Does the Legendre-Hadamard condition imply a generalized Gårding inequality?

Question

For simplicity, we restrict to constant coefficients. Let $A^{ij}_{ab} \in \mathbb{R}$, $1 \le i, j \le n$ and $1 \le a, b\le m$, satisfy the Legendre-Hadamard condition: $$ A^{ij}_{ab}\xi_i\xi_jv^av^b \ge \lambda |\xi|^2|v|^2 $$ for some $\lambda > 0$ and any $\xi \in \mathbb{R}^n$ and $v \in \mathbb{R}^m$. Let $B$ be the unit ball in $\mathbb{R}^n$.

It is straightforward to use the Fourier transform to prove that there exists $c > 0$ such that given any $u \in C^\infty_0(B,\mathbb{R}^m)$, $$ \int_{B} A^{ij}_{ab}\partial_iu^a\partial_ju^b \ge c\int_B |\partial u|^2. $$ If the (stronger) Legendre condition $$ A^{ij}_{ab}p_i^ap_j^b \ge \lambda\sum_{i,a}(p_i^a)^2 $$ for any $p_i^a \in \mathbb{R}$ holds, it is easy to use an extension operator to extend the inequality to any function $u \in C^\infty(\overline{B},\mathbb{R}^m)$.

Question: Does the Legendre-Hadamard condition imply a Gårding inequality of the form $$ \int_{B} A^{ij}_{ab}\partial_iu^a\partial_ju^b \ge \int_B c|\partial u|^2 - c'|u|^2, $$ for any $u \in C^\infty(\overline{B},\mathbb{R}^m)$? If not, what is a counterexample?

Answer 1

The answer is No, and this is an interesting question.

As Terry Tao commented, the test case is when $B$ is replaced by a half-space $H$, so let us consider the latter case for a moment and fields $u$ with compact support in $\bar B$ (i.e. not vanishing at the boundary).

Because $H$ is dilation invariant, a Garding inequality implies actually an inequality of the form $$ \int_H A^{ij}_{ab}\partial_iu^a\partial_ju^b \ge c\int_H |\partial u|^2. $$ So we are led to the question whether such an inequality holds true or not. I analysed this problem in my paper "Second-order initial-boundary value problems of variational type". J. of Functional Analysis, 236 (2006), pp 409-446. The existence of such an inequality is equivalent to a Lopatinskii-type property, which can be expressed in terms of a Lopatinskii determinant associated with the symbol $A$. The technique uses a Fourier transform in the $n-1$ coordinates along the boundary. You will find in Section 6 examples and counter-examples.

I do not resist to give the following counter-example: take the linear isotropic elasticity, satisfying the Legendre-Hadamard condition. It has two wave velocities, $c_S$ for shear (transverse) waves and $c_P$ for pressure (longitudinal) waves. In real media, we have $c_S<c_P$. However, L.-H. does not prevent from the possibility of $c_P<c_S$, in which case the inequality above does not hold. Actually, the lack of inequality reveals the ill-posedness of the Initial-boundary-value problem in the half-space when the boundary condition is Neumann-type (here free displacement).

Now, when the inequality above holds true in every half-space, I proved that the Garding inequality holds true in bounded domains. It is not contained in my article (it was in the submission) because it is essentially an exercise.

Link to my JFA paper. The main result is Theorem 3.5. Isotropic elasticity is treated in Paragraph 6.3, see Theorem 6.2.

Answer 2

Gui-Qiang Chen suggested me to look at your question.

I assume that your notation $\partial u$ means the gradient $\nabla u$. In the following I use $\bar\partial u$ to denote the standard first order operator for complex functions.

There is another simple counter-example in the $2\times 2$ case. Consider the convex functional $I(u)=\int_B|\bar\partial u|^2dx$ for $u=u_1+iu_2$ and write the quadratic form in terms of $u_1$ and $u_2$. Since $|\bar\partial u|^2=|P_{E_{\bar\partial}}\nabla u|^2$, where $P_{E_{\bar\partial}}$ is the orthogonal projection to the subspace of anti-conformal matrices and $\nabla u$ is the gradient of $(u_1,u_2)$, we have, for every rank-one matrix $\xi\otimes\eta$ that $|P_{E_{\bar\partial}} \xi\otimes\eta|^2=|\xi|^2|\eta|^2/2$. It is easy to see that the corresponding coefficient tensor $A$ satisfies the strong L-H condition. However, if one takes any holomorphic function $u$, then $I(u)=0$. This leads to easy counter-examples for every fixed pair of constants $c>0$ and $c^\prime>0$, e.g. $u_n=e^{nz}$. This type of constructions by using subspaces of conformal and anti-conformal matrices in the space of $2\times 2$ real matrices are commonly used in the vectorial calculus of variations related to quasiconvexity and material microstructure.

If $A$ is not a constant tensor, say, $A\in L^\infty$, even under the homogeneous Dirichlet boundary condition, Garding's inequality fails in general. I constructed such an example some years ago: `A counterexample in the theory of coerciveness for elliptic systems. J. Partial Differential Equations 2 (1989), no. 3, 79–82 (MR1026095)'. I also have some recent results on the so-called universal coercivity problem: 'On coercivity and regularity for linear elliptic systems. Calc. Var. PDEs 40 (2011), no. 1-2, 65–97 (MR2745197)'.