(X\left ( 0\right ) =0\) then the above gives \(0=A\) and the solution becomes\[ X\left ( x\right ) =B\sin \left ( \sqrt {\lambda }x\right ) \] Taking derivatives\[ X^{\prime }\left ( x\right ) =\sqrt {\lambda }B\cos \left ( \sqrt {\lambda }x\right ) \] Since \(X^{\prime }\left ( L\right ) =0\) the above becomes\[ 0=\sqrt {\lambda }B\cos \left ( \sqrt {\lambda }L\right ) \] Which implies \(\sqrt {\lambda }L=\frac {n\pi }{2}\) for \(n=1,3,5,\cdots \) or \[ \lambda _{n}=\left ( \frac {n\pi }{2L}\right ) ^{2}\qquad n=1,3,5,\cdots \] Hence the eigenfunctions are\[ \Phi _{n}\left ( x\right ) =\sin \left ( \sqrt {\lambda _{n}}x\right ) \qquad n=1,3,5,\cdots \] The time ODE now becomes\begin {align*} \frac {1}{c^{2}}\left ( \frac {T^{\prime \prime }}{T}+b\frac {T^{\prime }}{T}\right ) & =-\lambda _{n}\\ \frac {T^{\prime \prime }}{T}+b\frac {T^{\prime }}{T} & =-c^{2}\lambda _{n}\\ T^{\prime \prime }+bT^{\prime }+c^{2}\lambda _{n}T & =0 \end {align*}
The characteristic equation \(r^{2}+br+c^{2}\lambda _{n}=0\) has the roots \(r=\frac {-B}{2A}\pm \frac {1}{2A}\sqrt {B^{2}-4AC}\rightarrow r=\frac {-b}{2}\pm \frac {1}{2}\sqrt {b^{2}-4c^{2}\lambda _{n}}\) or\[ r=\frac {-b}{2}\pm \frac {1}{2}\sqrt {b^{2}-4c^{2}\lambda _{n}}\] Case \(b^{2}<4c^{2}\lambda _{n}\) for all \(n\). This is called the underdamped case, which generates damped oscillations. The roots becomes\[ r=\frac {-b}{2}\pm \frac {1}{2}i\sqrt {4c^{2}\lambda _{n}-b^{2}}\] Let \(\beta _{n}=\frac {1}{2}\sqrt {4c^{2}\lambda _{n}-b^{2}}\), then\[ r=\frac {-b}{2}\pm i\beta _{n}\] Hence the solution is\[ T_{n}\left ( t\right ) =e^{\frac {-b}{2}t}\left ( D_{n}\cos \left ( \beta _{n}t\right ) +E_{n}\sin \left ( \beta _{n}t\right ) \right ) \] Therefore the complete solution becomes\begin {equation} u\left ( x,t\right ) =\sum _{n=1,3,5,\cdots }^{\infty }e^{\frac {-b}{2}t}\left ( D_{n}\cos \left ( \beta _{n}t\right ) +E_{n}\sin \left ( \beta _{n}t\right ) \right ) \Phi _{n}\left ( x\right ) \tag {1} \end {equation} At \(t=0\) the above becomes\[ f\left ( x\right ) =\sum _{n=1,3,5,\cdots }^{\infty }D_{n}\Phi _{n}\left ( x\right ) \] Applying orthogonality gives\begin {align*} \int _{0}^{L}f\left ( x\right ) \Phi _{n}\left ( x\right ) dx & =D_{n}\int _{0}^{L}\Phi _{n}^{2}\left ( x\right ) dx\\ & =\frac {L}{2}D_{n} \end {align*}
Hence\begin {equation} D_{n}=\frac {2}{L}\int _{0}^{L}f\left ( x\right ) \Phi _{n}\left ( x\right ) dx\tag {2} \end {equation} Taking time derivative of (1) gives\[ u_{t}\left ( x,t\right ) =\sum _{n=1,3,5,\cdots }^{\infty }\frac {-b}{2}e^{\frac {-b}{2}t}\left ( D_{n}\cos \left ( \beta _{n}t\right ) +E_{n}\sin \left ( \beta _{n}t\right ) \right ) \Phi _{n}\left ( x\right ) +e^{\frac {-b}{2}t}\left ( -\beta _{n}D_{n}\sin \left ( \beta _{n}t\right ) +E_{n}\beta _{n}\cos \left ( \beta _{n}t\right ) \right ) \Phi _{n}\left ( x\right ) \] At \(t=0\) since \(g\left ( x\right ) =0\) then the above becomes\begin {align} 0 & =\sum _{n=1,3,5,\cdots }^{\infty }\left ( \frac {-b}{2}D_{n}+E_{n}\beta _{n}\right ) \Phi _{n}\left ( x\right ) \nonumber \\ 0 & =\frac {-b}{2}D_{n}+E_{n}\beta _{n}\nonumber \\ E_{n} & =\frac {bD_{n}}{2\beta _{n}}\tag {3} \end {align}
Using (2,3), the solution (1) now becomes\begin {align*} u\left ( x,t\right ) & =\sum _{n=1,3,5,\cdots }^{\infty }e^{\frac {-b}{2}t}\left ( D_{n}\cos \left ( \beta _{n}t\right ) +\frac {bD_{n}}{2\beta _{n}}\sin \left ( \beta _{n}t\right ) \right ) \Phi _{n}\left ( x\right ) \\ & =\sum _{n=1,3,5,\cdots }^{\infty }D_{n}e^{\frac {-b}{2}t}\left ( \cos \left ( \beta _{n}t\right ) +\frac {b}{2\beta _{n}}\sin \left ( \beta _{n}t\right ) \right ) \Phi _{n}\left ( x\right ) \end {align*}
Or\[ u\left ( x,t\right ) =\sum _{n=1,3,5,\cdots }^{\infty }\left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \Phi _{n}\left ( s\right ) ds\right ) e^{\frac {-b}{2}t}\left ( \cos \left ( \beta _{n}t\right ) +\frac {b}{2\beta _{n}}\sin \left ( \beta _{n}t\right ) \right ) \Phi _{n}\left ( x\right ) \] But \(\beta _{n}=\frac {1}{2}\sqrt {4c^{2}\lambda _{n}-b^{2}}\),\(\lambda _{n}=\left ( \frac {n\pi }{2L}\right ) ^{2}\) and \(\Phi _{n}\left ( x\right ) =\sin \left ( \frac {n\pi }{2L}x\right ) \), hence the above becomes
\begin {align*} u\left ( x,t\right ) & =\sum _{n=1,3,5,\cdots }^{\infty }\left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {n\pi }{2L}s\right ) ds\right ) e^{\frac {-b}{2}t}\cos \left ( \frac {1}{2}\sqrt {4c^{2}\left ( \frac {n\pi }{2L}\right ) ^{2}-b^{2}}t\right ) \sin \left ( \frac {n\pi }{2L}x\right ) \\ & +\sum _{n=1,3,5,\cdots }^{\infty }\left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {n\pi }{2L}s\right ) ds\right ) e^{\frac {-b}{2}t}\frac {b\sin \left ( \frac {1}{2}\sqrt {4c^{2}\left ( \frac {n\pi }{2L}\right ) ^{2}-b^{2}}t\right ) }{\sqrt {4c^{2}\left ( \frac {n\pi }{2L}\right ) ^{2}-b^{2}}}\sin \left ( \frac {n\pi }{2L}x\right ) \end {align*}
Or\begin {align*} u\left ( x,t\right ) & =\sum _{n=0}^{\infty }\left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac {-b}{2}t}\cos \left ( \frac {1}{2}\sqrt {4c^{2}\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}t\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) \\ & +\sum _{n=0}^{\infty }\left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac {-b}{2}t}\frac {b\sin \left ( \frac {1}{2}\sqrt {4c^{2}\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}t\right ) }{\sqrt {4c^{2}\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) \end {align*}
Case \(b^{2}=4c^{2}\left ( \frac {n\pi }{2L}\right ) ^{2}\). We see that for \(n=1\) it becomes critical, because then \(b=\frac {\pi c}{L}\) and now and the discriminant is zero in this case. This is called the critical damped case. For \(n>1\), it becomes underdamped, which is the above case. So we only need to find solution for \(n=1\). In this case, the solution to \(T^{\prime \prime }+bT^{\prime }+c^{2}\lambda _{1}T=0\) is\[ T_{1}\left ( t\right ) =D_{1}e^{\frac {-b}{2}t}+E_{1}te^{\frac {-b}{2}t}\] Therefore the complete solution becomes\begin {equation} u\left ( x,t\right ) =\left ( D_{1}e^{\frac {-b}{2}t}+E_{1}te^{\frac {-b}{2}t}\right ) \sin \left ( \frac {\pi }{2L}x\right ) +\sum _{n=3,5,\cdots }^{\infty }\left ( D_{n}e^{\frac {-b}{2}t}+E_{n}te^{\frac {-b}{2}t}\right ) \sin \left ( \sqrt {\lambda _{n}}x\right ) \tag {4} \end {equation} For \(n=1\), At \(t=0\), from intial conditions (4) becomes\[ f\left ( x\right ) =D_{1}\sin \left ( \frac {\pi }{2L}x\right ) \] By orthognailty the above gives\begin {equation} D_{1}=\frac {2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {\pi }{2L}x\right ) dx\tag {5} \end {equation} Taking time derivative of (4) for \(n=1\), gives\[ u\left ( x,t\right ) =\left ( \frac {-b}{2}D_{1}e^{\frac {-b}{2}t}+E_{1}\left ( e^{\frac {-b}{2}t}-\frac {b}{2}te^{\frac {-b}{2}t}\right ) \right ) \sin \left ( \frac {\pi }{2L}x\right ) \] At \(t=0\) and since \(g\left ( x\right ) =0\), then the above becomes\begin {align} 0 & =\left ( \frac {-b}{2}D_{1}+E_{1}\right ) \sin \left ( \frac {\pi }{2L}x\right ) \nonumber \\ \frac {-b}{2}D_{1}+E_{1} & =0\nonumber \\ E_{1} & =\frac {b}{2}D_{1}\tag {6} \end {align}
Using (5,6) then (4) becomes\begin {align*} u\left ( x,t\right ) & =D_{1}\left ( e^{\frac {-b}{2}t}+\frac {b}{2}te^{\frac {-b}{2}t}\right ) \sin \left ( \frac {\pi }{2L}x\right ) +\sum _{n=3,5,\cdots }^{\infty }\left ( D_{n}e^{\frac {-b}{2}t}+E_{n}te^{\frac {-b}{2}t}\right ) \sin \left ( \sqrt {\lambda _{n}}x\right ) \\ & =\left ( \frac {2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {\pi }{2L}x\right ) dx\right ) \left ( e^{\frac {-b}{2}t}+\frac {b}{2}te^{\frac {-b}{2}t}\right ) \sin \left ( \frac {\pi }{2L}x\right ) \\ & +\sum _{n=3,5,\cdots }^{\infty }\left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \sqrt {\lambda _{n}}s\right ) ds\right ) \left ( e^{\frac {-b}{2}t}+\frac {bt}{2}e^{\frac {-b}{2}t}\right ) \sin \left ( \sqrt {\lambda _{n}}x\right ) \end {align*}
For the case of \(n>1\), the solution was found in the above underdamped case. Putting all these together, gives the solution as\begin {align*} u\left ( x,t\right ) & =\left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {\pi }{2L}s\right ) ds\right ) \left ( e^{\frac {-b}{2}t}+\frac {b}{2}te^{\frac {-b}{2}t}\right ) \sin \left ( \frac {\pi }{2L}x\right ) \\ & +\sum _{n=2}^{\infty }\left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac {-b}{2}t}\cos \left ( \frac {1}{2}\sqrt {4c^{2}\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}t\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) \\ & +\sum _{n=2}^{\infty }\left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac {-b}{2}t}\frac {b\sin \left ( \frac {1}{2}\sqrt {4c^{2}\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}t\right ) }{\sqrt {4c^{2}\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) \end {align*}
