Added June 21, 2019
Solve the heat equation for \(u(x,t)\) \[ u_t = k u_{xx} \] For \(0<x<L\) and \(t>0\). The boundary conditions are \begin {align*} u_x(0,t) &= A(t) \\ u(L,t) &= 0 \\ \end {align*}
And initial condition is \(u(x,0)=0\)
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], t] == k*D[u[x, t], {x, 2}]; bc = {Derivative[1,0][u][0, t] == A[t], u[L, t] == 0}; ic = u[x, 0] == 0; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t},Assumptions->{k>0,L>0}], 60*10]];
\[\left \{\left \{u(x,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {\sqrt {2} \cos \left (\frac {\pi x (2 K[1]-1)}{2 L}\right ) \left (\frac {4 \sqrt {2} e^{-\frac {k \pi ^2 t (1-2 K[1])^2}{4 L^2}} A(0) L^{3/2}}{\pi ^2 (1-2 K[1])^2}+\int _0^t \frac {4 \sqrt {2} \exp \left (-\frac {k \pi ^2 (1-2 K[1])^2 (t-K[2])}{4 L^2}\right ) L^{3/2} A'(K[2])}{\pi ^2 (1-2 K[1])^2} \, dK[2]\right )}{\sqrt {L}}+A(t) (x-L)\right \}\right \}\]
Maple ✓
restart; pde := diff(u(x,t),t)= k*diff(u(x,t),x$2); ic := u(x,0)=0; bc := D[1](u)(0,t) =A(t),u(L,t)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde,ic,bc],u(x,t)) assuming k>0,L>0),output='realtime'));
\[u \left (x , t\right ) = \left (-L +x \right ) A \left (t \right )+\int _{0}^{t}\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {8 L \cos \left (\frac {\left (2 n +1\right ) \pi x}{2 L}\right ) \left (\frac {d}{d \tau }A \left (\tau \right )\right ) {\mathrm e}^{-\frac {\left (t -\tau \right ) \left (n +\frac {1}{2}\right )^{2} \pi ^{2} k}{L^{2}}}}{\left (2 n +1\right )^{2} \pi ^{2}}\right )d \tau +8 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {A \left (0\right ) L \cos \left (\frac {\left (2 n +1\right ) \pi x}{2 L}\right ) {\mathrm e}^{-\frac {\pi ^{2} \left (2 n +1\right )^{2} k t}{4 L^{2}}}}{\left (2 n +1\right )^{2} \pi ^{2}}\right )\]
Hand solution
This 1D heat PDE has left one end with boundary condition that is time dependent.\begin {align*} u_{t} & =ku_{xx}\qquad 0<x<L,t>0\\ u_{x}\left ( 0,t\right ) & =A\left ( t\right ) \\ u\left ( L,t\right ) & =0\\ u\left ( x,0\right ) & =0 \end {align*}
Solution
Since the boundary condition is not homogeneous, we need to first find a reference function \(r\left ( x,t\right ) \). Let\[ r\left ( x,t\right ) =A\left ( t\right ) \left ( x-L\right ) \] This function only needs to satisfy the nonhomogeneous boundary conditions given. i.e. \(\frac {\partial r}{\partial x}\left ( 0,t\right ) =A\left ( t\right ) ,r\left ( L,t\right ) =0\). Now we can write\begin {equation} u\left ( x,t\right ) =v\left ( x,t\right ) +r\left ( x,t\right ) \tag {1} \end {equation} Since \(r\left ( x,t\right ) \) satisfies the nonhomogeneous B.C’s, then \(v\left ( x,t\right ) \) satisfies the homogeneous boundary conditions. Substituting the above back into the original PDE gives\begin {align} v_{t}+r_{t} & =k\left ( v_{xx}+r_{xx}\right ) \nonumber \\ v_{t}+A^{\prime }\left ( t\right ) \left ( x-L\right ) & =kv_{xx}\nonumber \\ v_{t} & =kv_{xx}+A^{\prime }\left ( t\right ) \left ( L-x\right ) \nonumber \\ & =kv_{xx}+Q\left ( x,t\right ) \tag {2} \end {align}
