Added January 10, 2020.
Solve \[ u_{tt} = c^4 \left ( u_{xx}+ u_{yy} \right ) \]
\begin {align*} 0 & <x<L\\ 0 & <y<H \end {align*}
Boundary conditions on \(x\)
\begin {align*} u\left ( 0,y,t\right ) & =0\\ u\left ( L,y,t\right ) & =0 \end {align*}
And boundary conditions on \(y\)\begin {align*} u\left ( x,0,t\right ) & =0\\ u\left ( x,H,t\right ) & =0 \end {align*}
Initial conditions\begin {align*} u\left ( x,y,0\right ) & =f\left ( x,y\right ) \\ \frac {\partial u}{\partial t}\left ( x,y,0\right ) & =g\left ( x,y\right ) \end {align*}
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 {align*} \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 {align*}
Therefore\begin {align*} \frac {1}{c^{2}}\frac {T^{\prime \prime }}{T} & =-\lambda \\ \frac {X^{\prime \prime }}{X}+\frac {Y^{\prime \prime }}{Y} & =-\lambda \end {align*}
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 {align*} \frac {X^{\prime \prime }}{X} & =-\mu \\ -\lambda -\frac {Y^{\prime \prime }}{Y} & =-\mu \end {align*}
Or\begin {align*} X^{\prime \prime }+\mu X & =0\\ Y^{\prime \prime }+Y\left ( \lambda -\mu \right ) & =0 \end {align*}
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 {align*} \sqrt {\mu }L & =n\pi \\ \mu & =\left ( \frac {n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \end {align*}
Therefore the \(X_{n}\left ( x\right ) \) eigenfunctions are\[ X_{n}\left ( x\right ) =B_{n}\sin \left ( \frac {n\pi }{L}x\right ) \qquad n=1,2,3,\cdots \] Now, solving the \(Y\left ( y\right ) \) ODE above\[ Y^{\prime \prime }+Y\left ( \lambda -\left ( \frac {n\pi }{L}\right ) ^{2}\right ) =0 \] 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 {align*} \sqrt {\lambda -\left ( \frac {n\pi }{L}\right ) ^{2}}H & =m\pi \qquad 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}\qquad n=1,2,3,\cdots ,m=1,2,3,\cdots \end {align*}
Hence the \(Y_{nm}\left ( y\right ) \) solution is\[ Y_{nm}=B_{nm}\sin \left ( \frac {m\pi }{H}y\right ) \qquad n=1,2,3,\cdots ,m=1,2,3,\cdots \] The time ode \(T\left ( t\right ) \) is now solved\begin {align*} 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 {align*}
Combining all solutions, and merging all constants into two results in\begin {align} u_{nm}\left ( x,y,t\right ) & =X_{n}\left ( x\right ) Y_{nm}\left ( y\right ) T_{nm}\left ( t\right ) \nonumber \\ u\left ( x,y,t\right ) & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }X_{m}\left ( x\right ) Y_{mn}\left ( y\right ) T_{mn}\left ( t\right ) \nonumber \\ & =\sum _{n=1}^{\infty }\sum _{m=1}^{\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 ) \nonumber \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\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 ) \tag {1} \end {align}
Initial conditions are now used to find \(A_{nm},B_{nm}\). At \(t=0\)\begin {align*} u\left ( x,y,0\right ) & =f\left ( x,y\right ) \\ \frac {\partial u}{\partial t}\left ( x,y,0\right ) & =g\left ( x,y\right ) \end {align*}
