Two PDE Questions

Question

1. Solve the equation

$$ u_x^3-u_y=0~, $$ with $u(x,0)=2x^\frac{3}{2}$

2. Solve the equation

$$ u=xu_x+yu_y+\frac{1}{2}(u_x^2+u_y^2)~, $$ with $u(x,0)=\dfrac{1}{2}(1-x^2)$

How I solve these problems if I learnt PDE three weeks ago?

Answer 1

$1.$

$u_x^3-u_y=0$

$u_{xy}-3u_x^2u_{xx}=0$

Let $v=u_x$ ,

Then $v_y-3v^2v_x=0$ with $v(x,0)=3\sqrt x$

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

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

$\dfrac{dv}{dt}=0$ , letting $v(0)=v_0$ , we have $v=v_0$

$\dfrac{dx}{dt}=-3v^2=-3v_0^2$ , letting $x(0)=f(v_0)$ , we have $x=-3v_0^2t+f(v_0)=-3v^2y+f(v)$ , i.e. $v=F(x+3v^2y)$

$v(x,0)=3\sqrt x$ :

$F(x)=3\sqrt x$

$\therefore v=3\sqrt{x+3v^2y}$

$v^2=9(x+3v^2y)$

$v^2=9x+27v^2y$

$v^2-27v^2y=9x$

$v^2(1-27y)=9x$

$v^2=\dfrac{9x}{1-27y}$

$v=\pm\dfrac{3\sqrt x}{\sqrt{1-27y}}$

$u_x=\pm\dfrac{3\sqrt x}{\sqrt{1-27y}}$

$u(x,y)=\pm\dfrac{2x^\frac{3}{2}}{\sqrt{1-27y}}+g(y)$

$u_y=\pm\dfrac{27x^\frac{3}{2}}{(1-27y)^\frac{3}{2}}+g_y(y)$

$\therefore\pm\dfrac{27x^\frac{3}{2}}{(1-27y)^\frac{3}{2}}\mp\dfrac{27x^\frac{3}{2}}{(1-27y)^\frac{3}{2}}-g_y(y)=0$

$g_y(y)=0$

$g(y)=C$

$\therefore u(x,y)=\pm\dfrac{2x^\frac{3}{2}}{\sqrt{1-27y}}+C$

$u(x,0)=2x^\frac{3}{2}$ :

$C=0$ and the negative part reject

$\therefore u(x,y)=\dfrac{2x^\frac{3}{2}}{\sqrt{1-27y}}$