Quasilinear Equation but with non-zero initial condition?

[LieToMe]

The way I was taught to solve many quasi-linear PDEs was by harnessing the initial condition in the characteristic method at $u(x,0) = f(x)$. What if however I need use alternative initial conditions such as $u(x,y=c) = f(x)$ for some constant $c$? Can the solution be propagated the same way?

 

[BvU]

What do you think of shifting the $y$-axis and solving two different initial value problems (one backwards equation) ?

 

[LieToMe]

What do you think of shifting the $y$-axis and solving two different initial value problems (one backwards equation) ?

I think that sounds reasonable but I want to make sure.

 

[pasmith]

The way I was taught to solve many quasi-linear PDEs was by harnessing the initial condition in the characteristic method at $u(x,0) = f(x)$. What if however I need use alternative initial conditions such as $u(x,y=c) = f(x)$ for some constant $c$? Can the solution be propagated the same way?

Yes - you can use $(x, v = y - c)$ as your independent variable instead of $(x,y)$. (Or use $(x, c - y)$ if you need to work back to $y = 0$.)