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Fick's first law gives the diffusive fluxLocal conservation says $C_t+\nabla \cdot \mathbf{J}=0$, and therefore

For $D(C)=kC$, conservation of the deposited amount requiresThe dimensions satisfy $[M]=[C]L^2$ and $[k]=L^2/([C]T)$. Thus $Mkt$ has dimension $L^4$, so the similarity length is $(Mkt{)}^{1/4}$. Requiring the concentration scale to have dimension $[C]$ givesHence

In polar coordinates, substitution into
$C_t=k{r}^{-1}(rC{C}_r{)}_r$ givesMultiplying by $\xi$ and integrating once, with regularity and zero radial flux at the origin, yieldsWhere $F>0$, this reduces to $F^{\prime }=-\xi /4$, soNonnegativity and zero flux at the moving front give the compactly supported profileThe normalization givesand hence $A=1/\sqrt{2}$ and ${\xi }_0^4=32$. This is the Two-dimensional Barenblatt solution with diffusivity proportional to concentration, and its support radius isso $N=32$.

Now add the linear reaction term and writeSubstitution cancels the terms $aC$ and leavesChoosingproduces $G_{\tau }=k\nabla \cdot (G\nabla G)$, as in the linear-reaction time change for quadratic nonlinear diffusion. Therefore

For $a=0$, the front grows like $t^{1/4}$ while the central concentration decays like $t^{-1/2}$. For $a>0$, $\tau \sim e^{at}/a$, so the front grows like $e^{at/4}$ and the central concentration like $e^{at/2}$. For $a<0$, $\tau \to 1/|a|$: the front approaches a finite limiting radius while the concentration decays exponentially to zero.

A state $q$ is accessible whenfor some word $w\in {\Sigma }^{\ast }$. States $p$ and $q$ are equivalent, or indistinguishable, whenfor every continuation $w$.

This equivalence is preserved by transitions: if $p\sim q$, then $\delta (p,a)\sim \delta (q,a)$. Hence define $D/\sim$ with states $[q]$, initial state $[q_0]$, transitionand accepting classes $[q]$ with $q\in F$. Induction on word length givesso $D$ and its quotient deterministic finite automaton by indistinguishable states accept the same language. If two quotient states were equivalent, their representatives would be equivalent in $D$, so the classes would be equal. Thus no two distinct quotient states are equivalent.

For the unary alphabet, accessibility means that all states occur on the orbitFiniteness makes this orbit a directed tail entering a directed cycle. The quotient retains exactly one accepting state and merges precisely those positions having the same future acceptance pattern.

More explicitly, if the unique accepting state lies before the cycle, every state after it can never reach acceptance and these states collapse to one rejecting sink. The minimal diagram is then a directed chain from the initial state through the unique accepting state and onward to that sink, which has a self-loop.

If the accepting state lies on the cycle, all cycle states are distinct because their next visits to the accepting state occur in different residue classes modulo the cycle length. Any remaining tail is a chain feeding the cycle. In a minimal diagram with a nonempty tail, the accepting state is the cycle vertex immediately preceding the entry vertex; otherwise the final tail state has the same future acceptance sequence as a cycle state and would be merged. A pure cycle with one accepting vertex is also possible. These are exactly the minimal accessible unary DFAs with one accepting state.

By the shell theorem, only the mass inside the particle's radius contributes:The conserved Newtonian energy per unit particle mass isWriting $r=a(t)r_0$ and using conservation of enclosed mass, so that $\rho a^3$ is constant, givesThis is the first Friedmann equation without a cosmological constant.

The derivation has two central defects. It uses instantaneous Newtonian gravity and omits the relativistic gravitational effect of pressure, so it is not valid for general cosmological matter. It also starts from a finite ball with a preferred centre and boundary, contrary to exact homogeneity and isotropy; the assumption $r\ll L$ only hides that defect locally.

Order the phase-space coordinates asThen Hamilton's equations areThe Poisson bracket is

For $X=X(x)$ with Jacobian $J_{ab}=\partial X_a/\partial x_b$, the chain rule givesThis has Hamiltonian form with the same canonical matrix for every $H$ exactly whenTaking determinants gives $(\det J{)}^2=1$; the symplectic condition fixes the orientation, so $\det J=1$. Hence the change-of-variables formula shows that the phase-space volume element is invariant. This is Liouville theorem in Hamiltonian mechanics.

Choose paths from $0$ to $z$ in the cut plane and deform them continuously with the endpoint, never crossing the specified cut. With $0\le \arg z<2\pi$, use the sheet reached from the positive real axis; when an endpoint is moved into the left half-plane, the path passes around the upper endpoint $i$ of the cut. The cut makes all such admissible paths homotopic with fixed endpoints, so the integral is single valued and its endpoint derivative is $(1+z^2{)}^{-1/2}$; hence it is analytic.

For $z=-\sinh u$, continue the endpoint counterclockwise from the positive real axis through the upper half-plane. Inthe continued square root equals $-\cosh u$ at $z=-\sinh u$. The logarithm is reached with argument $\pi$, and therefore

The identitiesshow directly why continuation without an argument restriction is multivalued. The complete set of values of the multivalued inverse hyperbolic sine at $\sinh u$ is