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The variational approach is the way to go. You know that the Laplacian corresponds to minimizing the Dirichlet energy $\int_\Omega |\nabla u|^2 $; the question is how to fit $\int_{\partial \Omega}\beta(u)v\,dH^{N-1}$ into the variational problem. Let's try a generic boundary energy term: $$\int_{\partial \Omega}\Phi(u) \,dH^{N-1}$$ with $\Phi$ to be ...


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The PDE of this form is typically posed in some Sobolev space $W^{1,p}$. At a minimum, we need the weak formulation $$\int (-a(x,u,\nabla u)\nabla \phi + c(x,u,\nabla u) \phi - g\phi)=0 \tag{1}$$ to make sense for $\phi\in C^\infty_0$. But often, to actually use $(1)$ one needs $\phi$ to be somehow based on $u$. So we'd like to allow $\phi$ in a Sobolev ...


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$\def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits}$ $$ \avint_{\partial B(x,t)} \, dS(y) = 1 $$ and therefore $$ \bigl| u(x) - \avint_{\partial B(x,t)} u(y) \, dS(y)\bigr| = \bigl| \avint_{\partial B(x,t)} (u(x) - u(y)) \, dS(y) \bigr| \le \avint_{\partial B(x,t)} |u(x) - u(y)| \, dS(y) \\ \le \max_{y \in \partial B(x,t)} \{ |u(x) - u(y)| \} \quad . $$ ...


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Counterexample: let $u_1(x,t) = 0$, $u_2(x,t) = t$.


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Given $\phi$, let $$ A=\{u\in W^{1,p};u-\phi\in W^{1,p}_0\} $$ be the admissible set. The solution $u$ is the unique minimizer of the Dirichlet energy $$ I(v)=\int_\Omega|\nabla v|^p $$ in the set $A$. (This is a standard result in calculus of variations.) Now, let $M=\sup\phi$ and $m=\inf\phi$ and define $$ v(x) = \begin{cases} M, & u(x)>M\\ m, ...


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What you have agrees with what I got. To increase the value of this answer, here's an animation of the solution:


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If you have $P(x)y''+Q(x)y'+R(x)y=0$ then it has a singularity when $\frac{Q(x)}{P(x)}$ and $\frac{R(x)}{P(x)}$ are not defined May be there is a simple and single definition for Degeneracy but i am not aware or able to word what i know properly hopefully the following helps. Degeneracy is a more complicated depending on equation i would say there are ...


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Here is a hint. You already noticed that we want to use Maximum principle and Maximum principle only happens on harmonic functions. Let's divide your problem into too cases. Namely, \begin{cases}-\Delta u_1=0 & \text{in }U\\ \quad \, \, \, u_1=g & \text{on } \partial U \end{cases} \begin{cases}-\Delta u_2=f & \text{in }U\\ \quad \, \, \, ...


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You can write the tangent plane explicitly as the graph of the function $$ T(x)=u(x_0)+Du(x_0)(x-x_0). $$ Now simply note that $T$ is a harmonic function (being a degree 1 polynomial). If the tangent plane only intersects the graph of $u$ at $x_0$ then the tangent plane is everywhere above (or below) the graph of $u$, i.e. $T(x)\geq u(x)$ for all $x\in ...


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The starting point is the general formula for harmonic functions in the disk as the sum of separated solutions: $$ u(r,\theta) = \frac{A_0}{2}+\sum_{n=1}^\infty (A_n r^n \cos n\theta+ B_n r^n \sin n\theta) \tag{1}$$ To find the coefficients from the Neumann condition one writes $$ u_r(r,\theta) = \sum_{n=1}^\infty ( n A_n r^{n-1} \cos n\theta+ n B_n ...


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From what I can gather from my old Graduate PDE course, what you want to do is use something called a Riemann function for the PDE. I refer to my text on the subject, Guenther & Lee, Partial Differential Equations of Mathematical Physics and Integral Equations, Prentice-Hall (1988), Sec. 4-6, pp. 114-121. A Riemann function is akin to a Green's ...


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The Helmholtz equation is separable only in ellipsoidal coordinates (and degenerations like polar coordinates, and cartesian of course). For Laplace, there are a couple more; see the MathWorld article about Laplace's equation. A good book source on this subject is Chapter 5 of Morse & Feshbach, part I.


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$y\dfrac{\partial f}{\partial r}+\tanh(r)\left(z\dfrac{\partial f}{\partial y}-y\dfrac{\partial f}{\partial z}\right)+\text{sech}(r)\left(y\dfrac{\partial f}{\partial x}-x\dfrac{\partial f}{\partial y}\right)=0$ $y\dfrac{\partial f}{\partial r}+y~\text{sech}(r)\dfrac{\partial f}{\partial x}+(z\tanh(r)-x~\text{sech}(r))\dfrac{\partial f}{\partial ...


