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

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)\,

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

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{d B_{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}k B_{m,n}(t)\,. The last integral around the curve is zero because the outward normal vector of u and φ are zero because both functions are zero there by the boundary conditions. And so

\frac{dB_{m,n}}{dt} = -\lambda_{m,n} k B_{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}{a b}\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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