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Let $u(x,t)$ be the temperature along a 1-D rod, from $x=0$ to $x=L$.

$\frac{\partial u}{\partial t} = A \frac{\partial^2 u}{\partial x^2}+Bu$, where A and B are constants.

Initial condition is $u(x,0)=0$.

Boundary conditions are:

$\frac{\partial u}{\partial x}|_{x=0}=f(t)$ (arbitrarily prescribed heat influx on the left side of the rod)

$\frac{\partial u}{\partial x}|_{x=L}=0$ (insulated on the other end).

I failed to solve this problem using separation of variables, and would like to know: is there any other analytic methods I can use before I try numerical solutions?

Thanks!

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  • $\begingroup$ You can use separation of variables. Try it one last time. $\endgroup$ – Chinny84 Sep 24 '14 at 19:28
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Sure, you should take try to take the inverse Laplace transform of $ f(t) $. Here is a sketch. for simplicity, I'll flip the conditions so that $ \partial_x u|_{x = 0} = 0 $ and $ \partial_x u_{x = L} = f$.

Let $ \psi_k(x,t) = \frac{-1}{k \sin(k)} \cos(k x) e^{-\lambda_k t} $ where $ \lambda = A k^2 + B $ (mod some negative signs). Then you see that $ \psi_k(x,t) $ solves the heat equation with boundary conditions $ \partial_x \psi_k|_{x=0} = 0 $ and $ \partial_x \psi_k|_{x=L} = e^{-\lambda_k t} $. So, if you can find $ g $ so that $ f(t) = \int_{0}^{\infty} g(k) \; e^{-\lambda_k t} \; dk $ then you're done: $ u = \int_{0}^{\infty} g(k) \; \psi_k(x,t) \; dk $. The way to do that is to take inverse laplace transform, which I'll let you look up.

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