The General Heun Equation: Standard and Normal Forms

The general Heun equation is the canonical scalar equation with four distinct regular singular points. It is the first Fuchsian model in which local exponents do not determine the global equation: after the singularity position and exponent data are fixed, one accessory parameter remains.

This page fixes the convention used throughout the book, derives its Riemann scheme, and converts it exactly to the book’s normal form. Only then do we identify the normalized local function called HeunG. Geometry and monodromy of the accessory parameter belong to the next page; transformations, software crosswalks, and polynomial sectors are treated later in the chapter.

The six-parameter canonical equation

Let

$$a\in\mathbb C\setminus\{0,1\}.$$

The house convention is

$$\begin{aligned} y''(z) &+ \left( \frac{\gamma}{z} + \frac{\delta}{z-1} + \frac{\epsilon}{z-a} \right)y'(z)\\ &+ \frac{ \alpha\beta z-q }{ z(z-1)(z-a) }y(z) =0, \end{aligned}$$

with the Fuchs relation

$$\gamma+\delta+\epsilon = \alpha+\beta+1.$$

We write $p(z)$ for the coefficient of $y'$ and $r(z)$ for the coefficient of $y$ in this equation.

Thus $\epsilon$ is usually suppressed:

$$\epsilon = \alpha+\beta-\gamma-\delta+1.$$

The six independent parameters have three different jobs:

Data	Parameters	Role
Marked four-point geometry	$a$	Places the fourth puncture after the other three are sent to $0$, $1$, and $\infty$
Local exponent data	$\alpha,\beta,\gamma,\delta$, with $\epsilon$ constrained	Fixes the four unordered exponent pairs
Global coefficient freedom	$q$	Accessory parameter not fixed by those exponent pairs

Up to a Möbius coordinate change and an exponent-shifting scalar gauge, every second-order Fuchsian equation with four distinct singular points on the Riemann sphere can be put in this form.

The exponent ledger

At a finite singular point, substituting a Frobenius ansatz gives

$$\begin{array}{c|c|c} \text{point}&\text{indicial equation}&\text{exponents}\\ \hline 0&\rho(\rho-1+\gamma)=0&0,\ 1-\gamma\\ 1&\rho(\rho-1+\delta)=0&0,\ 1-\delta\\ a&\rho(\rho-1+\epsilon)=0&0,\ 1-\epsilon \end{array}$$

At infinity, set $y(z)\sim z^{-\rho}$. Since

$$p(z) \sim \frac{\gamma+\delta+\epsilon}{z}, \qquad r(z) \sim \frac{\alpha\beta}{z^2},$$

the leading equation is

$$\rho^2 + \left( 1-\gamma-\delta-\epsilon \right)\rho + \alpha\beta =0.$$

It factors as $(\rho-\alpha)(\rho-\beta)$ precisely when the Fuchs relation holds. Without that relation infinity is still regular singular, but the symbols $\alpha,\beta$ are not its actual exponents. With the house convention, the Riemann scheme is

$$P\!\left\{ \begin{matrix} 0&1&a&\infty\\ 0&0&0&\alpha\\ 1-\gamma&1-\delta&1-\epsilon&\beta \end{matrix} ;z \right\}.$$

Here “exponents $\alpha,\beta$ at infinity” means that the corresponding solutions behave as $z^{-\alpha}$ and $z^{-\beta}$ in the $z$-coordinate. Their full ledger obeys

$$(1-\gamma) + (1-\delta) + (1-\epsilon) + \alpha+\beta =2.$$

The equation is unchanged by $\alpha\leftrightarrow\beta$.