Applying first initial condition to (1) gives\[ f\left ( x,y\right ) =\sum _{n=1}^{\infty }\left ( \sum _{m=1}^{\infty }A_{nm}\sin \left ( \frac {m\pi }{H}y\right ) \right ) \sin \left ( \left ( \frac {n\pi }{L}\right ) x\right ) \] Applying 2D orthogonality gives\begin {align*} \int _{0}^{L}\int _{0}^{H}f\left ( x,y\right ) \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\left ( x,y\right ) \sin \left ( \left ( \frac {n\pi }{L}\right ) x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \ dxdy \end {align*}
Taking time derivative of (1) gives\begin {align*} \frac {\partial u}{\partial t}\left ( x,y,t\right ) & =\sum _{n=1}^{\infty }\sum _{m=1}^{\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}^{\infty }\sum _{m=1}^{\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 {align*}
At \(t=0\) the above becomes\[ g\left ( x,y\right ) =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }c\sqrt {\lambda _{nm}}B_{nm}\sin \left ( \left ( \frac {n\pi }{L}\right ) x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \] Applying 2D orthogonality gives\begin {align*} \int _{0}^{L}\int _{0}^{H}g\left ( x,y\right ) \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\left ( x,y\right ) \sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \ dxdy \end {align*}
Summary of solution\begin {align*} u\left ( x,y,t\right ) =\sum _{n=1}^{\infty }\left ( \sum _{m=1}^{\infty }A_{nm}\sin \left ( \frac {m\pi }{H}y\right ) \cos \left ( c\sqrt {\lambda _{nm}}t\right ) \right ) \sin \left ( \frac {n\pi }{L}x\right ) & +\sum _{n=1}^{\infty }\left ( \sum _{m=1}^{\infty }B_{nm}\sin \left ( \frac {m\pi }{H}y\right ) \sin \left ( c\sqrt {\lambda _{nm}}t\right ) \right ) \sin \left ( \frac {n\pi }{L}x\right ) \\ A_{nm} & =\frac {4}{LH}\int _{0}^{L}\int _{0}^{H}f\left ( x,y\right ) \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\left ( x,y\right ) \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 {align*}
____________________________________________________________________________________
Added January 10, 2020.
Solve \[ u_{tt} = c^4 \left ( u_{xx}+ u_{yy} \right ) \]
\begin {align*} 0 & <x<L\\ 0 & <y<H \end {align*}
Boundary conditions on \(x\)
\begin {align*} u\left ( 0,y,t\right ) & =0\\ u\left ( L,y,t\right ) & =0 \end {align*}
And boundary conditions on \(y\)\begin {align*} u\left ( x,0,t\right ) & =0\\ u\left ( x,H,t\right ) & =0 \end {align*}
Initial conditions\begin {align*} u\left ( x,y,0\right ) & =f\left ( x,y\right ) \\ \frac {\partial u}{\partial t}\left ( x,y,0\right ) & =g\left ( x,y\right ) \end {align*}
Using \(L=1,H=2,c=\frac {1}{10},f(x,y)=x \cos (y),g(x,y)=0\).
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);
\[u \left ( x,y,t \right ) =\sum _{m=1}^{\infty } \left ( \sum _{n=1}^{\infty }4\,{\frac {\sin \left ( n\pi \,x \right ) \sin \left ( 1/2\,m\pi \,y \right ) \cos \left ( 1/20\,\pi \,\sqrt {{m}^{2}+4\,{n}^{2}}t \right ) m \left ( \cos \left ( 2 \right ) \left ( -1 \right ) ^{m+n}- \left ( -1 \right ) ^{n} \right ) }{n \left ( {\pi }^{2}{m}^{2}-4 \right ) }} \right ) \]
Hand solution
The basic solution for this type of PDE was already given in problem 5.2.1.1 on page 1492 as