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(a) Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example: $\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$ $\dfrac{dx}{ds}=-xt=-xs$ , letting $x(0)=x_0$ , we have $x=x_0e^{-\frac{s^2}{2}}=x_0e^{-\frac{t^2}{2}}$ $\dfrac{df}{ds}=0$ , letting $f(0)=F(x_0)$ , we have $f(x,t)=F(x_0)=F\left(xe^\frac{t^2}{2}\right)$ ...


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To me*, restriction estimate means a bound on the norm of an operator $T:X\to Y$ where $X$ is some space of functions in a domain $\Omega$ (often the entire Euclidean space) $Y$ is some space of functions on a subset $E\subset \Omega$ of strictly smaller dimension than $\Omega$ itself For smooth functions, the operator agrees with the restriction ...


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The (angular) frequency of a separated solution is the coefficient of $t$ inside the trigonometric functions involved in it. That is $\omega_{m,n}$ in your notation. The formula for $\omega_{m,n}$ tells you that it's smallest when $m=n=1$. By the way, the proper way to write the series you have is $$u(x,\,y,\,t) = \sum^{\infty}_{m=1}\sum^{\infty}_{n=1} ...


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Ian made two excellent points in comments: In the absence of nice smoothness properties in the equation, nice convergence properties of a numerical method can fail. Your numerical method is actually an accurate solver for a Poisson problem where the right side is the indicator function of a small square whose side length has order $h$. I think ...


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Rewrite $$\frac{{\rm R}''(x)}{{\rm R}(x)}+\frac{{\rm S}''(y)}{{\rm S}(y)}=\frac{k}{c^{2}}$$ as $$\frac{{\rm R}''(x)}{{\rm R}(x)}=\frac{k}{c^{2}}-\frac{{\rm S}''(y)}{{\rm S}(y)}$$ The left side is independent of $y$. The right hand side is independent of $x$. Since they are identically equal, they are in fact constant. Let the constant be $\lambda$. Then $$ ...


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A good introduction can be found on the book Differential Equations in Abstract Spaces of G. E. LADAS and V . LAKSHMIKANTHAM.


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You are correct that from the two boundary conditions, we get $v(x)=b$. But how to determine the value of that constant $b$? If you solve the initial-boundary value problem for $u(x,t)$ (by separation of variables) and take the limit as $t\to\infty$, you will see that every term in the solution vanishes except the leading (constant) term. So you have ...


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The characteristic curves would be $3x^2+2y^2 = C$. Hence, $u(x,y) = f(3x^2+2y^2)$ and $u(x,x) = f(5x^2) = e^{x^2}$. Hence, $f(x) = e^{x/5}$ and $u(x,y) = e^{\frac{3x^2+2y^2}{5}}$


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Your numerical methods will generally converge to a weak solution, which will also be a strong solution if one exists. Here the weak solution is the zero function, and there is no strong solution (which should be unsurprising, since the forcing on the right side is not continuous). You can check that the weak solution is the zero function quite easily, ...


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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} ...


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Start changing the orders $$\int_{x = 0}^{x = \lambda} \sum_{k=-\infty}^\infty e^{-\left( \frac{x-k \lambda}{\sigma} \right)^2} dx=\sum_{k=-\infty}^\infty\int_{x = 0}^{x = \lambda} e^{-\left( \frac{x-k \lambda}{\sigma} \right)^2} dx$$ and now use, from Gaussian distribution,$$\int e^{-\left( \frac{x-k \lambda}{\sigma} \right)^2} dx=\frac{1}{2} \sqrt{\pi } ...


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Define $u_k=x-k\lambda$. Then $$ \begin{align*} \int_{0}^\lambda \sum_{k=-\infty}^\infty e^{-\left( \frac{x-k\lambda}{\sigma}\right)^2}dx &= \sum_{k=-\infty}^\infty \int_{0}^\lambda e^{-\left( \frac{x-k\lambda}{\sigma}\right)^2}dx\\ &= \sum_{k=-\infty}^\infty \int_{-k\lambda}^{-(k-1)\lambda} e^{-\left( \frac{u_k}{\sigma}\right)^2}du_k\\ &= ...


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An outline: If $s-|a|>n/2$ or $s-|b|>n/2$ then we're done since, in the first case, we get $D^a u\in L^{\infty}$ with appropriate control by Sobolev embedding. If $s-|a|, s-|b|<n/2$, then again by Sobolev embedding we get that $D^a u\in L^p$ for every $p\in [2, 2^*_a]$, where $2^*_a:=2n/(n-2(s-|a|))$, and similarly for $b$. The desired inequality ...


