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1 \hypertarget{v11-coupled_inverse_full-when-the-pair-reduces-and-when-its-parameters-are-intrinsic}{%
2 \chapter[Reduction and intrinsic parameters]{When the pair reduces, and when its parameters are
3 intrinsic}\label{app:coupled-inverse}\label{v11-coupled_inverse_full-when-the-pair-reduces-and-when-its-parameters-are-intrinsic}}
4 
5 Throughout, \(q,r,a,b\in(0,1)\). Let the normalized coupled solution be
6 
7 \[
8 U'=1+U(qz)V(rz),\qquad V'=1+U(az)V(bz),\qquad U(0)=V(0)=0.
9 \]
10 
11 Write \(U(z)=z+u_1z^3+u_2z^5+u_3z^7+\cdots\) and
12 \(V(z)=z+v_1z^3+v_2z^5+v_3z^7+\cdots\). Put \(p=qr\), \(k=ab\).
13 Coefficient matching gives
14 
15 \[
16 u_1=\frac p3,\qquad v_1=\frac k3,
17 \] \[
18 u_2=\frac p{15}(p q^2+k r^2),\qquad
19 v_2=\frac k{15}(p a^2+k b^2),
20 \] \[
21 u_3=\frac p7(q^4u_2+p^2u_1v_1+r^4v_2),\qquad
22 v_3=\frac k7(a^4u_2+k^2u_1v_1+b^4v_2).
23 \]
24 
25 Define the fifth-order separation quantity
26 
27 \[
28 \boxed{D=k^2u_2-p^2v_2
29 =\frac{pk ar}{15}(q^2-b^2)(qb-ar).}
30 \]
31 
32 This factorization will distinguish equality of normalized functions up
33 to amplitude and argument rescaling from a change of coordinates in
34 their defining functional differential operator. These are different
35 equivalence relations.
36 
37 \hypertarget{v11-coupled_inverse_full-exact-classification-of-component-rescaling}{%
38 \section{Exact classification of component
39 rescaling}\label{v11-coupled_inverse_full-exact-classification-of-component-rescaling}}
40 
41 \textbf{Theorem G.1.} The following statements are equivalent:
42 
43 \begin{enumerate}
44 \def\labelenumi{\arabic{enumi}.}
45 \tightlist
46 \item
47   There are nonzero complex constants \(A,B\) such that \(V(z)=A U(Bz)\)
48   for every complex \(z\).
49 \item
50   \(D=0\).
51 \item
52   \(q=b\) or \(qb=ar\).
53 \end{enumerate}
54 
55 If these conditions hold, put \(s=\sqrt{k/p}>0\), and let \(T_{x,y}\)
56 denote the normalized scalar solution \(T'=1+T(xz)T(yz)\), \(T(0)=0\).
57 Then
58 
59 \[
60 \boxed{
61 U(z)=\sqrt{s}\,T_{q,sr}(z/\sqrt{s}),\qquad
62 V(z)=s^{-1/2}T_{q,sr}(\sqrt{s}\,z).
63 }
64 \]
65 
66 The effective scales belong to \((0,1)\): if \(q=b\), then
67 \(sr=\sqrt{ar}\), and if \(qb=ar\), then \(s=b/r\) and \(sr=b\).
68 
69 \textbf{Proof.} Normalization gives \(AB=1\). Comparing cubic
70 coefficients gives \(B^2=k/p\), and comparing fifth coefficients gives
71 \(v_2=(k/p)^2u_2\). Thus component rescaling implies \(D=0\), even when
72 complex rescalings are allowed. The displayed factorization, strict
73 positivity, and \(q,b>0\) show that \(D=0\) is equivalent to \(q=b\) or
74 \(qb=ar\).
75 
76 For the converse define the displayed pair using \(T=T_{q,sr}\). Direct
77 differentiation gives
78 
79 \[
80 U'(z)=1+T(qz/\sqrt{s})T(r\sqrt{s}\,z)
81 =1+U(qz)V(rz).
82 \]
83 
84 The second equation reduces to equality of the unordered pairs of
85 arguments
86 
87 \[
88 \{q\sqrt{s},\;s^{3/2}r\}
89 =\{a/\sqrt{s},\;b\sqrt{s}\}.