From standard form to the book’s normal form

Write the standard equation as

$$y''+p(z)y'+r(z)y=0.$$

On a simply connected patch with chosen branches, set

$$\begin{aligned} y(z) &= z^{-\gamma/2} (1-z)^{-\delta/2} \left( 1-\frac za \right)^{-\epsilon/2} \psi(z). \end{aligned}$$

Constant branch factors do not affect the transformed equation. The Liouville formula gives

$$\psi''(z)+T(z)\psi(z)=0, \qquad T=r-\frac12p'-\frac14p^2.$$

Introduce the full exponent differences

$$\begin{aligned} \theta_0&=1-\gamma, & \theta_1&=1-\delta,\\ \theta_a&=1-\epsilon, & \theta_\infty&=\alpha-\beta, \end{aligned}$$

and the centered double-pole coefficients

$$\Delta_j = \frac{1-\theta_j^2}{4}, \qquad j\in\{0,1,a,\infty\}.$$

Two combinations keep the result compact:

$$\Lambda = \Delta_\infty-\Delta_0-\Delta_1-\Delta_a, \qquad \kappa_{\mathrm H} = q-\frac{\gamma}{2} \left( a\delta+\epsilon \right).$$

Direct substitution yields the normal-form coefficient

$$\begin{aligned} T(z) &= \frac{\Delta_0}{z^2} + \frac{\Delta_1}{(z-1)^2} + \frac{\Delta_a}{(z-a)^2}\\ &\quad+ \frac{ \Lambda z-\kappa_{\mathrm H} }{ z(z-1)(z-a) }. \end{aligned}$$

This form exposes three facts at once:

1. each double-pole coefficient depends only on one local exponent difference;
2. the coefficient at infinity is $\Delta_0+\Delta_1+\Delta_a+\Lambda=\Delta_\infty$;
3. the accessory parameter enters linearly, but its normal-form coordinate is the shifted quantity $\kappa_{\mathrm H}$ rather than the raw $q$.

For geometry near the moving puncture, it is useful to make the residue at $z=a$ itself the accessory coordinate. With

$$D(z)=z(z-1)(z-a),$$

the same coefficient can be written

$$\begin{aligned} T(z) &= \frac{\Delta_0}{z^2} + \frac{\Delta_1}{(z-1)^2} + \frac{\Delta_a}{(z-a)^2}\\ &\quad+ \frac{\Lambda}{z(z-1)} + \frac{a(a-1)c_a}{D(z)}, \end{aligned}$$

where

$$c_a = \frac{a\Lambda-\kappa_{\mathrm H}}{a(a-1)}.$$

Indeed, $c_a$ is the residue of $T$ at $z=a$. The three accessory coordinates are affinely related:

$$\frac{\partial\kappa_{\mathrm H}}{\partial q}=1, \qquad \frac{\partial c_a}{\partial q} = -\frac1{a(a-1)}, \qquad \frac{\partial T}{\partial q} = -\frac1{D(z)}.$$

At a finite singular point $z_j$, the two normal-form powers are

$$\psi(z) \sim (z-z_j)^{(1\pm\theta_j)/2}.$$

The scalar gauge is generally multivalued. Thus standard and normal forms are locally equivalent after branches are chosen, but a global comparison must retain the gauge’s scalar monodromy.

Abel’s identity audits the gauge

For any standard-form basis, Abel’s identity gives

$$\Wr[y_1,y_2] = C\,z^{-\gamma} (1-z)^{-\delta} \left( 1-\frac za \right)^{-\epsilon}.$$

Here a logarithm of $z$ is fixed on the chosen lift, while the last two branch factors may be normalized to one at $z=0$. Since the square of the Liouville gauge is exactly the nonconstant factor in this expression,

$$\Wr[y_1,y_2] = \left[ z^{-\gamma} (1-z)^{-\delta} \left( 1-\frac za \right)^{-\epsilon} \right] \Wr[\psi_1,\psi_2].$$

Consequently $\Wr[\psi_1,\psi_2]=C$ is constant, as it must be for an equation without a first-derivative term. In the nonresonant normalization

$$y_0=1+O(z), \qquad y_1=z^{1-\gamma}\bigl(1+O(z)\bigr),$$

one has $C=1-\gamma$. This checks both the gauge direction and its normalization.

DLMF sign convention

DLMF §31.2(ii) writes the normal equation as $\psi''=U\psi$. This book writes $\psi''+T\psi=0$, so $T=-U$. Copying DLMF’s partial-fraction coefficients without changing this sign produces the wrong local exponents.

