Worked Case Files and Problems

This capstone applies the chapter’s full diagnostic workflow to four benchmark problems. The point is not merely to recognize named functions. Each case tracks singularities, local or sectorial bases, branches, Wronskians, formal data, analytic data, and exceptional parameters in one consistent ledger.

The examples also separate four increasingly refined questions:

1. What kind of singularity is present?
2. What is its formal type?
3. Which normalized analytic basis realizes that type?
4. How do two normalized bases compare?

A formula is considered checked only after its direction, determinant, and parameter domain agree.

The underlying tools were developed on the pages about scalar equations and systems, regular singularities, formal irregular classification, sectorial normalization, and confluence. Here they are used as an audit protocol rather than rederived.

Four benchmark data sheets

Model	Singular structure	Canonical lesson
Bessel	$0$ regular singular; $\infty$ of Poincaré rank one	Frobenius resonance and normalized oscillatory bases coexist
Airy	Every finite point ordinary; $\infty$ slope $3/2$	A turning point is not an ODE singularity; ramification and sectorial recessiveness matter
Gauss hypergeometric	$0,1,\infty$ regular singular	An exact gamma-function connection matrix can be audited by Wronskians
Diagonal versus triangular rank-two pair	One Poincaré-rank-one irregular point with the same formal exponential factors	Formal type does not determine Stokes data

Throughout, $\Wr[f,g]=fg'-f'g$, continuation acts on the right, and powers use the branch declared in each case.

Bessel: Frobenius at zero, Hankel at infinity

Consider

$$z^2y''+zy'+(z^2-\nu^2)y=0.$$

The origin is regular singular with exponents $\pm\nu$. Infinity is irregular: after $y=z^{-1/2}\psi$,

$$\psi'' +\left[ 1+\frac{\tfrac14-\nu^2}{z^2} \right]\psi=0,$$

so the exponential factors in $x=1/z$ are

$$q_\pm(x)=\pm\frac{\ii}{x}.$$

Unit-leading Frobenius basis

For $\nu\notin\mathbb Z$, define

$$\begin{aligned} f_+(z) &= 2^\nu\Gamma(\nu+1)J_\nu(z),\\ f_-(z) &= 2^{-\nu}\Gamma(1-\nu)J_{-\nu}(z). \end{aligned}$$

On the principal branch,

$$f_\pm(z)=z^{\pm\nu}\left(1+O(z^2)\right).$$

Their Wronskian can be read from the leading powers or from the standard Bessel identity:

$$\Wr[f_+,f_-]=-\frac{2\nu}{z}.$$

Thus the positive local monodromy in $\Phi_0=(f_+,f_-)$ is

$$M_0 = \begin{pmatrix} \ee^{2\pi\ii\nu}&0\\ 0&\ee^{-2\pi\ii\nu} \end{pmatrix}.$$

Unit-leading oscillatory basis

Put

$$\alpha_\nu = \frac{\pi\nu}{2}+\frac{\pi}{4}$$

and define, in a common sector about the positive real axis,

$$\begin{aligned} h_+(z) &= \sqrt{\frac{\pi}{2}}\, \ee^{\ii\alpha_\nu} H_\nu^{(1)}(z),\\ h_-(z) &= \sqrt{\frac{\pi}{2}}\, \ee^{-\ii\alpha_\nu} H_\nu^{(2)}(z). \end{aligned}$$

Then

$$\begin{aligned} h_+(z) &\sim z^{-1/2}\ee^{\ii z} \left(1+O(z^{-1})\right),\\ h_-(z) &\sim z^{-1/2}\ee^{-\ii z} \left(1+O(z^{-1})\right), \end{aligned}$$

and

$$\Wr[h_+,h_-]=-\frac{2\ii}{z}.$$

The Wronskian is not constant because the original Bessel equation has $p=1/z$.

