Consider the following integral equation
$$f(t) = A(t) + \int_0^t K(s,t)f'(s)ds,\quad \forall t\ge 0,\label{1}\tag{$\ast$}$$
where $A: \mathbb R_+\to [0,1]$ and $K: \{(s,t): 0\le s\le t\}\to[0,1]$ are Holder continuous functions so that
- $A$ and $K$ are infinitely differentiable restricted on $(0,\infty)$ and $\{(s,t): 0<s<t\}$;
- $A(0)=1$ and $K(t,t)=1/2$ for all $t\ge 0$;
- There exists $C>0$ s.t. $$|A'(t)|\le \frac{C}{\sqrt{t}}\quad \mbox{and}\quad |\partial_sK(s,t)|+|\partial_tK(s,t)| \le \frac{C}{\sqrt{t-s}},\quad \forall 0<s<t.$$
Does there exist a unique solution $f$ to \eqref{1} satisfying $f(0)=1$? I'm OK if additional conditions are required.
PS : Assuming $f$ is differentiable on $(0,\infty)$, differentiating both sides of \eqref{1} yields
$$f'(t) = 2A'(t) + 2\int_0^t \partial_t K(s,t)f'(s)ds,\quad \forall t>0.\label{2}\tag{$\diamond$}$$
\eqref{2} is a Volterra integral equation, while we have two singularities, i.e. $A(0)=\infty$ and $\partial_tK(t,t)=\infty$. I've no idea how to show the existence/uniqueness of the solution to \eqref{2}.
PS2 : Applying integration by parts to \eqref{1}, one gets another Volterra integral equation
$$f(t)=2A(t)-2K(0,t)-2\int_0^t \partial_s K(s,t)f(s)ds,\quad \forall t\ge 0.\label{3}\tag{$\star$}$$
I can prove that \eqref{3} admits a unique solution $\hat f$. This implies in particular that, if \eqref{1} has a solution, it must be unique. But I don't know how to show $\hat f$ is differentiable and thus the solution to \eqref{1}.
Any answer, references and comments are highly appreciated.
