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Take the undisturbed free surface as $z=0$, the bottom as $z=-h$, and the walls as $x,y=0,a$. For an incompressible flow which is an irrotational flow, write $\mathbf{u}=\nabla \varphi$ with a velocity potential satisfying ${\nabla }^2\varphi =0$. Impermeability gives ${\varphi }_z=0$ at the bottom, ${\varphi }_x=0$ at the $x$ walls, and ${\varphi }_y=0$ at the $y$ walls.

For the actual surface $z=\eta (x,y,t)$, the exact kinematic boundary condition for a free-surface graph isThe Unsteady Bernoulli equation, with atmospheric pressure at the surface and its time-dependent constant absorbed into $\varphi$, givesFor small disturbances about fluid at rest, $\varphi$ and $\eta$ are first order. Products of first-order quantities and Taylor corrections from evaluating at $z=\eta$ are second order. The linear surface conditions are thereforeor, after eliminating the elevation,

The separated normal modes of surface gravity waves in a rectangular tank havewhere $m,n$ are nonnegative integers not both zero. The horizontal factors satisfy the wall conditions, and the vertical factor satisfies both the Laplace equation and the bottom condition. At the top, ${\varphi }_z=k_{mn}\tanh (k_{mn}h)\varphi$; substitution into the surface condition givesThe elevation isReal physical standing waves are the real parts or linear combinations of these complex modes. The uniform $m=n=0$ mode has zero frequency and would change the mean depth; conservation of the fixed water volume excludes it from the disturbances. Modes with equal $m^2+n^2$ are degenerate and can be superposed.

For a variation $y+\epsilon \eta$ with $\eta (a)=0$ but arbitrary $\eta (b)$, differentiation and integration by parts giveFirst choose variations supported inside the interval. The fundamental lemma of the calculus of variations gives the Euler-Lagrange equation. Then allow arbitrary endpoint variation to obtain the natural boundary conditions for a free endpoint:Because $F$ has no explicit dependence on $x$,Thus the Beltrami identity is $F-y^{\prime }{F}_{y^{\prime }}=k$, constant along an extremal.

For the brachistochrone problem, put $u=-y>0$, so conservation of energy gives the bead speed $v=\sqrt{2gu}$. Its travel time isThe Beltrami identity now gives $u(1+u^{\prime 2})=2R$ for a positive constant $R$. A descending cycloid parametrizes this relation:The free endpoint has $F_{u^{\prime }}=0$, hence $u^{\prime }(x_0)=0$. On the first descending arc this occurs at $\theta =\pi$. ThereforeAlong this arc $ds=2R\sin (\theta /2)d\theta$ and $v=2\sqrt{gR}\sin (\theta /2)$, soThe initial vertical tangent causes no divergent travel time: the quotient $ds/v$ has the finite constant limit $\sqrt{R/g}d\theta$.

To verify global minimality, take any admissible wire graph of finite travel time below the release level, and let $H>0$ be its maximum depth. For $0<u\le H$, the Cauchy-Schwarz inequality givesIndeed the two coefficients on the right, after multiplication by $\sqrt{u}$, form a unit vector. Since the path starts at depth zero and reaches depth $H$, the weighted total variation is at least the integral over all depths from $0$ to $H$. ConsequentlyThe last inequality is minimized at $H=2x_0/\pi$. Our cycloid attains equality throughout: its depth increases to $H$ and $u^{\prime }=\sqrt{H/u-1}$. Thus it achieves the genuine minimum over the wire graphs, not merely a stationary time. Reflection handles a negative horizontal endpoint coordinate by replacing $x_0$ with $|x_0|$.

The contraction mapping theorem says that if a nonempty complete metric space $(X,d)$ has a map $T:X\to X$ satisfying $d(Tx,Ty)\le qd(x,y)$ for every $x,y$ and some fixed $0\le q<1$, then $T$ has a unique fixed point $p$, and $T^{n}x\to p$ for every starting point. In particular $d(T^{n}x,p)\le q^{n}d(x,p)$; an estimate using only consecutive iterates is $d(T^{n}x,p)\le q^{n}d(x,Tx)/(1-q)$.

Let $s=\sqrt{a}$ and $T(x)=(x+a/x)/2$ for $x>0$. Direct subtraction givesThus every iterate after the first lies in $[s,\infty )$. On this interval,The mean value theorem gives a uniform Lipschitz constant $1/2$, and $T$ maps the interval into itself. The interval is closed in $\mathbb{R}$, hence complete. By the contraction mapping theorem, its iterates converge to its unique fixed point, which is $s$. Therefore every positive initial guess converges to $\sqrt{a}$.

The local error $e_n=x_n-s\ge 0$ satisfies the sharper exact recursionIn relative error ${\delta }_n=e_n/s$, this is ${\delta }_{n+1}={\delta }_n^2/[2(1+{\delta }_n)]$. Thus when ${\delta }_n\approx 10^{-d}$, the next relative error is approximately $\frac{1}{2}10^{-2d}$: one step roughly doubles the number of correct decimal digits. In particular one further step suffices for one additional digit once the iteration is sufficiently close. For absolute-error accuracy, the precise sufficient condition $e_n\le s/5$ gives $e_{n+1}\le e_n/10$; this states explicitly how close is sufficient when $a$ has an arbitrary scale.

In the Poincare disc model, hyperbolic lines are diameters and circular arcs meeting the boundary circle orthogonally. In the upper half-plane model, they are vertical lines and semicircles with centres on the real axis. The metric is conformal, so hyperbolic perpendicularity agrees with Euclidean perpendicularity.

