The free surface is material, so applying the material derivative to $y-\eta (x,t)$ gives the exact kinematic boundary conditionFor an incompressible fluid with constant density, the Euler equations areSince the flow is irrotational, $\mathbf{u}=\nabla \varphi$ and $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\nabla \varphi {|}^2/2)$. Integrating the resulting gradient equation gives a function of time on the right of Bernoulli equation. Adding a time-dependent constant to $\varphi$ removes that function, without changing velocity, and choosing the atmospheric-pressure reference givesThe pressure immediately above the surface is atmospheric; the liquid-side pressure includes the stated surface tension jump.

At first order in wave amplitude, the products ${\varphi }_x{\eta }_x$ and $|\nabla \varphi {|}^2$, the slope-squared correction to curvature, and displacement corrections in evaluating $\varphi$ at $y=\eta$ are all higher order. Hence, on $y=0$,For $\eta =A\cos (kx-\omega t)$, a harmonic velocity potential decaying as $y\to -\infty$ and satisfying the first condition isIndeed $\Delta \varphi =0$ and ${\varphi }_y{|}_0=A\omega \sin (kx-\omega t)={\eta }_t$. Substituting in the second condition yields the dispersion relation for a gravity-capillary wave:Taking the positive-frequency branch, the phase velocity and group velocity areSubtracting givesSince $\omega >0$,The threshold is also the wavenumber at the minimum phase speed of a gravity-capillary wave.

On $\zeta =a{e}^{i\theta }$, the Joukowski map gives $z=2a\cos \theta$, so the image is the segment $S=[-2a,2a]$. Solving the resulting quadratic gives the inverse candidatesOn $D=\mathbb{C}\setminus [-2a,2a]$, choose the analytic square root $s(z)$ of $z^2-4a^2$ characterized by $s(z)/z\to 1$ at infinity. One explicit definition iswhere the square root on the right is the principal square root. Its argument never lies on the nonpositive real axis for $z\in D$, so this defines the required holomorphic function. Then $g(z)=(z+s(z))/2\sim z$.

The two quadratic roots have product $a^2$. For $z\notin S$, neither root lies on $|\zeta |=a$, since that circle maps into $S$. Consequently exactly one root lies outside the circle, and it is the branch $g$ chosen at infinity. Also, if two exterior points have the same image, thenwhose second factor cannot vanish. Thus the Joukowski map is a conformal map from $|\zeta |>a$ onto $D$, and $g$ is conformal throughout $D$. The two endpoints are branch points and are excluded.

The complex potential in the circle plane is the sum of the uniform flow term $U{e}^{-i\alpha }\zeta$, its circle-image term $U{a}^2{e}^{i\alpha }/\zeta$, and the point vortex term $-i\Gamma \log \zeta /(2\pi )$. On $\zeta =a{e}^{i\theta }$, the first two sum to $2Ua\cos (\theta -\alpha )$, which is real, while the imaginary part of the vortex term is the constant $-\Gamma \log a/(2\pi )$. Hence the circle is a streamline, giving no normal flow. At infinity the first term gives the required uniform flow; the logarithmic term supplies circulation $\Gamma$. The potential may be multivalued by a real constant, but its velocity is single-valued.

Conformal transport therefore givesIn the frame where the plate is horizontal, its length is $4a$ and the stream has angle $\alpha$; rotating the frame gives the equivalent incidence description. The complex velocity isAt the trailing edge $z=2a$, corresponding to $\zeta =a$, the denominator has a simple zero. Finite velocity requiresso the Kutta condition isWith this choice, cancellation of the common zero givesThus the limiting velocity has zero transverse component and is tangent to the plate.

The stationary Schrödinger equation isThe Laplacian represents the kinetic-energy operator. For a spherically symmetric function, $\Delta \psi ={\psi }^{\prime \prime }+2{\psi }^{\prime }/r$. The attractive Coulomb potential of the nucleus is $V=-q^2/(4\pi {ϵ}_{0}r)$; $E$ is the energy eigenvalue. Multiplication by $-2m/{\hslash }^2$ produces the displayed radial equation with Bohr radiusFor a normalizable bound state, write ${\kappa }^2=-2mE/{\hslash }^2$ and take $\lambda >0$. Substitution of $\psi =(A+Br)e^{-\lambda r}$ ingives, after division by the exponential, the coefficientsIf $B=0$ and $A\ne 0$, these force $\lambda =1/a$ and $\kappa =\lambda$. If $B\ne 0$, the first coefficient gives ${\kappa }^2={\lambda }^2$, the second gives $\lambda =1/(2a)$, and the third gives $B=-A/(2a)$. Up to nonzero scalar factors, the two radial bound states and their energies are consequentlyThey are linearly independent as functions and correspond to different energy eigenvalues. Strictly, they are not two solutions of one fixed-energy differential equation: each is a solution at its own energy.

For the higher-energy state, the spherical Born rule includes the radial factor $4\pi r^2$. Its normalization constant and the angular factor cancel in the ratioThe given integral formula makes the denominator $8a^5$ and the numerator $8a^5-21a^5/e$. Thus

For a regular isoperimetric constraint, introduce a constant Lagrange multiplier $\lambda$ and set $L=F-\lambda G$. With fixed endpoint values, the Euler-Lagrange equation isIf $F$ and $G$ have no explicit $x$ dependence, the Beltrami identity gives the first integralThe regularity assumption excludes the exceptional case in which the constraint itself has zero first variation for every admissible perturbation.

Parametrize the surface of revolution by $r(z,\theta )=(f(z)\cos \theta ,f(z)\sin \theta ,z)$. The magnitude of $r_z\times r_{\theta }$ is $f\sqrt{1+f^{\prime 2}}$. Integrating the resulting surface area element over $\theta$, and integrating the circular cross-sectional areas for the volume, givesHere $f\ge 0$ because it is a radius. For a regular connected closed surface, take $f>0$ in the interior; the closing endpoints can have unbounded $f^{\prime }$ even when the surface itself is smooth.