Case \(b^{2}>4c^{2}\left ( \frac {n\pi }{2L}\right ) ^{2}\). Will consider only the case when this is true for \(n=1\) only. If this is true for larger \(n\), then same solution needs to be summed for each mode. But for simplicity, will consider \(n=1\) here. In this case, the roots are\[ r=\frac {-b}{2}\pm \frac {1}{2}\sqrt {b^{2}-4c^{2}\lambda _{1}}\] Where now \(b^{2}-4c^{2}\lambda _{1}\) is positive. Hence we get \(r_{1}=\frac {-b}{2}+\frac {1}{2}\sqrt {b^{2}-4c^{2}\lambda _{1}},r_{2}=\frac {-b}{2}-\frac {1}{2}\sqrt {b^{2}-4c^{2}\lambda _{1}}\) or \begin {align*} r_{1} & =\frac {-b}{2}+\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}\\ r_{1} & =\frac {-b}{2}-\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}} \end {align*}
And the solution to \(T_{1}^{\prime \prime }+bT_{1}^{\prime }+c^{2}\lambda _{1}T=0\) is\[ T_{1}\left ( t\right ) =e^{\frac {-b}{2}t}\left ( D_{1}e^{\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}t}+E_{1}e^{\frac {-1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}t}\right ) \] For the rest of the modes, the solution is from above\[ T_{n}\left ( t\right ) =e^{\frac {-b}{2}t}\left ( D_{n}\cos \left ( \beta _{n}t\right ) +E_{n}\sin \left ( \beta _{n}t\right ) \right ) \qquad n=3,5,7,\cdots \] Hence the complete solution becomes\begin {align} u\left ( x,t\right ) & =e^{\frac {-b}{2}t}\left ( D_{1}e^{\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}t}+E_{1}e^{\frac {-1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}t}\right ) \sin \left ( \frac {\pi }{2L}x\right ) \tag {7}\\ & +\sum _{n=3,5,\cdots }^{\infty }e^{\frac {-b}{2}t}\left ( D_{n}\cos \left ( \beta _{n}t\right ) +E_{n}\sin \left ( \beta _{n}t\right ) \right ) \sin \left ( \frac {n\pi }{2L}x\right ) \nonumber \end {align}
At \(t=0\) and for \(n=1\), the above becomes\[ f\left ( x\right ) =\left ( D_{1}+E_{1}\right ) \sin \left ( \frac {\pi }{2L}x\right ) \] Hence\begin {equation} \left ( D_{1}+E_{1}\right ) =\frac {2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {\pi }{2L}x\right ) dx\tag {8} \end {equation} Taking time derivative of (7) and for \(n=1\) at \(t=0\) it gives\[ 0=\left ( \frac {-b}{2}\left ( D_{1}+E_{1}\right ) +\left ( \frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}D_{1}-E_{1}\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}\right ) \right ) \sin \left ( \frac {\pi }{2L}x\right ) \] Hence\begin {align} \frac {-b}{2}\left ( D_{1}+E_{1}\right ) +\left ( \frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}D_{1}-E_{1}\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}\right ) & =0\nonumber \\ -E_{1}\left ( \frac {b}{2}+\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}\right ) & =\left ( \frac {b}{2}-\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}\right ) D_{1}\nonumber \\ E_{1} & =\frac {-\frac {b}{2}+\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}{\frac {b}{2}+\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}D_{1}\tag {9} \end {align}
From (8,9)\begin {align*} D_{1} & =\frac {2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {\pi }{2L}x\right ) dx-E_{1}\\ D_{1}-\frac {\frac {b}{2}-\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}{\frac {b}{2}+\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}D_{1} & =\frac {2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {\pi }{2L}x\right ) dx\\ D_{1} & =\frac {\frac {2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {\pi }{2L}x\right ) dx}{1-\frac {\frac {b}{2}-\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}{\frac {b}{2}+\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}}\\ & =\frac {\frac {b}{2}+\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}{\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}\frac {2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {\pi }{2L}x\right ) dx \end {align*}
And therefore\begin {align*} E_{1} & =\frac {-\frac {b}{2}+\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}{\frac {b}{2}+\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}\left ( \frac {\frac {b}{2}+\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}{\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}\right ) \frac {2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {\pi }{2L}x\right ) dx\\ & =\frac {-\frac {b}{2}+\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}{\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}\frac {2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {\pi }{2L}x\right ) dx \end {align*}
For \(n>1\) the solution is the same as the underdamped case above. Hence the complete solution becomes from (7)\begin {align*} u\left ( x,t\right ) & =e^{\frac {-b}{2}t}\left ( D_{1}e^{\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}t}+E_{1}e^{\frac {-1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}t}\right ) \sin \left ( \frac {\pi }{2L}x\right ) \\ & +\sum _{n=3,5,\cdots }^{\infty }e^{\frac {-b}{2}t}\left ( D_{n}\cos \left ( \beta _{n}t\right ) +E_{n}\sin \left ( \beta _{n}t\right ) \right ) \sin \left ( \frac {n\pi }{2L}x\right ) \end {align*}
Or\begin {align*} u\left ( x,t\right ) & =e^{\frac {-b}{2}t}\left ( \frac {\frac {b}{2}+\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}{\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}\right ) \frac {2}{L}\left ( \int _{0}^{L}f\left ( s\right ) \sin \left ( \sqrt {\frac {\pi }{2L}}s\right ) ds\right ) e^{\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}t}\sin \left ( \frac {\pi }{2L}x\right ) \\ & +e^{\frac {-b}{2}t}\left ( \frac {-\frac {b}{2}+\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}{\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}\right ) \frac {2}{L}\left ( \int _{0}^{L}f\left ( s\right ) \sin \left ( \sqrt {\frac {\pi }{2L}}s\right ) ds\right ) e^{\frac {-1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}t}\sin \left ( \frac {\pi }{2L}x\right ) \\ & +\sum _{n=3,5,\cdots }^{\infty }\left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {n\pi }{2L}s\right ) ds\right ) e^{\frac {-b}{2}t}\cos \left ( \frac {1}{2}\sqrt {4c^{2}\left ( \frac {n\pi }{2L}\right ) ^{2}-b^{2}}t\right ) \sin \left ( \frac {n\pi }{2L}x\right ) \\ & +\sum _{n=3,5,\cdots }^{\infty }\left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {n\pi }{2L}s\right ) ds\right ) e^{\frac {-b}{2}t}\frac {b\sin \left ( \frac {1}{2}\sqrt {4c^{2}\left ( \frac {n\pi }{2L}\right ) ^{2}-b^{2}}t\right ) }{\sqrt {4c^{2}\left ( \frac {n\pi }{2L}\right ) ^{2}-b^{2}}}\sin \left ( \frac {n\pi }{2L}x\right ) \end {align*}
Or\begin {align*} u\left ( x,t\right ) & =e^{\frac {-b}{2}t}\left ( \frac {\frac {b}{2}+\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}{\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}\right ) \frac {2}{L}\left ( \int _{0}^{L}f\left ( s\right ) \sin \left ( \sqrt {\frac {\pi }{2L}}s\right ) ds\right ) e^{\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}t}\sin \left ( \frac {\pi }{2L}x\right ) \\ & +e^{\frac {-b}{2}t}\left ( \frac {-\frac {b}{2}+\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}{\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}\right ) \frac {2}{L}\left ( \int _{0}^{L}f\left ( s\right ) \sin \left ( \sqrt {\frac {\pi }{2L}}s\right ) ds\right ) e^{\frac {-1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}t}\sin \left ( \frac {\pi }{2L}x\right ) \\ & +\sum _{n=1}^{\infty }\left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac {-b}{2}t}\cos \left ( \frac {1}{2}\sqrt {4c^{2}\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}t\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) \\ & +\sum _{n=1}^{\infty }\left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac {-b}{2}t}\frac {b\sin \left ( \frac {1}{2}\sqrt {4c^{2}\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}t\right ) }{\sqrt {4c^{2}\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) \end {align*}
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Added July 9, 2019 \[ u_{tt} + b u_t = c^2 u_{xx} \] For \(t>0\) and \(0<x<L\) and boundary conditions \begin {align*} u(0,t) &= 0\\ u_x(L,t) &=0 \end {align*}
With initial conditions \begin {align*} u(x,0) &= f(x)\\ u_t(x,0) &= 0 \\ \end {align*}
Using the following values \begin {align*} L &=3 \\ c &= 4\\ f(x) &= \left \{ \begin {array} [c]{ccc}\frac {3h}{L}x & & 0<x<\frac {L}{3}\\ h & & \frac {L}{3}<x<L \end {array} \right .\\ h &=\frac {1}{10}\\ b &=\frac {1}{2} \frac {\pi c}{L} \end {align*}
Hence \(b = \frac {2 \pi }{3}\)
Mathematica ✓