\begin {align*} u\left ( x,y,t\right ) =\sum _{n=1}^{\infty }\left ( \sum _{m=1}^{\infty }A_{nm}\sin \left ( \frac {m\pi }{H}y\right ) \cos \left ( c\sqrt {\lambda _{nm}}t\right ) \right ) \sin \left ( \frac {n\pi }{L}x\right ) & +\sum _{n=1}^{\infty }\left ( \sum _{m=1}^{\infty }B_{nm}\sin \left ( \frac {m\pi }{H}y\right ) \sin \left ( c\sqrt {\lambda _{nm}}t\right ) \right ) \sin \left ( \frac {n\pi }{L}x\right ) \\ A_{nm} & =\frac {4}{LH}\int _{0}^{L}\int _{0}^{H}f\left ( x,y\right ) \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\left ( x,y\right ) \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}\qquad n=1,2,3,\cdots ,m=1,2,3,\cdots \end {align*}
In this problem \begin {align*} L & =1\\ H & =2\\ c & =\frac {1}{10}\\ f\left ( x,y\right ) & =x\cos y\\ g\left ( x,y\right ) & =0 \end {align*}
Hence the solution becomes
\begin {align*} u\left ( x,y,t\right ) =\sum _{n=1}^{\infty }\left ( \sum _{m=1}^{\infty }A_{nm}\sin \left ( \frac {m\pi }{2}y\right ) \cos \left ( \frac {1}{10}\sqrt {\lambda _{nm}}t\right ) \right ) \sin \left ( n\pi x\right ) & +\sum _{n=1}^{\infty }\left ( \sum _{m=1}^{\infty }B_{nm}\sin \left ( \frac {m\pi }{2}y\right ) \sin \left ( \frac {1}{10}\sqrt {\lambda _{nm}}t\right ) \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}\qquad n=1,2,3,\cdots ,m=1,2,3,\cdots \end {align*}
But \begin {align*} 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\left ( -1\right ) ^{n}m\left ( -1+\left ( -1\right ) ^{m}\cos \left ( 2\right ) \right ) }{n\left ( m^{2}\pi ^{2}-4\right ) } \end {align*}
Hence the solution simplifies to
\begin {align*} u\left ( x,y,t\right ) & =\sum _{n=1}^{\infty }\left ( \sum _{m=1}^{\infty }\frac {4\left ( -1\right ) ^{n}m\left ( -1+\left ( -1\right ) ^{m}\cos \left ( 2\right ) \right ) }{n\left ( m^{2}\pi ^{2}-4\right ) }\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 ) \right ) \sin \left ( n\pi x\right ) \\ & =\sum _{n=1}^{\infty }\left ( \sum _{m=1}^{\infty }\frac {4\left ( -1\right ) ^{n}m\left ( -1+\left ( -1\right ) ^{m}\cos \left ( 2\right ) \right ) }{n\left ( m^{2}\pi ^{2}-4\right ) }\cos \left ( \pi \frac {\sqrt {m^{2}+4n^{2}}}{20}t\right ) \sin \left ( \frac {m\pi }{2}y\right ) \right ) \sin \left ( n\pi x\right ) \end {align*}
Animation is below
Source code used for the above
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Added June 17, 2019
Solve for \(u(x,y,t)\) with \(0<x<L\) and \(0<y<H\) and \(t>0\).
Solve \[ u_{tt} = c^2 \nabla ^2 u(x,y) \] With boundary conditions \begin {align*} u(x,0,t) &=0 \\ u(0,y,t) &= 0 \\ u(L,y,t) &=0 \\ u(x,H,t) &= 0 \end {align*}
With initial conditions \begin {align*} u(x,y,0) &=3 f_1(x) f_2(y) \\ u_t(x,y,0) &= 0 \end {align*}
And \[ f_{1}\left ( x\right ) =\left \{ \begin {array} [c]{ccc}x & & 0<x<\frac {L}{2}\\ L-x & & \frac {L}{2}<x<L \end {array} \right . \] Where \[ f_{2}\left ( y\right ) =\left \{ \begin {array} [c]{ccc}y & & 0<y<\frac {H}{2}\\ H-y & & \frac {H}{2}<y<H \end {array} \right . \] And \(L=2,H=3\) and \(c=\frac {1}{3}\).
