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The answer is Yes and is a consequence of Holmgren's uniqueness Theorem. See for instance Theorem 1.1.4 in this text.. The ellipticity serves here to ensure that there does not exist a characteristic …

This is not yet an answer. Yet, two comments and one information. 1- I don't think that the sign of $E$ is an issue. It might even happen that $\int E\,ds\,dt=0$, and you would be entitled to search …

Yes indeed, the maximum principle extends to the non-smooth setting. I am not sure that there is a complete theory, because so many situations can occur. But at least let me mention the following situ …

This is true, as long as your domain depends smoothly upon one real parameter. Say that you are insterested in the $n$ first eigenvalues. Using a Lyapunov-Schmidt procedure, you may reduce to the situ …

This is too long for a comment. Something is wrong in your statement. Because $v\in H^1(\Omega)$, $v|_{\partial\Omega}$ is naturally in $H^{1/2}$, by the trace theorem. On the other hand, when in add …

In one space dimension, the answer is yes, and the eigenvalues are real and simple (Sturm-Liouville theory). In higher space dimension, the answer is negative, because the operator needs not be diagon …

That depends on the elastic model that you deal with. Is it linear (infinitesimal displacements) or non-linear ? Is it isotropic or not ? In general, because elasticity is a system, not an equation, …

According to J. Liouville Sur l'équation aux différences partielles $\partial^2\log\lambda/\partial u\partial v\pm\lambda/2a^2$ J. Maths. Pures & Appl. 18 (1853), pp 71-71, the general solution of thi …

I discussed this question at lunch with my colleague Jean-Claude Sikorav, and we came up with a solution when $d=2$. We first consider a matrix $P$ such that $A\sharp B=PP^T$, then form $A'=P^{-1}AP^ …

The theory of removable singularities began with Laurent Véron. It was continued by H. Brézis and others. When the elliptic equation is linear, the important ingredient is the codimension of the sub …

You can solve the problem with even less regularity than in Rekalo's answer. If $u\in W^{1,p}(\Omega)$, it does not have a normal trace in general. But if you assume in addition that $\Delta u\in L^p( …

Set $G$ a primitive of $g$. Then the solution is a critical point of the functional $E:H^1(\Omega)\rightarrow{\mathbb R}$ defined by $$E[u]:=\int_\Omega \left(\frac12|\nabla u|^2+fu\right)dx-\int_{\pa …

This is actually a system of first-order PDEs, of $2n$ (the dimension of $M$) equations. To see that it is elliptic, let us consider the symplest case of $M={\mathbb R}^{2n}$ with the standard symplec …

In matrix analysis, the Schur complement is an object that you obtain after eliminating a part of the unknowns. It works that way: you have to solve $Mx=b$ where $M$ is a square, invertible matrix. Yo …

To begn with, your Boundary-Value Problem (BVP) is under-determined, because it lacks one boundary condition: because the PDE is elliptic and fourth-order, you need two boundary conditions, not only o …