# Tag Info

1

Certainly not, if $C$ is supposed to be real. For instance, suppose that $\rho_1$ and $\rho_2$ are probability densities such that $\rho_1\rho_2=0$. Then the left-hand side of your inequality is $\infty$, whereas its right-hand side is $2C$.

3

I doubt this bound can be improved much, at least for even $n$. Indeed: set $z_1 = z_2 = \ldots = z_k = k^{-1/2}$ and $z_{k+1} = z_{k+2} = \ldots = 0$, so that the $\ell^2$ norm of $(z_n)$ is $1$. (Intuitively, this is the worst-case scenario.) Then $$g(\mu) = (E_1(k^{-1/2} \mu))^k = (1 - k^{-1/2} z)^k e^{z \sqrt k} .$$ As $k \to \infty$, the above ...

2

Miscellaneous results. If $A$ is strictly upper triangular, then $x\cdot\nabla$ consists only is terms $x_j\partial_k$ with $j<k$. The action of $L$ over homogenous polynomials of degree $d$ is described, in the basis of monomials written in lexicographic order, by a strictly upper triangular matrix. hence the only eigenvalue is $\lambda=0$. In general \...

5

It is true and well-known (assuming $F$ is e.g. uniformly elliptic). The idea is sketched in ch. 9 of the book by Caffarelli-Cabre, but I am not sure of a precise reference at this level of generality. Interior smoothness follows from interior Schauder estimates (applied to difference quotients of $u$ and its derivatives, successively). To extend to the ...

9

Calderon-Zygmund theory generalises without much difficulty to doubling metric measure spaces (or more generally to "spaces of homogeneous type"). See for instance Chapter 1 of Stein, Elias M., Harmonic analysis: Real-variable methods, orthogonality, and oscillatory integrals. With the assistance of Timothy S. Murphy, Princeton Mathematical Series....

5

If you apply the maximum principle, at a point $p$ where the function $v$ reaches its minimum, you get $-\lambda^2 v(p) \geq \lambda^2$ so $v(p) \leq -1$. In particular, the function $u$ is not globally defined as it has to go to $-\infty$ at least at $p$.

3

Say that $L=\partial_x^2$ (the heat equation). Your problem is ill-posed in all reasonnable context : it does not admit a solution for generic data taken in spaces $L^2, C^\infty$ or even in distributional spaces, although it does when the data are analytic (Cauchy-Kowalevska). Let us consider the simplified situation where $F\equiv0$ and $(0,T)$ is replaced ...

1

Partial answer (the compact case for $n=3,p=2/3$ only): Let $M=\mathbb T = \mathbb R/(2\pi\mathbb Z)$ and let $\phi$ be a solution [Footnote 1] of the Klein-Gordon equation at the top of the question with $n=3, p=2/3$ such that $\phi(t_0, \cdot),\frac{\partial\phi}{\partial t}(t_0,\cdot)\in C^{4,\gamma}(\mathbb T^3)$ for some $\gamma>0$ and some $t_0>... 1 You are using the wrong space, that's all. The correct space is $$X=\{u \in H^1_{\textrm{loc}}(\mathbb R^2): u(\cdot+n)=u(\cdot) \quad \forall n=(n_1,n_2)\in \mathbb Z^2\}.$$ Then you apply Lax-Milgram and you are good to go. Look at Asymptotic Analysis for Periodic Structures, by Bensoussan, Lions, Papanicolaou (1979) chapter 1, for example. Answering ... 9 The short answer is No. A major problem is that there is no single universal definition for what it means for an equation to be solvable by separation of variables. This defect in the theory, as well as the lack of general statements of the form that you would like to see, was highlighted a while ago in this BAMS book review: Tom H. Koornwinder. "... 2 $$\tilde{u}_{tt} - \frac{2}{t}\tilde{u}_{t}-\Delta \tilde{u} = g_t -\frac{2}{t^2}\tilde{u}$$ $$\dot{E} = 2 \int \tilde{u}_t (2 t^{-1} \tilde{u}_t + g_t - 2 t^{-2} \tilde{u} )$$ $$\dot{E} = 2 \int \tilde{u}_t (2 t^{-1} \tilde{u}_t + g_t) - 2 t^{-2} \frac{d}{dt} \int \tilde{u}^2$$ $$\dot{E} + \frac{d}{dt} (2 t^{-2} \int \tilde{u}^2 ) = 2 \int \tilde{u}... 5 Absolutely not! Taking the difference v=u_1-u_2, you see that v(x,t) solves$$ \begin{cases} (L-\partial_t) v=0 & \mbox{for }(x,t)\in(0,1/2)\times(0,T]; \\ v(0,t)=f(t) & \mbox{for }x=0,\,t\in(0,T]\\ v(1/2,t)=0 & \mbox{for }x=1/2,\,t\in(0,T]\\ v(x,0) =0 & \mbox{for }x\in (0,1/2),\,t=0 \end{cases} $$for some left boundary data f=f_1-f_2... 1 Let's focus on N=1 only, the case N>1 is just a tensorization of the argument below. The whole argument is actually unrelated to the specific (aggregation-diffusion) PDE or gradient flow: As soon as you have a curve of probability measures \mu_t satisfying the continuity equation$$ \partial_t\mu_t+\nabla\cdot(v_t\mu_t)=0 $$for some (smooth enough) ... 6 Parabolic regularity show that u is regular for all positive times; in particular u(t,\cdot) \in W^{1,2}(M) for all t > 0. Interior parabolic estimates additionally show that there is a constant C = C(g,T) > 0 so that \lvert u(t,\cdot) \rvert_{W^{1,2}} \leq C \lvert f \rvert_{L^2} \quad \text{ for all t \in [T/2,2T].} \end{... 0 No, we cannot. Formally, \varphi is the eigenfunction of the unbounded operator L_s on L^2(\Omega), defined initially by$$ L_s u(x) = (-\Delta)^s u(x) \qquad \text{for } x \in \Omega , $$where u \in C_c^\infty(\Omega) (and it is understood that u(x) = 0 for x \notin \Omega), and then extended to an appropriate domain (e.g. by means of ... 2 I you are not really interested about viscosity solutions, but just the "philosophical reason" why the solution of the problem with inhomogeneous exterior condition can be written in terms of the heat kernel (as suggested by the comments), here is an answer. Choose z \notin \overline\Omega and define$$ u_z(x) = \int_0^\infty \int_\Omega e^{-\... 0 There seems to be a weakness to the approach with your (generalized) differential equation for$v$, because the Laplacian of the indicator function$\Delta\mathbf{1}_{\partial\Omega}$is not actually singular enough to produce a meaningful boundary condition on the surface$\partial\Omega$. We have that$\Delta\mathbf{1}_{\partial\Omega}=0$wherever it is ... 4 In dimension two, the Rado-Kneser-Choquet theorem explains how to choose boundary data to obtain a non vanishing Jacobian in the interior. Lewy's Theorem shows that harmonic one-to-one mappings have non vanishing jacobians. J. C. Wood one page paper called "Lewy's Theorem Fails in Higher Dimensions" gives a counter-example in dimensions larger or ... 8 To rule out the trivial case that one of the$f_j$is a nonzero constant, I interpret the linear independence condition as saying that no nontrivial linear combination of the$f_j$is constant. The set$\{\det J = 0\}$is not necessarily finite. Consider for example$u_1 = r\sin(\theta)$and$u_2 = r^2\sin(2\theta)$in$B_1 \subset \mathbb{R}^2\$, which both ...

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