Answer to Question #160499 in Differential Equations for Suchita

Question

u_t -ku_xx = F(x, t) 0<x<l t>0

u(x, 0)= f(x) for 0<x<l and u(0, t)= u(l, t)= 0 for t >0

Answer 1

CORRECTED SOLUTION

We seek the solution of the form of the functions that are solutions for the homogenous equation:

$u(t,x)=\sum\limits_{n=1}^{+\infty}u_n(t)\sin\frac{\pi n}{l}x$

Then

$u_t = \sum\limits_{n=1}^{+\infty}u'_n(t)\sin\frac{\pi n}{l}x$

$u_{xx}=-\sum\limits_{n=1}^{+\infty}(\frac{\pi n}{l})^2 u_n(t)\sin\frac{\pi n}{l}x$

$u_t-ku_{xx} = \sum\limits_{n=1}^{+\infty}(u'_n(t)+k(\frac{\pi n}{l})^2 u_n(t))\sin\frac{\pi n}{l}x$

Let $F(t,x)=\sum\limits_{n=1}^{+\infty}F_n(t)\sin\frac{\pi n}{l}x$ (a Fourier series for the function F(t,x) on the interval $0\leq x\leq l$ )

then $f_n(t)=\frac{2}{l}\int\limits_0^l F(t,x)\sin\frac{\pi n}{l}x\, dx$

Substituting this to the equation $u_t-ku_{xx}=F(t,x)$ we have:

$\sum\limits_{n=1}^{+\infty}(u'_n(t)+k(\frac{\pi n}{l})^2 u_n(t)-F_n(t))\sin\frac{\pi n}{l}x=0$

At the left side there is a Fourier series for the zero function, hence, all the Fourier coefficients are zeros, that is,

$u'_n(t)+k(\frac{\pi n}{l})^2u_n(t)=F_n(t)$

The solution of the homogeneous equation (with $F_n(t)=0$ ) is

$u_n(t)=c_ne^{-k(\pi n/l)^2t}$

To solve the non-homogeneous equation we are varying the constant c_n .

$u_n(t)=c_n(t)e^{-k(\pi n/l)^2t}$

We have the equation (ODE) for c_n(t):

$c'_n(t)e^{-k(\pi n/l)^2t} = F_n(t)$

To find the initial conditions we consider the function

$f(x) = u(0,x) =\sum\limits_{n=1}^{+\infty}u_n(0)\sin\frac{\pi n}{l}x=\sum\limits_{n=1}^{+\infty}c_n(0)\sin\frac{\pi n}{l}x$

Therefore, $c_n(0)=\frac{2}{l}\int\limits_0^l f(x)\sin\frac{\pi n}{l}x\, dx$ - Fourier coefficients for the function $f(x)$ .

Integrating ODE we have

$c_n(t)=c_n(0)+\int\limits_0^t F_n(s)e^{ks(\pi n/l)^2}ds$

Answer. $u(t,x)=\sum\limits_{n=1}^{+\infty}c_n(t)e^{-k(\pi n/l)^2t}\sin\frac{\pi n}{l}x$

with

$c_n(t)=c_n(0)+\int\limits_0^t F_n(s)e^{ks(\pi n/l)^2}ds$

$F_n(t)=\frac{2}{l}\int\limits_0^l F(t,x)\sin\frac{\pi n}{l}x\, dx$

$c_n(0)=\frac{2}{l}\int\limits_0^l f(x)\sin\frac{\pi n}{l}x\, dx$