TITLE: Unbounded entropy solution to Burgers' equation
QUESTION [13 upvotes]: I need to solve Problem 3.5 - 11 p. 164 of the book Partial Differential Equations by Lawrence C. Evans (2nd ed., AMS, 2010):


Show that
  $$
   u(x,t) = \begin{cases}
       -\dfrac{2}{3}\left(t+\sqrt{3x+t^2}\right); & \text{if } 4x + t^2 >0\\
       0; & \text{if } 4x + t^2<0
     \end{cases}
$$
  is an (unbounded) entropy solution of $u_t + \left(\dfrac{u^2}{2}\right)_x=0$.


Clearly it is easy to see that this is both unbounded and a solution to the given PDE, however I am not sure how to gather any information to say that it satisfies the entropy condition from the solution alone. Any hints would be welcomed.
Edit.
I know the definition (Evans, §3.4.3.b p. 150). A weak solution $u\in L^\infty(\Bbb R\times (0,\infty))$ of the initial value problem $u_t + \left(\dfrac{u^2}{2}\right)_x = 0$ with data $u|_{t=0} = g$ is an entropy solution if

$$
u(x+z,t) - u(x,t) \leq C \left(1 + \frac{1}{t}\right) z \tag{ii}
$$
  for some constant $C>0$ and a.e. $x$, $z \in \Bbb R$, $t>0$, with $z>0$.

Is it the one needed to solve the problem? How to use it?

REPLY [5 votes]: As suggested by @Saad, setting $t=0$ in $u(x,t)$ gives the initial data
$$
u(x,0) = g(x) =
\left\lbrace
\begin{aligned}
&{-{2}}\sqrt{{x}/{3}} & &\text{if}\quad x>0, \\
&0 & &\text{if}\quad x<0 .
\end{aligned}
\right.
$$
Instead of proving directly that $u$ is an entropy-condition-satisfying weak solution, let us derive the entropy solution to this initial-value problem and compare with the proposed $u$. We first derive a classical solution by applying the method of characteristics for short times. As long as the method of characteristics is valid, we have $u=g(x-u t)$. According to the expression of $g$, two cases must be considered:

for $x$ sufficiently small, $g(x)=0$, so that $u(x,t)=0$;
for $x$ sufficiently large, $g(x) = -2\sqrt{x/3}$, so that $u(x,t)={-\frac{2}{3}}\big(t+\sqrt{3x+t^2}\big)$.

Below is a plot of the characteristic curves in the $x$-$t$ plane deduced from the initial data:

An intersection of the characteristic lines is observed around the origin, and the method of characteristics fails there. According to the Lax entropy condition, a shock wave is generated. Its position $x_s(t)$ must satisfy the Rankine-Hugoniot condition to ensure that such a wave is a weak solution. Therefore, the shock speed is given by
$$
x'_s(t) = \frac{1}{2}\left( 0 - \frac{2}{3}\big(t+\sqrt{3 x_s(t)+t^2}\big) \right)
$$
with initial position $x_s(0)=0$, i.e., $x_s(t) = -t^2/4$. By construction, the (unique) entropy solution is therefore
$$
u(x,t) =
\left\lbrace
\begin{aligned}
&{-\frac{2}{3}}\big(t+\sqrt{3x+t^2}\big) & &\text{if}\quad x>-t^2/4, \\
&0 & &\text{if}\quad x<-t^2/4 ,
\end{aligned}
\right.
$$
which ends the proof.