9

Let $A$ be a self-adjoint operator with domain $D(A)\subset\mathcal H$ ($\mathcal H$ is some Hilbert space). An operator $C$ with $D(A)\subset D(C)$ is called relatively compact with respect to $C$ if $C(A-zI)^{-1}$ is compact for some (hence all) $z\notin\sigma(A)$. Paraphrasing Corollary 2, page 113 Section XIII.4, in [1], we have
If $C$ is relatively ...

5

My impression is that a PDE, or rather a class of PDEs, is interesting to the analysts when it is relevant for some analytical tool. Let me take a few examples. The list is not exhaustive.
linear constant coefficient PDEs in the whole space ${\mathbb R}^n$ are treated with Fourier analysis.
Elliptic (scalar) PDEs are analyzed with the maximum principle. ...

5

They are not. First of all, the existence of a convex entropy is not meaningful for a system given in this quasi-linear form. The reason is that you might make a change $v=\phi(u)$ of unknown, but the convexity is not preserved by composition by the diffeomorphism $\phi$.
In addition, if $n\ge3$, a generic quasi-linear system does not admit conservation ...

4

No, entropy convexity and hyperbolicity are not equivalent conditions. A necessary and sufficient condition for the system of differential equations to possess a strictly convex entropy is that the system is symmetrizable and hence hyperbolic. The symmetrizability condition is stronger than the condition of hyperbolicity, a system may have real eigenvalues ...

4

Of course not, because this equation is far from being elliptic. Actually, it is even under-determined, in the sense that you have only one equation, for $n^2$ unknowns (where the matrix is $n\times n$).
Let me however give you a result in this direction, that I discovered two years ago, which has important consequences in various domains.
Let $A$ be ...

3

It is indeed elementary with some slight maneuvering:
Since $s' < r' \leq r$, there exists $\alpha \in (0,1]$ such that \begin{equation}\|f\|_{r'} \lesssim \|f\|_{s'}^{1-\alpha} \, \|f\|_r^\alpha.\label{1}\tag{1}\end{equation}
Hence (by the $S^\theta_{r,s}$ assumption)
\begin{equation}\|f\|_{r'} \lesssim \|f\|_{s'}^{1-\alpha} \, \|f\|_s^{\alpha(1-\theta)}...

3

The ABP estimate indeed holds in your setting. The key is that the concave envelope of $u$ is in $C^{1,\,1}$, so the area formula is valid for its gradient. Assuming for simplicity that $L = \Delta$, that $\Omega = B_1$ and that $\sup_{\partial B_1} u = 0$, the way I would argue is:
Let $\Gamma$ be the concave envelope (the infimum of linear functions ...

2

To make my comment an answer: No. To see this, fix a cube $K$ and look at $L^p_K=\{f\in L^p: \text{ ess-supp}(f) \subseteq K\}$. On this space, the $L^p_{comp}$-topology coincides with the $L^p$-topology (this follows from the stricness of the inductive limit) and the $W^{k,p}$-topology is strictly finer on this subspace.

fa.functional-analysis ap.analysis-of-pdes sobolev-spaces limits-and-colimits topological-vector-spaces

2

Yes, the inequality is true. Let $\lambda$ be large enough so $(L+\lambda I)$ is positive definite. We have $$((L+\lambda I)y,y)=((L+\lambda I)^{1/2}y,(L+\lambda I)^{1/2}y)\ge C\|y\|^2_{H^{s/2}},$$
since the domain of $(L+\lambda I)^{1/2}$ is $H^{s/2}$ by the general theory of interpolation spaces. It follows that
$$(Ly,y)\ge C\|y\|^2_{H^{s/2}}-\lambda (y,y)....

2

The problem is precisely that you should not fix a priori the parameter $c$, because it will be exactly the Lagrange multiplier (or rather, $\lambda=-c$ will).
Think of it like this: Let me introduce a new parameter $R>0$, which I think of as a prescribed $L^2$ energy level.
Then the minimization problem
$$
\min\limits_{u\in(\dots)}\Bigg\{E(u):\qquad V(u)=...

1

You can find an overview of methods to obtain conservation laws from a wave equation in On the structure of conservation laws of (3+1)-dimensional wave equation. Noether's method requires that the PDE follows from a variational principle for a Lagrangian (as pointed out by Willie Wong). A direct algorithmic method to obtain conservation laws from a PDE ...

1

You get almost everywhere convergence for both first and second derivatives. In general, no uniform convergence can be expected. Take for example $\Omega=(-1,1)\subset\mathbb{R}$ and $u(x)=|x|$. Then $Du=\frac{x}{|x|}$ is discontinuous and $D^2u=2\delta_0$ in the sense of measures. On the other hand, both $Du_k$ and $D^2u_k$ are continuous functions, hence ...

1

I believe that a zero eigenvalue of $A(u)$ implies that the vector field $\partial_t$ is a characteristic direction. For instance, the equation $u_s + u_y = 0$ takes the form $u_t=0$ after switching to the coordinates $(t,x)=(s,y-s)$, where the spatial surfaces $t=$const and $s=$const are the same, while $\partial_t = \partial_s + \partial_y$. So perhaps you ...

1

Starting from the heat equation with a diffusion constant $D$,
$$u_t - D u_{xx} = 0,\;\;0<x<L,$$
rescale length and time as $x'=x/\lambda$, $t'=t/\tau$, $D'=D\tau/\lambda^2$, to obtain the dimensionless equation
$$u_{t'} - D' u_{x'x'} = 0,\;\;0<x'<L/\lambda.$$
So if you wish the domain in $x'$ to scale to infinity and infinite time $t$ to be ...

1

If $M$ is compact this is obviously false: if we can write $\phi=u\sqrt{-1}\partial\overline{\partial}u$ then $\int_M \omega\wedge\phi\leq 0$ integrating by parts. So for every $\phi$ which does not satisfy this (for example $\phi=\omega$) it will not be possible to find such $u$.

1

There will certainly be a weak-* convergent subsequence because $L^\infty(0, \infty; H)$ is the dual of the separable Banach space $L^1(0, \infty; H)$ and so its bounded sets are weak-* metrizable and relatively compact.
You can't expect an a.e. convergent subsequence though; this is not even true for $L^\infty(0, \infty; \mathbb{R})$ (consider $f_n(x) = ...

1

I find that the condition $A\eta+f(\eta) \ge 0$ is necessary and sufficient. Let us assume that $f$ is globally Lipschitz and depends only on $y$, so that there is no problem about global existence. Let us have in mind that $X$ is an $L^p$ space or a space of continuous functions.
1) Assume first that $\eta=0$. If we take $y_0=0$, then $y'(0)=f(0)$; if $f(0)...

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