I encountered a (likely simple) Volterra-style integral equation as part of my work. For an arbitrary function $p(s)$ which is square integrable, I have the following equation relating $p(s)$ and its integral with respect to the real-to-real function $Q$:
$$p(x) - C\int_{0}^{x}p(s)dQ(s) = 0$$
where $C$ is a known real constant.
I would hopefully like to show, without further assumptions, that this system has only the solution $p(s) = 0$. I have very little training in differential equations, so I'm a bit lost as where to begin.
Thank you in advance for any guidance!
Hello and welcome to math.stackexchange.
Here is how you can prove this. Write the equation as $$ p(x) = \int_0^x p(s) q(s) ds $$ where $q(s) = C Q'(s)$. The equation shows that $p(x)$ is differentiable, since it is equal to an antiderivative. Therefore by differentiating $$ p'(x) = p(x) q(x) \, . $$ and also $p(0) = \int_0^0 p(s) q(s) ds = 0$.
To prove from this that $p(x) = 0$ for all $x$, one can use the following trick: Set $r(x) = p(x)e^{-CQ(x)}$. Then $$r'(x) = p'(x) e^{-CQ(x)} + p(x) e^{-CQ(x)}\cdot (-CQ'(x)) = e^{-CQ(x)} \cdot \left(p'(x) - p(x) q(x) \right) = 0 $$ So $r(x) = constant = r(0) = 0$ and therefore also $q(x) = 0$ for all $x$.