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Take $x$ down the plane and $y$ normally away from it, with the solid at $y=0$ and the free surface at $y=h$. A steady unidirectional falling film flow has velocity $\mathbf{u}=u(y){\mathbf{e}}_x$. The streamwise Navier-Stokes equation reduces towhere $\nu$ is the kinematic viscosity. The lower no-slip boundary condition and upper stress-free boundary condition areIntegrating gives the velocity profileThe volume flux per unit width is

If the upper surface is replaced by a stationary solid plane, the second boundary condition becomes $u(h)=0$. The resulting plane-channel profile and flux areThe free surface exerts no tangential shear stress, whereas the stationary upper wall enforces no slip and exerts a retarding shear. This additional drag accounts for the reduced flux.

Introduce the orthogonal coordinatesThen the constraint and objective becomeThe Lagrange multiplier equations for a stationary point areIf $\lambda =0$, then $u=y=0$ and $v=\pm 1$, givingIf $\lambda \ne 0$, then $v=0$. The first two equations and the constraint implyConverting back to $(x,y,z)$ gives the other four points. Therefore all stationary points arewith matching signs, together withThe middle pair has $\varphi =1/2$, the last pair has $\varphi =-1/2$, and the first pair has $\varphi =0$.

The Poincare disc model isThis Riemannian metric has constant Gaussian curvature $-1$.

For $a\in D$ and $|\eta |=1$, the Möbius transformationis an isometry of the disc, and every orientation-preserving disc isometry has this form. In particular, $T_{z_1,1}$ sends $z_1$ to zero and $z_2$ to a point of modulusRotations are also isometries, so it remains only to integrate the metric along a radial geodesic from $0$ to $\rho$:Thus the Hyperbolic distance in the Poincare disc is

A Laurent series about $a$ is an expansionthat converges to $f$ on an annulus $r<|z-a|<R$.

The partial fraction decomposition isFor $0<|z|<1$, expand both denominators by the geometric series:This is actually a Taylor series at zero, since the apparent puncture at zero contains no singularity.

For $1<|z|<2$, the $z+2$ term still expands in nonnegative powers, while the quadratic term must be expanded in negative powers:Hence

A basis of a vector space $V$ is a family of vectors that is both linearly independent and a spanning set of $V$.

Write the given finite basis as $\mathcal{B}=\{v_1,\dots ,v_n\}$. First, any other basis ${\mathcal{B}}^{\prime }$ must also be finite. Each $v_i$ is a finite linear combination of elements of ${\mathcal{B}}^{\prime }$, so the union $S$ of the finitely many elements of ${\mathcal{B}}^{\prime }$ occurring in these $n$ expressions is finite. Since $S$ spans every $v_i$, it spans $V$. If some $w\in {\mathcal{B}}^{\prime }\setminus S$ existed, then $w$ would lie in the span of $S$, contradicting the linear independence of ${\mathcal{B}}^{\prime }$. Hence ${\mathcal{B}}^{\prime }=S$; write $|{\mathcal{B}}^{\prime }|=m$.

It remains to prove the elementary Steinitz exchange lemma directly. If independent vectors $w_1,\dots ,w_r$ lie in the span of $v_1,\dots ,v_n$, express $w_1$ in terms of the $v_i$. Some coefficient is nonzero, so the corresponding $v_i$ can be solved for in terms of $w_1$ and the other $v_i$; replacing it by $w_1$ preserves the span. Inductively, after replacing $k$ of the $v_i$ by $w_1,\dots ,w_k$, the expression for $w_{k+1}$ must have a nonzero coefficient on one of the unreplaced $v_i$, since otherwise $w_{k+1}$ would be a linear combination of $w_1,\dots ,w_k$. That $v_i$ can again be replaced. There are only $n$ original vectors to replace, so $r\le n$.

Apply this argument first to the independent family ${\mathcal{B}}^{\prime }$ and the spanning family $\mathcal{B}$ to obtain $m\le n$, and then with the two bases interchanged to obtain $n\le m$. Therefore

For an incompressible flow, $\nabla \cdot \mathbf{u}=0$. If the flow is also irrotational, then locally $\mathbf{u}=\nabla \varphi$ for a velocity potential, and therefore

The boundary conditions areThe fluid region is not simply connected, so a potential may change by the constant $\kappa$ after one circuit while its gradient remains single-valued. Superposing uniform flow, the cylinder doublet, and the circulation gives the potential flow around a circular cylinder with circulationIts velocity components are

On $r=a$, $u_r=0$ and, with $\lambda =\kappa /(4\pi Ua)$,The Bernoulli equation gives $p=p_{\infty }+\rho (U^2-u_{\theta }^2)/2$. The pressure force per unit length on the cylinder isThe $x$ component vanishes by symmetry, while ${\int }_0^{2\pi }{\sin }^2\theta d\theta =\pi$ givesin agreement with the Kutta–Joukowski theorem.

