# A basic partial differential equation of heat transfer

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!

• You can use separation of variables. Try it one last time. – Chinny84 Sep 24 '14 at 19:28

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.