PDE and differentiating through the sum

  • #1
joshmccraney
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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!
 

Answers and Replies

  • #2
Orodruin
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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.
 

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