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There are basically two definitions of convolution of a distribution $F \in \mathcal{S}'$ with a Schwartz function $\psi$, which turn out to be equivalent (cf. Folland, Real Analysis, Proposition 9.10). The first definition is to define $F \ast \psi$ as a pointwise-defined function(!), $$F \ast \psi (x) = \langle F, \tau_x \widetilde{\psi}\rangle,$$ ...

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On page $1$, the authors say: The value of $Lu(x)$ is well defined as long as $u$ is bounded in $\mathbb R^n$ and $C^{ 1,1}$ at $x$. The boundedness assumption is not stated for a few pages after that, but reappears in Definition 3.1.

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So: Yes. $u$ is AC iff $\exists w\in L^1_{loc}$ such that $u=\int w dx$. No. Distributions are by definition functionals, not function. Some distributions can be identified with a function (such as linear functionals on $L^1_{loc}$), but in general it does not make sense to talk about the value of a distribution at a point in space. You can only talk about ...

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Another approach to set a regularity condition is to make a change of variables to switch the type of boundary condition. For this equation, if $f(r)$ is regular at $r=0$, then $\frac{\text{d}}{\text{d}\rho}f(\rho^2)=0$ at $\rho=0$. We can switch to $g(\rho)=f(\rho^2)$ and rewrite the equation for $g(\rho)$: ...

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Ok, let me elaborate on my comment. The point is that we don't need convergence of the derivative, but only a bound on the derivative (which - for differentiable functions - is the same as a Lipschitz estimate). Take any "mollifier" $\varphi \in C_c^\infty (\Bbb{R})$ with $\varphi \geq 0$, $\int \varphi \, dx = 1$ and $\rm{supp}(\varphi) \subset (-1,1)$. ...

