It is given that $u_t=2u_{xx}-3u ,\hspace{0.3cm} u_x(0,t)=0=u_x(1,t)$ Use the Young's inequality to show that the energy $$E[u,u_x](t):=\frac{1}{2}\int_0^1(|u|^2+|u_x|^2)dx$$ satisfies the differential inequality

$$\frac{\partial}{\partial t}E(t)\leq -2\int_0^1|u_{xx}|^2dx-3E(t)$$

My problem:Please help me to prove this. I tried it many times but I failed.If you can prove with another method other than youngs inequality then also please share it.Thanks in advance.


I don't think you need Young's inequality. Just differentiate under the integral sign, integrate by parts to get rid of $u_{xt}=u_{tx}$, and then use the differential equation to substitute for $u_t$. You will want to integrate by parts once again to rewrite the $uu_{xx}$ integrals.

  • $\begingroup$ You're forgetting to use the boundary conditions given in the question, so when you integrate by parts, those terms go away. So you correctly end up with $\int_0^1 (-5u_x^2-3u^2-2u_{xx}^2)\,dx \le -2\int_0^1 u_{xx}^2\,dx - 3E(t)$. $\endgroup$ – Ted Shifrin Jan 24 '15 at 17:38
  • $\begingroup$ Yes you are 100% correct $\endgroup$ – Flip Jan 24 '15 at 17:40
  • $\begingroup$ sorry but how I can make this $$-3\int_0^1u^2dx-5\int_0^1u_x^2\leq -3E(t)$$?? $\endgroup$ – Flip Jan 24 '15 at 17:42
  • $\begingroup$ $-5\int_0^1 u_x^2\,dx \le -3\int_0^1 u_x^2\,dx$ :P $\endgroup$ – Ted Shifrin Jan 24 '15 at 17:43
  • $\begingroup$ -:( Thanks a lot. $\endgroup$ – Flip Jan 24 '15 at 17:44

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