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In general when $f: \Bbb{R}^n \to \Bbb{R}$ is the euclidean norm, defined by $$f(x) = |x| = \sqrt{\sum_{i=1}^n x_i^2}$$ then away from the origin, $f$ is $C^{\infty}$ and its derivative is given by $$f'(x)(\eta) = \left\langle \dfrac{x}{|x|}, \eta \right\rangle$$ (for all $x \neq 0$, and for all $\eta \... 2 Briefly, the derivative$E'(u)$is a linear functional on an appropriate function space. Let$v\in C^\infty_c(\mathbb{R}), then $$E'(u)(v)=\lim_{t\to 0}\frac{E(u+tv)-E(u)}{t}$$ So \begin{align*} E'(u)(v)&=\int_{\mathbb{R}} [2u(x,t)v(x,t)+2u_x(x,t)v_x(x,t)]\,\mathrm{d}x\\ &=\int_{\mathbb{R}} 2(u(x,t)-u_{xx}(x,t))v(x,t)\,\mathrm{d}x \end{align*} So ... 2 The PDEu_t + u_x = f-u$is a non-homogeneous linear advection equation. Let us solve the initial value problem$u(x,0)=u_0(x)$by using the method of characteristics. We introduce the parametrization$x(t)$of$x$such that the total time-derivative of$u(x(t),t)\$ satisfies $$\frac{\text d u}{\text d t} = u_x \frac{\text d x}{\text d t} + u_t = f-u \, . ... 2 Let us introduce the vector {\bf U} = (\theta, \eta)^\top of unknowns. Since \partial_t {\bf U} is nonlinear in {\bf U}, the Crank-Nicolson method requires to invert the nonlinear algebraic system F_\Delta (\theta,\eta) = 0 from Eq. (30) at each time step. This step provides the updated values {\bf U}_m^{n+1} from the current values {\bf U}_m^{n}.... 1$$u(x,t)=f\left(t+\frac{x}{2}\right)+g\left(t-\frac{x}{2}\right)\qquad\text{ is OK.} \tag 1$$Next you are supposed to determine the functions f and g according to the specified conditions. First condition :$$\quad u(x,t=0)=0=f\left(0+\frac{x}{2}\right)+g\left(0-\frac{x}{2}\right)g\left(-\frac{x}{2}\right)=-f\left(\frac{x}{2}\right)Let X=-\... 1 It's due to continuity and connectedness. Let's write this a tiny bit cleaner and fill in the gaps explicitly. Suppose that there exists x_0\in \Omega, with u(x_0)=M=\max_{\bar{\Omega}}u, and let S=\{x\in \Omega: u(x)=M\}. We will show that this set is non-empty and both open and closed in \Omega. If we can show this, then connectedness will imply ... 1 Integrating by parts and using the divergence theorem, we can write \begin{align} (\operatorname{div}[(\mathbf{b}\cdot\nabla u)\mathbf{b}],v) &= \int_\Omega\operatorname{div}[(\mathbf{b}\cdot\nabla u)\mathbf{b}]v dx = \\ &= \int_\Omega\operatorname{div}[(\mathbf{b}\cdot\nabla u)\mathbf{b}v] dx - \int_\Omega(\mathbf{b}\... 1 In an axisymmetric flow without swirl, u^\theta = 0. Thus from your formulas, \omega^r = 0 = \omega^z, and hence \omega = \omega^\theta e_\theta. Now apply the Biot-Savart law Wikipedia link u = \nabla\times(-\Delta)^{-1} \omega.$$This comes from the fact that \omega = \nabla\times u and the double curl identity$$ \nabla\times(\nabla\times u) = \...