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*}


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 $(0,t_{*})$.

  • $\begingroup$ What do we know about $u$? Is it $C^1$? or is it in $H^1$? $\endgroup$ – Tomás Oct 15 '13 at 13:16
  • $\begingroup$ 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 $L^{1}$. $\endgroup$ – Andrew Oct 15 '13 at 14:07
  • $\begingroup$ But the inequality is almost everywhere? $\endgroup$ – Tomás Oct 15 '13 at 14:13
  • $\begingroup$ Yes, the inequality holds almost everywhere. $\endgroup$ – Andrew Oct 15 '13 at 16:20
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    $\begingroup$ If $u$ is only in $L^1$ then $u(0)$ is not defined. Then how can you say $u(t)\leq e^{Lt}u(0)$? $\endgroup$ – timur Oct 22 '13 at 13:33

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