Solving Second-Order Homogeneous Linear Differential Equations in Terms of the Tri-Confluent Heun’s Function

Abstract

In this paper, we state an algorithm that checks whether a given second-order linear differential equation can be reduced to the tri-confluent Heun’s equation. The algorithm provides a method for finding solutions of the form $\exp (\int r(x\left)dx\right)·\mathrm{HeunT}(q,\alpha ,\gamma ,\delta ,ϵ,f(x\left)\right),$ where the parameters $\alpha ,\beta ,\lambda \in \mathbb{C},$ the functions $r,f\in \mathbb{C}\left(x\right),$ and f are not constant.

1. Introduction

The tri-confluent Heun’s function, $\mathrm{HeunT}(q,\alpha ,\gamma ,\delta ,ϵ,x),$ is a solution for the tri-confluent Heun’s differential equation [1]

$$y^{\prime \prime }+(\gamma +\delta x+ϵx^2)y^{\prime }+(\alpha x-q)y=0,$$

(1)

where $\lambda ,\alpha ,\beta \in \mathbb{C}.$ The equation corresponds to the following differential operator:

$$L_{HT}={\partial }^2+(\gamma +\delta x+ϵx^2)\partial +(\alpha x-q),$$

(2)

where $\partial =\frac{d}{dx}.$ There are many applications for the Heun’s functions in different fields of science such as quantum mechanics, gravitational physics, astrophysics, nuclear physics, chemical dynamics, mathematical chemistry, mathematical physics, etc. [2,3,4].

Applying the change of variables transformation [5], which is the substitution

$$(x,\partial ){\stackrel{f}{\to }}_{CH}(f,\frac{1}{f^{\prime }}\partial )$$

(3)

where $f\in \mathbb{C}\left(x\right)$ and f is not constant, on $L_{\mathrm{HT}}$ gives a second-order differential operator $\tilde{L}\in \mathbb{C}\left(x\right)[\partial ].$ Note that $\tilde{L}$ has a solution of the form $\mathrm{HeunT}(q,\alpha ,\gamma ,\delta ,ϵ,f(x\left)\right).$ Also, applying the exp-product transformation [5], which is

$$\partial {\stackrel{r}{\to }}_{EX}\partial -r,$$

(4)

where $r\in \mathbb{C}\left(x\right),$ on $\tilde{L}$ produces a second-order differential operator $L\in \mathbb{C}\left(x\right)[\partial ].$ From the definition of the transformations, the operator L has a solution of the form

$$\exp (\int r(x)dx)·\mathrm{HeunT}(q,\alpha ,\gamma ,\delta ,ϵ,f(x)).$$

(5)

If L is obtained from $L_{\mathrm{HT}}$ by $\left(3\right)$ and $\left(4\right),$ we write $L_{\mathrm{HT}}{\stackrel{f}{\to }}_{CH}\tilde{L}{\stackrel{r}{\to }}_{EX}L$, and the operator L is denoted by $\tilde{L}Ⓢ(\partial -r).$

Consider the second-order linear differential equations with solutions of the form    

$$\exp (\int r(x)dx)·y(f(x)),$$

(6)

where $r,f\in \mathbb{C}\left(x\right)$, and y is a special function (e.g., Bessel, Kummer, and the hypergeometric functions ${ }_2{F}_1(\alpha ,\beta ;\lambda ;x),$${ }_1{F}_1(\alpha ;\lambda ;x),$ and ${ }_0{F}_1(\lambda ;x)$) that satisfies a second-order linear differential equation (see [6]). These kinds of equations are studied by many authors (e.g., [7,8,9,10,11]). In the general case, where y is not a special function, Algorithm 17 in [12] reduces the second-order linear differential equations with solutions of the form $y\left(f\right(x\left)\right)$ to the same order with solutions of the form y by detecting the change of variable transformation $\left(3\right).$ Our algorithm in [13] studies the second-order differential equations with solutions of the form

$$\exp (\int r(x)dx)·\mathrm{HeunC}(\alpha ,\beta ,\gamma ,\delta ,\eta ,f(x)),$$

(7)

where $\mathrm{HeunC}$ is the confluent Heun’s function.

