5.2.1 Cartesian coordinates

5.2.1.1 [394] Rectangular membrane. Fixed on all edges, General solution

problem number 394

Added January 10, 2020.

Solve $$u_{tt}=c^4(u_{xx}+u_{yy})$$

$$\begin{array}{rl}0& <x<L\\ 0& <y<H\end{array}$$

Boundary conditions on $x$

$$\begin{array}{rl}u(0,y,t)& =0\\ u(L,y,t)& =0\end{array}$$

And boundary conditions on $y$$$\begin{array}{rl}u(x,0,t)& =0\\ u(x,H,t)& =0\end{array}$$

Initial conditions$$\begin{array}{rl}u(x,y,0)& =f(x,y)\\ \frac{\partial u}{\partial t}(x,y,0)& =g(x,y)\end{array}$$

Figure 5.85: PDE specification

Mathematica ✗

ClearAll["Global`*"]; 
pde = D[u[x, y, t], {t, 2}] == c^2*Laplacian[u[x, y, t], {x, y}]; 
ic = {Derivative[0, 0, 1][u][x, y, 0] == g[x, y], u[x, y, 0] == f[x, y]}; 
bc = {u[0, y, t] == 0, u[0, H, t] == 0, u[x, 0, t] == 0, u[x, L, t] == 0}; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, y, t], {x, y, t},Assumptions -> {H > 0, L > 0, c > 0}], 60*10]];
 

Failed

Maple ✓

restart; 
pde := diff(u(x, y, t), t$2) = c^2*VectorCalculus:-Laplacian(u(x,y,t),[x,y]); 
bc  := u(0,y,t)=0, 
       u(L,y,t)=0, 
       u(x, 0, t) = 0, 
       u(x, H, t) = 0; 
ic  := u(x, y, 0) = f(x,y),(D[3](u))(x, y, 0) = g(x,y); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc,ic],u(x,y,t))assuming L>0,H>0),output='realtime')); 
sol := subs(n1=m,sol);
 

Failed to convert to latex

Hand solution

Assuming $u=X\left(x\right)Y\left(y\right)T\left(t\right)$ and substituting into the PDE gives$$\begin{array}{rl}\frac{1}{c^2}{T}^{\prime \prime }XY& =X^{\prime \prime }YT+Y^{\prime \prime }XT\\ \frac{1}{c^2}\frac{T^{\prime \prime }}{T}& =\frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y}\end{array}$$

Therefore$$\begin{array}{rl}\frac{1}{c^2}\frac{T^{\prime \prime }}{T}& =-\lambda \\ \frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y}& =-\lambda \end{array}$$

The time ODE becomes$$T^{\prime \prime }+c^2\lambda T=0$$ And the space ODE is$$\frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y}=-\lambda$$ Separating this again gives$$\frac{X^{\prime \prime }}{X}=-\lambda -\frac{Y^{\prime \prime }}{Y}$$ Let the second separation variable be $\mu$. This gives two new ODE’s to solve$$\begin{array}{rl}\frac{X^{\prime \prime }}{X}& =-\mu \\ -\lambda -\frac{Y^{\prime \prime }}{Y}& =-\mu \end{array}$$

Or$$\begin{array}{rl}{X}^{\prime \prime }+\mu X& =0\\ Y^{\prime \prime }+Y(\lambda -\mu )& =0\end{array}$$

Solving for $X\left(x\right)$ ODE first, and knowing that only $\mu >0$ will give non trivial solutions (from the nature of the boundary conditions), gives the solution as$$X\left(x\right)=A\cos \left(\sqrt{\mu }x\right)+B\sin \left(\sqrt{\mu }x\right)$$ Applying B.C. at $x=0$ results in$$0=A$$ Therefore $X\left(x\right)=B\sin \left(\sqrt{\mu }x\right)$. Applying the B.C. at $x=L$ gives$$0=B\sin \left(\sqrt{\mu }L\right)$$ For non trivial solution$$\begin{array}{rl}\sqrt{\mu }L& =n\pi \\ \mu & ={\left(\frac{n\pi }{L}\right)}^{2}n=1,2,3,\cdots \end{array}$$

Therefore the $X_n\left(x\right)$ eigenfunctions are$$X_n\left(x\right)=B_n\sin \left(\frac{n\pi }{L}x\right)n=1,2,3,\cdots$$ Now, solving the $Y\left(y\right)$ ODE above$$Y^{\prime \prime }+Y(\lambda -{\left(\frac{n\pi }{L}\right)}^2)=0$$ The solution is$$Y_n\left(y\right)=A\cos \left(\sqrt{\lambda -{\left(\frac{n\pi }{L}\right)}^2}y\right)+B\sin \left(\sqrt{\lambda -{\left(\frac{n\pi }{L}\right)}^2}y\right)$$ Applying first B.C. gives $$0=A$$ Hence$$Y_n\left(y\right)=B\sin \left(\sqrt{\lambda -{\left(\frac{n\pi }{L}\right)}^2}y\right)$$ Applying the second B.C. gives$$0=B\sin \left(\sqrt{\lambda -{\left(\frac{n\pi }{L}\right)}^2}H\right)$$ For non trivial solution$$\begin{array}{rl}\sqrt{\lambda -{\left(\frac{n\pi }{L}\right)}^2}H& =m\pi m=1,2,3,\cdots \\ {\lambda }_{nm}-{\left(\frac{n\pi }{L}\right)}^2& ={\left(\frac{m\pi }{H}\right)}^2\\ {\lambda }_{nm}& ={\left(\frac{m\pi }{H}\right)}^2+{\left(\frac{n\pi }{L}\right)}^{2}n=1,2,3,\cdots ,m=1,2,3,\cdots \end{array}$$

Hence the $Y_{nm}\left(y\right)$ solution is$$Y_{nm}=B_{nm}\sin \left(\frac{m\pi }{H}y\right)n=1,2,3,\cdots ,m=1,2,3,\cdots$$ The time ode $T\left(t\right)$ is now solved$$\begin{array}{rl}{T}_{nm}^{\prime \prime }+c^2{\lambda }_{nm}{T}_{nm}& =0\\ T_{nm}\left(t\right)& =A_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)+B_{nm}\sin \left(c\sqrt{{\lambda }_{nm}}t\right)\end{array}$$

Combining all solutions, and merging all constants into two results in$$\begin{array}{rl}{u}_{nm}(x,y,t)& =X_n\left(x\right)Y_{nm}\left(y\right)T_{nm}\left(t\right)\\ u(x,y,t)& =\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{X}_m\left(x\right)Y_{mn}\left(y\right)T_{mn}\left(t\right)\\ & =\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{A}_{nm}\sin \left(\frac{n\pi }{L}x\right)\sin \left(\frac{m\pi }{H}y\right)\cos \left(c\sqrt{{\lambda }_{nm}}t\right)\\ \text{(1)}& & +\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{B}_{nm}\sin \left(\frac{n\pi }{L}x\right)\sin \left(\frac{m\pi }{H}y\right)\sin \left(c\sqrt{{\lambda }_{nm}}t\right)\end{array}$$

Initial conditions are now used to find $A_{nm},B_{nm}$. At $t=0$$$\begin{array}{rl}u(x,y,0)& =f(x,y)\\ \frac{\partial u}{\partial t}(x,y,0)& =g(x,y)\end{array}$$

