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Consider the kernel: $$k(x,y,\xi) = \sqrt{\frac{1}{\pi} \frac{1}{y+a}} \exp \left(-\frac{\xi^2}{y+a }\right)$$

I am trying to find the asymptotic form of the solutions to the following homogeneous Fredholm equation of the second kind:

$$K_\epsilon [u_\epsilon] (x) = \frac{1}{\epsilon} \int_0^1 k\left(x,y, \frac{x-y}{\epsilon} \right) \mathrm d y = \lambda_\epsilon u_\epsilon (x)$$

for small $\epsilon > 0$. That is, I want to find asymptotic series of the form:

$$\lambda_\epsilon = \lambda_0 + \lambda_1\epsilon + \lambda_2 \epsilon^2 + \dots$$ $$u_\epsilon (x) = u_0 (x) + u_1(x)\epsilon + u_2(x)\epsilon^2 + \dots$$

following the approach of Refs. [1,2]. Although [2] makes the assumption that the kernel is symmetric (my kernel does not satisfy this) I am not sure why this is needed so I tried to apply the method anyway. Eventually [2] finds $\lambda_0 = 1, \lambda_1 = 0$, transforming the problem into the solution of a Sturm-Liouville problem:

$$L[u_0] = \lambda_2 u_0, \quad u_0(0) = u_0(1) = 0$$

where $L[f] = (p(x)f'(x))' + q(x)f(x)$ with

$$p(x) = \int_0^\infty k(x,x,\xi) \xi^2 \mathrm d \xi, \quad q(x) = \int_0^\infty \frac{\partial^2 k(x,x,\xi)}{\partial y^2} \xi^2 \mathrm d \xi $$

I obtain $p(x) = (x+a)/2$ and $q(x) = 0$, which in turn implies that $u_0$ must be a linear combination of modified Bessel functions of the first and second kind, which cannot satisfy the boundary conditions $u_0(0) = u_0(1) = 0$ non-trivially.

I have reasons to believe that this result is incorrect. I need to confirm if my approach is sound or not, or where is my mistake, ultimately finding the correct asymptotic expansion.

Note that I am only interested in the first few terms of the expansion. Specifically I am happy with obtaining $\lambda_2$, $u_0$ and $u_1$.

[1] Malinovskii, Ju.G., Asymptotic expansion of the solutions of integral equations with $\delta$-form kernels, U.S.S.R. Comput. Math. Math. Phys. 12 (1972), No.6, 227-243 (1973). ZBL0268.45019.

[2] Malinovskii, Ju.G., Asymptotic behaviour of the eigenfunctions and eigenvalues of integral operators with $\delta$-type kernels, U.S.S.R. Comput. Math. Math. Phys. 13(1973), No.5, 32-44 (1974). ZBL0289.45022.

Note: These are the only two papers (in English) that I could find on this topic. If there is any more literature related to these kind of asymptotic expansions please let me know.

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  • $\begingroup$ Erm... Are you claiming that the Sturm-Liouville operator $f\mapsto (pf')'$ with positive smooth $p$ (I assume that $a>0$) has no eigenfunctions with the Dirichlet boundary conditions on $[0,1]$? That sounds fishy, doesn't it? $\endgroup$
    – fedja
    Commented Jul 17, 2017 at 0:30
  • $\begingroup$ @fedja You are totally correct (as usual). But I cannot tame these eigenfunctions into satisfying the boundary conditions. According to Mathemtica, the solution of $(pf')' = \lambda_2 f$ is $f(x) = c_1 I_0(4\sqrt{(x+a)\lambda_2} + c_2 K_0(4\sqrt{(x+a)\lambda_2}$ (modified Besel functions), but I cannot find $\lambda_2,c_1,c_2$ so as to make $f(0)=f(1)=0$ while $f(x)$ is real. $\endgroup$
    – valle
    Commented Jul 21, 2017 at 11:57
  • $\begingroup$ $\lambda_2$ is necessarily negative, so your arguments in the formula written are purely imaginary (and the coefficients complex). $\endgroup$
    – fedja
    Commented Jul 21, 2017 at 13:53
  • $\begingroup$ @fedja Yes, I noticed that much. In particular $I_0(\mathrm i x) = J_0(x)$ which is real. However $K_0(\mathrm i x)$ is complex, and I don't see how a constant complex coefficient can make it real.... unless $c_2 = 0$, in which case we must require $J_0(4\sqrt{a|\lambda_2|}) = J_0(4\sqrt{(1 + a)|\lambda_2|}) = 0$, which I think cannot be simultaneously met. This is where I am stuck. $\endgroup$
    – valle
    Commented Jul 21, 2017 at 14:05
  • $\begingroup$ ... which I think cannot be simultaneously met by $\lambda_2$ unless $a$ is very specific $\endgroup$
    – valle
    Commented Jul 21, 2017 at 14:11

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