Let $L={\partial }^2+a\left(x\right)\partial +b\left(x\right)$ be an irreducible second-order differential operator with coefficients in $\mathbb{C}\left(x\right).$ The goal of this paper is to reduce L “if and only if it is possible” to $L_{\mathrm{HT}},$$\left(2\right)$, by detecting the parameters of the transformations $\left(3\right)$ and $\left(4\right).$ The main contribution of this paper is an algorithm that reduces L to $L_{\mathrm{HT}}$ if L is obtained from $L_{\mathrm{HT}}$ by $\left(3\right)$ and $\left(4\right).$ The algorithm will extend the ability of the existing software that solves the tri-confluent Heun’s equations to solve second-order differential equations in terms of the tri-confluent Heun’s function.

Example 1.

Given

$$\begin{array}{cc}\hfill L& ={\partial }^2-\frac{(9x^8+x^6-4x^5-16x^4-10x^3+x^2+9)}{3x^4(x^2+1)}\partial \hfill \\ & -\frac{(180x^7-9x^6+190x^5-17x^4-160x^3-77x^2-170x-9)}{45x^4(x^2+1)}.\hfill \end{array}$$

The algorithm in this paper reduces L to $L_{\mathrm{HT}}={\partial }^2-(3x^2+\frac{1}{3})\partial -2x+\frac{1}{5}$ by computing the parameter of the transformations

$$f\left(x\right)=\frac{x^2-1}{x}\mathit{and}r\left(x\right)=\frac{-2}{3x}.$$

By use of the command dsolve in Maple software 2021, the functions

$$y_1\left(x\right)=\mathrm{HeunT}(\frac{1}{5},1,\frac{1}{3},x)\mathit{and}{y}_2\left(x\right)=e^{x^3+\frac{x}{3}}\mathrm{HeunT}(\frac{1}{5},-1,\frac{1}{3},-x)$$

are solutions of $L_{\mathrm{HT}}.$ Therefore, the functions

$$Y_i\left(x\right)=\exp (\int \frac{-2}{3x}dx)·y_i\left(\frac{x^2-1}{x}\right),$$

where $i=1,2,$ are solutions of L.

The importance of this paper lies in the application of the exp-product and the change of variable transformations or choosing certain values of the parameters of a well-known differential equation that could lead to another well-known differential equation. For example, the differential equation $y^{\prime \prime }+2ϵ-2(a{x}^2+b{x}^4)y=0$ is obtained by substituting the quartic potential $v\left(x\right)=a{x}^2+b{x}^4,$ where $a\in \mathbb{R}$ and $b\in {\mathbb{R}}^{+},$ into $y^{\prime \prime }+2(ϵ-v\left(x\right))y=0$, which is the one-dimensional Schrödinger equation. The solutions of the equation can be expressed in terms of the tri-confluent Heun’s functions [14]. Also, some forms of the one-dimensional Schrödinger equation can be solved by transforming them to the confluent and bi-confluent Heun equations [15,16,17]. A class of the quantum two-state problem models can be solved in terms of the tri-confluent Heun’s functions, and the class is symmetric for some values of the parameters of the class and the change of variable transformation (some second-order linear differential equations have Lie point symmetries which can be interpreted as the change of variables transformations of the form $x\mapsto ax+b,$ where a and b are constants) [4,18].

2. Preliminaries

A differential operator of order n with coefficients in $\mathbb{C}\left(x\right)$ takes the form ${\sum }_{i=0}^n{a}_i\left(x\right){\partial }^i,$ where $a_i\left(x\right)\in \mathbb{C}\left(x\right)$ and $\partial =\frac{d}{dx}.$ It corresponds to the homogeneous linear differential equation

$$a_n\left(x\right)y^{\left(n\right)}+a_{n-1}\left(x\right)y^{(n-1)}+\cdots +a_0\left(x\right)y=0.$$

A function $y\left(x\right)$ is a solution of $L\in \mathbb{C}\left(x\right)[\partial ]$ if $L\left(y\right)=0.$

