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Math Help - Two dimensional, non-homogenous heat equation

  1. #1
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    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.

    Thanks in advance!
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    Quote Originally Posted by pfarnall View Post
    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.

    Thanks in advance!
    Are you on a finite domain (like a rectangle) or infinite domain - \infty \le x,y \le \infty ? It will make a difference.
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    Quote Originally Posted by Danny View Post
    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)
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    Quote Originally Posted by pfarnall View Post
    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) .
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    Quote Originally Posted by Danny View Post
    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?
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    Quote Originally Posted by pfarnall View Post
    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

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

    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)

    so your PDE becomes

     <br />
\frac{T'}{D T} = \frac{X''}{X} + \frac{Y''}{Y}<br />

    which gives

     <br />
\frac{X''}{X} = \lambda_1,\;\;\; \frac{Y''}{Y} = \lambda_2,\;\;\;\frac{T'}{D T} = \lambda_1 + \lambda_2<br />
, 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.
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    Quote Originally Posted by Danny View Post
    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

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

    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)

    so your PDE becomes

     <br />
\frac{T'}{D T} = \frac{X''}{X} + \frac{Y''}{Y}<br />

    which gives

     <br />
\frac{X''}{X} = \lambda_1,\;\;\; \frac{Y''}{Y} = \lambda_2,\;\;\;\frac{T'}{D T} = \lambda_1 + \lambda_2<br />
, 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.
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