# Tag Info

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Why not a substitution? $y=x-ct \implies dy=-cdt$, and $$\mathcal{L}[u(t,x)](s) = \frac{1}{c} \int_{-\infty}^x \mathrm{exp}\left(-\frac{s}{c}(x-y)\right)u_0(y)dy = \int_0^\infty \mathrm{exp}(-st)\,u_0(x-ct)\,dt,$$ So $u(t,x) = u_0(x-ct)$.

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I think you have misunderstood the nature of the solutions to Bessel's equation: the differential equation $$y'' + \frac{1}{x}y' + \left(\lambda^2-\frac{n^2}{x^2}\right) y = 0$$ (Bessel's equation, with eigenvalue $\lambda^2$, $\lambda>0$) has a regular singular point at $x=0$, an irregular singular point at $\infty$, and all other points are regular ...

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Yes. A corollary to the Rellich Kondrachov theorem says that for all $p\ge 1$, we have $W^{1,p}$ compactly contained in $L^p$. Evans PDE book has a nice treatment of this. EDIT: Actually, if I recall correctly, for $\Omega \subseteq \mathbb R^n$, to prove that $W^{1,p}(\Omega)$ is compactly contained in $L^p(\Omega)$, we only need the Rellich Kondrachov ...

3

The product rule tells us $$\nabla |\nabla u| = \frac{\nabla (g(\nabla u,\nabla u))}{2 |\nabla u|}=\frac{g(\nabla^2u,\nabla u)}{|\nabla u|}.$$ Taking the squared norm and estimating the numerator with Cauchy-Schwarz gives us $$|\nabla|\nabla u||^2 \le \frac{|\nabla^2 u|^2 |\nabla u|^2}{|\nabla u|^2} = |\nabla^2 u|^2$$ as desired.

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Using the Spectral Theorem, $$Ax=\int_{0}^{\infty}\lambda dE(\lambda)x \\ \mathcal{D}(A) = \left\{ x : \int_{0}^{\infty}\lambda^2 d\|E(\lambda)x\|^2 < \infty \right\}.$$ Then the positive square root $\sqrt{A}$ is $$\sqrt{A}x = \int_{0}^{\infty}\sqrt{\lambda}dE(\lambda)x \\ \mathcal{D}(\sqrt{A}) = \left\{ x : ... 1 Try rechecking your work. The derivative of \ln(f(x)) is f'/f . Observe$$\frac{\partial w}{\partial x}= \frac{2 u \frac{\partial u }{\partial x }+2v\frac{\partial v }{\partial x} } {u^2 + v^2} $$where \frac{\partial u }{\partial x} = e^{x^2 +y} (2x) and \frac{\partial v}{\partial x} = e^{x + y^2} When (x,y) = 0 we have that u = 1, v = ... 2 The Spectral Theorem for A is given in terms of a Borel Spectral measure E$$ Ax = \int_{-\infty}^{\infty}\lambda dE(\lambda)x, $$and x \in \mathcal{D}(A) iff$$ \int_{-\infty}^{\infty}\lambda^2 d\|E(\lambda)x\|^2 < \infty. $$The operator e^{iA^2} is defined through the functional calculus as$$ e^{iA^2}x = ...

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Let's assume three things. First, that the "appropriate boundary condition" is $u=0$ on $\partial \Omega$. Second that we write $L_k$ for the operator defined with $A_k,b_k,c_k, r_k$. Third, we need to avoid the possibility that $0$ is in the spectrum of the operator $L_k$ for any $k$ so that we can get estimates for solutions to $L_k u =f$ solely in ...