Let $L={\partial }^2+A\left(x\right)\partial +B\left(x\right)$, where $A\left(x\right)$ and $B\left(x\right)$ are in $\mathbb{C}\left(x\right).$ A point $c\in \mathbb{C}$ is a singular point of L if c is a pole of $A\left(x\right)$ or $B\left(x\right).$ If c is a pole of $(x-c)A\left(x\right)$ or ${(x-c)}^{2}B\left(x\right),$ it is an irregular singular point of $L;$ otherwise, it is a regular singular point. The point $x=\infty$ is a singular point of L if 0 is a singular point of $L_{\frac{1}{x}},$ which is obtained from L by applying $\left(3\right)$ on L and $f\left(x\right)=\frac{1}{x}.$ The local parameter of $c\in \mathbb{C}$ is $t_c=x-c$ if $c\ne \infty$ and $t_c=\frac{1}{x}$ if $c=\infty .$

The operator L has two generalized series solutions at $c\in \mathbb{C}\cup \{\infty \}$ of the form

$$Y_1=\exp (\int \frac{e_1}{t_c}d{t}_c)U_1,Y_2=\exp (\int \frac{e_2}{t_c}d{t}_c)U_2,$$

(8)

where $e_j\in \mathbb{C}\left[t_c^{\frac{-1}{mj}}\right]$ and $U_j={\sum }_{i=0}^{\infty }{Z}_{j,i}{t}_c^{\frac{i}{m_j}}$ such that $Z_{j,i}\in \mathbb{C}[log\left(t_c\right)]$ and $Z_{j,0}\ne 0$ for $j=1,2,$ as shown in Definition 1 in [10]. In $\left(8\right),$ $e_1$ and $e_2$ are called the generalized exponents of L at $c,$ and $Y_1$ and $Y_2$ are called the formal solutions of L at $c.$ If c is not a singular point of L, then the generalized exponents of L at c are $e_1=0$ and $e_2=1$ [19]. At the regular singularities of $L,$ the generalized exponents are constants [20]. If c is an irregular singular point of $L,$ then at least one of the generalized exponents of L at c is non-constant [11].

3. The Effects of the Transformations on the Singularities and the Generalized Exponents of the Tri-Confluent Heun’s Operator

Let L be a second-order differential operator with coefficients in $\mathbb{C}\left(x\right)$ such that

$$L_{\mathrm{HT}}{\stackrel{f}{\to }}_{CH}\tilde{L}{\stackrel{r}{\to }}_{EX}L.$$

(9)

where $r,f\in \mathbb{C}\left(x\right)$, and f is not constant. In $\left(9\right),$ the operator $\tilde{L}$ is called a proper pullback of $L_{\mathrm{HT}}$ by $f,$ and the operator L is called a weak pullback of $L_{\mathrm{HT}}$ by f (Definition $4.1$ in [9]). If $L={\partial }^2+A\left(x\right)\partial +B\left(x\right)$ and $\tilde{L}={\partial }^2+c_1\left(x\right)\partial +c_0\left(x\right),$ we can rewrite $\left(9\right)$ as follows:

$$L_{\mathrm{HT}}{\stackrel{f}{\to }}_{CH}\tilde{L}{\stackrel{\frac{c_1\left(x\right)}{2}}{\to }}_{EX}NF\left(L\right){\stackrel{\frac{-A\left(x\right)}{2}}{\to }}_{EX}L,$$

(10)

where

$$NF\left(L\right)=LⓈ(\partial -\frac{A\left(x\right)}{2})={\partial }^2+B\left(x\right)-\frac{{\left(A\left(x\right)\right)}^2}{4}-\frac{{\left(A\left(x\right)\right)}^{\prime }}{2},$$

and $L=\tilde{L}Ⓢ(\partial -r\left(x\right))$, where $r\left(x\right)=\frac{c_1\left(x\right)-A\left(x\right)}{2}.$ The operator $NF\left(L\right)$ is called the normal form of $L,$ and it is used to remove the singularities of L that are not singularities of $\tilde{L}.$

