PDEMOC5

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$u_x + xu_y = u^2\,$

The characteristics are:

$\frac{dx}{dt}=1\,$ , $\frac{dy}{dt}=x\,$ , $\frac{dz}{dt}=z^2\,$

The intial curve was not given so assume $\Gamma=(0,0,z_0)\,$ .

The solution to the first DE is:

$x(s,t) = t + c_1(s)\,$

$x(s,0) = c_1(s) = 0\,$

$x=t\,$

The second DE is:

$\frac{dy}{dt}=x=t\,$

$y(s,t) = \frac{1}{2}t^2 + c_2(s)\,$

$y(s,0) = c_2(s) = 0\,$

$y=\frac{1}{2}t^2=\frac{1}{2}x^2\,$

The solution to the third DE is:

$z(s,t) = \frac{-1}{t+c_3(s)}\,$

$z(s,0) = \frac{-1}{c_3(s)}=z_0\,$

$c_3(s) = -z_0^{-1}\,$

$z=\frac{z_0}{1-z_0x}\,$

We now must find two new functions of x,y,z that are constant along gamma.

Let $\phi(x,y,z) = y-\frac{1}{2}x^2\,$ and $\psi(x,y,z)=z^{-1}+x\,$ .

So the solution is $y-\frac{1}{2}x^2 = f(x+\frac{1}{z})\,$ where $f\,$ is an arbitrary function as long as $f(\frac{1}{z_0})=0\,$ because those are the values along $\Gamma\,$ .

Main Page : Partial Differential Equations : Method of Characteristics

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