Equation class versus the local HeunG germ

For

$$\gamma\notin\mathbb Z_{\leq0},$$

there is a unique exponent-zero solution analytic at $z=0$ and normalized to one. DLMF denotes it by $\mathit{H\!\ell}(a,q;\alpha,\beta,\gamma,\delta;z)$; on heun.xyz the same germ is

$$\begin{aligned} &\operatorname{HeunG} \left( a,q,\alpha,\beta,\gamma,\delta;z \right),\\ &\operatorname{HeunG} \left( a,q,\alpha,\beta,\gamma,\delta;0 \right)=1. \end{aligned}$$

The missing parameter is always $\epsilon=\alpha+\beta-\gamma-\delta+1$. Clearing denominators and reading the constant term gives the useful sign check

$$\operatorname{HeunG} \left( a,q,\alpha,\beta,\gamma,\delta;z \right) = 1+\frac{q}{a\gamma}z+O(z^2).$$

The Taylor series is guaranteed to converge for $|z|<\min\{1,|a|\}$. Generically a nearest singular point sets the exact radius, but an exceptional solution may continue through it. Thus HeunG denotes one normalized local germ—not the entire solution space, a connection problem, or a globally preferred branch.

If an exponent difference is integral, a second local solution may contain a logarithm. For $\gamma\in\mathbb Z_{\leq0}$, the singular recurrence can obstruct existence; when its compatibility condition holds, the normalized germ need not be unique. The later page on canonical local bases treats those resonant cases systematically.

Local does not mean principal branch

Continuation of the Taylor germ along a path defines a branch on the universal cover of $\mathbb P^1\setminus\{0,1,a,\infty\}$. Software cuts are additional implementation conventions; they are not part of the differential equation.

Worked ledger: one equation in both forms

Keep $q$ free and choose

$$\begin{gathered} a=2, \qquad \gamma=\delta=\epsilon=\frac12, \\ \alpha=\frac16, \qquad \beta=\frac13. \end{gathered}$$

The Fuchs relation is satisfied because both sides equal $3/2$. The one-parameter family in standard form is

$$\begin{aligned} y'' &+ \frac12 \left( \frac1z + \frac1{z-1} + \frac1{z-2} \right)y'\\ &+ \frac{ z/18-q }{ z(z-1)(z-2) }y =0. \end{aligned}$$

Its exponent data do not vary with $q$:

$$\theta_0 = \theta_1 = \theta_a = \frac12, \qquad \theta_\infty=-\frac16.$$

Therefore

$$\Delta_0 = \Delta_1 = \Delta_a = \frac3{16}, \qquad \Delta_\infty=\frac{35}{144},$$

and the two normal-form combinations are

$$\Lambda=-\frac{23}{72}, \qquad \kappa_{\mathrm H}=q-\frac38.$$

The residue coordinate at the moving puncture is

$$c_a = -\frac q2-\frac{19}{144}.$$

In the residue decomposition, the normal equation is

$$\begin{aligned} \psi''+\Biggl[ &\frac3{16z^2} + \frac3{16(z-1)^2} + \frac3{16(z-2)^2}\\ &- \frac{23}{72z(z-1)} - \frac{q+19/72}{z(z-1)(z-2)} \Biggr]\psi=0. \end{aligned}$$

The same $q$ appears immediately in the local germ:

$$\operatorname{HeunG} \left( 2,q,\frac16,\frac13,\frac12,\frac12;z \right) = 1+qz+O(z^2).$$

This family separates local from global data cleanly. Varying $q$ changes neither singular positions nor exponent differences, yet changes both $\kappa_{\mathrm H}$ and $c_a$. Even at $q=-19/72$, where $c_a=0$, the double pole $3/[16(z-2)^2]$ remains. A vanishing accessory residue is not the removal of the puncture.

Exact reductions that test the convention

Two short tests catch many transcription errors. First, the equation is invariant under $\alpha\leftrightarrow\beta$: its coefficients depend only on $\alpha+\beta$ and $\alpha\beta$, while this exchange merely reverses the sign convention for $\theta_\infty$.

