New answers tagged

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I figured it out on my own. You can arrive at the second by integrating by parts of both terms on the left side and then taking the derivative.


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Second characteristic curve : $\frac{dx}{x-y} = \frac{dy}{y-x-u} = \frac{du}{u} =\frac{dx-dy+du}{(x-y)-(y-x-u)+u} = \frac{d(x-y+u)}{2(x-y+u)} = \frac{du}{u} $ $\frac{1}{2}\ln|x-y+u|-\ln|u|=$constant $\quad\to\quad \frac{x-y+u}{u^2}=C_2$ The implicit solution is $\quad x+y+u=F\left(\frac{x-y+u}{u^2}\right) \quad $any differentiable function $F$. The ...


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As a rule, all the problems with the modules to be addressed by intervals: $$u=\begin{cases} \dot u,\text{ if }\dot u_x>0\\ \ddot u, \text{ if }\ddot u_x<0, \end{cases}$$ $$-\dot u_t= \dot u_x +\dfrac12\dot u_{xx},$$ $$-\ddot u_t= -\ddot u_x +\dfrac12\ddot u_{xx},$$ The peculiarity of PDE case is that at the points of transition through zero of the ...


1

$$ x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}+t \frac{\partial z}{\partial t}=xyt$$ Solving thanks to the method of characteristics : The characteristic equations are : $$\frac{dx}{x}=\frac{dy}{y}=\frac{dt}{t}=\frac{dz}{xyt}$$ From $\frac{dx}{x}=\frac{dt}{t}$ the first characteristic : $\frac{x}{t}=c_1$ From ...


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As noted in a comment, the boundary condition $f(1,t)=0$ can be accommodated using a negative image charge at $x=2$, leading to the solution $p(x,t)-p(x-2,t)$ (with $p$ the fundemantal solution you gave). To get $f'(1,t)=0$, you need a positive image charge at $x=2$, leading to the solution $f(x,t)=p(x,t)+p(x-2,t)$, with \begin{align} ...


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$$ x(u+y^2)\frac{\partial u}{\partial x}- y (u+x^{2})\frac{\partial u}{\partial y}= (x^{2}-y^{2})u.$$ Before to answer to the question, let us see where $xyu=C_{1}$ and $ x^{2}+ y^{2}-2u= C_{2}$ come from. The characteristic equations are : $$\frac{dx}{x(u+y^2)}=\frac{dx}{-y(u+x^2)}=\frac{du}{(x^{2}-y^{2})u}$$ The first characteristic curve is determined ...


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The underlying problem is that the formula $\int \frac1x \,dx = \ln |x| +C$ is not quite correct. The correct general form of the antiderivatives of $1/x$ is $$\begin{cases} \ln x + C_1,\quad & x>0 \\ \ln (-x) + C_2,\quad & x<0\end{cases}$$ There's no reason for $C_1$ and $C_2$ to be the same constant. I'll explain how this is relevant, ...


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Noting that by the equation$$p^2 + q^2 + 1 = z^{-2}$$the equations for $\dot{p}$ and $\dot{q}$ can be simplified. Furthermore, the equation for $z$ is independent from the rest, which means that we can solve for $z$ first, then for $p$ and $q$. Geometrically, our equation looks like an eikonal equation since it can be written as$$\left|\nabla u\right|^2 = ...


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Some information about it can be found here Proving that the eigenfunctions of the Laplacian form a basis of $L^2(\Omega)$ (and of $H_0^1(\Omega)$) Basically it is so because $(-\Delta)^{-1} : L^2(\Omega) \longrightarrow L^2(\Omega)$ is a compact injecting self-adjoint operator on $L^2(\Omega)$ and on $H_{0}^1(\Omega)$. More precisely, from this it follows ...


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If the method of separation of variables is required, first you have to transform the inhomogeneous PDE to an homogeneous PDE. Any particular solution can be used. For example, obviously $U=-4y$ is a particular solution. So, let $U(x,y)=-4y+V(x,y)$ $U_y=-4+V_y \quad\to\quad U_{xy}=V_{xy}$ $4+U_y-U_{xy}=0 \quad\to\quad 4+(-4+V_y)+V_{xy}=0$ ...


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Hint If we set $V := U_y$, we may rewrite the equation as an "upper triangular" system of what are effectively o.d.e.s: $$\left\{ \begin{array}{rcl} U_y &=& V \\ V_x &=& V + 4 \end{array} \right.$$ The second equation has general solution $$V(x, y) = g(y) e^x - 4 .$$


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Answer: 1. Physics lecturers abuse notation, so here partials are used interchangeably with full derivatives. 2. ...