The PDE \(v_{t}=kv_{xx}+Q\left ( x,t\right ) \) has now homogeneous B.C. \(v_{x}\left ( 0,t\right ) =0,v\left ( L,0\right ) =0\). Where \(Q\left ( x,t\right ) =A^{\prime }\left ( t\right ) \left ( L-x\right ) \). The method of eigenfunction expansion is now used to solve (2). Let \[ v\left ( x,t\right ) =\sum _{n=1}^{\infty }a_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \] Substituting this back into (2) gives\begin {align*} \sum _{n=1}^{\infty }a_{n}^{\prime }\left ( t\right ) \Phi _{n}\left ( x\right ) & =k\sum _{n=1}^{\infty }a_{n}\left ( t\right ) \Phi _{n}^{\prime \prime }\left ( x\right ) +Q\left ( x,t\right ) \\ & =k\sum _{n=1}^{\infty }a_{n}\left ( t\right ) \Phi _{n}^{\prime \prime }\left ( x\right ) +\sum _{n=1}^{\infty }q_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \end {align*}
Where \(Q\left ( x,t\right ) =\sum _{n=1}^{\infty }q_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \). Now, since \(\Phi _{n}^{\prime \prime }\left ( x\right ) =-\lambda _{n}\Phi _{n}\left ( x\right ) \) then the above reduces to\begin {align} \sum _{n=1}^{\infty }a_{n}^{\prime }\left ( t\right ) \Phi _{n}\left ( x\right ) +k\sum _{n=1}^{\infty }a_{n}\left ( t\right ) \lambda _{n}\Phi _{n}\left ( x\right ) & =\sum _{n=1}^{\infty }q_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \nonumber \\ a_{n}^{\prime }\left ( t\right ) +k\lambda _{n}a_{n}\left ( t\right ) & =q_{n}\left ( t\right ) \tag {3} \end {align}
The eigenfunctions \(\Phi _{n}\left ( x\right ) \) come from solving the eigenvalue problem in \(v_{t}=kv_{xx}\) with homogeneous boundary conditions \(v_{x}\left ( 0,t\right ) =0,v\left ( L,t\right ) =0\). This was solved before in problem 4.1.1.28 on page 527. The eigenfunctions were found to be \(\Phi _{n}\left ( x\right ) =\cos \left ( \sqrt {\lambda _{n}}x\right ) ,n=1,3,5,\cdots \) with eigenvalues \(\lambda _{n}=\left ( \frac {n\pi }{2L}\right ) ^{2},n=1,3,5,\cdots \). Before solving the ODE (3), we need to first find \(q_{n}\left ( t\right ) \). Orthogonality is now used to find \(q_{n}\left ( t\right ) \)\begin {align*} Q\left ( x,t\right ) & =\sum _{n=1,3,5,\cdots }^{\infty }q_{n}\left ( t\right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) \\ A^{\prime }\left ( t\right ) \left ( L-x\right ) & =\sum _{n=1,3,5,\cdots }^{\infty }q_{n}\left ( t\right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) \\ \int _{0}^{L}A^{\prime }\left ( t\right ) \left ( L-x\right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) dx & =\frac {L}{2}q_{n}\left ( t\right ) \\ q_{n}\left ( t\right ) & =\frac {2A^{\prime }\left ( t\right ) }{L}\int _{0}^{L}\left ( L-x\right ) \cos \left ( \frac {n\pi }{2L}x\right ) dx\\ & =-\frac {2A^{\prime }\left ( t\right ) }{L}\left ( \frac {4L^{2}\left ( \cos \left ( \frac {n\pi }{2}\right ) -1\right ) }{\pi ^{2}n^{2}}\right ) \qquad n=1,3,5,\cdots \end {align*}
But \(\cos \left ( \frac {n\pi }{2}\right ) -1=-1\) for \(n=1,3,5,\cdots \). Hence the above becomes\begin {align*} q_{n}\left ( t\right ) & =\frac {2A^{\prime }\left ( t\right ) }{L}\left ( \frac {4L^{2}}{\pi ^{2}n^{2}}\right ) \qquad n=1,3,5,\cdots \\ & =\frac {8A^{\prime }\left ( t\right ) L}{\pi ^{2}n^{2}} \end {align*}
