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Consider the initial boundary value problem for wave equation:

$$\begin{align} & {{u}_{tt}}-\Delta u+u=f\left( x,t \right)\text{ }\left( x,t \right)\in \Omega \times \left( 0,T \right], \\ & u=0\text{ }\left( x,t \right)\in \partial \Omega \times \left( 0,T \right], \\ & u\left( x,0 \right)={{u}_{t}}\left( x,0 \right)=0\text{ }x\in \Omega , \\ \end{align}$$
Derive the following estimates:

$$E\left( u \right)=\int_{\Omega }{\left( {{\left| {{u}_{t}} \right|}^{2}}+{{\left| \Delta u \right|}^{2}}+{{\left| u \right|}^{2}} \right)dx}\le C\left\| f \right\|_{{{L}^{2}}\left( \Omega \times \left[ 0,T \right] \right)}^{2}$$ where $C$ is a constant depending on $T$ and $\Omega$.

My attempt:

Multiply the given wave equation on ${{u}_{t}} $, integrating over $\Omega$, and using the Gauss-Green theorem together with the boundary condition, we obtain $$\begin{align} & \frac{d}{dt}E\left( t \right)=\frac{d}{dt}\int_{\Omega }{f{{u}_{t}}dx}, \\ & E\left( t \right)=\int_{\Omega }{\left( \frac{1}{2}{{\left| {{u}_{t}} \right|}^{2}}+\frac{1}{2}{{\left| \Delta u \right|}^{2}}+\frac{1}{2}{{\left| u \right|}^{2}} \right)dx}=\int_{\Omega }{f{{u}_{t}}dx}. \\ \end{align}$$
By Holder's inequality
$$\int_{\Omega }{f{{u}_{t}}dx}\le {{\left\| f \right\|}_{{{L}^{2}}}}{{\left\| {{u}_{t}} \right\|}_{{{L}^{2}}}}.$$
Need help to continue.

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  • $\begingroup$ Helpful question: have you heard of Gronwall's lemma / inequality? $\endgroup$ – Glitch Nov 24 '15 at 2:33
  • $\begingroup$ I post an answer following your guidence. Thank you, Glitch. $\endgroup$ – Patrick Windance Nov 26 '15 at 14:51
  • $\begingroup$ No problem, I'm glad to be of assistance. Note, though, that I corrected the last line of your inequality to reflect the fact that you need squares and to integrate in time. $\endgroup$ – Glitch Nov 26 '15 at 22:59
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$\begin{align} & \frac{d}{dt}E\left( u \right)=2\int_{\Omega }{{{u}_{t}}{{u}_{tt}}+{{u}_{x}}{{u}_{xt}}+u{{u}_{t}}dx} \\ & \text{ }=2\int_{\Omega }{{{u}_{t}}{{u}_{tt}}-{{u}_{xx}}{{u}_{t}}+u{{u}_{t}}dx}+2\int_{\partial \Omega }{{{u}_{x}}{{u}_{t}}dS} \\ & \text{ }=2\int_{\Omega }{{{u}_{t}}fdx}\le \int_{\Omega }{{{\left| {{u}_{t}} \right|}^{2}}dx}+\int_{\Omega }{{{\left| f \right|}^{2}}dx} \\ & \text{ }\le \int_{\Omega }{\left( {{\left| {{u}_{t}} \right|}^{2}}+{{\left| \nabla u \right|}^{2}}+{{\left| u \right|}^{2}} \right)dx}+\int_{\Omega }{{{\left| f \right|}^{2}}dx} \\ & \text{ }=E\left( u \right)+\int_{\Omega }{{{\left| f \right|}^{2}}dx}. \\ \end{align}$

By Gronwall’s inequality

$E\left( u\left( x,T \right) \right)\le {{e}^{T}}\left( E\left( u\left( x,0 \right) \right)+\int_0^T {{\left\| f(\cdot,t) \right\|}^2_{{{L}^{2}}}} dt \right)={{e}^{T}} \int_0^T {{\left\| f(\cdot,t) \right\|}^2_{{{L}^{2}}}}dt =C\int_0^T {{\left\| f(\cdot,t) \right\|}^2_{{{L}^{2}}}}dt.$

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