# Tag Info

0

Yes, since your equation does not depend on $x$.

1

A solution is deemed to be a classical solution if it's differentiable as many times as needed to be able to evaluate it in the PDE (see e.g. this and this question for more about this). For your $u$ not to be a classical solution of the wave-equation this means that we need to show that $u_{xx}$ and $u_{tt}$ fails to exist at some point $(x,t) \in(0,2\pi) ... 0 Note that the Fourier transform is given by $$\tilde u(k,t) = \mathcal F_x[u(x,t)] = \int_{\Bbb R} u(x,t)e^{-ikx}\,dx$$ Thus, applying Leibniz's rule, we have $$\mathcal[F_x u_t(x,t)] = \int_{\Bbb R} \frac{\partial }{\partial t}u(x,t)e^{-ikx}\,dx =\\ \frac{\partial }{\partial t}\int_{\Bbb R} u(x,t)e^{-ikx}\,dx = \tilde u_t(k,t)$$ Perhaps you could take ... 0 The solution$u$is given by an analytic semigroup (defined e.g. by means of the spectral theorem), more precisely:$u(t,x):=e^{t\Delta}u_0(x)$. Now, this semigroup (like all analytic semigroups) satisfy an estimate $$\sup_{t>0} \|t^k \Delta^k e^{t\Delta}\|<\infty$$ for all$k\in \mathbb N$, and again in view of the spectral theorem this is optimal. ... 1 If$\Omega$is regular enough ($C^1$or uniformly Lipschitz) then there exists a continuous linear operator $$\operatorname{Tr}\colon W^{1,p}(\Omega) \to L^p(\partial \Omega,\mathcal{H}^{N-1})$$ such that$\operatorname{Tr}(u) = u$for every$u \in W^{1,p}(\Omega)\cap C^1(\Omega)$, there is a constant$C = C(\Omega,N,p)$such that ... 0 For the first you're right: Since$\Phi_t$is a positive$L^1$function with$\int_{\mathbb{R}} \Phi_t(x)dx=1$for all$t> 0$, we get that the convolution integral is well defined and by Holder's inequality we get $$|u(t,x)|\leq \sup_{y\in\mathbb{R}}|f(y)|.$$ Using the Dominated Convergence Theorem you can also see that if$f\in ...

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$$\frac{X''}{X}+\frac{Y''}{Y}+\frac{Z''}{Z}=0.$$ $$\frac{X''}{X}+\frac{Y''}{Y}=-\frac{Z''}{Z}=\lambda$$ $$\lambda-\frac{X''}{X} = \frac{Y''}{Y} = \mu,$$ so you solve 3 auxiliary problem 2 of them are Sturm Liouville: $$(1)\;\begin{cases}Z''+ \lambda Z &= 0 \\ Z(0)=0=Z(c)\end{cases} \qquad (2)\;\begin{cases} Y'' + \mu Y&=0, \\ ... 1 It seems to me that you have some misunderstanding in the notation between X and x or Y and y! In fact, we have X=X(x) and Y=Y(y). Take a look at the following steps$$\eqalign{ & x{u_x} = u + y{u_y} \cr & u = XY \cr & xX'Y = XY + yXY' \cr & {{xX'} \over X} = 1 + {{yY'} \over Y} = \mu \cr & \left\{ ...