Applying first initial condition to (1)  gives$$f(x,y)=\sum _{n=1}^{\mathrm{\infty }}(\sum _{m=1}^{\mathrm{\infty }}{A}_{nm}\sin \left(\frac{m\pi }{H}y\right))\sin \left(\left(\frac{n\pi }{L}\right)x\right)$$ Applying 2D orthogonality gives$$\begin{array}{rl}{\int }_0^L{\int }_0^{H}f(x,y)\sin \left(\left(\frac{n\pi }{L}\right)x\right)\sin \left(\frac{m\pi }{H}y\right)dxdy& =A_{nm}\left(\frac{L}{2}\right)\left(\frac{H}{2}\right)\\ A_{nm}& =\frac{4}{LH}{\int }_0^L{\int }_0^{H}f(x,y)\sin \left(\left(\frac{n\pi }{L}\right)x\right)\sin \left(\frac{m\pi }{H}y\right)dxdy\end{array}$$

Taking time derivative of (1) gives$$\begin{array}{rl}\frac{\partial u}{\partial t}(x,y,t)& =\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}-c\sqrt{{\lambda }_{nm}}{A}_{nm}\sin \left(\frac{n\pi }{L}x\right)\sin \left(\frac{m\pi }{H}y\right)\sin \left(c\sqrt{{\lambda }_{nm}}t\right)\\ & +\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}c\sqrt{{\lambda }_{nm}}{B}_{nm}\sin \left(\frac{n\pi }{L}x\right)\sin \left(\frac{m\pi }{H}y\right)\cos \left(c\sqrt{{\lambda }_{nm}}t\right)\end{array}$$

At $t=0$ the above becomes$$g(x,y)=\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}c\sqrt{{\lambda }_{nm}}{B}_{nm}\sin \left(\left(\frac{n\pi }{L}\right)x\right)\sin \left(\frac{m\pi }{H}y\right)$$ Applying 2D orthogonality gives$$\begin{array}{rl}{\int }_0^L{\int }_0^{H}g(x,y)\sin \left(\frac{n\pi }{L}x\right)\sin \left(\frac{m\pi }{H}y\right)dxdy& =B_{nm}\left(\frac{L}{2}\right)\left(\frac{H}{2}\right)\\ B_{nm}& =\frac{4}{LH}{\int }_0^L{\int }_0^{H}g(x,y)\sin \left(\frac{n\pi }{L}x\right)\sin \left(\frac{m\pi }{H}y\right)dxdy\end{array}$$

Summary of solution$$\begin{array}{rl}u(x,y,t)=\sum _{n=1}^{\mathrm{\infty }}(\sum _{m=1}^{\mathrm{\infty }}{A}_{nm}\sin \left(\frac{m\pi }{H}y\right)\cos \left(c\sqrt{{\lambda }_{nm}}t\right))\sin \left(\frac{n\pi }{L}x\right)& +\sum _{n=1}^{\mathrm{\infty }}(\sum _{m=1}^{\mathrm{\infty }}{B}_{nm}\sin \left(\frac{m\pi }{H}y\right)\sin \left(c\sqrt{{\lambda }_{nm}}t\right))\sin \left(\frac{n\pi }{L}x\right)\\ A_{nm}& =\frac{4}{LH}{\int }_0^L{\int }_0^{H}f(x,y)\sin \left(\left(\frac{n\pi }{L}\right)x\right)\sin \left(\frac{m\pi }{H}y\right)dxdy\\ B_{nm}& =\frac{4}{LH}{\int }_0^L{\int }_0^{H}g(x,y)\sin \left(\left(\frac{n\pi }{L}\right)x\right)\sin \left(\frac{m\pi }{H}y\right)dxdy\\ {\lambda }_{nm}& ={\left(\frac{m\pi }{H}\right)}^2+{\left(\frac{n\pi }{L}\right)}^2\end{array}$$

____________________________________________________________________________________

5.2.1.2 [395] Rectangular membrane. Fixed on all edges, zero velocity. Specific example

problem number 395

Added January 10, 2020.

Solve $$u_{tt}=c^4(u_{xx}+u_{yy})$$

$$\begin{array}{rl}0& <x<L\\ 0& <y<H\end{array}$$

Boundary conditions on $x$

$$\begin{array}{rl}u(0,y,t)& =0\\ u(L,y,t)& =0\end{array}$$

And boundary conditions on $y$$$\begin{array}{rl}u(x,0,t)& =0\\ u(x,H,t)& =0\end{array}$$

Initial conditions$$\begin{array}{rl}u(x,y,0)& =f(x,y)\\ \frac{\partial u}{\partial t}(x,y,0)& =g(x,y)\end{array}$$

Using $L=1,H=2,c=\frac{1}{10},f(x,y)=x\cos (y),g(x,y)=0$.

Figure 5.86: PDE specification

Mathematica ✗

ClearAll["Global`*"]; 
L=1;H=2;c=1/10; 
f[x_,y_]:=x*Cos[y]; 
g[x_,y_]:=0; 
pde = D[u[x, y, t], {t, 2}] == c^2*Laplacian[u[x, y, t], {x, y}]; 
ic = {Derivative[0, 0, 1][u][x, y, 0] == g[x, y], u[x, y, 0] == f[x, y]}; 
bc = {u[0, y, t] == 0, u[0, H, t] == 0, u[x, 0, t] == 0, u[x, L, t] == 0}; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, y, t], {x, y, t}], 60*10]];
 

Failed

Maple ✓

restart; 
L:=1; 
H:=2; 
c:=1/10; 
f:=(x,y)->x*cos(y); 
g:=(x,y)->0; 
pde := diff(u(x, y, t), t$2) = c^2*VectorCalculus:-Laplacian(u(x,y,t),[x,y]); 
bc  := u(0,y,t)=0, 
       u(L,y,t)=0, 
       u(x, 0, t) = 0, 
       u(x, H, t) = 0; 
ic  := u(x, y, 0) = f(x,y),(D[3](u))(x, y, 0) = g(x,y); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc,ic],u(x,y,t))),output='realtime')); 
sol := subs(n1=m,sol);
 

$$u(x,y,t)=\sum _{m=1}^{\mathrm{\infty }}\left(\sum _{n=1}^{\mathrm{\infty }}4\frac{\sin \left(n\pi x\right)\sin \left(1/2m\pi y\right)\cos \left(1/20\pi \sqrt{m^2+4n^2}t\right)m(\cos \left(2\right){(-1)}^{m+n}-{(-1)}^n)}{n({\pi }^2{m}^2-4)}\right)$$

Hand solution

The basic solution for this type of PDE was already given in problem 5.2.1.1 on page 1492 as

$$\begin{array}{rl}u(x,y,t)=\sum _{n=1}^{\mathrm{\infty }}(\sum _{m=1}^{\mathrm{\infty }}{A}_{nm}\sin \left(\frac{m\pi }{H}y\right)\cos \left(c\sqrt{{\lambda }_{nm}}t\right))\sin \left(\frac{n\pi }{L}x\right)& +\sum _{n=1}^{\mathrm{\infty }}(\sum _{m=1}^{\mathrm{\infty }}{B}_{nm}\sin \left(\frac{m\pi }{H}y\right)\sin \left(c\sqrt{{\lambda }_{nm}}t\right))\sin \left(\frac{n\pi }{L}x\right)\\ A_{nm}& =\frac{4}{LH}{\int }_0^L{\int }_0^{H}f(x,y)\sin \left(\frac{n\pi }{L}x\right)\sin \left(\frac{m\pi }{H}y\right)dxdy\\ B_{nm}& =\frac{4}{LH}{\int }_0^L{\int }_0^{H}g(x,y)\sin \left(\frac{n\pi }{L}x\right)\sin \left(\frac{m\pi }{H}y\right)dxdy\\ {\lambda }_{nm}& ={\left(\frac{m\pi }{H}\right)}^2+{\left(\frac{n\pi }{L}\right)}^{2}n=1,2,3,\cdots ,m=1,2,3,\cdots \end{array}$$

