The material surface is $F(x,y,t)=y-\eta (x,t)=0$. Its material derivative vanishes, giving the exact kinematic conditionFor irrotational flow $u=\nabla \varphi$, the Euler equations and $(u\cdot \nabla )u=\nabla (|u{|}^2/2)$ integrate spatially to ${\varphi }_t+|\nabla \varphi {|}^2/2+gy+p/\rho =C(t)$. Adding a suitable function of time to $\varphi$ changes no velocity and makes the right side $p_a/\rho$. ThusTake $p_a$ as the air-side pressure and use the given capillary pressure jump on the fluid side. At first order in wave amplitude, the product ${\varphi }_x{\eta }_x$ and kinetic-energy term are quadratic, the curvature denominator is one to the required order, and replacing $y=\eta$ by $y=0$ changes first-order quantities only at second order. ConsequentlyFor a harmonic wave $\eta =A\cos (kx-\omega t)$, the harmonic potential decaying with depth isIt satisfies the kinematic condition. Substituting into the dynamic condition givesThe phase speed is $c=\omega /k$ and the group velocity of a narrow wave packet is $c_g=d\omega /dk$. For the positive-frequency branch and $\sigma =T/\rho$,For positive surface tension this ratio exceeds one exactly when $\sigma k^2>g$, proving

On the circle put $\zeta =a{e}^{i\theta }$. The Joukowski map gives $z=2a\cos \theta$, so its image is the slit $S=[-2a,2a]$. Away from that circle, the inverse equation is ${\zeta }^2-z\zeta +a^2=0$. For $z\notin S$, neither root has modulus $a$, and their product is $a^2$, so exactly one root has modulus greater than $a$. Selecting that root defines the exterior inverse.

More explicitly, define $s(z)=\sqrt{z^2-4a^2}$ to be the branch analytic on $\mathbb{C}\setminus S$ normalized by $s(z)/z\to 1$ at infinity. It is positive for real $z>2a$ and negative for real $z<-2a$. ThenExistence and analyticity of this branch also follow from the exterior inverse: $J^{\prime }(\zeta )=1-a^2/{\zeta }^2$ is nonzero for $|\zeta |>a$, so the selected simple root varies analytically; $s=2g-z$ has the stated normalization. The exterior map is injective because $J({\zeta }_1)=J({\zeta }_2)$ with distinct roots would imply ${\zeta }_1{\zeta }_2=a^2$, impossible for two exterior points. Thus $g$ is a conformal bijection throughout $\mathbb{C}\setminus S$.

The proposed potential is a uniform stream plus its cylinder doublet and the circulation term. On $\zeta =a{e}^{i\theta }$, the first two terms sum to the real quantity $2Ua\cos (\theta -\alpha )$, while the imaginary part of the logarithmic term is the constant $-\Gamma \log a/(2\pi )$. Thus the cylinder is a streamline. At infinity the complex velocity tends to $U{e}^{-i\alpha }$ and the circulation is $\Gamma$, verifying both far-field conditions. Logarithms may be chosen locally; their derivative gives a single-valued velocity.

Composition therefore gives the plate potentialAt the edge $z=2a$, corresponding to $\zeta =a$, the physical complex velocity is $W^{\prime }(\zeta )/J^{\prime }(\zeta )$. Since the denominator has a simple zero, finiteness requireswhich is exactlyWith this cancellation, the limiting ratio is $W^{\prime \prime }(a)/J^{\prime \prime }(a)$. The Kutta value of the circulation gives $W^{\prime \prime }(a)=2U\cos \alpha /a$ and $J^{\prime \prime }(a)=2/a$. Hence the finite edge velocity is

The general time-independent Schrödinger equation is $[-{\hslash }^2{\nabla }^2/(2m)+V]\psi =E\psi$. For a spherically symmetric wave function, ${\nabla }^2\psi ={\psi }^{\prime \prime }+2{\psi }^{\prime }/r$; the Coulomb attraction between the proton and electron is $V=-q^2/(4\pi {ϵ}_{0}r)$. These supply respectively the kinetic and potential terms in the displayed radial equation. Rearranging identifiesas the Bohr radius.