Hence (3) becomes\begin {align*} a_{n}^{\prime }\left ( t\right ) +k\lambda _{n}a_{n}\left ( t\right ) & =q_{n}\left ( t\right ) \qquad n=1,2,3,\cdots \\ a_{n}^{\prime }\left ( t\right ) +k\left ( \frac {n\pi }{2L}\right ) ^{2}a_{n}\left ( t\right ) & =\frac {8A^{\prime }\left ( t\right ) L}{\pi ^{2}n^{2}}\\ a_{n}^{\prime }\left ( t\right ) +k\lambda _{n}a_{n}\left ( t\right ) & =\frac {8A^{\prime }\left ( t\right ) L}{\pi ^{2}n^{2}} \end {align*}
Integrating factor is \(e^{k\lambda _{n}t}\). Hence the above becomes\[ \frac {d}{dt}\left ( a\left ( t\right ) e^{k\lambda _{n}t}\right ) =\frac {8A^{\prime }\left ( t\right ) L}{\pi ^{2}n^{2}}e^{k\lambda _{n}t}\] Integrating gives\begin {align*} a_{n}\left ( t\right ) e^{k\lambda _{n}t} & =\int _{0}^{t}\frac {8A^{\prime }\left ( \tau \right ) L}{\pi ^{2}n^{2}}e^{k\lambda _{n}\tau }d\tau +a_{n}\left ( 0\right ) \\ a_{n}\left ( t\right ) & =a_{n}\left ( 0\right ) e^{-k\lambda _{n}t}+e^{-k\lambda _{n}t}\int _{0}^{t}\frac {8A^{\prime }\left ( \tau \right ) L}{\pi ^{2}n^{2}}e^{k\lambda _{n}\tau }d\tau \end {align*}
Hence\begin {align*} v\left ( x,t\right ) & =\sum _{n=1}^{\infty }a_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \\ & =\sum _{n=1,3,5,\cdots }^{\infty }\left ( a_{n}\left ( 0\right ) e^{-k\lambda _{n}t}+\frac {8L}{\pi ^{2}n^{2}}e^{-k\lambda _{n}t}\int _{0}^{t}A^{\prime }\left ( \tau \right ) e^{k\lambda _{n}\tau }d\tau \right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) \end {align*}
Since \(u\left ( x,t\right ) =v\left ( x,t\right ) +r\left ( x,t\right ) \) then\begin {equation} u\left ( x,t\right ) =A\left ( t\right ) \left ( x-L\right ) +\sum _{n=1,3,5,\cdots }^{\infty }\left ( a_{n}\left ( 0\right ) e^{-k\lambda _{n}t}+\frac {8L}{\pi ^{2}n^{2}}e^{-k\lambda _{n}t}\int _{0}^{t}A^{\prime }\left ( \tau \right ) e^{k\lambda _{n}\tau }d\tau \right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) \tag {4} \end {equation} At \(t=0,u\left ( x,0\right ) =0\) and the above becomes \begin {align*} 0 & =A\left ( 0\right ) \left ( x-L\right ) +\sum _{n=1,3,5,\cdots }^{\infty }a_{n}\left ( 0\right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) \\ A\left ( 0\right ) \left ( L-x\right ) & =\sum _{n=1,3,5,\cdots }^{\infty }a_{n}\left ( 0\right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) \end {align*}
Applying orthogonality\begin {align*} \int _{0}^{L}\left ( L-x\right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) dx & =a_{n}\left ( 0\right ) \frac {L}{2}\\ a_{n}\left ( 0\right ) & =\frac {2A\left ( 0\right ) }{L}\int _{0}^{L}\left ( L-x\right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) dx\\ & =\frac {2A\left ( 0\right ) }{L}\left ( -\frac {\cos \left ( L\sqrt {\lambda _{n}}\right ) -1}{\lambda _{n}}\right ) \end {align*}
But \(\cos \left ( L\sqrt {\lambda _{n}}\right ) -1=\cos \left ( L\frac {n\pi }{2L}\right ) -1=\) \(\cos \left ( \frac {n\pi }{2}\right ) -1=-1\) for \(n=1,3,5,\cdots \), and the above becomes\[ a_{n}\left ( 0\right ) =\frac {2A\left ( 0\right ) }{L\lambda _{n}}\] Therefore the solution (4) is \[ u\left ( x,t\right ) =A\left ( t\right ) \left ( x-L\right ) +\sum _{n=1,3,5,\cdots }^{\infty }e^{-k\lambda _{n}t}\left ( \frac {2A\left ( 0\right ) }{L\lambda _{n}}+\frac {8L}{\pi ^{2}n^{2}}\int _{0}^{t}A^{\prime }\left ( \tau \right ) e^{k\lambda _{n}\tau }\right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) \] Where \(\lambda _{n}=\left ( \frac {n\pi }{2L}\right ) ^{2}\). Hence\[ u\left ( x,t\right ) =A\left ( t\right ) \left ( x-L\right ) +\sum _{n=1,3,5,\cdots }^{\infty }e^{-k\left ( \frac {n\pi }{2L}\right ) ^{2}t}\left ( \frac {2A\left ( 0\right ) }{L\left ( \frac {n\pi }{2L}\right ) ^{2}}+\frac {8L}{\pi ^{2}n^{2}}\int _{0}^{t}A^{\prime }\left ( \tau \right ) e^{k\left ( \frac {n\pi }{2L}\right ) ^{2}\tau }d\tau \right ) \cos \left ( \frac {n\pi }{2L}x\right ) \]
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Added June 22, 2019
Solve the heat equation for \(u(x,t)\) \[ u_t = k u_{xx} \] For \(0<x<L\) and \(t>0\). The boundary conditions are \begin {align*} u_x(0,t) &= A(t) \\ u(L,t) &= 0 \\ \end {align*}
And initial condition is \(u(x,0)=0\). Using the following values \begin {align*} L&=5\\ k&=\frac {1}{100}\\ A(t)=e^t \end {align*}
Mathematica ✓
ClearAll["Global`*"]; L=5; k=1/100; pde = D[u[x, t], t] == k*D[u[x, t], {x, 2}]; bc = {Derivative[1,0][u][0, t] == Exp[t], u[L, t] == 0}; ic = u[x, 0] == 0; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t}], 60*10]];
\[\left \{\left \{u(x,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {40 e^{-\frac {\pi ^2 t (1-2 K[1])^2}{10000}} \cos \left (\frac {1}{10} \pi x (2 K[1]-1)\right ) \left (\pi ^2 (1-2 K[1])^2+10000 e^{\frac {\pi ^2 t (1-2 K[1])^2}{10000}+t}\right )}{\left (\pi ^2 (1-2 K[1])^2+10000\right ) (\pi -2 \pi K[1])^2}+e^t (x-5)\right \}\right \}\]
Maple ✓
restart; k:=1/100; L:=5; pde := diff(u(x,t),t)= k*diff(u(x,t),x$2); ic := u(x,0)=0; bc := D[1](u)(0,t) =exp(t),u(L,t)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde,ic,bc],u(x,t))),output='realtime'));
\[u \left (x , t\right ) = \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {10 \left (2500 \,{\mathrm e}^{t}+\left (n +\frac {1}{2}\right )^{2} \pi ^{2} {\mathrm e}^{-\frac {\pi ^{2} \left (2 n +1\right )^{2} t}{10000}}\right ) \cos \left (\frac {\left (2 n +1\right ) \pi x}{10}\right )}{\left (2500+\left (n +\frac {1}{2}\right )^{2} \pi ^{2}\right ) \left (n +\frac {1}{2}\right )^{2} \pi ^{2}}\right )+\left (x -5\right ) {\mathrm e}^{t}\]
Hand solution
This PDE general solution was obtained in problem 4.1.3.1 on page 649 as\[ u\left ( x,t\right ) =A\left ( t\right ) \left ( x-L\right ) +\sum _{n=1,3,5,\cdots }^{\infty }e^{-k\left ( \frac {n\pi }{2L}\right ) ^{2}t}\left ( \frac {2A\left ( 0\right ) }{L\left ( \frac {n\pi }{2L}\right ) ^{2}}+\frac {8L}{\pi ^{2}n^{2}}\int _{0}^{t}A^{\prime }\left ( \tau \right ) e^{k\left ( \frac {n\pi }{2L}\right ) ^{2}\tau }d\tau \right ) \cos \left ( \frac {n\pi }{2L}x\right ) \] When \(A\left ( t\right ) =e^{t}\), the above becomes\[ u\left ( x,t\right ) =e^{t}\left ( x-L\right ) +\sum _{n=1,3,5,\cdots }^{\infty }e^{-k\left ( \frac {n\pi }{2L}\right ) ^{2}t}\left ( \frac {2}{L\left ( \frac {n\pi }{2L}\right ) ^{2}}+\frac {8L}{\pi ^{2}n^{2}}\int _{0}^{t}e\left ( \tau \right ) e^{k\left ( \frac {n\pi }{2L}\right ) ^{2}\tau }d\tau \right ) \cos \left ( \frac {n\pi }{2L}x\right ) \] But \[ \int _{0}^{t}e\left ( \tau \right ) e^{k\left ( \frac {n\pi }{2L}\right ) ^{2}\tau }d\tau =\frac {e^{k\left ( \frac {n\pi }{2L}\right ) ^{2}t+t}-1}{k\left ( \frac {n\pi }{2L}\right ) ^{2}+1}\] And the general solution becomes\begin {align*} u\left ( x,t\right ) & =e^{t}\left ( x-L\right ) +\sum _{n=1,3,5,\cdots }^{\infty }e^{-k\left ( \frac {n\pi }{2L}\right ) ^{2}t}\left ( \frac {2}{L\left ( \frac {n\pi }{2L}\right ) ^{2}}+\frac {8L}{\pi ^{2}n^{2}}\frac {e^{k\left ( \frac {n\pi }{2L}\right ) ^{2}t+t}-1}{k\left ( \frac {n\pi }{2L}\right ) ^{2}+1}\right ) \cos \left ( \frac {n\pi }{2L}x\right ) \\ & =e^{t}\left ( x-L\right ) +\sum _{n=1,3,5,\cdots }^{\infty }\left ( \frac {2e^{-k\left ( \frac {n\pi }{2L}\right ) ^{2}t}}{L\left ( \frac {n\pi }{2L}\right ) ^{2}}+\frac {8L}{\pi ^{2}n^{2}}\frac {e^{t}-e^{-k\left ( \frac {n\pi }{2L}\right ) ^{2}t}}{k\left ( \frac {n\pi }{2L}\right ) ^{2}+1}\right ) \cos \left ( \frac {n\pi }{2L}x\right ) \\ & =e^{t}\left ( x-L\right ) +\sum _{n=1,3,5,\cdots }^{\infty }e^{-k\left ( \frac {n\pi }{2L}\right ) ^{2}t}\left ( \frac {2}{L\left ( \frac {n\pi }{2L}\right ) ^{2}}-\frac {8L}{\left ( k\left ( \frac {n\pi }{2L}\right ) ^{2}+1\right ) \pi ^{2}n^{2}}\right ) +\frac {8Le^{t}}{\left ( k\left ( \frac {n\pi }{2L}\right ) ^{2}+1\right ) \pi ^{2}n^{2}}\cos \left ( \frac {n\pi }{2L}x\right ) \\ & =e^{t}\left ( x-L\right ) +\sum _{n=1,3,5,\cdots }^{\infty }\left ( \frac {8L}{n^{2}\pi ^{2}}-\frac {8L}{\left ( k\left ( \frac {n\pi }{2L}\right ) ^{2}+1\right ) \pi ^{2}n^{2}}\right ) e^{-k\left ( \frac {n\pi }{2L}\right ) ^{2}t}+\frac {8Le^{t}}{\left ( k\left ( \frac {n\pi }{2L}\right ) ^{2}+1\right ) \pi ^{2}n^{2}}\cos \left ( \frac {n\pi }{2L}x\right ) \end {align*}
In this problem \(L=5,k=\frac {1}{100}\), hence the above becomes\[ u\left ( x,t\right ) =e^{t}\left ( x-5\right ) +\sum _{n=1,3,5,\cdots }^{\infty }\left ( \left ( \frac {40}{n^{2}\pi ^{2}}-\frac {40}{\left ( k\left ( \frac {n\pi }{10}\right ) ^{2}+1\right ) \pi ^{2}n^{2}}\right ) e^{-\frac {n^{2}\pi ^{2}}{10000}t}+\frac {40e^{t}}{\left ( k\left ( \frac {n\pi }{10}\right ) ^{2}+1\right ) \pi ^{2}n^{2}}\right ) \cos \left ( \frac {n\pi }{10}x\right ) \] The following is an animation of the solution
Source code used for the above
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Added June 22, 2019
Solve the heat equation for \(u(x,t)\) \[ u_t = k u_{xx} \] For \(0<x<L\) and \(t>0\). The boundary conditions are \begin {align*} u_x(0,t) &= A(t) \\ u(L,t) &= 0 \\ \end {align*}
And initial condition is \(u(x,0)=0\). Using the following values \begin {align*} L&=5\\ k&=\frac {1}{100}\\ A(t)=\sin (t) \end {align*}
Mathematica ✓
ClearAll["Global`*"]; L=5; k=1/100; pde = D[u[x, t], t] == k*D[u[x, t], {x, 2}]; bc = {Derivative[1,0][u][0, t] == Sin[t], u[L, t] == 0}; ic = u[x, 0] == 0; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], {x, t}], 60*10]];
\[\left \{\left \{u(x,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {400000 \cos \left (\frac {1}{10} \pi x (2 K[1]-1)\right ) \left (\pi ^2 \cos (t) (1-2 K[1])^2-e^{-\frac {\pi ^2 t (1-2 K[1])^2}{10000}} \pi ^2 (1-2 K[1])^2+10000 \sin (t)\right )}{\left (\pi ^4 (1-2 K[1])^4+100000000\right ) (\pi -2 \pi K[1])^2}+(x-5) \sin (t)\right \}\right \}\]
Maple ✓