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It's primarily due to the similarity to the corresponding algebraic equations. For example, let $A \in \mathbb{R}^{n\times n}$ be a symmetric positive definite matrix. We say that the PDEs: $$\begin{cases} \sum_{i,j=1}^n A_{ij} \partial_i \partial_j u =f &\text{is elliptic} \\ \partial_t u - \sum_{i,j=1}^n A_{ij} \partial_i \partial_j u ... 0 The differential equation,$$(1-\alpha x)\partial_x^4 y +2\alpha\partial_x^3 y = 0,$$is despite the usage of partial derivatives an ordinary differential equation since the function y=y(x) to be determined depends solely on the variable x. Introducing the function,$$z=\partial_x^3y,$$the differential equation is recast in the form of a homogeneous ... 0$$ u_x+uu_y=0$$The first part is correct : The characteristic system is$$\dfrac{dx}{1}=\dfrac{dy}{u}=\dfrac{u}{"0"},$$so$$\dfrac{du}{"0"} \rightarrow u=C_1$$where C_1 is constant. And$$\dfrac{dy}{dx}=u \rightarrow ux-y = C_2$$where C_2 is constant. At this point we already have 2 characteristics equations. There is no need for more : The ... 0 This is not an answer to the question about numerical method. Nevertheless, the result below can be numericaly evaluated in order to compare to the direct numerical methods.$$-cu_x+uu_x+u_{xxx}=0=(u-c)u_x+u_{xxx}$$This is an ODE of the autonomous kind. The usual change of function is : u_x=f(u) \quad\to\quad u_{xx}=f'f \quad\to\quad ... 1$$u_{xy}-xu_x+u=0$$This elliptic PDE is already on canonical form. Seach for particular solutions. Method of separation of variables, with u=f(x)g(y)$$f'g'-xf'g+fg=0$$\frac{g'}{g}=x-\frac{1}{\frac{f'}{f}}=(function of y)=(function of x)=constant=\lambda$$\begin{cases} \frac{g'}{g}=\lambda \\ (x-\lambda)f'-f=0 \end{cases} \qquad\to\qquad ...

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The method that would be used by Fourier is to separate variables, discard all of the resulting separated functions that are unbounded in the upper half plane, and then form integral linear combinations of the others. Start by separating variables with $u(x,y)=X(x)Y(y)$; to do this, plug into the equation and divide by $XY$: $$... 3 This is studied in potential theory: the function u is the Newtonian potential of f,$$u(x)=\int_{\mathbb{R}^n} K(x-y)f(y)\,dy$$where K(x)=c_n|x|^{2-n} for n\ne 2 and K(x)=c_2\log|x| for n=2. In dimensions n\ge 3 the kernel K decays at infinity, so u(x)\to 0 as |x|\to\infty in this case, provided f is reasonable (integrable and ... 1 Apart from a few typos, your derivation looks correct. There should be e^{-\alpha^2 t} in the final equation and I get x + \tan(x) = 0 (which is \cot(x) = -\frac{1}{x}), as the defining equation for the \alpha_n's. This follows from$$X(x) = \sin(\alpha x) \implies 0 = X(1) + X'(1) = \sin(\alpha) + \alpha \cos(\alpha) \implies \tan(\alpha) + \alpha ...

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This seems to work: The hyperbolicity implies $$\theta \int_\Omega |Du|^2 \, dx \leq \int \sum_{i,j} a^{ij} u_i u_j \, dx = B[u,u] - \int_\Omega c u^2 \, dx \leq B[u,u]\,,$$ since $c \geq 0$. For $u \in H_0^1(\Omega)$ the Poincaré inequality implies that there exists $C \geq 0$ such that $$\|u\|_{H^1(\Omega)}^2 = \|u\|_{L^2(\Omega)}^2 + ... 0 always with the same problem, i want to calculate the solution with separate variable methode. That what i try: we put$$u(x,t)= X(x) T(t) \neq 0$$then we have$$\dfrac{T'(t)}{T(t)}= \dfrac{X"(x)}{X(x)}$$So we obtain twe equations$$ T'(t) + \lambda T(t)= 0 $$and$$ X''(x)+ \lambda X(x)=0, X(0)=0, X(1)+ X'(1)=0 $$this second equation admits eigenvalues ... 0 Uniqueness holds, due to the condition that u=o(1) at infinity (that is, u(x)\to 0 as |x|\to\infty). Your function 1-1/|x| does not satisfy that. To show uniqueness, let u be the difference of two solutions. Given \epsilon>0, there is R such that |u|<\epsilon on the sphere |x|=R. By the maximum principle applied in the domain \{x ... 1 Your approach is perfectly appropriate. There was, however, a flaw in the execution of that way forward. Note that we have$$\begin{align} F(k)&=\langle H(a-|x|),e^{-ikx}\rangle\\\\ &=\frac{1}{-ik}\langle H(a-|x|),\frac{de^{-ikx}}{dx}\rangle\\\\ &=\frac{1}{-ik}\langle \text{sgn}(x)H'(a-|x|),e^{-ikx})\\\\ &=-\frac{1}{ik}\langle ...