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Let $f(t) = \begin{cases}e^{-1/t} & t > 0 \\ 0 & t \le 0\end{cases}$, $g(t) = \dfrac{f(2-t)}{f(t-1)+f(2-t)}$, and $\phi(t) = g(|t|)$. It is well known that $f \in C^{\infty}$. Since $\max\{t-1,2-t\} \ge \frac{1}{2} > 0$, one of $f(t-1)$ and $f(2-t)$ is strictly positive. The other is non-negative. Hence $f(t-1)+f(2-t) > 0$. Thus, $g \in ... 0 The order of a differential equation is the highest ordered partial derivative in the equation. In your example, the orders of each term (from left to right) are 2, 3, 2 and 1. Therefore the order the equation is 3. A similar classification is the degree of the differential equation. This is the maximum exponent of the highest ordered term of the ... 4 Jose,the order of the differential equation is the maximum power that a partial derivative has, and obviously,this is equal to$3$. 3 Finding the general solution of the PDE (below) is not the more difficult part of the task : As usual, the arduous part is to determine, among the infinite number of solutions which one fits with the boundary conditions : 0 Lets assume you have a diffeomorphism$\phi:M_1 \mapsto M_2 $then I would map the solution on$M_2$back to$M_1$via the inverse mapping$\phi^{-1}$and compare the two solutions. I hope I understood your problem correctly ? 0 While it is true that$u_m'(t) \in H_0^1(U)$(since the basis is smooth in space), we only have a bound uniform in$m$on$u_m'$in the space$L^2(0,T;H^{-1}(U))$, so we only have the weak convergence in that space. If we had a bound on$u_m'$in$L^2(0,T;L^2(U))$or$L^2(0,T;H^1(U))$then we could say that$u_m' \rightharpoonup u$with the convergence in ... 0 Unless I'm missing something, the constraint (2)$\nabla \lambda \times \mathbf{A} = 0$means that the vector field must be of the form$\mathbf{A} = f \nabla \lambda$where f is a scalar function. Inserting into (1) yields$\nabla f \times \nabla \lambda = f \lambda \nabla \lambda$. The only possible solutions are$\nabla\lambda = 0$, so$\lambda$has ... 1 (Converted and expanded from a comment.) You can artificially add to your list of unknowns the coordinate functions$x$and$\mathbf{y}$. They solve first order PDEs of the form $$\partial_x x = 1$$ and $$\partial_x \mathbf{y} = 0.$$ So starting with a system$\mathbf{v}that solves $$\partial_x \mathbf{v} = \mathbf{g}(x,\mathbf{y}, \mathbf{v}, ... 1 No, the conjugate of a harmonic function that is continuous up to the boundary need not be continuous up to the boundary, or even bounded. This is related to the fact that the Hilbert transform does not preserve continuity (though it does preserve Hölder continuity of exponents \alpha\in (0,1)). Here is an example. Let F be a conformal map of the ... 0 I'm not sure what's confusing you. When you plug in i=1,j=5, because of how you wrote the formula, you are using the iteration to find the value of w_{1,6}, which is simply not a value of interest since it isn't in the mesh. The only thing I can think that might be confusing you is that your BC numerically says that w_{0,j}=0 for all j and ... 0 For zero and negative Yamabe cases the proof is relatively simple. For the full resolution one needs to use the positive mass theorem. 3 Why are you restricting yourself to x \in (0,\infty) when the domain is for the entire real line? In this case, it seems easier to go with Fourier transforms, i.e.,$$u(s,t) = \int_{-\infty}^{\infty} dx \, U(x,t) \, e^{i s x}$$Then the PDE becomes$$u_t + s^2 u = \sqrt{2 \pi} \, e^{-s^2/2} = e^{-s^2 t}\frac{d}{dt}\left [e^{s^2 t} u \right ]... 0 I do not think you can apply the first theorem to the second statement you want to prove, because there is an essential difficulty: The first theorem only applies to holomorphic functions. The classical Cauchy-Kovalevskaya Theorem also need to assume the functions are analytic. Otherwise I think there are counter-examples. For a correct statement of the ... 0 Lets check by direct computation. Let \phi be some test function and consider \partial_t P in the sense of distributions. \begin{align*} (\partial_t P, \phi) = & - (P , \partial_t \phi) \end{align*} Lets check the RHS first, since P = \delta^{(2)} ( \alpha - \beta (t) ), we have that - ( P, \partial_t \phi ) = - (\Im \partial_t \phi)( \alpha) ... 2 Of course it is! The method of characteristics for your linear 1st oder PDE reads: $$\frac{\mathrm{d}x}{1} = \frac{\mathrm{d}y}{1}= \frac{\mathrm{d}u}{u},$$ which leads tox-y = \xi$(from 1st and 2nd identities) and$u = \eta \, e^x$(from 1st and 3rd identities). Put$\eta = \eta(\xi ) = \eta(x-y)$and you are done. Note the following funny fact of ... 2 The change of variables should make one of the new variables constant on the characteristic lines. Thus try$w = x - y$,$z = y$. 0 If$u_{\varepsilon \eta} \times \text{something} = 0$, then$u_{\varepsilon \eta} = 0$or$\text{something} = 0$. If you go for the first option, you'll get the solution. $$u_{\varepsilon \eta} = 0 \implies u(\varepsilon,\eta) = f(\varepsilon) + g(\eta),$$ for some arbitrary functions$f$and$g$. Can you show us the intermediate steps? Cheers! 