Integrodifference equation

In mathematics, Gårding's inequality is a result that gives a lower bound for the bilinear form induced by a real linear elliptic partial differential operator. The inequality is named after Lars Gårding.

Statement of the inequality

Let Ω be a bounded, open domain in n-dimensional Euclidean space and let H^k(Ω) denote the Sobolev space of k-times weakly differentiable functions u : Ω → R with weak derivatives in L². Assume that Ω satisfies the k-extension property, i.e., that there exists a bounded linear operator E : H^k(Ω) → H^k(Rⁿ) such that (Eu)|_Ω = u for all u in H^k(Ω).

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

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

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

$\sum _{\left|\alpha \right|,\left|\beta \right|=k}{\xi }^{\alpha }{A}_{\alpha \beta }(x){\xi }^{\beta }>\theta |\xi {|}^{2k}\text{ for all }x\in \mathrm{\Omega },\xi \in {\mathbb{ℝ}}^n\setminus \{0\}.$

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

$A_{\alpha \beta }\in L^{\mathrm{\infty }}(\mathrm{\Omega })\text{ for all }\left|\alpha \right|,\left|\beta \right|\le k.$

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

$B[u,u]+G\Vert u{\Vert }_{L^2(\mathrm{\Omega })}^2\ge C\Vert u{\Vert }_{H^k(\mathrm{\Omega })}^2\text{ for all }u\in H_0^k(\mathrm{\Omega }),$

where

$B[v,u]=\sum _{0\le \left|\alpha \right|,\left|\beta \right|\le k}\underset{\mathrm{\Omega }}{\int }{A}_{\alpha \beta }(x){\mathrm{D}}^{\alpha }u(x){\mathrm{D}}^{\beta }v(x)\mathrm{d}x$

is the bilinear form associated to the operator L.

Application: the Laplace operator and the Poisson problem

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

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

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

$B[u,v]=⟨f,v⟩\text{ for all }v\in H_0^1(\mathrm{\Omega }),$

where

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

$⟨f,v⟩=\underset{\mathrm{\Omega }}{\int }f(x)v(x)\mathrm{d}x.$

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

$B[u,u]\ge C\Vert u{\Vert }_{H^1(\mathrm{\Omega })}^2-G\Vert u{\Vert }_{L^2(\mathrm{\Omega })}^2\text{ for all }u\in H_0^1(\mathrm{\Omega }).$

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

$B[u,u]\ge K\Vert u{\Vert }_{H^1(\mathrm{\Omega })}^2\text{ for all }u\in H_0^1(\mathrm{\Omega }),$

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.

References

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