I have an integral equation that has the following form:
$$f(x,t)=g(x,t)+\int_0^t h(u)f(x+u,t-u)du$$
for all $x,t\in \mathbb{R_+}$ and $f(x,0)=1, \forall x\in\mathbb{R_+}$ with $g$ and $h$ are given functions.
I would like to solve this equation and this is my approach:
Taking the derivative with $t$ for both sides, we have:
$$\frac{\partial f(x,t)}{\partial t}=\frac{\partial g(x,t)}{\partial t}+\frac{\partial}{\partial t}\int_0^t h(u)f(x+u,t-u)du$$ Let $T$ be the primitive function of $h\circ f(t)$, we have: $$\frac{\partial f(x,t)}{\partial t}=\frac{\partial g(x,t)}{\partial t}+\frac{\partial}{\partial t}(T(t)-T(0))$$ We have $\frac{\partial}{\partial t}T(t)=h(t)f(x+t,0)=h(t)$ and $\frac{\partial}{\partial t}T(0)=0$. Hence: $$\frac{\partial f(x,t)}{\partial t}=\frac{\partial g(x,t)}{\partial t}+h(t)$$ Thus by taking the integral for both sides we have, $$f(x,t)=g(x,t)+\int_0^t h(u)du$$ And it means that $$\int_0^t h(u)f(x+u,t-u)du=\int_0^t h(u)du$$
Is there something wrong in my approach?