restart; k:=1/100; L:=5; pde := diff(u(x,t),t)= k*diff(u(x,t),x$2); ic := u(x,0)=0; bc := D[1](u)(0,t) =sin(t),u(L,t)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde,ic,bc],u(x,t))),output='realtime'));
\[u \left (x , t\right ) = 400000 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {\left (\pi ^{2} \left (n +\frac {1}{2}\right )^{2} \cos \left (t \right )-\left (n +\frac {1}{2}\right )^{2} \pi ^{2} {\mathrm e}^{-\frac {\pi ^{2} \left (2 n +1\right )^{2} t}{10000}}+2500 \sin \left (t \right )\right ) \cos \left (\frac {\left (2 n +1\right ) \pi x}{10}\right )}{16 \pi ^{2} \left (6250000+\left (n +\frac {1}{2}\right )^{4} \pi ^{4}\right ) \left (n +\frac {1}{2}\right )^{2}}\right )+\left (x -5\right ) \sin \left (t \right )\]
Hand solution
This PDE general solution was obtained in problem 4.1.3.1 on page 649 as\[ u\left ( x,t\right ) =A\left ( t\right ) \left ( x-L\right ) +\sum _{n=1,3,5,\cdots }^{\infty }e^{-k\lambda _{n}t}\left ( \frac {2A\left ( 0\right ) }{L\lambda _{n}}+\frac {8L}{\pi ^{2}n^{2}}\int _{0}^{t}A^{\prime }\left ( \tau \right ) e^{k\lambda _{n}\tau }d\tau \right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) \] Where \(\lambda _{n}=\left ( \frac {n\pi }{2L}\right ) ^{2},n=1,3,5,\cdots \). When \(A\left ( t\right ) =\sin \left ( t\right ) \), the above becomes\[ u\left ( x,t\right ) =\sin \left ( t\right ) \left ( x-L\right ) +\sum _{n=1,3,5,\cdots }^{\infty }e^{-k\lambda _{n}t}\left ( \frac {8L}{\pi ^{2}n^{2}}\int _{0}^{t}\cos \left ( \tau \right ) e^{k\lambda _{n}\tau }d\tau \right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) \] But \[ \int _{0}^{t}\cos \left ( \tau \right ) e^{k\lambda _{n}\tau }d\tau =\frac {k\lambda _{n}e^{k\lambda _{n}t}\cos \left ( t\right ) +e^{k\lambda _{n}t}\sin \left ( t\right ) -k\lambda _{n}}{k^{2}\lambda _{n}^{2}+1}\] And the general solution becomes\begin {align*} u\left ( x,t\right ) & =\left ( x-L\right ) \sin \left ( t\right ) +\sum _{n=1,3,5,\cdots }^{\infty }e^{-k\lambda _{n}t}\left ( \frac {8L}{\pi ^{2}n^{2}}\left ( \frac {k\lambda _{n}e^{k\lambda _{n}t}\cos \left ( t\right ) +e^{k\lambda _{n}t}\sin \left ( t\right ) -k\lambda _{n}}{k^{2}\lambda _{n}^{2}+1}\right ) \right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) \\ & =\left ( x-L\right ) \sin \left ( t\right ) +\sum _{n=1,3,5,\cdots }^{\infty }\frac {8L}{\pi ^{2}n^{2}}\left ( \frac {k\lambda _{n}\cos \left ( t\right ) +\sin \left ( t\right ) -k\lambda _{n}e^{-k\lambda _{n}t}}{k^{2}\lambda _{n}^{2}+1}\right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) \end {align*}
In this problem \(L=5,k=\frac {1}{100}\), hence the above becomes\[ u\left ( x,t\right ) =\left ( x-5\right ) \sin \left ( t\right ) +\sum _{n=1,3,5,\cdots }^{\infty }\frac {40}{\pi ^{2}n^{2}}\left ( \frac {\frac {1}{100}\left ( \frac {n\pi }{10}\right ) ^{2}\cos \left ( t\right ) +\sin \left ( t\right ) -\frac {1}{100}\left ( \frac {n\pi }{10}\right ) ^{2}e^{-\frac {1}{100}\left ( \frac {n\pi }{10}\right ) ^{2}t}}{\left ( \frac {1}{100}\left ( \frac {n\pi }{10}\right ) ^{2}\right ) ^{2}+1}\right ) \cos \left ( \frac {n\pi }{10}x\right ) \] The following is an animation of the solution
Source code used for the above
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Added Nov 25, 2018.
