Originally Posted by

**Danny** I'll provide some guidence. First let $\displaystyle c = u + c_1$ so that the boundary conditions (I'm assuming that the square region is $\displaystyle L \times L$)

$\displaystyle c(0,y,t) = c(L,y,t) = c(x,0,t) = c(x,L,t) = c_1$

becomes

$\displaystyle u(0,y,t) = u(L,y,t) = u(x,0,t) = u(x,L,t) = 0$

and the PDE becomes

$\displaystyle

u_t = D\left( u_{xx} + u_{yy}\right) - f(x,y,t)

$

We'll first consider the PDE without a source term ($\displaystyle f$) then address it later.

Now assume separable solutions $\displaystyle u = X(x) Y(y) T(t)$

so your PDE becomes

$\displaystyle

\frac{T'}{D T} = \frac{X''}{X} + \frac{Y''}{Y}

$

which gives

$\displaystyle

\frac{X''}{X} = \lambda_1,\;\;\; \frac{Y''}{Y} = \lambda_2,\;\;\;\frac{T'}{D T} = \lambda_1 + \lambda_2

$, where $\displaystyle \lambda_1$ and $\displaystyle \lambda_2$ are constants

with the boundary condition $\displaystyle X(0) = X(L) = Y(0) = Y(L) = 0$. Now we are a point like solving the heat equation in one space dimension.

Your turn - You pick it up from here.