New answers tagged

1

It seems that $A$ dictates the evolution of the system over time. My best guess is that the relevant piece of information here is $A + A^T = 0 \implies$ $A$ has imaginary eigenvalues $\implies$ energy is conserved On the other hand, $A + A^T \preceq 0 \implies$ $A$ has eigenvalues with a non-positive real part $\implies $ energy decays with a steady state ...


0

You are on the right track, but the integrals involving $v(x)$ and $v_x(x)$ have to be expressed explicitly to go further :


2

No for $p=1$, yes for $1<p<\infty$. If $\phi\in \Delta C^\infty_c$ then $\int\phi=0$; this shows that $\Delta C^\infty_c$ is not dense in $L^1$. One might think at first that this shows the same thing for other $p$, but it doesn't, because the integral is not a bounded linear functional. Suppose from now on that $1<p<\infty$. Suppose that $K\...


1

The trace theorem you mentioned is an "if and only if" characterization, hence the question you asked is equivalent to: is it true that $f \in W^{1-\frac 1p,p}(D')$ extends to a function $\bar{f} \in W^{1-\frac 1p,p}(D)$ (here $D' \subseteq D = \partial \Omega$). As in the case of extension theorems for standard Sobolev spaces, the answer depends on $D$.


1

Assume that $v_1^2+v_2^2\neq 0$. Write $u(x,y)=U(z,w)$ with $$z:=v_1x+v_2y\text{ and }w:=v_2x-v_1y\,.$$ Hence, $$x=\frac{v_1z+v_2w}{v_1^2+v_2^2}\text{ and }y=\frac{v_2z-v_1w}{v_1^2+v_2^2}\,.$$ Observe that $z$ and $w$ are independent (by checking that $\frac{\partial z}{\partial w}=0$ and $\frac{\partial w}{\partial z}=0$). Then, you can see that $$\frac{\...


2

To be clear, you want to find $P_n(t)$ for a birth-death process with constant birth rate $q(i,i+1)=\lambda$ and constant death rate $q(i,i-1)=\mu$? This particular birth and death process is exactly the $M/M/1$ queue. As you can see, under "Transient solution", there is a solution for the probability mass function dependent on time for a particular state.


2

Since $\|f - f_n \|_p \to 0$, we can extract a subsequence $f_{n_j}$ so that $\| f - f_{n_j} \|_p \le \frac{1}{2^j}$. Put $$g = \lvert f \rvert + \sum^\infty_{j=1} \lvert f - f_{n_j} \rvert.$$ Then $g \in L^p$ and by the triangle inequality we have $$\lvert f_{n_j} \rvert \le g \,\,\, (\text{almost everywhere}).$$ From this subsequence, you can extract a ...


1

Differentiate both sides of the equation $f(\lambda x) = \lambda^{a}f(x)$ in $\lambda$ to obtain: \begin{equation} \sum_{i = 1}^{n}x_{i}(\partial_{i}f)(\lambda x) = a\lambda^{a-1}f(x) \end{equation} Setting $\lambda = 1$ yields $\sum_{i = 1}^{n}x_{i}(\partial_{i}f)(x) = af(x)$, known as Euler's identity. Differentiating the equation displayed above in $\...


1

Use the Co-area formula with $g=|\nabla u|^{-1} \chi_{\Omega_t}$ to get $$ \int_{\Omega_t} dA=\int_t^\infty \left(\int_{\partial \Omega_x} \frac{ds}{|\nabla u|}\right)dx.$$ Differentiating this with respect to $t$ gives the formula you're looking for.


1

You can find the general statement and proof in Chapter 6 of the book "Sobolev Spaces" by Robert A. Adams and John. J. F. Fournier.


0

The claim $$\int_{\partial B_r(x)} \frac \partial {\partial r}u(y)ds_y=\frac \partial {\partial r}\int_{\partial B_r(x)} u(y)ds_y$$ is not true since the domain of integration depends on $r$. You also later claim that $|\partial B_r(x)|$ is constant, when it is in fact proportional to $r^{n-1}$.


