Parameter Crosswalks among Common Heun Conventions

A parameter crosswalk is not a dictionary that replaces one tuple by another. It is a certificate that two differential operators agree after a declared coordinate change and gauge, together with a rule that transports the solution normalization. This distinction is essential in the Heun family: two packages can use the same function name for different equations, and two identical equations can still define different functions by choosing different base points or continuation paths.

The house reference on this page is the DLMF equation convention used in the confluence hierarchy. The symbols HeunG, HeunC, HeunD, HeunB, and HeunT are reserved for declared software interfaces. They are not synonyms for the mathematical equation classes GHE, CHE, DCHE, BHE, and THE.

The four-layer firewall

Never write only “the Heun equation” or “the solution is HeunD(...).” Record the equation class, the exact canonical ODE, the normalized germ or basis, and the software interface separately.

Layer	Example	What fixes it
Equation class	CHE	Singularity pattern and generic formal ranks
Canonical representative	DLMF 31.12.1	Exact coefficient functions and parameter order
Normalized solution	$C_0(0)=1$	Base point, exponent, leading term, and path
Software interface	Wolfram HeunC[...]	Package equation, argument order, and branch convention

A crosswalk is an operator identity

Let the source and target equations be

$$\mathcal L_{\mathrm S}u=0, \qquad \mathcal L_{\mathrm T}y=0.$$

A scalar crosswalk declares

$$\begin{aligned} z &= \phi(x),\\ y\bigl(\phi(x)\bigr) &= \chi(x)u(x),\\ \Theta_{\mathrm T} &= F(\Theta_{\mathrm S}), \end{aligned}$$

where $\Theta_{\mathrm S}$ and $\Theta_{\mathrm T}$ are the ordered parameter tuples. The decisive check is an identity

$$\left( \mathcal L_{\mathrm T}y \right)\bigl(\phi(x)\bigr) = m(x) \mathcal L_{\mathrm S}u(x),$$

after the substitution, with $m(x)\neq0$ on the working domain. Comparing solutions numerically is useful only after this symbolic identity has been established.

Three verbs keep the degree of equivalence visible:

• Rename means the variable and all coefficient functions are identical; only parameter labels or their order change.
• Rescale or specialize means a redundant coefficient is fixed by an affine coordinate change or by restricting a larger parameter family.
• Transform means a nontrivial coordinate or gauge is required. Base points, leading coefficients, branches, and possibly sectors must then be transported.

Crosswalk certificate

A reproducible crosswalk records the two ODEs, both ordered tuples, $\phi$ and $\chi$, their inverses and exclusions, the nonzero operator multiplier, the image of every base point, the transformed normalization, and all branch or sector choices.

The five DLMF reference cards

The canonical-basis page defines the book’s normalized objects. DLMF itself names the general-Heun germ $H\!\ell$ but, in §31.12, primarily fixes the four confluent equations rather than one universal software-style function for each.

Class	DLMF tuple used here	Book object
GHE	$(a,q,\alpha,\beta,\gamma,\delta)$	$H\!\ell$ at $0$
CHE	$(q,\alpha,\gamma,\delta,\epsilon)$	$C_0$ at $0$
DCHE	$(q,\alpha,\gamma,\delta)$	Ordinary-point or sectorial basis
BHE	$(q,\alpha,\gamma,\delta)$	$B_0$ at $0$
THE	$(q,\alpha,\gamma)$	$(T_0,T_1)$ at $0$

Here and below a subscript $\mathrm D$, $\mathrm W$, or $\mathrm M$ marks a DLMF, Wolfram, or Maple parameter whenever the same letter could be ambiguous.

Class	Wolfram relation to DLMF	Maple relation to DLMF
GHE	Direct	Direct
CHE	Direct	Parameter map
DCHE	Scale and moved base point	Coordinate, gauge, and moved base point
BHE	Direct specialization	Square-root rescaling
THE	Direct specialization	Affine and cubic-root normalization

The word “direct” applies to the displayed equation and tuple. Equality of principal values after analytic continuation still requires compatible paths and package branch conventions.

General Heun is direct—but only generically

The DLMF general Heun equation is

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

where

$$\epsilon=\alpha+\beta+1-\gamma-\delta, \qquad a\neq0,1.$$

Its exponent-zero germ at $z=0$ begins

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

For $\gamma\notin\mathbb Z_{\leq0}$, the normalized recurrence is unobstructed and the following three local objects use the same equation, parameter order, and unit value at the origin:

$$\begin{gathered} H\!\ell \left( a,q;\alpha,\beta,\gamma,\delta;z \right),\\ \text{Wolfram } \operatorname{HeunG} \left[ a,q,\alpha,\beta,\gamma,\delta,z \right],\\ \text{Maple } \operatorname{HeunG} \left( a,q,\alpha,\beta,\gamma,\delta,z \right). \end{gathered}$$

This is a germ identity near $0$. It becomes an identity of continued values only when the logarithm branches and continuation paths agree. At $\gamma\in\mathbb Z_{\leq0}$, the literal unit-normalized recurrence can be obstructed or nonunique; a limiting or compatibility prescription must be declared instead of substituting blindly.

