# Unique solution of Volterra integral equation of second kind

Dear Maths Stackexchange,

In the context of a physics problem, I am looking at a linear integral equation, more specifically a 2nd kind Volterra equation in the unknown $g(t)$: \begin{equation} \phantom{texttexttexttex}g(t) = f(t) + \int_0^t K(t,s) g(s) \mathrm{d}s \phantom{texttexttexttex}(1) \end{equation} The independent variable $t\geq0$ always.

Now, the problem I have is that in my case $K(t,s)$ has the form $K(t,s)=K(t)K(s)$ where $K(t)$ is a (complicated) known function that has a pole of order 2 at $t=0$ (i.e. $t^2K(t)\to K_o<\infty$ as $t\to0$) but is smooth everywhere else. By contrast, the function $K(s)=s$ and $f(t)$ is smooth everywhere on $[0,\infty]$ and satisfies $\lim\limits_{t\to 0} f(t) = 0$. Because of this, $K(t,s)$ is not in $L^2([0,\epsilon]\times[0,\epsilon])$.

Now, an easy way of solving an integral equation of type (1) is by converting it into an initial value problem. If I differentiate (1) w.r.t. after dividing by $K(t)$, then I find the first order linear ODE, \begin{equation} \phantom{texttext}g'(t) + \left(\frac{\mathrm{d}}{\mathrm{d}t} \log(\tilde{K}(t)) - tK(t)\right)g(t) = \tilde{f}'(t)K(t)\phantom{tetextxt}(2) \end{equation} where $\tilde{K}=\frac{1}{K(t)}$ and $\tilde{f}(t)=\frac{f(t)}{K(t)}$. This equation I can solve exactly, but I need an initial condition to do so. And here lies my problem. My first try would have been to require $g(0)=0$ as an intial condition. However, upon inspection of the coefficients at $t=0$, we see that they are singular there. Therefore, imposing any initial condition at $t=0$ fails.

Now this leads me to the question: Does (1) have a unique solution despite the pole in $K(t)$ and which initial condition for $g(t)$ does this unique solution correspond to in (2)? Since the differential equation (2) follows from the integral equation (1), any equation of (1) must satisfy (2), but if there was a unique solution of (1), then obviously only one single out of the many solutions of (2) would correspond to that unique solution of (1). Can you help me sort out my confusion?

I would like to use the ODE formulation of the problem because it allows an easy analytical solution.

Best regards

Edit: clarified some ambiguities.

I have found the solution while browsing the literature. The trick is to exploit the fact that the kernel of (1) is separable. I use the notation $K(t,s)=\alpha(t)\beta(s)$ in this post as well as the notation of (1). The separability allows us to define $\gamma(t)=\int_0^tg(s)\beta(s)\mathrm{d}s$ such that (1) becomes: $$\gamma'(t) = \alpha(t)\beta(t)\gamma(t) + f(t)\beta(t) (3)$$ Noting that by definition of $\gamma(t)$, $g(0)=0$. Therefore the unique solution of (3) is $$\gamma(t) = e^{\int_0^t\alpha(s)\beta(s)\mathrm{d}s}\int_0^tf(s)\beta(s)e^{-\int_0^s\alpha(s')\beta(s')\mathrm{d}s'}\mathrm{d}s$$ From the definition of $\gamma(t)$ then follows the solution for $g(t)$ $$g(t) = f(t)+ \int_0^t f(s)\beta(s) e^{\int_s^t\alpha(s')\beta(s')\mathrm{d}s'}$$ By the uniqueness of the solution for $\gamma(t)$ this is the unique solution of (1) for $g(t)$. QED.