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There could be several issues here. The first one is rather mathematical: your PDE + initial and boundary conditions could be ill-posed, that is, there could exist multiple solutions. It's not very easy to determine this by looking at the PDE; moreover, for physical applications, this is often regarded as a sign that the model is flawed. Assuming that your ...

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Setting $v=u_1-u_2$, define $I(t)=\frac{1}{2}\int_0^1v(x,t)^2dx$. Then $$\frac{dI}{dt}=\int_0^1v(x,t)v_t(x,t)dx=\int_0^1vv_{xx}dx$$ Integrating by parts, this becomes: $$\frac{dI}{dt}=\left[vv_x\right]_0^1-\int_0^1v_x^2dx$$ Now, note: $vv_x\vert_{x=0}=0,vv_x\vert_{x=1}=-v(1,t)^2$ by the conditions given, so: ...

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Anytime you see a finite domain, standard heat equation, think separation of variables. Try to use $u(x,y,t) = T(t)X(x)Y(y),$ and you'll get 3 ODE's to solve, something along the lines of \begin{align} T'(t) + c^2(\lambda_x + \lambda_y)T(t) &= 0\\ X'' - \lambda_x X &=0\\ Y'' - \lambda_y Y &=0. \end{align} This should be the motivated approach ...

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With respect to the new bounty: Here's a way to see this. Suppose $u$ is a compactly supported solution to $\bar{\partial}u=f$. Then we have, for large enough $R>0$ $$0=\int_{\partial D(0,R)} u(z)dz = 2i\int_{D(0,R)} \bar{\partial}u(z)\: d\bar{z}\wedge dz=2i\int_{D(0,R)} f(z)\:d\bar{z}\wedge dz$$ Taking $R\to\infty$, this implies that if a compactly ...

3

We can get ridd of the source term by defining $$f(x,y,z,t) = xy - ct + u(x,y,z,t)\tag{1}$$ so that the PDE for $u$ becomes $$ax^2u_{xx} + bx u_x + u_t = 0\tag{2}$$ We can then perform a change of variables $z = \log(x)$ to get $$au_{zz} + (b-a) u_z + u_t = 0\tag{3}$$ which is linear second order homogenous PDE with constant coefficients and you can ...

1

Let us write f as a multivariate polynomial or power series $$f(t,x,y) = \sum_{{k,l,m}\in{\mathbb N}^3}c_{klm}x^ky^lt^m$$ The partial differential operators will perform (sum omitted): $$\frac{\partial f} {\partial x} = kc_{klm}x^{k-1}y^lt^m \hspace{1cm} \frac{\partial f} {\partial y} = lc_{klm}x^{k}y^{l-1}t^m \hspace{1cm}\frac{\partial f} {\partial t} = ... 0 Wherever you have derivatives of \phi (first term of each of the integrals above), try to integrate by parts so that you pass the derivatives to other functions and you get just \phi. Then use the boundary conditions and factor out \phi to isolate the terms for the classical equation inside the integral. For instance:$$ 0= \int_{a}^{b} \big(f(x)u(x) ...

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The entire field of quantum mechanics is based on the Schrödinger equation which is a wave equation. Any other field that studies waves (like water waves in fluid dynamics or acoustics, signal theory, $\dots$) needs wave equations.

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One must convert the derivative operators first, \begin{align} \frac{\partial u}{\partial t} = \frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial t} + \frac{\partial u}{\partial \eta}\frac{\partial \eta}{\partial t} = -u_\xi + u_\eta \end{align} and likewise \begin{align} \frac{\partial u}{\partial x} = \frac{\partial u}{\partial \xi}\frac{\partial ...

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You can think of it like this: you can treat differential operators with constant coefficients like polynomials. So it boils down to the following computation: $$\frac{\partial \widetilde{u}}{\partial \xi} = \frac{\partial}{\partial \xi} u\left(\xi+\eta,\xi - \eta\right) = \frac{\partial u}{\partial x} + \frac{\partial u}{\partial t},$$and in the same way ...

