PDE7

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$u_{tt}=c^2(u_{xx}+u_{yy})\,$

$u(x,0,t) = 0\,$
$u(x,b,t) = 0\,$
$u(0,y,t) = 0\,$
$u(a,y,t) = 0\,$

$u(x,y,0) = f(x,y)\,$
$u_t(x,y,0) = g(x,y)\,$

$0<x<a,\,\,\,\,\, 0<y<b,\,\,\,\,\, t>0\,$

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 $ψ$ is:

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

$\psi(t) = c_1 \cos(\sqrt{\lambda}\,c t) + c_2 \sin(\sqrt{\lambda}\,c t)\,$

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

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

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

$A_{m,n} = \frac{4}{ab}\int_0^a \!\!\int_0^b f(x,y) \sin(\frac{m\pi x}{a})\sin(\frac{n\pi y}{b})\,dy\,dx\,$

$g(x,y) = u_t(x,y,0) = \sum_{m,n=1}^\infty B_{m,n} \sqrt{\lambda_{m,n}} \sin(\frac{m\pi x}{a})\sin(\frac{n\pi y}{b})\,$

$B_{m,n} = \frac{4}{ab\sqrt{\lambda_{m,n}}}\int_0^a\!\! \int_0^b g(x,y) \sin(\frac{m\pi x}{a})\sin(\frac{n\pi y}{b})\,dy\,dx\,$

Partial Differential Equations

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PDE7

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