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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(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(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\,

Partial Differential Equations

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