I am struggling with the similarity solution method laid out in Evan's PDE to find the entropy solution of a PDE. This is one example we briefly touched on in class but class ended and we never came back to it.
Compute explicitly the entropy solution of the Burger's equation $$ u_{t}+\left(\frac{u^{2}}{2}\right)_{x} = 0, \text{ in }\mathbb{R}\times(0,\infty)$$ $$u(x,0) = g(x), \text{ on }\mathbb{R}\times{t=0}$$ where $g(x)$ is given by $$0, \text{ for } x<0$$ $$1, \text{ for } 0\leq x\leq 1$$ $$0, \text{ for } x>1$$
I can come up with the characteristic ODEs which are $$\frac{dx}{dt}=u$$ $$\frac{du}{dt}=0$$
This gives us $u(x,t) = constant$ and that implies (with the initial data) that $u(x,t) = g(x_{o})$ where $x_{o} = x(0)$. This in turn implies that $x(t)=x_{o}+g(x_{o})t$ as characteristics go. I can then look specifically at the three regions and obtain
$$x_{0} < 0 \Longrightarrow g(x_{0}) = 0 \Rightarrow x(t) = x_{0} $$ $$ 0 < x_{0} < 1 \Longrightarrow g(x_{0}) = 1 \Rightarrow x(t) =x_{0} + t $$ $$ x_{0} > 1 \Longrightarrow g(x_{0}) = 0 \Rightarrow x(t) = x_{0} $$
Now we notice that projected characteristics intersect at $x=0$, therefore we need to put in a shock curve. We want $u_{r} = 1$ and $u_l = 0$ and by utilizing the RH jump condition we have the shock curve $\xi_{1}(t)$ given by $$\xi_{1}'(t) = \frac{\frac{(u_{l}^{2})}{2}-\frac{(u_{r}^{2})}{2}}{u_{l}-u_{r}} = \frac{0 - \frac{1}{2}}{0-1} = \frac{1}{2}$$
Thus our shock curve is given by $\xi_{1}(t) = \frac{1}{2}t$, from here I have no idea what to do anymore. . .