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};
\[\left \{\left \{u(x,y,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {n=1}{\overset {\infty }{\sum }}\underset {K[3]=1}{\overset {\infty }{\sum }}\frac {288 \cos \left (\frac {1}{18} \pi t \sqrt {9 n^2+4 K[3]^2}\right ) \sin \left (\frac {n \pi }{2}\right ) \sin \left (\frac {n \pi x}{2}\right ) \sin \left (\frac {1}{2} \pi K[3]\right ) \sin \left (\frac {1}{3} \pi y K[3]\right )}{n^2 \pi ^4 K[3]^2} & (n|K[3])\in \mathbb {Z}\land n\geq 1\land K[3]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
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);
\[u \left ( x,y,t \right ) =\sum _{m=1}^{\infty } \left ( \sum _{n=1}^{\infty }288\,{\frac {\sin \left ( 1/3\,m\pi \,y \right ) \sin \left ( 1/2\,n\pi \,x \right ) \cos \left ( 1/18\,\pi \,\sqrt {4\,{m}^{2}+9\,{n}^{2}}t \right ) \sin \left ( 1/2\,n\pi \right ) \sin \left ( 1/2\,m\pi \right ) }{{n}^{2}{\pi }^{4}{m}^{2}}} \right ) \]
Hand solution
The basic solution for this type of PDE was already given in problem 5.2.1.8 on page 1533 as\begin {align*} u\left ( x,y,t\right ) & =\sum _{n=1}^{\infty }\sum _{m=1}^{\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\left ( x,y\right ) \sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \ dxdy \end {align*}
In this problem \begin {align*} L & =2\\ H & =3\\ c & =\frac {1}{3}\\ f\left ( x,y\right ) & =3f_{1}\left ( x\right ) f_{2}\left ( y\right ) \end {align*}
And\[ f_{1}\left ( x\right ) =\left \{ \begin {array} [c]{ccc}x & & 0<x<\frac {L}{2}\\ L-x & & \frac {L}{2}<x<L \end {array} \right . \] And\[ f_{2}\left ( y\right ) =\left \{ \begin {array} [c]{ccc}y & & 0<y<\frac {H}{2}\\ H-y & & \frac {H}{2}<y<H \end {array} \right . \] 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
The following shows selected modes. For example, for \(n=1,m=1\) the solution becomes\[ u\left ( x,y,t\right ) =A_{1,1}\sin \left ( \frac {\pi }{L}x\right ) \sin \left ( \frac {\pi }{H}y\right ) \cos \left ( c\sqrt {\left ( \frac {\pi }{H}\right ) ^{2}+\left ( \frac {1\pi }{L}\right ) ^{2}}t\right ) \] And for \(n=1,m=5\) then the solution becomes\[ u\left ( x,y,t\right ) =A_{1,5}\sin \left ( \frac {\pi }{L}x\right ) \sin \left ( \frac {5\pi }{H}y\right ) \cos \left ( c\sqrt {\left ( \frac {5\pi }{H}\right ) ^{2}+\left ( \frac {\pi }{L}\right ) ^{2}}t\right ) \] And so on.
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Added June 18, 2019
Solve for \(u(x,y,t)\) with \(0<x<L\) and \(0<y<H\) and \(t>0\).
Solve \[ u_{tt} = c^2 \nabla ^2 u(x,y) \] With boundary conditions \begin {align*} u(x,0,t) &=0 \\ u(0,y,t) &= 0 \\ u(L,y,t) &=0 \\ u(x,H,t) &= 0 \end {align*}
With initial conditions \begin {align*} u(x,y,0) &=f(x,y) \\ u_t(x,y,0) &= 0 \end {align*}
And \begin {align*} L & =20\\ H & =30\\ c & =\frac {1}{3}\\ f\left ( x,y\right ) & =f_{1}\left ( x\right ) f_{2}\left ( y\right ) \end {align*}
Where \(f\left ( x,y\right ) \) is an approximation of delta in the middle of the membrane \[ f_{1}\left ( x\right ) =\left \{ \begin {array} [c]{ccc}1 & & \frac {45}{100}L<x<\frac {55}{100}L\\ 0 & & \text {otherwise}\end {array} \right . \] And\[ f_{2}\left ( y\right ) =\left \{ \begin {array} [c]{ccc}1 & & \frac {45}{100}H<y<\frac {55}{100}H\\ 0 & & \text {otherwise}\end {array} \right . \]