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Let $\tilde{u}=v$. $$\dfrac{\partial^2 v(\xi,\eta)}{\partial t^2} - \dfrac{\partial^2 v(\xi,\eta)}{\partial x^2} = 0$$ Note that: $$\xi=\frac{x+t}{2}$$ $$\eta=\frac{x-t}{2}$$ Now use chain rule $$\dfrac{\partial v(\xi,\eta)}{\partial t} =\dfrac{\partial v(\xi,\eta)}{\partial \xi}\dfrac{\partial \xi}{\partial t}+\dfrac{\partial v(\xi,\eta)}{\partial ... 1$$\frac{du}{dx} + \frac{du}{dy} + u = e^{x+2y}$$Let : \quad u(x,y)=v(x,y)+\frac{1}{4}e^{x+2y}$$\frac{dv}{dx} + \frac{dv}{dy} = -v$$System of characteristic differential equations :$$\frac{dx}{1}=\frac{dy}{1}=\frac{dv}{-v}$$First characteristic equation from \frac{dx}{1}=\frac{dy}{1}\qquad\to\qquad x-y=c_1 Second characteristic equation from ... 1$$ y(t) = L^{-1} [ F(s) e^{-s} - F(s) e^{-2s}] = (t-1)u(t-1) - f(t-2)u(t-2)$$u(t-t_0) is the unit step function. It has the value 0 for all values of t<t_0 and has the value 1 for all values of t>t_0. Using this definition you can see that for 0<t<1 all unit step functions vanish. Resulting in y(t)=0. For 1<t<2 only ... 2 For t<1 we have u(t-1)=0 and for t<2 we have u(t-2)=0. So for t<1 we have f(t-1)u(t-1)=0 and f(t-2)u(t-2)=0 and then y(t)=0. For t> 1 we have u(t-1)=1 and for t<2 we have u(t-2)=0. So for 1<t<2 we have f(t-1)u(t-1)=f(t-1) and f(t-2)u(t-2)=0 and then$$y(t)=f(t-1)={1 \over 2} - e^{-(t-1)} + {1 \over 2} ...

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This isn't an answer to the question of sketching the solution. This is the analytical solution of the EDP : $$u_t+\cos(t)u_x=-u$$ Solving with the method of characteristics : The characteristic system of differential equations is : $$\frac{dt}{1}=\frac{dx}{\cos(t)}=\frac{du}{-u}$$ First characteristic equation : From ...

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They are not equivalent on an infinite-dimensional Hilbert space. The weak convergence for operators in this case is in the usually called "weak operator topology": $$A_n\xrightarrow{wot} A\ \ \iff\ \ \langle A_nx,y\rangle\to\langle Ax,y\rangle,\ \ \forall x,y\in H.$$ The weak operator topology is known to be coarser than the $\sigma$-weak operator ...

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Continuous but not in any Sobolev space: let $f(x)=g(x_1)$ where $g$ is a continuous but not absolutely continuous: Cantor staircase, or a Weierstrass-type nowhere differentiable function. Since Sobolev functions are absolutely continuous on almost every line segment parallel to a coordinate axis, $f$ is not in $W^{1,p}$. In $W^{1,p}(U)$ but without a ...

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Assume a solution of the form $$u(t, x, y) = c_{000} + c_{100}t + c_{010}x + c_{001}y + c_{200}t^{2} + c_{020}x^{2} + c_{002}y^{2} + c_{110}tx + c_{011}xy + c_{101}ty$$ Since $u(0, x, y) = x + y - 2x^{2}$, we find that $$u(t, x, y) = c_{100}t + x + y + c_{200}t^{2} - 2x^{2} + c_{110}tx + c_{101}ty$$ Write out partial derivatives: \begin{align} ...

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(1) and (2) are not the same in the sense that functions are defined in different domains unless $U=\mathbb{R}^d$. The notation $C(\overline{U})$ is seldom defined as in (2) in PDE books (I have never seen one), although the space $C(X)$ (which denotes the set of continuous complex or real functions) where $X$ is a compact Hausdorff space is well discussed ...

