# PDE7

 $u_{{tt}}=c^{2}(u_{{xx}}+u_{{yy}})\,$ $u(x,0,t)=0\,$ $u(x,b,t)=0\,$ $u(0,y,t)=0\,$ $u(a,y,t)=0\,$ $u(x,y,0)=f(x,y)\,$ $u_{t}(x,y,0)=g(x,y)\,$ $00\,$

First, notice that there are two derivates of $u$ w.r.t $t$ in the differential equation and two inital conditions. There are also two derivatives on $x$ and $y$ and two boundary conditions on each. Now, separate variables.

$u(x,y,t)=\psi (t)\phi (x,y)\,$

The new boundary conditions are:

$\phi (0,y)=0\,$

$\phi (a,y)=0\,$

$\phi (x,0)=0\,$

$\phi (x,b)=0\,$

Plug back into the original DE.

$\psi ''\phi =c^{2}\psi (\phi _{{xx}}+\phi _{{yy}})\,$

Separate variables. For simplicity, keep higher derivatives on top, keep the $c^{2}\,$ term with the psi function, and let the constant be $-\lambda \,$.

${\frac {\psi ''}{c^{2}\psi }}={\frac {\phi _{{xx}}+\phi _{{yy}}}{\phi }}=-\lambda \,$

The solution for the DE involving $\psi$ is:

$\psi ''+\lambda c^{2}\psi =0\,$

• $\psi (t)=c_{1}\cos({\sqrt {\lambda }}\,ct)+c_{2}\sin({\sqrt {\lambda }}\,ct)\,$

Don't worry about the inital conditions until later. Work is done on this function for now.

$\phi _{{xx}}+\phi _{{yy}}+\lambda \phi =0\,$

Separate variables. Keep higher derivatives on top.

$\phi (x,y)=X(x)Y(y)\,$

$X''(x)Y(y)+X(x)Y''(y)+\lambda X(x)Y(y)=0\,$

$Y(y)(X''(x)+\lambda X(x))=-X(x)Y''(y)\,$

The choice of $-\mu \,$ is for convenience.

${\frac {X''(x)+\lambda X(x)}{-X(x)}}={\frac {Y''(y)}{Y(y)}}=-\mu \,$

Write the new ODEs and BCs.

$Y''(y)+\mu Y(y)=0,\,\,\,Y(0)=0,\,\,\,Y(b)=0\,$

$X''(x)+(\lambda -\mu )X(x)=0,\,\,\,X(0)=0,\,\,\,X(a)=0\,$

The solution of $X$ is:

$X(x)=c_{3}\cos({\sqrt {\lambda -\mu }}\,x)+c_{4}\sin({\sqrt {\lambda -\mu }}\,x)\,$

$X(0)=c_{3}=0\,$

$X(a)=c_{4}\sin({\sqrt {\lambda -\mu }}\,a)=0\,$

Letting $c_{4}=0\,$ would give no interesting solutions.

${\sqrt {\lambda _{m}-\mu _{m}}}={\frac {m\pi }{a}},\,\,\,m=1,2,3,...\,$

The $m=0$ case would not add any information to the sum in the final answer.

$X(x)=c_{4}\sin({\frac {m\pi x}{a}}),\,\,\,m=1,2,3,...\,$

$Y(y)=c_{5}\cos({\sqrt {\mu }}\,y)+c_{6}\sin({\sqrt {\mu }}\,y)\,$

$Y(0)=c_{5}=0\,$

$Y(b)=c_{6}\sin({\sqrt {\mu }}\,b)=0\,$

Letting $c_{6}=0\,$ would give no interesting solutions.

${\sqrt {\mu _{n}}}={\frac {n\pi }{b}},\,\,\,n=1,2,3,...\,$

$Y(y)=c_{6}\sin({\frac {n\pi y}{b}}),\,\,\,n=1,2,3,...\,$

• The eigenvalues are $\lambda _{{m,n}}={\frac {m^{2}\pi ^{2}}{a^{2}}}+{\frac {n^{2}\pi ^{2}}{b^{2}}},\,\,\,m,n=1,2,3,...\,$

The eigenfunctions $\phi (x,y)=X(x)Y(y)\,$ are

• $\phi (x,y)=\sin({\frac {m\pi x}{a}})\sin({\frac {n\pi y}{b}})\,$

No coefficients are needed in the formula for the eigenfunctions. They are absorbed by the other constants $A$ and $B$. The solution is

• $u(x,y,t)=\psi (t)\phi (x,y)=\sum _{{m,n=1}}^{\infty }\sin({\frac {m\pi x}{a}})\sin({\frac {n\pi y}{b}})\left[A_{{m,n}}\cos({\sqrt {\lambda _{{m,n}}}}\,ct)+B_{{m,n}}\sin({\sqrt {\lambda _{{m,n}}}}\,ct)\right]\,$

The first initial condition gives

$f(x,y)=u(x,y,0)=\sum _{{m,n=1}}^{\infty }A_{{m,n}}\sin({\frac {m\pi x}{a}})\sin({\frac {n\pi y}{b}})\,$

This is the sine-sine double Fourier series for f(x,y), so the coefficients are given by

• $A_{{m,n}}={\frac {4}{ab}}\int _{0}^{a}\!\!\int _{0}^{b}f(x,y)\sin({\frac {m\pi x}{a}})\sin({\frac {n\pi y}{b}})\,dy\,dx\,$

$g(x,y)=u_{t}(x,y,0)=\sum _{{m,n=1}}^{\infty }B_{{m,n}}{\sqrt {\lambda _{{m,n}}}}\sin({\frac {m\pi x}{a}})\sin({\frac {n\pi y}{b}})\,$

• $B_{{m,n}}={\frac {4}{ab{\sqrt {\lambda _{{m,n}}}}}}\int _{0}^{a}\!\!\int _{0}^{b}g(x,y)\sin({\frac {m\pi x}{a}})\sin({\frac {n\pi y}{b}})\,dy\,dx\,$