An interpolation inequality

[gugo82]

It's an exercise from the classic Gilbarg&Trudinger's book on elliptic PDE (i.e. Elliptic Partial Differential Equations of Second Order, Springer)*; I've just slightly modified it and added some hints.

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Exercise:

Let \( \displaystyle \Omega \subseteq \mathbb{R}^N \) be a bounded open set with \( \displaystyle C^1 \) boundary and \( \displaystyle u\in C^2(\overline{\Omega}) \) with \( \displaystyle u=0 \) on \( \displaystyle \partial \Omega \) ; moreover let \( \displaystyle \text{D} u \) , \( \displaystyle \Delta u \) denote respectively the gradient and the Laplacian of \( \displaystyle u \) , i.e. \( \displaystyle \text{D} u=(\partial_1 u,\ldots ,\partial_N u) \) and \( \displaystyle \Delta u =\sum_{i}^N \partial^2_{i,i} u \) .

1. Prove that for each \( \displaystyle \varepsilon >0 \) there exists \( \displaystyle C(\varepsilon) >0 \) s.t.:

\( \displaystyle \lVert \text{D} u\rVert_2^2 \leq \varepsilon \lVert u\rVert_2^2+C(\varepsilon) \lVert \Delta u\rVert_2^2 \)

or, more explicitly, prove that the following inequality holds:

\( \displaystyle \int_\Omega |\text{D} u|^2\ \text{d} x \leq \varepsilon \int_\Omega u^2\ \text{d} x+C(\varepsilon) \int_\Omega (\Delta u)^2 \ \text{d} x \) .

2. Which is the optimal value of \( \displaystyle C(\varepsilon) \) in the previous inequality?

Testo nascosto, fai click qui per vederlo

Hints: Remember that \( \displaystyle |\text{D} u|^2 =\langle \text{D} u,\text{D} u\rangle \) and apply Green's first identity to evaluate \( \displaystyle \lVert \text{D} u\rVert_2^2 \) in terms of \( \displaystyle u \) and \( \displaystyle \Delta u \) ; then prove that it is possible to choose \( \displaystyle C(\varepsilon) >0 \) s.t. the inequality:

(*) \( \displaystyle -t\ s\leq \varepsilon \ t^2+C(\varepsilon)\ s^2 \)

holds for all \( \displaystyle (t,s)\in \mathbb{R}^2 \) and find the optimal value for \( \displaystyle C(\varepsilon) \) ; finally use (*) to conclude.

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* To be precise, it's an exercise from chapter 2.

[dissonance]

In what follows \( \displaystyle \int = \int_{\Omega} \) .

• Writing \( \displaystyle \int \lvert \nabla u \rvert ^2\, dx=\int \nabla u \cdot \nabla u\, dx \) and integrating by parts we get
  
  \( \displaystyle \int \lvert \nabla u \rvert ^2\, dx=\int u(-\Delta u)\,dx \)
  
  since \( \displaystyle u \equiv 0 \) on \( \displaystyle \partial \Omega \) .
  
• Now we fix \( \displaystyle \epsilon \) and turn to the inequality
  
  \( \displaystyle -ts \le \epsilon t^2 + C_{\epsilon} s^2 \) , which we claim being true for all \( \displaystyle t, s \in \mathbb{R} \) if \( \displaystyle C_{\epsilon} \ge (4\epsilon)^{-1} \) .
  
  To prove this we put \( \displaystyle f(t, s)=-\frac{\epsilon t^2+ts}{s^2} \) for all \( \displaystyle t, s \in \mathbb{R}, s\ne 0 \) and apply standard calculus techniques to find the maximum value it attains:
  
  for every fixed \( \displaystyle s \ne 0 \) the partial function \( \displaystyle f_s(t) \) has the only critical point \( \displaystyle t=- {s \over 2 \epsilon} \) , which is its global maximum (it's a downward parabola); evaluating \( \displaystyle f_s(- {s \over 2 \epsilon})=(4 \epsilon)^{-1} \) we find that it is independent of \( \displaystyle s \) .
  So \( \displaystyle (4\epsilon)^{-1} \) is the global maximum of \( \displaystyle f(t, s) \) .
  
• From the previous inequality we get
  
  \( \displaystyle (-\Delta u)u \le \epsilon u^2 + C_{\epsilon} \Delta u^2 \) for any \( \displaystyle C_{\epsilon} \ge (4\epsilon)^{-1} \) ;
  
  the thesis follows by integration. Also, the optimal value for \( \displaystyle C_{\epsilon} \) is \( \displaystyle (4\epsilon)^{-1} \) .

P.S.: @Gugo - This is the second time I work on one of your exercises while having a walk with the dog! The dog isn't very happy, though.

[gugo82]

Correct!


P.S.: Please, give my most sincere apologies to your dog.