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I have solved the heat equation and have gotten to the stage of getting a general solution $$u(x,t)=x+\sum^\infty_{n=1} c_n \sin(\pi n x)e^{-\pi^2 n^2 t}$$

And I have the initial condition $$u(x,0)=x+\sin(\pi x)$$

How do I incorporate the initial condition to get the particular solution of $$u(x,t)=x+\sin(\pi x)e^{-\pi^2 t}$$

Al jabra
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1 Answers1

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Set $t=0$ and you get

$$u(x,0)=x+\sum_{n=1}^\infty c_n\sin(\pi nx)=x+\sin(\pi x)\\ g(x):=u(x,0)-x=\sum_{n=1}^\infty c_n\sin(\pi nx)=\sin(\pi x)$$

But since $\int_0^1 \sin(\pi nx)\sin(\pi mx)\,dx=\begin{cases}\frac 1 2&\text{if }n=m\\0 &\text{if }n\neq m\end{cases}$

And since $g(x)$ is a $L^2[0,1]$ function, it must hold

$$c_n=2\int_0^1 g(x)\sin(\pi nx)\,dx=2\int_0^1 \sin(\pi x)\sin(\pi nx)\,dx=\begin{cases}1&\text{if }n=1\\0&\text{if not}\end{cases}$$

  • ok thanks, it seems I need to learn much more about fourier expansions etc... – Al jabra May 27 '15 at 10:47
  • Roughly speaking,@Aljabra , you want to compare the terms of the Fourier expansion in sines and say they must be the same. This is the case for functions in $L^2[0,1]$. I took a little time to leave a quick (yet incomplete) explanation of this. –  May 27 '15 at 10:53
  • SO basically we use the fourier expansion to find the constants? – Al jabra May 27 '15 at 11:07
  • The one in ${\sin(\pi nx)}_{n\ge 1}$ over $[0,1]$, for this case. –  May 27 '15 at 11:09