Gårding's inequality

Let ${\displaystyle \Omega }$ be a bounded, open domain in ${\displaystyle n}$-dimensional Euclidean space and let ${\displaystyle H^{k}(\Omega )}$ denote the Sobolev space of ${\displaystyle k}$-times weakly differentiable functions ${\displaystyle u\colon \Omega \rightarrow \mathbb {R} }$ with weak derivatives in ${\displaystyle L^{2}(\Omega )}$. Assume that ${\displaystyle \Omega }$ satisfies the ${\displaystyle k}$-extension property, i.e., that there exists a bounded linear operator ${\displaystyle E\colon H^{k}(\Omega )\rightarrow H^{k}(\mathbb {R} ^{n})}$ such that ${\displaystyle Eu\vert _{\Omega }=u}$ for all ${\displaystyle u\in H^{k}(\Omega )}$.

Let L be a linear partial differential operator of even order 2k, written in divergence form

${\displaystyle (Lu)(x)=\sum _{0\leq |\alpha |,|\beta |\leq k}(-1)^{|\alpha |}\mathrm {D} ^{\alpha }\left(A_{\alpha \beta }(x)\mathrm {D} ^{\beta }u(x)\right),}$

and suppose that L is uniformly elliptic, i.e., there exists a constant θ > 0 such that

${\displaystyle \sum _{|\alpha |,|\beta |=k}\xi ^{\alpha }A_{\alpha \beta }(x)\xi ^{\beta }>\theta |\xi |^{2k}{\mbox{ for all }}x\in \Omega ,\xi \in \mathbb {R} ^{n}\setminus \{0\}.}$

Finally, suppose that the coefficients A_αβ are bounded, continuous functions on the closure of Ω for |α| = |β| = k and that

${\displaystyle A_{\alpha \beta }\in L^{\infty }(\Omega ){\mbox{ for all }}|\alpha |,|\beta |\leq k.}$

Then Gårding's inequality holds: there exist constants C > 0 and G ≥ 0

${\displaystyle B[u,u]+G\|u\|_{L^{2}(\Omega )}^{2}\geq C\|u\|_{H^{k}(\Omega )}^{2}{\mbox{ for all }}u\in H_{0}^{k}(\Omega ),}$

where

${\displaystyle B[v,u]=\sum _{0\leq |\alpha |,|\beta |\leq k}\int _{\Omega }A_{\alpha \beta }(x)\mathrm {D} ^{\alpha }u(x)\mathrm {D} ^{\beta }v(x)\,\mathrm {d} x}$

is the bilinear form associated to the operator L.

Be careful, in this application, Garding's Inequality seems useless here as the final result is a direct consequence of Poincaré's Inequality, or Friedrich Inequality. (See talk on the article).

As a simple example, consider the Laplace operator Δ. More specifically, suppose that one wishes to solve, for f ∈ L²(Ω) the Poisson equation

${\displaystyle {\begin{cases}-\Delta u(x)=f(x),&x\in \Omega$ ;\\u(x)=0,&x\in \partial \Omega ;\end{cases}}}

where Ω is a bounded Lipschitz domain in Rⁿ. The corresponding weak form of the problem is to find u in the Sobolev space H₀¹(Ω) such that

${\displaystyle B[u,v]=\langle f,v\rangle {\mbox{ for all }}v\in H_{0}^{1}(\Omega ),}$

where

${\displaystyle B[u,v]=\int _{\Omega }\nabla u(x)\cdot \nabla v(x)\,\mathrm {d} x,}$

${\displaystyle \langle f,v\rangle =\int _{\Omega }f(x)v(x)\,\mathrm {d} x.}$

The Lax–Milgram lemma ensures that if the bilinear form B is both continuous and elliptic with respect to the norm on H₀¹(Ω), then, for each f ∈ L²(Ω), a unique solution u must exist in H₀¹(Ω). The hypotheses of Gårding's inequality are easy to verify for the Laplace operator Δ, so there exist constants C and G ≥ 0

${\displaystyle B[u,u]\geq C\|u\|_{H^{1}(\Omega )}^{2}-G\|u\|_{L^{2}(\Omega )}^{2}{\mbox{ for all }}u\in H_{0}^{1}(\Omega ).}$

Applying the Poincaré inequality allows the two terms on the right-hand side to be combined, yielding a new constant K > 0 with

${\displaystyle B[u,u]\geq K\|u\|_{H^{1}(\Omega )}^{2}{\mbox{ for all }}u\in H_{0}^{1}(\Omega ),}$

which is precisely the statement that B is elliptic. The continuity of B is even easier to see: simply apply the Cauchy–Schwarz inequality and the fact that the Sobolev norm is controlled by the L² norm of the gradient.

• Renardy, Michael; Rogers, Robert C. (2004). An introduction to partial differential equations. Texts in Applied Mathematics 13 (Second ed.). New York: Springer-Verlag. p. 356. ISBN 0-387-00444-0. (Theorem 9.17)