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I came across the following weak differential inequality while looking through F.Otto's paper on $L^{1}$ contraction and uniqueness of quasilinear elliptic-parabolic equation:

\begin{align*} - \int_{0}^{T} \alpha'(t) u(t) \; dt \leq \alpha(0) u(0) + L \int_{0}^{T} \alpha(t) u(t) \; dt \end{align*} holds for all non-negative $\alpha \in C^{\infty}_{0}((-\infty, T))$. Additional information on $u$ are: (1) $u \in L^{1}(0,T)$ and (2) the paper extends $u$ to negative times by $u(t) := u(0)$ for $t < 0$.

Can someone show me how one can derive a Gronwall-type result: \begin{align*} u(t) \leq \exp(Lt) u(0). \end{align*}

Thanks.

I have made some progress, but it is not quite there yet. My inspiration comes from a post about weak ODEs (Finding Weak Solutions to ODEs): We can rewrite the differntial inequality as \begin{align*} - \int_{0}^{T} e^{-Lt} (e^{Lt}\alpha)' u \; dt \leq \alpha(0) u(0). \end{align*} Define $\psi := e^{Lt} \alpha \in C^{\infty}_{0}((-\infty, T))$ and since $\alpha$ is non-negative, $\psi$ is also non-negative. Moreover, $\psi(0) = \alpha(0)$. So we have \begin{align*} \int_{0}^{T} e^{-Lt} \psi'u \; dt \geq - \alpha(0) u(0) = - \psi(0) u(0) = \int_{0}^{T} u(0) \psi' \; dt. \end{align*} Hence, \begin{align*} \int_{0}^{T} (e^{-Lt} u - u(0))\psi' \; dt \geq 0 \end{align*} holds for all non-negative $\psi \in C^{\infty}_{0}((-\infty, T))$. If $\psi'$ was non-positive over $(0,T)$, then one can deduce that $e^{-Lt} u - u(0) \leq 0$ and then we obtain the desired inequality for $u$ for almost all $t$.

However, my approach only works when $\psi' \leq 0$ over $(0,T)$. I don't see how this can be true for all test function $\psi$ of the form $e^{Lt}\alpha$ where $\alpha \in C^{\infty}_{0}((-\infty,T))$. Since $\alpha \in C^{\infty}_{0}((-\infty, T))$ its derivative $\alpha'$ must be non-positive for $\alpha$ to have compact support. But if $\alpha$ is a smooth function with a maximum at some $0 < t_{*} < T$, then $\alpha'$ would be positive in $(0,t_{*})$.

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  • What do we know about $u$? Is it $C^1$? or is it in $H^1$? – Tomás Oct 15 '13 at 13:16
  • If the derivative $u'$ exists, then we can put the derivative on $(e^{-Lt}u - u(0))$ to conclude that $(e^{-Lt}u - u(0))' \leq 0$ and then use the Gronwall inequality to deduce $e^{-Lt}u - u(0) \leq 0$. But in this paper, I believe $u$ is only in $L^{1}$. – Andrew Oct 15 '13 at 14:07
  • But the inequality is almost everywhere? – Tomás Oct 15 '13 at 14:13
  • Yes, the inequality holds almost everywhere. – Andrew Oct 15 '13 at 16:20
  • Why do you need it to hold "for all test function $\psi$" of that form? You just need one right. – hopo2 Oct 19 '13 at 18:01
  • My attempt merely translated the inequality that holds for all non-negative $\alpha \in C^{\infty}{0}$ to an equivalent inequality that holds for all non-negative $\psi \in C^{\infty}{0}$. From this "variational" inequality I want to deduce the conclusion of the classical Gronwall inequality. But it seems that I cannot make the assertion that $$\int_{0}^{T} (e^{-Lt}u - u(0))\psi' ; dt \geq 0$$ for all non-negative $\psi \in C^{\infty}_{0}(-\infty, T)$ implies $(e^{-Lt}u - u(0)) \leq 0$ for almost all $t$. – Andrew Oct 20 '13 at 11:33
  • Is the following meaningful? First forget about $u(0)$ in your last inequality because $\int_{-\infty}^T \psi'=0$. Then put $v(t)=e^{-Lt}u(t)$. The function $v$ satisfies $\int v\psi'\geq 0$ for all nonnegative test functions $\psi$. This means that the distributional derivative of $v$ is a non-positve measure, say $-\mu$. Hopefully, $v$ and $\widetilde v(t)=-\int_{]-\infty ,t)} d\mu$ have the same distributional derivative, i.e. $-\mu$. So $v$ and $\widetilde v$ are almost everywhere equal, and hence $v$ is equal a.e. to a non-increasing function. This should give the required result. – Etienne Oct 20 '13 at 21:24
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    If $u$ is only in $L^1$ then $u(0)$ is not defined. Then how can you say $u(t)\leq e^{Lt}u(0)$? – timur Oct 22 '13 at 13:33

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