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Formally (and by this I mean: applied in smooth functions) the $p$-Laplace operator is defined by $$\Delta_pu=\operatorname{div}(|\nabla u|^{p-2}\nabla u).$$ If $h(x)=\frac{u(x_0+dx)}{d}$, we see from the previous equality that $$\Delta_p h(x)=\frac{\operatorname{div}(|\nabla u(x_0+dx)|^{p-2}\nabla u(x_0+dx)}{d^{p-1}},$$ so if $\Delta_p u=f$, it is ...


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It's basically an exercise in distributional notation: Take $u(x)=-\frac{|x|^2}{8\pi}\ln|x|$ and $v(x)=-(2\pi)^{-1}(1+\ln|x|)$, then we have $$ \int_{\mathbb{R}^n}u\Delta^2 \varphi dx := \langle \Delta^2 u, \varphi \rangle = \langle \Delta u, \Delta \varphi \rangle =\langle v, \Delta \varphi \rangle. $$ (where the brackets denote the dual pairing of a ...


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Notice $u_t\cong 0$ on $\partial \Omega$!


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You will need to specify a lot more information about the problem. Are your sensors reporting the heat at steady state? At some finite time after heat started flowing from the point? Is that time known? What is a heat source? Is it a single infinitesimal point? Several discrete points? A curve/area? Is it a source in that it is keeping the temperature ...


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In my experience as a theoretical physicist one rarely encounters situations where the dielectric constant is a (continuously varying) function of spatial coordinates. Therefore there is no great need to be able to solve this Maxwell equation analytically for the case of an arbitrarily varying dielectric constant. On the other hand there are many ...


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Let $B=B(0,1/2)$ and we recall the definition of weak partial derivative. We say a function $g\in L^1_{\text{loc}}(B)$ is weak derivative of $f$ if for any $\phi\in C_c^\infty (B)$ we have $$\int_B f\partial_i \phi\,dx=-\int_B g\,\phi\,dx $$ So we need to find out what is $g$ in your question. Let us suppose $x\neq 0$ and we compute the classical ...


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Any partial differential equation can be written as a first order system, so basically what you are asking is how to generalize method of characteristics to general PDEs and systems. As you have already observed, in general it is hopeless. However, there are some things that can be said. For general PDEs and systems, the notion of characteristic surfaces ...


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$y^{\prime}$ is an argument of $L$, so it makes perfect sense to (partial) differentiate with respect to it. Go through your question and mentally replace every instance of $y^{\prime}$ with $z$ and see if it makes more sense.


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The original equation can be rewritten as $$u_{t}-u_{xx}=\int_{0}^{t}\int_{-\infty }^{\infty }p(\zeta,\tau) \delta(x-\zeta)\delta(t-\tau)d\zeta d\tau, -\infty<x<\infty,t>0$$ and then for the first procedure the correct expression is $$u_{1t}-u_{1xx}=p(\zeta,\tau) \delta(x-\zeta)\delta(t-\tau), -\infty<x<\infty,t>0$$ and identical ...


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This equation, coupled with the additional condition that $\nabla \cdot u = 0$ and some boundary conditions (otherwise the equation in underdetermined) is exactly the Navier-Stokes equation for incompressible fluid dynamics. Due to its importance in theoretical and computational study of fluids, existence and smoothness of its solutions is extremely ...


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I'll introduce a bit of notation to make the explanation smoother. Let's start with a coordinate system $u^i=(u^1,u^2,u^3)$. The vector field with respect to this notation can be written $\mathbf x = \mathbf x (u^1,u^2,u^3) = \mathbf x (u^i)$. I'll assume that we work in Euclidean space, it's just the coordinate system that is curvilinear. So we can express ...


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Let $\begin{cases}p=x^2\\q=y^2\end{cases}$ , Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial x}=2x\dfrac{\partial u}{\partial p}$ $\dfrac{\partial u}{\partial y}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial y}+\dfrac{\partial u}{\partial ...


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Your approach is right, although it wasn't necessary to write out (3) with both $c_1$ and $c_2$ in there. When we need a function with constant Laplacian, the quadratic polynomials are to look for: so, $\varphi(x)=(R^2-|x|^2)/(2n)$ is the test function to use. And you used it correctly: $$ -\int_{B_R} u =\frac{1}{2n} \int_{B_R} f (x) (R^2-|x|^2) $$ which ...