A fully normalized connection matrix

Set

$$A_\nu=2^\nu\Gamma(\nu+1), \qquad B_\nu=2^{-\nu}\Gamma(1-\nu),$$

and

$$d_\pm = \sqrt{\frac{\pi}{2}}\, \ee^{\pm\ii\alpha_\nu}.$$

The elementary identities

$$\begin{aligned} J_\nu &= \frac12 \left( H_\nu^{(1)}+H_\nu^{(2)} \right),\\ J_{-\nu} &= \frac12 \left( \ee^{\ii\pi\nu}H_\nu^{(1)} +\ee^{-\ii\pi\nu}H_\nu^{(2)} \right) \end{aligned}$$

give

$$\Phi_0 = \Phi_\infty C_{\infty0}, \qquad \Phi_\infty=(h_+,h_-),$$

with

$$C_{\infty0} = \frac12 \begin{pmatrix} d_+^{-1}&0\\ 0&d_-^{-1} \end{pmatrix} \begin{pmatrix} A_\nu&B_\nu\ee^{\ii\pi\nu}\\ A_\nu&B_\nu\ee^{-\ii\pi\nu} \end{pmatrix}.$$

The determinant audit is immediate:

$$\det C_{\infty0} = \frac{\Wr[f_+,f_-]}{\Wr[h_+,h_-]} =-\ii\nu.$$

This detects a missing phase or reversed column before any numerical test.

Resonance is not synonymous with a logarithm

When $\nu=n\in\mathbb Z$, $J_{-n}=(-1)^nJ_n$ and the displayed Frobenius basis collapses; $Y_n$ supplies the logarithmic second solution. When $\nu\in\mathbb Z+\tfrac12$, the exponent difference is an integer, but $J_\nu$ and $J_{-\nu}$ remain independent and no logarithm is forced.

Airy: an ordinary turning point and ramified infinity

The Airy equation

$$y''-zy=0$$

has no finite singularity. Its zero of the WKB coefficient at $z=0$ is a simple turning point, but the exact equation and all its solutions are holomorphic there.

At infinity, set $x=1/z$ and then $x=t^2$. The formal exponential factors are

$$q_\pm(t) = \pm\frac{2}{3t^3},$$

so the slope is $3/2$ in the original $x$-coordinate. With the natural normalization from formal irregular classification,

$$M_{\mathrm f} = \ii \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}, \qquad M_{\mathrm f}^2=-I.$$

This is formal monodromy at infinity, not monodromy around the ordinary point $z=0$.

Let

$$\omega=\ee^{2\pi\ii/3}.$$

The three rotated functions

$$\operatorname{Ai}(z), \qquad \operatorname{Ai}(\omega z), \qquad \operatorname{Ai}(\omega^2z)$$

all solve the same equation. The $k$th one is recessive where the chosen lift of $\arg(\omega^kz)$ lies between $-\pi/3$ and $\pi/3$. No two-dimensional solution space can contain three independent recessive solutions, and the exact relation is

$$\operatorname{Ai}(z) +\omega\operatorname{Ai}(\omega z) +\omega^2\operatorname{Ai}(\omega^2z) =0.$$

With the book’s Wronskian sign,

$$\Wr \left[ \operatorname{Ai}(z), \operatorname{Ai}(\omega z) \right] = \frac{\ee^{-\pi\ii/6}}{2\pi}.$$

These raw rotations can be normalized so that the lateral transition has no hidden phase. Define, on the lifted angular sectors,

$$A_k(z) = 2\sqrt{\pi}\, \ee^{-\pi\ii k/6} \operatorname{Ai}(\omega^{-k}z), \qquad k\in\mathbb Z.$$

The cyclic identity and the rotated Wronskian give

$$A_{k+2}=A_k-\ii A_{k+1}, \qquad \Wr[A_0,A_1]=2.$$

In the central sector, introduce the familiar formal-ordering basis

$$\phi_+ = \sqrt{\pi}\operatorname{Bi}(z), \qquad \phi_- = 2\sqrt{\pi}\operatorname{Ai}(z).$$

