# Thread: Another form of diffusion equation

1. ## Another form of diffusion equation

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

X,T,U here are all dimensionless.

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

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

find the analytic solution of this problem.

2. ## what i got

what i got here,

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

but i have no idea how to solve.

3. Originally Posted by silversand
what i got here,

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

but i have no idea how to solve.
If you were to solve

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

you would use a separation of variables and ultimately lead to

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

where

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

Since you have a source term in your new equation

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

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

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

and seek a solution of your porblem in the form

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

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

4. ## how

how to determine the coefficients : $\displaystyle c_{n}, k_{n}$ ?

5. Originally Posted by silversand
how to determine the coefficients : $\displaystyle c_{n}, k_{n}$ ?
The $\displaystyle c_n$ is found from the initial condition. For the $\displaystyle k_n$, it's $\displaystyle k_n = \frac{2}{1} \int_0^1 (x-1) \sin n \pi x\, dx$.