Another form of diffusion equation

**silversand** · January 19th 2009, 03:25 PM

now by defining several dimensionless variables, we got another form of diffusion equation:
$U_{T}=U_{XX} \ \ \ \ T>0, \ \ 0<X<1$

X,T,U here are all dimensionless.

U with initial conditions:
$U(X,0)=0, \ \ 0\leq X\leq1$ and boundary conditions :
$U(0,T)=T, \ \ U(1,T)=0, \ \ T\geq0$

$U(X,T)=T(1-X)+V(X,T)$

find the analytic solution of this problem.

**silversand** · January 19th 2009, 03:26 PM

what i got here,

$V(X,0)=V(0,T)=V(1,T)=0 \ \ \ \ (1-X)+V_{T}=V_{XX}$

but i have no idea how to solve.

**Danny** · January 19th 2009, 03:48 PM

Originally Posted by silversand

what i got here,

$V(X,0)=V(0,T)=V(1,T)=0 \ \ \ \ (1-X)+V_{T}=V_{XX}$

but i have no idea how to solve.

If you were to solve

$v_t = v_{xx}$ subject to $v(0,t) = v(1,t) = 0$

you would use a separation of variables and ultimately lead to

$v = \sum_{n=1}^\infty T_n(t) \sin n \pi x$

where

$T_n(t) = c_n e^{- n^2 \pi^2 t}$

Since you have a source term in your new equation

$v_t = v_{xx} + \underbrace{x-1}_{source}$

then find a Fourier sine series for the source term of the form

$x - 1 = \sum_{n=1}^\infty k_n \sin n \pi x$

and seek a solution of your porblem in the form

$v = \sum_{n=1}^\infty T_n(t) \sin n \pi x$

where $T_n(t)$ is to be determined (this should give rise to an ODE for $T_n$ ).