I got stuck with some strange point, solving Volterra integral equation:
$$ \int_0^t (t-s)f(s) ds =\sqrt{t}. $$
The solution can be obtained by ssuccessive differnetiation $$ \int_0^t f(s)ds=\frac{1}{2\sqrt{t}}, \quad \mbox{and then} $$ $$ f(t)=-\frac{1}{4t\sqrt{t}} $$
But, when I substitute this solution to the original equation $$ \int_0^t(t-s)\left[-\frac{1}{4s\sqrt{s}}\right]ds=\left[\frac{s+t}{2\sqrt{s}}\right]_0^t=\sqrt{t}-\infty $$
I can't figure out where I was wrong. Please explain.
On
Notice that integration by part gives $$\int_{0}^{t}(t-s)\,f(s) \, ds = \left[(t-s)\int_{0}^{s}f(u)du \right]_{s=0}^{s=t} +\int_{0}^{t}\int_{0}^{s}f(u)\,du\,ds = V^{2}(f)(t)$$
where $V^{2}(f)$ is understood as the composition of Volterra operatorer $$V(f)(t):= \int_{0}^{t}f(u) \, du$$ with itself.
Now for the Volterra operator to make sense, we must at least have that $f$ is locally integrable in a neighbourhood of $0$, say $f\in L^{1}(0,1))$. But then $V(f)$ lands into the space of continuous functions on $[0,1]$ (This is a straightforward appliction of Dominated convergence theorem).
Consequently composing with $V$ again we obtain $V^{2}(f)\in C^{1}([0,1])$.
Going back to your initial equation, we wished to solve $$V^{2}(f)(t) = \sqrt{t}$$ The previous argument shows that we cannot even find an $L^{1}(0,1)$-solution $f$ to this equation, since $t\rightarrow \sqrt{t}$ fails to be continuously differentiable on $[0,1]$. What I meant to explain was that if you cannot even find a solution with minimal to no restrictions, then you are pretty much out of luck.
I am not an expert on integral equations or the like, but the origin of the trouble is clear -- the differentiation process loses sense of a constant. The infinity comes because your function is not integrable near zero. Being only mildly acquainted with integral equations, I would try two other approaches -- Neumann series and Laplace transform -- and see if they give something reasonable and get some intuition from the result.