This problem is from Sandro Salsa's book (1), page 201.
Example 4.4. Consider the initial value problem $$ \left\{\begin{array}{l} u_{t}+(1-2 u) u_{x}=0 \\ u(x, 0)=\frac{1}{2} \arctan (\pi x) . \end{array}\right. \tag{4.45} $$ We have $q(u)=u-u^{2}$, $q'(u)=1-2 u$, $q''(u)=-2$, and $g(\xi)=\frac{1}{2} \arctan (\pi \xi)$, $g'(\xi)=\pi / 2\left(1+\pi^{2} \xi^{2}\right)$. Therefore, the function $$ z(\xi)=-q''(g(\xi)) g^{\prime}(\xi)=\frac{\pi}{\left(1+\pi^{2} \xi^{2}\right)} $$ has a maximum at $\xi_{M}=0$ and $z(0)=\pi$. The breaking-time is $t_{s}=1 / \pi$ and $$ x_{s}=q'(g(\xi_{M})) t_{s}+\xi_{M}=1 / \pi . $$ Thus, the shock curve starts from $(1 / \pi, 1 / \pi)$. For $0 \leq t<1 / \pi$ the solution $u$ is smooth and implicitly defined by the equation $$ G(x, t, u)=u-\frac{1}{2} \arctan [\pi x-\pi(1-2 u) t]=0 . \tag{4.46} $$ After $t=1 / \pi$, equation (4.46) defines $u$ as a multivalued function of $(x, t)$ and does not define a solution anymore.
The book says that:
We have to insert a shock wave into the multivalued graph [in Fig. 4.16], in a way that the conservation law is preserved. We will see that the correct insertion point is prescribed by the Rankine-Hugoniot condition. It turns out that this corresponds to cutting off from the multivalued profile the two equal area lobes $A$ and $B$ [as described in Fig. 4.17], at time $t=0.7$ (G. B. Whitham equal area rule).
"Indeed, due to the odd symmetry in the variables $x, u$ of $G$ with respect to the point $(t, 0)$, we have here $x=s(t)=t$."
I have one question:
- I do not understand How do you get $x=s(t)=t$ with Whitham equal area rule?
I would be grateful for your help.
(1) S. Salsa, Partial Differential Equations in Action: From Modelling to Theory, 3rd ed., Springer, 2016. doi:10.1007/978-3-319-31238-5
