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I'm looking to solve the PDE below for a function $f(p,R,t)$.

$\frac{\delta f(p,R,t)}{\delta t}-D(R,t)\,p\,\frac{\delta f(p,R,t)}{\delta p}=0 $

I'm aware that the method of characteristics is supposed to be used to make progress, but I have no idea how to apply it in this case.

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  • $\begingroup$ The domain of $f$, the boundary conditions and the nature of $D$ are important to find out the better approach to solving the problem. $\endgroup$ – rafa11111 Nov 9 '18 at 18:05
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$$\frac{\partial f(p,R,t)}{\partial t}-D(R,t)\,p\,\frac{\partial f(p,R,t)}{\partial p}=0 \tag 1$$ There is no partial differential with respect to $R$. Thus, $R$ can be considered as a parameter, or a dummy variable if you prefer. In a first step, we forget it in order to simplify the typing. $$\frac{\partial f(p,t)}{\partial t}-D(t)\,p\,\frac{\partial f(p,t)}{\partial p}=0 \tag 2$$ The system of characteristic ODEs is : $$\frac{dt}{1}=\frac{dp}{-D(t)p}=\frac{df}{0}$$ A first family of characteristic curves comes from $\frac{df}{0}=$differential of finite function of $t$ and $p$, which implies $df=0$ $$f=c_1$$ A second family of characteristic curves comes from $\frac{dt}{1}=\frac{dp}{-D(t)p}$ which is a separable ODE. $$\ln|p|+\int D(t)dt=c_2$$ General solution of PDE $(2)$, expressed on the form of implicit equation : $$\Phi\left(f\:,\:\left(\ln|p|+\int D(t)dt\right)\: \right)=0$$ where $\Phi$ is an arbitrary function of two variables.

Or equivalently on explicit form : $$f(p,t)=F\left(\ln|p|+\int D(t)dt \right)$$ where $F$ is an arbitrary function.

Coming back to the PDE $(1)$ in re-introducing the dummy variable $R$ gives the general solution of PDE $(1)$ :

$$f(p,R,t)=F\left(\:\left(\ln|p|+\int D(R,t)dt \right)\:,\:R\:\right)$$ where $F$ is an arbitrary function of two variables.

No boundary and no initial condition are specified in the question of the OP. So, one cannot go further. If some viable boundary and initial conditions where specified one could try to determine a specific function $F$ of two variables fitting the specified conditions. Often this is the most difficult part of the job.

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