Let $L\in \mathbb{C}\left(x\right)[\partial ]$ be as in $\left(9\right).$ To reduce L to $L_{\mathrm{HT}},$ we need to compute f and $c_1\left(x\right)$ in $\left(10\right).$ Algorithm 17 in [12] reduces $\tilde{L}$ to $L_{\mathrm{HT}}$ by computing the pullback function $f\in \mathbb{C}\left(x\right)$ if $\mathbb{C}⊊\mathbb{C}\left(f\right(x\left)\right)⊊\mathbb{C}\left(x\right).$ Therefore, we need to compute $c_1\left(x\right)$ in $\left(10\right)$ to have a complete method for reducing weak pullback operators of $L_{\mathrm{HT}}.$ For this, we study the effects of the transformations on the singularities and the generalized exponents of $L_{\mathrm{HT}}.$

The operator $L_{\mathrm{HT}}$ has only one singular point, which is $x=\infty ,$ and the singular point is irregular. The generalized exponents (we use the command gen_exp in Maple software to compute them) of $L_{\mathrm{HT}}$ at ∞ are $e_1=\frac{\alpha }{ϵ}$ and $e_2=ϵt_{\infty }^{-3}+\delta t_{\infty }^{-2}+\gamma t_{\infty }^{-1}+2-\frac{\alpha }{ϵ},$ and the generalized exponents of $L_{\mathrm{HT}}$ at any other point in $\mathbb{C}$ are 0 and $1.$

Lemma 1.

Let $\tilde{L}$ be a proper pullback of $L_{\mathrm{HT}}$ by $f\in \mathbb{C}\left(x\right)$ and f be not constant.

• If c is a root of f with multiplicity $m\in \mathbb{N}\mathit{,}$ then c is a regular singular point of L, and the generalized exponents of  $\tilde{L}$ at c are $e_1=0$ and $e_2=m;$
• If c is a pole of f of order$m\in \mathbb{N}\mathit{,}$ then c is an irregular singular point of $\tilde{L}$, and the generalized exponents of L at c are $e_1=m\left(\frac{\alpha }{ϵ}\right)$ and $e_2={\sum }_{i=-3m}^{-1}{a}_i{t}_c^i+m(2-\frac{\alpha }{ϵ})$, where  $a_{-3m},\dots ,a_{-1}\in \mathbb{C}.$

The proof is similar to the proof of Theorem 2 [5].

Let $\tilde{L}={\partial }^2+c_1\left(x\right)\partial +c_0\left(x\right)$ be as in $\left(9\right),$ and let $\{e_1,e_2\}$ be the set of the generalized exponents of $\tilde{L}$ at $c.$ If e is the generalized exponent of $\partial -r$ at $c,$ then $\{e_1+e,e_2+e\}$ is the set of the generalized exponents of $\tilde{L}Ⓢ(\partial -r)$ at c (Lemma 4 in [5]). If $\{e_1={\sum }_{i=j_1}^0{a}_{1,i}{t}_c^i,e_2={\sum }_{i=j_2}^0{a}_{2,i}{t}_c^i\}$ where $j_1,j_2\in {\mathbb{Z}}^{-}\cup \left\{0\right\}$ is the set of the generalized exponents of $\tilde{L}$ at $c,$ then the Laurent series of $c_1\left(x\right)$ at c is $\frac{-e_1-e_2-j_1-j_2+1}{t_c}+{\sum }_{i=0}^{\infty }{a}_i{\left(t_c\right)}^i$ (see Section 3.5 in [21] and Lemma $3.4$ in [22]). Therefore, the set of the generalized exponents of $\tilde{L}Ⓢ(\partial -\frac{c_1\left(x\right)}{2})$ at c is

$$\{\frac{e_1-e_2-j_1-j_2+1}{2},\frac{-e_1+e_2-j_1-j_2+1}{2}\}$$

(11)

because the generalized exponent of $\partial -\frac{c_1\left(x\right)}{2}$ at c is $\frac{-e_1-e_2-j_1-j_2+1}{2}$.

Lemma 2.