Confluent Heun exposes Maple’s reparameterization

The DLMF CHE is

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

Wolfram’s call

$$\operatorname{HeunC} \left[ q,\alpha,\gamma,\delta,\epsilon,z \right]$$

uses this equation literally and selects the analytic germ

$$C_0(z) = 1-\frac q\gamma z+O(z^2).$$

As for the GHE, the displayed germ requires $\gamma\notin\mathbb Z_{\leq0}$ unless an exceptional prescription is supplied.

Maple’s five parameters

Maple writes

$$\operatorname{HeunC} \left( \alpha_{\mathrm M}, \beta_{\mathrm M}, \gamma_{\mathrm M}, \delta_{\mathrm M}, \eta_{\mathrm M}, z \right)$$

for the exponent-zero solution of

$$\begin{aligned} U'' &+ \left[ \alpha_{\mathrm M} + \frac{\beta_{\mathrm M}+1}{z} + \frac{\gamma_{\mathrm M}+1}{z-1} \right]U'\\ &+ \left( \frac{\mu}{z} + \frac{\nu}{z-1} \right)U =0, \end{aligned}$$

where

$$\begin{aligned} \mu = \frac12 \bigl( &\alpha_{\mathrm M} -\beta_{\mathrm M} -\gamma_{\mathrm M} +\alpha_{\mathrm M}\beta_{\mathrm M}\\ &-\beta_{\mathrm M}\gamma_{\mathrm M} \bigr) -\eta_{\mathrm M}, \end{aligned}$$

and

$$\begin{aligned} \nu = \frac12 \bigl( &\alpha_{\mathrm M} +\beta_{\mathrm M} +\gamma_{\mathrm M} +\alpha_{\mathrm M}\gamma_{\mathrm M}\\ &+\beta_{\mathrm M}\gamma_{\mathrm M} \bigr) +\delta_{\mathrm M} +\eta_{\mathrm M}. \end{aligned}$$

Combining the last two simple fractions gives

$$\frac{\mu}{z} + \frac{\nu}{z-1} = \frac{ (\mu+\nu)z-\mu }{ z(z-1) }.$$

Coefficient comparison therefore yields the forward map

$$\begin{aligned} \epsilon&=\alpha_{\mathrm M},\\ \gamma&=\beta_{\mathrm M}+1,\\ \delta&=\gamma_{\mathrm M}+1,\\ q&=\mu,\\ \alpha&=\mu+\nu. \end{aligned}$$

The inverse map is

$$\begin{aligned} \alpha_{\mathrm M}&=\epsilon,\\ \beta_{\mathrm M}&=\gamma-1,\\ \gamma_{\mathrm M}&=\delta-1,\\ \delta_{\mathrm M} &= \alpha-\frac{\epsilon(\gamma+\delta)}2,\\ \eta_{\mathrm M} &= \frac12 + \frac{\gamma(\epsilon-\delta)}2 -q. \end{aligned}$$

The normalization crosses as well:

$$U'(0) = -\frac{\mu}{\beta_{\mathrm M}+1} = -\frac q\gamma = C_0'(0).$$

Thus this is a genuine parameter reparameterization of the same normalized local germ, not merely an equivalence of equation classes. Principal values away from the disk of the local series still require compatible continuation conventions.

Doubly confluent scaling moves the base point

Wolfram’s five-parameter DCHE is

$$\begin{aligned} z^2y'' &+ \left( \gamma_{\mathrm W} +\delta_{\mathrm W}z +\epsilon_{\mathrm W}z^2 \right)y'\\ &+ \left( \alpha_{\mathrm W}z-q_{\mathrm W} \right)y =0. \end{aligned}$$

Its named function

$$\operatorname{HeunD} \left[ q_{\mathrm W}, \alpha_{\mathrm W}, \gamma_{\mathrm W}, \delta_{\mathrm W}, \epsilon_{\mathrm W}, z \right]$$

is normalized at the ordinary point $z=1$ by

$$y(1)=1, \qquad y'(1)=0.$$

The DLMF DCHE is instead scale-fixed:

$$\begin{aligned} Y'' &+ \left( \frac{\delta_{\mathrm D}}{t^2} + \frac{\gamma_{\mathrm D}}{t} +1 \right)Y'\\ &+ \frac{\alpha_{\mathrm D}t-q_{\mathrm D}}{t^2}Y =0. \end{aligned}$$

For $\epsilon_{\mathrm W}\neq0$, set

$$t=\epsilon_{\mathrm W}z, \qquad y(z)=Y(t).$$

Direct substitution gives

$$\begin{aligned} q_{\mathrm D} &= q_{\mathrm W},\\ \alpha_{\mathrm D} &= \frac{\alpha_{\mathrm W}}{\epsilon_{\mathrm W}},\\ \gamma_{\mathrm D} &= \delta_{\mathrm W},\\ \delta_{\mathrm D} &= \epsilon_{\mathrm W}\gamma_{\mathrm W}. \end{aligned}$$

The transformed initial data are

$$Y(\epsilon_{\mathrm W})=1, \qquad Y'(\epsilon_{\mathrm W})=0.$$

The base point moved from $z=1$ to $t=\epsilon_{\mathrm W}$. This is why the equation crosswalk does not imply equality with a hypothetical DLMF function normalized at $t=1$.

