solve partial differential equation

Question

$$y^2u_x + xu_y = \sin(u^2) \\ u(x,0)=x$$

I get the projected characteristic curve on xy plane easily. However, cannot get the other one.

actually the problem is getting the value of $U_{xx}, U_{xy}, U_{yy}, U_x, U_y$ on x-axis. How can I do this?

Answer 1

$y^2u_x+xu_y=\sin u^2$

$\dfrac{u_x}{x}+\dfrac{u_y}{y^2}=\dfrac{\sin u^2}{xy^2}$

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

$\dfrac{dy}{dt}=\dfrac{1}{y^2}$ , letting $y(0)=0$ , we have $y^3=3t$

$\dfrac{dx}{dt}=\dfrac{1}{x}$ , letting $x(0)=x_0$ , we have $x^2=2t+x_0^2$

$\dfrac{du}{dt}=\dfrac{\sin u^2}{xy^2}=\dfrac{\sin u^2}{3^\frac{2}{3}~t^\frac{2}{3}\sqrt{2t+x_0^2}}$ , we have $\int^u\csc u^2~du=\int^t\dfrac{dt}{3^\frac{2}{3}~t^\frac{2}{3}\sqrt{2t+x_0^2}}+f(x_0^2)$ , i.e. $\int^u\csc u^2~du=\int^\frac{y^3}{3}\dfrac{dt}{3^\frac{2}{3}~t^\frac{2}{3}\sqrt{2t+x_0^2}}+f\left(x^2-\dfrac{2y^3}{3}\right)$