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The heat conduction problem

Posted October 13th, 2007 by vic777
in

What is a solution for the initial boundary-value problem (using separation of variables method)?
$ u_t=4u_x_x, 0<x<1, t>0; $
$ u(x,0)=x^2(1-x), 0\le x\le1; $
$ u(0,t)=0 ; u(1,t)=0 ;t\ge0 $

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Comments

Answer

On October 18th, 2007 Isoscel says:

Look for a nonzero solution of the form u(x,t)=X(x)T(t).
We get T'(t)X(x)-4T(t)X"(x)=0 or
$ \frac{T'(t)}{4T(t)}=\frac{X"(x)}{X(x)}=-\lambda^2 $
(*)We get $ T'(t)+4\lambda^2T(t)=0 $
(**)$ X"(x)+\lambda^2X(x)=0\: X(0)=X(1)=0 $
Second equation has a solution of the form
$ X(x)=a\cos(\lambda x)+b\sin(\lambda x) $
$ X(0)=a=0\: X(1)=b\sin\lambda =0 $
$ \lambda_n=n\pi $
So $ X_n(x)=\sin(n\pi x) $
Equation (*) has solution $ T_n(t)=c_ne^{-4n^2\pi ^2t} $
Look now for a solution of our equation
$ u(x,t)=\sum_{n=1}^{+\infty}T_n(t)X_n(x)=\sum_{n=1}^{+\infty}c_ne^{-4n^2\pi ^2t}\sin(n\pi x) $
From $ u(x,0)=\sum_{n=1}^{+\infty}c_n\sin(n\pi x)=x^2(1-x) $ we have
$ c_n=2\int_{0}^{1}x^2(1-x)\sin(n\pi x)dx=-\frac{4}{m^3\pi ^3}(1+2(-1)^n) $
So finally solution can be written
$ u(x,t)=\sum_{n=1}^{+\infty}-\frac{4}{m^3\pi ^3}(1+2(-1)^n)e^{-4n^2\pi ^2t}\sin(n\pi x) $

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