At the direct specialization $\epsilon_{\mathrm W}=1$, the DLMF equation is Wolfram’s equation with

$$\left( q_{\mathrm W}, \alpha_{\mathrm W}, \gamma_{\mathrm W}, \delta_{\mathrm W}, \epsilon_{\mathrm W} \right) = \left( q_{\mathrm D}, \alpha_{\mathrm D}, \delta_{\mathrm D}, \gamma_{\mathrm D}, 1 \right).$$

Notice the $\gamma$–$\delta$ swap in the function call. The resulting ordinary-point solution is useful, but it is not one of the sectorially normalized endpoint solutions defined by the irregular formal type.

Maple’s symmetric double-confluent equation

Maple’s four-parameter HeunD uses an ordinary point at $x=0$ and irregular points at $x=\pm1$. Writing its parameters as $(A_{\mathrm M},B_{\mathrm M},C_{\mathrm M},D_{\mathrm M})$, define

$$\begin{aligned} \mathcal D(x) &= (x-1)^3(x+1)^3,\\ P_{\mathrm M}(x) &= A_{\mathrm M}x^4-2x^5+4x^3\\ &\quad -A_{\mathrm M}-2x,\\ Q_{\mathrm M}(x) &= -B_{\mathrm M}x^2\\ &\quad +(-C_{\mathrm M}-2A_{\mathrm M})x -D_{\mathrm M}. \end{aligned}$$

Its equation is

$$U'' - \frac{P_{\mathrm M}(x)}{\mathcal D(x)}U' - \frac{Q_{\mathrm M}(x)}{\mathcal D(x)}U =0,$$

with

$$U(0)=1, \qquad U'(0)=0.$$

This cannot be translated by moving four entries between slots. There is, however, an exact coordinate-and-gauge crosswalk.

Choose a square root

$$k^2=\delta_{\mathrm D}, \qquad k\neq0,$$

and set

$$\begin{aligned} z &= k\frac{x+1}{x-1}, \\ y\bigl(z(x)\bigr) &= \left( \frac{1-x}{1+x} \right)^{(\gamma_{\mathrm D}-1)/2} U(x). \end{aligned}$$

Near $x=0$, define the power using the logarithm branch

$$\left. \Log\left( \frac{1-x}{1+x} \right) \right|_{x=0} =0.$$

This fixes the gauge to have value $1$ at the Maple base point. Another logarithm lift multiplies both transported initial data below by the same nonzero constant.

The DLMF equation transforms into Maple’s equation with

$$\begin{aligned} A_{\mathrm M} &= 4k,\\ B_{\mathrm M} &= -1+2\gamma_{\mathrm D}-\gamma_{\mathrm D}^2-4q_{\mathrm D}\\ &\quad +4k \left( 1+\alpha_{\mathrm D}-\gamma_{\mathrm D} \right),\\ C_{\mathrm M} &= 8k \left( \alpha_{\mathrm D}-1 \right),\\ D_{\mathrm M} &= 1-2\gamma_{\mathrm D}+\gamma_{\mathrm D}^2+4q_{\mathrm D}\\ &\quad +4k \left( 1+\alpha_{\mathrm D}-\gamma_{\mathrm D} \right). \end{aligned}$$

The two choices of $k$ are different coordinate-and-gauge charts. The Maple base point maps as

$$x=0 \longmapsto z=-k.$$

After transporting the Maple initial data, the DLMF solution satisfies

$$y(-k)=1, \qquad y'(-k) = \frac{\gamma_{\mathrm D}-1}{2k}.$$

This last derivative is the simplest proof that Maple HeunD is not Wolfram HeunD with renamed slots: even after the singularities are moved to $0$ and $\infty$, their defining ordinary-point data are generically different. When $\delta_{\mathrm D}=0$, this chart degenerates and the endpoint must be reclassified before choosing another reduction.

