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For constant mass density and incompressible flow, use $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times \mathbf{\omega }$ in the Euler equations for an inviscid fluid. Taking the curl removes the pressure and kinetic energy gradients, giving ${\mathbf{\omega }}_t=\nabla \times (\mathbf{u}\times \mathbf{\omega })$. The supplied vector identity, with $\nabla \cdot \mathbf{u}=0$ and $\nabla \cdot \mathbf{\omega }=0$, yields the vorticity equationIn a two-dimensional flow $\mathbf{u}=(u,v,0)$ is independent of $z$, and $\mathbf{\omega }=\omega {\mathbf{e}}_z$. The right side vanishes, so $D\omega /Dt=0$: each material element retains its vorticity. The vortex lines are parallel to $z$ and are carried by the planar velocity. Also planar incompressibility preserves the area of a material cross-section; integrating its conserved vorticity over that area shows that its circulation, or vortex strength, stays constant. Thus the vortex lines move with the fluid without changing strength.

Place a line vortex of circulation $\Gamma$ at $(0,b)$ above the wall $y=0$. Its velocity field is $\Gamma {\mathbf{e}}_z\times (\mathbf{r}-{\mathbf{r}}_0)/(2\pi |\mathbf{r}-{\mathbf{r}}_0{|}^2)$. The method of images puts a vortex of circulation $-\Gamma$ at $(0,-b)$. With $\mathbf{u}=({\psi }_y,-{\psi }_x)$, the resulting stream function isIt is identically zero on the wall, so $u_y=-{\psi }_x=0$ there. In the fluid half-plane it has just the required vortex singularity, with no other singularities, and its velocity decays at infinity. This is the no-background-flow solution with the wall: the difference from another such solution is a regular irrotational, divergence-free decaying field with zero normal velocity. Reflect its tangential component evenly and its normal component oddly across the wall. The reflected components are bounded harmonic functions on the whole plane, so they are constant by harmonic Liouville theorem, and decay forces both constants to be zero. Thus replacing the wall by the image gives the same flow. The real vortex moves under the image velocity, excluding its own singular velocity, soIts signed velocity parallel to the wall is $\Gamma /(4\pi b)$, and its speed is $|\Gamma |/(4\pi b)$.

For the quarter-plane, place images of circulation $-\Gamma$ at $(-x,y)$ and $(x,-y)$, and $+\Gamma$ at $(-x,-y)$. The paired stream functions make both walls streamlines. Summing their velocities at the real vortex givesIn polar coordinates,It follows directly thatThus the line-vortex trajectory in a quarter-plane is $r\sin 2\theta =\text{constant}$.

For fixed endpoints, the Euler-Lagrange equation is $F_y-\frac{d}{dx}{F}_{y^{\prime }}=0$. Since $F$ has no explicit dependence on $x$, differentiating givesThus $F-y^{\prime }{F}_{y^{\prime }}=C$, the Beltrami identity.

With $r^{\prime }=dr/d\theta$, the arc length element and the prescribed speed give the travel-time functionalFor the circular road $r=a$, this is $T_{\mathrm{circle}}=\pi /(2ka)$.

For the integrand $F=\sqrt{r^{\prime 2}+r^2}/r^2$, the Beltrami identity simplifies toThe candidate $r=a(\cos \theta +\sin \theta )$ has $r^{\prime 2}+r^2=2a^2$ and the required endpoints. To prove it really minimizes the travel time, rather than merely satisfies a first integral, put $u=1/r$. ThenThis is the Euclidean arc length integrand for the inverted curve $(X,Y)=(u\cos \theta ,u\sin \theta )$, whose endpoints are $(1/a,0)$ and $(0,1/a)$. Every connecting curve has length at least their Euclidean distance $\sqrt{2}/a$, by the triangle inequality. Equality is attained by the straight line segment $X+Y=1/a$. In polar form it gives $u(\cos \theta +\sin \theta )=1/a$, hence the proposed $r$. It stays strictly positive throughout the interval. ConsequentlyThis travel-time minimization with speed proportional to squared radius is globally solved by the inversion: its minimum is strictly less than $\pi /(2ka)$.

The contraction mapping theorem says that if $(X,d)$ is a nonempty complete metric space and $T:X\to X$ satisfies $d(Tx,Ty)\le qd(x,y)$ for a constant $0\le q<1$, then $T$ has exactly one fixed point.

For the proof, start at $x_0\in X$ and set $x_n=T^n{x}_0$. Iterating the contraction bound gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$. The triangle inequality then gives, for $m>n$,Thus $(x_n)$ is a Cauchy sequence and has a limit $x_{\ast }$ by completeness. Since the contraction is continuous, $T{x}_{\ast }=\lim T{x}_n=\lim x_{n+1}=x_{\ast }$. If $y_{\ast }$ is another fixed point, then $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing distance zero. This proves existence and uniqueness.