ClearAll["Global`*"]; L=3; c=4; h=1/10; b=1/2*(Pi*c/L); f=Piecewise[{{3*h/L*x,0<x<L/3},{h,L/3<x<L}}]; pde = D[u[x, t], {t, 2}] + b*D[u[x,t],t] == c^2*D[u[x, t], {x, 2}]; bc = {u[0, t] == 0, Derivative[1, 0][u][L, t] == 0}; ic = {Derivative[0, 1][u][x, 0] == 0, u[x, 0] == f}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}], 60*10]];
\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}\sqrt {\frac {2}{3}} e^{-\frac {\pi t}{3}} \sin \left (\frac {1}{6} \pi x (2 K[1]-1)\right ) \left (-\frac {6 \sqrt {6} \cos \left (\frac {1}{3} \pi (K[1]+1)\right ) \cos \left (\frac {1}{2} t \sqrt {\frac {16}{9} \pi ^2 (2 K[1]-1)^2-\frac {4 \pi ^2}{9}}\right )}{5 \pi ^2 (1-2 K[1])^2}-\frac {6 \sqrt {6} \cos \left (\frac {1}{3} \pi (K[1]+1)\right ) \sin \left (\frac {1}{2} t \sqrt {\frac {16}{9} \pi ^2 (2 K[1]-1)^2-\frac {4 \pi ^2}{9}}\right )}{5 \pi ^2 (1-2 K[1])^2 \sqrt {16 K[1]^2-16 K[1]+3}}\right ) & K[1]\in \mathbb {Z}\land K[1]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
Maple ✓
restart; L:=3; c:=4; h:=1/10; b:=1/2*(Pi*c/L); f:=piecewise(0<x and x<L/3,3*h/L*x,L/3<x and x<L,h); pde := diff(u(x,t),t$2) + b*diff(u(x,t),t) = c^2*diff(u(x,t),x$2); bc := u(0,t)=0,D[1](u)(L,t)=0; ic := D[2](u)(x,0)=0,u(x,0)=f; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc,ic],u(x,t)) ),output='realtime'));
\[u \left ( x,t \right ) =\sum _{n=0}^{\infty }{\frac {-{\frac {3\,i}{5}}}{{\pi }^{2} \left ( 1+2\,n \right ) ^{2}}\sin \left ( {\frac { \left ( 1+2\,n \right ) \pi \,x}{6}} \right ) \left ( \sqrt {3}\sin \left ( {\frac {n\pi }{3}} \right ) +\cos \left ( {\frac {n\pi }{3}} \right ) \right ) \left ( \left ( i\sqrt {16\,{n}^{2}+16\,n+3}+1 \right ) {{\rm e}^{{\frac {\pi \,t}{3} \left ( i\sqrt {16\,{n}^{2}+16\,n+3}-1 \right ) }}}+ \left ( i\sqrt {16\,{n}^{2}+16\,n+3}-1 \right ) {{\rm e}^{-{\frac {\pi \,t}{3} \left ( i\sqrt {16\,{n}^{2}+16\,n+3}+1 \right ) }}} \right ) {\frac {1}{\sqrt {16\,{n}^{2}+16\,n+3}}}}\]
Hand solution
Solving the wave PDE on string underdamped case \(t>0,0<x<L\)\[ u_{tt}+bu_{t}=c^{2}u_{xx}\qquad 0<x<L,t>0 \] Boundary conditions\begin {align*} u\left ( 0,t\right ) & =0\\ \left . \frac {\partial u}{\partial x}\right \vert _{x=L} & =0 \end {align*}
Initial conditions, \(t=0\)\begin {align*} u_{t}\left ( x,0\right ) & =0\\ u\left ( x,0\right ) & =f\left ( x\right ) \end {align*}
Using \begin {align*} f\left ( x\right ) & =\left \{ \begin {array} [c]{ccc}\frac {3h}{L}x & & 0<x<\frac {L}{3}\\ h & & \frac {L}{3}<x<L \end {array} \right . \\ b & =\frac {2\pi }{3}\\ c & =4\\ L & =3 \end {align*}
Hence the PDE becomes \(u_{tt}+\frac {2\pi }{3}u_{t}=16u_{xx}\). The general solution to the above PDE was given in problem 5.1.1.23 on page 1318. The eigenvalues are given as \[ \lambda _{n}=\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}\qquad n=0,1,2,\cdots \] And the discriminant is \(b^{2}-4c^{2}\lambda _{n}=b^{2}-4c^{2}\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}=b^{2}-\left ( 4\right ) \left ( 16\right ) \left ( \frac {\left ( 2n+1\right ) ^{2}\pi ^{2}}{36}\right ) \). For \(n=0\) this gives\(\ b^{2}-\frac {16}{9}\pi ^{2}\). But\(\ b=\frac {2\pi }{3}\). Hence discriminant is \(\left ( \frac {2\pi }{3}\right ) ^{2}-\frac {16}{9}\pi ^{2}=-\frac {4}{3}\pi ^{2}\). Since discriminant is negative, then this is underdamped wave with damped oscillations as the solution given from the above problem as\begin {align*} u\left ( x,t\right ) & =\sum _{n=0}^{\infty }\left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac {-b}{2}t}\cos \left ( \frac {1}{2}\sqrt {4c^{2}\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}t\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) \\ & +\sum _{n=0}^{\infty }\left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac {-b}{2}t}\frac {b\sin \left ( \frac {1}{2}\sqrt {4c^{2}\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}t\right ) }{\sqrt {4c^{2}\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) \end {align*}
Replacing given values in the above solution results in\begin {align} u\left ( x,t\right ) & =\sum _{n=0}^{\infty }\left ( \frac {2}{3}\int _{0}^{3}f\left ( s\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac {-\pi }{3}t}\left ( \cos \left ( \frac {1}{2}\sqrt {64\left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\left ( \frac {2\pi }{3}\right ) ^{2}}t\right ) \right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}x\right ) \tag {1}\\ & +\sum _{n=0}^{\infty }\left ( \frac {2}{3}\int _{0}^{3}f\left ( s\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}s\right ) ds\right ) e^{\frac {-\pi }{3}t}\left ( \frac {2\pi }{3}\frac {\sin \left ( \frac {1}{2}\sqrt {64\left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\left ( \frac {2\pi }{3}\right ) ^{2}}t\right ) }{\sqrt {64\left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\left ( \frac {2\pi }{3}\right ) ^{2}}}\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}x\right ) \nonumber \end {align}
But \begin {align*} \int _{0}^{3}f\left ( s\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}s\right ) & =\frac {1}{10}\int _{0}^{1}x\sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}x\right ) dx+\frac {1}{10}\int _{1}^{3}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}x\right ) dx\\ & =\frac {18}{5\pi ^{2}\left ( 2n+1\right ) ^{2}}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) \end {align*}
Hence the solution (1) becomes\begin {align} u\left ( x,t\right ) & =\sum _{n=0}^{\infty }\left ( \frac {2}{3}\frac {18}{5\pi ^{2}\left ( 2n+1\right ) ^{2}}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) \right ) e^{\frac {-\pi }{3}t}\left ( \cos \left ( \frac {1}{2}\sqrt {64\left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\left ( \frac {2\pi }{3}\right ) ^{2}}t\right ) \right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}x\right ) \tag {2}\\ & +\sum _{n=0}^{\infty }\left ( \frac {2}{3}\frac {18}{5\pi ^{2}\left ( 2n+1\right ) ^{2}}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) \right ) e^{\frac {-\pi }{3}t}\left ( \frac {2\pi }{3}\frac {\sin \left ( \frac {1}{2}\sqrt {64\left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\left ( \frac {2\pi }{3}\right ) ^{2}}t\right ) }{\sqrt {64\left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\left ( \frac {2\pi }{3}\right ) ^{2}}}\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}x\right ) \nonumber \end {align}
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Source code used for the above
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Added July 10, 2019 \[ u_{tt} + b u_t = c^2 u_{xx} \] For \(t>0\) and \(0<x<L\) and boundary conditions \begin {align*} u(0,t) &= 0\\ u_x(L,t) &=0 \end {align*}
With initial conditions \begin {align*} u(x,0) &= f(x)\\ u_t(x,0) &= 0 \\ \end {align*}
Using the following values \begin {align*} L &=3 \\ c &= 4\\ f(x) &= \left \{ \begin {array} [c]{ccc}\frac {3h}{L}x & & 0<x<\frac {L}{3}\\ h & & \frac {L}{3}<x<L \end {array} \right .\\ h &=\frac {1}{10}\\ b &=\frac {\pi c}{L} \end {align*}
Hence \(b = \frac {4 \pi }{3}\)
Mathematica ✓
ClearAll["Global`*"]; L=3; c=4; h=1/10; b=Pi*c/L; f=Piecewise[{{3*h/L*x,0<x<L/3},{h,L/3<x<L}}]; pde = D[u[x, t], {t, 2}] + b*D[u[x,t],t] == c^2*D[u[x, t], {x, 2}]; bc = {u[0, t] == 0, Derivative[1, 0][u][L, t] == 0}; ic = {Derivative[0, 1][u][x, 0] == 0, u[x, 0] == f}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}], 60*10]];
\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}\sqrt {\frac {2}{3}} e^{-\frac {2 \pi t}{3}} \sin \left (\frac {1}{6} \pi x (2 K[1]-1)\right ) \left (-\frac {6 \sqrt {6} \cos \left (\frac {1}{3} \pi (K[1]+1)\right ) \cos \left (\frac {1}{2} t \sqrt {\frac {16}{9} \pi ^2 (2 K[1]-1)^2-\frac {16 \pi ^2}{9}}\right )}{5 \pi ^2 (1-2 K[1])^2}-\frac {3 \sqrt {6} \cos \left (\frac {1}{3} \pi (K[1]+1)\right ) \sin \left (\frac {1}{2} t \sqrt {\frac {16}{9} \pi ^2 (2 K[1]-1)^2-\frac {16 \pi ^2}{9}}\right )}{5 \pi ^2 \sqrt {(K[1]-1) K[1]} (2 K[1]-1)^2}\right ) & K[1]\in \mathbb {Z}\land K[1]\geq 2 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
Maple ✓
restart; L:=3; c:=4; h:=1/10; b:=Pi*c/L; f:=piecewise(0<x and x<L/3,3*h/L*x,L/3<x and x<L,h); pde := diff(u(x,t),t$2) + b*diff(u(x,t),t) = c^2*diff(u(x,t),x$2); bc := u(0,t)=0,D[1](u)(L,t)=0; ic := D[2](u)(x,0)=0,u(x,0)=f; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc,ic],u(x,t)) ),output='realtime'));