Biconfluent forms differ by one scale

The DLMF BHE is

$$y'' - \left( \frac{\gamma_{\mathrm D}}z +\delta_{\mathrm D} +z \right)y' + \frac{\alpha_{\mathrm D}z-q_{\mathrm D}}z\,y =0.$$

Wolfram’s five-parameter equation is

$$\begin{aligned} z y'' &+ \left( \gamma_{\mathrm W} +\delta_{\mathrm W}z +\epsilon_{\mathrm W}z^2 \right)y'\\ &+ \left( \alpha_{\mathrm W}z-q_{\mathrm W} \right)y =0. \end{aligned}$$

Therefore the direct specialization is

$$\begin{aligned} q_{\mathrm W}&=q_{\mathrm D},\\ \alpha_{\mathrm W}&=\alpha_{\mathrm D},\\ \gamma_{\mathrm W}&=-\gamma_{\mathrm D},\\ \delta_{\mathrm W}&=-\delta_{\mathrm D},\\ \epsilon_{\mathrm W}&=-1. \end{aligned}$$

For $\gamma_{\mathrm D}\notin\mathbb Z_{\geq0}$, where the analytic recurrence is unobstructed,

$$B_0(z) = \operatorname{HeunB} \left[ q_{\mathrm D}, \alpha_{\mathrm D}, -\gamma_{\mathrm D}, -\delta_{\mathrm D}, -1, z \right].$$

Indeed Wolfram’s first derivative is $q_{\mathrm W}/\gamma_{\mathrm W}$, which becomes $-q_{\mathrm D}/\gamma_{\mathrm D}$ as required by

$$B_0(z) = 1-\frac{q_{\mathrm D}}{\gamma_{\mathrm D}}z+O(z^2).$$

The three minus signs are part of the convention.

An arbitrary nondegenerate Wolfram scale also reduces to DLMF. For $\epsilon_{\mathrm W}\neq0$, choose a square root

$$\lambda^2=-\epsilon_{\mathrm W}$$

and set $t=\lambda z$. The target parameters are

$$\begin{aligned} \gamma_{\mathrm D} &= -\gamma_{\mathrm W},\\ \delta_{\mathrm D} &= -\frac{\delta_{\mathrm W}}{\lambda},\\ \alpha_{\mathrm D} &= \frac{\alpha_{\mathrm W}}{\lambda^2},\\ q_{\mathrm D} &= \frac{q_{\mathrm W}}{\lambda}. \end{aligned}$$

The origin is fixed. Where the analytic recurrence is unobstructed, the unit-leading germ—including its forced first derivative—is preserved under $w(z)=B_0(\lambda z)$. The choice of $\lambda$ is part of the crosswalk. When $\epsilon_{\mathrm W}=0$, the generic rank-$2$ normalization has degenerated and this scaling does not apply.

Maple and Ronveaux’s four-parameter form

Maple uses the Ronveaux-style equation. To keep its accessory combination visible, set

$$Q_{\mathrm M} = \frac{ D_{\mathrm M} +(1+A_{\mathrm M})B_{\mathrm M} }{2}.$$

Then

$$\begin{aligned} xU'' &+ \left( 1+A_{\mathrm M} -B_{\mathrm M}x -2x^2 \right)U'\\ &+ \left[ \left( C_{\mathrm M}-A_{\mathrm M}-2 \right)x -Q_{\mathrm M} \right]U =0, \end{aligned}$$

Where its analytic recurrence is unobstructed, Maple selects the unit-leading germ at $x=0$. Choose $c^2=2$ and put

$$z=cx.$$

The Maple-to-DLMF map is

$$\begin{aligned} \gamma_{\mathrm D} &= -(1+A_{\mathrm M}),\\ \delta_{\mathrm D} &= \frac{B_{\mathrm M}}c,\\ \alpha_{\mathrm D} &= \frac{ C_{\mathrm M}-A_{\mathrm M}-2 }{2},\\ q_{\mathrm D} &= \frac{ D_{\mathrm M} +(1+A_{\mathrm M})B_{\mathrm M} }{2c}. \end{aligned}$$

The inverse is

$$\begin{aligned} A_{\mathrm M} &= -\gamma_{\mathrm D}-1,\\ B_{\mathrm M} &= c\delta_{\mathrm D},\\ C_{\mathrm M} &= 2\alpha_{\mathrm D}-\gamma_{\mathrm D}+1,\\ D_{\mathrm M} &= c \left( 2q_{\mathrm D} +\gamma_{\mathrm D}\delta_{\mathrm D} \right). \end{aligned}$$

Both square roots of $2$ are valid; changing $c$ rotates the canonical coordinate. Because $x=0$ and $z=0$ coincide and the scalar gauge is trivial, the unit-leading germs map directly when $\gamma_{\mathrm D}\notin\mathbb Z_{\geq0}$:

$$U(x)=B_0(cx).$$

At an obstructed value, the same statement requires a declared compatible limit or a separately constructed local solution.

The derivative audit reads

$$\begin{aligned} U'(0) &= cB_0'(0)\\ &= \frac{ D_{\mathrm M} +(1+A_{\mathrm M})B_{\mathrm M} }{ 2(1+A_{\mathrm M}) }, \end{aligned}$$

which is exactly the derivative forced by Maple’s equation.

