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I have the following integral equation:

$\phi(x, y) = \frac{a}{x-y} \int_y^x \phi(s, y) ds + \frac{b}{x-y} \int_y^x \phi(x, s) ds$

where $a > 1$ and $b> 1$ are constants, and $x \geq y$. The unknown function $\phi : \Delta \rightarrow \mathbb{R}$ (where $\Delta = \{(x, y) \in [0,1]^2 : x \geq y \}$) satisfies $\phi(x,x)=0$ for every $x \in [0,1]$, and is continuously differentiable.

I know one non-trivial solution is $\phi(x,y) = C (x-y)^{a+b-1}$, where $C$ is a constant. I am wondering if there is any other non-trivial solution?

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    $\begingroup$ It looks interesting. What is the motivation? $\endgroup$
    – Lev Borisov
    Commented Dec 19, 2013 at 13:58
  • $\begingroup$ Thanks for the reply! This integral equation comes from auction theory. The equation is satisfied by the bidding strategy of a Nash equilibrium, and I would like to know if this equilibrium is unique. Appreciate any leads! $\endgroup$
    – Songzi Du
    Commented Dec 19, 2013 at 18:01
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    $\begingroup$ Have you tried looking for solutions in the form $\phi(x,y)=(x-y)^{a+b-1} g(x,y)$ where $g(x,y)$ is a convergent power series in $x-y,x+y$? By rewriting your equation as a recursive relation on the coefficients of the power series, you may get a sense of how unique it is. $\endgroup$
    – Lev Borisov
    Commented Dec 19, 2013 at 22:22
  • $\begingroup$ Thank you! Following your suggestion I wrote down the recursive relation, which is very useful. For example, when $a=2$ and $b=3$, $\phi(x, y) = (x-y)^4 + (x-y)^4 (x+y) + \frac{1}{5} (x-y)^5$ is a solution. $\endgroup$
    – Songzi Du
    Commented Dec 20, 2013 at 12:43

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