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  How can we deal with a singular integral equation where the singular kernel is sum of Cauchy and Hilbert kernels

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In the context of studying the eigenvalue distribution of Matrix Models, one encounters integral equations which needs to be solved in order to obtain the eigenvalue density in the Large $N$ limit. To be more precise, referring to: https://arxiv.org/pdf/hep-th/0410165.pdf , what I am talking about is essentially Eqn. (2.47). Given the potential of the matrix model i.e $W(\lambda)$, one needs to solve for $\rho(\lambda)$. This is done by defining the resolvent at zero genus according to Eqn (2.56) and subsequently evaluating Eqn (2.60) to get a closed form expression for the genus zero resolvent ($\omega_0$) . Finally using Eqn (2.58), it is easy to determine the discontinuity in the resolvent across the relevant branch cut. This gives the eigenvalue density.

However, the case I'm looking at does not have an effective action of the form given in Eqn (2.41). This changes things significantly. Instead the singular kernel that I have is the sum of the Cauchy kernel (just discussed above) and the Hilbert kernel ($~\cot (\frac{\lambda-\lambda'}{2})$). Overall, the eigenvalue density can be determined from the equation:

\[\Pr \int_a^b d\lambda' \rho(\lambda')\left[ \frac{1}{\lambda-\lambda'}+\frac{t}{2}\cot\left(\frac{t(\lambda-\lambda')}{2}\right)\right]=G(\lambda)\]where Pr denotes the principal value of the integral. $G(\lambda)$ is a known function and $t>0$ is a constant.

My questions are the following:

1) Is there an expression for the "resolvent" for the above class of singular integral equations?

2) The Hilbert kernel seems to be defined when the range of the integral $[a,b] \equiv [0,2\pi]$. Does this put some kind of bound on $a$ and $b$.

Thanks in advance. 

asked Oct 7 in Mathematics by DebangshuMukherjee (160 points) [ revision history ]
edited Oct 7 by DebangshuMukherjee

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