PDE and differentiating through the sum

1. Jan 28, 2015

joshmccraney

Hi PF!

I'm reading my math text and am looking at the heat eq $u_t = u_{xx}$, where we are are given non-homogenous boundary conditions. We are solving using the method of eigenfunction expansion.

Evidently we begin by finding the eigenfunction $\phi (x)$ related to the homogenous boundary conditions. From here we say the solution takes the form $u(x,t) = \sum a(t) \phi(x)$. When plugging this result into the heat equation we are not allowed to differentiate w.r.t. x under the sum. The reason evidently is because the boundary conditions for the actual problem and $\phi(x)$ do not agree (the problem has non-homogenous B.C. yet the eigenfunction satisfies homogenous B.C.).

Can anyone tell me why not satisfying the same B.C. conditions implies we cannot differentiate w.r.t. that variable $x$?

Thanks so much!

2. Jan 30, 2015

Orodruin

Staff Emeritus
The point is that you can approximate the solution arbitrarily well (in a distribution sense) with a function that does not fulfill the BC, i.e., the sum will converge to the solution everywhere but at the boundary.

A different way of taking care of your boundary conditions is to make a change of variables to transfer your inhomogeneity to the PDE rather than the BC. The resulting PDE can be solved using series expansion of the resulting inhomogeneity.