1

You can do this. Let $v_\lambda(x)=u(\lambda x)\,,$ then $v_\lambda$ is also harmonic if $u$ is, so: $\displaystyle \Delta v_\lambda=0\implies\int_{B_r(x)}\Delta v_\lambda \,dV=0\implies \int_{\partial B_r(x)}\frac{\partial v_\lambda}{\partial r} \,r^{n-1}\,dS=0\implies \int_{\partial B_r(x)}\frac{\partial v_\lambda}{\partial \lambda} \,r^{n-1}\,dS=0\implies ...


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If $LF=\delta$ and $Lu=0\,,$ then $L(F+u)=\delta\,,$ and it would seem that $L$ has nonzero kernel. Are there conditions on $F$?


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These bump functions, or test functions, are extremely important in distribution theory. They can be constructed using partitions of unity. I didn't find any good references online in this context, but if you can get hold of Hörmanders ``The Analysis of Linear Partial Differential Operators I'', then it's an excellent reference.


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Define $A:H^s_0 \to (H^s_0)^*$ by $\langle Au, v \rangle = (u,v)_{H^s_0}$ and consider the equation $\langle Au, v \rangle = (f,v){L^2}.$ $u$ exists uniquely, and the solution map $T(f) = u$, $T:L^2 \to L^2$ is compact. Then one can apply the Hilbert-Schmidt theorem.


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here is a scheme you can try to see if it conserves the mass better. i have not tried the scheme so i cant be sure. $$\frac{{(dx)}^2}{dt}(u_j^{n+1} - u_j^n) = (1+au^n_{j+1/2})(u_{j+1}^n - u_j^n) + (1+au_{j-1/2}^n)(u_j^n - u_{j-1}^n), j = 0,1,\cdots M$$ where $u_{j+1/2}^n = {u_j^n + u_{j+1}^n \over 2}, u_{-1}^n = u_{1}^n, u_{M-1}^n = u_{M+1}^n.$


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The setup For completeness I want to state the parts from the book that are needed: We are given a (in general nonlinear) operator $$A: W^{1,p}(\Omega) \rightarrow W^{1,p}(\Omega)^*,$$ which is coercive (see Lemma 2.35 in the given literature). At that, coercivity means the existence of a mapping $\xi: \mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that $\xi$ ...


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As I've mentioned in the comment, just use the SDE that you derived to obtain ODEs for the moments of $X_T$. That is, let $m(t) = \mathsf EX_t$, then $\mathrm dm_t = \mu\mathrm dt$. A similar equation you can derive for $v(t) = X^2_t$ using Ito's lemma. That's pretty much it. Edit: you will get the following SDE $$ \mathrm dX^2_t = (2\mu X_t +\sigma^2 ...


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The separation of variables method leads to solutions of the form $\sin(\omega x)\sinh(\omega y)$ or $\sin(\omega x)\cosh(\omega y)$ or $\cos(\omega x)\sinh(\omega y)$ or $\cos(\omega x)\cosh(\omega y)$ or other equivalent forms with exp fonctions, where $\omega$ is any constant. The conditions $\frac{\partial u}{\partial x}(0,y)=0$ and $\frac{\partial ...


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The computation of the characteristic curves is wrong. Its differential equation is $$ \frac{dx}{x\,y}=\frac{dy}{2\,y^2-x^6}\implies\frac{dy}{dx}=\frac{2}{x}\,y-\frac{x^5}{y}. $$ This is a Bernoulli equation, whose solution is $$ y=\pm\, x^2\,\sqrt{C-x^2\,},\quad C>0. $$


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$$ \partial_t C = \frac{1}{r}\partial_r \left(r\partial_r C\right) - \frac{k}{D}C $$ Seperation of variables $C(r,t) = R(r)T(t)$ $$ \frac{\dot{T}}{T} = \frac{1}{R}\frac{1}{r}\frac{d}{dr} \left(r\frac{dR}{dr}\right) - \frac{k}{D} = -\lambda_n^2 $$ this leads to $$ \frac{\dot{T}}{T} = -\lambda_n^2\\ \frac{1}{R}\frac{1}{r}\frac{d}{dr} ...


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Define $T:W^{s,p}(\Omega)\to L^p(\Omega)\times L^p(\Omega\times\Omega)$ by $$T(u)=\left(u,\frac{|u(x)-u(y)|}{|x-y|^{N/p+s}}\right),$$ where $W^{s,p}(\Omega)$ is equipped with the norm given by you and $L^p(\Omega)\times L^p(\Omega\times\Omega)$ is equipped with the norm $$\|(u,v)\|=\left(\int_\Omega |u|^p+\int_{\Omega\times\Omega}|v(x,y)|^p\right)^{1/p}.$$ ...



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