In this problem $$\begin{array}{rl}L& =1\\ H& =2\\ c& =\frac{1}{10}\\ f(x,y)& =x\cos y\\ g(x,y)& =0\end{array}$$

Hence the solution becomes

$$\begin{array}{rl}u(x,y,t)=\sum _{n=1}^{\mathrm{\infty }}(\sum _{m=1}^{\mathrm{\infty }}{A}_{nm}\sin \left(\frac{m\pi }{2}y\right)\cos \left(\frac{1}{10}\sqrt{{\lambda }_{nm}}t\right))\sin \left(n\pi x\right)& +\sum _{n=1}^{\mathrm{\infty }}(\sum _{m=1}^{\mathrm{\infty }}{B}_{nm}\sin \left(\frac{m\pi }{2}y\right)\sin \left(\frac{1}{10}\sqrt{{\lambda }_{nm}}t\right))\sin \left(n\pi x\right)\\ A_{nm}& =2{\int }_0^1{\int }_0^{2}x\cos y\sin \left(n\pi x\right)\sin \left(\frac{m\pi }{2}y\right)dxdy\\ B_{nm}& =0\\ {\lambda }_{nm}& ={\left(\frac{m\pi }{2}\right)}^2+{\left(n\pi \right)}^{2}n=1,2,3,\cdots ,m=1,2,3,\cdots \end{array}$$

But $$\begin{array}{rl}{A}_{nm}& =2{\int }_0^1{\int }_0^{2}x\cos y\sin \left(n\pi x\right)\sin \left(\frac{m\pi }{2}y\right)dxdy\\ & =\frac{4{(-1)}^{n}m(-1+{(-1)}^m\cos \left(2\right))}{n(m^2{\pi }^2-4)}\end{array}$$

Hence the solution simplifies to

$$\begin{array}{rl}u(x,y,t)& =\sum _{n=1}^{\mathrm{\infty }}(\sum _{m=1}^{\mathrm{\infty }}\frac{4{(-1)}^{n}m(-1+{(-1)}^m\cos \left(2\right))}{n(m^2{\pi }^2-4)}\sin \left(\frac{m\pi }{2}y\right)\cos \left(\frac{1}{10}\sqrt{{\left(\frac{m\pi }{2}\right)}^2+{\left(n\pi \right)}^2}t\right))\sin \left(n\pi x\right)\\ & =\sum _{n=1}^{\mathrm{\infty }}(\sum _{m=1}^{\mathrm{\infty }}\frac{4{(-1)}^{n}m(-1+{(-1)}^m\cos \left(2\right))}{n(m^2{\pi }^2-4)}\cos \left(\pi \frac{\sqrt{m^2+4n^2}}{20}t\right)\sin \left(\frac{m\pi }{2}y\right))\sin \left(n\pi x\right)\end{array}$$

Animation is below

Source code used for the above

Figure 5.87: Source code

Figure 5.88: Source code

____________________________________________________________________________________

5.2.1.3 [396] All 4 edges fixed, zero initial velocity, Specific example

problem number 396

Added June 17, 2019

Solve for $u(x,y,t)$ with $0<x<L$ and $0<y<H$ and $t>0$.

Solve $$u_{tt}=c^2{\mathrm{\nabla }}^{2}u(x,y)$$ With boundary conditions $$\begin{array}{rl}u(x,0,t)& =0\\ u(0,y,t)& =0\\ u(L,y,t)& =0\\ u(x,H,t)& =0\end{array}$$

With initial conditions $$\begin{array}{rl}u(x,y,0)& =3f_1(x)f_2(y)\\ u_t(x,y,0)& =0\end{array}$$

And $$f_1\left(x\right)=\{\begin{array}{ccc}x& & 0<x<\frac{L}{2}\\ L-x& & \frac{L}{2}<x<L\end{array}$$ Where $$f_2\left(y\right)=\{\begin{array}{ccc}y& & 0<y<\frac{H}{2}\\ H-y& & \frac{H}{2}<y<H\end{array}$$ And $L=2,H=3$ and $c=\frac{1}{3}$.

Figure 5.89: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
L=2; 
H=3; 
c=1/3; 
f1[x_] :=Piecewise[{{x, x < L/2}, {L - x, x > L/2}}]; 
f2[y_] := Piecewise[{{y, y < H/2}, {H - y, y > H/2}}]; 
pde =  D[u[x, y, t], {t, 2}] == c^2 * Laplacian[u[x, y, t], {x, y}]; 
ic  = {u[x, y, 0] == 3*f1[x]*f2[y], Derivative[0, 0, 1][u][x, y, 0] == 0}; 
bc  = {u[x, 0, t] == 0, u[0, y, t] == 0, u[L, y, t] == 0, u[x, H, t] == 0}; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, y, t], {x, y, t}], 60*10]]; 
sol =  sol /. {K[1] -> n, K[2] -> m};
 

$$\left\{\{u(x,y,t)\to \begin{array}{cc}\{& \begin{array}{cc}\underset{n=1}{\sum ^{\mathrm{\infty }}}\underset{K[3]=1}{\sum ^{\mathrm{\infty }}}\frac{288\cos \left(\frac{1}{18}\pi t\sqrt{9n^2+4K[3{]}^2}\right)\sin \left(\frac{n\pi }{2}\right)\sin \left(\frac{n\pi x}{2}\right)\sin (\frac{1}{2}\pi K[3])\sin (\frac{1}{3}\pi yK[3])}{n^2{\pi }^{4}K[3{]}^2}& (n|K[3])\in \mathbb{Z}\wedge n\ge 1\wedge K[3]\ge 1\\ \text{Indeterminate}& \text{True}\end{array}\end{array}\}\right\}$$

Maple ✓

restart; 
L   := 2; 
H   := 3; 
c   := 1/3; 
f1  := x-> piecewise(x < L/2,x, x > L/2,L - x); 
f2  := y-> piecewise(y < H/2,y, y > H/2,H - y); 
pde := diff(u(x, y, t), t$2) = c^2* VectorCalculus:-Laplacian(u(x, y, t), 'cartesian'[x, y]); 
ic  := u(x,y,0)=3*f1(x)*f2(y),D[3](u)(x,y,0)=0; 
bc  := u(x,0,t)=0,u(0,y,t)=0,u(L,y,t)=0,u(x,H,t)=0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic,bc], u(x, y, t))),output='realtime')); 
sol := subs(n1=m,sol);
 

$$u(x,y,t)=\sum _{m=1}^{\mathrm{\infty }}\left(\sum _{n=1}^{\mathrm{\infty }}288\frac{\sin \left(1/3m\pi y\right)\sin \left(1/2n\pi x\right)\cos \left(1/18\pi \sqrt{4m^2+9n^2}t\right)\sin \left(1/2n\pi \right)\sin \left(1/2m\pi \right)}{n^2{\pi }^4{m}^2}\right)$$

