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I am trying to solve the heat equation for a semi infinite rod with lateral surfaces insulated and $u(x,0)$ = $u_0$ for $x>0$, $u(0,t)=u_1$ for $t>0$, and the $\lim_{t\to\infty} u(x,t)=u_0$.

I think I need to start with:

$$\frac{d^2u}{dx^2}=\frac{1}{\alpha^2}\frac{du}{dt}$$

Then I take Laplace Transform:

$$\frac{d^2U}{ds^2}-\frac{s}{\alpha^2}U=0$$

Can someone help me with the following steps?

JohnD
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Jackson Hart
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  • First, there is a mistake in the way the Laplace Transform is applied, the result is $$\frac{d^2U}{dx^2}-\frac{s}{\alpha^2}U + \frac1{\alpha^2}u(x,0)=0$$ which is still a differential equation in $x$. Also, it seems that you are saying that the initial conditions and the final conditions are the same, is that correct? The next step is to solve the differential equation – kleineg Sep 08 '15 at 17:16
  • The last term comes because $$\mathcal{L}\left({\frac d {dt}}u\right) = s \mathcal{L}(u) - u(x,0)$$ – kleineg Sep 08 '15 at 17:21

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