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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?

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    $\begingroup$ The key test case is probably when the ball $B$ is replaced by a half-space (and the lower order term is deleted); this test case has to be true if your question is to hold, by a scaling argument. One should probably try a reflection argument in the half-space case; if it works, it should be adaptable to the ball setting. $\endgroup$
    – Terry Tao
    Jan 22, 2015 at 1:48
  • $\begingroup$ Terry, thanks! Obvious after you said it. I'll try that. $\endgroup$
    – Deane Yang
    Jan 22, 2015 at 2:13
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    $\begingroup$ In the half-space case one can also try taking a Fourier transform in the tangential directions, this appears to reduce things to a 1D problem on a half-line which should be directly solvable. (To adapt this sort of method to the ball though may require some Fourier integral operators or something to replace the Fourier transform.) $\endgroup$
    – Terry Tao
    Jan 22, 2015 at 2:46
  • $\begingroup$ Terry, perhaps I'm trying to do the reflection too naively but I can't get it to work. If I reflect the function about the hyperplane $x^n = 0$, then the argument seems to work only if $A^{in}_{ab} = 0$ for $1 \le i \le n-1$. Or if $u=0$ along the hyperplane. $\endgroup$
    – Deane Yang
    Jan 23, 2015 at 0:32
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    $\begingroup$ In addition to my answer: the integral inequality, when stated for fields vanishing at the boundary, is called quasi-convexity. When stated for every fields, John Ball call it quasi-convexity up to the boundary. $\endgroup$ Jan 24, 2015 at 13:29

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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.

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  • $\begingroup$ Many thanks! I haven't had a chance to study your paper carefully, but it looks good. I've learned that elliptic systems of PDE's can have much more subtle behavior than scalar elliptic PDE's. $\endgroup$
    – Deane Yang
    Jan 24, 2015 at 14:57
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    $\begingroup$ Yes indeed. There is also an old paper, with too a wide scope to answer to your question, by Agmon, Douglis & Nirenberg. There is also the seminal work of Lopatinskii. $\endgroup$ Jan 24, 2015 at 15:09
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    $\begingroup$ @Denis: for those of us less well-versed, can you indicate which of the works of Lopatinskii is the seminal one you refer to? Thanks in advance. $\endgroup$ Jan 24, 2015 at 21:21
  • $\begingroup$ @Willy. Lopatinskii worked on elliptic equations and systems, as did later ADN. Actually, I am not an elliptic, but an hyperbolic PDEist. Nevertheless, Lopatinskii's idea applies also to the hyperbolic context, as quoted by R. Sakamoto and H.-O. Kreiss in the early 70'. It is amazing that when a conference was organized in Russia for the Lopatinskii's centenary, no hyperbolist was invited ! All this to say that I don't have a precise answer to your query. Sometimes, for hyperbolic boundary value problem, we speak of a KSL condition, instead of just a Lopatinskii condition. $\endgroup$ Jan 25, 2015 at 7:11
  • $\begingroup$ Complement about elasticity: genuine materials obey to nonlinear equations, because linearity is not compatible with objectivity and the fact that energy tends to infinity as $\det\nabla u\rightarrow0^+$. It might be possible that for some state, quasi-convexity holds in the classical sense, but not "up to the boundary". Such states can happen in the bulk of the elastic body, if it is suitably deformed ; but they cannot be observed on its surface, where they would be Hadamard unstable. $\endgroup$ Jan 26, 2015 at 10:32
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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)'.

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  • $\begingroup$ Thanks! That is indeed a nice and simple counterexample. $\endgroup$
    – Deane Yang
    Feb 1, 2015 at 23:20

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