Substitute $\psi =(A+Br)e^{-\lambda r}$ into the rearranged equation and put $ϵ=2mE/{\hslash }^2$. Equating the coefficients of $r,1,r^{-1}$ after cancelling the exponential givesA nonzero state has $A\ne 0$. If $B=0$, these identities give $\lambda =1/a$ and $ϵ=-1/a^2$. If $B\ne 0$, they give $\lambda =1/(2a)$, $B=-A/(2a)$, and $ϵ=-1/(4a^2)$. Thus the two independent eigenfunctions and their corresponding energies areThey are eigenfunctions at different energies, rather than two independent solutions at one fixed energy.

For the higher-energy state, radial probabilities include the spherical volume factor $4\pi r^2$. The normalization constant and $4\pi$ cancel in the ratio. Using the supplied integral at $b=a$ and at infinity,HenceThis calculation fixes the denominator as $8e$ in the source's unparenthesized expression; interpreting it as a product with $e$ would give a negative probability.

For symmetric quadratic coefficient matrices, the Euler-Lagrange equations give $T\ddot{q}+Vq=0$, where $T$ is the positive-definite mass matrix. Substitute a normal-mode oscillation $q=v\cos (\omega t+\varphi )$ to getThe positive roots ${\omega }^2$ determine the squared normal frequencies and their nonzero null vectors give mode shapes. Repeated eigenvalues give multidimensional mode spaces.

The particle's kinetic energy is $m|\dot{q}{|}^2/2$. For a taut elastic string at length $r=\sqrt{x^2+y^2+z^2}$, Hooke's law gives tension $\lambda (r-a)/a$ and elastic energy $\lambda (r-a{)}^2/(2a)$. With $\lambda =mg$ and gravitational potential $mgz$, this gives exactlyThis is the taut-string regime containing the equilibrium. Differentiating the potential givesAt $(0,0,-2a)$, $r=2a$ and all three accelerations vanish.

With $z=-2a+\zeta$, expansion to quadratic order gives $r=2a-\zeta +(x^2+y^2)/(4a)+O(3)$. Substitution cancels the linear terms, leavingThe mass matrix remains $mI$, and the stiffness matrix is $\mathrm{diag}(mg/(2a),mg/(2a),mg/a)$. ThereforeThe horizontal frequency is doubly degenerate: a linearly polarized oscillation in any horizontal direction is a normal mode. The remaining mode is vertical stretching and contraction of the string, with horizontal coordinates zero.

For a normal constrained extremal, introduce a constant Lagrange multiplier and use $L=F-\lambda G$. The Euler-Lagrange equation isWhen both integrands have no explicit $x$ dependence, the Beltrami identity gives $L-y^{\prime }{L}_{y^{\prime }}=\text{constant}$, as follows by differentiating this expression and using the Euler-Lagrange equation.

Parametrize the surface by $(f(z)\cos \theta ,f(z)\sin \theta ,z)$. The cross product of its two coordinate tangents has magnitude $f\sqrt{1+(f^{\prime }{)}^2}$, so integration in $\theta$ gives $A=2\pi {\int }_a^{b}f\sqrt{1+(f^{\prime }{)}^2}dz$. Integrating the cross-sectional disk areas gives $V=\pi {\int }_a^b{f}^{2}dz$.

Dividing the augmented functional $A-\lambda V$ by $\pi$, useThe Euler-Lagrange equation isThe Beltrami identity becomes $2f/\sqrt{1+(f^{\prime }{)}^2}-\lambda f^2=C$. At an endpoint $f\to 0$, the absolute value of the first term is at most $2f$, so $C=0$, even if the profile has an infinite endpoint slope. For a nondegenerate regular closed surface take $f>0$ on $(a,b)$. Then $\lambda >0$, and, with $R=2/\lambda$,Combining this identity with the Euler-Lagrange equation yields $f^{\prime \prime }=-R^2/f^3$. Consequently $(f^2{)}^{\prime \prime }=2((f^{\prime }{)}^2+f{f}^{\prime \prime })=-2$. The two zero endpoint values uniquely determineThe positive radius profile is therefore a semicircle and generates a sphere, centred at height $(a+b)/2$, with radius $R=(b-a)/2$. In particularThe graph is smooth in the interior; its vertical slopes at the two poles are compatible with a smooth spherical surface. Its volume is $4\pi R^3/3$, the volume associated with this stationary candidate.