\[u \left ( x,t \right ) =\sum _{n=0}^{\infty }\begin {cases}{\frac {4}{5\,{\pi }^{2}}{{\rm e}^{-{\frac {2\,\pi \,t}{3}}}} \left ( \pi \,t+{\frac {3}{2}} \right ) \sin \left ( {\frac {\pi \,x}{6}} \right ) } & n=0\\{\frac {-{\frac {3\,i}{10}}}{ \left ( 1+2\,n \right ) ^{2}{\pi }^{2}}\sin \left ( {\frac { \left ( 1+2\,n \right ) \pi \,x}{6}} \right ) \left ( \sqrt {3}\sin \left ( {\frac {n\pi }{3}} \right ) +\cos \left ( {\frac {n\pi }{3}} \right ) \right ) \left ( 2\,i{{\rm e}^{{\frac {2\,i}{3}}\pi \, \left ( -2\,\sqrt {n}\sqrt {n+1}+i \right ) t}}\sqrt {n}\sqrt {n+1}+2\,i{{\rm e}^{{\frac {2\,i}{3}}\pi \, \left ( 2\,\sqrt {n}\sqrt {n+1}+i \right ) t}}\sqrt {n}\sqrt {n+1}-{{\rm e}^{{\frac {2\,i}{3}}\pi \, \left ( -2\,\sqrt {n}\sqrt {n+1}+i \right ) t}}+{{\rm e}^{{\frac {2\,i}{3}}\pi \, \left ( 2\,\sqrt {n}\sqrt {n+1}+i \right ) t}} \right ) {\frac {1}{\sqrt {n}}}{\frac {1}{\sqrt {n+1}}}} & \text {otherwise}\end {cases}\]
Hand solution
Solving the wave PDE on string underdamped case \(t>0,0<x<L\)\[ u_{tt}+bu_{t}=c^{2}u_{xx}\qquad 0<x<L,t>0 \] Boundary conditions\begin {align*} u\left ( 0,t\right ) & =0\\ \left . \frac {\partial u}{\partial x}\right \vert _{x=L} & =0 \end {align*}
Initial conditions, \(t=0\)\begin {align*} u_{t}\left ( x,0\right ) & =0\\ u\left ( x,0\right ) & =f\left ( x\right ) \end {align*}
Using \begin {align*} f\left ( x\right ) & =\left \{ \begin {array} [c]{ccc}\frac {3h}{L}x & & 0<x<\frac {L}{3}\\ h & & \frac {L}{3}<x<L \end {array} \right . \\ b & =\frac {4\pi }{3}\\ c & =4\\ L & =3\\ h & =\frac {1}{10} \end {align*}
Hence the PDE becomes \(u_{tt}+\frac {4\pi }{3}u_{t}=16u_{xx}\). The general solution to the above PDE was given in problem 5.1.1.23 on page 1318. The eigenvalues are given as \[ \lambda _{n}=\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}\qquad n=0,1,2,\cdots \] And the discriminant is \(b^{2}-4c^{2}\lambda _{n}=b^{2}-4c^{2}\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}=b^{2}-\left ( 4\right ) \left ( 16\right ) \left ( \frac {\left ( 2n+1\right ) ^{2}\pi ^{2}}{36}\right ) \). For \(n=0\) this gives\(\ b^{2}-\left ( \frac {4}{3}\pi \right ) ^{2}\). But\(\ b=\frac {4\pi }{3}\). Hence discriminant is zero for \(n=0\). This means this is critically damped in first mode. Using the the solution for this case from the above general solution as
\begin {align*} u\left ( x,t\right ) & =\left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {\pi }{2L}s\right ) ds\right ) \left ( e^{\frac {-b}{2}t}+\frac {b}{2}te^{\frac {-b}{2}t}\right ) \sin \left ( \frac {\pi }{2L}x\right ) \\ & +\sum _{n=1}^{\infty }\left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac {-b}{2}t}\cos \left ( \frac {1}{2}\sqrt {4c^{2}\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}t\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) \\ & +\sum _{n=1}^{\infty }\left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac {-b}{2}t}\frac {b\sin \left ( \frac {1}{2}\sqrt {4c^{2}\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}t\right ) }{\sqrt {4c^{2}\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) \end {align*}
Replacing given values in the above solution results in\begin {align} u\left ( x,t\right ) & =\left ( \frac {2}{3}\int _{0}^{3}f\left ( x\right ) \sin \left ( \frac {\pi }{6}x\right ) dx\right ) \left ( e^{\frac {-2\pi t}{3}}+\frac {b}{2}te^{\frac {-2\pi t}{3}}\right ) \sin \left ( \frac {\pi }{6}x\right ) \tag {1}\\ & +\sum _{n=1}^{\infty }\left ( \frac {2}{3}\int _{0}^{3}f\left ( s\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}s\right ) ds\right ) e^{\frac {-2\pi t}{3}}\cos \left ( \frac {1}{2}\sqrt {64\left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\left ( \frac {4\pi }{3}\right ) ^{2}}t\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) \nonumber \\ & +\sum _{n=1}^{\infty }\left ( \frac {2}{3}\int _{0}^{3}f\left ( s\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}s\right ) ds\right ) e^{\frac {-b}{2}t}\frac {\left ( \frac {4\pi }{3}\right ) \sin \left ( \frac {1}{2}\sqrt {64\left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\left ( \frac {4\pi }{3}\right ) ^{2}}t\right ) }{\sqrt {64\left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\left ( \frac {4\pi }{3}\right ) ^{2}}}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}x\right ) \nonumber \end {align}
But
\[ \int _{0}^{3}f\left ( x\right ) \sin \left ( \frac {\pi }{6}x\right ) dx=\frac {9}{5\pi ^{2}}\]
And\begin {align*} \int _{0}^{3}f\left ( s\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}s\right ) & =\frac {1}{10}\int _{0}^{1}x\sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}x\right ) dx+\frac {1}{10}\int _{1}^{3}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}x\right ) dx\\ & =\frac {18}{5\pi ^{2}\left ( 2n+1\right ) ^{2}}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) \end {align*}
Hence the solution (1) becomes
\begin {align} u\left ( x,t\right ) & =\frac {6}{5\pi ^{2}}\left ( e^{\frac {-2\pi t}{3}}+\frac {2}{3}\pi te^{\frac {-2\pi t}{3}}\right ) \sin \left ( \frac {\pi }{6}x\right ) \tag {1}\\ & +\sum _{n=1}^{\infty }\frac {12}{5\pi ^{2}\left ( 2n+1\right ) ^{2}}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) e^{\frac {-2\pi t}{3}}\cos \left ( \frac {1}{2}\sqrt {64\left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\left ( \frac {4\pi }{3}\right ) ^{2}}t\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}x\right ) \nonumber \\ & +\sum _{n=1}^{\infty }\frac {12}{5\pi ^{2}\left ( 2n+1\right ) ^{2}}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) e^{\frac {-2\pi t}{3}}\frac {\left ( \frac {4\pi }{3}\right ) \sin \left ( \frac {1}{2}\sqrt {64\left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\left ( \frac {4\pi }{3}\right ) ^{2}}t\right ) }{\sqrt {64\left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\left ( \frac {4\pi }{3}\right ) ^{2}}}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}x\right ) \nonumber \end {align}
Animation is below
Source code used for the above
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Added July 11, 2019 \[ u_{tt} + b u_t = c^2 u_{xx} \] For \(t>0\) and \(0<x<L\) and boundary conditions \begin {align*} u(0,t) &= 0\\ u_x(L,t) &=0 \end {align*}
With initial conditions \begin {align*} u(x,0) &= f(x)\\ u_t(x,0) &= 0 \\ \end {align*}
Using the following values \begin {align*} L &=3 \\ c &= 4\\ f(x) &= \left \{ \begin {array} [c]{ccc}\frac {3h}{L}x & & 0<x<\frac {L}{3}\\ h & & \frac {L}{3}<x<L \end {array} \right .\\ h &=\frac {1}{10}\\ b &=\frac {3}{2} \frac {\pi c}{L} \end {align*}
Hence \(b = 2 \pi \)
Mathematica ✓
ClearAll["Global`*"]; L=3; c=4; h=1/10; b=3/2*Pi*c/L; f=Piecewise[{{3*h/L*x,0<x<L/3},{h,L/3<x<L}}]; pde = D[u[x, t], {t, 2}] + b*D[u[x,t],t] == c^2*D[u[x, t], {x, 2}]; bc = {u[0, t] == 0, Derivative[1, 0][u][L, t] == 0}; ic = {Derivative[0, 1][u][x, 0] == 0, u[x, 0] == f}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}], 60*10]];
\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}\sqrt {\frac {2}{3}} e^{-\pi t} \sin \left (\frac {1}{6} \pi x (2 K[1]-1)\right ) \left (-\frac {6 \sqrt {6} \cos \left (\frac {1}{3} \pi (K[1]+1)\right ) \cos \left (\frac {1}{2} t \sqrt {\frac {16}{9} \pi ^2 (2 K[1]-1)^2-4 \pi ^2}\right )}{5 \pi ^2 (1-2 K[1])^2}-\frac {18 \sqrt {6} \cos \left (\frac {1}{3} \pi (K[1]+1)\right ) \sin \left (\frac {1}{2} t \sqrt {\frac {16}{9} \pi ^2 (2 K[1]-1)^2-4 \pi ^2}\right )}{5 \pi ^2 (1-2 K[1])^2 \sqrt {16 K[1]^2-16 K[1]-5}}\right ) & K[1]\in \mathbb {Z}\land K[1]\geq 2 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
Maple ✓
restart; L:=3; c:=4; h:=1/10; b:=3/2*Pi*c/L; f:=piecewise(0<x and x<L/3,3*h/L*x,L/3<x and x<L,h); pde := diff(u(x,t),t$2) + b*diff(u(x,t),t) = c^2*diff(u(x,t),x$2); bc := u(0,t)=0,D[1](u)(L,t)=0; ic := D[2](u)(x,0)=0,u(x,0)=f; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc,ic],u(x,t)) ),output='realtime'));
\[u \left ( x,t \right ) =\sum _{n=0}^{\infty }{\frac {3}{5\,{\pi }^{2} \left ( 1+2\,n \right ) ^{2}}\sin \left ( {\frac { \left ( 1+2\,n \right ) \pi \,x}{6}} \right ) \left ( \sqrt {3}\sin \left ( {\frac {n\pi }{3}} \right ) +\cos \left ( {\frac {n\pi }{3}} \right ) \right ) \left ( \left ( \sqrt {-16\,{n}^{2}-16\,n+5}+3 \right ) {{\rm e}^{{\frac {\pi \,t}{3} \left ( \sqrt {-16\,{n}^{2}-16\,n+5}-3 \right ) }}}+{{\rm e}^{-{\frac {\pi \,t}{3} \left ( \sqrt {-16\,{n}^{2}-16\,n+5}+3 \right ) }}} \left ( \sqrt {-16\,{n}^{2}-16\,n+5}-3 \right ) \right ) {\frac {1}{\sqrt {-16\,{n}^{2}-16\,n+5}}}}\]
Hand solution
Solving the wave PDE on string underdamped case \(t>0,0<x<L\)\[ u_{tt}+bu_{t}=c^{2}u_{xx}\qquad 0<x<L,t>0 \] Boundary conditions\begin {align*} u\left ( 0,t\right ) & =0\\ \left . \frac {\partial u}{\partial x}\right \vert _{x=L} & =0 \end {align*}
Initial conditions, \(t=0\)\begin {align*} u_{t}\left ( x,0\right ) & =0\\ u\left ( x,0\right ) & =f\left ( x\right ) \end {align*}
Using \begin {align*} f\left ( x\right ) & =\left \{ \begin {array} [c]{ccc}\frac {3h}{L}x & & 0<x<\frac {L}{3}\\ h & & \frac {L}{3}<x<L \end {array} \right . \\ b & =2\pi \\ c & =4\\ L & =3\\ h & =\frac {1}{10} \end {align*}
