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I have a problem with the following first-order PDE: \begin{equation} \frac{\partial u}{\partial x} - a(x,t)\frac{1}{u}\frac{\partial u}{\partial t} = 0 \end{equation} where $a(x,t)$ is a smooth and well-behaved function. I am looking for its solution, or even better, the Green's function to the differential operator, as I am trying to solve the inhomogeneous version of it (replace the zero on the RHS by $b(x,t)$).

What I got so far (for different properties of $a(x,t)$):

  • $a(x,t) = 1 \Rightarrow u(x,t) = - \frac{x}{t}$
  • $a(x,t) = a(t) \Rightarrow u(x,t) = - \frac{x}{\int^t dt' / a(t')}$ (product Ansatz and separation of variables)
  • $a(x,t) = a(x) \Rightarrow u(x,t) = - \frac{\int^x dx' \, a(x')}{t}$ (product Ansatz and separation of variables)
  • $a(x,t) = a_1(x) \cdot a_2(t) \Rightarrow u(x,t) = - \frac{\int^x dx' \, a_1(x')}{\int^t dt' / a_2(t')}$ (product Ansatz and separation of variables)

What, however, if $a(x,t)$ cannot be factorized? I assumed that $a(x,t)$ could be written as the product of two functions only dependent on $x$ or $t$, respectively: \begin{equation} u(x,t) = - \frac{\int^x dx' \, a_1(x')}{\int^t dt' / a_2(t')} = -\frac{1}{a_1(x)a_2(t)} \frac{\int^x dx' \, a_1(x')a_2(t)}{\int^t dt' / (a_1(x)a_2(t'))} = -\frac{1}{a(x,t)} \frac{\int^x dx' \, a(x',t)}{\int^t dt' / a(x,t')} \end{equation} But testing my solution revealed that the assumption of two factors, even if I do not need to know their explicit functional forms, was wrong.

Simon Krug
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    The solution via the method of characteristics comes from solving $$\frac{dt}{1} = - \frac{u dx}{a(x,t)} = \frac{du}{0}$$ The last expression implies $u = C_{2} = f(C_{1})$ where $f$ is an arbitrary differentiable function and $C_{1}$ is the characteristic curve that comes from solving $$\frac{dt}{1} = - \frac{C_{2} dx}{a(x,t)} \implies x + \frac{1}{C_{2}} \int a(x,t) dt = C_{1}$$ Without knowing the explicit form of $a$ this is as far as you can go. – Matthew Cassell Sep 13 '22 at 03:38

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