0

$$(1-2f_{x}^{2})f_{t}f_{xx} + 2f_{x}f_{tx}(1+f_{x}^{2})=0$$ Try Separation of variables: $$f_{t}f_{xx} - 2f_{t}f_{xx}f_{x}^{2} + 2f_{x}f_{tx} + 2 f_{tx} f_{x}^{3}=0$$ write $f = X(x)T(t)$ Then you have: $$X\dot{T}\ddot{X} T - 2X\dot{T}\ddot{X} \dot{X}^2 T^3 + 2\dot{T} \dot{X}^2 T + 2\dot{T} \dot{X}^4 T^3=0$$ $$(X\ddot{X} + 2 \dot{X}^2)\dot{T}T = (2X\ddot{X}...


1

Let $L$ be a $k^{th}$-order linear differential operator, i.e. one which satisfies $L(\alpha u + \beta v) = \alpha L u + \beta Lv$ for all $u,v\in C^k$ and constants $\alpha,\beta$ (of course this notion can be weakened past $C^k$, but this will do for here). We say that the equation $L u =f$, for some given function $f$, is linear. A semilinear equation ...


1

If we set $$x(\xi, t) = f(\xi + \beta t, t) = f(z, t)$$ we find (by the chain rule) \begin{align} x_{t} &= f_{z} \cdot z_{t} + f_{t} \\ &= \beta f_{z} + f_{t} \\ x_{\xi} &= f_{z} \cdot z_{\xi} \\ &= f_{z} \\ x_{\xi \xi} &= f_{zz} \cdot z_{\xi} \\ &= f_{zz} \end{align} Substituting into our original PDE, we get \begin{align} x_{t}...


1

Another particular solution you missed can be found as: $\frac{1}{(D+D')^2}x=\frac{1}{D'^2}$$[1+\frac{D}{D'}]^{-2}x=\frac{1}{D'^2}(1-2D/D'+....]x=\frac{1}{D'^2}(x-2y)=\frac{xy^2}{2}-\frac{y^3}{3}$


0

Here some general issues that can come up when solving this numerically and how one can resolve them. Consider the equation $u_{rr} + \frac{2}{r}u_r = f(u,r)$. Discretized on a grid with $n$ points in the radial direction from $[R_{\rm min},R_{\rm max}]$ this becomes (an example; there are other possible discretizations) $$\frac{u_{i+1}+u_{i-1}-2u_i}{h^2} +...


1

The only way to have a spatially smooth solution is for $\partial u/\partial r$ go to zero as $r$ goes to zero. This means you can use L'Hopitals's rule in evaluating that second term: $\displaystyle \frac{1}{r}\frac{\partial u}{\partial r} \rightarrow \frac{\partial_r(\partial u/\partial r)}{\partial_r(r)} = \frac{\partial^2 u}{\partial r^2}$


1

You can transform to Sturm-Liouville form by dividing by $x$: $$ (xM')'+\frac{\lambda}{x}M = 0. $$ Whenever you have such a form, you can get rid of the multiplier in the derivative. There is a standard trick when you change the independent variable. In this case, let $M(x) = P(\int\frac{1}{x}dx)$. Then $$ xM'(x) = ...


0

I tried to prove the Part 1 as follows. Please check if it is valid. We prove Part 1 by contradiction. Suppose the conclusion is not true, then $\max_{\overline{U}} u\neq \max_{\partial U} u.$ But since $\partial U\subset \overline{U},$ we have $\max_{\overline{U}} u>\max_{\partial U} u.$ Because $u\in C(\overline{U}),$ with $U$ bounded, it follows ...


0

$y''=Ay'+B\sin(2Cy)$ E.g. $y(0):=a$ and $y'(0):=b$. => $y''(0)=Ab+B\sin(2Ca)$, $y'''(0)=Ay''(0)+2CB\cos(2Cy(0))y'(0)=A^2 b+AB\sin(2Ca)+2CBb\cos(2Ca)$ and so on. You can construct a Taylor series for $y(t)$, if you know $y^{(n)}(0)$.