Use an isometry of the hyperbolic plane to send an arbitrary line to the positive imaginary axis. For $P=x+iy$ with $x\ne 0$, a point on that axis is $Q=iv$ with $v>0$. The hyperbolic distance formula givesPut $R=\sqrt{x^2+y^2}$. The last expression is $(v+R^2/v)/(2y)$, uniquely minimized at $v=R$, since $v+R^2/v\ge 2R$ with equality only there. Hence the unique closest point is $Q_0=iR$. The semicircle $|z|=R$ passes through $P$ and $Q_0$ and meets the imaginary axis at a right angle. Conversely a semicircle centred at a real number $b$ can meet that axis orthogonally only if $b=0$; a vertical line cannot furnish such a crossing. Thus the perpendicular is unique. Isometries transfer both uniqueness and minimization back to the original line.

For $P$ on the specified second semicircle, $a-r\le x\le a+r$ and $R=|P|\le a+r$. Its perpendicular to the axis is the semicircle just found, with radius $R$. MoreoverSince $t\le \mathrm{artanh}t$ for $0\le t<1$, we obtainThis is uniform over the whole semicircle, including points arbitrarily close to the ideal endpoints.

Finally send any first line of an ultraparallel pair to the axis by an isometry of the hyperbolic plane. The second cannot be a vertical line, since that would share the ideal point at infinity. Its two finite endpoints lie on the same side of zero; otherwise its semicircle would meet the axis, or share the endpoint zero. Reflect horizontally if necessary, so those endpoints are $a-r,a+r$ with $a>r>0$. The uniform estimate above then givesThus ultraparallel hyperbolic lines have strictly positive separation; disjointness in the interior alone would not suffice.

An orthogonal transformation preserves the real inner product. If $W$ is an invariant subspace, its orthogonal complement is invariant too: $\alpha W=W$ in finite dimensions, and for $v\in W^{\perp }$, $w\in W$, we have $⟨\alpha v,w⟩=⟨v,{\alpha }^{-1}w⟩=0$.

Complexify the space and the transformation. There is a complex eigenvalue $\lambda$, and preservation of the Hermitian norm gives $|\lambda |=1$. If $\lambda$ is real, it is $1$ or $-1$, and the real kernel of $\alpha -\lambda I$ contains a nonzero vector. Its span is an invariant line on which $\alpha$ acts as $+1$ or $-1$.

If $\lambda$ is nonreal, write an eigenvector as $u+iv$. Its real and imaginary parts are linearly independent: otherwise a nonzero real vector would have the nonreal eigenvalue. The real plane $W=\mathrm{span}(u,v)$ is invariant, because comparison of real and imaginary parts in $\alpha (u+iv)=\lambda (u+iv)$ gives real combinations of $u,v$. The restriction has determinant $\lambda \bar{\lambda }=1$. In an oriented orthonormal basis of $W$, it is therefore a rotationThere is no invariant real line in this plane: such a line would supply a real eigenvalue, whereas its two eigenvalues are nonreal.

Apply the argument repeatedly to the invariant orthogonal complement. Thus the smallest mutually orthogonal invariant blocks are lines with action $\pm 1$, and planes with a genuine rotation. Planes with angle $0$ or $\pi$ can be decomposed further into lines.

In dimension three a suitably chosen orthonormal basis therefore gives eitherorChanging the orientation of the plane reverses the angle, so this angle range loses no cases. These exhaust rotations, reflections and their possible compositions in three dimensions.

Order the states as sandwich $N$, Hall $H$, and own cooking $C$. The transition matrix, with current state indexing the row, isThe two probabilities $1/4$ in the first row are unconditional alternatives to the probability $1/2$ of returning. Let $a_n=(P^n{)}_{HH}$, so $n$ counts days after the initial Hall lunch. Expanding the determinant gives the characteristic polynomialThe three distinct eigenvalues imply that the sequence has the formusing the specified $0^0=1$. Directly, $a_0=1$, $a_1=1/3$ and $a_2=7/36$. The last two equations give $A=3/14$ and $B=-5/7$; then $C=3/2$. ConsequentlyThe limiting value is also the Hall entry of the stationary distribution $(4/7,3/14,3/14)$. The tiny finite-time correction is retained in the boxed exact probability. Sixty days later means sixty transitions, rather than the sixtieth day including the initial day.

In the position representation, $\hat{p}=-i\hslash \nabla$. The orbital angular momentum operator and its squared magnitude areFor example, with $D=x\cdot \nabla$, the latter is $-{\hslash }^2[r^2\Delta -D^2-D]$. Operator products here act on smooth wavefunctions in a common invariant domain.

To calculate the angular momentum commutation relations, use $[x_a,p_b]=i\hslash {\delta }_{ab}$ and the vanishing position-position and momentum-momentum commutators. Expanding the products givesMultiplication by ${ϵ}_{iab}{ϵ}_{jcd}$ then givesFinally, the product commutator rule yieldsThe final cancellation pairs the antisymmetric Levi-Civita symbol with an expression symmetric in $j,k$. Thus

Choose the coordinate vectors $b=e_i$ and $c=e_j$ in the assumed bilinear identity. It gives $T_{ij}=0$ for each pair of indices, hence the tensor is zero.

Let $\Omega$ be the enclosed region. Apply the divergence theorem to the vector field $(b\cdot x)c$. Since $b,c$ are constant,ThereforeWriting the left side as $b_i{c}_j{\int }_S{x}_i{n}_{j}dS$ and using the initial tensor argument gives the boundary integral of position and normal componentsContract with the Kronecker delta to obtain ${\int }_{S}x\cdot ndS=3V$. Contract with the Levi-Civita symbol to obtainThus the requested integrals are $3V$ and the zero vector, respectively. These applications require the usual sufficiently regular closed boundary and outward-normal orientation of the divergence theorem.