The Heun Triconfluent function

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HeunT - The Heun Triconfluent function

HeunTPrime - The derivative of the Heun Triconfluent function

	Calling Sequence
	HeunT(, , , z) HeunTPrime(, , , z)

Parameters

	-	algebraic expression
	-	algebraic expression
	-	algebraic expression
z	-	algebraic expression

Description

•	The HeunT function is the solution of the Heun Triconfluent equation. Following the first reference (at the end), the equation and the conditions at the origin satisfied by HeunT are

>	FunctionAdvisor(definition, HeunT);

$HeunT(alpha, beta, gamma, z) = DESol({diff(_Y(z), `$`(z, 2))-(3*z^2+gamma)*(diff(_Y(z), z))-((3-beta)*z-alpha)*_Y(z)}, {_Y(z)}, {_Y(0) = 1, (D(_Y))(0) = 0})$

(1)

•	The HeunT(,,,z) function is a local solution to Heun's Triconfluent equation, computed as a standard power series expansion around the origin, a regular point. Because the single singularity is located at , this series converges in the whole complex plane.

•

The Triconfluent Heun Equation (THE) above is obtained from the Doubleconfluent Heun Equation (DHE) through a confluence process, that is, a process where two singularities coalesce, performed by redefining parameters and taking limits. In this case the two irregular singularities of the DHE are coalesced into one irregular singularity at . The resulting Heun Triconfluent equation, thus, has the structure of singularities f of the 0F1 hypergeometric equation and so can be related to the Airy functions.

•

A special case happens when in HeunT(,,,z), the second parameter satisfies , where is a positive integer. In this case the th+1, th+2 and th+3 coefficients form a polynomial system for the remaining parameters and ; when this system is identically satisfied all the subsequent coefficients cancel too and the series truncates, resulting in a polynomial form of degree for HeunT. Remark: for this situation leads to a constant, for HeunT will also be a constant since its series expansion satisfies at 0 = 0 and for the polynomial system for and is inconsistent. So the non-trivial polynomial forms of HeunT are of degree .

Examples

Heun's Triconfluent equation,

THE := diff(y(z), z, z) = (3*z^2+gamma)*(diff(y(z), z))+((3-beta)*z-alpha)*y(z)

THE := diff(y(z), `$`(z, 2)) = (3*z^2+gamma)*(diff(y(z), z))+((3-beta)*z-alpha)*y(z)

(2)

can be transformed into another version of itself, that is, an equation with one regular and one irregular singularities respectively located at 0 and through transformations of the form

z = kappa*t, y(z) = exp(t*(kappa-1)*(kappa^2+kappa+1)*(gamma+t^2)/(2*kappa^3))*u(t)

z = kappa*t, y(z) = exp((1/2)*t*(kappa-1)*(kappa^2+kappa+1)*(gamma+t^2)/kappa^3)*u(t)

(3)

where are new variables and . Under this transformation, the HeunT parameters transform according to -> , -> , -> . These transformations form a group of six elements and imply on identities, among which you have

[[HeunT(alpha, beta, gamma, z) = HeunT(j*alpha, beta, j^2*gamma, j*z), And(j^3 = 1)], [HeunT(alpha, beta, gamma, z) = HeunT(alpha, -beta, gamma, -z)*exp(z^3), And(gamma = 0)]]

(4)

When, in HeunT(,,,z), , where is a positive integer, the th+1, th+2 and th+3 coefficients form a polynomial system for the remaining parameters and . When this system is identically satisfied all the subsequent coefficients cancel too and the series truncates, resulting in a polynomial form of degree for HeunT. For example, this is the necessary condition for a polynomial form

(5)

Considering the first non-trivial case, for , the function is

(6)

So the coefficients of for equal to 4, 5, and 6 in the series expansion are

Q := series(1-((1/2)*alpha)*z^2+(-3/2-(1/6)*gamma*alpha)*z^3+((1/24)*alpha^2-(3/8)*gamma-(1/24)*gamma^2*alpha)*z^4+((1/60)*(9+gamma*alpha)*(-(1/2)*gamma^2+alpha))*z^5+(-(1/720)*alpha^3+(1/240)*gamma^2*alpha^2+(1/720)*(-gamma^4+27*gamma)*alpha-(1/80)*gamma^3)*z^6+O(z^7),z,7)

(7)

c4, c5, c6 := (1/24)*alpha^2-(3/8)*gamma-(1/24)*gamma^2*alpha, (1/60)*(9+gamma*alpha)*(-(1/2)*gamma^2+alpha), -(1/720)*alpha^3+(1/240)*gamma^2*alpha^2+(1/720)*(-gamma^4+27*gamma)*alpha-(1/80)*gamma^3

(8)

solving for and , requesting from solve to return using RootOf, you have

(9)

$[{alpha = 0, gamma = 0}, {alpha = (1/2)*RootOf(36+_Z^3)^2, gamma = RootOf(36+_Z^3)}]$

(10)

substituting for instance the first of these two solutions in HT we have

>	$HT_polynomial := subs([{alpha = 0, gamma = 0}, {alpha = (1/2)*RootOf(36+_Z^3)^2, gamma = RootOf(36+_Z^3)}][1], HT)$

(11)

When the function admits a polynomial form, as is the case of HT_polynomial by construction, to obtain the actual polynomial of degree n (in this case n=3) use

(12)

	See Also
	FunctionAdvisor, Heun, HeunB, HeunC, HeunD, HeunG, hypergeom

References

Decarreau, A.; Dumont-Lepage, M.C.; Maroni, P.; Robert, A.; and Ronveaux, A. "Formes Canoniques de Equations confluentes de l'equation de Heun". Annales de la Societe Scientifique de Bruxelles. Vol. 92 I-II, (1978): 53-78.