Coupled contraction functions and their Perron growth law
Definition and exact coefficient construction
Fix real parameters
Consider the normalized pair
Theorem F.1.
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There is a unique pair of entire solutions. Both functions are odd and have strictly positive coefficients at every odd degree. Write
Then \(u_0=v_0=1\), and, for \(n\ge1\),
Proof.
Comparing successive coefficients in the differential equations determines the coefficient of degree \(k+1\) from coefficients of degrees at most \(k\). Thus the formal pair is unique. The displayed recurrence constructs an odd formal pair satisfying both equations. Formal uniqueness excludes any additional even coefficients. Strict positivity follows by induction.
The elementary bounds
hold by simultaneous induction. If they hold below \(n\), each product in either recurrence is at most \(2^{-(n-1)}\), and there are \(n\) products. The resulting bound is \(n2^{-(n-1)}/(2n+1)<2^{-n}\). Hence the two series converge on \(|z|<\sqrt2\), and define a holomorphic solution there.
Put \(Q=\max(q,r,a,b)<1\). If the pair is already defined on \(|z|<R\), then
define holomorphic extensions on \(|z|<R/Q\). The integrands are holomorphic on that disk. On the old disk the new functions agree with the old pair because their derivatives and values at zero agree. On the larger disk, every contracted argument still lies in the old disk, so the extensions satisfy the same equations. Iterating exhausts the complex plane. Any other entire solution agrees with the unique germ and therefore everywhere. \(\square\)
In particular,
The factorial coefficients \(A_n=(2n+1)!u_n\) and \(B_n=(2n+1)!v_n\) satisfy
Thus \(A_n,B_n\) are polynomials with nonnegative integer coefficients in the four scales. These formulas are exact algebraic constructions. Finite checks of them do not replace the analytic proof.
The characteristic exponent and its positive vector
For \(s>0\), let
There is a unique \(\rho>0\) with
Indeed, the left side strictly increases continuously from zero to one, while the right side strictly decreases continuously from one to zero. This equation is precisely \(\det(I-P(\rho))=0\). Define
Both coordinates of \(w\) are positive, \(\max(w_1,w_2)=1\), and direct multiplication gives
The other eigenvalue is \(q^\rho+b^\rho-1\), whose absolute value is strictly less than one. Consequently
Conversely, a positive matrix of this form having spectral radius one has eigenvalue one and satisfies the determinant equation. This proves the proposed spectral characterization without requiring a numerical eigensolver. A positive left eigenvector is
The notation \(w\) for the growth vector is deliberately different from \(v_n\), the coefficients of \(V\).
Exact order and explicit positive type bounds
For \(R>0\), positivity gives
Set \(F_1=U\), \(F_2=V\), and \(H_i(R)=\log F_i(R)\). Since \(F_i(R)\ge R\), these logarithms are positive for \(R>1\). The type convention used here is
Theorem F.2.
Both functions have order \(\rho\), positive lower type and finite upper type. More precisely, the explicit constants below satisfy
Moreover,
This radial exponent limit is stronger than the limsup used to define order. It does not assert that either type has a limit or that growth is uniform in the angular variable.
Explicit lower comparison
Let
We prove the stronger statement
On \(2\le R\le B\), its right side is at most \(\log2\), because \(w_i\le1\), whereas \(H_i(R)\ge\log R\ge\log2\).
For the first equation, integrate only over the final unit interval to obtain, when \(R\ge B\),
The second equation gives the same inequality with \((q,r)\) replaced by \((a,b)\) and \(U(R)\) replaced by \(V(R)\). All four child radii are at least two. Every scale \(p\in\{q,r,a,b\}\) satisfies
because \((1-p)C\ge p\). If the comparison already holds at the child radii, then
The second row of \(P(\rho)w=w\) proves the second component in exactly the same way.
For a formal propagation, put
Every child of \(R\ge B\) is at most \(Q(R-1)\le R-d\). Simultaneous induction over the intervals \([B+kd,B+(k+1)d]\), starting with the established seed interval, therefore proves both comparisons for every \(R\ge2\). Continuity handles shared endpoints. The argument uses only monotonicity of positive powers and works unchanged when \(0<\rho<1\).
Explicit upper comparison
Set
Choose any integer \(N\ge0\) satisfying \(Q^{-N}\ge3/m\), and define
These constants require only the parameters and the uniquely defined root \(\rho\). They require no evaluation of an unknown solution.
To bound a seed interval, the coefficient majorant from Theorem F.1 gives \(F_i(1)\le2\). Put \(R_k=Q^{-k}\), \(Z_0=2\), and
Monotonicity of the positive functions and the integral equations give \(F_i(R_k)\le Z_k\) simultaneously. As \(Z_k\ge1\),
Summing this elementary recurrence gives \(\log Z_N\le W_N\). Therefore \(H_i(e^t)\le W_N\) throughout \(1\le t\le1+\Delta\), since \(e^t\le e/m<3/m\le R_N\).
