Regular Singularities: Frobenius, Resonance, and Monodromy

At a regular singular point, every solution has at most power-logarithmic growth, but the logarithm is not determined by the indicial roots alone. The Frobenius recurrence contains a finite resonant obstruction. Its vanishing separates a log-free integer exponent difference from a nontrivial unipotent part of local monodromy.

This page derives that obstruction with fixed coefficient indices, then places the scalar calculation inside the system-level Levelt description.

Fuchsian coefficients and a logarithm branch

Let $x=z-z_0$ and consider

$$y''+p(x)y'+q(x)y=0$$

with

$$\begin{aligned} p(x)&=\sum_{j=-1}^{\infty}p_jx^j,\\ q(x)&=\sum_{j=-2}^{\infty}q_jx^j. \end{aligned}$$

These convergent Laurent series express the regular-singular conditions $xp$ and $x^2q$ holomorphic at $x=0$.

Choose a simply connected sector in the punctured disk and a branch

$$x^\rho=\exp\bigl(\rho\operatorname{Log}x\bigr).$$

A Frobenius ansatz is

$$y(x)=x^\rho\sum_{n=0}^{\infty}a_nx^n, \qquad a_0\neq0.$$

The branch is essential. A positive circuit changes $x^\rho$ by $\ee^{2\pi\ii\rho}$ and changes $\operatorname{Log}x$ by $2\pi\ii$.

The indicial polynomial and full recurrence

Define

$$I(s) =s(s-1)+p_{-1}s+q_{-2}.$$

After substituting the Frobenius ansatz, the coefficient of $x^{\rho+n-2}$ is

$$\begin{aligned} 0={}&I(\rho+n)a_n\\ &+\sum_{k=0}^{n-1} \left[ (\rho+k)p_{n-k-1} +q_{n-k-2} \right]a_k. \end{aligned}$$

At $n=0$, this gives the indicial equation

$$I(\rho)=0.$$

For $n\geq1$, the recurrence is

$$\begin{aligned} a_n =-\frac{1}{I(\rho+n)} \sum_{k=0}^{n-1} \left[ (\rho+k)p_{n-k-1}\right.\\ \left. +q_{n-k-2} \right]a_k, \end{aligned}$$

provided the denominator is nonzero.

The recurrence is worth retaining in this indexed form. It makes a resonant denominator and its numerator visible without relying on a named special function.

Convergence

The Frobenius series obtained by this recurrence is not merely formal. Under the regular-singular hypotheses it converges in a punctured disk at least up to the nearest other singularity of the coefficients, with the chosen branch of $x^\rho$. The power-logarithmic form is therefore an analytic local classification, unlike the generally divergent formal series at an irregular singularity.

Nonresonant roots

Let the indicial roots be $\rho_1$ and $\rho_2$. If

$$\rho_1-\rho_2\notin\mathbb Z,$$

then neither recurrence encounters the other root. Setting $a_0=1$ in each series gives two independent local solutions

$$y_j(x) =x^{\rho_j} \left( 1+\sum_{n=1}^{\infty}a_{n,j}x^n \right).$$

In this ordered Frobenius basis,

$$M_0= \begin{pmatrix} \ee^{2\pi\ii\rho_1} & 0\\ 0 & \ee^{2\pi\ii\rho_2} \end{pmatrix}$$

for a counterclockwise loop. The basis normalization matters, but the eigenvalues and their ratio do not.

Integer-separated roots and the obstruction

Now order the roots so that

$$\rho_1-\rho_2=N\in\mathbb Z_{>0}.$$

The larger-root recurrence is nonsingular for all $n\geq1$, so a solution $y_1=x^{\rho_1}(1+O(x))$ always exists.

For the smaller root $\rho_2$, the denominator first vanishes at $n=N$:

$$I(\rho_2+N)=I(\rho_1)=0.$$

Normalize $a_0=1$ and compute $a_1,\ldots,a_{N-1}$. The resonant obstruction is

$$\begin{aligned} \mathcal R_N =\sum_{k=0}^{N-1} \left[ (\rho_2+k)p_{N-k-1}\right.\\ \left. +q_{N-k-2} \right]a_k. \end{aligned}$$

There are two cases.

