Help with characteristic method technicality

Question

I'd like to derive a slightly more general result than the transport equation.

I have the PDE $0 = u_y(x,y) - g(y)f(x)u_x(x,y)$ with $u(x,0) = h(x),$

and in imploring the characteristic method, I arrive at

$$ 1 = \frac{dy}{ds}, \ g(y)f(x) = \frac{dx}{ds}, \ 0 = z(x(s),y(s)) .$$

Now, when I'm integrating these ODE, where I'm confused is whether I treat the $g(y)$ term as a constant or if I can integrate $\int_{s_0}^{s} g(v)dv = \int_{s_0}^{x(s)}\frac{dx/ds}{f(x)}ds.$

I'm sure this solution has some kind of ascertainable result.

Answer 1

$$u_y(x,y) - g(y)f(x)u_x(x,y)=0$$ Charpit-Lagrange characteristic ODEs : $$\frac{dy}{1}=\frac{dx}{- g(y)f(x)}=\frac{du}{0}$$ A first characteristic equation comes from $\frac{du}{0}\quad\implies\quad du=0\quad $ thus : $$u=c_1$$ A second characteristic equation comes from solving $\frac{dy}{1}=\frac{dx}{- g(y)f(x)}\quad\implies\quad g(y)dy+\frac{dx}{f(x)}=0$ $$\int g(y)dy+\int\frac{dx}{f(x)}=c_2$$ The general solution of the PDE on the form of implicit equation $c_1=\Phi(c_2)$ is : $$\boxed{u(x,y)=\Phi\left(\int_{y_0}^y g(\mu)d\mu+\int_{x_0}^x\frac{d\nu}{f(\nu)}\right)}$$ $\Phi$ is an arbitrary function (to be determined according to the specified condition).

Condition : $u(x,0)=h(x)$

We set $y_0=0$ into the above general solution without loss of generality. $$h(x)=\Phi\left(\int_{0}^0 g(\mu)d\mu+\int_{x_0}^x\frac{d\nu}{f(\nu)}\right)$$ $$h(x)=\Phi\left(\int_{x_0}^x\frac{d\nu}{f(\nu)}\right)$$ Let $X(x)=\int_{x_0}^x\frac{d\nu}{f(\nu)}$

Supposing that $\frac{d\nu}{f(\nu)}$ is integrable and that the inverse function of $X(x)$ can be expressed on the form of $x=\Psi(X)$, then : $$\Phi(X)=h\bigg(\Psi(X)\bigg)$$

Now $\Phi(X)$ is a known function since $\Psi(X)$ has been expressed. We put $\Phi(X)$ into the above general solution where $X=\int_{0}^y g(\mu)d\mu+\int_{x_0}^x\frac{d\nu}{f(\nu)}$ $$\boxed{u(x,y)=h\left(\Psi\left(\int_{0}^y g(\mu)d\mu+\int_{x_0}^x\frac{d\nu}{f(\nu)}\right)\right)}$$ This is the particular solution which satisfies both the PDE and the condition.

Note that obtaining the function $\Psi$ from the function $1/f(x)$ involves an integration and then an inversion which is the most difficult part of the task. The most often this is problematic. Certainly few functions $f(x)$ are nice enough to finally get an analytical result.

Comment:

One might be surprised that an arbitrary constant $x_0$ seems to remains in the final equation. As a matter of fact the final result doesn't depends on the choice of starting point $x_0$ for the integration. Try with some examples of explicit functions $f(x)$. You will obseved that $x_0$ vanishes along the process of inversion.