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The intermediate value theorem states that if $f:[\alpha ,\beta ]\to \mathbb{R}$ is continuous and $y$ lies between $f(\alpha )$ and $f(\beta )$, then some $c\in [\alpha ,\beta ]$ satisfies $f(c)=y$.

It is enough to prove the caseLetChoose $x_n\in E$ with $x_n\to c$. Continuity gives $f(c)\le y$. If $f(c)<y$, continuity would make $f(x)<y$ for some $x>c$, contradicting that $c$ is an upper bound of $E$. Thus $f(c)=y$. The case with reversed endpoint inequalities follows by replacing $f$ with $-f$.

The flatness problem is that the observed cosmological density parameter is close to $\Omega =1$, although in an ordinary decelerating universe any departure from one grows with time. Extrapolation backward therefore requires implausibly precise early cancellation of the curvature term.

The cosmological perfect-fluid continuity equation isConsequentlyFor an expanding universe $H>0$. Ifthen $\rho a^2$ increases. The Friedmann equation impliesIt therefore decreases during this period, driving $\Omega$ toward one. By the Friedmann acceleration equation, the same condition gives $\ddot{a}>0$ when the cosmological constant is included in the effective fluid. This is the inflationary solution of the flatness problem.

For the scalar field, the slow-roll approximation meansThe equations reduce toTaking the positive-field branch gives $H\simeq m\varphi /\sqrt{3}$, and thereforeWith $\varphi (0)={\varphi }_i$,Now integrate $H=\dot{a}/a\simeq m\varphi /\sqrt{3}$:Thus the quadratic-potential slow-roll solution isFinally,so equivalently

An adiabatic invariant remains constant to leading order when a system parameter changes on a timescale much longer than one period. For a periodic one-degree-of-freedom system, the adiabatic invariance of the action supplies such an invariant.

The momentum magnitude between collisions is $p_0=\sqrt{2mE}$. A complete orbit crosses the gap once in each direction, so the unnormalized action used in the question isConsequentlybecause the speed is $\sqrt{2E/m}$ and the round-trip distance is $2L$.

When $L(t)$ varies adiabatically, conservation of the adiabatic particle between moving parallel walls action givesA moving wall does work at each collision: an approaching wall raises the particle's energy and a receding wall lowers it. Elasticity holds in the instantaneous rest frame of the wall.

The Papperitz symbolspecifies a second-order Fuchsian differential equation. The first row lists its three distinct regular singular points $0,1,\infty$; $z$ is the independent variable. The two entries below each singular point are its characteristic exponents. Thus local solutions have leading behaviorsand, with the usual convention at infinity,When an exponent difference is an integer, a logarithmic second solution may replace the naive second Frobenius power. The entries obey the Fuchs relationFor a second-order equation with exactly three regular singular points, these exponent data determine the equation up to multiplication by a nonzero function; there is no accessory parameter. The symbol is therefore the one for the Gauss hypergeometric equation. Its distinguished solution $F(a,b;c;z)$ is the exponent-zero solution analytic at zero and normalized to one there. For the ordinary power-series definition one assumes $c\notin \{0,-1,-2,\dots \}$, with exceptional parameter values handled separately or by continuation.

Now putThe hypergeometric equation in $u$ has exponent pairsThe Möbius transformation of a Papperitz symbol sendsso $Y$ has symbol

On the other hand, $F(a,b;c;z)$ has exponent pairsat $0,1,\infty$. Multiplication by $(1-z{)}^a$ applies the dependent-variable rescaling of a Papperitz symbol: it adds $a$ to both exponents at $z=1$ and subtracts $a$ from both at infinity. Hencehas exactly the same three exponent pairs as $Y$.

Both functions are analytic near $z=0$ and equal one at $z=0$, so uniqueness of the normalized exponent-zero hypergeometric solution gives the Pfaff transformationThe identity first holds near zero with the branch of $(1-z{)}^a$ equal to one there, and then extends by analytic continuation on any domain where compatible branches are chosen.

Put $a=f^{\prime }(0)>0$. Linearizing the morphogen reaction-diffusion equation at $C=0$ givesSeek separated modes $C(x,t)=X(x)e^{\lambda t}$. ThenThe mixed Dirichlet-Neumann modes on an interval areIndeed, the Dirichlet condition selects sine functions, while the Neumann condition requires $\cos (k_{n}L)=0$. Their growth rates areThe largest growth rate is the lowest mode,Linear stability requires ${\lambda }_0<0$, after which all higher modes also decay. Therefore the critical length for a linearly growing morphogen isEquality gives a neutral lowest mode, while a larger domain is linearly unstable.