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Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example: $\dfrac{dx}{dt}=x$ , letting $x(0)=1$ , we have $x=e^t$ $\dfrac{dy}{dt}=-y$ , letting $y(0)=y_0$ , we have $y=y_0e^{-t}=\dfrac{y_0}{x}$ $\dfrac{du}{dt}=u$ , letting $u(0)=f(y_0)$ , we have $u(x,y)=e^tf(y_0)=xf(xy)$ $u(x,0)=\sin\dfrac{\pi x}{4}$ : xf(0)=\sin\dfrac{\pi ... 2 The Laplacian is the trace of the Hessian, which is positive. However the Hessian is symmetric, and symmetric plus negative semidefinite implies nonpositive trace (this can be seen easily through diagonalization). 0 Hint Use the following orthogonality \eqalign{ & {\varphi _{mn}}(r,\theta ) = {J_{3n}}\left( {\sqrt {{\lambda _{nm}}} r} \right)\cos \left( {3n\theta } \right) \cr & \int_{r = 0}^R {\int_{\theta = 0}^{2\pi } {{\varphi _{mn}}{\varphi _{pq}}rdrd\theta } } = {\delta _{mp}}{\delta _{nq}}\int_{r = 0}^R {\int_{\theta = 0}^{2\pi } {\varphi ... 1 \newcommand{\M}{M} \newcommand{\til}{\tilde} \newcommand{\ep}{\epsilon} \newcommand{\Ga}{\Gamma} \newcommand{\brk}[1]{\left(#1\right)} \newcommand{\R}{\mathbb{R}} \newcommand{\pd}[2]{\frac{\partial#1}{\partial#2}} Here is an alternative proof, using local existence-and-uniqueness result on Linear First-Order Cauchy problems, which I think ... 0 I don't know if the following example is nontrivial enough for you, but I'll let you decide that. Let \psi be a Schwartz function such that \widehat{\psi} has support in the annulus \{2^{-1/2}\leq|\xi|\leq 2^{1/2}\}. Consider the symbol a given bya(x,\xi)=\sum_{j=1}^{\infty}a_{j}(x)\widehat{\psi}(2^{-j}\xi), \tag{1}$$where the a_{j} are ... 2 For differential equations, the formal adjoint sets up an efficient way to integrate by parts, whether in one dimension or in many dimensions. For example, when you integrate by parts, derivatives are stripped from the left and are moved to the right with a negative.$$ \int \left(a\frac{d}{dx}f\right)gdx =-\int f\left(\frac{d}{dx}(ag)\right)dx $$So the ... 1 This holds if and only if \frac{\partial L}{\partial m} = 0. In this case, you obtain the equation $$0 = 2\cdot\Sigma m + \lambda\cdot \mathbf{1},$$ which is equivalent with your expression for m after shuffeling some terms around. 2 Since you want to speak about derivatives in C ^\infty (M), you have to endow this with a topology, and the usual requirement of topological completeness will force you to restrict yourself to only a subspace of it. Let us introduce the notation: V ^n f = \underbrace {V (V ( \dots (V} _{n \ \text{times}} f \underbrace {) \dots ))} _{n \ \text{times}} ... 1 The most basic preconditioner for$$A= \begin{pmatrix} A_{11} \ A_{12} \ A_{13}\\ A_{21} \ A_{22} \ A_{23}\\ A_{31} \ A_{32} \ A_{33}\\ \end{pmatrix}$$is a Jacobi (or diagonal) preconditioner, i.e. \mathrm{diag}(A). Now if you have suitable pre-conditioner for A_{22}, say P_{22}, then I would try to use the following preconditioner$$P= ... 0 Yes, Taylor expansions are your friends here. Your local truncation error is given by \begin{align} & y_{i+1} - y_i - \Delta x (3/2 f_i - 1/2 f_{i-1}) \\ & = \left( y_i + \Delta x y_i' + \frac{\Delta x^2}{2} y_i'' + \mathcal{O}(\Delta x^3) \right) - y_i \\ & + \Delta x \left( \frac{3}{2}f_i - \frac{1}{2} \left( f_i - \Delta x f_i' + ... 0 Not sure why you call this PDE, but you can rewrite equation as $${px^2+z^2 \over x-z} = {qy^2+z^2 \over y-z}$$ (unlessx=z$or$y=z$which gives additional solutions$x=y=z$,$x=\pm \sqrt{-q \over p+1}y=z$and$y=\pm \sqrt{-p \over q+1}x=z$) From${px^2+z^2 \over x-z}=c$we have$x(c,z)$and likewise$y(c,z)$(for$c$not between$2z(p \pm \sqrt{p^2+p})$... 0 How to relate the Laplacian of composition$u\circ \phi$to the Laplacian of$u$and some derivatives of map$\phi$? In general this is a messy formula involving the second derivatives of$\phi$; it does not preserve the structure of Poisson's equation. But if the Jacobian matrix of$\phi$is a multiple of an orthogonal matrix, i.e.,$(D\phi)^T (D\phi) =c ...

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What mathematical software tools do you have available? It usually doesn't pay to reinvent the wheel. In Maple you could do something like this: pde:= diff(u(x,t),t,t) = diff(u(x,t),x,x) + u(x,t)^2; ibc:= {u(0,t) = 1, D[2](u)(x,0) = 0, u(x,0)=1,u(1,t) = 1}; solution:= pdsolve(pde,ibc, numeric,time=t,range = 0..1);

