23 Wave PDE in 2D Cartesian coordinates

23.1 In square, given initial position but with zero initial velocity
23.2 In square with damping. Given zero initial position but with non-zero initial velocity
23.3 In rectangle. All 4 edges are fixed and given non-zero initial position with zero initial velocity
23.4 In rectangle. All 4 edges are fixed and given non-zero initial position with zero initial velocity (Haberman 8.5.5 (a))

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23.1 In square, given initial position but with zero initial velocity

problem number 160

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 \[ \frac {\partial ^2 u}{\partial t^2} = \frac {1}{4} \left ( \frac {\partial ^2 u}{\partial x^2}+ \frac {\partial ^2 u}{\partial y^2} \right ) \]

With boundary conditions

\begin {align*} \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 {align*}

With initial conditions

\begin {align*} \frac {\partial u}{\partial t}(x,y,0) &=0 \\ u(x,0) &= x y (\pi -y) \end {align*}

Mathematica

ClearAll[u, t, y, x]; 
 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]];
 

\[ \text {Failed} \]

Maple

 
x:='x'; t:='t'; y:='y'; u:='u'; 
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; 
sol:=pdsolve([pde,bc,ic],u(x,y,t)); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc,ic],u(x,y,t))),output='realtime')); 
sol:=subs(n1=m,sol);
 

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

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23.2 In square with damping. Given zero initial position but with non-zero initial velocity

problem number 161

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 \[ \frac {\partial ^2 u}{\partial t^2} = \frac {1}{4} \left ( \frac {\partial ^2 u}{\partial x^2}+ \frac {\partial ^2 u}{\partial y^2} \right ) -\frac {1}{10} \frac {\partial u}{\partial t} \]

With boundary conditions

\begin {align*} 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 {align*}

With initial conditions

\begin {align*} 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 {align*}

Mathematica

ClearAll[u, t, y, x]; 
 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]];
 

\[ \text {Failed} \]

Maple

 
x:='x'; t:='t'; y:='y'; u:='u'; 
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; 
sol:=pdsolve([pde, ic,bc], u(x, y, t)); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic,bc], u(x, y, t))),output='realtime')); 
sol:=subs(n1=m,sol);
 

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

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23.3 In rectangle. All 4 edges are fixed and given non-zero initial position with zero initial velocity

problem number 162

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 {align*} u(x,0,t) &=0 \\ u(0,y,t) &= 0 \\ u(1,y,t) &=0 \\ u(x,2,t) &= 0 \end {align*}

With initial conditions

\begin {align*} u(x,y,0) &=\frac {1}{10} (x-x^2)(2 y-y^2) \\ \frac {\partial u}{\partial t}(x,y,0) &= 0 \end {align*}

Mathematica

ClearAll[u, t, y, x, n, m]; 
 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]];
 

\[ \text {Bad latex generated} \]

Maple

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

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

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23.4 In rectangle. All 4 edges are fixed and given non-zero initial position with zero initial velocity (Haberman 8.5.5 (a))

problem number 163

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 \(u=0\).

\[ \frac {\partial ^2 u(x,y,t)}{\partial t^2} = c^2 \nabla (u) + Q(x,y,t) \]

If the memberane is rectangle \((0<x<L,0<y<H)\).

With initial conditions

\begin {align*} u(x,y,0) &=f(x,y) \\ \frac {\partial u}{\partial t}(x,y,0) &= 0 \end {align*}

See my HW9, Math 322, UW Madison.

Mathematica

ClearAll[u, t, y, x, n, m, L, H, Q, f]; 
 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]];
 

\[ \text {Failed} \]

Maple

 
x:='x'; t:='t'; L:='L'; c:='c';u:='u';Q:='Q'; 
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'));
 

\[ \text { sol=() } \]