Hand solution

The basic solution for this type of PDE was already given in problem 5.2.1.8 on page 1533 as$$\begin{array}{rl}u(x,y,t)& =\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{A}_{nm}\sin \left(\frac{n\pi }{L}x\right)\sin \left(\frac{m\pi }{H}y\right)\cos \left(c\sqrt{{\left(\frac{m\pi }{H}\right)}^2+{\left(\frac{n\pi }{L}\right)}^2}t\right)\\ A_{nm}& =\frac{4}{LH}{\int }_0^L{\int }_0^{H}f(x,y)\sin \left(\frac{n\pi }{L}x\right)\sin \left(\frac{m\pi }{H}y\right)dxdy\end{array}$$

In this problem $$\begin{array}{rl}L& =2\\ H& =3\\ c& =\frac{1}{3}\\ f(x,y)& =3f_1\left(x\right)f_2\left(y\right)\end{array}$$

And$$f_1\left(x\right)=\{\begin{array}{ccc}x& & 0<x<\frac{L}{2}\\ L-x& & \frac{L}{2}<x<L\end{array}$$ And$$f_2\left(y\right)=\{\begin{array}{ccc}y& & 0<y<\frac{H}{2}\\ H-y& & \frac{H}{2}<y<H\end{array}$$ This is animation of the above solution using these specific values for for $40$ seconds. (Animation will only show in the HTML version)

Source code used for the above

Figure 5.90: Source code

Figure 5.91: Code used for animation

The following shows selected modes. For example, for $n=1,m=1$ the solution becomes$$u(x,y,t)=A_{1,1}\sin \left(\frac{\pi }{L}x\right)\sin \left(\frac{\pi }{H}y\right)\cos \left(c\sqrt{{\left(\frac{\pi }{H}\right)}^2+{\left(\frac{1\pi }{L}\right)}^2}t\right)$$ And for $n=1,m=5$ then the solution becomes$$u(x,y,t)=A_{1,5}\sin \left(\frac{\pi }{L}x\right)\sin \left(\frac{5\pi }{H}y\right)\cos \left(c\sqrt{{\left(\frac{5\pi }{H}\right)}^2+{\left(\frac{\pi }{L}\right)}^2}t\right)$$ And so on.

$n$	$m$	animation
$1$	$1$
$1$	$3$
$1$	$5$
$3$	$1$
$3$	$5$
$5$	$1$
$5$	$3$

____________________________________________________________________________________

5.2.1.4 [397] All 4 edges fixed, zero initial velocity, Specific example, delta in center

problem number 397

Added June 18, 2019

Solve for $u(x,y,t)$ with $0<x<L$ and $0<y<H$ and $t>0$.

Solve $$u_{tt}=c^2{\mathrm{\nabla }}^{2}u(x,y)$$ With boundary conditions $$\begin{array}{rl}u(x,0,t)& =0\\ u(0,y,t)& =0\\ u(L,y,t)& =0\\ u(x,H,t)& =0\end{array}$$

With initial conditions $$\begin{array}{rl}u(x,y,0)& =f(x,y)\\ u_t(x,y,0)& =0\end{array}$$

And $$\begin{array}{rl}L& =20\\ H& =30\\ c& =\frac{1}{3}\\ f(x,y)& =f_1\left(x\right)f_2\left(y\right)\end{array}$$

Where $f(x,y)$ is an approximation of delta in the middle of the membrane $$f_1\left(x\right)=\{\begin{array}{ccc}1& & \frac{45}{100}L<x<\frac{55}{100}L\\ 0& & \text{otherwise}\end{array}$$ And$$f_2\left(y\right)=\{\begin{array}{ccc}1& & \frac{45}{100}H<y<\frac{55}{100}H\\ 0& & \text{otherwise}\end{array}$$

Figure 5.92: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
L  = 20; 
H  = 30; 
c  = 1/3; 
f1[x] := Piecewise[{{1,45/100*L <= x<= 55/100*L},{0,True}}]; 
f2[y] := Piecewise[{{1,45/100*H<= y<= 55/100*H},{0,True}}]; 
pde =  D[u[x, y, t], {t, 2}] == c^2 * Laplacian[u[x, y, t], {x, y}]; 
ic  = {u[x, y, 0] == f1[x]*f2[y], Derivative[0, 0, 1][u][x, y, 0] == 0}; 
bc  = {u[x, 0, t] == 0, u[0, y, t] == 0, u[L, y, t] == 0, u[x, H, t] == 0}; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, y, t], {x, y, t}], 60*10]]; 
sol =  sol /. {K[1] -> n, K[2] -> m};
 

$$\left\{\{u(x,y,t)\to \begin{array}{cc}\{& \begin{array}{cc}\underset{n=1}{\sum ^{\mathrm{\infty }}}\underset{K[3]=1}{\sum ^{\mathrm{\infty }}}\frac{4(\cos \left(\frac{9n\pi }{20}\right)-\cos \left(\frac{11n\pi }{20}\right))(\cos (\frac{9}{20}\pi K[3])-\cos (\frac{11}{20}\pi K[3]))\cos \left(\frac{1}{180}\pi t\sqrt{9n^2+4K[3{]}^2}\right)\sin \left(\frac{n\pi x}{20}\right)\sin (\frac{1}{30}\pi yK[3])}{n{\pi }^{2}K[3]}& (n|K[3])\in \mathbb{Z}\wedge n\ge 1\wedge K[3]\ge 1\\ \text{Indeterminate}& \text{True}\end{array}\end{array}\}\right\}$$

Maple ✓

restart; 
L   := 20; 
H   := 30; 
c   := 1/3; 
f1  := x-> piecewise(x>45/100*L and x< 55/100*L,1, true,0); 
f2  := y-> piecewise(y>45/100*H and y< 55/100*H,1, true,0); 
pde := diff(u(x, y, t), t$2) = c^2* VectorCalculus:-Laplacian(u(x, y, t), 'cartesian'[x, y]); 
ic  := u(x,y,0)=f1(x)*f2(y),D[3](u)(x,y,0)=0; 
bc  := u(x,0,t)=0,u(0,y,t)=0,u(L,y,t)=0,u(x,H,t)=0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic,bc], u(x, y, t))),output='realtime')); 
sol := subs(n1=m,sol);
 

$$u(x,y,t)=\sum _{m=1}^{\mathrm{\infty }}(\sum _{n=1}^{\mathrm{\infty }}-16\frac{\sin \left(1/20n\pi x\right)\sin \left(1/30m\pi y\right){(\sin \left(1/20n\pi \right))}^2\cos \left(1/20n\pi \right)(-1-2\cos \left(1/5n\pi \right)+2\cos \left(1/10n\pi \right))(1+2\cos \left(1/10n\pi \right)-2\cos \left(1/20n\pi \right))(1+2\cos \left(1/10n\pi \right)+2\cos \left(1/20n\pi \right))}{n{\pi }^{2}m}\cos \left(\frac{\pi \sqrt{4m^2+9n^2}t}{180}\right)(-\cos \left(\frac{11m\pi }{20}\right)+\cos \left(\frac{9m\pi }{20}\right)))$$