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For any function $\phi$ and coordinate variable $w$, we will use $\partial_w$ and $\phi_w$ as a shorthand of the differential operator $\frac{\partial}{\partial w}$ and the partial derivative $\partial_w \phi = \frac{\partial\phi}{\partial w}$. Consider following change of variables $$(x,y) = \left(\frac{\sqrt{3}u+v}{2}, ...


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HINT (partial answer) : Search for a particular linear solution on the form $z=ax+by+c$ Puting into the PDE leads to $$z=ax+by+a^2+ab+b^2 \quad\text{any }a\:,\: b.$$


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The answer to the question raised by "sequence" is given in the comments. So, my answer is only a different form (but equivalent) of the method of characteristics with advantage of clearness : $$u_t+cu_x=e^{2x}$$ The characteristic equations are : $\quad \frac{dt}{1}=\frac{dx}{c}=\frac{du}{e^{2x}}$ From $\frac{dt}{1}=\frac{dx}{c}$ , the first ...


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There are many ways to show that there is no unique solution to this PDE. One of the simplest we can use to argue is: "the derivative doesn't read constants" So if $u$ is a particoular integral of your PDE, then also $u+constant$ is


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Compute the characteristics through the points $(0,\tau)$, $\tau>0$. On these characteristics, the value of $U$ is constant and equal to $\tau$, the value on $(0,\tau)$.


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Have you checked if the proof to theorem A. uses weak lower-semicontinuity or sequential weak lower-semicontinuity? I would guess it is the latter, especially after reading @user127096 response in the link you provided. As for your questions, (i) Note that for $t\in\mathbb R$, $$ I(tu)=\frac{|t|^2}{2}\int|\nabla u|^2-\frac{|t|^q}{q}\int|u|^q\sim_{|t|\to0} ...


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I figure out that the solution of this problem \begin{align*} &\text{PDE:}~~u_{tt}=u_{xx}+u_{yy}+x^3-3xy^2, \qquad (x,y)\in\mathbb{R}^2, t>0,\\ &\text{IC:}~~ u(x,y,0)=\operatorname{e}^x\cos(y),\quad u_t(x,y,0)=\operatorname{e}^y\sin(x),\qquad (x,y)\in\mathbb{R}^2 \end{align*} is \begin{gather*} u(x,y,t)=\operatorname{e}^x\cos(y)+t\cdot ...


1

Consider $$\left\{\begin{array}{c} −\Delta u = f & \text{on }\Omega \\ u(x)=0, \partial\Omega \end{array}\right.,$$ where $\Omega \subset \mathbb{R}^{N}$ is open, bounded and $f \in L^{2}(\Omega)$ See that $u \in H^{1}_{0}(\Omega)$ is a weak solution for the problem above if $(u,v)_{H^{1}_{0}(\Omega)} = \displaystyle\int_{\Omega} \nabla u \nabla v = ...


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For the sake of an answer: This usually means that the solutions are less regular than the functions in the equations or the initial data.


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I put some model solutions online today, you can see it there (Solutions 9 exercise nr 3) https://wiki.helsinki.fi/display/mathstatKurssit/Sobolev+spaces%2C+spring+2016


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Not sure why you would want to find spatial periodicity separately for each cos factor. However, a general expression for a traveling wave is $U(x,t) = a\cos(2\pi)(kx-nt)$. The expression $U(x,t) = a\cos(2\pi(kx-nt)+g)$ represents another wave with phase difference $g$ relative to the first expression (Waves, C.A. Coulson). If you recast your first cos term ...


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The footnote 31 on page 305 explains the logic of the choice of $h$: the support of $w$ is at some distance from the top and side surface of the cylinder $ Q_+$, but it need not be separated from the bottom surface, which is $x_N=0$. This is how the chosen partition of unity works: in order to cover the domain by finitely many sets, we need these sets to ...


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You have found one particular solution, and since the PDE is linear any other solution will differ from that one by a solution of the homogeneous equation: $$ x \frac{\partial z}{\partial x}+t \frac{\partial z}{\partial t}+y \frac{\partial z}{\partial y}=0 . $$ This equation says that the directional derivative of $z$ in the radial direction is zero; in ...


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The result is wrong. You cannot even get pointwise convergence on a sequence $(m_k,n_k)\to (\infty,\infty)$ if, e.g., the $u_n$ are a typical highly oscillatory sequence of functions with a joint compact support. In that case you just have $f_m u_n = u_n$ for $k$ large enough and the $u_n$ do not converge pointwise (on any subsequence).


