# PDEMOC14

Consider $\displaystyle uu_x+u_y=0\,$ with the IC $\displaystyle u(x,0)=h(x)=\begin{cases} u_0>0, & x\le 0 \\ u_0(1-x), & 0 .

Show that a shock develops at a finite time and describe the weak solution.

The chracteristic line is $\displaystyle x=uy+x_0\,$ .

To see this, note that the characteristics are $\displaystyle \frac{dx}{dy}=u\,$ so $\displaystyle x(y)=uy+c_1, x(0)=c_1=x_0\,$ and $\displaystyle x=uy+x_0\,$ .

Examine the middle of the region.

For $\displaystyle 0 ,

$\displaystyle u=u_0(1-x_0)=u_0(1-x+uy)\implies u=\frac{u_0(1-x)}{1-u_0y}\,$

Now we know

$\displaystyle u(x,y) = \begin{cases} u_0>0, & x\le 0 \\ \frac{u_0(1-x)}{1-u_0y}, & 0 .

So $\displaystyle y=\frac{1}{u_0}\,$ is the shock and the characteristics intersect at $\displaystyle (1,\frac{1}{u_0})\,$ . Next we'd like to define $\displaystyle G(u)=\frac{1}{2}u^2\,$ (in this case) so that the pde is in the form $\displaystyle [G(u)]_x+u_y=0\,$ .

The jump condition taken from the far left side to the far right side is:

$\displaystyle \xi'(y) = \frac{G(u_r)-G(u_l)}{u_r-u_l} = \frac{u_0}{2}\,$

$\displaystyle \xi(y) = \frac{u_0}{2}y + c_2\,$

Plug in the shock $\displaystyle (1,\frac{1}{u_0})\,$ to get $\displaystyle 1=\xi(y)=\frac{1}{2}+c_2\implies c_2=\frac{1}{2}\,$ .

Finally, $\displaystyle \xi(y)=\frac{u_0}{2}y+\frac{1}{2}\,$ .

The weak solution is $\displaystyle u(x,y) = \begin{cases} u_0, & x<\frac{u_0}{2}y+\frac{1}{2} \\ 0, & x>\frac{u_0}{2}y+\frac{1}{2}\end{cases}\,$ .