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The brachistochrone problem asks for the curve along which a particle released from rest reaches a specified target fastest under a uniform gravitational field, without friction. With downward depth $u$, conservation of energy gives speed $\sqrt{2gu}$ and travel time $\int \sqrt{(1+u^{\prime 2})/(2gu)}dx$. The Beltrami identity makes $u(1+u^{\prime 2})$ constant, leading to a cycloid. If only the horizontal endpoint coordinate $x_0>0$ is fixed, the natural boundary conditions for a free endpoint select the half arch ending at its deepest point, with depth $2x_0/\pi$ and minimum time $\sqrt{\pi x_0/g}$.

A positively curved pressureless matter universe with zero cosmological constant has the cycloid parametrization displayed above. In natural units, if the Friedmann equation is ${\dot{a}}^2=C/a-k$ with $C,k>0$, then $A=C/(2k)$ and $u=\sqrt{k}\tau$ for conformal time measured from the Big Bang. Substituting in $a^{\prime 2}=Ca-k{a}^2$ verifies the scale factor; integrating $dt=ad\tau$ gives cosmic time. The universe reaches maximum size at $u=\pi$ and recollapses into a Big Crunch at $u=2\pi$.

For pressureless homologous spherical collapse, let $\varphi$ be its cycloid parameter. The density contrast equation becomes ${\delta }_{\varphi \varphi }-\tan (\varphi /2){\delta }_{\varphi }-3\delta /(1+\cos \varphi )=0$. One solution is ${\delta }_g=\sin \varphi /(1+\cos \varphi {)}^2$. Reduction of order gives an independent solution that decays at collapse,For $ϵ=\pi -\varphi$, the integral is ${ϵ}^5/40+O({ϵ}^7)$, while ${\delta }_g\sim 4{ϵ}^{-3}$. Therefore ${\delta }_d\sim {ϵ}^2/10$. Since $R\sim R_0{ϵ}^2/4$, the displayed powers follow. The growing branch is unique up to scale and addition of the decaying branch; many mixed solutions grow, so it is not literally the only function with growth. Finite perturbations eventually leave the linear regime.

Write $H=\dot{R}/R$. A fluid element in the homologous spherical flow obeys $\dot{\mathbf{r}}=H\mathbf{r}$, so $\mathbf{r}/R$ remains constant. Also $\nabla \cdot {\mathbf{u}}_0=3H$. Along every element, the continuity equation therefore gives $D{\rho }_0/Dt=-3H{\rho }_0$. Every element starts with the same mass density, and hence has the same mass density at subsequent times:This proves preservation of uniformity, rather than assuming it when conserving the total mass.

By the shell theorem, only the mass interior to radius $r$ contributes to its gravitational acceleration. That mass is $4\pi {\rho }_0{r}^3/3$. The same result follows directly from the Poisson equation for Newtonian gravity: spherical symmetry gives $(r^2{\Phi }_r{)}_r=4\pi G{\rho }_0{r}^2$, and regularity at the centre sets the integration constant to zero. Hence ${\Phi }_r=4\pi G{\rho }_{0}r/3$ andThe material derivative of the background velocity is $(\dot{H}+H^2)\mathbf{r}=(\ddot{R}/R)\mathbf{r}$. There is no pressure force, and the momentum equation consequently reduces toMultiplication by $\dot{R}$ and integration, using collapse from rest, givesChoose the negative square root for contraction. Put $C^2=2K/R_0$ and $R=R_0{\cos }^2(\varphi /2)$. For $0<\varphi <\pi$, the energy integral then gives $\dot{R}=-C\tan (\varphi /2)$, while $dR/d\varphi =-R_0\sin \varphi /2$. Their ratio isIntegrating from the initial turning point proves the cycloid parametrization of pressureless homologous spherical collapse:It covers the entire contracting branch, with complete collapse at $\varphi =\pi$ and $t_c=\sqrt{3\pi /(32G{\rho }_{00})}$. At $R=0$ the uniform dust configuration is singular, so the smooth calculation is restricted to $t<t_c$.

Each Fourier mode among the comoving perturbations of a homologously collapsing cloud has its phase fixed on each fluid element. In fact, for $\mathbf{k}=\mathbf{q}/R$,Its physical wavelength contracts in proportion to $R$, while its wavelength in the comoving coordinate $\mathbf{r}/R$ stays fixed. This is the physical reason to use a time-dependent wavenumber; holding the physical wavenumber fixed would not follow the contracting pattern. Boundary effects are being excluded in this bulk Fourier mode calculation.

Let ${\rho }^{\prime }={\rho }_1(t)e^{i\mathbf{k}\cdot \mathbf{r}}$ and let the perturbed velocity and Newtonian gravitational potential have amplitudes ${\mathbf{u}}_1(t)$ and ${\Phi }_1(t)$. Linearizing the continuity equation, the uniform background eliminates ${\mathbf{u}}_1\cdot \nabla {\rho }_0$. The advected phase eliminates its own time derivative, leavingFor the density contrast $\delta ={\rho }_1/{\rho }_0$, use ${\dot{\rho }}_0=-3H{\rho }_0$ to obtain $\dot{\delta }=-i\mathbf{k}\cdot {\mathbf{u}}_1$. The linear momentum equation and Poisson equation for Newtonian gravity areDifferentiating the density contrast equation must include $\dot{\mathbf{k}}=-H\mathbf{k}$:Thus the compressive Fourier mode with $k\ne 0$ satisfiesThe pressureless approximation removes the usual acoustic restoring term of Jeans instability. A transverse velocity perturbation instead obeys ${\dot{\mathbf{u}}}_{1\perp }+H{\mathbf{u}}_{1\perp }=0$ and does not drive this density contrast.

