Search Results

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 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 …

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 …

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 …

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: …

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 …

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 …

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 …

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, …

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( …

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 …

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 ${ …

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 …

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 …

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 …