Let L be a weak pullback of  $L_{\mathrm{HT}}$ by $f\in \mathbb{C}\left(x\right)\setminus \mathbb{C}.$ Then, the set of the generalized exponents of  $NF\left(L\right)$ at a singular point $c\in \mathbb{C}\cup \{\infty \}$ is of the form $\{\frac{-m+1}{2},\frac{m+1}{2}\}$ or $\{{\sum }_{i=-3m}^{-1}\frac{a_i{t}_c^i}{2}-\frac{m\alpha }{ϵ}+\frac{5m+1}{2},-\left({\sum }_{i=-3m}^{-1}\frac{a_i{t}_c^i}{2}-\frac{m\alpha }{ϵ}\right)+\frac{m+1}{2}\}.$

Proof. 

Let $\tilde{L}={\partial }^2+c_1\left(x\right)\partial +c_0\left(x\right)\in \mathbb{C}\left(x\right)[\partial ]$ be the proper pullback of $L_{\mathrm{HT}}$ by $f.$ Then, $NF\left(L\right)=NF\left(\tilde{L}\right)=\tilde{L}Ⓢ(\partial -\frac{c_1\left(x\right)}{2}).$ Since the set of the generalized exponents of $\tilde{L}$ at a singular point c is of the form $\{0,m\}$ or $\{m\left(\frac{\alpha }{ϵ}\right),{\sum }_{i=-3m}^{-1}{a}_i{t}_c^i+m(2-\frac{\alpha }{ϵ})\}$ (Lemma $1),$ the generalized exponents of $NF\left(L\right)$ at c are of the form $\{\frac{-m+1}{2},\frac{m+1}{2}\}$ or $\{{\sum }_{i=-3m}^{-1}\frac{a_i{t}_c^i}{2}-\frac{m\alpha }{ϵ}+\frac{5m+1}{2},-\left({\sum }_{i=-3m}^{-1}\frac{a_i{t}_c^i}{2}-\frac{m\alpha }{ϵ}\right)+\frac{m+1}{2}\}$, and it is computed by substituting the generalized exponents of $\tilde{L}$ at c in $\left(11\right).$    □

4. Reducing a Weak Pullback of the Tri-Confluent Heun’s Operator

Let L be a weak pullback of $L_{\mathrm{HT}}$ by f such that

$$L_{\mathrm{HT}}{\stackrel{f}{\to }}_{CH}\tilde{L}{\stackrel{r}{\to }}_{EX}L.$$

(12)

The generalized exponents of $NF\left(L\right)$ at its singularities can be used to compute the parameter of the exp-product transformation that transfers $NF\left(L\right)$ to $\tilde{L}$ (Lemmas 1 and 2). If c is a regular singular point of $NF\left(L\right)$ and $\{\frac{-m+1}{2},\frac{m+1}{2}\}$ is the set of the generalized exponents of $NF\left(L\right)$ at $c,$ the generalized exponents of the operators $NF\left(L\right)Ⓢ(\partial +\frac{\frac{-m+1}{2}}{t_c})$ and $\tilde{L}$ at c are 0 and $m.$ Also, if c is an irregular singular point of $NF\left(L\right)$ and $\{{\sum }_{i=-3m}^{-1}\frac{a_i{t}_c^i}{2}-\frac{\alpha }{ϵ}+\frac{5m+1}{2},-\left({\sum }_{i=-3m}^{-1}\frac{a_i{t}_c^i}{2}-\frac{m\alpha }{ϵ}\right)+\frac{m+1}{2}\}$ is the set of the generalized exponents of $NF\left(L\right)$ at $c,$ and the operators $NF\left(L\right)Ⓢ(\partial +\frac{-\left({\sum }_{i=-3m}^{-1}\frac{a_i{t}_c^i}{2}\right)+\frac{m+1}{2}}{t_c})$ and $\tilde{L}$ have the same set of generalized exponents at $c.$

