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Consider uu_{x}+u_{y}=0\, with the IC u(x,0)=h(x)={\begin{cases}u_{0}>0,&x\leq 0\\u_{0}(1-x),&0<x<1\\0,&x\geq 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_{0}y}}\,

Now we know

u(x,y)={\begin{cases}u_{0}>0,&x\leq 0\\{\frac  {u_{0}(1-x)}{1-u_{0}y}},&0<x<1\\0,&x\geq 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}}\,.

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