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.

***

Exercise:

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

1. Prove that for each [tex]$\varepsilon >0$[/tex] there exists [tex]$C(\varepsilon) >0$[/tex] s.t.:

[tex]$\lVert \text{D} u\rVert_2^2 \leq \varepsilon \lVert u\rVert_2^2+C(\varepsilon) \lVert \Delta u\rVert_2^2$[/tex]

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

[tex]$\int_\Omega |\text{D} u|^2\ \text{d} x \leq \varepsilon \int_\Omega u^2\ \text{d} x+C(\varepsilon) \int_\Omega (\Delta u)^2 \ \text{d} x$[/tex].

2. Which is the optimal value of [tex]$C(\varepsilon)$[/tex] in the previous inequality?

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

(*) [tex]$-t\ s\leq \varepsilon \ t^2+C(\varepsilon)\ s^2$[/tex]

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

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

[gugo82]

Correct!


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

[dissonance]

In what follows [tex]\int = \int_{\Omega}[/tex].

[*:1fvtnn9c]Writing [tex]\int \lvert \nabla u \rvert ^2\, dx=\int \nabla u \cdot \nabla u\, dx[/tex] and integrating by parts we get

[tex]\int \lvert \nabla u \rvert ^2\, dx=\int u(-\Delta u)\,dx[/tex]

since [tex]u \equiv 0[/tex] on [tex]\partial \Omega[/tex].
[/*:m:1fvtnn9c]
[*:1fvtnn9c]Now we fix [tex]\epsilon[/tex] and turn to the inequality

[tex]-ts \le \epsilon t^2 + C_{\epsilon} s^2[/tex], which we claim being true for all [tex]t, s \in \mathbb{R}[/tex] if [tex]C_{\epsilon} \ge (4\epsilon)^{-1}[/tex].

To prove this we put [tex]f(t, s)=-\frac{\epsilon t^2+ts}{s^2}[/tex] for all [tex]t, s \in \mathbb{R}, s\ne 0[/tex] and apply standard calculus techniques to find the maximum value it attains:

for every fixed [tex]s \ne 0[/tex] the partial function [tex]f_s(t)[/tex] has the only critical point [tex]t=- {s \over 2 \epsilon}[/tex], which is its global maximum (it's a downward parabola); evaluating [tex]f_s(- {s \over 2 \epsilon})=(4 \epsilon)^{-1}[/tex] we find that it is independent of [tex]s[/tex].
So [tex](4\epsilon)^{-1}[/tex] is the global maximum of [tex]f(t, s)[/tex].
[/*:m:1fvtnn9c]
[*:1fvtnn9c]From the previous inequality we get

[tex](-\Delta u)u \le \epsilon u^2 + C_{\epsilon} \Delta u^2[/tex] for any [tex]C_{\epsilon} \ge (4\epsilon)^{-1}[/tex];

the thesis follows by integration. Also, the optimal value for [tex]C_{\epsilon}[/tex] is [tex](4\epsilon)^{-1}[/tex].[/*:m:1fvtnn9c][/list:u:1fvtnn9c]

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.