Obstruction	Second solution	Projective monodromy
$\mathcal R_N=0$	A pure smaller-root Frobenius series exists	Identity
$\mathcal R_N\neq0$	The second solution contains $y_1\operatorname{Log}x$	Nontrivially unipotent after removing the scalar eigenvalue

When $\mathcal R_N=0$, the coefficient $a_N$ is free because adding a multiple of the larger-root series preserves the smaller leading term. One normally sets $a_N=0$ to choose a representative modulo $y_1$.

When $\mathcal R_N\neq0$, a convenient second solution has the form

$$y_2(x) =x^{\rho_2}h_2(x) +C\,y_1(x)\operatorname{Log}x, \qquad C\neq0.$$

If $y_1=x^{\rho_1}(1+O(x))$ and $h_2(0)=1$, substitution at the resonant power gives the exact coefficient

$$C =-\frac{\mathcal R_N}{I'(\rho_1)} =-\frac{\mathcal R_N}{N}.$$

The coefficient $a_N$ in $h_2$ remains free: changing it adds a multiple of $y_1$. One may again set $a_N=0$.

The numerical value of $C$ changes when either solution is rescaled, but the statement $C=0$ or $C\neq0$ is invariant.

Resonance is not a logarithm

An integer exponent difference makes one recurrence singular. A logarithm is present only when the corresponding compatibility condition fails.

Repeated roots

If $\rho_1=\rho_2=\rho$, the first Frobenius solution exists, but a second pure series with the same leading exponent cannot be independent. The second solution is logarithmic:

$$y_2 =y_1\operatorname{Log}x +x^\rho h_2(x),$$

after a suitable normalization.

The inevitability of the logarithm can also be seen from Abel’s identity. Two log-free solutions with the same leading power would have a Wronskian vanishing to higher order than the nonzero Wronskian prescribed by the equation.

Reduction of order as a residue test

Given one nonzero solution $y_1$, a second solution is

$$y_2 =y_1 \int^x \frac{ \exp\left(-\int^\xi p(s)\,\dd s\right) }{ y_1(\xi)^2 } \,\dd\xi.$$

Suppose $y_1=x^{\rho_1}h_1(x)$ is the larger-root solution and $\rho_1-\rho_2=N\geq0$. Since

$$\rho_1+\rho_2=1-p_{-1},$$

the integrand begins at order

$$x^{-N-1}.$$

A logarithm appears exactly when the Laurent expansion of the integrand has a nonzero $x^{-1}$ coefficient. For $N=0$, the leading coefficient already has this order and is nonzero, so the repeated-root logarithm is unavoidable. For $N>0$, the residue can vanish; this is the reduction-of-order version of $\mathcal R_N=0$.

Local monodromy with a logarithm

Use the ordered basis

$$\Phi= \begin{pmatrix} y_1 & y_2\\ y_1' & y_2' \end{pmatrix},$$

where $y_1$ is the larger-root solution and

$$y_2=x^{\rho_2}h_2+C\,y_1\operatorname{Log}x.$$

Because $\rho_1-\rho_2\in\mathbb Z$, both powers have the same monodromy eigenvalue

$$\lambda=\ee^{2\pi\ii\rho_1} =\ee^{2\pi\ii\rho_2}.$$

Counterclockwise continuation gives

$$y_1^\gamma=\lambda y_1, \qquad y_2^\gamma =\lambda y_2+2\pi\ii C\lambda y_1.$$

With the book’s right-action convention,

$$M_0 =\lambda \begin{pmatrix} 1 & 2\pi\ii C\\ 0 & 1 \end{pmatrix}.$$

Its determinant supplies an Abel-identity check:

$$\begin{aligned} \det M_0 &=\exp\left[ 2\pi\ii(\rho_1+\rho_2) \right]\\ &=\exp(-2\pi\ii p_{-1}). \end{aligned}$$

Reversing the loop inverts this matrix. Reversing the basis order moves the off-diagonal entry to the opposite triangular position.

The eigenvalues alone cannot detect $C$. This is why an exponent difference or a trace of monodromy is insufficient at resonance.

