The standard form of the Heun equation is
d 2 y d z 2 + ( γ z + δ z − 1 + ϵ z − t ) d y d z + α β z − q z ( z − 1 ) ( z − t ) y = 0 , \frac{d^{2}y}{dz^{2}}+\left(\frac{\gamma}{z}+\frac{\delta}{z-1}+\frac{\epsilon}{z-t}\right)\frac{dy}{dz}+\frac{\alpha\beta z-\mathsf{q}}{z(z-1)(z-t)}y=0\,, d z 2 d 2 y + ( z γ + z − 1 δ + z − t ϵ ) d z d y + z ( z − 1 ) ( z − t ) α β z − q y = 0 , where ϵ = α + β − γ − δ + 1 \epsilon=\alpha+\beta-\gamma-\delta+1 ϵ = α + β − γ − δ + 1 q \mathsf{q} q { 0 , 1 , t , ∞ } \{0,1,t,\infty\} { 0 , 1 , t , ∞ }
y = P { 0 1 t ∞ 0 0 0 α ; z 1 − γ 1 − δ 1 − ϵ β } . y=P
\begin{Bmatrix}
0 & 1 & t & \infty & {}\\
0 & 0 & 0 & \alpha & ;z\\
1-\gamma & 1-\delta & 1-\epsilon & \beta &
\end{Bmatrix}. y = P ⎩ ⎨ ⎧ 0 0 1 − γ 1 0 1 − δ t 0 1 − ϵ ∞ α β ; z ⎭ ⎬ ⎫ . The two linearly independent solutions near z = 0 z=0 z = 0
y 1 [ 0 ] = H ℓ ( t , q ; α , β , γ , δ ; z ) , y 2 [ 0 ] = z 1 − γ H ℓ ( t , ( t δ + ϵ ) ( 1 − γ ) + q ; α + 1 − γ , β + 1 − γ , 2 − γ , δ ; z ) . \begin{aligned}
y^{[0]}_{1} &= H\!\ell \bigl(t,\mathsf{q};\alpha,\beta,\gamma,\delta;z\bigr),\\
y^{[0]}_{2} &= z^{1-\gamma}\,\mathit{H\!\ell}\bigl(t, (t\delta+\epsilon)(1-\gamma) + \mathsf{q}; \alpha+1-\gamma, \beta+1-\gamma, 2-\gamma, \delta; z\bigr).
\end{aligned} y 1 [ 0 ] y 2 [ 0 ] = H ℓ ( t , q ; α , β , γ , δ ; z ) , = z 1 − γ H ℓ ( t , ( t δ + ϵ ) ( 1 − γ ) + q ; α + 1 − γ , β + 1 − γ , 2 − γ , δ ; z ) . where H ℓ H\!\ell H ℓ z = 0 z=0 z = 0 z = 1 z=1 z = 1 z = 1 z=1 z = 1
y 1 [ 1 ] = H ℓ ( 1 − t , α β − q ; α , β , δ , γ ; 1 − z ) , y 2 [ 1 ] = ( 1 − z ) 1 − δ H ℓ ( 1 − t , ( ( 1 − t ) γ + ϵ ) ( 1 − δ ) + α β − q ; α + 1 − δ , β + 1 − δ , 2 − δ , γ ; 1 − z ) . \begin{aligned}
y^{[1]}_{1} &= H\!\ell \bigl(1-t, \alpha\beta - \mathsf{q}; \alpha, \beta, \delta, \gamma; 1-z\bigr),\\
y^{[1]}_{2} &= (1-z)^{1-\delta}\,\mathit{H\!\ell}\bigl(1-t, ((1-t)\gamma+\epsilon)(1-\delta) + \alpha\beta - \mathsf{q}; \alpha+1-\delta, \beta+1-\delta, 2-\delta, \gamma; 1-z\bigr).