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You are confusing the multivariant and single-variant chain rules in your fomula. You don't get the multivariant version simply by replacing "$d$" with "$\partial$". $f$ is function of two variables, not just one, and $\zeta$ and $\eta$ are functions of both $x$ and $y$, not just one or the other. The chain rule for two variables is: \frac{\partial ... 1 \begin{align} & \frac{d}{dt}E\left( u \right)=2\int_{\Omega }{{{u}_{t}}{{u}_{tt}}+{{u}_{x}}{{u}_{xt}}+u{{u}_{t}}dx} \\ & \text{ }=2\int_{\Omega }{{{u}_{t}}{{u}_{tt}}-{{u}_{xx}}{{u}_{t}}+u{{u}_{t}}dx}+2\int_{\partial \Omega }{{{u}_{x}}{{u}_{t}}dS} \\ & \text{ }=2\int_{\Omega }{{{u}_{t}}fdx}\le \int_{\Omega }{{{\left| ... 0 A helpful concept is the domain of influence. The initial position at (x_0,0) influences the values of u at (x_0\pm ct, t). The initial velocity at (x_0,0) influences the values of u on the interval from (x_0 - ct, t) to (x_0 + ct, t). Both these statements are a consequence of d'Alembert's formula. Case 1: g\equiv 0, f is supported on ... 1 If you apply maximum principle on \Omega \setminus B_{\epsilon}(\mathbf{x}_0), there are two boundaries. As you have said Green function takes zero on \partial \Omega. For another boundary, note that H, being harmonic, is a twice continuously differentiable function in \Omega, so |H(\mathbf{x}_0)|<\infty, which implies that \mathbf{x}_0 is a ... 0 u is a given function of x and t, so the triplet (x,t,u) is really (x,t,u(x,t)) which leaves two degrees of freedom, namely, x and t. The capital T means "transpose": exchanging rows and columns of a matrix. In this case it means re-interpreting (x,t,u) as a column of three numbers rather than a horizontal row. Many authors prefer to express the coordinates ... 0 I accidentally find the answer U_t+A(U)U_x=0 $$where A=F' referring to page 2 in https://www.math.psu.edu/bressan/PSPDF/clawtut09.pdf 0 Ignoring weights, multiply both sides of the original DE by a test function and "formally" integrate by parts. You get$$\int_\Omega -\nabla u \cdot \nabla v dx + \int_{\Gamma_e} v \nabla u \cdot n d \Gamma + \int_{\Gamma_n} v \nabla u \cdot n d \Gamma = \int_\Omega g v dx.$$Now we apply the boundary conditions in the weak form. On the Neumann part of ... 3 For the first question, take something like$$ f(t) = \sin(e^{t^2}). $$Then f is bounded (so in particular of exponential growth), but$$ f'(t) = 2t e^{t^2} \cos( e^{t^2} ) $$is not of exponential growth. (Look at the values for z=2\pi k, k \in \mathbb{Z}.) For the second question, see Jyrki's answer. 1 Seek a solution of the form$$u(x,t)=\sum_{\lambda}C_{\lambda}(t)X_{\lambda}$$where X_{\lambda} are eigenfunctions of the operator \frac{d^2}{dx^2} with homogeneous Neumann boundary conditions. We know the eigenfunctions are 1 and \cos(nx), n \in \mathbb{N}. Thus, we have$$u(x,t)=\sum_{n=0}^{\infty}C_n(t)\cos(nx)$$Substituting the trial ... 3 Answering the second question with a "standard" example of a function that is not of exponential order, but does have a Laplace transform in the region \Re s>0. Build a function out of spiky triangles$$ \Delta_{H,A}(x)= \begin{cases} H-\frac{H^2}{A}|x|,&\ \text{if $|x|\le A/H$, and}\\ 0,&\ \text{otherwise.} \end{cases} $$Here H>0,A>0 ... 2 EDIT: This proves the wrong thing; ignore this answer. If f' is of exponential order, so is f. Consider the integral$$\int_a^x f'(t) dt$$From the fundamental theorem of calculus, this differs from f by at most a constant. If |f(x)| ≤ |g(x)| for sufficiently large x, \int_a^x |f(t)| dt ≤ \int_a^x |g(t)| dt for sufficiently large x and a.$$ ...

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$$F[f](\xi) = F[(-\Delta+\lambda)((-\Delta+\lambda)^{-1}f)](\xi) = (|\xi|^2+\lambda)F[(-\Delta+\lambda)^{-1}f](\xi),$$ hence $$F[(-\Delta+\lambda)^{-1}f](\xi)=\frac{1}{|\xi|^2+\lambda}F[f](\xi)$$

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Let $0<s<1$ be fixed, and let $\alpha>0$ be such that $\alpha>2s$. Suppose $f\in C^{0,\alpha}(\mathbb{R}^{n})$ and $$|f(x)|\lesssim (1+|x|)^{\delta},\quad x\in\mathbb{R}^{n}$$ for $0<\delta<2s$. Then I claim that the integral above is absolutely convergent. Indeed, for \begin{align*} &\int_{1\geq ...

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Assuming the correct regularity on $f$, this is an application of the divergence theorem. Indeed notice that $$\int_{\Omega}f\,dx = \int_{\Omega}\Delta u\, dx = \int_{\partial \Omega}\nabla u \cdot n\, d\mathcal{H}^{N-1} = 0,$$ where the last equality follows from the Neumann condition. This is usually referred to as a compatibility condition.

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If you want to avoid the measure theoretic framework, there is a quick and easy way to build a useful integral for Banach-valued functions. It's called the regulated integral / Cauchy integral. A good reference is Dieudonne's Foundations of Modern Analysis, but I think you can also find it in Bourbaki. In any case, it avoids measure theory and gives an ...

