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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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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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