Letting $D:=\{(x,y):\ 0\leq x\leq y\leq1 \text{ and } y>0\}$, I have a continuous function $k:D\to[0,\infty)$ that satisfies some properties that I list below. I'm interested in continuous and bounded-variation functions $\varphi:[0,1]\to\mathbb R$ that solve the Volterra equation $$0=\int_0^y \varphi(x)k(x,y) \text{ d}x\ \ \ \forall y\in[0,1].$$ For brevity, let me call any such $\varphi$ a solution for $k$.
My question: Does my $k$ necessarily have the zero function as its unique solution? And if not, are there smoothness/boundedness conditions I could put on $k$ to ensure it has a unique solution?
My $k$ satisfies the following properties (the first of which seems important):
- Every $y\in(0,1]$ has $\int_0^y k(x,y) \text{ d}x=1.$
- The partial derivative $k_2$ with respect to the second argument exists and is continuous on $D$.
- Every $(x,y)\in D$ with $x>0$ has $k(x,y)>0$.
- The function $x\mapsto \tfrac{k(x,\hat y)}{k(x,y)}$ is weakly increasing on $(0,y]$ whenever $0<y<\hat y\leq1$. Equivalently, the function $\tfrac{k_2}{k}$ is weakly increasing in its second argument wherever the latter is strictly positive.
We can also convert to a Volterra equation of the second kind (with a boundary condition) in the usual way. Defining the continuous function $\tilde k:D\to\mathbb R$ via $\tilde k(x,y):=\tfrac{-k_2(x,y)}{k(y,y)}$, it's easy to see that $\varphi$ is a solution if and only if it has $\varphi(0)=0$ (given the first bullet above) and $$\varphi(y)=\int_0^y \varphi(x)\tilde k(x,y) \text{ d}x\ \ \ \forall y\in(0,1].$$
I have no idea which of the above listed properties is useful for establishing uniqueness. However, note that the first condition tells us $k$ cannot be extended continuously to the closure of $D$, and (loosely) tells us how quickly $k(x, y)$ explodes as $y\to0$.