Hand solution

The basic solution for this type of PDE was already given in problem 5.2.1.8 on page 1533 as$$\begin{array}{rl}u(x,y,t)& =\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{A}_{nm}\sin \left(\frac{n\pi }{L}x\right)\sin \left(\frac{m\pi }{H}y\right)\cos \left(c\sqrt{{\left(\frac{m\pi }{H}\right)}^2+{\left(\frac{n\pi }{L}\right)}^2}t\right)\\ A_{nm}& =\frac{4}{LH}{\int }_0^L{\int }_0^{H}f(x,y)\sin \left(\frac{n\pi }{L}x\right)\sin \left(\frac{m\pi }{H}y\right)dxdy\end{array}$$

In this problem $$\begin{array}{rl}L& =20\\ H& =30\\ c& =\frac{1}{3}\\ f(x,y)& =f_1\left(x\right)f_2\left(y\right)\end{array}$$

Where $f(x,y)$ is an approximation of delta in the middle of the membrane $$f_1\left(x\right)=\{\begin{array}{ccc}1& & \frac{45}{100}L<x<\frac{55}{100}L\\ 0& & \text{otherwise}\end{array}$$ And$$f_2\left(y\right)=\{\begin{array}{ccc}1& & \frac{45}{100}H<y<\frac{55}{100}H\\ 0& & \text{otherwise}\end{array}$$ This is animation of the above solution using these specific values for for $40$ seconds. (Animation will only show in the HTML version)

Source code used for the above

Figure 5.93: Source code

Figure 5.94: Code used for animation

____________________________________________________________________________________

5.2.1.5 [398] All 4 edges fixed

problem number 398

Taken from Mathematica helps pages on DSolve

Solve for $u(x,y,t)$ with $0<x<1$ and $0<y<2$ and $t>0$.

Solve $$\frac{{\partial }^{2}u}{\partial t^2}=\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}$$ With boundary conditions $$\begin{array}{rl}u(x,0,t)& =0\\ u(0,y,t)& =0\\ u(1,y,t)& =0\\ u(x,2,t)& =0\end{array}$$

With initial conditions $$\begin{array}{rl}u(x,y,0)& =\frac{1}{10}(x-x^2)(2y-y^2)\\ \frac{\partial u}{\partial t}(x,y,0)& =0\end{array}$$

Figure 5.95: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x, y, t], {t, 2}] == Laplacian[u[x, y, t], {x, y}]; 
ic  = {u[x, y, 0] == (1/10)*(x - x^2)*(2*y - y^2), Derivative[0, 0, 1][u][x, y, 0] == 0}; 
bc  = {u[x, 0, t] == 0, u[0, y, t] == 0, u[1, y, t] == 0, u[x, 2, t] == 0}; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, y, t], {x, y, t}], 60*10]]; 
sol =  sol /. {K[1] -> n, K[2] -> m}; 
sol =  Assuming[Element[{n, m}, Integers], FullSimplify[sol]];
 

$$\left\{\{u(x,y,t)\to \begin{array}{cc}\{& \begin{array}{cc}\underset{n=1}{\sum ^{\mathrm{\infty }}}\underset{K[3]=1}{\sum ^{\mathrm{\infty }}}\frac{32(-1+(-1{)}^n)(-1+(-1{)}^{K[3]})\cos \left(\frac{1}{2}\pi t\sqrt{4n^2+K[3{]}^2}\right)\sin (n\pi x)\sin (\frac{1}{2}\pi yK[3])}{5n^3{\pi }^{6}K[3{]}^3}& K[3]\in \mathbb{Z}\wedge n\ge 1\wedge K[3]\ge 1\\ \text{Indeterminate}& \text{True}\end{array}\end{array}\}\right\}$$

Maple ✓

restart; 
pde := diff(u(x, y, t), t$2) =  VectorCalculus:-Laplacian(u(x, y, t), 'cartesian'[x, y]); 
ic  := u(x,y,0)=(1/10)*(x-x^2)*(2*y-y^2),(D[3](u))(x,y,0)=0; 
bc  := u(x,0,t)=0,u(0,y,t)=0,u(1,y,t)=0,u(x,2,t)=0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic,bc], u(x, y, t))),output='realtime')); 
sol := subs(n1=m,sol);
 

$$u(x,y,t)=\sum _{m=1}^{\mathrm{\infty }}(\sum _{n=1}^{\mathrm{\infty }}-\frac{32\sin \left(n\pi x\right)\sin \left(1/2m\pi y\right)\cos \left(1/2\pi \sqrt{m^2+4n^2}t\right)(-{(-1)}^{m+n}+{(-1)}^m+{(-1)}^n-1)}{5n^3{\pi }^6{m}^3})$$

____________________________________________________________________________________

5.2.1.6 [399] All edges fixed (Haberman 8.5.5 (a)

problem number 399

Added Nov 27, 2018.

This is problem 8.5.5 part(a) from Richard Haberman applied partial differential equations 5th edition.

Solve the initial value problem for membrane with time-dependent forcing and fixed boundaries $u=0$. $$u_{tt}=c^2{\mathrm{\nabla }}^{2}u+Q(x,y,t)$$ If the memberane is rectangle $(0<x<L,0<y<H)$. With initial conditions $$\begin{array}{rl}u(x,y,0)& =f(x,y)\\ \frac{\partial u}{\partial t}(x,y,0)& =0\end{array}$$

See my HW9, Math 322, UW Madison.

Figure 5.96: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x, y, t], {t, 2}] == c^2*Laplacian[u[x, y, t], {x, y}] + Q[x, y, t]; 
ic  = {u[x, y, 0] == f[x, y], Derivative[0, 0, 1][u][x, y, 0] == 0}; 
bc  = {u[0, y, t] == 0, u[L, y, t] == 0, u[x, 0, t] == 0, u[x, H, t] == 0}; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, y, t], {x, y, t}, Assumptions -> {L > 0, H > 0, t > 0, c > 0}], 60*10]];
 

$$\left\{\{u(x,y,t)\to \begin{array}{cc}\{& \begin{array}{cc}\underset{K[1]=1}{\sum ^{\mathrm{\infty }}}\underset{K[3]=1}{\sum ^{\mathrm{\infty }}}2\sqrt{\frac{1}{H}}\sqrt{\frac{1}{L}}({\int }_0^t\frac{2({\int }_0^L{\int }_0^{H}Q(x,y,K[4])\sin \left(\frac{\pi xK[1]}{L}\right)\sin \left(\frac{\pi yK[3]}{H}\right)dydx)\sin (c\pi \sqrt{\frac{K[1{]}^2}{L^2}+\frac{K[3{]}^2}{H^2}}(t-K[4]))}{c\pi \sqrt{\frac{HK[1{]}^2}{L}+\frac{LK[3{]}^2}{H}}}dK[4]+\frac{2\cos \left(c\pi t\sqrt{\frac{K[1{]}^2}{L^2}+\frac{K[3{]}^2}{H^2}}\right){\int }_0^L{\int }_0^{H}f(x,y)\sin \left(\frac{\pi xK[1]}{L}\right)\sin \left(\frac{\pi yK[3]}{H}\right)dydx}{\sqrt{HL}})\sin \left(\frac{\pi xK[1]}{L}\right)\sin \left(\frac{\pi yK[3]}{H}\right)& (K[1]|K[3])\in \mathbb{Z}\wedge K[1]\ge 1\wedge K[3]\ge 1\\ \text{Indeterminate}& \text{True}\end{array}\end{array}\}\right\}$$

Maple ✗

restart; 
interface(showassumed=0); 
pde := diff(u(x,y,t),t$2)=c^2*(diff(u(x,y,t),x$2)+diff(u(x,y,t),y$2))+Q(x,y,t); 
bc  := u(0,y,t)=0,u(L,y,t)=0,u(x,0,t)=0,u(x,H,t)=0; 
ic  := u(x,y,0)=f(x,y), eval( diff(u(x,y,t),t),t=0)=0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc,ic],u(x,y,t)) assuming L>0,H>0,c>0,t>0),output='realtime'));
 

sol=()

____________________________________________________________________________________

5.2.1.7 [400] 2 edgs fixed, 2 free, zero initial velocity

problem number 400

Taken from Maple PDE help pages. This wave PDE inside square with free to move on left edge and right edge, and top and bottom edges are fixed. It has zero initial velocity, but given a non-zero initial position. Where $0<x<\pi$ and $0<y<\pi$ and $t>0$.

