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See Theorem 1.5 in C. Brauner & B. Nicolaenko

Not necessarily. I mean, it depends upon the torus you consider. Notice that in the case of the standard one ${\mathbb T}^d={\mathbb R}^d/{\mathbb Z}^d$, the answer is positive. But if you torus is ${ …

I apologize for having posted this question too early. I realize that the answer to the first question is negative. Actually suppose that $D=D(0;R)$ is a disk and $f=f(r)$ is a radial function. If a c …

To begin with, the ellipticity condition is useless if you don't ask also that $$\sum_{i,j}a_{ij}\xi_i\xi_j\le M|\xi|^2$$ for some finite constant $M$. Now the resolvant estimate is used to define an …

The fact that both $u_j$ solve the PDE is not an issue. What matters is that both $u_j$ are smooth over the closure $\overline\Omega$, positive in $\Omega$, vanish on $\partial\Omega$ and their norma …

I think not. Set the dimension $n=2$ and consider the elliptic equation $$u_{,11}+u_{,12}+u_{,22}=0$$ in $B_+$. The extension being $u(x)=-u(x_1,-x_2)$, it satifies in $B_-$ the equation $$u_{,11}-u_{ …

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 …

Here is a construction. It elaborates from perturbation analysis of eigenvalues. However it starts from the situation of a non-simple eigenvalue. So, let me start with the standard self-adjoint $L_0=- …

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 …

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 …

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 …

The notion of viscosity solution is based on the comparison principle (an application of the maximum principle), against smooth sub-/super-solutions. This is a not-at-all-straightforward extension of …

Let $A$ be this matrix. Because of the formula $$\int_D\sigma\nabla u\cdot\nabla v\, dx=\sum_{i,j}a_{ij}U_iI_j,$$ ($U$ for voltages of $u$, $I$ for currents of $v$), we see three necessary conditions: …

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 …

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 …