Triconfluent forms move an ordinary base point

The DLMF THE is

$$y'' + \left( \gamma_{\mathrm D}+z \right)z\,y' + \left( \alpha_{\mathrm D}z-q_{\mathrm D} \right)y =0.$$

It is the direct Wolfram specialization

$$\begin{aligned} q_{\mathrm W}&=q_{\mathrm D},\\ \alpha_{\mathrm W}&=\alpha_{\mathrm D},\\ \gamma_{\mathrm W}&=0,\\ \delta_{\mathrm W}&=\gamma_{\mathrm D},\\ \epsilon_{\mathrm W}&=1. \end{aligned}$$

Since Wolfram’s HeunT is normalized by $y(0)=1$ and $y'(0)=0$,

$$T_0(z) = \operatorname{HeunT} \left[ q_{\mathrm D}, \alpha_{\mathrm D}, 0, \gamma_{\mathrm D}, 1, z \right].$$

Advanced: reduce a general Wolfram triconfluent scale

Start from Wolfram’s equation

$$\begin{aligned} w'' &+ \left( \gamma_{\mathrm W} +\delta_{\mathrm W}z +\epsilon_{\mathrm W}z^2 \right)w'\\ &+ \left( \alpha_{\mathrm W}z-q_{\mathrm W} \right)w =0. \end{aligned}$$

For $\epsilon_{\mathrm W}\neq0$, choose

$$\lambda^3=\epsilon_{\mathrm W}$$

and one root $h$ of

$$h^2 - \frac{\delta_{\mathrm W}}{\lambda^2}h + \frac{\gamma_{\mathrm W}}{\lambda} =0.$$

With

$$t=\lambda z+h, \qquad w(z)=Y(t),$$

the Wolfram equation becomes the DLMF THE with

$$\begin{aligned} \gamma_{\mathrm D} &= \frac{\delta_{\mathrm W}}{\lambda^2} -2h,\\ \alpha_{\mathrm D} &= \frac{\alpha_{\mathrm W}}{\epsilon_{\mathrm W}},\\ q_{\mathrm D} &= \frac{q_{\mathrm W}}{\lambda^2} + \frac{\alpha_{\mathrm W}h}{\epsilon_{\mathrm W}}. \end{aligned}$$

The Wolfram origin maps to $t=h$, so its named function becomes the DLMF solution with

$$Y(h)=1, \qquad Y'(h)=0,$$

not generally $T_0(t)$. The cube-root and quadratic-root choices are part of the coordinate chart. If $\epsilon_{\mathrm W}=0$, this rank-$3$ normalization is unavailable and the equation must be reclassified before a different reduction is chosen.

Maple instead uses

$$\begin{aligned} U'' &- \left( \Gamma_{\mathrm M}+3x^2 \right)U'\\ &+ \left[ A_{\mathrm M} +(B_{\mathrm M}-3)x \right]U =0, \end{aligned}$$

with $U(0)=1$ and $U'(0)=0$. In the same variable this is exactly the Wolfram specialization

$$\begin{aligned} q_{\mathrm W}&=-A_{\mathrm M},\\ \alpha_{\mathrm W}&=B_{\mathrm M}-3,\\ \gamma_{\mathrm W}&=-\Gamma_{\mathrm M},\\ \delta_{\mathrm W}&=0,\\ \epsilon_{\mathrm W}&=-3. \end{aligned}$$

Thus Maple and Wolfram name the same origin-normalized function on that five-parameter slice.

To map the DLMF reduced form to Maple’s reduced form, choose one cube root

$$c^3=-3$$

and set

$$z = cx-\frac{\gamma_{\mathrm D}}2, \qquad y(z)=U(x).$$

Coefficient comparison gives

$$\begin{aligned} \Gamma_{\mathrm M} &= \frac{c\gamma_{\mathrm D}^2}{4},\\ A_{\mathrm M} &= -c^2 \left( q_{\mathrm D} +\frac{ \alpha_{\mathrm D}\gamma_{\mathrm D} }{2} \right),\\ B_{\mathrm M} &= 3 \left( 1-\alpha_{\mathrm D} \right). \end{aligned}$$

The three choices of $c$ are residual rotations of the canonical coordinate. More importantly,

$$x=0 \longmapsto z=-\frac{\gamma_{\mathrm D}}2.$$

Consequently this equation map does not identify the two functions normalized at their respective origins. A DLMF solution normalized at $z=0$ must first be evaluated, with its derivative, at $z=-\gamma_{\mathrm D}/2$ before it can be expanded in Maple’s origin-normalized basis. Locally the derivative transforms as

$$U'(0) = c\,y'\left( -\frac{\gamma_{\mathrm D}}2 \right).$$

Normalizations and derivatives must cross too

Suppose again that

$$z=\phi(x), \qquad y\bigl(\phi(x)\bigr)=\chi(x)u(x),$$

and $z_0=\phi(x_0)$. At an ordinary point,

$$u(x_0) = \frac{y(z_0)}{\chi(x_0)},$$

and differentiation gives

$$u'(x_0) = \frac{ \phi'(x_0)y'(z_0) -\chi'(x_0)u(x_0) }{ \chi(x_0) }.$$

Even if $\chi(x_0)=1$, a nonzero $\chi'(x_0)$ changes the derivative. This is the ordinary-point version of a general rule:

• At a regular singular point, transport the exponent, the branch of the local power, and the Frobenius leading coefficient.
• At an irregular point, transport the ordered sectorial basis, the angular lift, the asymptotic normalization, and the lateral summation choice.
• After analytic continuation, transport the path. Equality of local germs does not force two package principal values to agree across different cuts.