0 If you'll abide the strangeness of it, group theory (Lie theory) provides an answer. Use this stretching group: $$G(x,t,u)=(\lambda x,\lambda^\beta t, \lambda^\alpha u)\lambda_o=1$$$\lambda_o=1$is the unit transformation without which no group is complete. Plug these transformed variables into the equation and you get $$\lambda x \frac{\lambda^\alpha ... 0 You've got it so far. For the T equation, you have that T'' = \alpha^{2} \lambda T (you should really put the \alpha^{2} in this equation to make your life easier). You need a function that, when differentiated twice, gives you back the same function times a constant. You seem like you know what that is. For the X equation, you have X'' = \lambda ... 0 That looks correct. I think there are product solutions for each case. See the way it is done for the heat equation... ...or check out what I just found. 0 Multiply (40) by u_t and integrate:$$\int_{\mathbb{R}^N}u_t(u_t)_t-\int_{\mathbb{R}^N}u_t\Delta u_t=\int_{\mathbb{R}^N}f_tu_t.\tag{1}$$Use integration by parts in (1) combined with (u_t^2)_t/2=u_t(u_t)_t to conclude that$$\frac{1}{2}\int_{\mathbb{R}^N} (u_t^2)_t+\int_{\mathbb{R}^N} |Du_t|^2=\int_{\mathbb{R}^N} f_tu_t. \tag{2}$$Now integrate ... 1 In order to derive the Black-Scholes equation, you can do it in a more intuitive way (with economics). Also, this needs understanding of Brownian Motions and stochastic differential equations (SDEs), Îto's Lemma, and some basical assumptions that I will not list here (they're already available in Wikipedia and other references). Let's see it in two steps: ... 2 Let me answer a different more general question$$ v(t,S) = \mathrm{e}^{-rt}U\left(t,g(x,t)\right).\tag{1}  v_t = -r\mathrm{e}^{-rt}U\left(t,g(x,t)\right) + \mathrm{e}^{-rt}U_t.\tag{2} $$now you must be happy with Eq.(2)? so lets compute U_t,$$ U_t = \dfrac{\partial U}{\partial t} + \dfrac{\partial U}{\partial g}\dfrac{\partial g(x,t)}{\partial ... 1$ \newcommand{\pd}[2]{ \frac{\partial #1}{\partial #2} } $Attending to the positive answer to my request from the OP, consider the non-linear 1st order PDE: $$F(x_i,u,p_i) = 0, \quad i = 1,\ldots, n, \quad p_i = \pd{u}{x_i}, \quad u = u(x_1,\ldots,x_n). \tag{1}$$ Assume$F \in \mathcal{C}^1_{x_i}$and compute the partial derivative of eq.$(1)$with ... 0 You can proceed as @Chinny84 points out in his answer or in a more conventional way. The characteristic curves for the PDE: $$\square u = u_{tt} - u_{xx} = 0,$$ where we can define$A = 1, B = 0$and$C=-1$, so$\Delta = \sqrt{B^2-4AC} > 0classifies the PDE as hyperbolic; are given by the equations: \begin{align} A \xi_t ^2 + B \xi_t \xi_x + C ... 0 Let\alpha=c_1$,$\beta=c_2$and$U=f$(completely for shortening the latex solution, no real need to do that)$\frac{d}{dx}(f(x, y)e^{-(c_1 x)-c_2 y)}\cdot \frac{d}{dx}(uv) = v \frac{du}{ dx}+u\frac{dv}{dx}$, where$u = e^{c_1 (-x)-c_2 y}$and$v = f(x, y)=f(x, y) (\frac{d}{dx}(e^{-(x c_1)-y c_2}))+e^{-x c_1-y c_2} (\frac{d}{dx}(f(x, y)))\cdot$... 1 You remembered to use the chain rule, but apparently forgot the product rule. Use it once: $$V_x=e^{-ax-by}U_x-a e^{-ax-by}U$$ then use it again: $$V_{xx}=e^{-ax-by}U_{xx}- a e^{-ax-by}U_x - a e^{-ax-by}U_x + a^2 e^{-ax-by}U$$ You may notice the same term appearing in the middle twice; it saves a bit of time to remember the product formula for second ... 0 HINT Let$u = u(r(t,x),s(t,x))$. Use the chain rule to find$u_{tt}-u_{xx}$and compare coefficients. $$\begin{eqnarray*} \frac{\partial u}{\partial t} &=& \frac{\partial u}{\partial r}\frac{\partial r}{\partial t}+\frac{\partial u}{\partial s}\frac{\partial s}{\partial t} &=& c_{1,1}\frac{\partial u}{\partial r}+c_{2,1}\frac{\partial ... 0 \textbf{Hint:}$$ L = \partial_{tt} - \partial_{xx} = \left(\dfrac{\partial}{\partial t} + \dfrac{\partial}{\partial x}\right)\left(\dfrac{\partial}{\partial t} - \dfrac{\partial}{\partial x}\right) $$Now we can choose r and s such that the operators reduce to \partial_r\partial_s 1 In fact, \alpha\beta is optimal. To prove it, let f,g:[0,1]\to\mathbb{R} with f(x)=x^\alpha and g(x)=x^\beta. Note that f\circ g\in C^{\alpha\beta}, however, for all \gamma>\alpha\beta, f\circ g does not belong to C^\gamma. 1 There's a simpler explanation of why R'(0) = 0: You assumed your solution is radially symmetric, and performed a separation of variables. If R'(0) \neq 0 you have a cusp at the origin, where the function u is in fact not differentiable. So you require R'(0) = 0. The power series explanation is basically the same: you need$$ R'' + R'/r = c R $$... 0 Google these if not already done: Breather, Solilton And also check out Richard Palais's home page: http://vmm.math.uci.edu And in math overflow.. 1 Using abstract index notation: There are two types of indices here: I will use i,j,\ldots for indices relative to the domain (which appears to be \mathbb{R}^2) and A,B,\ldots for indices relative to the co-domain (basically from the number of equations) which is \mathbb{R}^N. We can write \mathbf{c} as the rank 4 tensor in index notation$$ ... 2 The Lagrange-Charpit equations tell us that: $$\frac{\mathrm{d}x }{3p^2} = \frac{\mathrm{d} y}{-1} = \frac{\mathrm{d}u }{3p^3-1} = -\frac{\mathrm{d} p}{0} = -\frac{\mathrm{d} q}{0}.$$ Thus, from the last two fractions we can conclude that: $$p = A, \quad q = B, \quad A,B \in \mathbb{R}.$$ If you separately integrate both equations$^1$, since$p = ...

