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How to solve a highly nonlinear PDE

GROUPS:
I have a highly nonlinear PDE, in a form
D[f[x,y],x]^2==G
where f[x,y] is the function I wanna solve. G is some function of f[x,y], D[f[x,y],x], x and y.
So,there are two solutions:
D[f[x,y],x]=-Sqrt[G],and D[f[x,y],x]=+Sqrt[G]
after specify some initial conditions, I can first solve the first one (- sign), numerically. then at some point, this solution terminates because of singularies or something else (the solution, which is a interpolating function, will turn from a real value to a complex value). The terminating point is a line y0=y0. At this point I want to change the equation to the (+ sign) one, with the new initial conditions given by f[x,y0[x]].
How can I do the above steps?
Especially, first, how can I get this terminating line y0=y0; second, how can I specify f[x,y0] as the new initial conditions?

By the way, how to make Mathematica solve PDE only in real number field?
POSTED BY: Yiyang Zhang
Answer
1 year ago
Because there is in general no closed form solution for non-linear PDE/ODE, you want to be very specific about the actual function that you are dealing with. A good idea is to give a set of real values for demonstration. 
POSTED BY: Shenghui Yang
Answer
1 year ago
By default, NDSolve, or NDSolveValue, will solve numerically in complex numbers. how can I restrict in real numbers?
POSTED BY: Yiyang Zhang
Answer
1 year ago
This is not the same as FindRoot, which only works on the real domain. Differential equation by default needs complex numbers, even for simple second-order linear systems, I do not understand why you want to work on real domain only.  
POSTED BY: Shenghui Yang
Answer
1 year ago
Without any more information, I think the best way forward is to extract the solutions with both roots. You can then write a short function which chooses the correct root based on whether it is returning a real result or not. 

A simple example of the kind of problem you are looking to solve would be helpful in demonstrating how to work with it.
POSTED BY: Sean Clarke
Answer
1 year ago