I'm not very familiar with the Method of Characteristics for PDEs but have been reading up with several *.pdfs and YouTubes in the past few days. Despite that I got stuck on a relatively simple semi-linear Real World case.
The equation is for a heat transfer problem:
$$\frac{1}{v}u_t+u_x=-\alpha u$$
Initial data: $$u(0,x)=u_0$$
where $u(t,x)$ is a reduced temperature ($u>0$), $t$ is time, $x$ is distance travelled through a pipe, $v$ is a flow velocity (constant) and $\alpha$ a constant.
The characteristics are:
$$\mathrm{d}x=v\mathrm{d}t=-\frac{\mathrm{d}u}{\alpha u}$$
Now I need two find two functions, say $\phi$ and $\psi$, along the characteristics, so that:
$$\mathrm{d}\phi = \mathrm{d}\psi=0$$
I chose:
$$\phi=x-vt$$
and:
$$\psi=x+\frac{\ln u}{\alpha}$$
The implicit solution is then supposed to be (with $F$ an arbitrary function):
$$F(x-vt,x+\frac{\ln u}{\alpha})=0$$
But how to render this explicit? And how to apply the initial data?
Any help is much appreciated.
Making use of the implicit function theorem, we can write too $f(x-vt)=x+\dfrac{\ln u}{\alpha}$ an expression making easier the imposing of the initial conditions, $u(0,x)=u_0$
$$f(x)=x+\dfrac{\ln u_0}{\alpha}$$
$$x-vt+\dfrac{\ln u_0}{\alpha}=x+\dfrac{\ln u}{\alpha}$$
So is, $u(t,x)=u_0e^{-\alpha vt}$