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I have to know how we can show the following inequality:

$\|u\|_{2}\leq\|u_{0}\|_{2}+\int_{0}^{t}\|u_{t}(t,x)\|_{2}dt$

where

$\|u\|_{2}=\Big(\int_{\Omega}u^{2}dx\Big)^{1/2}$, $u_{0}=u(x,0)$, $u_{t}=\frac{\partial u}{\partial t}$, $\Omega$ is an open bounded domain in $R^{n}$ and $u=u(x,t)$.

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  • $\begingroup$ What is $u_t$? What space does $u$ belongs? Have you tried something? $\endgroup$ – Tomás Sep 29 '13 at 13:06
  • $\begingroup$ Notation is inconsistent. Especially the $\|\cdot\|_2$ seems to have two different meanings (which would be ok if defined properly). Also, what should $u_t(s)$ be? $\endgroup$ – Dirk Sep 29 '13 at 13:08
  • $\begingroup$ Dear friends, please note that I have revised the problem according your questions. $\endgroup$ – Albert Sep 29 '13 at 15:18
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Assuming that $u \in H^1(0, T; L^2(\Omega))$, you have $$u(T) = u(0) + \int_0^T u_t(t) \, dt.$$ By the triangle inequality, we find $$\| u(T) \|_{L^2(\Omega)} \le \| u(0)\|_{L^2(\Omega)} + \left\| \int_0^T u_t(t) \, dt\right\|_{L^2(\Omega)} \le \| u(0)\|_{L^2(\Omega)} + \int_0^T \| u_t(t) \|_{L^2(\Omega)} \, dt. $$

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  • $\begingroup$ Thanks a lot. But how did you write the last inequality? $\endgroup$ – Albert Sep 30 '13 at 11:38
  • $\begingroup$ This follows from simple properties of the Bochner integral, see en.wikipedia.org/wiki/Bochner_integral. $\endgroup$ – gerw Sep 30 '13 at 13:00

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