Does this linear integral equation defined on $[-a,a]$ have any nonconstant solution? I am more interested in the exsistence of solution rather than the solution itself. The boundary condition has two cases:

Case $\bf(1)$ : $~y(-a)=y(a)=b~~$ and

Case $\bf(2)$ : $~-y(-a)=y(a)=c~$. $$\int_{-a}^{a}{ \frac{y(x)-y(x')}{(x-x')^2} ~~d x'}=0$$

I feel the solution, it exists at all, could be something simple. But I am not sure how to (dis)prove the exsistence or solve it.

  • $\begingroup$ $y(x)=b$ for all $x$ solves case 1). $\endgroup$ – Kavi Rama Murthy Feb 14 at 7:42
  • $\begingroup$ @KaviRamaMurthy Sorry, I mean nontrivial solution. Edited. $\endgroup$ – xiaohuamao Feb 14 at 7:44
  • $\begingroup$ How do you treat the singularity at $x=x'$? Especially at the interval ends $\pm a$ where you can not apply the Cauchy principal value? $\endgroup$ – Lutz Lehmann Feb 15 at 10:22
  • $\begingroup$ @LutzLehmann I think we can assume the nominator should cancel out the singularity for any valid smooth solution. $\endgroup$ – xiaohuamao Feb 15 at 19:19
  • $\begingroup$ How so? You can expand $y(x')=y(x)+y'(x)(x'-x)+\frac12y''(c_{x'})(x'-x)^2$, but what happens to the $y'(x)$ term? $\endgroup$ – Lutz Lehmann Feb 15 at 23:21

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