Let ${\mathrm{S}}_{\mathrm{reg}}=\{c_1,c_2,\dots ,c_n\}$ and ${\mathrm{S}}_{\mathrm{irr}}=\{v_1,v_2,\dots ,v_k\}$ be the sets of the regular and irregular singularities of $NF\left(L\right),$ respectively. If $c\in {\mathrm{S}}_{\mathrm{reg}}$ and $\{\frac{-m+1}{2},\frac{m+1}{2}\}$ is the set of the generalized exponents of $NF\left(L\right)$ at $c,$ we take $r_c\left(x\right)=\frac{-m+1}{2t_c}$ if $c\ne \infty$ and $r_c\left(x\right)=0$ if $c=\infty$ (if the constant parts of the generalized exponents of $NF\left(L\right)Ⓢ(\partial -r)$ are equal to the constant parts of the generalized exponents $\tilde{L}$ at each finite singular point, the constant parts of the generalized exponents of the operators at ∞ are equal (Fuchs’s relation, shown in [21,23])). If $v\in {\mathrm{S}}_{\mathrm{irr}}$ and $\{{\sum }_{i=j}^{-1}{a}_i{t}_v^i+a_{0,1},-{\sum }_{i=j}^{-1}{a}_i{t}_v^i+a_{0,2}\}$ are the set of the generalized exponents of $NF\left(L\right)$ at $v,$ we take

$$r_{v,1}\left(x\right)=\frac{{\sum }_{i=j}^{-1}{a}_i{t}_v^i+\frac{\frac{j}{3}+1}{2}}{t_v}\mathrm{and}{r}_{v,2}\left(x\right)=\frac{-{\sum }_{i=j}^{-1}{a}_i{t}_v^i+\frac{\frac{j}{3}+1}{2}}{t_v}$$

if $v\ne \infty ,$ and

$$r_{v,1}\left(x\right)=\frac{{\sum }_{i=j}^{-1}{a}_i{t}_v^i}{t_v}\mathrm{and}{r}_{v,2}\left(x\right)=\frac{-{\sum }_{i=j}^{-1}{a}_i{t}_v^i}{t_v}$$

if $v=\infty .$ The parameter of the exp-product transformation that reduces $NF\left(L\right)$ to $\tilde{L}$ is a function of the form

$$Y_{z_1,\dots ,z_k}=-(r_{c_1}\left(x\right)+\cdots +r_{c_n}\left(x\right)+r_{v_1,z_1}\left(x\right)+\cdots +r_{v_k,z_k}\left(x\right)),$$

(13)

for $z_1,\dots ,z_k\in \{1,2\}.$ Therefore, the operator $\tilde{L}$ in $\left(12\right)$ is equal to one of the operators

$$L_{z_1,\dots ,z_k}=NF\left(L\right)Ⓢ(\partial -Y_{z_1,\dots ,z_k}),$$

(14)

for $z_1,\dots ,z_k\in \{1,2\}.$

Notation 1.

Let L be in $C\left(x\right)[\partial ]$, where  $C\subseteq \mathbb{C}$ is a field of characteristic $0.$ Let $P\left(x\right)\in C\left[x\right]$ be the denominator of L such that $P\left(x\right)={\left(P_1\left(x\right)\right)}^{s_1}\cdots {\left(P_j\left(x\right)\right)}^{s_j}$ and $P_i\left(x\right)$ is an irreducible in $C\left[x\right].$ If $q_1,\dots ,q_{s-1}$ and $q_s$ are the roots of $P_i\left(x\right)$, the sets of the generalized exponents of L at $q_1,\dots ,q_{s-1}$ and $q_s$ are the same. Also, the sets of the generalized exponents of $LⓈ(\partial -\mathrm{Tr}(\frac{e}{x-q_j}))$ at $q_1,\dots ,q_{s-1}$, and $q_s$ are the same $(\mathrm{Tr}(\frac{e}{x-q_j})$ is the trace of $\frac{e}{x-q_j}$ over the field extension $C(x,q_j)/C\left(x\right),$ and it is the sum of $\frac{e}{x-q_j}$ and its conjugates. Note that $\mathrm{Tr}(\frac{e}{x-q_j})\in C(x)).$

5. The Algorithm

Given a monic second-order differential operator, $L\in C\left(x\right)[\partial ],$ where $C\subseteq \mathbb{C}.$ In this section, we state an algorithm that reduces L to $L_{\mathrm{HT}}$ if L is a weak pullback of $L_{\mathrm{HT}}$ by f and $C⊊C\left(f\right(x\left)\right)⊊C\left(x\right).$

Example 2.

