How Can I Solve This Non-Linear PDE in 2D?

[L0r3n20]

Hi all. I'm trying to solve this PDE but I really can't figure how. The equation is
<br /> f(x,y) + \partial_x f(x,y) - 4 \partial_x f(x,y) \partial_y f(x,y) = 0<br />
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?

 

[pasmith]

Hi all. I'm trying to solve this PDE but I really can't figure how. The equation is
<br /> f(x,y) + \partial_x f(x,y) - 4 \partial_x f(x,y) \partial_y f(x,y) = 0<br />
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?

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,
<br /> \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}<br />
which by comparison with your equation gives the following system:
<br /> \frac{\mathrm{d}x}{\mathrm{d}t} = 1 \\<br /> \frac{\mathrm{d}y}{\mathrm{d}t} = -4\frac{\partial f}{\partial x} \\<br /> \frac{\mathrm{d}f}{\mathrm{d}t} = - f<br />
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
<br /> \frac{\mathrm{d}y}{\mathrm{d}t} = 4f \\<br />
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.