0

Hints Notice the general identity $$\psi \nabla ^2 \phi=\nabla \cdot (\psi \nabla \phi)-\nabla \psi \cdot \nabla \phi \tag{1}$$ Integrate over the domain Use the divergence theorem Put $\psi=\phi$


1

$$x^2\frac{d^2M}{dx^2}+x\frac{dM}{dx}+\lambda M(x)=0$$ The change of variable $\quad z=e^x\quad\to\quad x=\ln|z[\quad\to\quad \frac{dz}{dx}=z \quad$ isn't a good idea because it leads to a more complicated form of ODE : $\frac{dM}{dx}=\frac{dM}{dz}\frac{dz}{dx}=z\frac{dM}{dz}$ $\frac{d^2M}{dx^2}=\left(\frac{d}{dz}\frac{dM}{dx}\right)\frac{dz}{dx}=\frac{d}{...


2

The "Maximum Principle" is a very useful tool to answer to such a question (as John Barber rightly did). I will not come back with this principle to repeat what was already said and which is more general for all times from $t=0$ to $t\to\infty$. In the present case, since the wording of the question concerns only $t\to\infty$ and where it is not question of ...


3

There is a property of certain PDEs known as the Maximum Principle. (See these notes (pdf warning) for its application to the heat equation.) The maximum principle states that any solution $u(x,t)$ of the heat equation in a region of $(x,t)$ must have its extrema (minimum and maximum values) on the boundary of that region. In this case, the region is $(x,t)...


3

Using the integrating factor method borrowed from first order linear ODEs, your equation becomes $\frac{\partial}{\partial t}(e^{\lambda^2 t}u)=k \frac{\partial^2}{\partial x^2} \left ( e^{\lambda^2 t} u \right )$. Thus $e^{\lambda^2 t} u$ solves the ordinary heat equation.


0

If $\lambda=0$ is an eigenvalue, then $X(x)$ will be of the form $X(x) = mx + c$. Plug this into the boundary conditions, solve for $m$ and $c$, and voilà! The converse for part (a) is just as easy - if the condition is satisfied, then show there exist $m$ and $c$ so that $X(x)$ satisfies the boundary conditions.


0

This is nothing more than integration by parts. Multiply your equation by $u$, integrate and take the derivative w.r.t. to t out (I let the formal verification to you): $$ \frac{1}{2}\frac{d}{dt}\int_\Omega |u(x,t)|^2 dx = \int_\Omega ( u\Delta u )(x,t)dx + \int_\Omega f(x,t) dx. $$ Using integration by parts on the form $u\Delta u$ and integrating in time ...


1

Why not $L^1$: In studying elliptic equation, it is most convenient to consider $L^p$ space for $1<p<\infty$. The reason is that one does not have nice $L^p$-estimates $$\|u\|_{W^{2,p}(\Omega)}\le C (\|f\|_{L^p(\Omega)} + \|u\|_{L^p(\Omega)}$$ for $p=1$ (Here we assume that $p(x)$ is nice). Note that the above estimates is crucial in establishing ...


0

You want an operator that is symmetric. Suppose $$ Lf = a\frac{d^2}{dx^2}f+b\frac{d}{dx}f+cf $$ You want $a$ and $b$ to be such that $$ \int_{a}^{b}(Lf)gdx = \int_{a}^{b}f(Lg)dx+\mbox{evaluation terms}. $$ Symmetry of operators plays a critical role in Sturm-Liouville problems. It turns out that $L$ needs to have the form $$...


0

Let $S$ be the graph of the function $x^2-y^2$ in the neigbourhood of $(0,0)$ positive normals in the $(x,z)$ plane ($y=0$) are not disjoint, and neagtive normal in the $(z,y)$ plane are not disjoint.


