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This post is mainly to ask good reference that explains all details needed to solve a (solvable by separation of variables) non homogeneous PDE by using separation of variables. Take, for example

$ \left\{\begin{matrix} u_{tt}(x,t) + u_{xx}(x,t) =f(x,t) & 0<x <L, t>0 \\ u(0,t)=u(L,t)=0 & t>0 \\ u(x,0)=u(x,K)=0 & 0<x<L \end{matrix}\right. $

Every notes I have been given (including the ones I link above) say something like:

"We look for solutions u in the form u(x, t) = T(t)X(x). As before we look at the eigenvalue problem $\left\{\begin{matrix} X'' +\lambda X =0 \\ X(0) =0=X(L) & \end{matrix}\right.$"

Why? I understand this is valid for an homogeneous PDE where the basic principle is to asume $u(x,t)=X(x)T(t)$, so then $u_{tt} +u_{xx}=0=XT''+X''T$, and then

$$ \frac{X''}{X}=-\frac{T''}{T}= - \lambda \quad \quad (^*)$$

But this does not (seem to) work when the RHS of the equation is an arbitrary function, as $ (^*)$ would be something like

$$\frac{X''}{X}=-\frac{T''}{T}-\frac{f}{XT}$$

and I see no way to conclude anything similar from this last equation.

Note this was already asked here, but the answer was not an actual answer.

D1X
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  • Remember that a general solution is not $X(x)T(t)$, but $\sum_i c_i X_i(x) T_i(t)$. Thus the source term $f$ needs to be split across this sum - I'd start by assuming $f(x,t) = \sum_i b_i(t) X_i(x)$ and equating coefficients of $X_i$ in the PDE, which should give you a family of ODEs for $T_i$ with source terms $b_i$. – Anthony Carapetis Feb 12 '17 at 07:49
  • @AnthonyCarapetis Thanks for the answer, but still it does not answer why the system $X'' +\lambda X =0, X(0)=0=X(L) $ is even considered in the first place. This comes from asuming the PDE is homogeneous, but it is not. – D1X Feb 12 '17 at 09:11
  • The eigenfunctions of the spatial operator (i.e. the solutions of the homogeneous problem $X_i'' + \lambda_i X_i = 0$) provide a natural basis with which to separate the (inhomogeneous) PDE, since after making the substitution for $u$ in terms of $X_i$ and $T_i$, we can replace $X_i''$ with $-\lambda_i X_i$ and thus equate coefficients of the $X_i$. – Anthony Carapetis Feb 12 '17 at 09:17

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