Solve $$u_{tt}=\frac{1}{4}(\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2})$$ With boundary conditions $$\begin{array}{rl}\frac{\partial u}{\partial x}u(0,y,t)& =0\\ \frac{\partial u}{\partial x}u(\pi ,y,t)& =0\\ u(x,0,t)& =0\\ u(x,\pi ,0)& =0\end{array}$$

With initial conditions $$\begin{array}{rl}\frac{\partial u}{\partial t}(x,y,0)& =0\\ u(x,0)& =xy(\pi -y)\end{array}$$

Figure 5.97: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x, y, t], {t, 2}] == (1*(D[u[x, y, t], {x, 2}] + D[u[x, y, t], {y, 2}]))/4; 
ic  = {Derivative[0, 0, 1][u][x, y, 0] == 0, u[x, y, 0] == x*y*(Pi - y)}; 
bc  = {Derivative[1, 0, 0][u][0, y, t] == 0, Derivative[1, 0, 0][u][Pi, y, t] == 0, u[x, 0, t] == 0, u[x, Pi, t] == 0}; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, y, t], {x, y, t}], 60*10]];
 

$$\left\{\{u(x,y,t)\to \begin{array}{cc}\{& \begin{array}{cc}\underset{K[3]=1}{\sum ^{\mathrm{\infty }}}-\frac{2(-1+(-1{)}^{K[3]})\cos (\frac{1}{2}tK[3])\sin (yK[3])}{K[3{]}^3}+\underset{K[1]=1}{\sum ^{\mathrm{\infty }}}\underset{K[3]=1}{\sum ^{\mathrm{\infty }}}-\frac{8(-1+(-1{)}^{K[1]})(-1+(-1{)}^{K[3]})\cos (xK[1])\cos \left(\frac{1}{2}t\sqrt{K[1{]}^2+K[3{]}^2}\right)\sin (yK[3])}{{\pi }^{2}K[1{]}^{2}K[3{]}^3}& (K[1]|K[3])\in \mathbb{Z}\wedge K[1]\ge 1\wedge K[3]\ge 1\\ \text{Indeterminate}& \text{True}\end{array}\end{array}\}\right\}$$

Maple ✓

restart; 
pde := diff(u(x, y, t), t, t) = (1/4)*(diff(u(x, y, t), x, x))+(1/4)*(diff(u(x, y, t), y, y)); 
bc  := (D[1](u))(0, y, t) = 0, 
       (D[1](u))(Pi, y, t) = 0, 
       u(x, 0, t) = 0, 
       u(x, Pi, t) = 0; 
ic  := u(x, y, 0) = x*y*(Pi-y),(D[3](u))(x, y, 0) = 0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc,ic],u(x,y,t))),output='realtime')); 
sol := subs(n1=m,sol);
 

$$u(x,y,t)=\sum _{n=1}^{\mathrm{\infty }}-2\frac{({(-1)}^n-1)\sin \left(ny\right)\cos \left(1/2nt\right)}{n^3}+\sum _{n=1}^{\mathrm{\infty }}\left(\sum _{m=1}^{\mathrm{\infty }}8\frac{(-1+{(-1)}^m+{(-1)}^n-{(-1)}^{n+m})\cos \left(mx\right)\sin \left(ny\right)\cos \left(1/2\sqrt{m^2+n^2}t\right)}{{\pi }^2{m}^2{n}^3}\right)$$

____________________________________________________________________________________

5.2.1.8 [401] All 4 edges fixed, zero initial velocity, general solution

problem number 401

Added June 16, 2019

Solve for $u(x,y,t)$ with $0<x<L$ and $0<y<H$ and $t>0$.

Solve $$u_{tt}=c^2{\mathrm{\nabla }}^{2}u(x,y)$$ With boundary conditions $$\begin{array}{rl}u(x,0,t)& =0\\ u(0,y,t)& =0\\ u(L,y,t)& =0\\ u(x,H,t)& =0\end{array}$$

With initial conditions $$\begin{array}{rl}u(x,y,0)& =f(x,y)\\ u_t(x,y,0)& =0\end{array}$$

Figure 5.98: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x, y, t], {t, 2}] == c^2 * Laplacian[u[x, y, t], {x, y}]; 
ic  = {u[x, y, 0] == f[x,y], Derivative[0, 0, 1][u][x, y, 0] == 0}; 
bc  = {u[x, 0, t] == 0, u[0, y, t] == 0, u[L, y, t] == 0, u[x, H, t] == 0}; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[x, y, t], {x, y, t}], 60*10]]; 
sol =  sol /. {K[1] -> n, K[2] -> m};
 

$$\left\{\{u(x,y,t)\to \begin{array}{cc}\{& \begin{array}{cc}\underset{n=1}{\sum ^{\mathrm{\infty }}}\underset{K[3]=1}{\sum ^{\mathrm{\infty }}}\frac{4\cos \left(\pi t\sqrt{c^2(\frac{n^2}{L^2}+\frac{K[3{]}^2}{H^2})}\right)({\int }_0^L{\int }_0^{H}f(x,y)\sin \left(\frac{n\pi x}{L}\right)\sin \left(\frac{\pi yK[3]}{H}\right)dydx)\sin \left(\frac{n\pi x}{L}\right)\sin \left(\frac{\pi yK[3]}{H}\right)}{HL}& (n|K[3])\in \mathbb{Z}\wedge n\ge 1\wedge K[3]\ge 1\wedge c^2{H}^2{L}^2(H^2{n}^2+L^{2}K[3{]}^2)>0\\ \text{Indeterminate}& \text{True}\end{array}\end{array}\}\right\}$$

Maple ✗

restart; 
pde := diff(u(x, y, t), t$2) = c^2* VectorCalculus:-Laplacian(u(x, y, t), 'cartesian'[x, y]); 
ic  := u(x,y,0)=f(x,y),D[3](u)(x,y,0)=0; 
bc  := u(x,0,t)=0,u(0,y,t)=0,u(L,y,t)=0,u(x,H,t)=0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic,bc], u(x, y, t))),output='realtime')); 
sol := subs(n1=m,sol);
 

time expired

Hand solution

Solve for $u(r,\theta ,t)$$$u_{tt}=c^2{\mathrm{\nabla }}^{2}u(x,y)$$ With boundary conditions such that all edges are fixed, and initial conditions $u(x,y,0)=f(x,y)$ and initial velocity $g(x,y)=0.$

