PDEMOC4

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$u_x + 2u_y = u^2, u(x,0)=h(x)\,$

The parameterized curve at $t=0\,$ is $\Gamma(s,0,h(s))\,$ . This comes from the symbols $x,0,h(x)\,$ in the IC.

The characteristics are $\frac{dx}{dt}=1\,$ , $\frac{dy}{dt}=2\,$ , $\frac{dz}{dt}=z^2\,$ .

These differential equations give $x(s,t)=t+c_1(s)\,$ , $y(s,t)=2t+c_2(s)\,$ , $-z(s,t)^{-1}=t-c_3(s)^{-1}\,$ .

At $t=0\,$ , $x=c_1(s)=s\,$ , $y=c_2(s)=0\,$ , and $z=-c_3(s)^{-1}=h(s)\,$ .

So the solutions are $x=t+s\,$ , $y=2t\,$ , and $z=-1/(t-h(s)^{-1})\,$ .

Therefore $s=x-t\,$ , $t=y/2\,$ , and $z=u(x,y)=\frac{h(x-y/2)}{1-\frac{1}{2} y h(x-y/2)}\,$ .

Main Page : Partial Differential Equations : Method of Characteristics

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