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Differentiation of the Hubble parameter gives $\dot{H}+H^2=\ddot{a}/a$. The strong energy condition and the acceleration equation therefore implyAssume the present universe is expanding, $H_0>0$, and normalize the scale factor by $a(t_0)=1$. Since $\ddot{a}\le 0$, the past expanding branch has $\dot{a}(t)\ge \dot{a}(t_0)>0$. On this branch,Integrating from a past time $t$ to $t_0$ gives $H(t{)}^{-1}\le H_0^{-1}+t-t_0$. Thus, where the denominator is positive,The printed inequality needs this past-time qualification: for future times with $H>0$, the inequality reverses. Independently, a concave function lies below its tangent, soPositivity of $a$ prevents the past branch extending before the zero of this tangent. If $t_B$ is the beginning of the expanding universe, its age satisfies $\boxed{t_0-t_B\le H_0^{-1}}$. Without the normalization $a(t_0)=1$, the tangent bound instead reads $a(t)\le a(t_0)[1+H_0(t-t_0)]$.

For variations vanishing at both endpoints, differentiating the action and using integration by parts givesThe fundamental lemma of the calculus of variations therefore yields the Euler-Lagrange equationsFor the specified Lagrangian, the gauge transformation changes it byFor arbitrary $F$, it is a total derivative precisely when $\boxed{\alpha =-1}$. The corresponding change in action is then an endpoint term whose variation is zero. In general time-dependent notation, the equations areThe electric field $-\nabla \varphi -{\partial }_t\mathbf{A}$ and magnetic field $\nabla \times \mathbf{A}$ are unchanged by the gauge transformation with $\alpha =-1$. Although the original potentials have no time dependence, their transformed versions generally do; allowing this is essential to the stated invariance.

The given Euler product for the gamma function has the finite approximantsaway from the poles. Split the denominator of ${\Gamma }_{2m+1}(2z)$ into even and odd factors. The resulting exact identity isThe right side does not depend on $z$, so its limit is a constant $C$. Evaluate the ratio at $z=1/2$: $\Gamma (1)=1$, and the gamma reflection formula, together with positivity on the positive real axis, gives $\Gamma (1/2)=\sqrt{\pi }$. HenceThis is the gamma duplication formula. The argument first applies where the factors are finite; the identity then holds as an identity of meromorphic functions, with values at poles interpreted accordingly.

For this chemical reaction network, the law of mass action gives the reaction rates are $k_1{s}^{2}e$, $k_{2}c$ and $k_{3}c$. Their stoichiometry givesHere the convention for $k_1$ absorbs any combinatorial factor for two identical reactants. Direct differentiation gives the two conservation lawsEliminating $e$ and $p$ leavesFor positive $s_0,e_0,k_1$, use the dimensionless variables $u=s/s_0$, $v=c/e_0$ and $\tau =k_1{e}_0{s}_{0}t$. With prime marks denoting differentiation in $\tau$, this becomeswhere

Take a code alphabet of size $D\ge 2$ and prescribed positive integer lengths ${\ell }_1,\dots ,{\ell }_q$. A decipherable code means that concatenation of codewords determines the entire finite sequence of codewords uniquely. First prove the necessary Kraft inequality. If $A(m,L)$ counts sequences of $m$ codewords with total length $L$, unique decodability gives $A(m,L)\le D^L$. ThereforeTaking $m$th roots and letting $m\to \infty$ gives $K\le 1$.

Conversely, order the lengths increasingly and construct a prefix code in the rooted $D$-ary tree. At depth ${\ell }_j$, previously selected codewords forbid exactly ${\sum }_{i<j}{D}^{{\ell }_j-{\ell }_i}$ vertices. The partial Kraft inequality impliesso an available vertex exists. Select it as the next codeword. Earlier lengths are no greater, so this preserves the prefix code property. A prefix code has unique decodability: in two purported parsings, the first unequal codewords would force one to be a prefix of the other. This proves the equivalence of decipherable and prefix code lengths: the two existence conditions are equivalent.

For a one-symbol alphabet, at most one positive-length codeword can be decipherable: two different unary codewords commute under concatenation. At most one is also exactly the prefix-code condition, so the equivalence still holds. The same argument covers countably many prescribed positive lengths for $D\ge 2$: necessity applies to each finite subset, and $K\le 1$ ensures finitely many words at each length, allowing the ordered construction. Empty codewords are excluded, as required for unique decodability of arbitrary sequences.

For isentropic gas with $P=K{\rho }^{\gamma }$, $\gamma >1$, the sound speed obeys $c^2=P^{\prime }(\rho )$. The one-dimensional mass and momentum equations areSince $c\propto {\rho }^{(\gamma -1)/2}$, these implyAdding and subtracting yields the Riemann invariantsA simple wave has one invariant constant throughout. For a right-moving wave into a resting state with sound speed $c_0$, $R_{-}=0$, so $c=c_0+(\gamma -1)u/2$ and the right characteristic speed is $c_0+bu$, $\boxed{b=(\gamma +1)/2}$.

For the damped scalar equation, characteristics from $\xi$ satisfyAssume smooth bounded initial data with bounded derivative, and the stated maximal compressive slope exists. ThenFor the damping threshold for a simple wave, if $m={\max }_{\xi }[-u_0^{\prime }(\xi )]>0$, put $\boxed{{\alpha }_c=bm}$. For $0<\alpha <{\alpha }_c$, a gradient blow-up first occurs atFor $\alpha >{\alpha }_c$, $J$ stays uniformly positive. At $\alpha ={\alpha }_c$ it still satisfies $J\ge e^{-\alpha t}>0$ for every finite time. At a point attaining the minimum slope, $u_x=u_0^{\prime }$ for all time, rather than blowing up. Hence the precise conclusion is no finite-time shock occurs when $\alpha \ge {\alpha }_c$, not the printed strict “if and only if $\alpha >{\alpha }_c$”. Equality only makes the limiting characteristic map degenerate as $t\to \infty$. The profile $u_0(\xi )=-\sin \xi$ directly verifies this endpoint distinction. If there is no compressive slope, take ${\alpha }_c=0$; no finite-time shock occurs for positive damping. A nonattained maximal slope is interpreted with a supremum and the corresponding infimum of breakdown times. The gas interpretation also requires $c_0+(\gamma -1)u/2>0$.

Work per unit axial length, with $y$ vertically upward and $x$ across the thin gap. In the frame of the downward-translating cylinder, the wall moves upward at speed $U$ and the cylinder surface is stationary. Let $\eta$ be viscosity and use the leading lubrication theoryThe transverse pressure gradient is negligible; solving $\eta u_{xx}=p_y$ with $u(0)=U$, $u(h)=0$ givesThe flux $q$ is constant by incompressibility. Equal ambient pressure at both ends implies $\int p_{y}dy=0$. With $I_j={\int }_{-\infty }^{\infty }{h}^{-j}dy$ and $I_2/I_3=4h_0/3$, this yieldsThe tangential traction on the cylinder isSince $I_2=I_1/(2h_0)$, its integrated force is $-2\eta U{I}_1+4\eta U{h}_0{I}_2=0$. Pressure acts radially on a circular cylinder and contributes no torque; the tangential traction has moment arm $a$ to leading order. HenceFor clarity, the total vertical hydrodynamic force is not zero. Its leading pressure contribution isupward, while the leading shear contribution just calculated vanishes. Thus this model predicts no induced rolling direction: a freely rotating cylinder can remain nonrotating, rather than rolling like a wheel in contact with the wall. This is the torque-free cylinder in a lubrication gap cancellation; finite ends or different boundary conditions would require a different model.