A stagnation point satisfies $u_r=u_{\theta }=0$. On the cylinder this means $\sin \theta =\lambda$. Thus $|\lambda |<1$ gives two surface stagnation points, $|\lambda |=1$ gives one coincident surface point, and $|\lambda |>1$ gives none on the surface. Away from the cylinder, $u_r=0$ requires $\cos \theta =0$. Solving $u_{\theta }=0$ then gives one physical exterior root when $|\lambda |>1$:At $|\lambda |=1$ this root lies on $r=a$ and agrees with the single surface point.

Use $\mathbf{E}=-\nabla \varphi$ and Gauss's law $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$. The divergence theorem and the decay at infinity giveMultiplying by $1/2$ proves that the two expressions for the electrostatic energy agree.

The uniform volume charge density in the thick shell isSpherical symmetry and the integral form of Gauss's law giveThus $|\mathbf{E}|$ is zero inside, rises continuously through the charged region, reaches $Q/(4\pi {ϵ}_0{b}^2)$ at $b$, and then decays as $r^{-2}$. Taking $\varphi (\infty )=0$ and integrating $E=-d\varphi /dr$ givesThe electric potential is continuous, constant inside, decreases smoothly across the charge, and continues as $1/r$ outside.

As $b\to a$, the charge becomes a surface shell. ThenThe potential remains continuous, but the normal electric field jumps by $Q/(4\pi {ϵ}_0{a}^2)$ across the surface charge density.

The field-energy expression givesThe charge-potential expression gives the same result directly because the entire charge lies where the potential equals $\varphi (a)$:This is the electrostatic energy of a uniformly charged spherical shell.

Since $\varphi (a)=Q/(4\pi {ϵ}_{0}a)$ is proportional to $Q$,to first order. This is the work required to bring the additional charge $\delta Q$ from infinity to the potential created by the charge already present.

The Time-dependent Schrödinger equation for the one-dimensional Quantum harmonic oscillator isA stationary state has $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$, so cancellation of the common time factor gives

SetThe equation becomesWith $\psi =f(y)e^{-y^2/2}$,and hence $f$ obeys the Hermite differential equation

If $f$ is monic of degree $N$, the coefficient of $y^N$ in this equation is $ϵ-1-2N$. It must vanish, soWriting $f={\sum }_{k=0}^N{a}_k{y}^k$ gives the recurrenceThe coefficient of $y^{N-1}$ first gives $a_{N-1}=0$, and the recurrence preserves parity. If $N$ is even, every odd coefficient vanishes, so $f$ is an even function. The Gaussian factor is also even; therefore the stationary state $\psi$ has even parity.

Since $x=r\cos \theta$, the function on the left of the defining expansion is $e^{ix}$. Thereforewhich is the two-dimensional Helmholtz equation.

In polar coordinates,Substitute the defining Fourier cosine series and equate the coefficient of each $\cos (n\theta )$. Its angular second derivative contributes $-n^2$, so every Bessel function $J_n$ satisfiesor equivalently the Bessel differential equation

Expand the exponential through cubic order and useThis givesComparing with $J_0+2{\sum }_{n\ge 1}{i}^n{J}_n\cos (n\theta )$ yields

Jordan lemma states that if $a>0$ and $g$ is analytic in the upper half-plane outside a fixed circle, withthen on the upper semicircle $C_R$,Indeed, $|e^{iaR{e}^{i\theta }}|=e^{-aR\sin \theta }$. Using symmetry and $\sin \theta \ge 2\theta /\pi$ on $0\le \theta \le \pi /2$ gives

The residue of $f$ at an isolated singularity $a$ is the coefficient of $(z-a{)}^{-1}$ in its Laurent series, equivalentlyIf $f(z)=g(z)/(z-a{)}^k$, the Taylor expansion of the analytic function $g$ shows that

For the real integral, apply contour integration in the upper half-plane toJordan's lemma removes the semicircular arc. The only enclosed singularity is the double pole at $z=i$, whose residue isThe residue theorem therefore givesIts real part vanishes because $x^3\cos x/(1+x^2{)}^2$ is an odd function, while its imaginary part is the requested integral. Hence