Since $d/dt=(C/R)d/d\varphi$, the density contrast equation becomesFor a direct verification, put $s=\tan (\varphi /2)$. As $d/d\varphi =(1+s^2)d/(2ds)$, the first-derivative terms cancel, reducing the equation toThe function ${\delta }_g=s(1+s^2)/2$ has second derivative $3s$ and satisfies this equation. Transforming back gives the required growing branch:To establish what the other solution does, use reduction of order, not just the growth of this example. The coefficient of ${\delta }_{\varphi }$ is $d\log R/d\varphi$, so an independent solution is proportional to ${\delta }_g\int d\varphi /(R{\delta }_g^2)$. Choose the integration endpoint at complete collapse to select the decaying branch:For example, evaluating the integral with $a=\tan (s/2)$ gives integrand $4/[a^2(1+a^2{)}^2]$ and antiderivative $-4/a-6\arctan a-2a/(1+a^2)$. Its endpoint value gives the displayed expression. Its Wronskian with ${\delta }_g$ is $-(1+\cos \varphi {)}^{-1}$, so the two solutions are independent on $0<\varphi <\pi$.

For $ϵ=\pi -\varphi \to 0$, $R\sim R_0{ϵ}^2/4$, ${\delta }_g\sim 4{ϵ}^{-3}$ and the integral in ${\delta }_d$ is ${ϵ}^5/40+O({ϵ}^7)$. Consequently the two density modes of a collapsing pressureless sphere haveEquivalently, with $\tau =t_c-t$, the local powers are ${\tau }^{-1}$ and ${\tau }^{2/3}$. There is one independent growing branch: every solution is $A{\delta }_g+B{\delta }_d$, and only the branch with $A=0$ decays at collapse. The uniqueness is up to normalization and addition of the decaying solution; it does not mean that only one literal function can grow. A fixed nonzero initial perturbation will eventually violate $|\delta |\ll 1$, so the divergent linear result is not a nonlinear solution through collapse.

Use natural units with $c=1$, write $H=\dot{a}/a$, and measure cosmic time from the Big Bang. The cosmological continuity equation follows by differentiating the Friedmann equation and using the Friedmann acceleration equation; for pressureless matter it gives $\dot{\rho }+3H\rho =0$, hence $\rho ={\rho }_0{a}^{-3}$ when the present scale factor is $a_0=1$. The present critical density and cosmological density parameter giveConsequently, with $C=H_0^2{\Omega }_0$, the Friedmann equation becomes ${\dot{a}}^2=C/a-k$.

Introduce conformal time through $dt=ad\tau$. A prime denotes a derivative with respect to $\tau$, so $a^{\prime }=a\dot{a}$ andSet $u=\sqrt{k}\tau$, with $u=0$ at the Big Bang. On the expanding branch the substitution $a=C(1-\cos u)/(2k)$ satisfies this equation. Integrating $dt=adu/\sqrt{k}$, with $t(0)=0$, gives the closed matter-dominated Friedmann solution:The parametrization continues smoothly through the maximum scale factor, reached at $u=\pi$, onto the contracting branch. Its curve in the $(t,a)$ plane is a rescaled cycloid.

For a presently expanding universe, $H_0>0$ selects $0<u_0<\pi$. The normalization $a(u_0)=1$ givesSubstitution yields the age of a closed dust universe:The first later zero of the scale factor occurs at $u=2\pi$, so the lifetime of a closed dust universe gives the Big Crunch timeThis is the total lifetime measured from the Big Bang; the remaining cosmic time is $t_{\mathrm{BC}}-t_0$. As a check, the limit ${\Omega }_0\downarrow 1$ gives $H_0{t}_0\to 2/3$, while the Big Crunch recedes to infinite time, as expected for a flat matter-dominated universe with zero cosmological constant.

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 $|x_0|$.

A uniform spherical cloud with negligible pressure and velocity $\mathbf{u}=(\dot{R}/R)\mathbf{r}$ remains uniform: the continuity equation gives $\rho R^3={\rho }_{00}{R}_0^3$. The shell theorem gives interior gravitational acceleration $-(4\pi G\rho /3)\mathbf{r}$, so its outer radius satisfies the displayed equation, with fixed mass $M=4\pi {\rho }_{00}{R}_0^3/3$. Collapse from rest has ${\dot{R}}^2=2GM(1/R-1/R_0)$. Writing $C^2=2GM/R_0$, its cycloid parametrization is $R=R_0(1+\cos \varphi )/2$ and $Ct=R_0(\varphi +\sin \varphi )/2$, for $0\le \varphi \le \pi$. Complete collapse occurs after $\sqrt{3\pi /(32G{\rho }_{00})}$. This Newtonian dust solution ceases to be a regular fluid configuration at zero radius.