Let $u=X\left(x\right)Y\left(y\right)T\left(t\right)$. Substituting into the above PDE gives$$\begin{array}{rl}\frac{1}{c^2}{T}^{\prime \prime }XY& =X^{\prime \prime }YT+Y^{\prime \prime }XT\\ \frac{1}{c^2}\frac{T^{\prime \prime }}{T}& =\frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y}\end{array}$$

Hence$$\begin{array}{rl}\frac{1}{c^2}\frac{T^{\prime \prime }}{T}& =-\lambda \\ \frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y}& =-\lambda \end{array}$$

The time ODE becomes$$T^{\prime \prime }+c^2\lambda T=0$$ And the space ODE is$$\begin{array}{rl}\frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y}& =-\lambda \\ \frac{X^{\prime \prime }}{X}& =-\lambda -\frac{Y^{\prime \prime }}{Y}\end{array}$$

Using a new separation variable $\mu$ gives the following two ODE’s$$\begin{array}{rl}\frac{X^{\prime \prime }}{X}& =-\mu \\ -\lambda -\frac{Y^{\prime \prime }}{Y}& =-\mu \end{array}$$

Or$$\begin{array}{rl}{X}^{\prime \prime }+\mu X& =0\\ X\left(0\right)& =0\\ X\left(L\right)& =0\end{array}$$

And$$\begin{array}{rl}{Y}^{\prime \prime }+Y(\lambda -\mu )& =0\\ Y\left(0\right)& =0\\ Y\left(H\right)& =0\end{array}$$

Solving first for the $X\left(x\right)$ ODE, and knowing that $\mu$ must be positive only here from the nature of the boundary conditions gives$$X=A\cos \left(\sqrt{\mu }x\right)+B\sin \left(\sqrt{\mu }x\right)$$ Applying B.C. at $x=0$$$0=A$$ Hence solution becomes $X\left(x\right)=B\sin \left(\sqrt{\mu }x\right)$. Applying the B.C. at $x=L$ gives$$0=B\sin \left(\sqrt{\mu }L\right)$$ Non trivial solution requires that$$\begin{array}{rl}\sqrt{\mu }L& =n\pi n=1,2,3,\cdots \\ {\mu }_n& ={\left(\frac{n\pi }{L}\right)}^2\end{array}$$

Therefore the eigenfunctions $X_n\left(x\right)$ are$$X_n\left(x\right)=\sin \left(\frac{n\pi }{L}x\right)n=1,2,3,\cdots$$ Solving the $Y\left(y\right)$ ODE$$Y_n^{\prime \prime }+(\lambda -{\left(\frac{n\pi }{L}\right)}^2)Y_n=0n=1,2,3,\cdots$$ The nature of the boundary conditions on $Y\left(y\right)$ suggests that $(\lambda -{\left(\frac{n\pi }{L}\right)}^2)$ must be positive (if $\lambda -{\left(\frac{n\pi }{L}\right)}^2=0$ or $\lambda -{\left(\frac{n\pi }{L}\right)}^2<0$, trivial solutions result).

Hence the solution for $Y_n\left(y\right)$ becomes$$Y_n\left(y\right)=A\cos \left(\sqrt{\lambda -{\left(\frac{n\pi }{L}\right)}^2}y\right)+B\sin \left(\sqrt{\lambda -{\left(\frac{n\pi }{L}\right)}^2}y\right)$$ Applying first B.C. $Y\left(0\right)=0$ gives$$0=A$$ The solution becomes$$Y_n\left(y\right)=B_n\sin \left(\sqrt{\lambda -{\left(\frac{n\pi }{L}\right)}^2}y\right)$$ Applying second B.C. $Y\left(H\right)=0$ gives$$0=B\sin \left(\sqrt{\lambda -{\left(\frac{n\pi }{L}\right)}^2}H\right)$$ Non trivial solution requires that$$\begin{array}{rl}\sqrt{\lambda -{\left(\frac{n\pi }{L}\right)}^2}H& =m\pi n=1,2,3,\cdots ,m=1,2,3,\cdots \\ {\lambda }_{nm}-{\left(\frac{n\pi }{L}\right)}^2& ={\left(\frac{m\pi }{H}\right)}^2\\ {\lambda }_{nm}& ={\left(\frac{m\pi }{H}\right)}^2+{\left(\frac{n\pi }{L}\right)}^2\end{array}$$

Hence the $Y_{nm}\left(y\right)$ eigenfunctions are$$Y_{nm}\left(y\right)=\sin \left(\frac{m\pi }{H}y\right)n=1,2,3,\cdots ,m=1,2,3,\cdots$$ Now the time $T\left(t\right)$ ode is solved, and since ${\lambda }_{nm}$ is positive, then $$\begin{array}{rl}{T}_{nm}^{\prime \prime }+c^2{\lambda }_{nm}{T}_{nm}& =0\\ T_{nm}\left(t\right)& =A_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)+B_{nm}\sin \left(c\sqrt{{\lambda }_{nm}}t\right)\\ & =A_{nm}\cos \left(c\sqrt{{\left(\frac{m\pi }{H}\right)}^2+{\left(\frac{n\pi }{L}\right)}^2}t\right)+B_{nm}\sin \left(c\sqrt{{\left(\frac{m\pi }{H}\right)}^2+{\left(\frac{n\pi }{L}\right)}^2}t\right)\end{array}$$

Combining all solution , and merging all constants into two results in

$$\begin{array}{rl}{u}_{nm}(x,y,t)& =X_n\left(x\right)Y_{nm}\left(y\right)T_{nm}\left(t\right)\\ u(x,y,t)& =\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{X}_n\left(x\right)Y_{nm}\left(y\right)T_{nm}\left(t\right)\\ \text{(1)}& & =\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{A}_{nm}\sin \left(\frac{n\pi }{L}x\right)\sin \left(\frac{m\pi }{H}y\right)\cos \left(c\sqrt{{\left(\frac{m\pi }{H}\right)}^2+{\left(\frac{n\pi }{L}\right)}^2}t\right)+\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{B}_{nm}\sin \left(\frac{n\pi }{L}x\right)\sin \left(\frac{m\pi }{H}y\right)\sin \left(c\sqrt{{\left(\frac{m\pi }{H}\right)}^2+{\left(\frac{n\pi }{L}\right)}^2}t\right)\end{array}$$

Initial conditions are now used to find $A_{nm},B_{nm}$. At $t=0$

$$\begin{array}{rl}u(x,y,0)& =f(x,y)\\ \frac{\partial u}{\partial t}(x,y,0)& =0\end{array}$$

Applying first initial condition to (1)  gives$$f(x,y)=\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{A}_{nm}\sin \left(\frac{n\pi }{L}x\right)\sin \left(\frac{m\pi }{H}y\right)$$ Applying 2D orthogonality gives$$\begin{array}{rl}{\int }_0^L{\int }_0^{H}f(x,y)\sin \left(\frac{n\pi }{L}x\right)\sin \left(\frac{m\pi }{H}y\right)dxdy& =A_{nm}\left(\frac{L}{2}\right)\left(\frac{H}{2}\right)\\ A_{nm}& =\frac{4}{LH}{\int }_0^L{\int }_0^{H}f(x,y)\sin \left(\frac{n\pi }{L}x\right)\sin \left(\frac{m\pi }{H}y\right)dxdy\end{array}$$