For a fundamental row frame

$$\mathcal B_y = \left( y_1,y_2 \right),$$

a scalar coordinate-and-gauge map acts on both columns. A further constant matrix is generally required if the target software basis uses different initial data:

$$\mathcal B_y\bigl(\phi(x)\bigr) = \chi(x)\, \mathcal B_u(x)\, N.$$

The matrix $N$ is fixed by matching initial or asymptotic normalization, not by the parameter tuple.

Physical parameters belong in a separate layer

A physical reduction often has the form

$$\begin{aligned} R(r) &= \exp(\sigma z) z^{\rho_0} (1-z)^{\rho_1} y(z),\\ z &= \phi(r). \end{aligned}$$

The map from physical data to a Heun tuple must retain every part of this formula. A robust pipeline is

$$\begin{gathered} \text{physical parameters}\\ \downarrow\\ \text{coefficient functions in }r\\ \downarrow\quad \left( z=\phi(r),\ \text{gauge} \right)\\ \text{declared canonical Heun equation}\\ \downarrow\\ \text{normalized basis or software call}. \end{gathered}$$

For a raw scalar equation

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

the Liouville substitution

$$y = \exp\left( -\frac12\int^z p(s)\,\dd s \right)\psi$$

gives the convention-resistant normal form

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

Comparing $T$ after aligning coordinates is often the fastest way to detect a missed scalar gauge. Under

$$z=\phi(x), \qquad \psi\bigl(\phi(x)\bigr) = \sqrt{\phi'(x)}\,u(x),$$

the projective coefficient becomes

$$T_x(x) = \bigl(\phi'(x)\bigr)^2 T\bigl(\phi(x)\bigr) + \frac12 \{\phi,x\},$$

where

$$\{\phi,x\} = \frac{\phi'''}{\phi'} - \frac32 \left( \frac{\phi''}{\phi'} \right)^2.$$

Thus $T$ is gauge-resistant in a fixed coordinate, not coordinate-invariant; even a Möbius map, whose Schwarzian vanishes, retains the $(\phi')^2$ factor. The normal form also separates invariant exponent differences and irregular exponential differences from the bookkeeping of first-derivative coefficients. The general-Heun page derives the corresponding standard-to-normal accessory map.

A frequency, energy, or separation constant should not be called “the accessory parameter” until its explicit relation to canonical $q$ has been written. That relation may be affine, nonlinear, or gauge dependent.

A crosswalk audit that fails loudly

Use this sequence before trusting a symbolic or numerical conversion:

1. Record the source ODE, variable, ordered parameter tuple, and normalization.
2. Classify its singular points and generic formal ranks.
3. Declare the exact target equation and parameter convention.
4. State the coordinate, gauge, inverse maps, branch choices, and excluded strata.
5. Clear denominators and compare every coefficient of $u''$, $u'$, and $u$.
6. Check local exponent differences or irregular exponential differences.
7. Transport the base point, leading coefficient, derivative, branch, path, and sector.
8. Check a first Taylor coefficient or an exact Wronskian.
9. Substitute the inverse parameter map back into the source operator.
10. Only then perform a generic, nonresonant numerical spot check.
11. Record software versions and the official definition pages used.

Version-aware computation

Record $Version in Wolfram Language and kernelopts(version) in Maple. As documented in SymPy 1.14 and mpmath 1.3, neither project exposes a documented built-in HeunG/HeunC/HeunD/HeunB/HeunT interface. Do not invent a Python parameter order. A generic IVP solver is a different object and requires an explicit complex-path strategy for analytic continuation.

Common pitfalls

A matching tuple is not a matching function. The Wolfram-to-DLMF DCHE rescaling maps the equation but generally moves the base point. Transport the two initial values before comparing numbers.

A software name is not an equation class. Maple and Wolfram both expose HeunD, but their canonical variables, parameter counts, singularity locations, and defining base points differ.

A missed scale changes several slots at once. In the BHE, the single choice $c^2=2$ changes the constant, linear, and accessory coefficients. Changing only the quadratic coefficient in $y'$ cannot be correct.

A direct germ identity is not automatically a principal-value identity. Wolfram and Maple can continue around cuts by different implicit paths. Attach a path and branch prescription whenever the argument leaves the initial series disk.

Exceptional parameters cannot be handled by canceled denominators. If a displayed first coefficient contains $1/\gamma$, return to the recurrence at the exceptional value. A compatible logarithmic or limiting solution may exist, but it requires a separate declaration.

Polynomiality is not invariant under an arbitrary gauge. A polynomial may become an exponential or power times a polynomial. The next page treats polynomial, algebraic, and quasi-exact sectors in declared gauges.

Exercises

1. Repair an incomplete HeunD statement

A calculation ends with “the solution is $\operatorname{HeunD}(q,\alpha,\gamma,\delta,\epsilon;z)$.” List the additional data needed to make the statement reproducible.

