# How to solve for $\Gamma(X,t)$ in $\Gamma_{t,X} = S(X,t) \Gamma_X$?

Given this equation

$$\frac{\partial^2 \Gamma}{\partial t \partial X} = S(X,t) \frac{\partial \Gamma}{\partial X},$$

how do you solve for $\Gamma(X,t)$?

$S(X,t)$ is unknown and we impose conditions (i) $\Gamma(X,0) = X$ and (ii) $\Gamma(0,t) = 0$ and $t,X \geq 0$.

Start with $$\frac{\partial^2\Gamma(x,t)}{\partial t \partial x} = S(x,t) \frac{\partial\Gamma(x,t)}{\partial x}$$
Assuming it is nonzero divide by ${\partial\Gamma(x,t)}/{\partial x}$ to get $$\frac{\partial}{\partial t}\log\frac{\partial \Gamma(x,t)}{\partial x} = S(x,t)$$ integrate, then exponentiate to get $$\frac{\partial \Gamma(x,t)}{\partial x} = B(x) \exp \left(\int_0^t S(x,\tau)d\tau\right)$$
where $B(x)$ is some function of $x$ only. Now $\Gamma(0,t)=0$, so $$\Gamma(x,t) = \Gamma(x,t)-\Gamma(x,0) = \int_0^xB(\xi) \exp \left(\int_0^t S(\xi,\tau)d\tau\right)\,d\xi$$ Now substitute $t=0$ to get $$x = \Gamma(x,0) = \int_0^x B(\xi)\,d\xi$$ for all $x$. Therefore $B(x) = 1$ for all $x$. Finally, $$\Gamma(x,t) = \int_0^x \exp \left(\int_0^t S(\xi,\tau)d\tau\right)\,d\xi$$