0

This is not possible. Take $r =1$ and let's work on the real line. Suppose there were such a function $f.$ Then $f(x) = 1, 0\le x \le 1,$ $f(x) = 0, x\ge 2,$ and $|f'(x)|\le 1$  everywhere. Note that $f'(1) = 0.$ Let $g(x)$ be the line connecting $(1,1)$ with $(2,0).$ Then the slope of $g$ is $-1.$ Because $f'(1) = 0,$ $f$ is above $g$ on $(1,1+\epsilon]$ ...


1

$\newcommand{\dd}{\partial}$If $u$ is a (differentiable) function of two real variables, then loosely, "$du$" isn't a number, but an ordered pair of numbers $$ \frac{\dd u}{\dd t}(t, s) = \lim_{h \to 0} \frac{u(t + h, s) - u(t, s)}{h},\qquad \frac{\dd u}{\dd s}(t, s) = \lim_{k \to 0} \frac{u(t, s + k) - u(t, s)}{k}. $$ Technically, at each point $(t, s)$, ...


1

Yes. The necessary and sufficient condition is that $A$ is bounded. Indeed, the domain of $AE_I$ is the whole Hilbert space, and the domain of $E_IA$ is the domain $\mathcal D(A)$ of $A$.


1

Your answer doesn't make much sense - complex analysis does not apply in any sense when the dimension is odd. I'm guessing $\partial/\partial n$ denotes the normal derivative along the boundary, in which case the divergence theorem provides the answer: since $u$ is harmonic, $\nabla u$ is divergence-free, and thus $$\iint_{\partial \Omega} \frac{\partial u}{\...


0

Hint: Use multipliers $1,-1,1$ and then equate to the third term. Edit: Using above you will get, $$\frac{dx-dy+du}{2(x-y+u)}=\frac{du}{u}$$ Now solve, and you will get $\frac{x-y+u}{u^2}=c_2$. Then combine with the one you have already found and use the initial condition.


1

In your first equation you can divide with $\cos (\sqrt{\lambda} L)$ and get $$ \tan (\sqrt{\lambda} L)= - \beta \sqrt{\lambda}, $$ Define $f(x) = \tan (\sqrt{x} L)+ \beta \sqrt{x}$ and notice, $$\lim _{x\rightarrow^{+} \frac{(\pi +4n\pi)^2}{4L^2} }f(x)=-\infty$$ and that $$\lim _{x\rightarrow^{-} \frac{(\pi +4(n+1)\pi)^2}{4L^2} }f(x)=\infty$$ ...


0

Yes, the tensor product is dense. To approximate a function $f$ in $C_c^{\infty}((a_1,b_1) \times (a_2,b_2))$ you can use functions of the form $p(x,y)\chi_1(x)\chi_2(y)$ where $p$ is any polynomial and $\chi_1,\chi_2$ are cut-off functions that are equal to $1$ on most of the interval, so that $\chi_1(x)\chi_2(y)\equiv 1$ on the support of $f$. Then the ...


2

First of all, the wrong thing that you think is wrong, is totally wrong! :) So, yes, you are right! If you take the divergence and curl of the following equation $$ \textbf{F}=-\nabla \phi+\nabla \times \textbf{A} $$ then you will get, respectively $$\boxed{ \begin{align} \nabla \cdot \textbf{F} &= - \nabla^2 \phi \\ \nabla \times \textbf{F} &= - ...


2

If ${\bf F} = - \nabla \phi + \nabla \times {\bf A}$, we get $\nabla \cdot {\bf F} = - \Delta \phi$. So we want to get $\phi$ by solving a Laplace equation. Then we get $\bf A$ by solving $\nabla \times {\bf A} = {\bf F} + \nabla \phi$. In your first example, ${\bf F}_1$ is indeed divergence-free, so you can take any $\phi$ whose Laplacian is $0$. You ...


-1

Take a look at the The Laplace Transform, Joel L. Schiff 1991 Springer Verlag, ISBN 0-387-98698-7, Examples on p182 onwards.