Hence the PDE becomes \(u_{tt}+2\pi u_{t}=16u_{xx}\). The general solution to the above PDE was given in problem 5.1.1.23 on page 1318. The eigenvalues are given as \[ \lambda _{n}=\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}\qquad n=0,1,2,\cdots \] And the discriminant is \(b^{2}-4c^{2}\lambda _{n}=b^{2}-4c^{2}\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}=b^{2}-\left ( 4\right ) \left ( 16\right ) \left ( \frac {\left ( 2n+1\right ) ^{2}\pi ^{2}}{36}\right ) \). For \(n=0\) this gives\(\ b^{2}-\left ( \frac {4}{3}\pi \right ) ^{2}\). But\(\ b=2\pi \). Hence discriminant is positive for \(n=0\). This means this is critically damped in first mode. Using the the solution for this case from the above general solution as
\begin {align*} u\left ( x,t\right ) & =e^{\frac {-b}{2}t}\left ( \frac {\frac {b}{2}+\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}{\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}\right ) \frac {2}{L}\left ( \int _{0}^{L}f\left ( s\right ) \sin \left ( \sqrt {\frac {\pi }{2L}}s\right ) ds\right ) e^{\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}t}\sin \left ( \frac {\pi }{2L}x\right ) \\ & +e^{\frac {-b}{2}t}\left ( \frac {-\frac {b}{2}+\frac {1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}{\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}}\right ) \frac {2}{L}\left ( \int _{0}^{L}f\left ( s\right ) \sin \left ( \sqrt {\frac {\pi }{2L}}s\right ) ds\right ) e^{\frac {-1}{2}\sqrt {b^{2}-4c^{2}\left ( \frac {\pi }{2L}\right ) ^{2}}t}\sin \left ( \frac {\pi }{2L}x\right ) \\ & +\sum _{n=1}^{\infty }\left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac {-b}{2}t}\cos \left ( \frac {1}{2}\sqrt {4c^{2}\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}t\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) \\ & +\sum _{n=1}^{\infty }\left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}s\right ) ds\right ) e^{\frac {-b}{2}t}\frac {b\sin \left ( \frac {1}{2}\sqrt {4c^{2}\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}t\right ) }{\sqrt {4c^{2}\left ( \frac {\left ( 2n+1\right ) \pi }{2L}\right ) ^{2}-b^{2}}}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) \end {align*}
Replacing given values in the above solution results in\begin {align} u\left ( x,t\right ) & =e^{-\pi t}\left ( \frac {\pi +\frac {1}{2}\sqrt {4\pi ^{2}-64\left ( \frac {\pi }{6}\right ) ^{2}}}{\sqrt {4\pi ^{2}-64\left ( \frac {\pi }{6}\right ) ^{2}}}\right ) \frac {2}{3}\left ( \int _{0}^{3}f\left ( s\right ) \sin \left ( \sqrt {\frac {\pi }{6}}s\right ) ds\right ) e^{\frac {1}{2}\sqrt {4\pi ^{2}-64\left ( \frac {\pi }{6}\right ) ^{2}}t}\sin \left ( \frac {\pi }{6}x\right ) \tag {1}\\ & +e^{-\pi t}\left ( \frac {-\pi +\frac {1}{2}\sqrt {4\pi ^{2}-64\left ( \frac {\pi }{6}\right ) ^{2}}}{\sqrt {4\pi ^{2}-64\left ( \frac {\pi }{6}\right ) ^{2}}}\right ) \frac {2}{3}\left ( \int _{0}^{3}f\left ( s\right ) \sin \left ( \sqrt {\frac {\pi }{6}}s\right ) ds\right ) e^{\frac {-1}{2}\sqrt {4\pi ^{2}-64\left ( \frac {\pi }{6}\right ) ^{2}}t}\sin \left ( \frac {\pi }{6}x\right ) \nonumber \\ & +\sum _{n=1}^{\infty }\left ( \frac {2}{3}\int _{0}^{3}f\left ( s\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}s\right ) ds\right ) e^{-\pi t}\cos \left ( \frac {1}{2}\sqrt {64\left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-4\pi ^{2}}t\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}x\right ) \nonumber \\ & +\sum _{n=1}^{\infty }\left ( \frac {2}{3}\int _{0}^{3}f\left ( s\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}s\right ) ds\right ) e^{-\pi t}\frac {2\pi \sin \left ( \frac {1}{2}\sqrt {64\left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-4\pi ^{2}}t\right ) }{\sqrt {64\left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-4\pi ^{2}}}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}x\right ) \nonumber \end {align}
But
\[ \int _{0}^{3}f\left ( x\right ) \sin \left ( \frac {\pi }{6}x\right ) dx=\frac {9}{5\pi ^{2}}\]
And\begin {align*} \int _{0}^{3}f\left ( s\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}s\right ) & =\frac {1}{10}\int _{0}^{1}x\sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}x\right ) dx+\frac {1}{10}\int _{1}^{3}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}x\right ) dx\\ & =\frac {18}{5\pi ^{2}\left ( 2n+1\right ) ^{2}}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) \end {align*}
Hence the solution (1) becomes
\begin {align} u\left ( x,t\right ) & =e^{-\pi t}\left ( \frac {\pi +\sqrt {\pi ^{2}-16\left ( \frac {\pi }{6}\right ) ^{2}}}{2\sqrt {\pi ^{2}-16\left ( \frac {\pi }{6}\right ) ^{2}}}\right ) \left ( \frac {6}{5\pi ^{2}}\right ) e^{\sqrt {\pi ^{2}-16\left ( \frac {\pi }{6}\right ) ^{2}}t}\sin \left ( \frac {\pi }{6}x\right ) \tag {2}\\ & +e^{-\pi t}\left ( \frac {-\pi +\sqrt {\pi ^{2}-16\left ( \frac {\pi }{6}\right ) ^{2}}}{2\sqrt {\pi ^{2}-16\left ( \frac {\pi }{6}\right ) ^{2}}}\right ) \left ( \frac {6}{5\pi ^{2}}\right ) e^{-\sqrt {\pi ^{2}-16\left ( \frac {\pi }{6}\right ) ^{2}}t}\sin \left ( \frac {\pi }{6}x\right ) \nonumber \\ & +\sum _{n=1}^{\infty }\frac {12}{5\pi ^{2}\left ( 2n+1\right ) ^{2}}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) e^{-\pi t}\cos \left ( \sqrt {16\left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\pi ^{2}}t\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}x\right ) \nonumber \\ & +\sum _{n=1}^{\infty }\frac {12}{5\pi ^{2}\left ( 2n+1\right ) ^{2}}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) e^{-\pi t}\frac {\pi \sin \left ( \sqrt {16\left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\pi ^{2}}t\right ) }{\sqrt {16\left ( \frac {\left ( 2n+1\right ) \pi }{6}\right ) ^{2}-\pi ^{2}}}\sin \left ( \frac {\left ( 2n+1\right ) \pi }{6}x\right ) \nonumber \end {align}
Animation is below
Source code used for the above
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Added July 2, 2018. This is Example 2 (pde 10) taken from Maple document What_is_New_after_Maple_2018.pdf
Solve \[ -u_{tt} + u(x,t)= u_{xx} + 2 e^{-t} \left ( x - \frac {1}{2} x^2 + \frac {1}{2} t - 1 \right ) \] With boundary condition \begin {align*} u(0,t) &= 0 \\ \frac {\partial u(1,t)}{\partial x} &= 0 \end {align*}
And initial conditions \begin {align*} u(x,0) &= x^2-2 x \\ u(x,1)&= u(x,\frac {1}{2}) + e^{-1} \left ( \frac {1}{2} x^2-x\right ) - \left ( \frac {3}{4} x^2- \frac {3}{2}x \right ) e^{\frac {-1}{2}} \end {align*}
Mathematica ✗
ClearAll["Global`*"]; pde = -D[u[x, t], {t, 2}] + u[x, t] == D[u[x, t], {x, 2}] + 2*Exp[-t]*(x - (1/2)*x^2 + (1/2)*t - 1); bc = {u[0, t] == 0, Derivative[1, 0][u][1, t] == 0}; ic = {u[x, 0] == x^2 - 2*x, u[x, 1] == u[x, 1/2] + ((1/2)*x^2 - x)*Exp[-1] - ((3*x^2)/4 - (3/2)*x)*Exp[-2^(-1)]}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], x, t], 60*10]];
Failed
Maple ✓
restart; pde := -diff(u(x, t), t, t) + u(x, t) = diff(u(x, t), x, x)+ 2*exp(-t)*(x-(1/2)*x^2+(1/2)*t-1); ic := u(x, 0) = x^2-2*x, u(x, 1) = u(x, 1/2)+((1/2)*x^2-x)*exp(-1)-(3/4*(x^2)-3/2*x)*exp(-1/2); bc := u(0, t) = 0, eval(diff(u(x, t), x), {x = 1}) = 0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic, bc],u(x,t))),output='realtime'));
\[u \left ( x,t \right ) =-{\frac {x{{\rm e}^{-t}} \left ( -2+x \right ) \left ( t-2 \right ) }{2}}\]
Hand solution
Solve \begin {equation} -u_{tt}+u=u_{xx}+2e^{-t}\left ( x-\frac {1}{2}x^{2}+\frac {1}{2}t-1\right ) \qquad t>0,0<x<1\tag {1} \end {equation} Boundary conditions\begin {align*} u\left ( 0,t\right ) & =0\\ \left . \frac {\partial u}{\partial x}\right \vert _{x=1} & =0 \end {align*}
Initial conditions, \(t=0\)\begin {align*} u\left ( x,0\right ) & =x^{2}-2x\\ u\left ( x,1\right ) & =u\left ( x,\frac {1}{2}\right ) +e^{-1}\left ( \frac {1}{2}x^{2}-x\right ) -\left ( \frac {3}{4}x^{2}-\frac {3}{2}x\right ) e^{\frac {-1}{2}} \end {align*}
Since boundary conditions are homogeneous, we can directly use eigenfunction expansion method. Let the solution be \begin {equation} u\left ( x,t\right ) =\sum _{n=1}^{\infty }c_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \tag {2} \end {equation} Where \(\Phi _{n}\left ( x\right ) \) are the eigenfunctions of the corresponding homogeneous PDE \(-u_{tt}+u=u_{xx}\). Using separation of variables, Let \(u=X\left ( x\right ) T\left ( t\right ) .\) Substituting this back in \(-u_{tt}+u=u_{xx}\) gives\begin {align*} -T^{\prime \prime }X+XT & =X^{\prime \prime }T\\ -\frac {T^{\prime \prime }}{T}+1 & =\frac {X^{\prime \prime }}{X}=-\lambda \end {align*}
The eigenvalue ODE is\begin {align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X^{\prime }\left ( 1\right ) & =0 \end {align*}
