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My trouble is in finding the solution $u = u(x,y)$ of the semilinear PDE $$x^2u_x +xyu_y = u^2$$ passing through the curve $u(y^2,y) = 1.$

So I started by using the method of characteristics to obtain the set of differential, by considering the curve $\Gamma = (y^2, y, 1)$. I then reparametrize $\Gamma$ by $r\in\mathbb{R}$, as $\Gamma = (r^2, r, 1)$.

Then, to my understanding, I need to solve the set of ODE's $$\frac{dx}{ds}(r,s) = x^2$$ $$\frac{dy}{ds}(r,s) = xy$$ $$\frac{dz}{ds} = z^2$$ with initial conditions $x(r,0) = r^2$, $y(r,0) = r$, and $z(r,0) =r^2$.

And this is where I run into a wall. How can I solve the system of ODE's, when they are in respect to $r$ and $s$? Also, I'm not even sure if I'm going about this in the correct way.

Your thoughts are appreciated, as always. ~Dom

madisonfly
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1 Answers1

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$x^2u_x+xyu_y=u^2$

$xu_x+yu_y=\dfrac{u^2}{x}$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=x$ , letting $x(0)=x_0$ , we have $x=x_0e^t$

$\dfrac{dy}{dt}=y$ , letting $y(0)=y_0$ , we have $y=y_0e^t$

$\dfrac{du}{dt}=\dfrac{u^2}{x}=\dfrac{u^2}{x_0e^t}=\dfrac{u^2e^{-t}}{x_0}$ , letting $u(0)=\dfrac{1}{f\left(\dfrac{x_0}{y_0}\right)}$ , we have $u(x,y)=\dfrac{x_0}{x_0f\left(\dfrac{x_0}{y_0}\right)+e^{-t}}=\dfrac{xe^{-t}}{xe^{-t}f\left(\dfrac{xe^{-t}}{ye^{-t}}\right)+e^{-t}}=\dfrac{x}{xf\left(\dfrac{x}{y}\right)+1}$

$u(y^2,y)=1$ :

$\dfrac{y^2}{y^2f(y)+1}=1$

$f(y)=1-\dfrac{1}{y^2}$

$\therefore u(x,y)=\dfrac{x}{x\left(1-\dfrac{y^2}{x^2}\right)+1}=\dfrac{x^2}{x^2+x-y^2}$

doraemonpaul
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