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On page 87 of the book Hyperbolic Conservation Laws in Continuum Physics by C. M. Dafermos, there is a theorem which I summarise as follows

Theorem. (Theorem 4.5.2 in the book.) Let $U$ be a weak solution to the conservation law $\partial_t U + \text{div }G(U) =0,$ with initial data $U_0$ and $U\in\mathcal O\subseteq \mathbb R^n, x\in \mathbb R^m, G$ an appropriate matrix-valued function. Suppose that $U_0 - \bar U \in L^2$ for some constant $\bar U,$ and $\bar U$ satisfy the entropy condition for entropy $\eta.$ If we normalise $\eta$ so that $\eta(\bar U ) =0, D\eta(\bar U) =0,$ then $$S(t) = \int_{\mathbb R^m} \eta (U(x,t)) dx$$ is a decreasing function of $t.$

I am worried about the statement of this theorem because the only assumption made on $G$ is that it is smooth in the image of $U.$ (This is stated at the beginning of the chapter.) No other technical assumptions are given. This seems insufficient to control the growth of $Q,$ and there are problems that integrals of $Q$ diverging.

If we look into the proof, there is a sentence: "fix $s$ sufficiently large so that $s\eta \geq |Q|$ for all output of the function $U.$" I do not know why such $s$ must exist. We know that $DQ_\alpha = D\eta \cdot DG_\alpha,$ by the definition of entropy flux. So if $G_\alpha$ grow very fast, we cannot expect $Q$ to be dominated by $\eta.$ Apparently some bound on $G$ need to be assumed; but nearly nothing is assumed in the book. (Of course we can normalise so that $Q(\bar U) =0$. But if we do so, there is no guarantee that $U$ will always stay close to $0$ at infinity.)

So, are we implicitly assuming some conditions on $G$, which are not stated in the book?

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I just quickly read the proof you mentioned, and I think what is meant is following:

  1. Note that in Section 4.3 it is noted that any weak solution $U$ may be renormalized to be a continuous (in weak* topology) mapping from $[0,T)\to L^\infty$.
  2. For the argument, one proves a statement to be true on $[0,T)$ by proving it true on $[0,\tau]$ for all $\tau\in [0,T)$. The interval $[0,\tau]$ is compact.
  3. This suggests that $\{Q(U(t,x)): (t,x)\in [0,\tau]\times \mathbb{R}^m\}$ should be essentially bounded. (We only need the comparison to hold a.e. since the desired result is integrated.)
  4. You should be more worried then about where $\eta$ is small, and not where $Q$ is large; this is resolved by the convexity of $\eta$.
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  • $\begingroup$ If point 3 holds then I have nothing to worry about. A weak* convergent sequence is automatically bounded due to uniformly bounded principle. From this, the weak* continuity of $U$ implies that $L^\infty$ norm for fixed times are locally bounded. Therefore, $U$ is essentially bounded on $[0,\tau] \times \mathbb R^n$ since $[0,\tau]$ is compact. Is this the way to prove it? $\endgroup$
    – Ma Joad
    Jul 14, 2021 at 14:06
  • $\begingroup$ More or less; I would start with the image of $U|_{[0,\tau]}$ being a compact set in $L^\infty$ with weak* topology and use essentially what you wrote to argue it must be norm bounded. But that's mostly stylistic. $\endgroup$ Jul 14, 2021 at 15:39
  • $\begingroup$ Thank you very much! $\endgroup$
    – Ma Joad
    Jul 14, 2021 at 23:52

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