I am solving that problem:
$$u_t+u_x+\sigma(x)u=g(x), x\in[0,l], t\ge0$$ $$u(x,0)=\phi(x)$$ $$u(0,t)=\gamma(t)=\int_{0}^{l}\beta(s)u(s,t)ds$$
I have already solved initial value problem with $\gamma(t)$ and got solution in general case to u(x,t) in two areas: $$u_1(x,t)=\phi(x-t)e^{-\int_{x-t}^{x}\sigma(\tau)d\tau} + \int_{x-t}^{x}g(\psi)e^{-\int_{\psi}^{x}\sigma(\tau)d\tau}d\psi, x-t>0$$
$$u_2(x,t)=\gamma(t-x)e^{-\int_{0}^{x}\sigma(\tau)d\tau} + \int_{0}^{x}g(\psi)e^{-\int_{\psi}^{x}\sigma(\tau)d\tau}d\psi, x-t<0$$
Now I think that I need to put this solution u(x,t) in $\gamma(t)=\int_{0}^{l}\beta(s)u(s,t)ds$.
But I also doesn't clearly understand how I need put it.
I think we need to split integral to two parts:
$\gamma(t)=\int_{0}^{l}\beta(s)u(s,t)ds=\int_{0}^{t}\beta(s)u_2(s,t)ds+\int_{t}^{l}\beta(s)u_1(s,t)ds$ (why $u_2$ in first integral?)
In this case we get integral equations of Volterra 2nd kind:
$\gamma(t) = \int_{0}^{t}\beta(s)\gamma(t-s)e^{-\int_{0}^{s}\sigma(\tau)d\tau}ds + H(t)$, where $H(t)$ is function, that we can calculate because all components are known.
So, please explain to me if is correct and why I can split gamma integral in this way into two parts. And also what is true way to solve this problem? Please give me some advices to get final solution in general case. More over please explain me how to use information about borders of t and x.