It satisfies

$$\begin{aligned} A_0&=\phi_-,\\ A_1&=\phi_+-\frac{\ii}{2}\phi_-,\\ A_{-1}&=\phi_++\frac{\ii}{2}\phi_-. \end{aligned}$$

Thus the two lateral dominant solutions differ by the shared recessive solution:

$$(A_{-1},A_0) = (A_1,A_0) \left( I+\ii E_{21} \right).$$

This is an exact right-acting Stokes transition for the displayed ordering. It occurs across the direction on which the exponential ratio $\exp(-\tfrac43z^{3/2})$ is maximally small. The equal-magnitude rays, defined instead by $\operatorname{Re}(z^{3/2})=0$, have $\arg z=\pi/3,\pi,5\pi/3$ modulo $2\pi$. Stating both phase conditions avoids the conflicting “Stokes” and “anti-Stokes” terminology in the literature.

The asymptotic normalization

$$\operatorname{Ai}(z) \sim \frac{1}{2\sqrt{\pi}}\, z^{-1/4} \exp\left(-\frac23z^{3/2}\right)$$

requires a branch and a sector. Rotating $z$ changes both. Since the exact solutions are entire, the ordered Stokes factors around infinity must combine with $M_{\mathrm f}$ to give the trivial analytic continuation of a fixed entire basis.

Gauss: a gamma connection matrix with a determinant check

Consider

$$z(1-z)y'' +\left[ c-(a+b+1)z \right]y' -ab\,y=0.$$

The exponent pairs at $0,1,\infty$ are

$$\{0,1-c\}, \qquad \{0,c-a-b\}, \qquad \{a,b\},$$

where the exponents at infinity refer to the local coordinate $t=1/z$. All three points are regular singular.

For comparison with the oper form, set

$$\begin{aligned} \theta_0&=1-c,\\ \theta_1&=c-a-b,\\ \theta_\infty&=a-b, \end{aligned} \qquad \Delta_j=\frac{1-\theta_j^2}{4}.$$

The Liouville-normal-form equation $\psi''+T(z)\psi=0$ has

$$\begin{aligned} T(z) ={}& \frac{\Delta_0}{z^2} +\frac{\Delta_1}{(z-1)^2}\\ &+ \frac{ \Delta_\infty-\Delta_0-\Delta_1 }{ z(z-1) }. \end{aligned}$$

Set

$$\delta=c-a-b.$$

Assume first that $c,\delta\notin\mathbb Z$. On the plane cut along $(-\infty,0]$ and $[1,\infty)$, take the branches positive on $0<z<1$. Define

$$\begin{aligned} f_0(z) &= {}_2F_1(a,b;c;z),\\ g_0(z) &= z^{1-c} {}_2F_1 \left( \begin{matrix} a-c+1,\ b-c+1\\ 2-c \end{matrix} ;z \right) \end{aligned}$$

and

$$\begin{aligned} f_1(z) &= {}_2F_1 \left( \begin{matrix} a,\ b\\ 1-\delta \end{matrix} ;1-z \right),\\ g_1(z) &= (1-z)^\delta {}_2F_1 \left( \begin{matrix} c-a,\ c-b\\ 1+\delta \end{matrix} ;1-z \right). \end{aligned}$$

Thus

$$\Phi_0=(f_0,g_0), \qquad \Phi_1=(f_1,g_1)$$

are unit-leading Frobenius bases at $0$ and $1$.

