$L^p$ contraction of scalar conservation law solution As we know, the following scalar conservation law:
$$\partial_t u+\partial_xf(u)=0$$
where $f:\mathbb{R}\to\mathbb{R}$ locally Lipschtz, its entropy solutions satisfies $L^1$ contraction:
If $u,v$ are bounded entropy solutions, then
$$\int_{-\infty}^{\infty}|u(t,x)-v(t,x)|dx\leq\int_{-\infty}^{\infty}|u(0,x)-v(0,x)|dx$$
This is a well-known conclusion, but does it also hold in $L^p$ sense? That is:
$$\int_{-\infty}^{\infty}|u(t,x)-v(t,x)|^pdx\leq\int_{-\infty}^{\infty}|u(0,x)-v(0,x)|^pdx$$
where $p>1$.
I saw someone write in his paper that it holds only if the equation is linear $(f''\equiv 0)$. Is this correct, and how to prove it?
 A: When $f$ is linear, we have $u_t + au_x + b = 0$, with solution $u(t, x) = u(0, x-at) - bt$. In this case $\|u(t,\cdot)-v(t,\cdot)\|_{L^p}$ is independent of $t$. 
When $f$ is nonlinear, instead of transport at constant speed $a=f'(u)$, we have transport at the speed $f'(u)$ that depends on $u$. Imagine $u$ and $v$ as two waves that initially occupy the same position, except $v$ is slightly higher. The maximum of the difference $|v-u|$ is small initially, but for positive times the location of the two waves is different, with one moving ahead of the other. So $\max|u-v|$ will increase. And since $L^\infty$ norm is the limit of $L^p$ norms as $p\to\infty$, at least some of those norms will increase too. 
The above is of course vague so let's consider an explicit counterexample. The easiest case to analyse is Burgers' equation $x_t+uu_x=0$, which corresponds to $f(u)=u^2/2$.  Let $a<b$ be two positive constants and consider initial conditions
$$
u(0, x) = \begin{cases} 0\quad & x < 0 \\ a\quad & 0< x< 1 \\ 0\quad & x > 1 \end{cases}
$$
$$
v(0, x) = \begin{cases} 0\quad & x < 0 \\ b\quad & 0< x< 1 \\ 0\quad & x > 1 \end{cases}
$$
For $t>0$ we have shock waves $x=1+at/2$ for $u$ and $x=1+bt/2$ for $b$. There are also rarefaction waves $0<x<at$ for $u$ and $0<x<bt$ for $b$. Consider positive time $t>0$ that is small enough that $bt < 1+bt/2$; that is, rarefactions do no catch up with shocks. Then 
$$
u(t, x) = \begin{cases} 0\quad & x < 0 \\ x/t\quad & 0< x< at \\ a\quad & at< x<1+at/2 \\ 0\quad & x > 1+at/2 \end{cases}
$$
$$
v(t, x) = \begin{cases} 0\quad & x < 0 \\ x/t\quad & 0< x< bt \\ b\quad & bt< x<1+bt/2 \\ 0\quad & x > 1+bt/2 \end{cases}
$$
The integral $\int |u-v|^p\,dx$ splits into three, 
$$
\int_{at}^{bt} (x/t - a)^p\,dx  = t\frac{(b-a)^{p+1}}{p+1}
$$
$$
\int_{bt}^{1+at/2} (b - a)^p\,dx  = (b-a)^p (1+at/2-bt)
$$
$$
\int_{1+at/2}^{1+bt/2} (b - 0)^p\,dx  = b^p t (b-a)/2
$$
This is a linear function of $t$; we want to know if its slope is positive. The slope (the coefficient of $t$) is 
$$
k(a, b) = \frac{(b-a)^{p+1}}{p+1} + (b-a)^p (a/2-b) + b^p (b-a)/2
$$ 
When $p=1$, we have $k(a,b)=0$. When $p>1$, it looks complicated but considering $a$ fixed and $b\to a^+$ we find that the first two terms are $o(b-a)$, and thus 
$$
k(a,b) = b^p (b-a)/2 + o(b-a) > 0
$$ 
when $b$ is sufficiently close to $a$. (Note that this means the bulk of the difference comes from the front edge of the wave, where $v$ moves ahead of $u$.) 
This demonstrates that for a suitable choice of $a,b$ the $L^p$ norm of $u-v$ increases with time, at least until the rarefaction catches the shock wave. 