After dividing $A-\lambda V$ by $\pi$, its integrand is $L=2f\sqrt{1+f^{\prime 2}}-\lambda f^2$. The Euler-Lagrange equation and Beltrami identity becomeAs $f$ tends to zero at an endpoint, $0\le 2f/\sqrt{1+f^{\prime 2}}\le 2f$ also tends to zero, so $C=0$. On the positive interior this implies $\lambda >0$ and, writing $R=2/\lambda$,The Euler-Lagrange equation also yields $f^{\prime \prime }=-R^2/f^3$. Thereforeincluding at the maximum where $f^{\prime }=0$. Integrating and using the first integral givesThus the regular extremal is a sphere. The two zeros fix its center and radius:Equivalently, the prescribed volume fixes $R=(3V/(4\pi ){)}^{1/3}$. If $a,b$ are fixed in advance, their separation and the prescribed volume must be compatible with these equations; there is no positive regular stationary profile of the stated type for an incompatible volume.

For minimality, the three-dimensional isoperimetric inequality states that every sufficiently regular closed surface enclosing volume $V$ haswith equality precisely for a sphere. Our extremal has $A=4\pi R^2$ and $V=4\pi R^3/3$, so it attains equality and is a global minimum of area at the fixed volume. This also makes clear the connected-surface assumption behind uniqueness: allowing several disconnected components introduces stationary collections of spheres, rather than the single positive generating profile used here.

A metric space is sequentially compact if every sequence has a convergent subsequence whose limit belongs to the space. A limit point $\ell$ of $S$ is a point for which every open ball about $\ell$ contains a point of $S$ other than $\ell$.

Let $(x_n)$ lie in a compact metric space. If one value occurs infinitely often, take a constant subsequence. Otherwise its set of values $S$ is infinite. If $S$ had no limit point, every point of the space would have a neighborhood containing at most one member of $S$. A finite subcover of these neighborhoods would force $S$ finite, a contradiction. Thus $S$ has a limit point; the stated subsequence result supplies a convergent subsequence. This proves sequential compactness.

Assume now that $X$ is nonempty, as is necessary for the requested minimum and fixed point. The function $r(x)=d(f(x),x)$ is continuous, as a composition of $x\mapsto (f(x),x)$ with the metric. The extreme value theorem on a compact space implies that $r$ attains its infimum, say at $p$. If $f(p)\ne p$, strict decrease of distances givescontradicting minimality. Thus $f(p)=p$. If $p,q$ were distinct fixed points, then $d(p,q)=d(f(p),f(q))<d(p,q)$, another contradiction. Hence there is exactly one fixed point. This is the fixed point theorem for a strictly distance-decreasing map on a compact metric space.

Compactness is essential. On $\mathbb{R}$, takeHere $f(x)-x=\log (1+e^{-x})>0$, so there is no fixed point. Nevertheless $0<f^{\prime }(x)=e^x/(1+e^x)<1$ everywhere, and the mean value theorem gives $|f(x)-f(y)|<|x-y|$ for distinct $x,y$. There is no uniform Lipschitz constant less than $1$, since $f^{\prime }(x)\to 1$ as $x\to \infty$.

A connected space has no expression as the union of two disjoint nonempty open subsets. A continuous surjection to the two-point discrete space would pull back its singleton sets to such a separation. Conversely, a separation $X=U\bigsqcup V$ defines a continuous surjection by taking the values $0$ on $U$ and $1$ on $V$. Thus these two conditions are equivalent, including for the empty space.

If a subset $S\subseteq \mathbb{R}$ contains $a<c$ but omits some $b$ between them, its intersections with $(-\infty ,b)$ and $(b,\infty )$ separate it. Hence every nonempty connected subset is an interval. Conversely, let an interval $I$ have a separation and choose $a<b$ in its opposite sides. Restrict the associated two-valued continuous function to $[a,b]\subseteq I$ and regard it as a real-valued function. The intermediate value theorem forces the value $1/2$, a contradiction. Singletons and the empty set are connected directly.

For the union assertion, let $u:X\to \{0,1\}$ be continuous. Each restriction $u{|}_{A_i}$ is constant because $A_i$ is connected. Since every pair of the $A_i$ intersects, these constants agree, so $u$ is constant on $X$. There is no continuous surjection, and $X$ is connected. The empty indexing case gives $X=\varnothing$ and is harmless.

To apply this to the plane, for each $t\in \mathbb{R}$ take the union of the horizontal line at height $t$ with the vertical axis. Its two lines are connected and intersect; all such unions share the vertical axis and cover ${\mathbb{R}}^2$. Thus the plane is connected. For the punctured plane, take, for every $r>0$, the circle of radius $r$ together with the positive horizontal ray. The circle is the continuous image of $[0,2\pi ]$, and intersects that connected ray. These unions share the ray and cover ${\mathbb{R}}^2\setminus \{0\}$, so the punctured plane is connected too. Translation gives the same conclusion after removing any one point.

Finally, suppose $g:{\mathbb{R}}^2\to \mathbb{R}$ were a continuous injection. Its image $I$ is a connected interval with at least three points. Choose an interior point $c$ of $I$ and its unique preimage $p$. The restriction of $g$ to ${\mathbb{R}}^2\setminus \{p\}$ is a continuous surjection onto $I\setminus \{c\}$. Its domain is connected, whereas its image is separated by the two half-lines on either side of $c$. This contradicts preservation of connectedness under continuous images. Therefore no continuous injection of the plane into the line exists.