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Separation of Variables - Fourier Series

The idea of Separation of Variables can be used to obtain the exact solution for the Diffusion Problem:
$\displaystyle \frac{\partial u(x,t)}{\partial t}$ $\textstyle =$ $\displaystyle D\frac{\partial^2u(x,t)}
{\partial x^2}$ (18)

in the region $\Omega = [0;L]\times[0;\infty]$ with the boundary conditions $u(x=0,t)=0$ and $u(x=L,t)=0$ with the initial condition $u(x,t=0)=f(x)$.

The solution is a family of solutions $u_k(x,t)$ with the form

$\displaystyle u_k(x,t)=X_k(x)\cdot T_k(t).$     (19)

Since the differential equation is linear, the exact solution can be determined as a linear combination of single solutions.
By substitution of (19) into (18) we obtain (derivatives are markes with $'$)

$\displaystyle X_k(x)T_k'(t)$ $\textstyle =$ $\displaystyle D\cdot X_k''(x)T_k(t)$ (20)
       
$\displaystyle \frac{X_k(x)T_k'(t)}{D\cdot X_k(x)T_k(t)}$ $\textstyle =$ $\displaystyle \frac{D\cdot X_k''(x)T_k(t)}
{D\cdot X_k(x)T_k(t)}$  
       
$\displaystyle \frac{T_k'(t)}{D\cdot T_k(t)}$ $\textstyle =$ $\displaystyle \frac{X_k''(x)}{X_k(x)}$ (21)

(21) can obviosly only be true for any arbitray $x$ and $t$ if the relation of $T$ and $T'$ equals the relation of $X$ and $X''$:
$\displaystyle \frac{T_k'(t)}{D\cdot T_k(t)}$ $\textstyle =$ $\displaystyle \frac{X_k''(x)}{X_k(x)} = -\lambda_k.$  

Thus, there are two separate ordinary differential equations of $x$ and $t$:
$\displaystyle T_k'(t)+\lambda_kD\cdot T_k(t)$ $\textstyle =$ $\displaystyle 0$ (22)
$\displaystyle \mbox{and}$      
$\displaystyle X_k''(x)+\lambda_kX_k(x)$ $\textstyle =$ $\displaystyle 0.$ (23)

(22) is the well known Exponential Decay Equation with its solutions:
$\displaystyle T_k(t)=e^{-\lambda_kDt}.$     (24)

(23) can be solved with Sine- and Cosine-approaches.
Considering the boundary conditions $u(x=0,t)=0$ and $u(x=L,t)=0$, the Cosine-approaches are not applicable and we get:

$\displaystyle X_k(x)=sin{\frac{k\pi x}{L}},$     (25)
       
$\displaystyle \lambda_k=\Big(\frac{k\pi}{L}\Big)^2, \quad k=1,2,\ldots \;.$      

By substituting both solutions in (19), we get:
$\displaystyle u_k(x,t)=\exp(-\Big(\frac{k\pi}{L}\Big)^2Dt)\cdot sin{\frac{k\pi x}{L}}.$     (26)

Since $u_k(x,t)$ is linear, all linear combinations are solutions, too:
$\displaystyle u_k(x,t)=\sum_{k=1}^\infty c_k\cdot\exp(-\Big(\frac{k\pi}{L}\Big)^2Dt)\cdot sin{\frac{k\pi x}{L}},$     (27)

for arbitrary coefficients $c_k$.
To determine the coefficients $c_k$ which fullfill the initial condition
$u(x,t=0)=f(x)$, $f(x)$ may be developed in a Fourier series.


next up previous
Next: Numerical solutions of PDEs Up: Solution of a PDE Previous: Solution of a PDE
Heiko Enderling 2003-11-13