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A) Use the first derivative test. Let $u_0$ be your function that minimizes the your functional. Let $w\in \Gamma$ be arbitrary. A computation shows \begin{align} E(u_0+tv) = \frac{1}{2} \int_\Omega |\nabla u_0|^2 dx + \frac{t^2}{2}\int_\Omega |v|^2 dx + t \int_\Omega \nabla u_0 \nabla v \,dx - \int_\Omega (u_0+tv) f \, dx \end{align} From which we deduce ...

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A good starting point with something like this (more item $ii$, but it can help with $i$ as well) is to figure out what the weak formulation is. Once you have that in hand, the functional analytic tools that you need will begin to become clearer. It can also sometimes be useful when you're setting up the weak formulation to initially ignore the subtle ...

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The solution is actually very simple:/, I did not realize that the term $\int \lVert \nabla u_n\rVert^2$ had a negative sign on the inequality that I have on lemma 7.5. Since it does, you can multiply everything by -1, and get the inequality that helps you to conclude just as 7.3.

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You correctly found the general solution : $$u(x,y,z)=F(X,Y)\quad\text{with}\quad\begin{cases}X=y-x\\Y=z-2x\end{cases}$$ Condition : $$u(x,0,z)=x+z=F(X,Y)\quad\text{with}\quad\begin{cases}X=-x\\Y=z-2x\end{cases}\quad\implies\quad \begin{cases}x=-X\\z=Y+2(-X)=Y-2X\end{cases}$$ $$F(X,Y)=x+z=(-X)+(Y-2X)=Y-3X$$ Now the function $F$ is known : $$F(X,Y)=Y-3X$$ We ...

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Hint Define $f(x) = f_1(x) + i f_2(x)$ as a complex map, find $f$ as the solution of a complex ODE. $f_1,f_2$ are then respectively the real and imaginary parts of $f$.