Meromorphic gauges and exponent shifts

A scalar meromorphic gauge

$$y=x^m\widetilde y, \qquad m\in\mathbb Z,$$

gives

$$\widetilde I(s)=I(s+m), \qquad \widetilde\rho_j=\rho_j-m.$$

Equivalently, multiplication of a solution by $x^m$ adds $m$ to its exponent. Since $m\in\mathbb Z$, the monodromy eigenvalues are unchanged. In a system, integer shears can shift individual exponents by integers. Consequently:

• exponents depend on a choice of logarithmic lattice or meromorphic gauge;
• their classes modulo integers determine monodromy eigenvalues;
• the nilpotent or Jordan part still requires the resonant extension data.

For $y=\exp(-\tfrac12\int p)\psi$,

$$\sigma_j=\rho_j+\frac{p_{-1}}2, \qquad M_\psi=\ee^{\pi\ii p_{-1}}M_y$$

in corresponding bases. Hence $\det M_\psi=1$ and the projective class is unchanged. If $\theta=\rho_1-\rho_2$, then

$$\sigma_{1,2}=\frac{1\pm\theta}{2},$$

giving eigenvalues $-\ee^{\pm\pi\ii\theta}$, exactly as fixed on the conventions page.

A scalar regular singularity as a Fuchsian system

The naive state $(y,y')^{\mathsf T}$ can contain a double pole through $q$. That does not make the underlying system irregular. Use the sheared state

$$Z= \begin{pmatrix} y\\ xy' \end{pmatrix}.$$

It satisfies

$$Z' =\frac1xB(x)Z, \qquad B(x)= \begin{pmatrix} 0 & 1\\ -x^2q(x) & 1-xp(x) \end{pmatrix},$$

and $B$ is holomorphic at the origin. Its residue is

$$B(0)= \begin{pmatrix} 0 & 1\\ -q_{-2} & 1-p_{-1} \end{pmatrix},$$

with

$$\det\!\left(\rho I-B(0)\right)=I(\rho).$$

Thus the scalar indicial roots are precisely the residue eigenvalues in this Fuchsian frame.

Fuchsian systems and Levelt form

Consider a Fuchsian system

$$Y' =\left( \frac{R}{x}+A_0+A_1x+\cdots \right)Y.$$

If no two eigenvalues of $R$ differ by a positive integer, a holomorphic gauge reduces the system locally to its residue equation. In an adapted frame,

$$\Phi(x)=H(x)x^R, \qquad H(0)\in GL(r,\mathbb C),$$

and the local monodromy is conjugate to $\ee^{2\pi\ii R}$.

At resonance, analytic terms can interact with the residue eigenspaces. Levelt form organizes the result as

$$\Phi(x)=H(x)x^D x^L,$$

where $H$ is holomorphic and invertible, $D=\operatorname{diag}(d_1,\ldots,d_r)$ with $d_i\in\mathbb Z$, and $L$ is a normalized logarithm whose spectrum lies in a fixed width-one strip. A Levelt ordering requires

$$x^DLx^{-D}$$

to be holomorphic at $x=0$; equivalently, $L_{ij}=0$ whenever $d_i<d_j$. The nilpotent part of $L$ is the resonant extension data. Indeed,

$$\begin{aligned} \Phi'\Phi^{-1} ={}&H'H^{-1}\\ &+\frac1xH \left( D+x^DLx^{-D} \right)H^{-1}. \end{aligned}$$

The order $x^Dx^L$ matters because $D$ and $L$ need not commute. Since $x^D$ is single-valued, a Levelt basis has

$$M_0=\ee^{2\pi\ii L}.$$

The holomorphic factor $H$ then matches this form to the given system. At resonance, the full monodromy Jordan form need not be obtained by naively exponentiating the residue of an arbitrary Fuchsian presentation.

The scalar Frobenius obstruction is the rank-two cyclic-vector version of this resonant extension data.

Worked examples

Euler equation: resonance without a logarithm

For

$$x^2y''+\alpha x y'+\beta y=0,$$

the equation is exactly homogeneous under scaling. Its indicial polynomial is

$$I(\rho) =\rho(\rho-1)+\alpha\rho+\beta.$$

If the roots are distinct, $x^{\rho_1}$ and $x^{\rho_2}$ are exact solutions, even when their difference is a nonzero integer. The resonant obstruction vanishes. If the root is repeated, the solutions are

$$x^\rho, \qquad x^\rho\operatorname{Log}x.$$

The same family therefore exhibits both log-free resonance and the unavoidable repeated-root logarithm.

