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In my research I've come across a handful of double integral equations, and I'm nearly at a total loss for how to derive anything useful from such things.

I've been lead to believe that even single integral equations are quite difficult to handle, so I would imagine problems escalate considerably when considering double integral equations.

That being said, the most compelling equation I've derived is $$ 0=\int_0^\ell\int_0^\ell f(s,t)\mu(t)\mu^\prime(s)\,ds\,dt; $$ where I'm attempting to solve for $\mu$.

The particular $f$ I'm dealing with has a variety of nice properties: $\ell$-periodic, nonnegative, symmetric, continuous. The $\mu$ I'm looking for will be those that are $\ell$-periodic, positive, and continuous. For the sake of this discussion you may assume $f$ and $\mu$ have any properties that you desire them to have.

Ultimately, the question is this: what information can I gather from the above statements? It would be incredible if there was some way to solve for solutions, but I doubt such a method exists. Obviously any constant $\mu$ will be a solution, but I'm trying to find other results.

Any advice or relevant references are highly appreciated.

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  • $\begingroup$ Isnt $\mu'$ supposed to be $\mu$? $\endgroup$
    – BigM
    Commented Nov 19, 2015 at 1:08
  • $\begingroup$ I suspect it's supposed to be the derivative of $\mu$, otherwise constant $\mu$ wouldn't be "obviously" a solution. $\endgroup$
    – Robert Israel
    Commented Nov 19, 2015 at 1:10
  • $\begingroup$ Yes, $\mu^\prime$ is indeed correct. For context, this equation arose in an attempt to use calculus of variations. $\endgroup$
    – AegisCruiser
    Commented Nov 19, 2015 at 1:13

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Take some nice "trial functions" $\mu_j(t)$, and consider the $d$-parameter family $\mu_\lambda(t) = \sum_{j=1}^d \lambda_j \mu_j(t)$. The right side of your equation is a quadratic form in $\lambda_1, \ldots, \lambda_d$. If it doesn't happen to be positive definite or negative definite, you can solve the quadratic form and get nontrivial solutions.

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