Second, impose

$$\epsilon=0, \qquad q=a\alpha\beta,$$

then both coefficient functions lose their pole at $z=a$, and the equation reduces exactly to

$$y'' + \left( \frac{\gamma}{z} + \frac{\delta}{z-1} \right)y' + \frac{\alpha\beta}{z(z-1)}y =0.$$

Because the Fuchs relation now gives $\delta=\alpha+\beta-\gamma+1$, this is the Gauss equation and

$$\operatorname{HeunG} \left( a,a\alpha\beta, \alpha,\beta,\gamma, \alpha+\beta-\gamma+1; z \right) = {}_2F_1(\alpha,\beta;\gamma;z)$$

as normalized germs wherever both sides are defined. The hypergeometric connection benchmark then supplies an exact global check.

By contrast, merely substituting $a=0$ or $a=1$ collides singularities and leaves the canonical equation’s stated domain. A meaningful confluence may require a gauge, a rescaling, and coordinated parameter limits.

What has—and has not—been fixed

Object	Fixed by this page?
Differential equation in the house convention	Yes, once $(a,q,\alpha,\beta,\gamma,\delta)$ are given
Exponent-zero germ at $z=0$	Yes generically, after normalization to one
Second local basis element at resonance	Not without a logarithmic or limiting prescription
Analytic-continuation branch away from zero	Not without a path or cut convention
Connection matrix between two singular points	Not by the local symbol HeunG
Distinguished values of $q$	Not until boundary, monodromy, regularity, or apparency conditions are supplied

The next page explains geometrically why four marked singularities leave one accessory parameter. Later pages construct canonical bases, transformation rules, software crosswalks, and polynomial sectors.

Common pitfalls

Treating all seven symbols as independent. The Fuchs relation removes one degree of freedom. A coefficient match that violates it does not have the advertised exponents $\alpha,\beta$ at infinity.

Calling HeunG the general solution. It is one normalized local germ. The general local solution requires a second basis element, and a global solution requires continuation data.

Using the raw accessory parameter in normal form. The Liouville gauge shifts $q$ to $\kappa_{\mathrm H}$. Different normal-form conventions may shift or rescale the accessory coordinate again.

Ignoring resonant normalization failure. Integer exponent differences affect the second solution, while $\gamma\in\mathbb Z_{\leq0}$ can obstruct the normalized exponent-zero germ itself. These are distinct phenomena.

Exercises

1. Recover the Fuchs relation

Starting from the standard equation, substitute $y(z)\sim z^{-\rho}$ as $z\to\infty$. Derive the indicial polynomial and determine the condition under which its roots are $\alpha$ and $\beta$.

Solution

At infinity,

$$y' \sim -\rho z^{-\rho-1}, \qquad y'' \sim \rho(\rho+1)z^{-\rho-2}.$$

The leading coefficient functions are

$$p(z) \sim \frac{\gamma+\delta+\epsilon}{z}, \qquad r(z) \sim \frac{\alpha\beta}{z^2}.$$

After dividing by $z^{-\rho-2}$, the indicial polynomial is

$$\rho^2 + \left( 1-\gamma-\delta-\epsilon \right)\rho + \alpha\beta.$$

It equals

$$(\rho-\alpha)(\rho-\beta) = \rho^2-(\alpha+\beta)\rho+\alpha\beta$$

if and only if

$$\gamma+\delta+\epsilon = \alpha+\beta+1.$$

2. Invert the exponent dictionary

Suppose the ordered full exponent differences $\theta_0,\theta_1,\theta_a,\theta_\infty$ are given in the book’s convention. Recover $\gamma,\delta,\epsilon,\alpha,\beta$. Which sign changes merely relabel local exponents?

