8 votes

Wick rotation for Laplace and wave equations

The transformation to imaginary time is used to relate the Green's function of the Laplacian to the Green's function of the d'Alembertian (therefore relating Laplace equation and heat equation). See ...
Carlo Beenakker's user avatar
7 votes
Accepted

Regularity of solution of $(-\Delta + w)f = 0$

As discussed in the comments, I interpret the question as asking about the asymptotics of $f'(r)$, $r\to 0+$, for solutions of $$ -\frac{d^2f}{dr^2} -\frac{2}{r} \frac{df}{dr} + w(r)f(r) = 0 ; \quad\...
Christian Remling's user avatar
6 votes

Weak convergence + convergence of the norm implies strong convergence in Orlicz spaces

There's a famous trick attributed to Minty (sometimes also to Browder and Lions) which allows to recover at least a.e. convergence when you're able to pass to the limit through an increasing non-...
Ayman Moussa's user avatar
  • 2,505
5 votes

Possible research directions in analysis?

I find that students often come in with these kinds of ideas, that almost everything is known, it's hard to do anything, the problems are mostly solved. Then they ask how to best position themselves ...
user378654's user avatar
5 votes
Accepted

A fractional weighted Poincaré inequality

It is not true. Start with a function $u$ which vanishes for $x<0$ and is equal to $1$ for $0 \leq x \leq \frac 12$ and then smooth from $x \geq \frac 12$. The Fourier coefficients behave like $1/n$...
Giorgio Metafune's user avatar
5 votes
Accepted

Stochastic representation of Laplace equation with Neumann boundary condition

Yes, see Section 4.4.2 in "Stochastic Differential Equations, Backward SDEs, Partial Differential Equations" by Pardoux and Rascanu.
Luke's user avatar
  • 280
4 votes
Accepted

Can we approximate a Hölder pdf by higher-order Hölder pdf's?

No, take any $p$ such that $p(x)=|x|^\alpha$ in some neighbourhood of the origin.
Martin Hairer's user avatar
4 votes
Accepted

Reference request: Uniformly elliptic partial differential operator generates positivity preserving semigroup

Check Chapter 4, more specifically Chapter 4.2 in Ouhabaz: Analysis of heat equations on domains (2005). ZBL1082.35003. The approach there goes by the form method which I personally perceive to be ...
Hannes's user avatar
  • 2,100
3 votes

A priori estimates to $u_t - \Delta u = u^2$

I figured out a way to show it, basically by separating the space and consider where $A^+ := \{u>0\}$ and $A^- := \{u<0\}$. Then the RHS reads \begin{equation} \int u^2 \min(u,c) = \int_{\{u\geq ...
mathdoge's user avatar
  • 159
2 votes
Accepted

Decompositions of $\partial_i$ to the radial direction and rotations in higher dimensions

The formula you gave in the beginning is a special case of the vector triple product formula in $\mathbb{R}^3$ $$ a\times (b\times c) = (a\cdot c) b - (a\cdot b) c $$ Let $v$ be an arbitrary vector ...
Willie Wong's user avatar
  • 35.2k
2 votes
Accepted

A question on biharmonic functions

No. Let $w = (1 - |x|^{2 - n})_+$, with $n > 2$. This is globally Lipschitz. It is harmonic when positive, so therefore also biharmonic. it is subharmonic everywhere (max of two harmonic functions ...
user378654's user avatar
2 votes

Regularity of solution of $(-\Delta + w)f = 0$

regarding the case $f\in L^n$ I can give you some suggestion. Since $L^n$ is contained in the Morrey space $L^{1,n-1}$ and you have Laplace operator you can say that $\nabla f $ belongs to BMO and ...
Giuseppe Di Fazio's user avatar
1 vote

$L^2$ regularity theory for elliptic equations: Is there another method other that the difference quotient method? Reference request

You need more than continuous. Take $d=1$ and $f=0$. Then $$ D(a D u)=0 \Rightarrow Du = \frac{C}{a}, $$ therefore $a\in C^0 \Leftrightarrow Du \in C^0$ (and that's it). To convince yourself that $u\...
username's user avatar
  • 2,454
1 vote

Justification for uniqueness of solutions to dispersive PDE

In fact, in exactly the same book as you referred to, exercise 2.24 gives a counter example. Tao himself has commented above. Maybe he is too busy to give more details. Roughly speaking, the reason ...
Qiuye Jia's user avatar

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