For minimality, the three-dimensional isoperimetric inequality says that every closed surface enclosing volume $V$ has $A^3\ge 36\pi V^2$, with equality for a sphere. The displayed extremal attains equality, proving that it minimizes area at its fixed enclosed volume. The uniqueness above concerns the regular connected positive profile with the two stated endpoint heights.

A sequentially compact space is one in which every sequence has a subsequence converging to a point of the space. A limit point $\ell$ of $S$ means every open ball around $\ell$ meets $S\setminus \{\ell \}$.

In a compact metric space, a sequence with a finite set of values has a constant subsequence. If its value set is infinite and had no limit point, each point of the space would have a neighbourhood containing at most one point of that value set. A finite subcover would then make the value set finite, a contradiction. It therefore has a limit point in the space, and the permitted subsequence result gives a convergent subsequence. This proves sequential compactness.

For the remaining assertions assume the compact space is nonempty, as required for a fixed-point claim. The displacement function $q(x)=d(f(x),x)$ is continuous by the stated continuity of the metric and of $f$. The extreme value theorem gives a minimum $\delta =q(p)$. If $\delta >0$, then $p\ne f(p)$, and strict decrease of distances impliescontradicting minimality. Thus $\delta =0$ and $p$ is a fixed point. Two distinct fixed points would satisfy $d(p,q)=d(f(p),f(q))<d(p,q)$, also impossible. Hence the fixed point is unique.

On the noncompact real line, takeIts derivative $e^x/(1+e^x)$ is strictly between zero and one. The mean value theorem therefore gives $|f(x)-f(y)|<|x-y|$ for each distinct pair. Yet $f(x)-x=\log (1+e^{-x})>0$, so it has no fixed point. Compactness, rather than merely pointwise strict shrinkage, supplies attainment of a zero displacement in the preceding proof.

A connected space has no separation into two disjoint nonempty open subsets whose union is the space. A continuous surjection to the discrete space $\{0,1\}$ produces such a separation by the two point preimages. Conversely, from a separation define a function equal to zero on one open piece and one on the other; it is continuous and surjective. Thus these formulations of connectedness are equivalent.

An interval in $\mathbb{R}$ is connected: if a continuous map from it to $\{0,1\}$ took both values, the intermediate value theorem would force value $1/2$, impossible. Conversely, a subset of $\mathbb{R}$ that is not an interval has points $a<b$ in it and a point $c\in (a,b)$ outside it. Intersecting it with $(-\infty ,c)$ and $(c,\infty )$ separates it. The empty set is connected, and singleton intervals are included.

For the union assertion, any separation of $X$ must put each nonempty connected $A_i$ entirely on one side, since intersecting the separation with $A_i$ would otherwise disconnect it. Pairwise nonempty intersections force all the $A_i$ onto the same side. Their union is $X$, contradicting a nonempty second side. The empty-union case is already connected.

The plane is the union of all lines through the origin. Each line is homeomorphic to $\mathbb{R}$ and connected, and every pair meets at the origin, so ${\mathbb{R}}^2$ is connected. For the punctured plane, let $R_{+}$ be the positive horizontal ray, and for each $r>0$ let $C_r$ be the circle of radius $r$. Each circle is a continuous image of $[0,2\pi ]$ and hence connected; $C_r\cup R_{+}$ is connected because its two connected pieces meet at $(r,0)$. These unions all contain $R_{+}$ and cover the punctured plane, proving it connected as well.

Finally suppose a continuous injection $g:{\mathbb{R}}^2\to \mathbb{R}$ existed. Its image is a connected interval, since a continuous image of a connected space is connected. It contains at least three distinct values; choose $p$ whose value lies strictly between two others. The punctured plane ${\mathbb{R}}^2\setminus \{p\}$ is connected by translation of the preceding result. By injectivity its image is precisely $g({\mathbb{R}}^2)\setminus \{g(p)\}$, which is disconnected by the missing interior point. This contradicts connectedness of a continuous image. Therefore