\end{aligned} y 1 [ 1 ] y 2 [ 1 ] = H ℓ ( 1 − t , α β − q ; α , β , δ , γ ; 1 − z ) , = ( 1 − z ) 1 − δ H ℓ ( 1 − t , (( 1 − t ) γ + ϵ ) ( 1 − δ ) + α β − q ; α + 1 − δ , β + 1 − δ , 2 − δ , γ ; 1 − z ) . Solutions to boundary value problems between z = 0 z=0 z = 0 z = 1 z=1 z = 1
The normal form of the Heun equation is
( d 2 d z 2 + 1 4 − θ 0 2 z 2 + 1 4 − θ 1 2 ( z − 1 ) 2 + 1 4 − θ t 2 ( z − t ) 2 + θ 0 2 + θ 1 2 + θ t 2 − θ ∞ 2 − 1 2 z ( z − 1 ) + ( t − 1 ) ( w 2 + θ t 2 − θ ∞ 2 − 1 4 ) z ( z − 1 ) ( z − t ) ) ψ ( z ) = 0 , \begin{aligned}
\Biggl(\frac{d^2}{dz^2}+\frac{\tfrac14-\theta_0^2}{z^2}+\frac{\tfrac14-\theta_1^2}{(z-1)^2}+\frac{\tfrac14-\theta_t^2}{(z-t)^2}+\frac{\theta_0^2+\theta_1^2+\theta_t^2-\theta_\infty^2-\tfrac12}{z(z-1)}
+ \frac{(t-1)\bigl(\mathsf{w}^2+\theta_t^2-\theta_\infty^2-\tfrac14\bigr)}{z(z-1)(z-t)}\Biggr)\psi(z)=0\,,
\end{aligned} ( d z 2 d 2 + z 2 4 1 − θ 0 2 + ( z − 1 ) 2 4 1 − θ 1 2 + ( z − t ) 2 4 1 − θ t 2 + z ( z − 1 ) θ 0 2 + θ 1 2 + θ t 2 − θ ∞ 2 − 2 1 + z ( z − 1 ) ( z − t ) ( t − 1 ) ( w 2 + θ t 2 − θ ∞ 2 − 4 1 ) ) ψ ( z ) = 0 , which is related to the standard form by ψ ( z ) = z 1 / 2 − θ 0 ( 1 − z ) 1 / 2 − θ 1 ( 1 − z / t ) 1 / 2 − θ t y ( z ) \psi(z)=z^{1/2-\theta_0}(1-z)^{1/2-\theta_1}(1-z/t)^{1/2-\theta_t}y(z) ψ ( z ) = z 1/2 − θ 0 ( 1 − z ) 1/2 − θ 1 ( 1 − z / t ) 1/2 − θ t y ( z )
θ 0 = 1 2 ( 1 − γ ) , θ 1 = 1 2 ( 1 − δ ) , θ t = 1 2 ( 1 − ϵ ) , θ ∞ = 1 2 ( β − α ) , ( t − 1 ) w 2 + q + θ 0 2 + θ ∞ 2 − t ( θ 0 + θ 1 − 1 2 ) 2 − ( θ 0 + θ t − 1 2 ) 2 = 0 . \begin{aligned}
& \theta_0 = \tfrac{1}{2}(1-\gamma),\qquad \theta_1 = \tfrac{1}{2}(1-\delta),\qquad \theta_t = \tfrac{1}{2}(1-\epsilon),\qquad \theta_\infty = \tfrac{1}{2}(\beta-\alpha),\\
& (t-1)\,\mathsf{w}^2+\mathsf{q}+\theta_0^2+\theta_\infty^2 - t\bigl(\theta_0+\theta_1-\tfrac{1}{2}\bigr)^2 - \bigl(\theta_0+\theta_t-\tfrac{1}{2}\bigr)^2 = 0\,.
\end{aligned} θ 0 = 2 1 ( 1 − γ ) , θ 1 = 2 1 ( 1 − δ ) , θ t = 2 1 ( 1 − ϵ ) , θ ∞ = 2 1 ( β − α ) , ( t − 1 ) w 2 + q + θ 0 2 + θ ∞ 2 − t ( θ 0 + θ 1 − 2 1 ) 2 − ( θ 0 + θ t − 2 1 ) 2 = 0 . The four regular singular points and the corresponding exponents are represented by the Riemann scheme as
ψ = P { 0 1 t ∞ 1 2 − θ 0 1 2 − θ 1 1 2 − θ t 1 2 − θ ∞ ; z 1 2 + θ 0 1 2 + θ 1 1 2 + θ t 1 2 + θ ∞ } . \psi=P
\begin{Bmatrix}
0 & 1 & t & \infty & {}\\
\tfrac{1}{2}-\theta_0 & \tfrac{1}{2}-\theta_1 & \tfrac{1}{2}-\theta_t & \tfrac{1}{2}-\theta_\infty & ;z\\
\tfrac{1}{2}+\theta_0 & \tfrac{1}{2}+\theta_1 & \tfrac{1}{2}+\theta_t & \tfrac{1}{2}+\theta_\infty &
\end{Bmatrix}. ψ = P ⎩ ⎨ ⎧ 0 2 1 − θ 0 2 1 + θ 0 1 2 1 − θ 1 2 1 + θ 1 t 2 1 − θ t 2 1 + θ t ∞ 2 1 − θ ∞ 2 1 + θ ∞ ; z ⎭ ⎬ ⎫ . The two linearly independent solutions near z = 0 z=0 z = 0
ψ ± [ 0 ] = z 1 2 ∓ θ 0 ( 1 − z ) 1 2 − θ 1 ( 1 − z t ) 1 2 − θ t H ℓ ( t , 1 4 + ( 1 − t ) w 2 + t ( 1 2 ∓ θ 0 − θ 1 ) 2 + θ t 2 − θ ∞ 2 ∓ θ 0 ( 1 − 2 θ t ) − θ t ; 1 ∓ θ 0 − θ 1 − θ t − θ ∞ , 1 ∓ θ 0 − θ 1 − θ t + θ ∞ , 1 ∓ 2 θ 0 , 1 − 2 θ 1 , z ) . \begin{aligned}
\psi^{[0]}_\pm &= z^{\frac{1}{2}\mp\theta_0}(1-z)^{\frac{1}{2}-\theta_1}\left(1-\frac{z}{t}\right)^{\frac{1}{2}-\theta_t}\, H\!\ell\Bigl(t, \tfrac14 + (1-t)\mathsf{w}^2 + t\bigl(\tfrac12\mp\theta_0-\theta_1\bigr)^2 + \theta_t^2 - \theta_\infty^2\\
&\quad \mp \theta_0(1-2\theta_t) - \theta_t; \; 1\mp\theta_0-\theta_1-\theta_t-\theta_\infty,\, 1\mp\theta_0-\theta_1-\theta_t+\theta_\infty,\, 1\mp 2\theta_0,\, 1-2\theta_1,\, z\Bigr).