Euler’s continuation formula gives the book-direction relation

$$\Phi_0=\Phi_1C_{10},$$

where

$$C_{10} = \begin{pmatrix} A&A'\\ B&B' \end{pmatrix}$$

and

$$\begin{aligned} A &= \frac{\Gamma(c)\Gamma(\delta)} {\Gamma(c-a)\Gamma(c-b)},\\ B &= \frac{\Gamma(c)\Gamma(-\delta)} {\Gamma(a)\Gamma(b)},\\ A' &= \frac{\Gamma(2-c)\Gamma(\delta)} {\Gamma(1-a)\Gamma(1-b)},\\ B' &= \frac{\Gamma(2-c)\Gamma(-\delta)} {\Gamma(a-c+1)\Gamma(b-c+1)}. \end{aligned}$$

The exact Wronskians are

$$\begin{aligned} \Wr[f_0,g_0] &= (1-c) z^{-c}(1-z)^{\delta-1},\\ \Wr[f_1,g_1] &= -\delta z^{-c}(1-z)^{\delta-1}. \end{aligned}$$

Therefore

$$\det C_{10} = \frac{1-c}{-\delta} = \frac{1-c}{a+b-c}.$$

The same result follows from the four gamma entries using $\Gamma(z)\Gamma(1-z)=\pi/\sin(\pi z)$. This independent determinant check is the quickest way to catch a swapped basis direction.

The local monodromy matrices in their own Frobenius bases are

$$D_0 = \operatorname{diag} \left( 1,\ee^{2\pi\ii(1-c)} \right), \qquad D_1 = \operatorname{diag} \left( 1,\ee^{2\pi\ii\delta} \right).$$

Because $\Phi_0=\Phi_1C_{10}$, a positively oriented loop around $1$ is represented in the $0$-basis by

$$M_1^{(0)} = C_{10}^{-1}D_1C_{10}.$$

This conjugation formula is path-specific: winding the continuation path before comparing the bases appends the corresponding monodromy factor.

A reproducible numerical checkpoint

Take

$$a=\frac13, \qquad b=\frac25, \qquad c=\frac76, \qquad z=\frac12.$$

Then $\delta=13/30$ and

$$C_{10} \approx \begin{pmatrix} 1.3957761521550798411& 1.1444361673723044296\\ -0.56597550632696216556& -0.18850261502591602281 \end{pmatrix}.$$

The displayed entries are rounded. Recomputing the gamma quotients without rounding, and then evaluating both functions and their first derivatives at 80-digit working precision, gives

$$\left\| \Phi_0-\Phi_1C_{10} \right\|_{\max} <3\times10^{-80},$$

where $\|A\|_{\max}=\max_{i,j}|A_{ij}|$ is the entrywise maximum norm. Using the rounded matrix printed above would, of course, give only a roughly 20-digit check. The independent exact check is

$$\det C_{10} = \frac5{13}.$$

The quoted residual is numerical evidence, not an additional proof; the Wronskian identity supplies the exact determinant test.

The displayed formulas are generic, not universal without limits:

• if $c\in\mathbb Z$ or $\delta\in\mathbb Z$, a local basis may contain a logarithm and must be obtained by resonant recombination;
• if $a-b\in\mathbb Z$, the basis at infinity is resonant even when the $0$-to-$1$ formula is regular;
• if $a$ or $b$ is a nonpositive integer, a polynomial solution appears and reciprocal gamma factors can make a connection coefficient vanish;
• if $c$ is a nonpositive integer, the standard ${}_2F_1(a,b;c;z)$ normalization is singular although the differential equation itself still makes sense.

Same formal type, different analytic class

Compare

$$\frac{\dd Y}{\dd x} = \begin{pmatrix} -x^{-2}&0\\ 0&0 \end{pmatrix}Y$$

with

$$\frac{\dd Y}{\dd x} = \begin{pmatrix} -x^{-2}&x^{-1}\\ 0&0 \end{pmatrix}Y.$$

Both have formal exponential factors

$$q_1(x)=\frac1x, \qquad q_2(x)=0,$$

zero formal powers, and $M_{\mathrm f}=I$. The first is already diagonal:

$$\Phi_{\mathrm d} = \begin{pmatrix} \ee^{1/x}&0\\ 0&1 \end{pmatrix},$$

so all its Stokes matrices and its actual monodromy are trivial.