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Your system \begin{aligned} \|u\|_{k,\infty:\Omega} & = \max_{|a|\leq k}\operatorname{ess-sup}_{x \in \Omega} \|D^{\alpha} u\|\\ & = \max_{|a|\leq k}\max_{i=1,\dots, n}\operatorname{ess-sup}_{x \in \Omega_i} \|D^{\alpha} u\|\\ & = \max_{i=1,\dots, n}\max_{|a|\leq k}\operatorname{ess-sup}_{x \in \Omega_i} \|D^{\alpha} u\|\\ & = \max_{i=1,\... 1 Since \theta>2, one gets from the upper bound of DJ(u_n)u_n (\frac\theta2-1) \|\nabla u\|_{L^2}^2 \le Cm(\Omega) + \theta J(u_n) - DJ(u_n)u_n. $$Using the estimate as in Prop 7.3$$ (\frac\theta2-1) \|\nabla u\|_{L^2}^2 \le Cm(\Omega) + \theta J(u_n) +c\|DJ(u_n)\|_{H^{-1}}^2 +\frac12(\frac\theta2-1) \|\nabla u\|_{L^2}^2 $$with c = \frac12 (\frac\... 0 From what I understand (and recall from working in this area), the constructions are equivalent. It is therefore more convenient to do both steps at the same time. 0 From e^{x}\geq x+1\Rightarrow e^{k^2t}\geq k^2t+1\Rightarrow e^{-k^2t}\leq\frac{1}{k^2t+1}. Hence if t>0 we have$$ \left|\sum^{\infty}_{k=1}\frac{2(-1)^{k+1}e^{-k^2 t}}{k}\sin(kx)\right|= 2\sum^{\infty}_{k=1}\frac{\left|\sin(kx)\right|}{k^3t+k}<\infty\textrm{ , }\forall x\in\textbf{R}\textrm{ and }t>0. $$Also for t=0 we have that the series ... 0 Let F(x,y)=16(x^2+y^2-1). Then F vanishes on x^2+y^2=1, and F_{xx}+F_{yy}=64.$$ F(1/4,1/\sqrt{2})=16(1/16+1/2-1)=1+8-16=-7. $$The answer is number 3. 1 Instead of the weak maximum principle you can apply the strong maximum principle. Assume that there exists a point x_0\in\Omega such that u(x_0)\leq 0, then there exists a point \tilde{x}_0 \in \Omega such that$$u(\tilde{x}_0) = \min_{\bar{\Omega}} usince g\geq 0. The strong maximum priciple now states that u is constant within \Omega, hence ... 1 Let z=x+y , Then y=z-x \dfrac{dy}{dx}=\dfrac{dz}{dx}-1 \therefore\dfrac{dz}{dx}-1=\sin z-e^x \dfrac{dz}{dx}=\sin z+1-e^x Which is similar as Solving ode similar to Adler's equation and follow the method in http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%... 1 \newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\... 1 Just to be clear, extending an aspect of @QiaochuYuan's answer: there is the very basic equality-of-mixed-partials criterion, that if F(x,y,z)=(f_x,f_y,f_z) for some scalar-valued f, then (because under mild hypotheses (f_x)_y=(f_y)_x), etc.), the partial with respect to y of the first component of F must be equal to the partial with respect to x ... 1 This is false; for example p, \phi could be constants and then \rho could be arbitrary. I imagine the intended interpretation is that it generally has no solutions if \rho is not constant, because gradients must be conservative and if \rho is not constant \rho times a conservative vector field generally won't be conservative. 1 Using what I explained in the commentary, you have : \begin{aligned} \int_{\Omega} V \cdot \nabla u & = \sum_{i=1}^n \int_{\Omega} V_i \partial_i u \\ & = \sum_{i=1}^n (-\int_{\Omega} \partial_i V_i u + \underbrace{\int_{\partial \Omega} V_i \ u \ n_i}_{=0} ) \text{ using integration by parts}\\ & = - \int_{\Omega} \sum_{i=1}^n \partial_i V_i u\... 1 You are absolutely on the right track. To clarify, all you really need to do is remember that you should be using definite integrals here in order to precisely wield the fundamental theorem of calculus. Indeed, if you have a solution then you can fix an x and integrate in time from 0 to t to see that (using the FTC) u_t(t,x) = u_t(0,x) + \int_0^t ...

