Question #160499

ut -kuxx = 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


1
Expert's answer
2021-02-04T08:07:37-0500

CORRECTED SOLUTION

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

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

Then

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

uxx=n=1+(πnl)2un(t)sinπnlxu_{xx}=-\sum\limits_{n=1}^{+\infty}(\frac{\pi n}{l})^2 u_n(t)\sin\frac{\pi n}{l}x

utkuxx=n=1+(un(t)+k(πnl)2un(t))sinπnlxu_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)=n=1+Fn(t)sinπnlxF(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 0xl0\leq x\leq l )

then fn(t)=2l0lF(t,x)sinπnlxdxf_n(t)=\frac{2}{l}\int\limits_0^l F(t,x)\sin\frac{\pi n}{l}x\, dx

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

n=1+(un(t)+k(πnl)2un(t)Fn(t))sinπnlx=0\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,

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

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

un(t)=cnek(πn/l)2tu_n(t)=c_ne^{-k(\pi n/l)^2t}

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

un(t)=cn(t)ek(πn/l)2tu_n(t)=c_n(t)e^{-k(\pi n/l)^2t}

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

cn(t)ek(πn/l)2t=Fn(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)=n=1+un(0)sinπnlx=n=1+cn(0)sinπnlxf(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, cn(0)=2l0lf(x)sinπnlxdxc_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)f(x) .

Integrating ODE we have

cn(t)=cn(0)+0tFn(s)eks(πn/l)2dsc_n(t)=c_n(0)+\int\limits_0^t F_n(s)e^{ks(\pi n/l)^2}ds


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

with

cn(t)=cn(0)+0tFn(s)eks(πn/l)2dsc_n(t)=c_n(0)+\int\limits_0^t F_n(s)e^{ks(\pi n/l)^2}ds

Fn(t)=2l0lF(t,x)sinπnlxdxF_n(t)=\frac{2}{l}\int\limits_0^l F(t,x)\sin\frac{\pi n}{l}x\, dx

cn(0)=2l0lf(x)sinπnlxdxc_n(0)=\frac{2}{l}\int\limits_0^l f(x)\sin\frac{\pi n}{l}x\, dx


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