\end{aligned} ψ ± [ 0 ] = z 2 1 ∓ θ 0 ( 1 − z ) 2 1 − θ 1 ( 1 − t z ) 2 1 − θ t H ℓ ( t , 4 1 + ( 1 − t ) w 2 + t ( 2 1 ∓ θ 0 − θ 1 ) 2 + θ t 2 − θ ∞ 2 ∓ θ 0 ( 1 − 2 θ t ) − θ t ; 1 ∓ θ 0 − θ 1 − θ t − θ ∞ , 1 ∓ θ 0 − θ 1 − θ t + θ ∞ , 1 ∓ 2 θ 0 , 1 − 2 θ 1 , z ) . The two linearly independent solutions near z = 1 z=1 z = 1
ψ ± [ 1 ] = ( 1 − z ) 1 2 ∓ θ 1 z 1 2 − θ 0 ( t − z t − 1 ) 1 2 − θ t H ℓ ( 1 − t , 3 4 − ( 1 − t ) w 2 − t ( 1 2 − θ 0 ∓ θ 1 ) 2 + θ 0 2 + θ 1 2 ∓ 2 θ 1 ( 1 − θ 0 − θ t ) − θ 0 − θ t ; 1 − θ 0 ∓ θ 1 − θ t − θ ∞ , 1 − θ 0 ∓ θ 1 − θ t + θ ∞ , 1 ∓ 2 θ 1 , 1 − 2 θ 0 ; 1 − z ) . \begin{aligned}
\psi^{[1]}_\pm &= (1-z)^{\frac{1}{2}\mp\theta_1}\, z^{\frac{1}{2}-\theta_0}\,\Bigl(\frac{t-z}{t-1}\Bigr)^{\frac{1}{2}-\theta_t}\, H\!\ell\Bigl(1-t, \tfrac34 - (1-t)\mathsf{w}^2 - t\bigl(\tfrac12-\theta_0\mp\theta_1\bigr)^2 + \theta_0^2 + \theta_1^2\\
&\quad \mp 2\theta_1\bigl(1-\theta_0-\theta_t\bigr) - \theta_0 - \theta_t; \; 1-\theta_0\mp\theta_1-\theta_t-\theta_\infty,\, 1-\theta_0\mp\theta_1-\theta_t+\theta_\infty,\, 1\mp 2\theta_1,\, 1-2\theta_0; \; 1-z\Bigr).
\end{aligned} ψ ± [ 1 ] = ( 1 − z ) 2 1 ∓ θ 1 z 2 1 − θ 0 ( t − 1 t − z ) 2 1 − θ t H ℓ ( 1 − t , 4 3 − ( 1 − t ) w 2 − t ( 2 1 − θ 0 ∓ θ 1 ) 2 + θ 0 2 + θ 1 2 ∓ 2 θ 1 ( 1 − θ 0 − θ t ) − θ 0 − θ t ; 1 − θ 0 ∓ θ 1 − θ t − θ ∞ , 1 − θ 0 ∓ θ 1 − θ t + θ ∞ , 1 ∓ 2 θ 1 , 1 − 2 θ 0 ; 1 − z ) . What makes Heun problems hard is that, for generic parameters, the connection coefficients between local bases (e.g. the basis near z = 0 z=0 z = 0 z = 1 z=1 z = 1
[1] J. Ren and Z. Yu, Holographic thermal correlators from recursions ,
JHEP 06 (2025) 183
[arXiv:2412.02608 ].
