PDEMOC14

From Exampleproblems

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

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

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

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

Examine the middle of the region.

For $0<x_0<1\,$ ,

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

Now we know

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

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

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

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

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

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

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

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

Main Page : Partial Differential Equations : Method of Characteristics

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