This is known to have the eigenvalues are \(\lambda _{n}=\left ( \frac {n\pi }{2L}\right ) ^{2}=\left ( \frac {n\pi }{2}\right ) ^{2}\,\), since \(L=1\). This is for \(n=1,3,5,\cdots \) and the eigenfunctions are \(\Phi _{n}\left ( x\right ) =\sin \left ( \sqrt {\lambda _{n}}x\right ) =\sin \left ( \frac {n\pi }{2}x\right ) \). Therefore the solution (2) is\begin {align*} u\left ( x,t\right ) & =\sum _{n=1,3,5,\cdots }^{\infty }c_{n}\left ( t\right ) \sin \left ( \frac {n\pi }{2}x\right ) \\ & =\sum _{n=1,3,5,\cdots }^{\infty }c_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \end {align*}
Substituting this back into (1) gives\begin {equation} -\sum _{n=1,3,5,\cdots }^{\infty }c_{n}^{\prime \prime }\left ( t\right ) \Phi _{n}\left ( x\right ) +\sum _{n=1,3,5,\cdots }^{\infty }c_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) =\sum _{n=1,3,5,\cdots }^{\infty }c_{n}\left ( t\right ) \Phi _{n}^{\prime \prime }\left ( x\right ) +\sum _{n=1,3,5,\cdots }^{\infty }b_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \tag {3} \end {equation} Where \(\sum _{n=1,3,5,\cdots }^{\infty }b_{n}\left ( t\right ) \Phi \left ( x\right ) =2e^{-t}\left ( x-\frac {1}{2}x^{2}+\frac {1}{2}t-1\right ) \). By orthogonality this becomes\begin {align*} \int _{0}^{1}2e^{-t}\left ( x-\frac {1}{2}x^{2}+\frac {1}{2}t-1\right ) \Phi _{n}\left ( x\right ) dx & =b_{n}\left ( t\right ) \int _{0}^{1}\Phi _{n}^{2}\left ( x\right ) dx\\ 2e^{-t}\int _{0}^{1}\left ( x-\frac {1}{2}x^{2}+\frac {1}{2}t-1\right ) \sin \left ( \frac {n\pi }{2}x\right ) dx & =\frac {1}{2}b_{n}\left ( t\right ) \end {align*}
But \(2e^{-t}\int _{0}^{1}\left ( x-\frac {1}{2}x^{2}+\frac {1}{2}t-1\right ) \sin \left ( \frac {n\pi }{2}x\right ) dx=\frac {2e^{-t}\left ( 8+\left ( t-2\right ) n^{2}\pi ^{2}\right ) }{n^{3}\pi ^{3}}\). Hence the above gives\[ b_{n}\left ( t\right ) =\frac {4e^{-t}\left ( 8+\left ( t-2\right ) n^{2}\pi ^{2}\right ) }{n^{3}\pi ^{3}}\] Substituting the above into (3) gives\[ -\sum _{n=1,3,5,\cdots }^{\infty }c_{n}^{\prime \prime }\left ( t\right ) \Phi _{n}\left ( x\right ) +\sum _{n=1,3,5,\cdots }^{\infty }c_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) =\sum _{n=1,3,5,\cdots }^{\infty }c_{n}\left ( t\right ) \Phi _{n}^{\prime \prime }\left ( x\right ) +\sum _{n=1,3,5,\cdots }^{\infty }\frac {4e^{-t}\left ( 8+\left ( t-2\right ) n^{2}\pi ^{2}\right ) }{n^{3}\pi ^{3}}\Phi _{n}\left ( x\right ) \] But \(\Phi _{n}^{\prime \prime }\left ( x\right ) =-\lambda _{n}\Phi _{n}\left ( x\right ) \), hence the above simplifies to\begin {align*} -c_{n}^{\prime \prime }\left ( t\right ) +c_{n}\left ( t\right ) & =-\lambda _{n}c_{n}\left ( t\right ) +\frac {4e^{-t}\left ( 8+\left ( t-2\right ) n^{2}\pi ^{2}\right ) }{n^{3}\pi ^{3}}\\ -c_{n}^{\prime \prime }\left ( t\right ) +\left ( 1+\lambda _{n}\right ) c_{n}\left ( t\right ) & =\frac {4e^{-t}\left ( 8+\left ( t-2\right ) n^{2}\pi ^{2}\right ) }{n^{3}\pi ^{3}}\\ c_{n}^{\prime \prime }\left ( t\right ) -\left ( 1+\frac {n^{2}\pi ^{2}}{4}\right ) c_{n}\left ( t\right ) & =-\frac {4e^{-t}\left ( 8+\left ( t-2\right ) n^{2}\pi ^{2}\right ) }{n^{3}\pi ^{3}} \end {align*}
The solution to this second order ODE can be found to be\[ c_{n}\left ( t\right ) =A_{n}e^{\frac {-\sqrt {n^{2}\pi ^{2}+4}t}{2}}+B_{n}e^{\frac {\sqrt {n^{2}\pi ^{2}+4}t}{2}}+\frac {16\left ( t-2\right ) e^{-t}}{n^{3}\pi ^{3}}\] Hence (2) becomes\begin {equation} u\left ( x,t\right ) =\sum _{n=1,3,5,\cdots }^{\infty }\left ( A_{n}e^{\frac {-\sqrt {n^{2}\pi ^{2}+4}t}{2}}+B_{n}e^{\frac {\sqrt {n^{2}\pi ^{2}+4}t}{2}}+\frac {16\left ( t-2\right ) e^{-t}}{n^{3}\pi ^{3}}\right ) \sin \left ( \frac {n\pi }{2}x\right ) \tag {4} \end {equation} At \(t=0\) the above becomes\[ x^{2}-2x=\sum _{n=1,3,5,\cdots }^{\infty }\left ( A_{n}+B_{n}-\frac {32}{n^{3}\pi ^{3}}\right ) \sin \left ( \frac {n\pi }{2}x\right ) \] Applying orthogonality gives\[ \int _{0}^{1}\left ( x^{2}-2x\right ) \sin \left ( \frac {n\pi }{2}x\right ) dx=\left ( A_{n}+B_{n}-\frac {32}{n^{3}\pi ^{3}}\right ) \frac {1}{2}\] But \(\int _{0}^{1}\left ( x^{2}-2x\right ) \sin \left ( \frac {n\pi }{2}x\right ) dx=-\frac {16}{n^{3}\pi ^{3}}\), hence the above gives\begin {align} -\frac {32}{n^{3}\pi ^{3}} & =A_{n}+B_{n}-\frac {32}{n^{3}\pi ^{3}}\tag {5}\\ A_{n} & =-B_{n}\nonumber \end {align}
Therefore the solution (4) now becomes\begin {equation} u\left ( x,t\right ) =\sum _{n=1,3,5,\cdots }^{\infty }A_{n}\left ( e^{\frac {-\sqrt {n^{2}\pi ^{2}+4}t}{2}}-e^{\frac {\sqrt {n^{2}\pi ^{2}+4}t}{2}}+\frac {16\left ( t-2\right ) e^{-t}}{n^{3}\pi ^{3}}\right ) \sin \left ( \frac {n\pi }{2}x\right ) \tag {6} \end {equation} The second initial conditions is \(u\left ( x,1\right ) =u\left ( x,\frac {1}{2}\right ) +e^{-1}\left ( \frac {1}{2}x^{2}-x\right ) -\left ( \frac {3}{4}x^{2}-\frac {3}{2}x\right ) e^{\frac {-1}{2}}\). At \(t=1\) the above gives\[ u\left ( x,1\right ) =\sum _{n=1,3,5,\cdots }^{\infty }A_{n}\left ( e^{\frac {-\sqrt {n^{2}\pi ^{2}+4}}{2}}-e^{\frac {\sqrt {n^{2}\pi ^{2}+4}}{2}}-\frac {16e^{-1}}{n^{3}\pi ^{3}}\right ) \sin \left ( \frac {n\pi }{2}x\right ) \] At \(t=\frac {1}{2}\) Eq (6) gives\[ u\left ( x,\frac {1}{2}\right ) =\sum _{n=1,3,5,\cdots }^{\infty }A_{n}\left ( e^{\frac {-\sqrt {n^{2}\pi ^{2}+4}}{4}}-e^{\frac {\sqrt {n^{2}\pi ^{2}+4}}{4}}-\frac {24e^{-\frac {1}{2}}}{n^{3}\pi ^{3}}\right ) \sin \left ( \frac {n\pi }{2}x\right ) \] Hence the second initial conditions implies \begin {multline*} \sum _{n=1,3,5,\cdots }^{\infty }A_{n}\left ( e^{\frac {-\sqrt {n^{2}\pi ^{2}+4}}{2}}-e^{\frac {\sqrt {n^{2}\pi ^{2}+4}}{2}}-\frac {16}{n^{3}\pi ^{3}}e^{-1}\right ) \sin \left ( \frac {n\pi }{2}x\right ) \\ -\sum _{n=1,3,5,\cdots }^{\infty }A_{n}\left ( e^{\frac {-\sqrt {n^{2}\pi ^{2}+4}}{4}}-e^{\frac {\sqrt {n^{2}\pi ^{2}+4}}{4}}-\frac {24}{n^{3}\pi ^{3}}e^{-\frac {1}{2}}\right ) \sin \left ( \frac {n\pi }{2}x\right ) =e^{-1}\left ( \frac {1}{2}x^{2}-x\right ) -\left ( \frac {3}{4}x^{2}-\frac {3}{2}x\right ) e^{\frac {-1}{2}} \end {multline*} Or\begin {multline*} \sum _{n=1,3,5,\cdots }^{\infty }A_{n}\left ( -2\sinh \left ( \frac {\sqrt {n^{2}\pi ^{2}+4}}{2}\right ) -\frac {16}{n^{3}\pi ^{3}}e^{-1}\right ) \sin \left ( \frac {n\pi }{2}x\right ) \\ -\sum _{n=1,3,5,\cdots }^{\infty }A_{n}\left ( -2\sinh \left ( \frac {\sqrt {n^{2}\pi ^{2}+4}}{4}\right ) -\frac {24}{n^{3}\pi ^{3}}e^{-\frac {1}{2}}\right ) \sin \left ( \frac {n\pi }{2}x\right ) =e^{-1}\left ( \frac {1}{2}x^{2}-x\right ) -\left ( \frac {3}{4}x^{2}-\frac {3}{2}x\right ) e^{\frac {-1}{2}} \end {multline*}Simplifying gives\begin {multline*} \sum _{n=1,3,5,\cdots }^{\infty }A_{n}\left ( 2\sinh \left ( \frac {\sqrt {n^{2}\pi ^{2}+4}}{4}\right ) -2\sinh \left ( \frac {\sqrt {n^{2}\pi ^{2}+4}}{2}\right ) -\frac {16}{n^{3}\pi ^{3}}e^{-1}+\frac {24}{n^{3}\pi ^{3}}e^{-\frac {1}{2}}\right ) \sin \left ( \frac {n\pi }{2}x\right ) =\\ e^{-1}\left ( \frac {1}{2}x^{2}-x\right ) -\left ( \frac {3}{4}x^{2}-\frac {3}{2}x\right ) e^{\frac {-1}{2}} \end {multline*}Applying orthogonality gives\[ A_{n}\left ( 2\sinh \left ( \frac {\sqrt {n^{2}\pi ^{2}+4}}{4}\right ) -2\sinh \left ( \frac {\sqrt {n^{2}\pi ^{2}+4}}{2}\right ) -\frac {16}{n^{3}\pi ^{3}}e^{-1}+\frac {24}{n^{3}\pi ^{3}}e^{-\frac {1}{2}}\right ) \frac {1}{2}=\int _{0}^{1}\left ( e^{-1}\left ( \frac {1}{2}x^{2}-x\right ) -\left ( \frac {3}{4}x^{2}-\frac {3}{2}x\right ) e^{\frac {-1}{2}}\right ) \sin \left ( \frac {n\pi }{2}x\right ) dx \]But \(\int _{0}^{1}\left ( e^{-1}\left ( \frac {1}{2}x^{2}-x\right ) -\left ( \frac {3}{4}x^{2}-\frac {3}{2}x\right ) e^{\frac {-1}{2}}\right ) \sin \left ( \frac {n\pi }{2}x\right ) dx=-\frac {8e^{-\frac {1}{2}}-12e^{-1}}{n^{3}\pi ^{3}}\), hence the above becomes\begin {align*} A_{n}\left ( 2\sinh \left ( \frac {\sqrt {n^{2}\pi ^{2}+4}}{4}\right ) -2\sinh \left ( \frac {\sqrt {n^{2}\pi ^{2}+4}}{2}\right ) -\frac {16}{n^{3}\pi ^{3}}e^{-1}+\frac {24}{n^{3}\pi ^{3}}e^{-\frac {1}{2}}\right ) & =-\frac {16e^{-1}}{n^{3}\pi ^{3}}+\frac {24e^{-\frac {1}{2}}}{n^{3}\pi ^{3}}\\ 2A_{n}\left ( \sinh \left ( \frac {\sqrt {n^{2}\pi ^{2}+4}}{4}\right ) -\sinh \left ( \frac {\sqrt {n^{2}\pi ^{2}+4}}{2}\right ) \right ) +A_{n}\left ( -\frac {16e^{-1}}{n^{3}\pi ^{3}}+\frac {24e^{-\frac {1}{2}}}{n^{3}\pi ^{3}}\right ) & =-\frac {16e^{-1}}{n^{3}\pi ^{3}}+\frac {24e^{-\frac {1}{2}}}{n^{3}\pi ^{3}} \end {align*}
Since this is true for all \(n=1,3,5,\cdots \) then\begin {align*} A_{n}\left ( \sinh \left ( \frac {\sqrt {n^{2}\pi ^{2}+4}}{4}\right ) -\sinh \left ( \frac {\sqrt {n^{2}\pi ^{2}+4}}{2}\right ) \right ) & =0\\ A_{n} & =1 \end {align*}
Which implies \(\sinh \left ( \frac {\sqrt {n^{2}\pi ^{2}+4}}{4}\right ) -\sinh \left ( \frac {\sqrt {n^{2}\pi ^{2}+4}}{2}\right ) =0\) but this is not possible for \(n=1,3,5,\cdots \). Something went wrong. I need to look at this again.