Now let
For \(t\ge1+\Delta\), all child radii are at least \(e\), so their function values are at least one. The full integral gives
The definition of \(D\) converts this into
Similarly,
On the seed interval, \(G_i(t)\le S\le c_+w_i e^{\rho t}\). The Perron equation propagates this comparison: substituting the bounds at the children in the first inequality gives \(c_+e^{\rho t}(q^\rho w_1+r^\rho w_2)=c_+w_1e^{\rho t}\), and the second row works likewise. All logarithmic delays belong to \([\log(1/Q),\Delta]\). Induction in intervals of length \(\log(1/Q)>0\), after the seed interval, therefore proves
The lower and upper comparisons imply the asserted positive, finite types. Taking logarithms of these bounds for sufficiently large \(R\) gives the claimed radial exponent limits and hence exact orders. \(\square\)
The two type vectors have exactly the Perron proportions
The explicit constants above are conservative. There is nevertheless an exact statement about the relative types of the two components.
Theorem F.3.
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There exist positive finite constants \(\underline\tau\le\overline\tau\) such that
In particular,
Either both ordinary type limits exist or neither does. When they exist, their vector is a positive multiple of \(w\).
Proof.
Use the final-unit integral inequality from the lower proof and divide by \(R^\rho\). Taking the liminf gives
Here \(p(R-1)\) tends to infinity continuously and \([p(R-1)]^\rho/R^\rho\to p^\rho\), which justifies each scaled liminf. The liminf of a sum is at least the sum of the liminfs. All quantities are positive and finite by Theorem F.2.
The full-integral inequality
and its second-row counterpart imply, on taking the limsup,
The terms \((\log R+\log2)/R^\rho\) vanish. Dotting either componentwise inequality with the strictly positive left eigenvector \(\ell\) from Section The characteristic exponent and its positive vector shows that its nonnegative difference vector has zero positive weighted sum. Every coordinate of that difference is therefore zero. Thus both type vectors are fixed by \(P(\rho)\). Its eigenvalue-one space is one dimensional, so each is a positive multiple of \(w\).
If one component has an ordinary type limit, its lower and upper types are equal. Because that coordinate of \(w\) is positive, this forces \(\underline\tau=\overline\tau\), and the other component also has a type limit. \(\square\)
This theorem does not assert a limit for \(\log U(R)/\log V(R)\) when type oscillations persist. It also does not identify the scalar values of either type or prove their equality.
Reductions, a worked unequal pair, and parameter structure
Exact scalar reduction. The components coincide if and only if the unordered pairs of scales coincide:
For sufficiency, the scalar solution \(T_{q,r}\) from routes/asymmetric_dilation.md makes \((T_{q,r},T_{q,r})\) solve the coupled equations. Theorem F.1 gives uniqueness. For necessity, equality at degree three gives \(qr=ab=p>0\). Equality at degree five then gives \(q^2+r^2=a^2+b^2\). The two positive unordered pairs have the same product and the same sum \(\sqrt{q^2+r^2+2p}\), so they coincide. This characterizes equality of the normalized components. A stronger assertion about all possible changes of variables requires its own nonreduction theorem.
On either scalar reduction locus, the characteristic equation is equivalent to \(q^\rho+r^\rho=1\), recovering the scalar order theorem. At \(q=r=a=b=1/2\),
A genuinely unequal order-one pair. Set
The determinant identity at \(\rho=1\) is \((1-1/5)(1-7/10)=(3/5)(2/5)=6/25\). Thus the exact order is one, and
The first terms are
Both lower and upper exponential types of \(V\) are exactly twice the corresponding types of \(U\). The row sums of \(P(1)\) are \(3/5\) and \(13/10\). Consequently neither an unweighted average row condition nor the maximum of the two scalar row orders yields this example's coupled order. The positive eigenvector accounts for the exchange of growth between the components.
For this example the elementary lower constants are \(m=1/5\), \(Q=7/10\), \(C=7/3\), \(B=11\), and \(c_-=3\log2/40\). Hence, for every \(R\ge2\),
Dependence of the order on the scales. The determinant equation shows that the order depends on the self scales \(q,b\) and the product \(ar\) of the two cross scales. Changing \(a\) and \(r\) while keeping their product fixed leaves the order fixed, provided both remain in \((0,1)\). The Perron vector and the normalized functions generally change.
For any prescribed \(s>0\) and any \(q,r\in(0,1)\), choices satisfying
lie on the order-\(s\) surface. For instance, one may select \(q,r,b\) with \((1-q^s)(1-b^s)<r^s\), then define
This constructs exact order surfaces without fitting growth from truncated series.