0

$(\partial_x-4\partial_t)(\partial_x+\partial_t)u=0$ and set $v:=(\partial_x+\partial_t)u$ then $v_x-4v_t=0$ is transport equation with solution $f\left(x+\frac{1}{4}t\right)$ Hence the problem is reduced to $u_x+u_t=f\left(x+\frac{1}{4}t\right)$. This is the nonhomogeneous transport equation with solution $u(x,t)=\int_0^t f(x+s-t+\frac 14 s)ds+g(x-t)\...


2

$$\frac{dx}{a}=\frac{dy}{b}=\frac{du}{1}$$ Only two characteristic curves are sufficient because this is a set of two characteristic equations (they are two sign =). For example, we chose those characteristic curves : From $\frac{dx}{a}=\frac{dy}{b}$ a first characteristic curve is : $ay-bx=c_1$ From $\frac{dx}{a}=\frac{du}{1}$ a second characteristic ...


0

There are multiple interpretations. The usual scenario is that you have a single heavy particle (like a pollen grain) in a sea of light particles (like water molecules). The heavy particle experiences collisions with the light particles. These light particles endow the heavy particle with force, and the heavy particle is also under the influence of a spatial ...


0

you can write it in a similar way as stated in wikipedia as $M \ddot{X} = - \nabla U(X) - \gamma \dot{X} + \sqrt{2\gamma k_{B}T}R(t)$, where $M$ are the masses of $N$ particles and coordinates $X=X(t)$. $U(X)$ is the particle interaction potential, so $- \nabla U(X)$ is the force calculated from the particle interaction potentials. $\gamma$ is a small ...


0

With regards to the 'missing' Detla Function: Recall that the Dirac Delta Function only has a non-zero value when its argument is zero, hence the product of any function and the Dirac Delta Function will only be non-zero when the Delta function's argument is set to be zero in the product (Note that the integral of 0 is 0, hence when the product of these two ...


2

$\newcommand{\pd}[2]{\frac{\partial#1}{\partial#2}}$ $\newcommand{\pdk}[3]{\frac{\partial^#3#1}{\partial#2^#3}}$ $\newcommand{\pdd}[3]{\frac{\partial^2#1}{\partial#2\ \partial #3}}$ Let $Q(\theta,t) = e^{\alpha t} \pd\kappa\theta$ and compute the time derivative: $$\pd Q t = \alpha e^{\alpha t}\pd \kappa \theta+e^{\alpha t} \pdd\kappa\theta t.$$ ...


0

We have $$ \lim_{t \to \infty}\left(\frac1{e^t}\right)=0 $$ thus, there exists a $t_0>0$ such that $$ \left|\frac1{e^t}\right|\leq \color{red}{1}, \quad \text{for all}\quad t\geq t_0, $$ giving, for any fixed $x$, as $t \to \infty$: $$ |u(x,t)|=\left|x+\dfrac{x}{e^t}\right| \leq \left|x\right|+\left|\dfrac{x}{e^t}\right|\leq\left|x\right|+|x|\times \...


1

The previous theorem (on page 355) shows that $w_k \in H^1_0(U)$. Identities $(6)$ and $(7)$ tell you that $\{ w_k \}_{k=1}^{\infty}$ is an orthogonal collection and the length of each $w_k$ with respect to $B$ is $||w_k||_{B} = \sqrt{B[w_k,w_k]} = \sqrt{\lambda_k}$. Thus, if we normalize $w_k$ we get an orthonormal collection of elements in $H^1_0(U)$ with ...


1

The set of characteristic equations is : $$\frac{dx}{x}=\frac{dy}{y}=\frac{dz}{z}$$ With the condition $z=1$ on the curve $x^2+y^2=1$ So, it should be easy to satisfy this condition if this curve would be a characteristic curve. So, it is of interest to make appear the terme $x^2+y^2$ in a combination of the characteristic equations : $$\frac{dx}{x}=\frac{...


1

One interpretation could be that it solves the homogeneous ODE given by $\dot{u}(t)-i\omega u(t) = 0$



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