The triangular system has the divergent formal normalizer

$$\widehat H = \begin{pmatrix} 1&\widehat h\\ 0&1 \end{pmatrix}, \qquad \widehat h = \sum_{n=1}^{\infty} (-1)^{n-1}(n-1)!x^n.$$

With the lateral convention already fixed,

$$S=I-2\pi\ii E_{12}, \qquad M_0=S^{-1} =I+2\pi\ii E_{12}.$$

Indeed,

$$\mathcal B\widehat h(\xi) = \frac1{1+\xi}, \qquad h(x) = -\ee^{1/x}\operatorname{Ei}(-1/x).$$

The Borel pole at $\xi=-1$ fixes the jump, and the logarithmic branch of $\operatorname{Ei}$ fixes the positive actual monodromy. The parameter-limit calculation shows how the corresponding parameter family recombines into $\operatorname{Log}x$ when the exponential scale collapses.

The comparison isolates the analytic modulus:

Datum	Diagonal system	Triangular system
Exponential factors	$1/x,\ 0$	$1/x,\ 0$
Formal powers	$0,\ 0$	$0,\ 0$
Formal monodromy	$I$	$I$
Formal normalizer	Convergent	Gevrey-$1$, divergent
Nonidentity Stokes factor	None	$I-2\pi\ii E_{12}$
Positive actual monodromy	$I$	$I+2\pi\ii E_{12}$

Formal classification answers which asymptotic types are possible. Analytic classification also remembers how their sectorial realizations glue.

A compact audit matrix

Case	First check	Independent check	Main failure mode
Bessel	Leading Frobenius and Hankel coefficients	Wronskian ratio for $\det C_{\infty0}$	Using $J_{-\nu}$ at integer $\nu$
Airy	Rotated solution relation	Constant Wronskian at $z=0$	Calling the turning point singular
Gauss	Euler continuation coefficients	Abel Wronskians and gamma reflection	Omitting branch or resonance data
Poincaré-rank-one pair	Same formal exponential factors	Borel residue and monodromy product	Equating formal and analytic type

Problems

1. Rebuild the Bessel determinant. Starting from the two Hankel identities, derive $C_{\infty0}$ and show directly from its entries that $\det C_{\infty0}=-\ii\nu$.

Solution

Insert

$$J_\nu=\frac12(H_\nu^{(1)}+H_\nu^{(2)}), \qquad J_{-\nu} =\frac12 \left( \ee^{\ii\pi\nu}H_\nu^{(1)} +\ee^{-\ii\pi\nu}H_\nu^{(2)} \right)$$

into the unit-leading bases. This gives the displayed matrix. Its determinant is

$$\begin{aligned} \det C_{\infty0} &= \frac{ -2\ii A_\nu B_\nu\sin(\pi\nu) }{ 4d_+d_- }\\ &= -\ii\nu, \end{aligned}$$

because

$$d_+d_-=\frac{\pi}{2}, \qquad A_\nu B_\nu = \Gamma(\nu+1)\Gamma(1-\nu) = \frac{\pi\nu}{\sin(\pi\nu)}.$$

2. Diagnose two resonant Bessel values. Compare $\nu=1$ and $\nu=1/2$. Does a logarithmic solution occur at the origin in each case?

Solution

Both exponent differences are integers:

$$2\nu=2 \quad\text{and}\quad 2\nu=1.$$

For $\nu=1$, $J_{-1}=-J_1$, so the pure Frobenius pair collapses and the second standard solution $Y_1$ contains a logarithm. For $\nu=1/2$, $J_{1/2}$ and $J_{-1/2}$ are independent and reduce to sine and cosine:

$$J_{1/2}(z) = \sqrt{\frac{2}{\pi z}}\sin z, \qquad J_{-1/2}(z) = \sqrt{\frac{2}{\pi z}}\cos z.$$

The resonant obstruction vanishes in the half-integer case, so no logarithm is required.

3. Prove the Airy rotation relation. Show that

$$F(z) = \operatorname{Ai}(z) +\omega\operatorname{Ai}(\omega z) +\omega^2\operatorname{Ai}(\omega^2z)$$

vanishes identically.

