Eigenvalue problem for integral equation

Question

We need to solve the eigenvalue problem for the integral operator $$ \frac1{1-x}\int_0^{+\infty} K\left(\frac{x}{t}\right) u(t)dt = \lambda u(x). $$ We assume that integral kernel $K$ is analytic and decays sufficiently fast at $+\infty$. I tried to construct a differential equation for $u$ as follows. First, I consider more general case $$ \frac1{1-x}\int_0^{+\infty} K\left(\frac{xa}{t}\right) u(t)dt = \lambda u(x), $$ now $u=u(x,a)$. Then I differentiated this equation in $x$ and in $a$, replacing the integral by $(1-x)\lambda u$: $$ (1-x)^{-1}\lambda u + (1-x)^{-1}\int_0^{+\infty} \frac{a}{t} K'\left(\frac{xa}{t}\right) u(t)dt = \lambda u_x; $$ $$ (1-x)^{-1}\int_0^{+\infty} \frac{x}{t} K'\left(\frac{xa}{t}\right) u(t)dt = \lambda u_a. $$ From the last two equations, we obtain $$ (1-x)^{-1} u + \frac{a}{x} u_a = u_x. $$ Thus, $xu_x-au_a=\frac{x}{1-x} u$. Then we introduce function $v$ such that $u=v e^{\int\frac{x}{1-x} dx}$, and we obtain for $v(x,a)$ the following equation $$ xv_x-av_a=0. $$ I use the method of characteristics but failed because I do not know "initial" starting values. Maybe I should change the approach in the very beginning?

Answer 1

$$x\frac{\partial v}{\partial x}-a\frac{\partial v}{\partial a}=0$$ Charpit-Lagrange characteristic ODEs : $$\frac{dx}{x}=\frac{da}{-a}=\frac{dv}{0}$$ A first characteristic equation comes from $dv=0$ $$v=c_1$$ A second characteristic equation comes from $\frac{dx}{x}=\frac{da}{-a}$ $$xa=c_2$$ General solution of the PDE on the form of implicit equation $c_1=F(c_2)$ $$v(x,a)=F(xa)$$ $F$ is an arbitray function. $$u=ve^{\int\frac{x}{1-x}dx}=\frac{e^{-x}}{1-x}F(xa)$$