A one-parameter resonant obstruction

Consider

$$y''+\frac{a}{x}y=0.$$

Here $p_{-1}=q_{-2}=0$, so the exponents are $0$ and $1$. For the smaller root $\rho_2=0$, the recurrence at $n=1$ has

$$I(1)=0, \qquad \mathcal R_1=q_{-1}a_0=a.$$

Thus:

• if $a=0$, the equation is $y''=0$ with log-free solutions $1$ and $x$;

• if $a\neq0$, the second local solution contains a logarithm. With $y_1=x(1+O(x))$ and $h_2(0)=1$,

  $$C=-a, \qquad M_0= \begin{pmatrix} 1 & -2\pi\ii a\\ 0 & 1 \end{pmatrix}.$$

Thus $a=0$ is the vanishing locus of the resonant obstruction, not a singular locus of the coefficient family in parameter space.

Bessel: two kinds of resonance

The Bessel equation has exponents $\pm\nu$ at the origin. Resonance occurs when $2\nu\in\mathbb Z$.

• For $\nu\in\mathbb Z$, the standard second solution $Y_\nu$ is logarithmic.
• For $\nu\in\mathbb Z+\tfrac12$, the solutions $J_\nu$ and $J_{-\nu}$ are independent and log-free.

The exponent difference issues the warning; the recurrence resolves it.

Apparent singularity from a cyclic vector

If a regular rank-two system is reduced through a coefficient $b(x)$ with a zero of order $m$, the scalar equation has exponents

$$0, \qquad m+1.$$

Both component solutions remain holomorphic and the scalar monodromy is the identity. This is a resonant but log-free apparent singularity produced by the scalar presentation.

Parameter limits at resonance

Away from resonance, a normalized Frobenius basis is often meromorphic in the exponents or other parameters because its recurrence contains $I(\rho+n)^{-1}$. As a parameter approaches an integer exponent difference, one basis vector can develop a pole or become proportional to the other.

The pole-to-logarithm mechanism is already visible in the resonant coefficient. Put

$$\delta=\rho_1-\rho_2, \qquad t=N-\delta.$$

Because $I(s)=(s-\rho_1)(s-\rho_2)$,

$$I(\rho_2+N)=Nt.$$

If the obstruction tends to a nonzero value, then

$$a_N =-\frac{\mathcal R_N}{Nt} \sim\frac Ct.$$

The singular multiple becomes finite only after it is recombined with the larger-root solution:

$$\frac Ct \left( x^{\rho_1+t}-x^{\rho_1} \right) \longrightarrow C\,x^{\rho_1}\operatorname{Log}x.$$

This does not imply that the normalized initial-value fundamental matrix or the local system is singular. A stable procedure is:

1. retain an initial-value basis at an ordinary match point;
2. express the Frobenius basis through a parameter-dependent change of basis;
3. subtract the singular multiple of the larger-root solution;
4. take the finite limit of the recombined parameter-dependent solution; a parameter derivative is valid only when it denotes the limit of a difference quotient such as the recombination displayed above;
5. only then take the limit of connection or monodromy matrices.

Gamma-function poles in a connection formula often cancel against this singular basis change. Substituting the resonant parameter into individual nonresonant coefficients before recombining the basis can create a false divergence.

A practical resonance decision

Given a scalar regular singularity:

1. compute $I(\rho)$ and order its roots;
2. if their difference is not an integer, build two Frobenius series;
3. if the difference is $N>0$, build the larger-root series and the smaller-root coefficients through $a_{N-1}$;
4. evaluate $\mathcal R_N$;
5. if it vanishes, choose a log-free smaller-root representative;
6. if it does not, construct the power-logarithmic solution and record its normalization;
7. compute monodromy only after the basis order and logarithm branch are fixed.

Common pitfalls

Stopping at the indicial equation. The roots determine candidate powers and monodromy eigenvalues. At resonance, the recurrence obstruction determines the logarithm and Jordan part.

