# Two dimensional, non-homogenous heat equation

• June 29th 2009, 02:45 AM
pfarnall
Two dimensional, non-homogenous heat equation
Could someone please point me in the right direction for finding an analytical solution to the following (if one exists):

$\frac{\partial c(x,y,t)}{\partial t}=D\nabla^{2}c(x,y,t)-f(x,y,t)$

f is probably going to be a constant function and I will probably have constant initial and boundary conditions but they're not finalised yet, I'm just looking for a gentle nudge in the right direction or for someone to tell me that it's unsolvable so that I can use a numerical method.

• June 29th 2009, 05:45 AM
Jester
Quote:

Originally Posted by pfarnall
Could someone please point me in the right direction for finding an analytical solution to the following (if one exists):

$\frac{\partial c(x,y,t)}{\partial t}=D\nabla^{2}c(x,y,t)-f(x,y,t)$

f is probably going to be a constant function and I will probably have constant initial and boundary conditions but they're not finalised yet, I'm just looking for a gentle nudge in the right direction or for someone to tell me that it's unsolvable so that I can use a numerical method.

Are you on a finite domain (like a rectangle) or infinite domain $- \infty \le x,y \le \infty$? It will make a difference.
• June 29th 2009, 05:47 AM
pfarnall
Quote:

Originally Posted by Danny
Are you on a finite domain (like a rectangle) or infinite domain $- \infty \le x,y \le \infty$? It will make a difference.

Oh yes, forgot to mention that, finite (like a rectangle ... a square in fact)
• June 29th 2009, 05:54 AM
Jester
Quote:

Originally Posted by pfarnall
Oh yes, forgot to mention that, finite (like a rectangle ... a square in fact)

Here's some ideas based on constant boundary conditions. First transform you BC's to zero. Then try a double Fourier series. You'll also need a Fourier series for your source term $f(x,y,t)$ .
• June 29th 2009, 06:28 AM
pfarnall
Quote:

Originally Posted by Danny
Here's some ideas based on constant boundary conditions. First transform you BC's to zero. Then try a double Fourier series. You'll also need a Fourier series for your source term $f(x,y,t)$ .

Ok I think I might be needing more than a gentle nudge as I don't really follow you and have never heard of a double Fourier series. I'm fairly sure all the boundary conditions will be the same constant value (lets say c1) and the initial condition will be another constant (c2). Could you guide me through it if it's not too much trouble?
• June 29th 2009, 07:55 AM
Jester
Quote:

Originally Posted by pfarnall
Ok I think I might be needing more than a gentle nudge as I don't really follow you and have never heard of a double Fourier series. I'm fairly sure all the boundary conditions will be the same constant value (lets say c1) and the initial condition will be another constant (c2). Could you guide me through it if it's not too much trouble?

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

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

becomes

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

and the PDE becomes

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

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

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

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

which gives

$
\frac{X''}{X} = \lambda_1,\;\;\; \frac{Y''}{Y} = \lambda_2,\;\;\;\frac{T'}{D T} = \lambda_1 + \lambda_2
$
, where $\lambda_1$ and $\lambda_2$ are constants

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

Your turn - You pick it up from here.
• June 29th 2009, 08:45 AM
pfarnall
Quote:

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

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

becomes

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

and the PDE becomes

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

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

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

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

which gives

$
\frac{X''}{X} = \lambda_1,\;\;\; \frac{Y''}{Y} = \lambda_2,\;\;\;\frac{T'}{D T} = \lambda_1 + \lambda_2
$
, where $\lambda_1$ and $\lambda_2$ are constants

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

Your turn - You pick it up from here.

Ah fantastic, thanks very much, I wasn't convinced that separation of variables was possible with three but that's probably because I've only ever done it with two. I'll give it a go from here though, thanks very much for all your help.