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It is a quotient space! $\mathbb{R}$ here should be thought of as the one-dimensional subspace of $L^2(\Omega)$ which consists of the (a.e.) constant functions. The norm given is the canonical norm on the quotient of a Banach space mod a closed subspace. This construction is discussed, for instance, in section III.4 of Conway's A Course in Functional ...

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I guess there are other typos in the answer. In the third equation $u_n(t)$ and not $b_n$ shall be defined. Further in the fourth and fifth equation $\sum_{n=1}^\infty$ is missed after the equal sign, unless the Einstein's summation rule is assumed. Probably, renaming $\lambda_n \rightarrow \lambda_n^2$ would simplify the understanding of the answer ...

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if $u_1$ and $u_2$ are two solutions, then $u = u_1 - u_2$ are is a solution of the homogenous problem $$\begin{cases} u_t-u_{xx}=0&x\in \mathbb{R},t>0\\ u(x,0)=0&x\in\mathbb{R} \end{cases}$$ So prove that $u = 0$ is the only solution to the homogenous equation.

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I know I'm giving you bad news, but what you in fact have is this (where obviously $A_{jk}$ are block matrices): $$A_{11}x_{1}+A_{12}x_{2}+A_{13}x_{3}=b_{1}$$ $$A_{21}x_{1}+A_{22}x_{2}+A_{23}x_{3}=b_{2}$$ $$A_{31}x_{1}+A_{32}x_{2}+A_{33}x_{3}=b_{3}$$ Now...unfortunately if you don't have any other restriction I guess that's as far as you can get. If you ...

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$u$ is a two variable function $u(x,t)$, $$u_{xx}=\frac{\partial^2x}{\partial x^2}$$.

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subscript is a derivative with respect to this variable. $$u_{xx} = \frac{\partial^2u}{\partial x \partial x}$$

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Hint: Introduce $\mathbf{y}=\frac{\mathbf{x}}{|x|^{2}}$. Then \begin{equation*} (\frac{\partial }{\partial \mathbf{y}}\cdot \frac{\partial }{\partial \mathbf{y}})u(\mathbf{y})=0 \end{equation*} Now express $\frac{\partial }{\partial \mathbf{y}}\cdot \frac{\partial }{ \partial \mathbf{y}}$ in terms of $\mathbf{x}$ and $\frac{\partial }{ \partial \mathbf{x}}$. ...

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Let $g$ be the restriction of $\Phi$ to the boundary of the square. Your boundary data determines $g$ up to a constant. You can then take this $g$ and solve the corresponding Dirichlet problem; standard theory tells that solutions exist uniquely. Therefore you can conclude that solutions are unique up to shifting by constants. (The fact that shifting a ...

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Once you reduced the problem to half-plane, the argument function can be used to produce a harmonic function with any piecewise constant boundary values you want. Specifically, to have $v_0$ on $[a,b]$ and zero value elsewhere, you would use $$h(z) = \frac{v_0}{\pi}(\arg(z-b)-\arg(z-a))$$ where the argument is understood as taking values in $[0,\pi]$ in ...

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Hint: Just solve the equation using the standard separate of variable approach with initial condition as a sinusoidal function. Be sure you can generalize this approach to dealing with any other initial conditions using Fourier series.

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The underlying problem is that the plate is subjected to distributed load with density $f$, and is fixed along the boundary in such a way that the boundary can neither mode nor rotate (clamping condition). The function $u$ represents vertical displacement of the plate when it has reached an equilibrium. Since an equilibrium is being studied, no time is ...

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Yes, solutions of ODE $u'=f(u)$ with $C^\infty$-smooth $f$ are themselves $C^\infty$-smooth in the domain of existence. This follows from "bootstrapping" argument: $u\in C^0$ $\implies$ $f\circ u\in C^0$ $\implies$ $u\in C^1$ $\implies$ $f\circ u\in C^1$ $\implies$ $u\in C^2$ $\implies \dots$ (It should be noted that the domain of existence $I$ may be ...

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After the basic shock waves $x=t/2$ and $x=t+2$ (shown in red) are accounted for, the characteristics look like this. This picture is correct for times $t<1$, but needs two adjustments after that. Note that the solution within rarefaction wave is $u(x,t) = (x-1)/t$. Rarefaction catches up with shock on the right: $x=3$, $t=1$. From this point on, ...

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In spherical coordinates, the Laplacian of $v$ is given by $$\Delta v = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial v}{\partial r}\right) + \frac{1}{r^2\sin^2 \psi}\frac{\partial^2 v}{\partial \theta^2} + \frac{1}{r^2\sin\psi}\frac{\partial}{\partial \psi}\left(\sin\psi \frac{\partial v}{\partial \psi}\right).$$ Let \$\rho = ...

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