Analytic solution to a Volterra integral equation with kernel divided by continous differentiable function

Question

I have been struggling with the following Volterra integral equation for days: $$f(x)=A_{1}-\frac{A_{2}}{h(x)}+\lambda \int_{a}^{x}\frac{h^{\prime }(s)}{h(x)}f(s)ds$$ , where a, A1 and A2 are positive constants, $\lambda$ is a negative constant, and $h(x)$ is a continuously differentiable function. It is a Volterra integral equation of second kind, and I have tried Adomian Decomposition Method or converting the problem to an ODE of initial value, but I am not able to solve it even if I consider the simplest case: h(x)=x. Does this type of integral equation have analytic solution? May Laplace transformation help to solve it? If I have to resort to numerical methods, are there available libraries for Mathematica, Matlab or other software I could use? Any help will be greatly appreciated.

Answer 1

One can brute force this using functional analysis techniques (which I just pray you know). I assume that $h$ does not vanish on $[a, b]$ where $b>a$ and is continuous on that domain. Some steps you will have to calculate for yourself, I do not want the answer to be too long...

We equip the space $C([a, b])$ with the usual norm $\lVert f \rVert_{\infty} := \sup_{x \in [a, b]} \lvert f(x) \rvert$. Consider the continuous linear operator $T: C([a, b]) \rightarrow C([a, b])$, $$ (Tf)(x) := \int^x_a k(s, x) f(s)~\mathrm{d}s $$ where $k \in C([a, b]^2)$ is defined as $k(s,x) := \lambda\frac{h'(s)}{h(x)}$. Let $K := \sup_{(s, x) \in [a, b]^2} \lvert k(s, x) \rvert$.

Then let $n \in \mathbb{N}$ be very large. We consider $T^n$, the $n$-time composition of $T$ with itself. Then, for any $f \in C([a, b])$ and any $x \in [a, b]$ we can prove that $$ \left \lvert (T^nf)(x) \right \rvert \leq \lVert f \rVert_\infty K^n \int^x_a \int^{x_1}_a ... \int^{x_{n-1}}_a~\mathrm{d}x_n~\mathrm{d}x_{n-1}...~\mathrm{d}x_1 = \lVert f \rVert_\infty\frac{K^n (x-a)^n}{n!}\leq \lVert f \rVert_\infty\frac{K^n(b-a)^n}{n!}. $$ So for $n$ very large, we have that $\lVert T^n \rVert < 1$, where $\lVert \cdot \rVert$ denotes the operator norm. Therefore the Neumann-Series $\displaystyle \sum_{n = 1}^\infty T^n$ converges in the operator norm. Thus $T- \mathrm{Id}$ has an inverse. In other words: For every $v \in C([a, b])$ the equation $$ v = Tf - f $$ has a unique solution $f \in C([a, b])$. Choose $v(x) := \frac{A_2}{h(x)}-A_1$ and you get a solution for your integral equation that is continuous on $[a, b]$.

So yes, there exists an analytic solution on compact intervals $[a, b]$, no doubt. Just iterating this argument for larger and larger intervals should give us a continuous solution on $[a, \infty)$... Maybe one can allow $x \in [\beta, b]$ where $\beta < a < b$ - which I don't expect to do any harm, but I would have to compute that separately. Then we would get a solution in $[\beta, b]$