Solution

Each term solves $y''=zy$, so $F$ does as well. At the origin,

$$F(0) = \left( 1+\omega+\omega^2 \right)\operatorname{Ai}(0) =0.$$

Differentiation supplies an extra rotation factor:

$$F'(0) = \left( 1+\omega^2+\omega^4 \right)\operatorname{Ai}'(0) =0.$$

Uniqueness for the ordinary initial-value problem gives $F\equiv0$.

4. Compute the rotated Airy Wronskian. Use

$$\operatorname{Ai}(0) = \frac{1}{3^{2/3}\Gamma(2/3)}, \qquad \operatorname{Ai}'(0) = -\frac{1}{3^{1/3}\Gamma(1/3)}$$

to compute $\Wr[\operatorname{Ai}(z),\operatorname{Ai}(\omega z)]$.

Solution

There is no first-derivative term, so the Wronskian is constant. At $z=0$,

$$\Wr = \operatorname{Ai}(0)\operatorname{Ai}'(0) (\omega-1).$$

Using

$$\Gamma(1/3)\Gamma(2/3)=\frac{2\pi}{\sqrt3}$$

and

$$1-\omega=\sqrt3\,\ee^{-\pi\ii/6},$$

one obtains

$$\Wr \left[ \operatorname{Ai}(z), \operatorname{Ai}(\omega z) \right] = \frac{\ee^{-\pi\ii/6}}{2\pi}.$$

5. Audit the Gauss connection determinant. Use Abel’s identity to derive $\det C_{10}$ without simplifying any gamma functions.

Solution

The coefficient of $y'$ is

$$p(z) = \frac cz +\frac{\delta-1}{1-z}.$$

Hence every Wronskian is a constant multiple of

$$z^{-c}(1-z)^{\delta-1}.$$

The unit-leading behavior at $0$ gives the constant $1-c$. At $1$,

$$f_1\sim1, \qquad g_1\sim(1-z)^\delta,$$

so

$$\Wr[f_1,g_1] \sim -\delta(1-z)^{\delta-1}.$$

Since $\Phi_0=\Phi_1C_{10}$,

$$\det C_{10} = \frac{\Wr[f_0,g_0]}{\Wr[f_1,g_1]} = \frac{1-c}{-\delta}.$$

6. Locate the analytic modulus. Prove that the diagonal and triangular rank-two systems of Poincaré rank one are formally equivalent but not analytically equivalent under a single-valued meromorphic gauge preserving the stated normalization.

Solution

For the triangular system, the off-diagonal entry of the formal normalizing gauge satisfies

$$x^2\widehat h'+\widehat h=x$$

and therefore

$$\widehat h = \sum_{n=1}^{\infty} (-1)^{n-1}(n-1)!x^n.$$

This formal gauge removes the off-diagonal entry, so both systems have the same formal normal form. The series diverges, however. Its two lateral sums across $\arg x=\pi$ differ by $-2\pi\ii\ee^{1/x}$, producing $S=I-2\pi\ii E_{12}$. The diagonal system has $S=I$.

Stokes data are invariants of analytic meromorphic equivalence once the formal normalization is fixed. Since the two Stokes cocycles differ, no such analytic gauge can identify the systems.

References

• NIST Digital Library of Mathematical Functions, §10.2, Bessel definitions, §10.5, Wronskians, and §10.17, Hankel asymptotics.
• NIST Digital Library of Mathematical Functions, §9.2, Airy differential equation, Wronskians, and connection formulas and §9.7, asymptotic expansions.
• NIST Digital Library of Mathematical Functions, §15.10, hypergeometric local bases and connection formulas.
• F. W. J. Olver, Asymptotics and Special Functions, AKP Classics, 1997 reprint, for sectorial asymptotics and error bounds.
• Y. Sibuya, Linear Differential Equations in the Complex Domain: Problems of Analytic Continuation, AMS Translations of Mathematical Monographs 82, 1990, for normalized sectorial solutions and Stokes data.