method of characteristics implicit equation for u?

Question

Trying to understand how method of characterists is implemented here[ Method of Characteristics, Quasilinear pde as given in the second solution

why does $x=f(u^t)+u(e^t-1)$

imply $u=e^{-t}F(x+u(1-e^t))$ ?

Answer 1

The notation in that answer is indeed a little confusing. $f$ is a function which takes the initial value of the function $u_0$ to the $x$-coordinate where that value begins propagating, $x_0$. The initial condition is $u_0=-\frac{x}{2}$, so there is a bijection between the starting points $x_0$ and the point-wise initial values $u_0$. $f$ is the bijection $f:u_0\mapsto x_0$.

Once that is clear, they compute $$ x = f(ue^t)+u[e^t-1], $$ which becomes $$ f(ue^t)=x-u[e^t-1]=x+u[1-e^t]. $$ Now, because $f$ is a bijection, there exists a bijective inverse, $f^{-1}$. Call this inverse $F$, and take $F$ of the last line, obtaining $$ ue^t=F(x+u[1-e^t]), $$ or $$ u=e^{-t}F(x+u[1-e^t]). $$

I hope that helps!