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};
\[\left \{\left \{u(x,y,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {n=1}{\overset {\infty }{\sum }}\underset {K[3]=1}{\overset {\infty }{\sum }}\frac {4 \left (\cos \left (\frac {9 n \pi }{20}\right )-\cos \left (\frac {11 n \pi }{20}\right )\right ) \left (\cos \left (\frac {9}{20} \pi K[3]\right )-\cos \left (\frac {11}{20} \pi K[3]\right )\right ) \cos \left (\frac {1}{180} \pi t \sqrt {9 n^2+4 K[3]^2}\right ) \sin \left (\frac {n \pi x}{20}\right ) \sin \left (\frac {1}{30} \pi y K[3]\right )}{n \pi ^2 K[3]} & (n|K[3])\in \mathbb {Z}\land n\geq 1\land K[3]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
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);
\[u \left ( x,y,t \right ) =\sum _{m=1}^{\infty } \left ( \sum _{n=1}^{\infty }-16\,{\frac {\sin \left ( 1/20\,n\pi \,x \right ) \sin \left ( 1/30\,m\pi \,y \right ) \left ( \sin \left ( 1/20\,n\pi \right ) \right ) ^{2}\cos \left ( 1/20\,n\pi \right ) \left ( -1-2\,\cos \left ( 1/5\,n\pi \right ) +2\,\cos \left ( 1/10\,n\pi \right ) \right ) \left ( 1+2\,\cos \left ( 1/10\,n\pi \right ) -2\,\cos \left ( 1/20\,n\pi \right ) \right ) \left ( 1+2\,\cos \left ( 1/10\,n\pi \right ) +2\,\cos \left ( 1/20\,n\pi \right ) \right ) }{n{\pi }^{2}m}\cos \left ( {\frac {\pi \,\sqrt {4\,{m}^{2}+9\,{n}^{2}}t}{180}} \right ) \left ( -\cos \left ( {\frac {11\,m\pi }{20}} \right ) +\cos \left ( {\frac {9\,m\pi }{20}} \right ) \right ) } \right ) \]
Hand solution
The basic solution for this type of PDE was already given in problem 5.2.1.8 on page 1533 as\begin {align*} u\left ( x,y,t\right ) & =\sum _{n=1}^{\infty }\sum _{m=1}^{\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\left ( x,y\right ) \sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \ dxdy \end {align*}
In this problem \begin {align*} L & =20\\ H & =30\\ c & =\frac {1}{3}\\ f\left ( x,y\right ) & =f_{1}\left ( x\right ) f_{2}\left ( y\right ) \end {align*}
Where \(f\left ( x,y\right ) \) is an approximation of delta in the middle of the membrane \[ f_{1}\left ( x\right ) =\left \{ \begin {array} [c]{ccc}1 & & \frac {45}{100}L<x<\frac {55}{100}L\\ 0 & & \text {otherwise}\end {array} \right . \] And\[ f_{2}\left ( y\right ) =\left \{ \begin {array} [c]{ccc}1 & & \frac {45}{100}H<y<\frac {55}{100}H\\ 0 & & \text {otherwise}\end {array} \right . \] 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
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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["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]];
\[\left \{\left \{u(x,y,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {n=1}{\overset {\infty }{\sum }}\underset {K[3]=1}{\overset {\infty }{\sum }}\frac {32 \left (-1+(-1)^n\right ) \left (-1+(-1)^{K[3]}\right ) \cos \left (\frac {1}{2} \pi t \sqrt {4 n^2+K[3]^2}\right ) \sin (n \pi x) \sin \left (\frac {1}{2} \pi y K[3]\right )}{5 n^3 \pi ^6 K[3]^3} & K[3]\in \mathbb {Z}\land n\geq 1\land K[3]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
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);
