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)sinlπnx
Then
ut=n=1∑+∞un′(t)sinlπnx
uxx=−n=1∑+∞(lπn)2un(t)sinlπnx
ut−kuxx=n=1∑+∞(un′(t)+k(lπn)2un(t))sinlπnx
Let F(t,x)=n=1∑+∞Fn(t)sinlπnx (a Fourier series for the function F(t,x) on the interval 0≤x≤l )
then fn(t)=l20∫lF(t,x)sinlπnxdx
Substituting this to the equation ut−kuxx=F(t,x) we have:
n=1∑+∞(un′(t)+k(lπn)2un(t)−Fn(t))sinlπnx=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(lπn)2un(t)=Fn(t)
The solution of the homogeneous equation (with Fn(t)=0 ) is
un(t)=cne−k(πn/l)2t
To solve the non-homogeneous equation we are varying the constant cn .
un(t)=cn(t)e−k(πn/l)2t
We have the equation (ODE) for cn(t):
cn′(t)e−k(πn/l)2t=Fn(t)
To find the initial conditions we consider the function
f(x)=u(0,x)=n=1∑+∞un(0)sinlπnx=n=1∑+∞cn(0)sinlπnx
Therefore, cn(0)=l20∫lf(x)sinlπnxdx - Fourier coefficients for the function f(x) .
Integrating ODE we have
cn(t)=cn(0)+0∫tFn(s)eks(πn/l)2ds
Answer. u(t,x)=n=1∑+∞cn(t)e−k(πn/l)2tsinlπnx
with
cn(t)=cn(0)+0∫tFn(s)eks(πn/l)2ds
Fn(t)=l20∫lF(t,x)sinlπnxdx
cn(0)=l20∫lf(x)sinlπnxdx
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