Given

$$\begin{array}{cc}\hfill L=& {\partial }^2+\frac{(8x^7-19x^6-4x^5+10x^4-36x^3+78x^2-72x+24)}{2x^7(x-2)}\partial \hfill \\ & +\frac{(8x^7-22x^6-6x^5-3x^4+29x^3-58x^2+76x-40)}{4x^8(x-2)}.\hfill \end{array}$$

The commands dsolve and Heunsols in Maple software 2021 do not find solutions for L. However, Algorithm 1 finds that L is a weak pullback of $\overline{L}={\partial }^2+\frac{4x^3+x^2+6}{2x^4}\partial +\frac{(2x-11)}{4x^5}$ by $f\left(x\right)=\frac{x^2}{x-1}$ and finds $r\left(x\right)=\frac{-1}{x}$ such that

$$\overline{L}{\stackrel{\frac{x^2}{x-1}}{\to }}_{CH}\tilde{L}{\stackrel{\frac{-1}{x}}{\to }}_{EX}L,$$

so the solutions of L can be computed by using the solutions of $\overline{L}$ and the parameters of the transformations. Since the functions

$$y_1\left(x\right)=\mathrm{HeunT}(\frac{1}{2},\frac{1}{4},\frac{1}{2},\frac{1}{x})\mathit{and}{y}_2\left(x\right)=\exp \left(\frac{x^2+2}{2x^3}\right)\mathrm{HeunT}(\frac{1}{2},\frac{-1}{4},\frac{1}{2},\frac{-1}{x})$$

are solutions of $\overline{L},$ the functions

$$Y_1=\frac{1}{x}{y}_1\left(\frac{x^2}{x-1}\right)\mathit{and}{Y}_2=\frac{1}{x}{y}_2\left(\frac{x^2}{x-1}\right)$$

are solutions of L.

Algorithm 1 Reduce weak proper pullback of tri-confluent Heun’s operator to LHT and find the pullback function f.

