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I need to solve $u_t = 3u_{xx}$ with $u(x,0) = 17\sin(\pi x)$ and $u(0,t) = u(4,t) = 0$ using the Laplace Transform.

So taking the Laplace transform, do I hold the terms with only $x$'s in them constant?

If so, here's what I've got:

$sU(x,s) -\dfrac {17\sin(\pi x)}{s} = 3U_{xx}(x,s)$ with $U(x,0) = 17\sin(\pi x)$ and $U(0,t) = U(4,t) = 0$.

So the solution to the ODE should be $U(x,s) = Ae^{\sqrt{\frac s3}x} + Be^{-\sqrt{\frac s3}x} + \dfrac {17\sin{\pi x}}{s^2 + 3\pi^2 s^2}$.

But when I try to apply my condition for $U(x,0)$, I see that the rightmost term is infinite. What did I do wrong?

Bob Dylan
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  • You need to take the inverse misplaced transform. The other condition doesn't apply to U. – abnry Mar 09 '15 at 16:55
  • So you're saying I need to take the inverse transform of $U(x,s)$ before trying to apply the conditions? How do I take the inverse transform of $e^{\sqrt{\frac s3}x}$? I haven't learned that one (or even how to take the inverse Laplace transform -- we just use a table). – Bob Dylan Mar 09 '15 at 17:04
  • $e^{cx}$ is constant with respect to $s$, so it's inverse laplace transform is the inverse laplace transform of a constant. A table should be fine for such a problem. – abnry Mar 09 '15 at 18:12
  • Except the $c$ in this case is $\sqrt{s/3}$. I just used $U(0,t) = U(4,t) = 0$ to solve it though. – Bob Dylan Mar 09 '15 at 18:21
  • My bad, didn't see that. – abnry Mar 09 '15 at 18:32

1 Answers1

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$L(u_t)=sL(u)-u(x,0)=sL(u)-17sin(\pi x)$ (There does not appear an $s$ in the denominator!)

kryomaxim
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