# Solve Quasilinear PDE with Initial Condition

Here is the question I am having trouble with:

Consider the quasilinear equation $u(x,t)$,

$u_t+uu_x=-\dfrac{1}{2u}$ with the initial condition $u(x,0)=\sin x$:

This is what I have so far:

Assume $u(x,t)$ is a solution.

Consider $h(t)=u(\xi(t),t)$, $h(0)=\sin x_0$

$h'(t)=u_t(\xi(t),t)+u_x(\xi(t),t)\cdot\xi'(t)$

Let $\xi'(t)=u$, then $h'(t)=-\dfrac{1}{2u}=-\dfrac{1}{2h(t)}$ and therefore

$h(t)=e^{-\frac{t}{2}}\sin x_0$.

I'm not really sure where to go from here. Any help would be greatly appreciated!

$u_t+uu_x=-\dfrac{1}{2u}$

$2uu_t+2u^2u_x=-1$

$\dfrac{du}{ds}=-1$ , letting $u(0)=0$ , we have $u=-s$

$\dfrac{dt}{ds}=2u=-2s$ , letting $t(0)=t_0$ , we have $t=t_0-s^2=t_0-u^2$

$\dfrac{dx}{ds}=2u^2=2s^2$ , letting $x(0)=f(t_0)$ , we have $x=f(t_0)+\dfrac{2s^3}{3}=f(u^2+t)+\dfrac{2u^3}{3}$

$u(x,0)=\sin x$ :

$x=f(\sin^2x)+\dfrac{2\sin^3x}{3}$

$f(\sin^2x)=x-\dfrac{2\sin^3x}{3}$

$f(x^2)=\sin^{-1}x-\dfrac{2x^3}{3}$

$f(x)=\sin^{-1}\sqrt{x}-\dfrac{2x^\frac{3}{2}}{3}$

$\therefore x=\sin^{-1}\sqrt{u^2+t}-\dfrac{2(u^2+t)^\frac{3}{2}}{3}+\dfrac{2u^3}{3}$