Problem 8.3.6 from Richard Haberman applied partial differential equations book, 5th edition
Solve the heat equation for \(u(x,t)\) \[ u_t = u_{xx} + \sin (5 x) e^{-2 t} \] For \(0<x<\pi \) and \(t>0\). The boundary conditions are \begin {align*} u(0,t) &= 1 \\ u(\pi ,t) &= 0 \\ \end {align*}
Initial condition is \(u(x,0)=0\)
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], t] == D[u[x, t], {x, 2}] + Sin[5*x]*Exp[-2*t]; bc = {u[0, t] == 1, u[Pi, t] == 0}; ic = u[x, 0] == 0; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, t], x, t], 60*10]];
\[\left \{\left \{u(x,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\left (\sqrt {\frac {2}{\pi }} \int _0^t e^{-K[1]^2 (t-K[2])} \left (\begin {array}{cc} \{ & \begin {array}{cc} e^{-2 K[2]} \sqrt {\frac {\pi }{2}} & K[1]=5 \\ 0 & \text {True} \\\end {array} \\\end {array}\right ) \, dK[2]-\frac {2 e^{-t K[1]^2}}{\pi K[1]}\right ) \sin (x K[1])-\frac {x}{\pi }+1\right \}\right \}\]
Maple ✓
restart; pde := diff(u(x,t),t)= diff(u(x,t),x$2)+ sin(5*x)*exp(-2*t); ic := u(x,0)=0; bc := u(0,t) =1,u(Pi,t)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde,ic,bc],u(x,t))),output='realtime'));
\[u \left (x , t\right ) = \frac {-23 x -46 \pi \left (\moverset {\infty }{\munderset {n =1}{\sum }}\frac {{\mathrm e}^{-n^{2} t} \sin \left (n x \right )}{\pi n}\right )+\pi \left (-{\mathrm e}^{-25 t}+{\mathrm e}^{-2 t}\right ) \sin \left (5 x \right )+23 \pi }{23 \pi }\]
Hand solution
This problem has nonhomogeneous B.C. and non-homogenous in the PDE itself (source present). First step is to use reference function to remove the nonhomogeneous B.C. then use the method of eigenfunction expansion on the resulting problem. Let \[ r\left ( x\right ) =c_{1}x+c_{2}\] At \(x=0,r\left ( x\right ) =1\), hence \(1=c_{2}\) and at \(x=\pi ,r\left ( x\right ) =0\), hence \(0=c_{1}\pi +1\) or \(c_{1}=-\frac {1}{\pi }\), therefore\[ r\left ( x\right ) =1-\frac {x}{\pi }\] Therefore\[ u\left ( x,t\right ) =v\left ( x,t\right ) +r\left ( x\right ) \] Where \(v\left ( x,t\right ) \) solution for the given PDE but with homogeneous B.C., therefore \begin {align} v_{t} & =v_{xx}+e^{-2t}\sin 5x\tag {1}\\ v\left ( 0,t\right ) & =0\nonumber \\ v\left ( \pi ,t\right ) & =0\nonumber \\ v\left ( x,0\right ) & =f\left ( x\right ) \nonumber \\ & =u\left ( x,0\right ) -r\left ( x\right ) \nonumber \\ & =0-\left ( 1-\frac {x}{\pi }\right ) \nonumber \\ & =\frac {x}{\pi }-1\nonumber \end {align}
We now solve (1). This is a PDE with homogeneous B.C. of the form \(v_{t}=v_{xx}+Q\left ( x,t\right ) \). The general solution to above PDE was solved in 4.1.6.4 on page 806 and the solution is
\begin {equation} v\left ( x,t\right ) =\sum _{n=1}^{\infty }e^{-k\lambda _{n}t}\Phi _{n}\left ( x\right ) \left ( \frac {2}{L}\int _{0}^{L}f\left ( s\right ) \Phi _{n}\left ( s\right ) ds\right ) +\sum _{n=1}^{\infty }e^{-k\lambda _{n}t}\Phi _{n}\left ( x\right ) \left ( \int _{0}^{t}\frac {2}{L}e^{k\lambda _{n}\tau }\left ( \int _{0}^{L}Q\left ( s,\tau \right ) \Phi _{n}\left ( s\right ) ds\right ) d\tau \right ) \tag {2} \end {equation}
Where \begin {align} \Phi _{n}\left ( x\right ) & =\sin \left ( \sqrt {\lambda _{n}}x\right ) \tag {3}\\ \lambda _{n} & =\left ( \frac {n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \nonumber \end {align}
Replacing \(L=\pi ,f\left ( x\right ) =\frac {x}{\pi }-1,Q\left ( x,t\right ) =e^{-2t}\sin \left ( 5x\right ) \) into (3,2) gives
\begin {align} \Phi _{n}\left ( x\right ) & =\sin \left ( nx\right ) \tag {3A}\\ \lambda _{n} & =n^{2}\qquad n=1,2,3,\cdots \nonumber \end {align}
And