\[u \left ( x,y,t \right ) =\sum _{m=1}^{\infty } \left ( \sum _{n=1}^{\infty }-{\frac {32\,\sin \left ( n\pi \,x \right ) \sin \left ( 1/2\,m\pi \,y \right ) \cos \left ( 1/2\,\pi \,\sqrt {{m}^{2}+4\,{n}^{2}}t \right ) \left ( - \left ( -1 \right ) ^{m+n}+ \left ( -1 \right ) ^{m}+ \left ( -1 \right ) ^{n}-1 \right ) }{5\,{n}^{3}{\pi }^{6}{m}^{3}}} \right ) \]
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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 \(u=0\). \[ u_{tt} = c^2 \nabla ^2 u + Q(x,y,t) \] 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["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]];
\[\left \{\left \{u(x,y,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}\underset {K[3]=1}{\overset {\infty }{\sum }}2 \sqrt {\frac {1}{H}} \sqrt {\frac {1}{L}} \left (\int _0^t \frac {2 \left (\int _0^L\int _0^HQ(x,y,K[4]) \sin \left (\frac {\pi x K[1]}{L}\right ) \sin \left (\frac {\pi y K[3]}{H}\right )dydx\right ) \sin \left (c \pi \sqrt {\frac {K[1]^2}{L^2}+\frac {K[3]^2}{H^2}} (t-K[4])\right )}{c \pi \sqrt {\frac {H K[1]^2}{L}+\frac {L K[3]^2}{H}}} \, dK[4]+\frac {2 \cos \left (c \pi t \sqrt {\frac {K[1]^2}{L^2}+\frac {K[3]^2}{H^2}}\right ) \int _0^L\int _0^Hf(x,y) \sin \left (\frac {\pi x K[1]}{L}\right ) \sin \left (\frac {\pi y K[3]}{H}\right )dydx}{\sqrt {H L}}\right ) \sin \left (\frac {\pi x K[1]}{L}\right ) \sin \left (\frac {\pi y K[3]}{H}\right ) & (K[1]|K[3])\in \mathbb {Z}\land K[1]\geq 1\land K[3]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
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=()
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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 \[ u_{tt} = \frac {1}{4} \left ( \frac {\partial ^2 u}{\partial x^2}+ \frac {\partial ^2 u}{\partial y^2} \right ) \] 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["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]];
\[\left \{\left \{u(x,y,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[3]=1}{\overset {\infty }{\sum }}-\frac {2 \left (-1+(-1)^{K[3]}\right ) \cos \left (\frac {1}{2} t K[3]\right ) \sin (y K[3])}{K[3]^3}+\underset {K[1]=1}{\overset {\infty }{\sum }}\underset {K[3]=1}{\overset {\infty }{\sum }}-\frac {8 \left (-1+(-1)^{K[1]}\right ) \left (-1+(-1)^{K[3]}\right ) \cos (x K[1]) \cos \left (\frac {1}{2} t \sqrt {K[1]^2+K[3]^2}\right ) \sin (y K[3])}{\pi ^2 K[1]^2 K[3]^3} & (K[1]|K[3])\in \mathbb {Z}\land K[1]\geq 1\land K[3]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
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);
\[u \left ( x,y,t \right ) =\sum _{n=1}^{\infty }-2\,{\frac { \left ( \left ( -1 \right ) ^{n}-1 \right ) \sin \left ( ny \right ) \cos \left ( 1/2\,nt \right ) }{{n}^{3}}}+\sum _{n=1}^{\infty } \left ( \sum _{m=1}^{\infty }8\,{\frac { \left ( -1+ \left ( -1 \right ) ^{m}+ \left ( -1 \right ) ^{n}- \left ( -1 \right ) ^{n+m} \right ) \cos \left ( mx \right ) \sin \left ( ny \right ) \cos \left ( 1/2\,\sqrt {{m}^{2}+{n}^{2}}t \right ) }{{\pi }^{2}{m}^{2}{n}^{3}}} \right ) \]
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Added June 16, 2019
Solve for \(u(x,y,t)\) with \(0<x<L\) and \(0<y<H\) and \(t>0\).