Solution

First declare the software or source, because Maple and Wolfram use different HeunD equations. Then write the exact ODE and ordered tuple. State the canonical variable and singularity locations. State the base point and both initial conditions—or, for a sectorial solution, the sector, angular lift, and asymptotic normalization. Finally record the continuation path, branch choices, software version, and parameter exclusions. Only this complete passport selects a reproducible function.

2. Derive the Maple confluent map

Starting from Maple’s coefficient

$$\frac{\mu}{z}+\frac{\nu}{z-1},$$

derive $q=\mu$ and $\alpha=\mu+\nu$. Verify the first Taylor coefficient at $z=0$.

Solution

Combining the fractions gives

$$\frac{\mu}{z} + \frac{\nu}{z-1} = \frac{ (\mu+\nu)z-\mu }{ z(z-1) }.$$

Comparison with $(\alpha z-q)/[z(z-1)]$ gives

$$q=\mu, \qquad \alpha=\mu+\nu.$$

The coefficient of $U'$ at the origin has residue $\beta_{\mathrm M}+1=\gamma$. Keeping the coefficient of $z^{-1}$ in the ODE gives

$$\left( \beta_{\mathrm M}+1 \right)U'(0) +\mu U(0) =0.$$

For $U(0)=1$,

$$U'(0) = -\frac{\mu}{\beta_{\mathrm M}+1} = -\frac q\gamma,$$

which agrees with $C_0(z)=1-qz/\gamma+O(z^2)$.

3. Reduce the five-parameter DCHE

Apply $t=\epsilon_{\mathrm W}z$ to Wolfram’s DCHE for $\epsilon_{\mathrm W}\neq0$. Derive the four DLMF parameters and transport the defining initial data.

Solution

Set $y(z)=Y(t)$. Then

$$y'=\epsilon_{\mathrm W}Y', \qquad y''=\epsilon_{\mathrm W}^2Y''.$$

Substitution and division by the transformed leading coefficient give

$$\begin{aligned} Y'' &+ \left[ \frac{ \epsilon_{\mathrm W}\gamma_{\mathrm W} }{t^2} + \frac{\delta_{\mathrm W}}t +1 \right]Y'\\ &+ \frac{ (\alpha_{\mathrm W}/\epsilon_{\mathrm W})t -q_{\mathrm W} }{t^2}Y =0. \end{aligned}$$

Hence

$$\begin{aligned} q_{\mathrm D}&=q_{\mathrm W},\\ \alpha_{\mathrm D} &= \alpha_{\mathrm W}/\epsilon_{\mathrm W},\\ \gamma_{\mathrm D}&=\delta_{\mathrm W},\\ \delta_{\mathrm D} &= \epsilon_{\mathrm W}\gamma_{\mathrm W}. \end{aligned}$$

Since $z=1$ maps to $t=\epsilon_{\mathrm W}$,

$$Y(\epsilon_{\mathrm W})=1.$$

Also $y'(1)=\epsilon_{\mathrm W}Y'(\epsilon_{\mathrm W})=0$, so $Y'(\epsilon_{\mathrm W})=0$. The target is not normalized at $t=1$ unless $\epsilon_{\mathrm W}=1$.

4. Scale the Maple biconfluent equation

Let $c^2=2$ and $z=cx$. Derive the Maple-to-DLMF BHE map and verify both the exponent difference and the first derivative. Assume $A_{\mathrm M}\notin\{-1,-2,\ldots\}$ so that the unit-leading analytic recurrence is unobstructed.

Solution

Write $U(x)=B(z)$ with $z=cx$. Then

$$U'=cB', \qquad U''=c^2B''.$$

Substitution into Maple’s equation, followed by division by $x$ and use of $c^2=2$, gives the DLMF equation with

$$\begin{aligned} \gamma_{\mathrm D} &= -(1+A_{\mathrm M}),\\ \delta_{\mathrm D} &= B_{\mathrm M}/c,\\ \alpha_{\mathrm D} &= (C_{\mathrm M}-A_{\mathrm M}-2)/2,\\ q_{\mathrm D} &= \frac{ D_{\mathrm M} +(1+A_{\mathrm M})B_{\mathrm M} }{2c}. \end{aligned}$$

The DLMF exponents are $0$ and $1+\gamma_{\mathrm D}$, so the second exponent is

$$1+\gamma_{\mathrm D}=-A_{\mathrm M},$$

which matches Maple’s indicial equation. The derivative check is

$$\begin{aligned} U'(0) &= c \left( -\frac{q_{\mathrm D}}{\gamma_{\mathrm D}} \right)\\ &= \frac{ D_{\mathrm M} +(1+A_{\mathrm M})B_{\mathrm M} }{ 2(1+A_{\mathrm M}) }. \end{aligned}$$

This is exactly the slope imposed by Maple’s constant-order equation.

