I am trying to solve the homogeneous Fredholm equation of the second kind:
$w(r) = \frac{c}{2}\frac{1}{\frac{\gamma + a}{v \sigma b} + 1 - \frac{\gamma}{f\sigma} r} \int_{-1}^1 d r' w(r'),$ $\quad\quad\quad\quad$ (1)
where $\sigma$, $c$, and $v$ are real and I have not been able to determine any restrictions on $\gamma$, $a$, $b$ nor $f$. This equation was derived from the homogeneous neutron transport equation in 1-D planar geometry, isotropic scattering, and constant material properties. The quantities $\gamma$ and $a$ are scaling parameters while $b$ and $f$ are translation parameters.
From what I have read, the kernel in this case is not a function of the integration variable $r'$ and therefore is not symmetric (i.e., $k(r) \neq k(r')$). I found a paper ("Elementary Solutions of the Transport Equation and Their Applications" by K.M. Case) that has an equation similar to this and the solution is given without much explanation. The equation and solution of this paper appear to be:
$\phi_{\alpha,k} (\mu) = \frac{ic}{2k} \cdot \frac{1}{\alpha - \mu} \int_{-1}^1 d \mu' \phi_{\alpha,k} (\mu')$ $\quad\quad\quad\quad$ (2)
and
$\phi_{\alpha,k}(r) = \frac{ic}{2k} P \left(\frac{1}{\alpha - r}\right) + \lambda(\alpha) \delta (r-\alpha)$, $\quad\quad\quad\quad$ (3)
where $P$ refers to the principle values are to be used during integration. Also $k$ is a fixed real number and
$\lambda (\alpha) = 1 - \frac{ic}{k} tanh^{-1} (\alpha)$. $\quad\quad\quad\quad$ (4)
The author, K.M. Case, does not explain how they got this solution to be complex or how they arrived at many of the conditions on variables like $k$ and $\alpha$. Both of these Fredholm equations are derived from the exact same neutron transport equation, just from different solution methods (mine being a Lie group analysis while K.M. Case makes an ansatz and finds the Fredholm equation). My goal is to arrive at his solution, which requires solving the above in detail to get the eigenvalues and eigenfunctions of the solution.
I believe I need to apply Hilbert-Schmidt theory, but I am not striking any luck on finding info for non-symmetric kernels.
Any insight into how to solve this equation would be very helpful. I want to understand how to solve this type of equation. Thanks.
Let's simplify the notation, and rewrite your integral equation as $$ w(r)=\frac{1}{A-Br}\int_{-1}^1 w(r')\,dr', \tag{1} $$ where $A=\frac{2}{c}\left(\frac{\gamma + a}{v \sigma b} + 1\right)$ and $B=\frac{2\gamma}{cf\sigma}$. If we assume that $A$ and $B$ are real and $A>B$, so that $A-Br\neq 0$ in the interval $-1\leq r\leq 1$, we may integrate both sides of $(1)$ over that interval: \begin{align} \int_{-1}^1w(r)\,dr&=\int_{-1}^1 w(r')\,dr'\int_{-1}^1\frac{dr}{A-Br} \\ &=-\frac{1}{B}\ln\left(\frac{A-B}{A+B}\right)\int_{-1}^1 w(r')\,dr'. \tag{2} \end{align} Eq. $(2)$ implies that $A$ and $B$ must satisfy the following relation: $$ B=\ln\left(\frac{A+B}{A-B}\right). \tag{3} $$ Therfore, the solution to $(1)$ is given by $$ w(r)=\frac{W}{A-Br}, \tag{4} $$ where $W=\int_{-1}^1w(r')\,dr'$, and $A$ and $B$ are related by Eq. $(3)$.