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Added December 20, 2018.
Left end fixed, right end oscillates, initially at rest. With source that depends on time and space.
Example 19, Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018
Solve for \(u(x,t)\) with \(0<x<\pi \) and \(t>0\) \[ \frac {\partial ^2 u}{\partial t^2} = 4 \frac {\partial ^2 u}{\partial x^2} + (1+t) x \] With boundary conditions \begin {align*} u(0,t) &= 0\\ u(\pi ,0) &=\sin (t) \end {align*}
With initial conditions \begin {align*} u(x,0) &= 0\\ \frac {\partial u}{\partial t}(x,0) &=0 \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], {t, 2}] == 4*D[u[x, t], {x, 2}] + (1 + t)*x; ic = {u[x, 0] == 0, Derivative[0, 1][u][x, 0] == 0}; bc = {u[0, t] == 0, u[Pi, t] == Sin[t]}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}], 60*10]];
\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {x \sin (t)}{\pi }+\underset {K[1]=1}{\overset {\infty }{\sum }}\sqrt {\frac {2}{\pi }} \sin (x K[1]) \left (\frac {(-1)^{K[1]} \left (-8 \sin (t) K[1]^3+2 \pi (-t+\cos (2 t K[1])-1) \left (4 K[1]^2-1\right ) K[1]+\left (4 (1+\pi ) K[1]^2-\pi \right ) \sin (2 t K[1])\right )}{4 \sqrt {2 \pi } K[1]^4 \left (4 K[1]^2-1\right )}+\frac {(-1)^{K[1]} \sin \left (2 t \sqrt {K[1]^2}\right )}{\sqrt {2 \pi } | K[1]| K[1]}\right ) & K[1]\in \mathbb {Z}\land K[1]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
Maple ✓
restart; pde := diff(u(x, t), t$2) = 4*diff(u(x, t), x$2)+(1+t)*x; bc := u(0,t)=0,u(Pi,t)=sin(t); ic := u(x,0)=0,eval(diff(u(x,t),t),t=0)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc,ic],u(x,t))),output='realtime'));
\[u \left ( x,t \right ) =\sum _{n=1}^{\infty }4\,{\frac { \left ( \left ( {n}^{4}+1/4\,\pi \,{n}^{2}-\pi /16 \right ) \sin \left ( 2\,nt \right ) -1/2\,n \left ( \left ( -\pi \,{n}^{2}+\pi /4 \right ) \cos \left ( 2\,nt \right ) +\sin \left ( t \right ) {n}^{2}+ \left ( t+1 \right ) \left ( 1/2+n \right ) \pi \, \left ( n-1/2 \right ) \right ) \right ) \left ( -1 \right ) ^{n}\sin \left ( nx \right ) }{\pi \,{n}^{4} \left ( 4\,{n}^{2}-1 \right ) }}+{\frac {x\sin \left ( t \right ) }{\pi }}\]
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Added May 26, 2019.
Taken from midterm 2 sample exam. UMN Math 5587, Fall 2016. Problem 8
Solve for \(u(x,t)\) with \(-\pi <x<\pi \) and \(t>0\) \[ u_{tt} = u_{xx} \] With boundary conditions \begin {align*} u(-\pi ,t) &= u(\pi ,t)\\ u_x(-\pi ,0) &=u_x(\pi ,t) \end {align*}
With initial conditions \begin {align*} u(x,0) &= x\\ u_t(x,0) &=0 \end {align*}
Mathematica ✗
ClearAll["Global`*"]; pde = D[u[x, t], {t, 2}] == D[u[x, t], {x, 2}] ; ic = {u[x, 0] == x, Derivative[0, 1][u][x, 0] == 0}; bc = {u[-Pi, t] == u[Pi,t], Derivative[1, 0][u][-Pi, t] ==Derivative[1, 0][u][Pi, t] }; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}], 60*10]];
Failed
Maple ✗
restart; pde := diff(u(x, t), t$2) = diff(u(x, t), x$2); bc := u(-Pi,t)=u(Pi,t),eval(diff(u(x,t),x),x=-Pi)=eval(diff(u(x,t),x),x=Pi); ic := u(x,0)=x,eval(diff(u(x,t),t),t=0)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc,ic],u(x,t))),output='realtime'));
sol=()
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Added May 26, 2019.
Taken from midterm 2 sample exam. UMN Math 5587, Fall 2016. Problem 10
Solve for \(u(x,t)\) with \(0<x<\pi \) and \(t>0\) \[ u_{tt} = u_{xx} \] With boundary conditions \(u(0,t) = u_t(\pi ,t)\) and initial conditions \begin {align*} u(x,0) &= 0\\ u_t(x,0) &=1 \end {align*}
Mathematica ✗
ClearAll["Global`*"]; pde = D[u[x, t], {t, 2}] == D[u[x, t], {x, 2}] ; ic = {u[x, 0] == 0, Derivative[0, 1][u][x, 0] == 1}; bc = u[0, t] == Derivative[1, 0][u][Pi, t] ; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}], 60*10]];
Failed
Maple ✗
restart; pde := diff(u(x, t), t$2) = diff(u(x, t), x$2); bc := u(0,t)=eval(diff(u(x,t),x),x=Pi); ic := u(x,0)=0,eval(diff(u(x,t),t),t=0)=1; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc,ic],u(x,t))),output='realtime'));
sol=()
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Added January 12, 2020.
Solve for \(u(x,t)\) with \(0<x<L\) and \(t>0\) \[ u_{tt} = c^2 u_{xx} \] With boundary conditions \(u(0,t) = 0, u_x(L,t)=C\) and zero initial conditions \begin {align*} u(x,0) &= 0\\ u_t(x,0) &=0 \end {align*}
For animations use \(L=10,c=1,C=5\)
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], {t, 2}] == c^2*D[u[x, t], {x, 2}]; ic = {u[x, 0] == 0, Derivative[0, 1][u][x, 0] == 0}; bc = {u[0, t] == 0, Derivative[1, 0][u][L, t] == C0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t},Assumptions->L>0], 60*10]];
\[\left \{\left \{u(x,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \text {C0} x+\underset {K[1]=1}{\overset {\infty }{\sum }}\frac {8 (-1)^{K[1]} \text {C0} L \cos \left (\frac {\pi t \sqrt {c^2 (2 K[1]-1)^2}}{2 L}\right ) \sin \left (\frac {\pi x (2 K[1]-1)}{2 L}\right )}{\pi ^2 (1-2 K[1])^2} & K[1]\in \mathbb {Z}\land ((c<0\land K[1]\geq 1)\lor (c>0\land K[1]\geq 1)) \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
Maple ✓
restart; pde := diff(u(x, t), t$2) = c^2*diff(u(x, t), x$2); bc := u(0,t)=0, D[1](u)(L,t)=C0; ic := u(x,0)=0,D[2](u)(x,0)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc,ic],u(x,t)) assuming L>0),output='realtime'));
\[u \left ( x,t \right ) =\sum _{n=0}^{\infty }-8\,{\frac {L \left ( -1 \right ) ^{n}{\it C0}}{ \left ( 1+2\,n \right ) ^{2}{\pi }^{2}}\sin \left ( 1/2\,{\frac { \left ( 1+2\,n \right ) \pi \,x}{L}} \right ) \cos \left ( 1/2\,{\frac {c \left ( 1+2\,n \right ) \pi \,t}{L}} \right ) }+x{\it C0}\]
Hand solution
Let \begin {equation} u\left ( x,t\right ) =v\left ( x,t\right ) +u_{E}\left ( x\right ) \tag {2} \end {equation} \(u_{E}\left ( x\right ) \) is the steady state solution which only needs to satisfy the non-homogeneous boundary conditions. At equilibrium \(\frac {\partial ^{2}u\left ( x,t\right ) }{\partial t^{2}}=0\) and the PDE becomes \(\frac {\partial ^{2}u_{E}\left ( x,t\right ) }{\partial t^{2}}=0\) or the ODE \(\frac {d^{2}u_{E}\left ( x,t\right ) }{dx^{2}}=0\) with B.C. \(u_{E}\left ( 0\right ) =0,u_{E}^{\prime }\left ( L\right ) =C\). The solution to this ODE is\[ u_{E}\left ( x\right ) =c_{1}x+c_{2}\] At first B.C.\[ 0=c_{2}\] Solution becomes \(u_{E}\left ( x\right ) =c_{1}x\). At second B.C. \(u_{E}^{\prime }\left ( x\right ) =c_{1}=C\). Therefore solution is\[ u_{E}\left ( x\right ) =Cx \] Hence\[ u\left ( x,t\right ) =v\left ( x,t\right ) +Cx \] \(v\left ( x,t\right ) \) is the solution to the PDE but with homogeneous B.C. Plugging (2) into (1) gives\[ \frac {\partial ^{2}v\left ( x,t\right ) }{\partial t^{2}}+\frac {\partial ^{2}u_{E}\left ( x,t\right ) }{\partial t^{2}}=c\left ( \frac {\partial ^{2}v\left ( x,t\right ) }{\partial x^{2}}+\frac {\partial ^{2}u_{E}\left ( x,t\right ) }{\partial x^{2}}\right ) \] But \(\frac {\partial ^{2}u_{E}\left ( x,t\right ) }{\partial x^{2}}=0\) and also \(\frac {\partial ^{2}u_{E}\left ( x,t\right ) }{\partial t^{2}}=0\), hence above becomes\[ \frac {\partial ^{2}v\left ( x,t\right ) }{\partial t^{2}}=c\frac {\partial ^{2}v\left ( x,t\right ) }{\partial x^{2}}\] With \(v\left ( x,t\right ) \) having now homogeneous B.C.\begin {align*} v\left ( 0,t\right ) & =0\\ \frac {\partial v\left ( L,t\right ) }{\partial x} & =0 \end {align*}
