I am trying to find an approximate solution to an eigenvalue equation using techniques from perturbation theory. Roughly speaking, the problem is as follows

$L^s \phi_q^s = \lambda_q^s \phi_q^s$

The operator $L^s$ can be written

$L^s = L^0 + s L^1$

where $s>0$ is small and $L^0$ is self-adjoint acting on some Hilbert space. I am trying to find solutions of the form

$\phi_q^s = \phi_q^0 + s \phi_q^1 + \ldots$

$\lambda_q^s = \lambda_q^0 + s \lambda_q^1 + \ldots$

Inserting the above expansions into the eigenvalue equation, and collecting terms of like orders in $s$ yields

$( L^0 - \lambda_q^0 ) \phi_q^0 = 0$ (1)

$( L^0 - \lambda_q^0 ) \phi_q^1 = (\lambda_q^1 - L^1 ) \phi_q^0 = 0$ (2)

When the Hilbert space is $L^2((a,b))$ ($-\infty < a < b < \infty$) the spectrum of $L^0$ is simple and purely discrete. Solving (1) yields a complete set of orthonormal basis functions with the orthogonality relation $<\phi_n^0,\phi^0_k>=\delta_{n,k}$. By the Fredholm alternative, in order for a solution $\phi_n^1$ of (2) to exist, the RHS of (2) must satisfy

$< \phi_n^0, (\lambda_n^1 - L^1 ) \phi_n^0 > = 0$

Thus $\lambda_n^1$ is given by

$\lambda_n^1 = < \phi_n^0, L^1 \phi_n^0 >$

And $\phi_n^1$ is given by applying the resolvent operator to the RHS of (2), which yields

$\phi_n^1 = \sum_{k \neq n} \frac{< \phi_k^0, L^1 \phi_n^0 >}{\lambda_n - \lambda_k} \phi_k^0$

When the Hilbert space is $L^2(-\infty,\infty)$ the spectrum of $L^0$ is purely absolutely continuous. Solving (1) still yeilds a complete set of orthonormal basis functions with the orthogonality relation $<\phi_p^0,\phi^0_q>=\delta(p-q)$. However, solving (2) is now more complicated. When the operator $L^1$ is such that

$< \phi_p^1 , L^1 \phi_q^0 > = \delta(p-q) f^1(q) + g^1(p,q)$ (3)

Trying a solution analgous to the discrete case works works

$\lambda_q^1 = f^1(q)$

$\psi_q^1 = \int \frac{g^1(p,q)}{\lambda_q - \lambda_p} \phi_p^0 dp$ (the integral converges). (4)

However, I am looking at various $L^1$. And, for certain $L^1$ I do not have (3). Rather, I have

$< \phi_p^1 , L^1 \phi_q^0 > = \delta(p-q) f^1(q) + h(q) \delta'(p-q) + g^1(p,q)$

(yes, that's a derivative of a delta function). Frankly, at this point, I am totally stuck. I've tried a solution of the form (4) with

$g^1(p,q) \to h(q) \delta'(p-q) + g^1(p,q)$.

But that solution blows up. My sense is that I should be looking for some sort of condition on $\lambda_q^1$ which would guarantee that a solution $\phi_q^1$ of (2) exists. But, I know of no such condition.

If it helps, you can think of $L^0$ as $-d^2/dx^2$ so that the eigenfunctions are $\phi_q^0(x)=e^{iqx}/\sqrt{2\pi}$. And $L_1 \phi_q^0 = x \phi_k^0(x)$. If you're wondering, the derivative of the delta function comes about as follows

$< \phi_q^0, x \phi_p^0 > = (1/2 \pi) \int x e^{i(p-q)x} dx = (1/2 \pi i) (d/dp) \int e^{i(p-q)x} dx = (1/2 \pi i)\delta'(p-q)$.

Any guidance on solving this problem would be greatly greatly appreciated.

asymptoticexpansion, in various regimes. Poincare and others already found examples in various scenarios in celestial mechanics, where "approximate" solutions were eventually shown tonotbe parts of convergent infinite series... but, nevertheless, were perfectly fine asymptotic expansions (which is how they'd been used all along, in reality). Just a thought... – paul garrett Jul 29 '11 at 20:23