5. Center the DLMF triconfluent equation

Use

$$z = cx-\frac{\gamma_{\mathrm D}}2, \qquad c^3=-3,$$

to derive $(A_{\mathrm M},B_{\mathrm M},\Gamma_{\mathrm M})$. Explain why the map is not an identity between the two origin-normalized functions.

Solution

With $y(z)=U(x)$, multiply the transformed equation by $c^2$. The first-derivative coefficient becomes

$$\begin{aligned} c\,z(z+\gamma_{\mathrm D}) &= c \left( c^2x^2-\frac{\gamma_{\mathrm D}^2}{4} \right)\\ &= -3x^2-\frac{c\gamma_{\mathrm D}^2}{4}. \end{aligned}$$

Thus

$$\Gamma_{\mathrm M} = \frac{c\gamma_{\mathrm D}^2}{4}.$$

The coefficient of $U$ becomes

$$\begin{aligned} c^2 \left( \alpha_{\mathrm D}z-q_{\mathrm D} \right) &= -c^2 \left( q_{\mathrm D} +\frac{\alpha_{\mathrm D}\gamma_{\mathrm D}}2 \right)\\ &\quad -3\alpha_{\mathrm D}x. \end{aligned}$$

Therefore

$$\begin{aligned} A_{\mathrm M} &= -c^2 \left( q_{\mathrm D} +\frac{\alpha_{\mathrm D}\gamma_{\mathrm D}}2 \right),\\ B_{\mathrm M} &= 3(1-\alpha_{\mathrm D}). \end{aligned}$$

But $x=0$ maps to $z=-\gamma_{\mathrm D}/2$, not to $z=0$. The equations are equivalent while their named origin-normalized solutions are generally different linear combinations.

6. Transport initial data through a gauge

Derive the transformation of $(y(z_0),y'(z_0))$ under

$$z=\phi(x), \qquad y\bigl(\phi(x)\bigr)=\chi(x)u(x).$$

What happens when $\chi(x_0)=1$ but $\chi'(x_0)\neq0$?

Solution

At $z_0=\phi(x_0)$,

$$u(x_0) = \frac{y(z_0)}{\chi(x_0)}.$$

Differentiate the defining relation:

$$\phi'(x)y'\bigl(\phi(x)\bigr) = \chi'(x)u(x)+\chi(x)u'(x).$$

Solving for the target derivative gives

$$u'(x_0) = \frac{ \phi'(x_0)y'(z_0) -\chi'(x_0)u(x_0) }{ \chi(x_0) }.$$

If $\chi(x_0)=1$ but $\chi'(x_0)\neq0$, the function value is unchanged while the derivative is shifted by $-\chi'(x_0)u(x_0)$. A unit gauge value therefore does not imply preserved initial data.

7. Catch the biconfluent sign error

Suppose someone maps the DLMF BHE to Wolfram using

$$\gamma_{\mathrm W}=\gamma_{\mathrm D}, \qquad \delta_{\mathrm W}=\delta_{\mathrm D}, \qquad \epsilon_{\mathrm W}=1.$$

Use both the ODE and the first Taylor coefficient to find the error.

Solution

Multiplying the DLMF equation by $z$ gives

$$\begin{aligned} z y'' &- \left( \gamma_{\mathrm D} +\delta_{\mathrm D}z +z^2 \right)y'\\ &+ \left( \alpha_{\mathrm D}z-q_{\mathrm D} \right)y =0. \end{aligned}$$

Wolfram places a plus sign before its entire $y'$ coefficient. Therefore

$$\begin{aligned} \gamma_{\mathrm W} &= -\gamma_{\mathrm D},\\ \delta_{\mathrm W} &= -\delta_{\mathrm D},\\ \epsilon_{\mathrm W} &= -1. \end{aligned}$$

The local check reaches the same conclusion. Wolfram’s analytic solution has

$$y'(0) = \frac{q_{\mathrm W}}{\gamma_{\mathrm W}},$$

which becomes

$$-\frac{q_{\mathrm D}}{\gamma_{\mathrm D}}$$

only with $\gamma_{\mathrm W}=-\gamma_{\mathrm D}$. This agrees with the DLMF germ $B_0(z)=1-q_{\mathrm D}z/\gamma_{\mathrm D}+O(z^2)$.

References

• NIST Digital Library of Mathematical Functions, §31.2, Differential Equations, §31.3, Basic Solutions, and §31.12, Confluent Forms. These sections fix the DLMF equations and the general-Heun local notation used here.
• Wolfram Language official documentation for HeunG, HeunC, HeunD, HeunB, and HeunT. These pages define the five software calls, equations, and normalizations.
• Maplesoft official help for Heun functions, HeunG, HeunC, HeunD, HeunB, and HeunT. The coefficient maps on this page follow by direct substitution into these defining equations.
• A. Ronveaux, ed., Heun’s Differential Equations, Oxford University Press, 1995, especially Parts B–E for the four confluent forms and their canonical solutions.
• SymPy, complete API index, and mpmath, complete index and ODE documentation, for the version-aware Python note.