Search Results

This operator has been studied in the paper by Miloslav Znojil, PT-symmetric square well, Phys. Lett. A, 285 (2001) 7-10, and then by the same and many other authors. However I did not find out whethe …

Since we are in the plane I use complex notation. The general solution of your equation is the sum of the potential and an arbitrary harmonic function: $$u(z)=\frac{1}{2\pi}\int\int\log|z-\zeta| f(\ze …

This is not true in dimension 2. Function $f(z)=ze^z$ is entire, and $f'(z)=0$ at one point, $z=-1$. It follows that the function $$u(x,y)=\mathrm{Re}f(x+iy)=e^x(x\cos y-y\sin y)$$ is harmonic, has on …

First, some general background. For a bounded domain, the boundary value problem is solved by the Poisson formula, however an explicit form of the Poisson kernel for a cylinder of finite length is pro …

To separate contribution of the "ends" and the lateral surface, write $w=u+v$, where $u$ has zero boundary conditions on the lateral surface, and $v$ is zero on the ends. The estimate $|u|\leq Ce^{-kt …

This question was addressed by Hansen and Nadirashvili in a series of papers, see, for example: MR1315353 Hansen, W., Nadirashvili, N., On Veech's conjecture for harmonic functions. Ann. Scuola Norm. …

You have to say what is the meaning of $\Delta u$, and of $\int\Delta u$. For the integral to have a meaning, $u$ has to be a distribution and $\Delta u$ has to be a (signed, Radon) measure. Such dist …

This follows from the so-called (Eberhard) Hopf Minimum Principle. If you have a positive (super-) harmonic function $u$ in a ball, and $u(z_0)=0$ for some boundary point $z_0$, then the normal deriva …

Just prove it yourself: Take $r=1$. Then $$w(x_0,2^{-n})\leq \lambda^n w(x_0,1)=:C\lambda^n.$$ To estimate $|u(x_0)-u(y)|$, where $y$ is close to $x_0$, choose $n$ so that $|x_0-y|\in[2^{-n-1},2^{-n} …

This is correct, and the same proof as in McOwen and Troyanov should work. In fact they were not the first who proved this result. The story begins with E. Picard, who wrote several papers on this (al …

The answer is no. Let $M$ be a region in the upper half-space $x_1>0$ in $R^n$, (you can take $n=2$) and $\partial M$ contains an open piece $U$ of the plane $x_1=0$. Take $f=0$ in $U$. Then your harm …

If they are talking about the Laplace operator, this statement is true only in dimension 2. And this is only sufficient, not necessary. In general, for solvability of the classical Dirichlet problem, …

No. Take any $u$ which is not zero, but compactly supported in $\Omega$. Then define $f=\Delta u$; it will be also compactly supported, and non-zero.

You must specify more exactly what do you mean by "singular analytic" (what singularities are allowed). Some version of this problem was investigated in arXiv:1309.5483, and in the literature mention …

Equation $\Delta u=0$ is called the Laplace equation, btw. Edit. The answer to your question is no. Consider $f(z)=1/(z+i)$. On the real line it belongs to $L^p$ with any $p>1$. In the upper half-pl …