# Non Linear PDE in 2 dimensions

1. Dec 18, 2012

### L0r3n20

Hi all. I'm trying to solve this PDE but I really can't figure how. The equation is
$$f(x,y) + \partial_x f(x,y) - 4 \partial_x f(x,y) \partial_y f(x,y) = 0$$
As a first approximation I think it would be possible to consider $$\partial_y f$$ a function of only y and $$\partial_x f$$ a function of only x but even in this case I couldn't find a general solution.
Any idea?

Last edited: Dec 18, 2012
2. Dec 18, 2012

### pasmith

Your equation needs to be supplemented by a boundary condition: say $f(x,g(x)) = h(x)$ for suitable $g(x)$.

The method of characteristics looks like a good bet.

By the chain rule,
$$\frac{\mathrm{d}f}{\mathrm{d}t} = \frac{\partial f}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t} + \frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}t}$$
which by comparison with your equation gives the following system:
$$\frac{\mathrm{d}x}{\mathrm{d}t} = 1 \\ \frac{\mathrm{d}y}{\mathrm{d}t} = -4\frac{\partial f}{\partial x} \\ \frac{\mathrm{d}f}{\mathrm{d}t} = - f$$
subject to the initial conditions $f(0) = f_0 = h(x_0)$, $x(0) = x_0$, $y(0) = y_0 = g(x_0)$ so that $f(x_0,g(x_0)) = h(x_0)$.

Solving the first equation gives $x = t + x_0$, and the third gives $f = f_0e^{-t} = f_0e^{x_0-x}$. Substituting these into the second gives
$$\frac{\mathrm{d}y}{\mathrm{d}t} = 4f \\$$
so that $y = y_0 + 4f_0(1 - e^{-t})$.

Therefore given a characteristic starting at $(x_0,g(x_0))$, the value of the function at $(x,y) = (x_0, g(x_0) + 4h(x_0)(1 - e^{-t}))$ is $h(x_0)e^{-t}$.

It is of vital importance that the curve $(x,g(x))$ on which the boundary condition is given is not a characteristic (ie a curve $(x(t),y(t)$) for some $(x_0,y_0)$). There may also be a problem if characteristics intersect.