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$\newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\... 1 Hint: You can suppose that $$u(x,y)=X(x)Y(y)$$so, you can see that $$u_{x}=YX' \quad u_{yy}=XY''$$Then, $$u_{x}=2u_{yy} \iff YX'=2XY'' \iff \frac{X'}{2X}=\frac{Y''}{Y}=-\lambda$$ and you can the use method of separation of variables that you know in ODE. 1 Let's think in terms of characteristics, and assume a parametrization$(x(t), t)$of the coordinates such that$x'(t) = f(t)$. Using the chain rule for the time-derivative of$u(x(t), t)$, we find $$\frac{\text d}{\text d t} u = u_t + f(t) u_x = h \, ,$$ i.e.$u'(t) = h\big(t, u(t), x(t)\big)$. Local existence and uniqueness results follow from the study ... 0 There are solution$x^m$of$(x^2 f')'+\lambda f=0$, where $$m(m+1)+\lambda=0 \\ m^2+m+\lambda = 0 \\ (m+1/2)^2=1/4-\lambda \\ m=-1/2\pm\sqrt{1/4-\lambda}.$$ For convenience, normalize this solution at$x=1$by imposing$f(1)=0$,$f'(1)=1$, which gives $$... 2 Via the boundary conditions given, we can restrict the solution set of the PDE to a single function (which depends on f(x)). This function will always have a saddle point at (x,y)=(0,0). Let u = p(x)q(y) the PDE becomes$$\Delta u = u_{xx}+u_{yy} = p''q + pq'' = 0 $$which gives$$-\frac{p''}{p} = \frac{q''}{q} = k$$for some constant k\in \mathbb{R} ... 1 Use$$2 \eta |u| |Du||D\eta| =2(\eta |Du|)(|u||D\eta)\le\epsilon \eta ^2 |Du|^2 +|u|^2|D\eta|^2/\epsilon.$$Now choose \epsilon so that$$ \frac{M\epsilon}{\theta} =\frac 12.$$Then$$\int _U \eta^2 |Du|^2 \le \frac {M}{\theta \epsilon} \int _U u^2 |D\eta |^2 +\frac 12 \int _U \eta^2|Du|^2$$and your inequality is obtained by moving the second term on the ... 3 The +1 is time. Thus the independent variables considered together in, say, a 2+1 dimensional model live in a 3 dimensional space, but when considering time separate from space it makes sense to write it as 2+1, to emphasize that the spatial dimension is 2 but the dynamical problem is under investigation. I find this terminology is most common in ... 1 So far you have$$ u(x, t) = \sum_n \sin\left(\frac{n\pi x}{l}\right)\left[ A_n e^{\lambda_+ t} + B_n e^{\lambda_- t} \right] \tag{1} $$where$$ \lambda_\pm = \frac{1}{2}\left[1 \pm \sqrt{1 - \left(\frac{2n\pi}{l}\right)^2}\right] $$Now consider the initial conditions u(x, 0) Replacing that in (1) you get$$ \sin x = \sum_n \sin\left(\frac{n\pi x}{l}\... 0 This is an error as noted in the errata for the second edition, maintained on Evans' homepage: Chapter 3 [...] page 132, line 5: Change “C.4” to “C.5” C.5 is about mollification; in particular you will want Theorem 7 of C.5. 2 $$\frac{\partial^2 u}{\partial x \partial y} =0$$ integrating with respect to$y$we get $$\frac{\partial u}{\partial x} = f(x)$$ and similaty integrating with respect to$x$we get $$u(x,y) =F(x) +G(y)$$ We have $$f(x)=\frac{\partial u}{\partial x} (x, x^2 ) =\sqrt{|x|}$$ therefore $$F(x) = \int \sqrt{|x|} dx$$ and hence $$F(x) =\frac{2}{3}\text{sign} (x)|... 