\begin {equation} v\left ( x,t\right ) =\sum _{n=1}^{\infty }e^{-kn^{2}t}\sin \left ( nx\right ) \left ( \frac {2}{\pi }\int _{0}^{\pi }\left ( \frac {s}{\pi }-1\right ) \sin \left ( ns\right ) ds\right ) +\sum _{n=1}^{\infty }e^{-kn^{2}t}\sin \left ( nx\right ) \left ( \int _{0}^{t}\frac {2}{\pi }e^{kn^{2}\tau }e^{-2\tau }\left ( \int _{0}^{\pi }\sin \left ( 5s\right ) \sin \left ( ns\right ) ds\right ) d\tau \right ) \tag {2A} \end {equation}
But \(\int _{0}^{\pi }\left ( \frac {s}{\pi }-1\right ) \sin \left ( ns\right ) ds=\frac {-1}{n}\) since \(n\) is integer. And \(\int _{0}^{\pi }\sin 5s\sin \left ( ns\right ) ds=0\) when \(n\neq 5\) and for \(n=5\) it becomes \(\frac {\pi }{2}\). Using these values in the above gives
\begin {align} v\left ( x,t\right ) & =\sum _{n=1}^{\infty }e^{-kn^{2}t}\sin \left ( nx\right ) \left ( \frac {-2}{\pi n}\right ) +e^{-k\left ( 5\right ) ^{2}t}\sin \left ( 5x\right ) \left ( \int _{0}^{t}\frac {2}{\pi }e^{k\left ( 5\right ) ^{2}\tau }e^{-2\tau }\left ( \frac {\pi }{2}\right ) d\tau \right ) \tag {2C}\\ & =-\frac {2}{\pi }\sum _{n=1}^{\infty }e^{-kn^{2}t}\frac {\sin \left ( nx\right ) }{n}+e^{-25kt}\sin \left ( 5x\right ) \left ( \int _{0}^{t}e^{25k\tau }e^{-2\tau }d\tau \right ) \end {align}
But \(\int _{0}^{t}e^{25k\tau }e^{-2\tau }d\tau =\frac {-1+e^{25kt-2t}}{25k-2}\) and the above becomes
\begin {align*} v\left ( x,t\right ) & =-\frac {2}{\pi }\sum _{n=1}^{\infty }e^{-kn^{2}t}\frac {\sin \left ( nx\right ) }{n}+e^{-25kt}\sin \left ( 5x\right ) \left ( \frac {-1+e^{25kt-2t}}{25k-2}\right ) \\ & =-\frac {2}{\pi }\sum _{n=1}^{\infty }\frac {1}{n}e^{-kn^{2}t}\sin \left ( nx\right ) +\sin \left ( 5x\right ) \left ( \frac {-e^{-25kt}+e^{-2t}}{25k-2}\right ) \end {align*}
Since \(u\left ( x,t\right ) =v\left ( x,t\right ) +r\left ( x\right ) \) then the final solution is
\[ u\left ( x,t\right ) =\left ( 1-\frac {x}{\pi }\right ) -\left ( \frac {2}{\pi }\sum _{n=1}^{\infty }\frac {1}{n}e^{-kn^{2}t}\sin \left ( nx\right ) \right ) +\sin \left ( 5x\right ) \left ( \frac {-e^{-25kt}+e^{-2t}}{25k-2}\right ) \]
Animation is below using \(k=1\), the solution becomes
\[ u\left ( x,t\right ) =\left ( 1-\frac {x}{\pi }\right ) -\left ( \frac {2}{\pi }\sum _{n=1}^{\infty }\frac {1}{n}e^{-n^{2}t}\sin \left ( nx\right ) \right ) +\sin \left ( 5x\right ) \left ( \frac {e^{-2t}-e^{-25t}}{23}\right ) \]
Source code used for the above
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added March 8, 2018. Exam problem
Solve the heat equation \[ u_t= u_{xx} \] For \(0<x<\pi \) and \(t>0\). The boundary conditions are \begin {align*} u(0,t) &= t \\ u(\pi ,t) &= 0 \end {align*}
Initial condition is \(u(x,0)=0\).
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], {t, 1}] == D[u[x, t], {x, 2}]; bc = {u[0, t] == t, u[Pi, t] == 0}; ic = u[x, 0] == 0; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}], 60*10]]; sol = sol /. {K[1] -> n};
\[\left \{\left \{u(x,t)\to \underset {n=1}{\overset {\infty }{\sum }}-\frac {\left (2-2 e^{-n^2 t}\right ) \sin (n x)}{n^3 \pi }-\frac {t x}{\pi }+t\right \}\right \}\]
Maple ✓
restart; interface(showassumed=0); pde := diff(u(x,t),t)=diff(u(x,t),x$2); bc := u(0,t)=t,u(Pi,t)=0; ic := u(x,0)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde,bc,ic],u(x,t))),output='realtime'));
\[u \left (x , t\right ) = \frac {12 \pi \left (\moverset {\infty }{\munderset {n =1}{\sum }}\frac {{\mathrm e}^{-n^{2} t} \sin \left (n x \right )}{\pi \,n^{3}}\right )-6 \left (\frac {1}{6} x^{2}+t -\frac {1}{3} \pi x \right ) \left (x -\pi \right )}{6 \pi }\]
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