For the integral map, the fundamental theorem of calculus shows that $Tf$ is continuous whenever $f$ is continuous. In the supremum norm,Hence $\Vert Tf-Tg{\Vert }_{\infty }\le \frac{3}{2}\max (a^2,b^2)\Vert f-g{\Vert }_{\infty }$. We can chooseThe space $C[a,b]$ with this supremum norm is complete: a uniformly Cauchy sequence has pointwise real limits, the Cauchy bound passes to those limits to give uniform convergence, and a uniform limit of continuous functions is continuous. The contraction mapping theorem therefore gives a unique continuous $y$ with $y(x)=1+{\int }_0^{x}3ty(t)dt$. The fundamental theorem of calculus makes this $y$ differentiable, with $y^{\prime }=3xy$ and $y(0)=1$. Conversely every solution of that initial-value problem obeys the integral equation, so uniqueness transfers to the ordinary differential equation. Explicitly $y(x)=e^{3x^2/2}$ on $[-1/2,1/2]$, as direct differentiation confirms.

In the Poincaré half-plane model, $\mathbb{H}=\{x+iy:y>0\}$ has Riemannian metricIts orientation-preserving isometries are the Möbius transformations $z\mapsto (az+b)/(cz+d)$ with real coefficients and $ad-bc=1$, with a simultaneous sign change representing the same map. The full isometry group also contains their compositions with $z\mapsto -\overline{z}$, which reverse orientation. The hyperbolic lines are vertical Euclidean lines and upper semicircles centred on the real axis; each meets the boundary orthogonally.

A hyperbolic line with finite endpoints $p<q$ is mapped to the imaginary axis byIts coefficient determinant is $q-p>0$, so rescaling the coefficients makes this an orientation-preserving isometry. The endpoints map to $0$ and $\infty$, which characterize the imaginary-axis hyperbolic line. For a vertical line with endpoints $p,\infty$, use $T(z)=z-p$. If $T_1,T_2$ send the two prescribed lines to the imaginary axis, $T_2^{-1}{T}_1$ sends the first to the second. This proves transitivity on hyperbolic lines.

For two ultraparallel hyperbolic lines, choose their common perpendicular and map it to the imaginary axis. A hyperbolic line perpendicular to that axis is a semicircle centred at zero: orthogonality at $iy$ forces the circle centre to have real coordinate zero. Thus the reflecting lines become semicircles of distinct radii $r_1,r_2>0$. The hyperbolic reflection in the radius-$r$ semicircle is $S_r(z)=r^2/\overline{z}$; it fixes that semicircle pointwise and reverses orientation. Their composition isIts $k$th power is $z\mapsto {\lambda }^{k}z$, and no positive power is the identity since $\lambda >0$ and $\lambda \ne 1$. Conjugation by the chosen isometry preserves order of a group element, so the original composition has infinite order. Reversing the order of the reflections replaces $\lambda$ by ${\lambda }^{-1}$ and gives the same conclusion.

Write $u_n=\mathbb{P}(X_n=0\mid X_0=0)$. The Markov property, conditioning on the state at time $n$, givesPut $r=\alpha +\beta -1$ and ${\pi }_0=(1-\beta )/(2-\alpha -\beta )$. Then ${\pi }_0=r{\pi }_0+1-\beta$, so subtraction yields $u_{n+1}-{\pi }_0=r(u_n-{\pi }_0)$. Iterating this linear recurrence givesFor $r=0$, read the $n=0$ expression using $r^0=1$, or state $u_0=1$ separately; for $n\ge 1$, $u_n={\pi }_0$. Since $|r|<1$, the result also shows convergence in a metric space to the stationary distribution probability ${\pi }_0$.

Under the null hypothesis, the defect count in each packet has a binomial distribution with parameters $3,\theta$. The likelihood, apart from factors independent of $\theta$, isThe maximum-likelihood estimate is thereforeThe fitted binomial distribution probabilities for counts $0,1,2,3$ are $27/64,27/64,9/64,1/64$, yielding expected packet counts $108,108,36,4$. The Pearson chi-squared goodness-of-fit test usesFor binomial goodness-of-fit with an estimated parameter, there are $4-1-1=2$ statistical degrees of freedom: the category counts have a fixed total, and one parameter was estimated. At the conventional $5\%$ level the chi-squared distribution with two statistical degrees of freedom has upper critical value $5.99$, so the data do not reject the independent common-defect-probability model. The approximate p-value is $e^{-X^2/2}\simeq 0.146$; even at $10\%$, the critical value $4.61$ exceeds the observed statistic.

This is a goodness-of-fit conclusion rather than proof that individual bulbs are independent. The last expected count is only $4$, so the chi-squared asymptotic approximation is somewhat marginal in that cell. The stated calibration is the usual asymptotic test with all four cells retained; if more accurate small-count inference is required, calibrate the fitted statistic under the binomial distribution model. Pooling cells would require a corresponding treatment of parameter fitting and test calibration.