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$$\dfrac{\partial{\phi}}{\partial{i}}=0$$ $$\dfrac{\partial{\phi}}{\partial{v}}=E-v-i R_0$$ Where E,$R_0$ are constants. Then (1) $\dfrac{\partial{\phi}}{\partial{i}}=0$ => ${\phi}=f(v)$ (2) $\dfrac{\partial{\phi}}{\partial{v}}=E-v-i R_0$ => ${\phi}=Ev-\dfrac{1}{2}v^2-i R_0v + f(i)$ Taking partial of (2) wrt $i$ yields: (3) ...

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If you want to do applied math without theory, then respectfully, you shouldn't go into applied math. Even applied mathematicians care about where things come from and how to justify them, so you won't be able to avoid proofs and theorems. With that said, a few of my favorite resources are as follows: Finite Difference Methods for Ordinary and Partial ...

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Anyway, here go some thoughts. Since you are using a Crank-Nicolson scheme, which is $\mathcal{O}(\Delta x^2, \Delta t^2)$ precise, your boundary conditions should also be discretize using a second order approximation, i.e.: \begin{align} \mu \varphi_x(0,t) - u \, \varphi(0,t) = 0 \implies & \mu \frac{\varphi_{1}- \varphi_{-1}}{2 \Delta x} - u \, ... 0 Do I understand your problem correctly if I state you want to solve the following problem (assumption A(x)=Id)  \hat U = \arg \min \{ \| U - Y \|^2_2 + \lambda \mathcal R( U) \}  where the data Y is given on a N \times N  grid and \mathcal R(.) is a convex regularization functional? Or is your data given only at certain points on the grid? 3 I suppose that i is a variable and not the square root of -1. On the one hand {∂_v}{∂_i}ϕ=0 and on the other {∂_i}{∂_v}ϕ=-R_0. So, if the two are equal, R_0 must vanish in order to avoid a contradiction. 0 You're solving the initial value problem: \dfrac{\partial}{\partial \sigma} x(\sigma,n) = x(\sigma,n),\ x(0,n) = n $$The variable n is not important here: this is just the ordinary differential equation$$ \dfrac{d}{d\sigma} x = x, \ x(0) = n You do remember how to solve constant-coefficient first-order linear equations, don't you?

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I'm not sure that your original question is well posed. If only $g$ is specified, then the boundary conditions are not fully-specified and the problem of solving for $\psi$ is ill-posed. I will continue, assuming that both $A$ and $g$ are specified and that you want to solve the problem for $\psi$ given the fully-specified boundary conditions. We have the ...

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Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example: $\dfrac{dx}{dt}=2x$ , letting $x(0)=1$ , we have $x=e^{2t}$ $\dfrac{dy}{dt}=x+y=e^{2t}+y$ , letting $y(0)=y_0+1$ , we have $y=e^{2t}+y_0e^t=x+y_0\sqrt x$ $\dfrac{du}{ds}=2u$ , letting $u(0)=f(y_0)$ , we have $u(x,y)=f(y_0)e^{2t}=f\left(\dfrac{y-x}{\sqrt x}\right)x$ ...

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I'll work in index notation with summation over doubled indices. In that same vein, I'll take $v_i$ rather than $v_{x_i}$ to be the the $x_i$-derivative of $v$. We are given that $\Delta$ has an eigenfunction $u$, i.e. $u_{ii}+\lambda u=0$ for some real $\lambda$. We also have some function $w$ such that $w_{ii}+\beta w<0$ for some real $\beta$. We now ...

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steady state solution? Does it mean that $u_t=0$? Yes, this is exactly what it means. The equation simplifies to $(u^2)_x=-2\sin x$, which integrates to $u=\pm \sqrt{2\cos x+B}$ for some constant $B$. For this to make sense, we need $B\ge 2$. This leads to a lower bound on $\int |u|$, attained when $B=2$. Show that the limiting solution when ...

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We start with a theorem: Theorem. Let $H$ be a Hilbert space and $T$ a compact self adjoint operator. Then there exist a Hilbert basis composed of eigenvectors of $T$. Proof. See Brezis chapter 6, in particular, theorem 6.11. Remark: Let $(\cdot,\cdot)$ denote inner product. In the conditions of the theorem, if $(Tu,u)> 0$ for all $u\neq 0$ ...

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