And initial conditions given by\begin {align*} v\left ( x,0\right ) & =u\left ( x,0\right ) -u_{E}\left ( x\right ) \\ & =0-Cx\\ & =-Cx \end {align*}
And\begin {align*} \frac {\partial v\left ( x,0\right ) }{\partial t} & =\frac {\partial u\left ( x,0\right ) }{\partial t}-\frac {\partial u_{E}\left ( x\right ) }{\partial t}\\ & =0 \end {align*}
In summary, the PDE to solve for \(v\left ( x,t\right ) \) is\begin {align} \frac {\partial ^{2}v\left ( x,t\right ) }{\partial t^{2}} & =c\frac {\partial ^{2}v\left ( x,t\right ) }{\partial x^{2}}\tag {3}\\ v\left ( 0,t\right ) & =0\nonumber \\ \frac {\partial v\left ( L,t\right ) }{\partial x} & =0\nonumber \\ v\left ( x,0\right ) & =-Cx\nonumber \\ \frac {\partial v\left ( x,0\right ) }{\partial t} & =0\nonumber \end {align}
Now we solve for PDE (3) for \(v\left ( x,t\right ) \) using separation of variables since the boundary conditions in space are now homogeneous. Let \(v\left ( x,t\right ) =X\left ( x\right ) T\left ( t\right ) \) and the PDE becomes\[ \frac {1}{c}T^{\prime \prime }X=X^{\prime \prime }T \] Dividing by \(XT\neq 0\) gives\begin {equation} \frac {1}{c}\frac {T^{\prime \prime }}{T}=\frac {X^{\prime \prime }}{X}=-\lambda \tag {4} \end {equation} Where \(\lambda \) is some real positive constant. The space ODE becomes\begin {align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X^{\prime }\left ( L\right ) & =0 \end {align*}
Case \(\lambda <0\): Let The solution is\[ X\left ( x\right ) =c_{1}\cosh \left ( \sqrt {\lambda }x\right ) +c_{2}\sinh \left ( \sqrt {\lambda }x\right ) \] At \(x=0\)\[ 0=c_{1}\] Hence solution becomes \[ X\left ( x\right ) =c_{2}\sinh \left ( \sqrt {\lambda }x\right ) \] Taking derivative\[ X^{\prime }\left ( x\right ) =\sqrt {\lambda }c_{2}\cosh \left ( \sqrt {\lambda }x\right ) \] Using second boundary conditions gives\[ 0=\sqrt {\lambda }c_{2}\cosh \left ( \sqrt {\lambda }L\right ) \] Since \(\cosh \) is zero only when its argument is zero. But we assumed \(\sqrt {\lambda }\) not zero here, then \(c_{2}=0\) in only other choice. Hence this gives trivial solution. Therefore \(\lambda <0\) is not possible.
Case \(\lambda =0\)\begin {align*} X^{\prime \prime } & =0\\ X\left ( 0\right ) & =0\\ X^{\prime }\left ( L\right ) & =0 \end {align*}
Solution is \(X\left ( x\right ) =c_{1}x+c_{2}\). First B.C. gives \(0=c_{2}\). Solution becomes \(X\left ( x\right ) =c_{1}x\). Second B.C. gives \(c_{1}=0\). This gives trivial solution again. Hence \(\lambda =0\) is not possible eigenvalue.
Case \(\lambda >0\): The solution becomes\[ X\left ( x\right ) =B_{1}\cos \left ( \sqrt {\lambda }x\right ) +B_{2}\sin \left ( \sqrt {\lambda }x\right ) \] AT first B.C.\[ 0=B_{1}\] Hence solution becomes\[ X\left ( x\right ) =B_{2}\sin \left ( \sqrt {\lambda }x\right ) \] Taking derivative\[ X^{\prime }\left ( x\right ) =\sqrt {\lambda }B_{2}\cos \left ( \sqrt {\lambda }x\right ) \] At second B.C.\[ 0=\sqrt {\lambda }B_{2}\cos \left ( \sqrt {\lambda }L\right ) \] To avoid trivial solution, take \(\cos \left ( \sqrt {\lambda }L\right ) =0\) or \(\sqrt {\lambda }L=\frac {\pi }{2},\frac {3\pi }{2},\frac {5\pi }{2},\cdots \) or \begin {align} \sqrt {\lambda } & =\frac {\pi }{2L},\frac {3\pi }{2L},\frac {5\pi }{2L},\cdots \nonumber \\ \sqrt {\lambda _{n}} & =\left ( \frac {n\pi }{2L}\right ) \qquad n=1,3,5,\cdots \nonumber \\ & =\frac {\left ( 2n+1\right ) \pi }{2L}\qquad n=0,1,2,\cdots \tag {5} \end {align}
Hence the space solution is\begin {equation} X_{n}\left ( x\right ) =B_{n}\sin \left ( \sqrt {\lambda _{n}}x\right ) \qquad n=1,3,5,\cdots \tag {6} \end {equation} Now we solve the time ODE \(T\left ( t\right ) \) from (4), which is\[ T^{\prime \prime }+\lambda cT=0 \] The solution is\[ T_{n}\left ( t\right ) =D_{n}\cos \left ( \sqrt {\lambda _{n}c}t\right ) +E_{n}\sin \left ( \sqrt {\lambda _{n}c}t\right ) \] Therefore\begin {align*} v\left ( x,t\right ) & =\sum _{n=0}^{\infty }T_{n}\left ( t\right ) X_{n}\left ( x\right ) \\ & =\sum _{n=0}^{\infty }\left ( D_{n}\cos \left ( \sqrt {\lambda _{n}c}t\right ) +E_{n}\sin \left ( \sqrt {\lambda _{n}c}t\right ) \right ) \sin \left ( \sqrt {\lambda _{n}}x\right ) \end {align*}
Where constant \(B_{n}\) merged with the other constants. Now At \(t=0\)\[ -Cx=\sum _{n=0}^{\infty }D_{n}\sin \left ( \sqrt {\lambda _{n}}x\right ) \] Applying orthogonality\begin {align*} -\int _{0}^{L}Cx\sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) dx & =D_{n}\int _{0}^{L}\sin ^{2}\left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) dx\\ -C\int _{0}^{L}x\sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) dx & =D_{n}\frac {L}{2}\\ -C\left ( \frac {4\left ( -1\right ) ^{n}L^{2}}{\left ( \pi +2n\pi \right ) ^{2}}\right ) & =D_{n}\frac {L}{2}\\ D_{n} & =-C\frac {8\left ( -1\right ) ^{n}L}{\left ( \pi +2n\pi \right ) ^{2}} \end {align*}
Therefore\[ v\left ( x,t\right ) =\sum _{n=0}^{\infty }\left ( -C\frac {8\left ( -1\right ) ^{n}L}{\left ( \pi +2n\pi \right ) ^{2}}\cos \left ( \sqrt {\lambda _{n}c}t\right ) +E_{n}\sin \left ( \sqrt {\lambda _{n}c}t\right ) \right ) \sin \left ( \sqrt {\lambda _{n}}x\right ) \] Taking time derivative\[ \frac {\partial v\left ( x,t\right ) }{\partial t}=\sum _{n=0}^{\infty }\left ( C\frac {8\left ( -1\right ) ^{n}L}{\left ( \pi +2n\pi \right ) ^{2}}\frac {\left ( 2n+1\right ) \pi }{2L}\sin \left ( \sqrt {\lambda _{n}c}t\right ) +E_{n}\sqrt {\lambda _{n}c}\cos \left ( \sqrt {\lambda _{n}c}t\right ) \right ) \sin \left ( \sqrt {\lambda _{n}}x\right ) \] At \(t=0\)\[ 0=\sum _{n=0}^{\infty }E_{n}\sqrt {\lambda _{n}c}\sin \left ( \sqrt {\lambda _{n}}x\right ) \] Hence \(E_{n}=0\). Therefore solution becomes\[ v\left ( x,t\right ) =\sum _{n=0}^{\infty }-C\frac {8\left ( -1\right ) ^{n}L}{\left ( \pi +2n\pi \right ) ^{2}}\cos \left ( \sqrt {\lambda _{n}c}t\right ) \sin \left ( \sqrt {\lambda _{n}}x\right ) \] Therefore, since \(u\left ( x,t\right ) =v\left ( x,t\right ) +u_{E}\left ( x\right ) \) then\begin {equation} u\left ( x,t\right ) =Cx-C\sum _{n=0}^{\infty }\frac {8\left ( -1\right ) ^{n}L}{\left ( \pi +2n\pi \right ) ^{2}}\cos \left ( \frac {\left ( 2n+1\right ) \pi }{2L}\sqrt {c}t\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{2L}x\right ) \tag {7} \end {equation} This animation runs for 40 seconds for \(L=10,c=1,C=5\). The solution becomes\begin {equation} u\left ( x,t\right ) =5x-5\sum _{n=0}^{\infty }\frac {80\left ( -1\right ) ^{n}}{\left ( \pi +2n\pi \right ) ^{2}}\cos \left ( \frac {\left ( 2n+1\right ) \pi }{20}\sqrt {10}t\right ) \sin \left ( \frac {\left ( 2n+1\right ) \pi }{20}x\right ) \tag {7} \end {equation}
Code used for the above is
L=10;c=1; C0=5; sqrtLam= (2 n +1) Pi/(2 L); Manipulate[ mysol=C0 x-C0 Sum[(8.0(-1)^n L)/(Pi+2n \[Pi])^2 Sin[sqrtLam x] Cos[sqrtLam Sqrt[c] t],{n,0,numberOfTerms}% ]; Grid[{{Row[{"Solution at time = ",i}]}, {Plot[mysol/.t->i,{x,0,10},PlotRange->{{0,10},{-10,100}}% ,ImageSize->500,GridLines->Automatic,GridLinesStyle->LightGray,PlotStyle->Red]}% }], {{i,0,"time"},0,100,.1,Appearance->"Labeled"}, {{numberOfTerms,1,"n"},1,1000,1,Appearance->"Labeled"} ]
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