When the pair reduces, and when its parameters are intrinsic
Throughout, \(q,r,a,b\in(0,1)\). Let the normalized coupled solution be
Write \(U(z)=z+u_1z^3+u_2z^5+u_3z^7+\cdots\) and \(V(z)=z+v_1z^3+v_2z^5+v_3z^7+\cdots\). Put \(p=qr\), \(k=ab\). Coefficient matching gives
Define the fifth-order separation quantity
This factorization will distinguish equality of normalized functions up to amplitude and argument rescaling from a change of coordinates in their defining functional differential operator. These are different equivalence relations.
Exact classification of component rescaling
Theorem G.1.
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The following statements are equivalent:
There are nonzero complex constants \(A,B\) such that \(V(z)=A U(Bz)\) for every complex \(z\).
\(D=0\).
\(q=b\) or \(qb=ar\).
If these conditions hold, put \(s=\sqrt{k/p}>0\), and let \(T_{x,y}\) denote the normalized scalar solution \(T'=1+T(xz)T(yz)\), \(T(0)=0\). Then
The effective scales belong to \((0,1)\): if \(q=b\), then \(sr=\sqrt{ar}\), and if \(qb=ar\), then \(s=b/r\) and \(sr=b\).
Proof.
Normalization gives \(AB=1\). Comparing cubic coefficients gives \(B^2=k/p\), and comparing fifth coefficients gives \(v_2=(k/p)^2u_2\). Thus component rescaling implies \(D=0\), even when complex rescalings are allowed. The displayed factorization, strict positivity, and \(q,b>0\) show that \(D=0\) is equivalent to \(q=b\) or \(qb=ar\).
For the converse define the displayed pair using \(T=T_{q,sr}\). Direct differentiation gives
The second equation reduces to equality of the unordered pairs of arguments
When \(qb=ar\), one has \(a=sq\) and \(b=sr\), giving equality in the displayed order. When \(q=b\), one has \(a=s^2r\), giving equality after interchanging the factors. The pair also has the required initial values and derivatives. Uniqueness of the formal coefficient recursion identifies it with the coupled solution. Finally, \(V(z)=s^{-1}U(sz)\), proving rescaling.
The positive parameter domain therefore contains two explicit reduction surfaces. Outside them the two component functions cannot be obtained from one another by amplitude and argument rescaling. This statement does not exclude every representation using previously known functions.
Derivatives through order seven recover the scales
Theorem G.2.
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If \(D\ne0\), the six coefficients \((u_1,v_1,u_2,v_2,u_3,v_3)\) uniquely determine the ordered tuple \((q,r,a,b)\) within \((0,1)^4\). Consequently, equality of the two labeled normalized germs implies equality of these parameters in this regime.
Proof.
First recover \(p=3u_1\), \(k=3v_1\). If \(x=q^2\), then \(r^2=p^2/x\), and the equation for \(u_2\) gives
This leaves at most two positive candidates. If there are two distinct positive roots \(x,x'\), their product is \(xx'=kp\). The corresponding seventh-coefficient values differ by
The difference is nonzero because \(D\ne0\) and the distinct positive roots have distinct squares. Thus \(u_3\) selects the unique root \(q^2\), giving \(q=\sqrt{x}>0\) and \(r=p/q\). A repeated quadratic root already gives a unique value and requires no selection.
Likewise \(y=a^2\) obeys
with root product \(yy'=k^3/p\). The two candidate seventh coefficients have difference
which is again nonzero for distinct positive candidates. Recover \(a=\sqrt{y}\) and \(b=k/a\). All selections must also respect the contraction domain. Since the supplied coefficients arise from an admissible pair, at least one admissible selection exists. The preceding differences give uniqueness.
This is an exact-data inverse theorem. The displayed differences become small as \(D\to0\), so the argument does not supply a uniform stable inverse near the reduction surfaces. It also does not identify unlabeled components without the additional component-permutation equivalence.
A worked pair with exact order one
Take
At \(\rho=1\), the characteristic matrix has positive eigenvector \((1/2,1)^T\) with eigenvalue one. Its other eigenvalue is \(-1/10\). The growth theorem therefore gives exact order one for both components. Here \(q\ne b\) and \(qb\ne ar\), so \(D\ne0\). The pair is outside both reduction surfaces and its four labeled parameters are recoverable from the coefficients through degree seven.