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Non Linear PDE in 2 dimensions 
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#1
Dec1812, 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? 


#2
Dec1812, 05:35 PM

HW Helper
Thanks
P: 990

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_0x}[/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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