Solve \[ u_{tt} = c^2 \nabla ^2 u(x,y) \] With boundary conditions \begin {align*} u(x,0,t) &=0 \\ u(0,y,t) &= 0 \\ u(L,y,t) &=0 \\ u(x,H,t) &= 0 \end {align*}
With initial conditions \begin {align*} u(x,y,0) &=f(x,y) \\ u_t(x,y,0) &= 0 \end {align*}
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};
\[\left \{\left \{u(x,y,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {n=1}{\overset {\infty }{\sum }}\underset {K[3]=1}{\overset {\infty }{\sum }}\frac {4 \cos \left (\pi t \sqrt {c^2 \left (\frac {n^2}{L^2}+\frac {K[3]^2}{H^2}\right )}\right ) \left (\int _0^L\int _0^Hf(x,y) \sin \left (\frac {n \pi x}{L}\right ) \sin \left (\frac {\pi y K[3]}{H}\right )dydx\right ) \sin \left (\frac {n \pi x}{L}\right ) \sin \left (\frac {\pi y K[3]}{H}\right )}{H L} & (n|K[3])\in \mathbb {Z}\land n\geq 1\land K[3]\geq 1\land c^2 H^2 L^2 \left (H^2 n^2+L^2 K[3]^2\right )>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
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 \(u\left ( r,\theta ,t\right ) \)\[ u_{tt}=c^{2}\nabla ^{2}u\left ( x,y\right ) \] With boundary conditions such that all edges are fixed, and initial conditions \(u\left ( x,y,0\right ) =f\left ( x,y\right ) \) and initial velocity \(g\left ( x,y\right ) =0.\)
Let \(u=X\left ( x\right ) Y\left ( y\right ) T\left ( t\right ) \). Substituting into the above PDE gives\begin {align*} \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 {align*}
Hence\begin {align*} \frac {1}{c^{2}}\frac {T^{\prime \prime }}{T} & =-\lambda \\ \frac {X^{\prime \prime }}{X}+\frac {Y^{\prime \prime }}{Y} & =-\lambda \end {align*}
The time ODE becomes\[ T^{\prime \prime }+c^{2}\lambda T=0 \] And the space ODE is\begin {align*} \frac {X^{\prime \prime }}{X}+\frac {Y^{\prime \prime }}{Y} & =-\lambda \\ \frac {X^{\prime \prime }}{X} & =-\lambda -\frac {Y^{\prime \prime }}{Y} \end {align*}
Using a new separation variable \(\mu \) gives the following two ODE’s\begin {align*} \frac {X^{\prime \prime }}{X} & =-\mu \\ -\lambda -\frac {Y^{\prime \prime }}{Y} & =-\mu \end {align*}
Or\begin {align*} X^{\prime \prime }+\mu X & =0\\ X\left ( 0\right ) & =0\\ X\left ( L\right ) & =0 \end {align*}
And\begin {align*} Y^{\prime \prime }+Y\left ( \lambda -\mu \right ) & =0\\ Y\left ( 0\right ) & =0\\ Y\left ( H\right ) & =0 \end {align*}
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 {align*} \sqrt {\mu }L & =n\pi \qquad n=1,2,3,\cdots \\ \mu _{n} & =\left ( \frac {n\pi }{L}\right ) ^{2} \end {align*}
Therefore the eigenfunctions \(X_{n}\left ( x\right ) \) are\[ X_{n}\left ( x\right ) =\sin \left ( \frac {n\pi }{L}x\right ) \qquad n=1,2,3,\cdots \] Solving the \(Y\left ( y\right ) \) ODE\[ Y_{n}^{\prime \prime }+\left ( \lambda -\left ( \frac {n\pi }{L}\right ) ^{2}\right ) Y_{n}=0\qquad n=1,2,3,\cdots \] The nature of the boundary conditions on \(Y\left ( y\right ) \) suggests that \(\left ( \lambda -\left ( \frac {n\pi }{L}\right ) ^{2}\right ) \) 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 {align*} \sqrt {\lambda -\left ( \frac {n\pi }{L}\right ) ^{2}}H & =m\pi \qquad 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 {align*}
Hence the \(Y_{nm}\left ( y\right ) \) eigenfunctions are\[ Y_{nm}\left ( y\right ) =\sin \left ( \frac {m\pi }{H}y\right ) \qquad n=1,2,3,\cdots ,m=1,2,3,\cdots \] Now the time \(T\left ( t\right ) \) ode is solved, and since \(\lambda _{nm}\) is positive, then \begin {align*} 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 {align*}
Combining all solution , and merging all constants into two results in
\begin {align} u_{nm}\left ( x,y,t\right ) & =X_{n}\left ( x\right ) Y_{nm}\left ( y\right ) T_{nm}\left ( t\right ) \nonumber \\ u\left ( x,y,t\right ) & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }X_{n}\left ( x\right ) Y_{nm}\left ( y\right ) T_{nm}\left ( t\right ) \nonumber \\ & =\sum _{n=1}^{\infty }\sum _{m=1}^{\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}^{\infty }\sum _{m=1}^{\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 ) \tag {1} \end {align}