Taking time derivative of (1) gives$$\begin{array}{rl}\frac{\partial u}{\partial t}(x,y,t)& =\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}-c\sqrt{{\left(\frac{m\pi }{H}\right)}^2+{\left(\frac{n\pi }{L}\right)}^2}{A}_{nm}\sin \left(\frac{n\pi }{L}x\right)\sin \left(\frac{m\pi }{H}y\right)\sin \left(c\sqrt{{\left(\frac{m\pi }{H}\right)}^2+{\left(\frac{n\pi }{L}\right)}^2}t\right)\\ & +\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}c\sqrt{{\left(\frac{m\pi }{H}\right)}^2+{\left(\frac{n\pi }{L}\right)}^2}{B}_{nm}\sin \left(\frac{n\pi }{L}x\right)\sin \left(\frac{m\pi }{H}y\right)\cos \left(c\sqrt{{\left(\frac{m\pi }{H}\right)}^2+{\left(\frac{n\pi }{L}\right)}^2}t\right)\end{array}$$

AT $t=0$ the above becomes$${\int }_0^{H}g(x,y)=\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}c\sqrt{{\left(\frac{m\pi }{H}\right)}^2+{\left(\frac{n\pi }{L}\right)}^2}{B}_{nm}\sin \left(\frac{n\pi }{L}x\right)\sin \left(\frac{m\pi }{H}y\right)$$ Applying 2D orthogonality gives$${\int }_0^L{\int }_0^{H}g(x,y)\sin \left(\left(\frac{n\pi }{L}\right)x\right)\sin \left(\frac{m\pi }{H}y\right)dxdy=B_{nm}\left(\frac{L}{2}\right)\left(\frac{H}{2}\right)$$ But the initial velocity $g(x,y)=0$. Hence $B_{nm}=0$ for all $n,m$.

Summary of solution$$\begin{array}{rl}u(x,y,t)& =\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{A}_{nm}\sin \left(\frac{n\pi }{L}x\right)\sin \left(\frac{m\pi }{H}y\right)\cos \left(c\sqrt{{\left(\frac{m\pi }{H}\right)}^2+{\left(\frac{n\pi }{L}\right)}^2}t\right)\\ A_{nm}& =\frac{4}{LH}{\int }_0^L{\int }_0^{H}f(x,y)\sin \left(\frac{n\pi }{L}x\right)\sin \left(\frac{m\pi }{H}y\right)dxdy\end{array}$$

____________________________________________________________________________________

5.2.1.9 [402] With damping

problem number 402

Taken from Maple PDE help pages. This wave PDE inside square with damping present.

Membrane is free to move on the right edge and also on top edge. But fixed at left edge and bottom edge.

It has zero initial position, but given a non-zero initial velocity. Where $0<x<1$ and $0<y<1$ and $t>0$. Solve $$u_{tt}+\frac{1}{10}{u}_t=\frac{1}{4}{\mathrm{\nabla }}^{2}u(x,y)$$ With boundary conditions $$\begin{array}{rl}u(0,y,t)& =0\\ \frac{\partial u}{\partial x}u(1,y,t)& =0\\ u(x,0,t)& =0\\ \frac{\partial u}{\partial y}u(x,1,t)& =0\end{array}$$

With initial conditions $$\begin{array}{rl}u(x,y,0)& =0\\ \frac{\partial u}{\partial t}(x,y,0)& =x(1-\frac{1}{2}x)(1-\frac{1}{2}y)y\end{array}$$

Figure 5.99: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x, y, t], {t, 2}] == (1*(D[u[x, y, t], {x, 2}] + D[u[x, y, t], {y, 2}]))/4 - (1*D[u[x, y, t], t])/10; 
ic  = {u[x, y, 0] == 0, Derivative[0, 0, 1][u][x, y, 0] == x*(1 - (1/2)*x)*(1 - (1/2)*y)*y}; 
bc  = {u[0, y, t] == 0, Derivative[1, 0, 0][u][1, y, t] == 0, u[x, 0, t] == 0, Derivative[0, 1, 0][u][x, 1, t] == 0}; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, y, t], {x, y, t}], 60*10]];
 

$$\left\{\{u(x,y,t)\to \begin{array}{cc}\{& \begin{array}{cc}\underset{K[1]=1}{\sum ^{\mathrm{\infty }}}\underset{K[3]=1}{\sum ^{\mathrm{\infty }}}\frac{5120e^{-t/20}\sin (\frac{1}{2}\pi x(2K[1]-1))\sin (\frac{1}{2}\pi y(2K[3]-1))\sin \left(\frac{1}{20}t\sqrt{50{\pi }^2(2K[1{]}^2-2K[1]+2K[3{]}^2-2K[3]+1)-1}\right)}{{\pi }^6(2K[1]-1{)}^3(2K[3]-1{)}^3\sqrt{50{\pi }^2(2K[1{]}^2-2K[1]+2K[3{]}^2-2K[3]+1)-1}}& (K[1]|K[3])\in \mathbb{Z}\wedge K[1]\ge 1\wedge K[3]\ge 1\\ \text{Indeterminate}& \text{True}\end{array}\end{array}\}\right\}$$

Maple ✓

restart; 
pde := diff(u(x, y, t), t$2) = 1/4*(diff(u(x, y, t), x$2)+diff(u(x, y, t), y$2))-(1/10)*(diff(u(x, y, t), t)); 
bc  := u(0, y, t) = 0, 
      (D[1](u))(1, y, t) = 0, 
      u(x, 0, t) = 0, 
      (D[2](u))(x, 1, t) = 0; 
ic  := u(x, y, 0) = 0, (D[3](u))(x, y, 0) = x*(1-(1/2)*x)*(1-(1/2)*y)*y; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic,bc], u(x, y, t))),output='realtime')); 
sol := subs(n1=m,sol);
 

$$u(x,y,t)=\sum _{m=0}^{\mathrm{\infty }}\left(\sum _{n=0}^{\mathrm{\infty }}5120\frac{{\mathrm{e}}^{-t/20}\sin \left(1/2(1+2m)\pi y\right)\sin \left(1/2(1+2n)\pi x\right)\sin \left(1/20t\sqrt{-1+(100m^2+100n^2+100m+100n+50){\pi }^2}\right)}{\sqrt{-1+(100m^2+100n^2+100m+100n+50){\pi }^2}{\pi }^6{(1+2m)}^3{(1+2n)}^3}\right)$$

____________________________________________________________________________________

5.2.1.10 [403] On the whole plane

problem number 403

From Mathematica DSolve help pages.

Hyperbolic partial differential equation with non-rational coefficients.

Solve for $u(x,y)$ $$u_{xx}-2\sin x{u}_{xy}-{\cos }^{2}x{u}_{yy}-\cos x{u}_y=0$$

Figure 5.100: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
ode = D[u[x, y], {x, 2}] - 2*Sin[x]*D[u[x, y], x, y] - Cos[x]^2*D[u[x, y], {y, 2}] - Cos[x]*D[u[x, y], y] == 0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[ode, u[x, y], {x, y}], 60*10]];
 

$$\{\{u(x,y)\to c_1(x-\cos (x)+y)+c_2(-x-\cos (x)+y)\}\}$$

Maple ✗

restart; 
interface(showassumed=0); 
ode := diff(u(x, y), x$2) - 2*sin(x)*diff(u(x, y),x,y)-cos(x)^2*diff(u(x, y), y$2) - cos(x)*diff(u(x, y), y) = 0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(ode, u(x, y))),output='realtime'));
 

sol=()

____________________________________________________________________________________