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Show that for weak solutions of \begin{aligned} &u_t + f(u)_x = 0 \quad\text{in}\quad \mathbb{R}\times (0,\infty) \\ &u=u_0 \quad\text{on}\quad t=0 \end{aligned} The entropy condition $$\dot{s}(|u_L-k|-|u_R-k|)\leq \text{sign}(u_L-k)(f(u_L)-f(k)) - \text{sign}(u_R-k)(f(u_R)-f(k))$$ for all $k \in \mathbb{R}$ is equivalent to the Oleinik condition $$\frac{f(k)-f(u_R)}{k-u_R}\leq \frac{f(u_L)-f(u_R)}{u_L-u_R} \leq \frac{f(k)-f(u_L)}{k-u_L}$$ for all $k$ between $u_L$ and $u_R$. ($\dot{s}=\frac{f(u_L)-f(u_R)}{u_L-u_R}$)

I have tried to solve this but I am stuck:

$\text{sign}(u_L-k)=-\text{sign}(u_R-k)$

$|u_L-k|-|u_R-k|=\text{sign}(u_L-k)(u_L-l)-\text{sign}(u_R-k)(u_R-k)=\text{sign}(u_L-k)(u_L+u_R-2k)$

$\text{sign}(u_L-k)(f(u_L)-f(k))-\text{sign}(u_R-k)(f(u_R)-f(k))=\text{sign}(u_L-k)(f(u_L)-f(k)+f(u_R)-f(k))$

Then with $h:=\text{sign}(u_L-k)$ the entropy condition becomes:

$\dot{s}(h(u_L+u_R-2k))\leq h(f(u_L)+f(u_R)-2f(k))$

For $h=-1$ (so $u_L<u_R$) I get:

$-\dot{s}(2u_R-2k)\leq -\dot{s}(u_L+u_R-2k)\leq -(f(u_L)+f(u_R)-2f(k))$ So

$\frac{f(u_L)+f(u_R)-2f(k)}{(2u_R-2k)}\leq \dot{s}$

This already looks kindof similar to the right inequality. I would need that $f(u_L)>f(u_R)$ to get the right inequality, but I don't that this has to be the case. Any advice?

EditPiAf
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Kekks
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1 Answers1

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One way to show this is going a step back into abstraction and realize that $\vert u - k \vert$ is exactly the Kruzkhov entropy function $\eta(u) \colon = \vert u - k \vert$, and that $\text{sign}(u - k) \big(f(u) - f(c)\big) $ is the corresponding Kruzkhov entropy flux $q(u)$. Then, you can rewrite your given statement as \begin{align} \dot s \big( \eta(u_L) - \eta(u_R) \big) &\leq q(u_L) - q(u_R)\\ -\dot s \big( \eta(u_R) - \eta(u_L) \big) q(u_R) - q(u_L) & \leq 0 \end{align}

Shamefully citing myself, I showed in this post how the formula above leads to the Oleiniks entropy condition. You can probably follow the steps along for your special type of entropy function & flux.

Dan Doe
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