PDE8

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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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PDE8

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