Input: $L={\partial }^2+A\left(x\right)\partial +B\left(x\right)\in C\left(x\right)[\partial ]$ where $C\subseteq \mathbb{C}$ is a field of characteristic $0.$ Output: A tri-confluent Heun’s differential operator, the pullback function $f,$ and the parameter of the exp-product transformation r or “failed” (failed means L is not a weak pullback of $L_{\mathrm{HT}}$ by non-trivial pullback function f (f is not a Möbius transformation)). 1. Compute the normal form of $L,$ which is $NF\left(L\right)=LⓈ(\partial -\frac{A\left(x\right)}{2}).$ 2. Let $P\left(x\right)\in C\left[x\right]$ be the denominator of $NF\left(L\right),$ and factor $P\left(x\right)$ over $C.$ 3. If $P\left(x\right)={\left(P_1\left(x\right)\right)}^{s_1}\cdots {\left(P_j\left(x\right)\right)}^{s_j},$ where $P_i\left(x\right)$ is an irreducible in $C\left[x\right],$ take $q_i$ to be a root of $P_i\left(x\right).$ 4. For each $q_i,$ compute the generalized exponent, $e_{i,1}$ and $e_{i,2},$ of $NF\left(L\right)$ at $q_i.$ Case $1:$ If $e_{i,1}$ and $e_{i,2}$ are in $\mathbb{Q}$ and $e_{i,1}-e_{i,2}\in \mathbb{Z},$ take $r_i\left(x\right)=\mathrm{Tr}\left(\frac{e_{i,1}}{t_{q_i}}\right)$ if $e_{i,1}\le e_{i,2}.$ If $e_{i,1}$ and $e_{i,2}$ are constants but not in $\mathbb{Q}$ or $e_{i,1}-e_{i,2}\notin \mathbb{Z},$ return “failed” (from Lemmas 1 and 2, it is clear that L is not obtained from $L_{\mathrm{HT}}$ by the transformations $\left(3\right)$ and $\left(4\right)$). Case $2:$ If $e_{i,1}={\sum }_{k=j}^{-1}{a}_k{t}_{q_i}^k+a_{0,1}$ and $e_{i,2}={\sum }_{k=j}^{-1}(-a_k)t_{q_i}^k+a_{0,2},$ then take $v_{i,1}\left(x\right)=\frac{{\sum }_{k=j}^{-1}{a}_k{t}_{q_i}^k+\frac{\frac{j}{3}+1}{2}}{t_{q_i}}$ and $v_{i,2}\left(x\right)=\frac{-{\sum }_{k=j}^{-1}{a}_k{t}_{q_i}^k+\frac{\frac{j}{3}+1}{2}}{t_{q_i}}$ otherwise return “failed”. 5. Put $r_{\infty }\left(x\right)=0$ if $\infty \notin {\mathrm{S}}_{\mathrm{irr}}.$ However, if $\infty \in {\mathrm{S}}_{\mathrm{irr}}$ and $$e_{i,1}=\sum _{k=j}^{-1}{a}_k{t}_{\infty }^k+a_{0,1}\mathrm{and}{e}_{i,2}=\sum _{k=j}^{-1}(-a_k)t_{\infty }^k+a_{0,2},$$ take $v_{\infty ,1}\left(x\right)=\frac{{\sum }_{i=j}^{-1}{a}_i{t}_{\infty }^i}{t_{\infty }}$ and $v_{v,2}\left(x\right)=\frac{-{\sum }_{i=j}^{-1}{a}_i{t}_{\infty }^i}{t_{\infty }}$ otherwise return “failed". 6. From Case 1 of Step $4,$ if $q_{s_1},\dots ,q_{s_k}$ are the regular singularities of $NF\left(L\right),$ take $R=r_{s_1}\left(x\right)+r_{s_2}\left(x\right)+\cdots +r_{s_k}\left(x\right)$ where $s_1,\dots ,s_k\in \{1,\dots ,j\}.$ 7. From Case 2 in Step 4, if $q_{z_1},\dots ,q_{z_t}$ are the irregular singularities of $NF\left(L\right),$ let $$H=\{\{v_{q_{z_1},1}\left(x\right),v_{q_{z_1},2}\left(x\right)\},\dots ,\{v_{q_{z_t},1}\left(x\right),v_{q_{z_t},t}\left(x\right)\},\{v_{\infty ,1}\left(x\right),v_{\infty ,2}\left(x\right)\}\},$$ and G is the set of the combinations of choosing one element from each set in $H.$ 8. For each $G_i=\{v_{q_{z_1},w_1}\left(x\right),v_{q_{z_2},w_2}\left(x\right),\dots ,v_{q_{z_t},w_t}\left(x\right),v_{\infty ,w_{\infty }}\left(x\right)\}\in G,$ where $w_1,w_2,$$\dots ,w_{\infty }\in \{1,2\},$ (a) Take $Y=R+v_{q_{z_1},w_1}\left(x\right)+v_{q_{z_2},w_2}\left(x\right)+\cdots +v_{q_{z_t},w_t}\left(x\right)+v_{\infty ,w_{\infty }}\left(x\right),$ where R is as in Step $6,$ and find $\tilde{L}=NF\left(L\right)Ⓢ(\partial +Y).$ (b) Apply Algorithm 17 in [12] on $\tilde{L}.$ If it reduces $\tilde{L}$ to $\overline{L}$ and finds non-trivial pullback function f$(f$ is not a Möbius transformation$),$ return $\overline{L},$f and $r\left(x\right)=\frac{Y-A\left(x\right)}{2},$ where $L=\tilde{L}Ⓢ(\partial -r\left(x\right)).$ (c) If Algorithm 17 in [12] does not reduce $\tilde{L},$ go to the next element of $G.$ 9. Return “failed” if the previous step does not reduce $L.$

6. Conclusions

We have provided an algorithm that reduces second-order differential operators, whose solutions are in the form

$$\exp (\int r(x)dx)·\mathrm{HeunT}(q,\alpha ,\gamma ,\delta ,ϵ,f(x))$$

where $r,f\in \mathbb{C}\left(x\right)$ and $\mathbb{C}⊊\mathbb{C}\left(f\right(x\left)\right)⊊\mathbb{C}\left(x\right),$ to the tri-confluent Heun’s operator, $\left(2\right).$ The algorithm detects the parameters of the transformations $\left(3\right)$ and $\left(4\right)$ by using the generalized exponents of the given differential operators. Related work for the other forms of Heun’s equations is in progress.