Solution

The three finite-point definitions give

$$\begin{aligned} \gamma&=1-\theta_0,\\ \delta&=1-\theta_1,\\ \epsilon&=1-\theta_a. \end{aligned}$$

The Fuchs relation and $\theta_\infty=\alpha-\beta$ then imply

$$\alpha+\beta = 2-\theta_0-\theta_1-\theta_a,$$

and hence

$$\begin{aligned} \alpha &= \frac{ 2-\theta_0-\theta_1-\theta_a+\theta_\infty }{2},\\ \beta &= \frac{ 2-\theta_0-\theta_1-\theta_a-\theta_\infty }{2}. \end{aligned}$$

An unordered exponent pair determines each $\theta_j$ only up to sign. At infinity, $\theta_\infty\mapsto-\theta_\infty$ simply exchanges $\alpha$ and $\beta$ and leaves the standard equation unchanged. At a finite point, a sign flip exchanges the two local powers; restoring the standard choice with one exponent equal to zero requires the corresponding exponent-shifting scalar gauge.

3. Derive the compact normal form

Use $T=r-p'/2-p^2/4$ to show that the double-pole coefficient at zero is $\Delta_0=(1-\theta_0^2)/4$. Then combine the remaining simple-pole terms into $(\Lambda z-\kappa_{\mathrm H})/D(z)$ and derive the residue coordinate $c_a$.

Solution

Near zero, the terms from $-p'/2-p^2/4$ are

$$\frac{\gamma}{2z^2} - \frac{\gamma^2}{4z^2}.$$

Their coefficient is

$$\frac{2\gamma-\gamma^2}{4} = \frac{1-(1-\gamma)^2}{4} = \Delta_0.$$

The same calculation gives $\Delta_1$ and $\Delta_a$. After removing those double poles, the remaining numerator over $D=z(z-1)(z-a)$ is

$$\begin{aligned} N(z) &= \alpha\beta z-q - \frac{\gamma\delta}{2}(z-a)\\ &\quad- \frac{\gamma\epsilon}{2}(z-1) - \frac{\delta\epsilon}{2}z\\ &= \Lambda z-\kappa_{\mathrm H}. \end{aligned}$$

The constant term gives

$$\kappa_{\mathrm H} = q-\frac{\gamma}{2}(a\delta+\epsilon).$$

Using the Fuchs relation, the coefficient of $z$ simplifies to

$$\alpha\beta - \frac{ \gamma\delta+\gamma\epsilon+\delta\epsilon }{2} = \Delta_\infty-\Delta_0-\Delta_1-\Delta_a = \Lambda.$$

Finally,

$$\frac{\Lambda z-\kappa_{\mathrm H}}{D(z)} = \frac{\Lambda}{z(z-1)} + \frac{a\Lambda-\kappa_{\mathrm H}}{D(z)}.$$

The residue at $z=a$ is therefore

$$c_a = \frac{a\Lambda-\kappa_{\mathrm H}}{a(a-1)},$$

and, with all exponent data and $a$ fixed,

$$\frac{\partial c_a}{\partial q} = -\frac1{a(a-1)}.$$

4. Normalize the local Wronskian

For $\gamma\notin\mathbb Z$, let $y_0=1+O(z)$ and $y_1=z^{1-\gamma}(1+O(z))$. Use Abel’s identity to determine their Wronskian on a local branch. Then show directly that the Liouville gauge makes the normal-form Wronskian constant.

Solution

Abel’s identity gives

$$\Wr[y_0,y_1]' = -p(z)\Wr[y_0,y_1].$$

Hence

$$\Wr[y_0,y_1] = C z^{-\gamma} (1-z)^{-\delta} \left( 1-\frac za \right)^{-\epsilon}.$$

Near zero,

$$\Wr[y_0,y_1] \sim (1-\gamma)z^{-\gamma},$$

so $C=1-\gamma$ when the last two branch factors are normalized to one at zero. If $y_j=s\psi_j$, where

$$s = z^{-\gamma/2} (1-z)^{-\delta/2} \left( 1-\frac za \right)^{-\epsilon/2},$$

then

$$\Wr[y_0,y_1] = s^2\Wr[\psi_0,\psi_1].$$

The factor $s^2$ is precisely the $z$-dependent Abel factor, so

$$\Wr[\psi_0,\psi_1]=1-\gamma$$

is constant. Had the scalar gauge been inverted, this cancellation would fail; the Wronskian is therefore a sensitive audit of the gauge direction.

