For the discretization of gradient, if I set $$ \mathcal \nabla := \begin{bmatrix}1&-1&\\&1&-1\end{bmatrix},$$ then $$ \mathcal \nabla^T\mathcal\nabla := \begin{bmatrix}1&0\\-1&1\\0&-1\end{bmatrix}\begin{bmatrix}1&-1&\\&1&-1\end{bmatrix}=\begin{bmatrix}1&-1&0\\-1&2&-1\\0&-1&1\end{bmatrix}$$ It seems that $\mathcal \nabla^T\mathcal\nabla=-\Delta$, not $\mathcal \nabla^T\mathcal\nabla=\Delta$. Is this true? If yes, could you explain why it holds?

I am supposed to prove the following: $$4\frac{\partial}{\partial z}\frac{\partial}{\partial\bar{z}}=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\,\,\,$$ Using the definitons $$\frac{\partial}{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x} + \frac{1}{i}\frac{\partial}{\partial y}\right)$$ and $$\frac{\partial}{\partial \bar{z}}=\frac{1}{2}\left(\frac{\partial}{\partial x} – \frac{1}{i}\frac{\partial}{\partial y}\right),$$ I get that $$4\frac{\partial}{\partial z}\frac{\partial}{\partial\bar{z}}=\frac{\partial^2}{\partial x^2}-\frac{1}{i}\frac{\partial^2}{\partial x\partial y}+\frac{1}{i}\frac{\partial^2}{\partial y\partial x}+\frac{\partial^2}{\partial y^2}.$$ This leaves me with a few questions. Assume $f:\Omega\to\mathbb{C}$, and let $u(x)=\textrm{Re}(f(x))$ and $v(x)=\textrm{Im}(f(x))$. Then if $\frac{\partial^2 […]

Knowing: $f(z)$ is analytical Prove: $$(\frac{\partial^2}{\partial^2 x} + \frac{\partial^2}{\partial^2 y})|f(z)|^2 = 4|f'(z)|^2$$ I have proved firstly that $\ln|f(z)|$ is harmonic function Let $$f(z) = u(x,y) + i v(x,y)$$ And I transformed what to prove into: $$u(\frac{\partial^2 u}{\partial^2 x} + \frac{\partial^2 u}{\partial^2 y}) + v(\frac{\partial^2 v}{\partial^2 x} + \frac{\partial^2 v}{\partial^2 y}) = 0$$ However, when I […]

How can I decide which are the connected components by looking at the eigenvectors of laplacian matrix ? I have the following adjacency matrix (all nodes are different), and have created the Laplacian matrix. I calculated the eigenvalues: $λ_1 = 0$, $λ_2 = 0$, $λ_3 = 1$ which mean there is 2 connected components. The […]

I would like to evaluate this integral over the surface of a sphere in 3D: $$ \int_{\partial D} \int_{\partial D} \frac{1}{|x-y|}dxdy. $$ It seems there is a lot of symmetry in this integral so I imagine there is a good chance there is an explicit solution. However, I usually deal with 2D Helmholtz problems so […]

In cartesian coordinates, the Laplacian is $$\nabla^2 = \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\qquad(1)$$ If it’s converted to spherical coordinates, we get $$\nabla^2=\frac{1}{r^2}\frac{\partial}{\partial r}\left( r^2 \frac{\partial}{\partial r}\right)+\frac{1}{r^2 sin\theta}\frac{\partial}{\partial \theta}\left(sin \theta \frac{\partial}{\partial \theta}\right)+\frac{1}{r^2 sin^2 \theta}\frac{\partial^2}{\partial \phi^2}\qquad(2)$$ I am following the derivation (i.e. the method of conversion from cartesian to spherical) in “Quantum physics of atoms, molecules, solids, nuclei […]

Question : How could I compute the (wave) kernel from the fact I have already found (wave) trace on unit circle? The definitions are related to the page $25$ of the following pdf. As the Spectrum$(S^1)=\{n^2 : n\ \in \mathbb{N}^*\}$, the trace (It this relevant for the question?) as distribution is simply $$w(t)=\sum_{k \geq 1} […]

In my previous question I asked about evaluating the following integral over the surface of a sphere in 3D: $$ \int_{\partial D} \int_{\partial D} \frac{1}{|x-y|}dxdy, $$ and it turned out that the answer is $(4\pi)^2$ for a unit sphere. Now what about the case where $x$ and $y$ are not on the surface of the […]

It is well known that for Dirichlet problem for Laplace equation on balls or half-space, we could use the green function to construct a solution based on the boundary data. For instance, one could find a nice proof in Evans PDE book, chapter 2.2, it is called the Poisson’s formula. Now, it comes to my […]

I am looking to find the radial part of Laplace’s operator in three dimensions. I looked up Laplace’s operator in spherical coordinates and from there I guess the radial part is: $\frac{\partial^2}{\partial r^2} + \frac{2\: \partial}{r\:\partial r}$. Is this correct and is there a better, quicker way to find the radial part?

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