Initial conditions are now used to find \(A_{nm},B_{nm}\). At \(t=0\)
\begin {align*} u\left ( x,y,0\right ) & =f\left ( x,y\right ) \\ \frac {\partial u}{\partial t}\left ( x,y,0\right ) & =0 \end {align*}
Applying first initial condition to (1) gives\[ f\left ( x,y\right ) =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }A_{nm}\sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \] Applying 2D orthogonality gives\begin {align*} \int _{0}^{L}\int _{0}^{H}f\left ( x,y\right ) \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\left ( x,y\right ) \sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \ dxdy \end {align*}
Taking time derivative of (1) gives\begin {align*} \frac {\partial u}{\partial t}\left ( x,y,t\right ) & =\sum _{n=1}^{\infty }\sum _{m=1}^{\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}^{\infty }\sum _{m=1}^{\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 {align*}
AT \(t=0\) the above becomes\[ \int _{0}^{H}g\left ( x,y\right ) =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }c\sqrt {\left ( \frac {m\pi }{H}\right ) ^{2}+\left ( \frac {n\pi }{L}\right ) ^{2}}B_{nm}\sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \] Applying 2D orthogonality gives\[ \int _{0}^{L}\int _{0}^{H}g\left ( x,y\right ) \sin \left ( \left ( \frac {n\pi }{L}\right ) x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \ dxdy=B_{nm}\left ( \frac {L}{2}\right ) \left ( \frac {H}{2}\right ) \] But the initial velocity \(g\left ( x,y\right ) =0\). Hence \(B_{nm}=0\) for all \(n,m\).
Summary of solution\begin {align*} u\left ( x,y,t\right ) & =\sum _{n=1}^{\infty }\sum _{m=1}^{\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\left ( x,y\right ) \sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{H}y\right ) \ dxdy \end {align*}
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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 \[ u_{tt} + \frac {1}{10} u_t = \frac {1}{4} \nabla ^2 u(x,y) \] 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["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]];
\[\left \{\left \{u(x,y,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}\underset {K[3]=1}{\overset {\infty }{\sum }}\frac {5120 e^{-t/20} \sin \left (\frac {1}{2} \pi x (2 K[1]-1)\right ) \sin \left (\frac {1}{2} \pi y (2 K[3]-1)\right ) \sin \left (\frac {1}{20} t \sqrt {50 \pi ^2 \left (2 K[1]^2-2 K[1]+2 K[3]^2-2 K[3]+1\right )-1}\right )}{\pi ^6 (2 K[1]-1)^3 (2 K[3]-1)^3 \sqrt {50 \pi ^2 \left (2 K[1]^2-2 K[1]+2 K[3]^2-2 K[3]+1\right )-1}} & (K[1]|K[3])\in \mathbb {Z}\land K[1]\geq 1\land K[3]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]
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);
\[u \left ( x,y,t \right ) =\sum _{m=0}^{\infty } \left ( \sum _{n=0}^{\infty }5120\,{\frac {{{\rm e}^{-t/20}}\sin \left ( 1/2\, \left ( 1+2\,m \right ) \pi \,y \right ) \sin \left ( 1/2\, \left ( 1+2\,n \right ) \pi \,x \right ) \sin \left ( 1/20\,t\sqrt {-1+ \left ( 100\,{m}^{2}+100\,{n}^{2}+100\,m+100\,n+50 \right ) {\pi }^{2}} \right ) }{\sqrt {-1+ \left ( 100\,{m}^{2}+100\,{n}^{2}+100\,m+100\,n+50 \right ) {\pi }^{2}}{\pi }^{6} \left ( 1+2\,m \right ) ^{3} \left ( 1+2\,n \right ) ^{3}}} \right ) \]
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From Mathematica DSolve help pages.
Hyperbolic partial differential equation with non-rational coefficients.
Solve for \(u(x,y)\) \[ u_{xx} -2 \sin x u_{x y} -\cos ^2 x u_{y y} -\cos x u_y=0 \]
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=()
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