5. Vary the accessory parameter with exponents fixed

For the worked family

$$a=2, \qquad \gamma=\delta=\epsilon=\frac12, \qquad \alpha=\frac16, \qquad \beta=\frac13,$$

compute $\Lambda$, $\kappa_{\mathrm H}$, and $c_a$. Find the value of $q$ for which $c_a=0$, and decide whether $z=2$ then ceases to be singular.

Solution

The full exponent differences and centered coefficients are

$$\begin{gathered} \theta_0=\theta_1=\theta_a=\frac12, \qquad \theta_\infty=-\frac16,\\ \Delta_0=\Delta_1=\Delta_a=\frac3{16}, \qquad \Delta_\infty=\frac{35}{144}. \end{gathered}$$

It follows that

$$\begin{aligned} \Lambda &= \frac{35}{144}-3\frac3{16} =-\frac{23}{72},\\ \kappa_{\mathrm H} &= q-\frac12\frac12 \left( 2\frac12+\frac12 \right) =q-\frac38,\\ c_a &= \frac{2\Lambda-\kappa_{\mathrm H}}{2} =-\frac q2-\frac{19}{144}. \end{aligned}$$

Thus $c_a=0$ at $q=-19/72$. Nevertheless,

$$T(z) = \frac{3}{16(z-2)^2} +O(1) \qquad (z\to2),$$

because $\Delta_a=3/16$. The simple accessory residue vanishes, but the double pole—and hence the regular singular point—remains.

6. Distinguish zero accessory residue from puncture removal

Starting from the standard equation, derive the conditions that make $z=a$ an ordinary point of its coefficient functions. Compare them with $c_a=0$ and explain why the two statements are not equivalent.

Solution

The residue of the $y'$ coefficient at $z=a$ is $\epsilon$, so removal requires

$$\epsilon=0.$$

The numerator of the $y$ coefficient must also vanish at $z=a$:

$$\alpha\beta a-q=0.$$

Hence the necessary and sufficient conditions are

$$\epsilon=0, \qquad q=a\alpha\beta.$$

They cancel the factor $z-a$ and leave the Gauss equation

$$y'' + \left( \frac{\gamma}{z} + \frac{\delta}{z-1} \right)y' + \frac{\alpha\beta}{z(z-1)}y =0.$$

By contrast, $c_a=0$ cancels only the simple-pole term assigned to the accessory coordinate in normal form. Unless $\Delta_a=0$ as well, the double pole at $a$ survives. Even $\Delta_a=0$ requires a further local check: it says $\theta_a=\pm1$, whereas standard-form pole removal selects the compatible gauge $\epsilon=0$, or $\theta_a=1$, together with $q=a\alpha\beta$.

References

• NIST Digital Library of Mathematical Functions, §31.2(i) fixes the standard Heun equation, its parameter roles, singularities, and exponents; §31.2(ii) gives its normal form, and §31.14 places it in the general Fuchsian parameter count.
• NIST Digital Library of Mathematical Functions, §31.3(i) defines the normalized local solution at zero, its recurrence, convergence disk, and resonant cases; §31.4 distinguishes solutions analytic at two singularities from a one-point local Heun germ.
• A. Ronveaux (ed.), Heun’s Differential Equations, Oxford University Press, 1995, Part A, develops the canonical equation, local functions, transformations, and polynomial sectors.
• F. M. Arscott, “Heun’s equation,” in Ronveaux (ed.), Heun’s Differential Equations, gives a concise classical account of the equation’s singularity and parameter structure.
• K. Heun, “Zur Theorie der Riemann’schen Functionen zweiter Ordnung mit vier Verzweigungspunkten”, Mathematische Annalen 33 (1888), 161–179, is the original source.
• G. Bonelli, C. Iossa, D. Panea Lichtig, and A. Tanzini, “Irregular Liouville correlators and connection formulae for Heun functions”, Communications in Mathematical Physics 397 (2023), 635–727, records a modern standard-to-normal-form dictionary used in CFT and gauge theory.