# ODE's & PDE's: Homogenous piecing vs Eigenexpansion vs Green functions

I don't know if i'm within rules of the forum to ask this question. If i'm not please comment before downvoting. If you know of a source that answers these questions, please suggest. It would be benificial if the source is concise, may be upto 70 pages max.

I'm getting confused repeatedly regarding this

Consider a inhomogenous linear operator $Lu = f$ for a boundary value problem(BVP). Now there seem to be three ways of solving these equations.

a) One solves for inhomogenous in terms of a particular solution & a complementary solution(i.e. solving $Lu=0$). If the boundary conditions are linear one solves for two equations $Lu = f$ for homogenous boundary & $Lu =0$ for inhomogenous boundary.

b)One solves the eigenvalue problem $Lu =\lambda u$, expands $f$ in terms of the eigenfunctions $u_n$ say $f = \Sigma (c_nu_n)$ and $u = \Sigma (a_nu_n)$ where $a_n$ are undetermined coefficients determined by the boundary conditions.

c) One solves the fundamental equation $Lg = \delta$ & then $u = g * f$ where '$*$' means convolution here.

My questions are as as follows

1) What is the justification behind b). Is $f$ representable in the form $f = \Sigma c_nu_n$. What happens when $\lambda _n$'s are continous & not discrete.

2) Are approaches a) & b) equivalent. What are the situations where one preferable over the other.

3) Does the Green function approach give all solutions including the particular & complementary. Is it equivalent to b), if so how?

• for (b) you don't need to expand $f$ in terms of eigenfunctions since it is known Commented Jun 16, 2015 at 10:33
• also you can expand $f$ as eigenfunctions if the set of eigenfunctions forms a basis for the space that $f$ lives in, an example is $Lu=-\Delta u$, and $f\in L^2$. Commented Jun 16, 2015 at 10:35

The first idea always works; this is basically a linear algebraic fact, since the BVP is essentially the system of linear equations $Lu=f,Ru=g$ where $R$ is the restriction to the boundary (or similar in the non-Dirichlet case) and $g$ is the given boundary function.
• Does the green function approach give particular & complementary solution as well since $Lg = \delta$ can also be broken into $Lg_1 = \delta$ and $Lg_2 = 0$ with $g=g_1+g_2$ Commented Jun 16, 2015 at 11:11
• @Ricky Technically yes, but part of the point is that you don't have to do that, at least if the solutions to the BVP are unique (i.e. $Lu=0$ with a zero boundary condition implies $u=0$). The convolution will satisfy both the equation and the boundary condition for free.