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For $u \in L^2$ one defines $\Delta u \in H^{-2} = (H_0^2)^*$ by $$ \langle \Delta u, v \rangle := \int u \, \Delta v \, \mathrm{d}x.$$ Using Cauchy-Schwarz, you find $$ |\langle \Delta u , v \rangle | \le \| u\|_{L^2} \, \|\Delta v\|_{L^2} \le C \, \|u\|_{L^2} \, \|v\|_{H_0^2}.$$ Hence, $\|\Delta u\|_{H^{-2}} \le C \, \|u\|_{L^2}$. This should also answer ...


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$V=B_{r/2}(x^0) ∩U$; Note that $\bigcup_{v∈ V}[ B_{ε}(v)+2ε e_1 ]$ is pushed up from $\partial U$ by about $ε$. The choice of $r/2$ stops this set from 'spilling out' of $B_r$.


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Regarding the differential equation: You have to differentiate the equation in direction of a testfunction $\phi\in C^\infty_0(\Omega)$ (or taken from some other suitable test function space): $$\begin{eqnarray} \frac{d}{dt }F(u+t\phi)|_{t=0} &=& \frac{d}{dt}|_{t=0}\int_\Omega\frac{1}{2}\langle \nabla (u+t\phi),\nabla (u+t\phi)\rangle + ...


1

From the absence of the boundary conditions I assume it's a Cauchy problem. Note that $u(x,0)\to1$, $x\to+\infty$ and $u(x,0)\to-1$, $x\to-\infty$. And the convergence is exponential in speed. So it's straightforward to check that $v(x)=u(x,0)-\text{sign}(x-1)\in L_1(\mathbb R)$. Denote $\Gamma(x,t)=(4\pi t)^{-1/2}e^{-x^2/(4t)}$ the fundamental solution of ...


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If $u (x,y) = f \left(\frac{x^3}{3} - \frac{y^4}{4}\right)$, then the boundary condition gives us $f ' \left(\frac{x^3}{3}\right) = x$. Hence, we have $$f ' (z) = \sqrt[3]{3z}$$ Integrating, we obtain $$f (z) = \frac{3\sqrt[3]{3}}{4} \, z^{\frac{4}{3}}$$ and, thus, $$u (x,y) = \frac{3\sqrt[3]{3}}{4} \left(\frac{x^3}{3} - ...


1

The Fourier Cosine series for $\sin(x)$ on $[0,\pi]$ is given by $$\sin(x)=\frac{a_0}{2}+\sum_{n=1}^\infty a_n \cos(nx)$$ where $a_n=\frac{2}{\pi}\int_0^\pi \sin(x)\cos(nx)\,dx$. Note that $a_0=\frac{4}{\pi}$ and $a_1=0$. For $n\ge 2$, we have $$\begin{align} a_n&=\frac{2}{\pi}\int_0^\pi \sin(x)\cos(nx)\,dx\\\\ &=\frac{1}{\pi}\int_0^\pi ...


0

For $n=0$ Your ode equation for $F(x)$ will become $$F''(x)=0$$ which gives $$F(x)=Ax+B$$ Now we do not know your boundary conditions but let say $F(0)=0=F(L)$, then $F(x)=0$, a trivial solution. For $n>0$ The ode for $F(x)$ will have the characteristics equation of the form $r^2-n=0$ with the roots $r=\pm n$ and thus the solution is ...


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Essentially, the problem is $u''+\lambda u=0$ on $[-\pi,\pi]$ with periodic boundary conditions. (Note that this interval has the same length as the circumference of the circle; this choice of parametrization ensures that the Laplace-Beltrami operator on the circle directly corresponds to the Laplacian on this interval.) The solutions to the DE itself are ...


1

Assume that $u(x,y)$ is a solution then $$du=\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy$$ $$\Rightarrow \frac{\partial u}{\partial y}=\frac{du-\frac{\partial u}{\partial x}dx}{dy}$$ The equation becomes $$a\frac{\partial u}{\partial x}+b\frac{du-\frac{\partial u}{\partial x}dx}{dy}=c$$ $$a\frac{\partial u}{\partial ...


1

You have two options, either do the product rule and then discretize, or discretize the equation as is. I'll derive them both: Product Rule $$\frac{du}{dt} = \frac{d}{dx}\left[\frac{1}{x^2+1} \frac{du}{dx}\right] $$ $$\frac{du}{dt} = \frac{d}{dx}\left[\frac{1}{x^2+1} \right]\frac{du}{dx} + \frac{1}{x^2+1}\frac{d^2u}{dx^2} $$ $$\frac{du}{dt} = ...