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You are talking about Fick's law of diffusion (https://en.wikipedia.org/wiki/Fick%27s_laws_of_diffusion). The negative sign comes from the fact, that diffusion always happens in the direction where concentration gets lesser (negative gradient). Imagine a situation where at $x=0$ you have no concentration of a particle and at $x=1$ you have 100% of one ...

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It does indeed hold. Well, if $\Omega$ is bounded, $u=0$ on $\Omega^c$ not just $\partial \Omega$, and with a slightly bigger integral on the right. On the plus side you only need one $k$. Your idea is already quite good. Now, you write $$u(x) = -h \sum_{l=0}^{N_h} D^h_k u(x+l h e_k),$$ where $N_h$ is such that $N_h h \geq diam(\Omega)$ and $N_h h \leq C$ ...

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In your original problem you need two derivatives of $u$, whereas for the weak formulation one suffices. So you have reduced the regularity required for $u$ by integrating by parts. Sobolev spaces have been created specifically to be the appropriate spaces where solutions of PDEs in work form live.

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initially we will be getting $\frac{ \mathrm dx + \mathrm dy + \mathrm dz } {\mathrm (x^2+y^2) + \mathrm (2xy) + \mathrm (z(x+y)^3 )}$ ----> eqn (1) we need to decide the factors to arrive at this condition $\mathrm A(x^2+y^2) + B(2xy) + C((x+y)^3)z = 0$ i.e. $\frac{ \mathrm A dx + \mathrm B dy + \mathrm C dz } {\mathrm A (x^2+y^2) + \mathrm B ... 1$\frac{dx}{x^2 + y^2} = \frac{dy}{2xy} = \frac{dz}{(x+y)^3z}$is not the general solution, but is the characteristic system of differential equations. HINT : To find the first characteristic equation observe that : $$\frac{dx}{x^2 + y^2} = \frac{dy}{2xy}=\frac{xdx-ydy}{x(x^2 + y^2)-2xy^2} =\frac{xdx-ydy}{x(x^2 - y^2)}$$ and solve the first ODE (simplify ... 0 Here is a simple proof of the mean-value theorem for harmonic functions of two variables. We start with Green's Third Identity $$u(\vec x_0)=\oint_C \left(u(\vec x')\frac{\partial G(\vec x_0,\vec x')}{\partial n'}-G(\vec x_0,\vec x')\frac{\partial u(\vec x')}{\partial n'}\right)\,d \ell' \tag 1$$ where$u(\vec x)$is harmonic in a region$Sbounded by ... 2 The most insightful way to study this equation is to write it as a two-dimensional dynamical system, like \begin{align} f' &= g, \\ g' &= g^2 -f. \tag{1} \end{align} Questions about the shape of solutions can best be answered by looking at the orbits of solutions in the(f,g)$phase plane. If you consider such an orbit as a graph$g(f)$, then ... 0 The ODE is on the "autonomous" kind :http://mathworld.wolfram.com/Autonomous.html The usual method to reduce the order is applied below. The result is the inverse function, on integral form. There is no closed form. Note : For easiness the symbol$f$is replaced by$y$. 0 The sign is wrong in the pressure term. The integration by parts in time gives $$\int_0^T (u_t,\varphi)=(u(T),\varphi(T))-(u(0),\varphi(0))-\int_0^T (u,\varphi_t)= -(u(0),\varphi(0))-\int_0^T (u,\varphi_t).$$ The correct formula should be$$\int_0^T[(u,\varphi_t)-(\nabla u,\nabla\varphi)-((\nabla\cdot u)u,\varphi)-(\nabla ... 1 Suppose$v_{xx}(x_0, t_0) < 0$for some$t_0 > 0$and$x_0$inside the spatial domain. This implies that$v_t(x_0, t_0) < 0$. Since$v$is$C^2$, this means that$v(x_0, t_0 - \delta) > v(x_0, t_0)$for a sufficiently small$\delta > 0$. Thus$t_0 = T$is not possible for a maximum. 1 It's OK for the chatacteristics system. Afterwards, it's a matter of combination of the differentials to put them on the form of exact differentials (below) :$F(X,Y)$is any differentiable function of two variables, with$X=(x+y)(u-1)$and$Y=(x-y)(u+1)\$.

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