90 \]
91 
92 When \(qb=ar\), one has \(a=sq\) and \(b=sr\), giving equality in the
93 displayed order. When \(q=b\), one has \(a=s^2r\), giving equality after
94 interchanging the factors. The pair also has the required initial values
95 and derivatives. Uniqueness of the formal coefficient recursion
96 identifies it with the coupled solution. Finally, \(V(z)=s^{-1}U(sz)\),
97 proving rescaling.
98 
99 The positive parameter domain therefore contains two explicit reduction
100 surfaces. Outside them the two component functions cannot be obtained
101 from one another by amplitude and argument rescaling. This statement
102 does not exclude every representation using previously known functions.
103 
104 \hypertarget{v11-coupled_inverse_full-seven-derivatives-recover-all-four-ordered-scales-away-from-reduction}{%
105 \section{Derivatives through order seven recover the scales}\label{v11-coupled_inverse_full-seven-derivatives-recover-all-four-ordered-scales-away-from-reduction}}
106 
107 \textbf{Theorem G.2.} If \(D\ne0\), the six coefficients
108 \((u_1,v_1,u_2,v_2,u_3,v_3)\) uniquely determine the ordered tuple
109 \((q,r,a,b)\) within \((0,1)^4\). Consequently, equality of the two
110 labeled normalized germs implies equality of these parameters in this
111 regime.
112 
113 \textbf{Proof.} First recover \(p=3u_1\), \(k=3v_1\). If \(x=q^2\), then
114 \(r^2=p^2/x\), and the equation for \(u_2\) gives
115 
116 \[
117 p x^2-\frac{15u_2}{p}x+kp^2=0.
118 \]
119 
120 This leaves at most two positive candidates. If there are two distinct
121 positive roots \(x,x'\), their product is \(xx'=kp\). The corresponding
122 seventh-coefficient values differ by
123 
124 \[
125 u_3(x)-u_3(x')
126 =\frac p7(x^2-x'^2)
127 \left(u_2-\frac{p^2}{k^2}v_2\right)
128 =\frac{pD}{7k^2}(x^2-x'^2).
129 \]
130 
131 The difference is nonzero because \(D\ne0\) and the distinct positive
132 roots have distinct squares. Thus \(u_3\) selects the unique root
133 \(q^2\), giving \(q=\sqrt{x}>0\) and \(r=p/q\). A repeated quadratic
134 root already gives a unique value and requires no selection.
135 
136 Likewise \(y=a^2\) obeys
137 
138 \[
139 p y^2-\frac{15v_2}{k}y+k^3=0,
140 \]
141 
142 with root product \(yy'=k^3/p\). The two candidate seventh coefficients
143 have difference
144 
145 \[
146 v_3(y)-v_3(y')
147 =\frac k7(y^2-y'^2)
148 \left(u_2-\frac{p^2}{k^2}v_2\right)
149 =\frac{D}{7k}(y^2-y'^2),
150 \]
151 
152 which is again nonzero for distinct positive candidates. Recover
153 \(a=\sqrt{y}\) and \(b=k/a\). All selections must also respect the
154 contraction domain. Since the supplied coefficients arise from an
155 admissible pair, at least one admissible selection exists. The preceding
156 differences give uniqueness.
157 
158 This is an exact-data inverse theorem. The displayed differences become
159 small as \(D\to0\), so the argument does not supply a uniform stable
160 inverse near the reduction surfaces. It also does not identify unlabeled
161 components without the additional component-permutation equivalence.
162 
163 \hypertarget{v11-coupled_inverse_full-a-worked-pair-with-exact-order-one}{%
164 \section{A worked pair with exact order
165 one}\label{v11-coupled_inverse_full-a-worked-pair-with-exact-order-one}}
166 
167 Take
168 
169 \[
170 (q,r,a,b)=\left(\frac15,\frac25,\frac35,\frac7{10}\right),\qquad
171 p=\frac2{25},\quad k=\frac{21}{50}.
172 \]
173 
174 At \(\rho=1\), the characteristic matrix has positive eigenvector
175 \((1/2,1)^T\) with eigenvalue one. Its other eigenvalue is \(-1/10\).
176 The growth theorem therefore gives exact order one for both components.
177 Here \(q\ne b\) and \(qb\ne ar\), so \(D\ne0\). The pair is outside both
178 reduction surfaces and its four labeled parameters are recoverable from
179 the coefficients through degree seven.
180 

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