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If we plug in $a=b=c=1$ into $$ a^{\frac1{80}}\le\frac1{20}\frac{a^{\frac14}}{b^{\frac34}c^6}+\frac6{20}b^{\frac1{16}}+\frac3{20}c^\alpha $$ we get that $1\le\frac12$. This doesn't seem promising.


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You can find a nice proof with lots of explanations and intermediate steps over here: https://vaughnclimenhaga.wordpress.com/2012/11/05/the-stable-manifold-theorem/


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Actually, you have $H_\perp^1 = \{0\}$, the set containing only the zero function. Indeed, suppose that $u \in H^1(\Omega)$ satisfies $\int_\Omega u \, v \, \mathrm{d}x = 0$ for all $v \in H_0^1(\Omega)$. Then, using the density of $H_0^1(\Omega)$ in $L^2(\Omega)$ you get $\int_\Omega u \, v \, \mathrm{d}x = 0$ for all $v \in L^2(\Omega)$ and this shows $u = ...


0

Actually it was (as you can check in the lecture notes you've referenced) $$\frac 1 {2\pi} \int_{\mathbb R} \int_{\mathbb R} e^{-ixt} g(x)\ dt\ dx.$$ If you integrate by $t$ first then it can be rewritten as (I'll assume below that integration interval is always $\mathbb R$) $$ \int dx\ g(x) \left(\frac 1 {\sqrt{2\pi}} \int \frac{1}{\sqrt{2\pi}} ...


0

Take g outside the inner integral on the left and you have the fourier transform of 1. This is the dirac delta, a distribution that integrates to 1 and is infinity at x. In this sense only the value of g where the dirac delta is non-zero becomes relevant. The point is then moving with x in your case.


0

Hint: Try guessing answers of the form $$ v = r^ \lambda$$ You'll find $$ r^2 v'' + r v ' -v = r^ \lambda P ( \lambda) = 0$$ if $r>0$, then $v$ is a solution if $P( \lambda) =0$.


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This is an Euler differential equation. Hint: look for a fundamental set of solutions of the form $v(r) = r^p$.


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I would recommend "An Introduction to Linear Analysis" which can be downloaded for free here: https://archive.org/details/AnIntroductionToLinearAnalysis. This was actually the book we used when I took a class on PDE's. It introduces all of the linear algebra needed for the classical theory on PDE's and in my opinion explains everything quite clearly. You can ...


2

The fundamental solution, as mentioned, satisfies $-u''+k^2 u = \delta_{y}(x)$. To the left or to the right of $y$, the fundamental solution satisfies $-u''+k^2u=0$. The fundamental solution needs to be continuous across $y$, and, in order to have the $\delta$ function behavior, there is a discontinuity in the first derivative of the solution with a jump of ...


0

What do you mean by 'solve'? If you're looking for an explicit expression for the solutions of this system, then you're out of luck, because there isn't any (for general $a$ and $b$). However, if you're interested in the stationary solutions of the Gray-Scott system, then there's a entire branch of research devoted to that, and related questions. To get a ...


0

While not an answer, I'll provide a few possible angles of attack in this post: We have that \begin{eqnarray} && 0 \leq \int_{U\subset\mathbb{R}^3}d^3r|\nabla \rho(\mathbf{r}) \times \nabla V(\mathbf{r})|^2 = \int_{U}d^3r\left(|\nabla \rho(\mathbf{r})|^2 |\nabla V(\mathbf{r})|^2 - |\nabla \rho(\mathbf{r}) \cdot \nabla V(\mathbf{r})|^2 \right)=\\ ...


0

You need a global Lipschitz condition with the same constant to apply a global version of the theorem. Your example $\dot x = x^3$ does not possess a global Lipschitz constant since its derivative $3x^2$ is unbounded. And indeed, the solution $x^{-2}=C-2t$ has a couple limitations on its domain. However, there is a theorem of Cauchy that tells you that if ...


2

Since the initial conditions are dependent only on $r$, we transform to spherical coordinates wherein $u=u(r,t)$. Then, we have in spherical coordinates $$\frac1{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial u(r,t)}{\partial r}\right)-\frac{\partial^2 u(r,t)}{\partial t^2}=0 \tag 1$$ We can rewrite the first term on the left-hand side of $(1)$ ...


4

First, you are missing the contribution from the $z-$component of the vortex convection term $\mathbb{u} \cdot \nabla \mathbb{\omega},$ which is $-\frac{ \alpha}{2}r \frac{d \xi}{d r}.$ The correct steady-state differential equation is $$\frac{d^2 \xi}{dr^2}+\frac 1r \frac{d \xi}{dr}+ \frac{\alpha}{ν} \xi= -\frac{\alpha}{2 \nu}r \frac{d \xi}{d r}.$$ This ...



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