I have a system of two coupled PDEs and I'm trying to understand what is going on:
$$ \frac{\partial u(x,t)}{\partial t} = \frac{\partial w(x,t)}{\partial x} + u(x,t)w(x,t)\\ \frac{\partial w(x,t)}{\partial t} = a(x)w(x,t) + b(x)u(x,t) + c(x) $$
There's a coupling between the $\frac{\partial u(x,t)}{\partial t}$ and $\frac{\partial w(x,t)}{\partial x}$, is there a way to decompose this relationship? I'm free to choose my boundary conditions if need be.
$$ \frac{\partial u(x,t)}{\partial t} = \frac{\partial w(x,t)}{\partial x} + u(x,t)w(x,t)\\ \frac{\partial w(x,t)}{\partial t} = a(x)w(x,t) + b(x)u(x,t) + c(x) $$ One can separate the functions $u(x,t)$ and $w(x,t)$ : $$u(x,t)=-\frac{1}{b(x)}\frac{\partial w(x,t)}{\partial t} + \frac{a(x)}{b(x)}w(x,t) + \frac{c(x)}{b(x)}$$ $$\boxed{-\frac{1}{b(x)}\frac{\partial^2 w(x,t)}{\partial t^2} + \frac{a(x)}{b(x)}\frac{\partial w(x,t)}{\partial t} = \frac{\partial w(x,t)}{\partial x} + \left(-\frac{1}{b(x)}\frac{\partial w(x,t)}{\partial t} + \frac{a(x)}{b(x)}w(x,t) + \frac{c(x)}{b(x)} \right)w(x,t)}$$ In this PDE there is only the function $w(x,t)$. This is a second order nonlinear PDE. One cannot expect an easy solving in general.
Better rely on numerical methods.
An alternative way leads to the same conclusion :
Considering the equation $\frac{\partial w(x,t)}{\partial t} = a(x)w(x,t) + b(x)u(x,t) + c(x)$ which is an ODE (since where is only one differential wrt $t$). Solving it for $w(x,t)$ leads to $$w(x,t)=e^{a(x)t}\left(\int e^{-a(x)t}\big(b(x)u(x,t)+c(x)\big)dt+F(x)\right)$$ $F(x)$ is an arbitrary function.
One put it into $\frac{\partial u(x,t)}{\partial t} = \frac{\partial w(x,t)}{\partial x} + u(x,t)w(x,t)$ . This leads to a first order nonlinear integro-differential equation where $u(x,t)$ is the only unknown. Again one cannot expect an easy solving in general.
Note: Without boundary and without initial condition one cannot go further. Possibly simplification might appear in particular case.