First order pde with characteristics [closed]

Question

Consider a first order pde of the type $$u_y+b(x)u_x=0$$ and suppose that the coefficient $b$ is not necessairly continuous (for instance with a jump in some point). Is it still possible to apply in some way the method of characteristics in order to solve the equation?

Answer 1

I once had to solve a similar PDE. In your notation, $b=u H(u_x)$ (where $H$ is the Heaviside (step) function), and so b was not a function of x, but rather of u and its x-derivative. It may be that this answer is thus geared towards a rather different question. (There are close votes on your question, so maybe when $b$ is merely a function of $x$, the problem is trivial in some way that is not coming to my mind immediately).

Anyways, while in that paper I proceeded "as a physicist" (ignoring all subtleties about existence, etc.), there is the following book, which contains some mathematical results along these lines: A. Melikyan, Generalized Characteristics of First Order PDEs (Birkhaüser, Boston, 1998). Apologies, I never had the chance to read it in detail (despite citing it) -- however, it might contain some results relevant to your problem too.

Answer 2

This pde can be solved formally as $$ u(x,y) = F(y - B(x)) $$ where $B(x)$ is an antiderivative of $1/b(x)$ and $F$ is an arbitrary differentiable function. Of course at points where $b(x)$ is discontinuous, $B(x)$ and $u(x,y)$ are likely not to be differentiable: the pde will only be satisfied there in some weak sense.