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u_{{t}}=k(\Delta u)+q(x,y,t)\,




Seperating variables for the homogeneous version of this equation leads to the same eigenvalue problem as in PDE7. Those eigenvalues and eigenfunctions are:

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

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

Now, represent the function u(x,y,t)\, in its eigenfunction expansion.

u(x,y,t)=\sum _{{m,n=1}}^{\infty }B_{{m,n}}(t)\phi _{{m,n}}(x,y)\,

B_{{m,n}}(t)={\frac  {4}{ab}}\int \!\!\!\int _{D}u(x,y,t)\phi _{{m,n}}(x,y)\,dx\,dy\,

Let t\rightarrow 0^{+}\, in the formula for B_{{m,n}}(t)\, above and use the inital condition to get

A_{{m,n}}={\frac  {4}{ab}}\int \!\!\!\int _{D}f(x,y)\phi _{{m,n}}(x,y)\,dx\,dy\,

D is the rectangular domain 0<x<a,\,\,\,0<y<b\,, and C is its boundary.

To evaluate the coefficients, take the derivative of the last equation with respect to t\,.

{\frac  {dB_{{m,n}}}{dt}}={\frac  {4}{ab}}\int \!\!\!\int _{D}u_{t}\phi _{{m,n}}\,dx\,dy\,

Substitute the DE for u

={\frac  {4}{ab}}\int \!\!\!\int _{D}k(\Delta u)\phi _{{m,n}}\,dx\,dy+{\frac  {4}{ab}}\int \!\!\!\int _{D}q\phi _{{m,n}}\,dx\,dy\,

Label the second integral Q\, and save it for later. This is a known function of t.

Q_{{m,n}}(t)={\frac  {4}{ab}}\int \!\!\!\int _{D}q(x,y,t)\phi _{{m,n}}(x,y)\,dx\,dy\,

Using Green's identities,

\int \!\!\!\int _{D}\phi \Delta u\,dx\,dy=-\int \!\!\!\int _{D}\nabla \phi \nabla u\,dx\,dy+\int _{C}\phi \nabla u\,ds\,

=-\int \!\!\!\int _{D}\nabla \phi \nabla u\,dx\,dy+\int _{C}\phi {\frac  {\partial u}{\partial \nu }}\,ds\,

\int \!\!\!\int _{D}\phi \Delta u\ -u\Delta \phi \,dx\,dy=\int _{C}\phi {\frac  {\partial u}{\partial \nu }}-u{\frac  {\partial \phi }{\partial \nu }}\,ds\,

and the fact that \Delta \phi _{{m,n}}=-\lambda _{{m,n}}\phi _{{m,n}}\,, the first integral can be transformed:

{\frac  {4}{ab}}\int \!\!\!\int _{D}k(\Delta u)\phi _{{m,n}}\,dx\,dy\,=\,-\lambda _{{m,n}}k{\frac  {4}{ab}}\int \!\!\!\int _{D}u\phi _{{m,n}}\,dx\,dy\,\,+\,\,{\frac  {4k}{ab}}\int _{C}\left({\frac  {\partial u}{\partial \nu }}\phi _{{m,n}}-u{\frac  {\partial \phi _{{m,n}}}{\partial \nu }}\right)\,ds\,

The first integral on the right is equal to -\lambda _{{m,n}}kB_{{m,n}}(t)\,. The last integral around the curve is zero because the outward normal vector of u and \phi are zero because both functions are zero there by the boundary conditions. And so

{\frac  {dB_{{m,n}}}{dt}}=-\lambda _{{m,n}}kB_{{m,n}}+Q_{{m,n}}(t)\,

So B_{{m,n}}\, is determined and gives the coefficient in the Fourier expansion of u(x,y,t)\,

  • B_{{m,n}}(t)=A_{{m,n}}e^{{-\lambda _{{m,n}}kt}}+\int _{0}^{t}e^{{-\lambda _{{m,n}}k(t-\tau )}}Q_{{m,n}}(\tau )\,d\tau \,

The problem is formally solved. But just for fun, the explicit solution is

  • u(x,y,t)=\sum _{{m,n=1}}^{\infty }{\frac  {4}{ab}}\int \!\!\!\int _{D}f(\xi ,\zeta )\sin({\frac  {m\pi \xi }{a}})\sin({\frac  {n\pi \zeta }{b}})\,d\xi \,d\zeta e^{{-\left[{\frac  {m^{2}\pi ^{2}}{a^{2}}}+{\frac  {n^{2}\pi ^{2}}{b^{2}}}\right]kt}}+\int _{0}^{t}\left(e^{{-\left[{\frac  {m^{2}\pi ^{2}}{a^{2}}}+{\frac  {n^{2}\pi ^{2}}{b^{2}}}\right]k(t-\tau )}}\left[{\frac  {4}{ab}}\int \!\!\!\int _{D}q(\xi ,\zeta ,t)\sin({\frac  {m\pi \xi }{a}})\sin({\frac  {n\pi \zeta }{b}})\,d\xi \,d\zeta \right]\right)\,d\tau \sin({\frac  {m\pi x}{a}})\sin({\frac  {n\pi y}{b}})\,

Which can probably be simplified.

Partial Differential Equations

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