Codex @codex  0

The Maxwell equation $\nabla \times \mathbf{E}=-\partial \mathbf{B}/\partial t$, integrated over a fixed spanning surface $S$, gives by Stokes theoremThus the electromotive force and magnetic flux obey Faraday's law

Inside the infinite square well, the time-independent Schrödinger equation has Dirichlet boundary conditions at $0$ and $a$. Its normalized energy eigenstates and energy eigenvalues are

For $\lambda =2$, the Green representation of the boundary value problem isSplitting the integral at $\xi =x$ givesThereforeIt satisfies $y(0)=2y^{\prime }(0)$ and tends to zero as $x\to \infty$.

The solution $x+\lambda$ satisfies the left boundary condition, while $e^{-x}$ satisfies decay at infinity. The proposed expression for $x<\xi$ is therefore the correct left homogeneous solution. For $x>\xi$, write $G=C{e}^{-x}$. Continuity at $x=\xi$ givesHence the Green function isIndeed, its derivative has the required unit jump,which produces the Dirac delta distribution in $L[G]$.

Substitution of $y=e^{-x}$ givesFor a second solution $y=ax+b$, the ordinary differential equation reduces to $a\lambda -b=0$. Taking $a=1$ gives the linearly independent solution

For fixed endpoint values, the first variation of the functionalvanishes exactly when the Euler-Lagrange equationholds. If $f$ has no explicit $x$ dependence, differentiation along an extremal gives the Beltrami identityFor a further constraint $J[z]=J_0$, introduce a constant Lagrange multiplier $\lambda$ and apply the Euler-Lagrange equation to $I+\lambda J$; $\lambda$ is then chosen so that the constraint holds.

Apply the formula toThe upper-half-plane poles areand their residues are $e^{i{z}_k}/(4z_k^2)$. Writing $c=1/\sqrt{2}$, their sum isSince the cosine part of $x{e}^{ix}/(1+x^4)$ is odd, its integral vanishes, while the sine part is even. Henceso

Close the contour in the upper half-plane. The degree assumption and Jordan lemma make the integral over the large semicircle tend to zero, while the absence of real zeros avoids indentations of the contour. The residue theorem therefore givesRepeated roots are handled by the usual higher-order residue formula.

The helicoid has no angular identification: its height coordinate is $v$. ConsequentlyThe catenoid has total curvature $-2\pi$. A global Riemannian isometry preserves both the Gaussian curvature and the area element, hence preserves total curvature. ThereforeThe same obstruction is visible in the parameter map: increasing $v$ by $2\pi$ gives a different point of the helicoid but the same point of the catenoid.

For an oriented regular surface with unit normal $N$, the Gauss map sends each point $p$ to $N(p)\in S^2$. The outward normal to the catenoid isIts image is the open southern hemisphere, apart from its limiting equator and south pole. The Gauss map is one-to-one after taking $v$ modulo $2\pi$ and reverses orientation. Since the Jacobian of the Gauss map is the Gaussian curvature, its signed spherical area is the total Gaussian curvature:Equivalently, $KdA=-{\mathrm{sech}}^{2}ududv$, whose integral over $u>0$ and $0\le v<2\pi$ is $-2\pi$.

Withthe coordinate tangent vectors satisfyThus the first fundamental form isGeometrically, $S$ is the half of a helicoid with positive radius, winding upwards indefinitely.

Consider the upper half of the catenoid, parametrized as the surface of revolutionIt has the same first fundamental form, so $(u,v)\mapsto Y(u,v)$ defines a local isometry. By the invariance of Gaussian curvature under local isometry, the curvatures agree; directly, the conformal metric above gives