0 I got the answer in the comments under my question. The correct code should be like this: weqn = 3D[u[x,y],{x,2}] + 8D[u[x,y],{x}, {y}] -3D[u[x,y],{y,2}] - 60D[u[x,y],x] +20D[u[x,y],y] == 0 ic = {u[x,0] ==-4E^(2x), Derivative[0,1][u][x,0] == -9x E^(2x)} sol = DSolve[{weqn, ic},u,{x,y}] You should use E instead e, because mathematica. 2 Solving xy \, u(x,y)=0: We know that x\,v(x)=0 has solutions v(x)=C\,\delta(x), where C is a constant. Generalizing this we get that y\,u(x,y) = A(y)\,\delta(x) for some distribution A(y). Then u(x,y) = B(y)\,\delta(x) + C(x)\,\delta(y), where A(y)=y\,B(y) and C(x) is some distribution. Thus,$$ u(x,y) = \delta(x) \otimes B(y) + C(x) \... 3 $$(t+1)u_t-u=(t+1)u_{xx}$$ $$\dfrac {(t+1)u_t-u}{(t+1)^2}=\dfrac {u_{xx}}{t+1}$$ $$\left(\dfrac u {t+1} \right)_t=\left (\dfrac {u}{t+1}\right)_{xx}$$ $$w_t(x,t)=w_{xx}(x,t)$$ Where$w(x,t)=\dfrac {u(x,t)} {t+1}$1 What you did is correct but the final result is missing. $$u(x,y)=f(y)+g(y-2x)$$ with arbitrary functions$f$and$g$. Another method to find the above result : $$u_{xx}+2u_{xy}=\frac{\partial}{\partial x}(u_x+2u_y)=0$$ Integrate : $$u_x+2u_y=\phi(y)\quad\text{any function}\quad \phi$$ First order linear PDE through method of characteristics : $$u(x,y)=\int \... 1 u(s,s) denotes the particular case when x=s and y=s for any values of s. Since the general solution is u(x,y)=f(y-x) thus$$u(s,s)=f(s-s)=f(0)=1$$This means that any function f which as the property f(0)=1 satisfies the condition. They are an infinity of functions such as f(0)=1. For examples f(X)=X+1 or f(X)=(X+1)^n or f(X)=\cos(X) or ... 1 We use that \left|\sum_{k=1}^{n}\frac{(-1)^ksin(kx)}{k}\right|\leq C which is a classic result; see for example this MSE post (edited per comments \sum_{k=1}^{n}\frac{(-1)^{k+1} sin(kx)}{k}= \sum_{k=1}^{n}\frac{sin(kx)}{k}- \sum_{k=1}^{[n/2]}\frac{sin(k(2x))}{k}, so the bound for the signed sum follows from the bound from the unsigned sum applied for x, ... 1 A simple example can be constructed as follows: consider the positive orthant$$ \Bbb R_+^n=\times_1^n[0,\infty[ $$and the associated characteristic function, i.e.$$ \chi_{\Bbb R_+^n} (x)= \begin{cases} 1 & x\in \Bbb R_+^n\\ 0 & x\notin \Bbb R_+^n \end{cases}. $$Then the function$$ g(x)= \chi_{\Bbb R_+^n} (x) e^{-\| x\|^2} $$belongs to L^p(\... 0 What you are adding is not an initial condition, but a boundary condition at x=0. The initial condition is given for x>0. The initial condition determines the solution only in the sector x\ge t. To determine the solution at a point with x<t you should consider the characteristic with slope =1 through that point. Using the same reasoning as ... 2 By changing the dependent variable u to a new dependent variable v by the substitution u(x,t) = v(x,t) + \phi (x) , the original problem can be reduced to two decoupled ones: precisely, by using we have that$$ \begin{split} \frac{\partial u}{\partial x} &= \frac{\partial v}{\partial x} + \phi '(x), \\ \frac{\partial^2 u}{\partial x^2} &= \... 1 We know that$u(x,t) = h(x-t)$. Notice that$x,t > 0$. The initial conditions give us$h(x) = f(x)$for$ x>0$, so$h(x-t) = f(x-t)$for$x>t$. Also,$h(-t) = g(t)$, so$h(x-t) = g(t-x)$for$t>x$. Thus$\begin{equation*} u(x,t) = \begin{cases} f(x-t) & x>t\\ g(t-x) & x<t ...

