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Non Linear PDE in 2 dimensions

by L0r3n20
Tags: dimensions, linear
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L0r3n20
#1
Dec18-12, 03:20 PM
P: 19
Hi all. I'm trying to solve this PDE but I really can't figure how. The equation is
[tex]
f(x,y) + \partial_x f(x,y) - 4 \partial_x f(x,y) \partial_y f(x,y) = 0
[/tex]
As a first approximation I think it would be possible to consider [tex] \partial_y f [/tex] a function of only y and [tex] \partial_x f [/tex] a function of only x but even in this case I couldn't find a general solution.
Any idea?
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pasmith
#2
Dec18-12, 05:35 PM
HW Helper
Thanks
P: 946
Quote Quote by L0r3n20 View Post
Hi all. I'm trying to solve this PDE but I really can't figure how. The equation is
[tex]
f(x,y) + \partial_x f(x,y) - 4 \partial_x f(x,y) \partial_y f(x,y) = 0
[/tex]
As a first approximation I think it would be possible to consider [tex] \partial_y f [/tex] a function of only y and [tex] \partial_x f [/tex] 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 [itex]f(x,g(x)) = h(x)[/itex] for suitable [itex]g(x)[/itex].

The method of characteristics looks like a good bet.

By the chain rule,
[tex]
\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}
[/tex]
which by comparison with your equation gives the following system:
[tex]
\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
[/tex]
subject to the initial conditions [itex]f(0) = f_0 = h(x_0)[/itex], [itex]x(0) = x_0[/itex], [itex]y(0) = y_0 = g(x_0)[/itex] so that [itex]f(x_0,g(x_0)) = h(x_0)[/itex].

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

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

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


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