Calling every integer difference logarithmic. Euler equations and half-integer Bessel functions give immediate counterexamples. Compute $\mathcal R_N$ or the reduction-of-order residue.

Exponentiating an arbitrary resonant residue. In a nonresonant Fuchsian gauge, $\exp(2\pi\ii R)$ is reliable. At resonance, use a Levelt-adapted exponent matrix and include nilpotent extension data.

Taking a resonant parameter limit coefficient by coefficient. A canonical basis can be singular while its span and the initial-value local system remain regular. Recombine or renormalize the basis before taking the limit.

Exercises

1. Derive the recurrence. Substitute the Frobenius ansatz into the equation and recover the displayed coefficient of $x^{\rho+n-2}$.

Solution

The derivative terms are

$$\begin{aligned} y'&=\sum_{k=0}^{\infty} (\rho+k)a_kx^{\rho+k-1},\\ y''&=\sum_{k=0}^{\infty} (\rho+k)(\rho+k-1)a_kx^{\rho+k-2}. \end{aligned}$$

In $py'$, the power $x^{\rho+n-2}$ occurs when $j+k=n-1$; in $qy$, it occurs when $j+k=n-2$. The terms with $k=n$ use $p_{-1}$ and $q_{-2}$ and combine with $y''$ to give $I(\rho+n)a_n$. The remaining terms have $0\leq k<n$ and give the stated sum.

2. Reduction-of-order residue. Prove that the integrand for a second solution begins as $x^{-N-1}$ when $N=\rho_1-\rho_2\geq0$. Explain the repeated-root case.

Solution

Abel’s factor behaves as

$$\exp\left(-\int p\,\dd x\right) \sim x^{-p_{-1}},$$

while $y_1^2\sim x^{2\rho_1}$. Vieta’s relation gives

$$-p_{-1}-2\rho_1 =\rho_2-\rho_1-1 =-N-1.$$

For $N=0$, the leading integrand is a nonzero multiple of $x^{-1}$, whose primitive is $\operatorname{Log}x$. For $N>0$, a logarithm depends on whether later terms produce a nonzero $x^{-1}$ coefficient.

3. Tune a logarithm on and off. Apply the Frobenius recurrence to $y''+(a/x)y=0$ through the resonant step. Compute the monodromy type for $a=0$ and $a\neq0$.

Solution

The indicial roots are $1$ and $0$. For the smaller root,

$$\mathcal R_1=a.$$

At $a=0$, the solutions $x$ and $1$ are single-valued and the monodromy is $I$. With $y_1=x(1+O(x))$ and the smaller-root analytic part normalized to one, $C=-a$. Thus in the ordered basis $(y_1,y_2)$,

$$M_0= \begin{pmatrix} 1 & -2\pi\ii a\\ 0 & 1 \end{pmatrix},$$

which is nontrivially unipotent when $a\neq0$.

4. A nilpotent Fuchsian residue. Solve

$$Y'=\frac{N}{x}Y, \qquad N= \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix},$$

and compute its monodromy.

Solution

Because $N^2=0$,

$$x^N =\exp(N\operatorname{Log}x) =I+N\operatorname{Log}x.$$

This is a fundamental matrix. Positive continuation adds $2\pi\ii$ to the logarithm, so

$$M_0 =\ee^{2\pi\ii N} =I+2\pi\ii N = \begin{pmatrix} 1 & 2\pi\ii\\ 0 & 1 \end{pmatrix}.$$

References

• NIST DLMF, Fuchs–Frobenius theory, for regular-singular classification, nonresonant series, logarithmic cases, and convergence.
• G. Teschl, Ordinary Differential Equations and Dynamical Systems, for a rigorous Frobenius construction and Fuchsian systems.
• P. Deligne, Équations différentielles à points singuliers réguliers, for regular connections, logarithmic lattices, and monodromy.
• K. Iwasaki, H. Kimura, S. Shimomura, and M. Yoshida, From Gauss to Painlevé, for Levelt form and the monodromy description of Fuchsian equations.
• NIST DLMF, integer-order Bessel series and half-odd-integer formulas, for the two resonance benchmarks.