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I finally managed to find a nice reference specifically for the three-dimensional situation: Cantarella, Jason; DeTurck, Dennis; Gluck, Herman (Mai 2002). „Vector Calculus and the Topology of Domains in 3-Space“. The American Mathematical Monthly 109.5, S. 409–442. doi: 10.1080/00029890.2002.11919870.

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Let $I = (a,b)\subset \mathbb{R}$, $a<b$, be an interval. Let $\varphi \in C_c^\infty(I)$, hence, $\phi$ is smooth and has compact support in $I$. Since $\varphi$ has compact support on an interval $[c,d] \subset (a,b)$, we can extend it to a function on $\mathbb{R}$ by setting $\varphi=0$ on $\mathbb{R}\setminus [c,d]$. Now we have $\varphi\in C_c^\infty(... 0 The two PDEs are transformed into one same PDE by convenient change of variable. $$\text{With}\quad x=e^X\qquad (\partial_t + {x}\, \partial_x) u = 0\quad\implies\quad (\partial_t + \partial_X) u = 0\tag 1$$ $$\text{With}\quad x=-\frac{1}{X}\qquad (\partial_t + {x^2}\, \partial_x) u = 0\quad\implies\quad (\partial_t + \partial_X) u = 0\tag 2$$ In both ... 1 If$h \in L^1_{loc}(\mathbb{R})$does not have a weak derivative, then you can define$f$via $$f(x) = h(x_j)$$ and$f$will not have a weak$j$-th derivative. 1 Since$\tilde u$is bounded in$B(x_0,r)$,$w=u-\tilde u=o(\Phi(x-x_0))$as$x\to x_0$: $$\lim\limits_{x\to x_0}\frac{w(x)}{\Phi(x-x_0)}=\lim\limits_{x\to x_0}\frac{u(x)-\tilde u(x)}{\Phi(x-x_0)}=0$$ since$\Phi(x-x_0)\to\infty$as$x\to x_0$and$\tilde u(x)$is bounded. 1 We have $$|-\langle dE_{\lambda}(u_m), u_m\rangle| \leq \|dE_{\lambda}(u_m)\| \cdot \|u_m\| \leq (1+\|u_m\| )\|dE_{\lambda}(u_m)\| = (1+\|u_m\| ) o(1),$$ where the last inequality is because of$dE_{\lambda}(u_m)\to 0$. 2 Actually, you were almost there: $$A'(\tau)B(\tau)=B'(\tau)A(\tau)\implies A'(\tau)B(\tau)-B'(\tau)A(\tau)=0,$$ and not the product rule, but the quotient rule is relevant, now: $$\frac{d}{d\tau}\,\frac{A(\tau)}{B(\tau)}=\frac{A'(\tau)B(\tau)-B'(\tau)A(\tau)}{B(\tau)^2}=0.\tag{quotient}$$ So$\displaystyle\frac{A(\tau)}{B(\tau)}$must be a constant, i.e.$0$... 1 In short, it is not necessary to use the sigmoid function at every layer. Check out Wikipedia for activation functions, sigmoid function and identity function are both activation functions. The main difference is that sigmoid function has range in$[0,1]$and is nonlinear, whereas identity function is linear and has no restriction for its output. From my ... 1 Mathematically, artificial neural networks are just mathematical functions. You can apply whatever function you want for each neuron it is still a function. If you want to apply the sigmoid function only in the last layer you can do that. However, activation functions have a certain purpose. They make a neural network more powerful. Observe that a ... 1 We first notice that proving the statement for real polynomials implies the result in complex case. Thus, I assume that$p \in \mathbb{R}[x_1, \ldots, x_n]$. The statement means that$\nabla p(x)$and$x$are not collinear everywhere. Arguing by contradiction, assume$\nabla p(x) \wedge x$vanishes everywhere. But $$0 = |\nabla p(x) \wedge x|^2 = \|\nabla p(... 0 Solution using the method of characteristics: We need to solve$$u_{x}+bu_{y}+cu=0 \implies \underbrace{\left(-\frac{1}{c}\right)}_{m}u_{x}+\underbrace{\left(-\frac{b}{c}\right)}_{n}u_{y}-u=0\iff \boxed{mu_{x}+nu_{y}-u=0} $$Now, we have a quasilinear partial differential equation and the relations between the differentials is$$\boxed{\frac{dx}{m}=\frac{dy}{... 1 $$u_x+bu_y=-cu$$ https://en.wikipedia.org/wiki/Method_of_characteristics $$\begin{cases} dx=ds\\ dy=b\:ds\\ du=-cu\:ds \end{cases}\quad\implies\quad \frac{dx}{1}=\frac{dy}{b}=\frac{du}{-cu}=ds$$ A first characteristic equation comes from solving$\quad dx=\frac{dy}{b}$$$bx-y=c_1$$ A second characteristic equation comes from solving$\quad\frac{dx}{1}=\frac{...

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Let $f$ be continuous on $[0,1]^k$ (hence uniformly continuous). Define (for fixed $n$ and $x=(x_1,...,x_k)$) $$B_n(f)(x) = \sum_{0 \le j_1,j_2,...,j_k \le n} f(\frac{j_1}{n},...,\frac{j_k}{n}) \prod_{i=1}^k {n \choose j_i}x_i^{j_i}(1-x_i)^{n-j_i}$$ Take $S_{n,i}(x) = \sum_{j=1}^n X_{j,i}(x)$, where $X_{j,i} \sim \mathcal B(1,x_i)$ (so that \$\mathbb P(X_{j,...

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