Skip to content

Local solutions

The standard form of the hypergeometric equation is

z(1z)y(z)+(c(a+b+1)z)y(z)aby(z)=0.z(1-z)y''(z)+\left(c-(a+b+1)z\right)y'(z)-aby(z)=0\,.

The three regular singular points {0,1,}\{0,1,\infty\} and the corresponding exponents are represented by the Riemann scheme as

y=P{0100a;z1ccabb}.y=P \begin{Bmatrix} 0 & 1 & \infty & {}\\ 0 & 0 & a & ;z\\ 1-c & c-a-b & b & {} \end{Bmatrix}.

The two linearly independent solutions near z=0z=0 are

y1[0]=2F1(a,b;c;z),y2[0]=z1c2F1(ac+1,bc+1;2c;z).\begin{aligned} y^{[0]}_1 &={}_2F_1(a,b;c;z)\,,\\ y^{[0]}_2 &= z^{1-c}\,{}_2F_1(a - c + 1, b - c + 1; 2 - c; z)\,. \end{aligned}

The two linearly independent solutions near z=1z=1 are

y1[1]=2F1(a,b;a+bc+1;1z),y2[1]=zcab2F1(ca,cb;cab+1;1z).\begin{aligned} y^{[1]}_1 &={}_2F_1(a, b; a + b - c + 1; 1 - z)\,,\\ y^{[1]}_2 &= z^{c-a-b}\,{}_2F_1(c - a, c - b; c - a - b + 1; 1 - z)\,. \end{aligned}

The two sets of solutions are connected by

(y1[0]y2[0])=(Γ(c)Γ(cab)Γ(ca)Γ(cb)Γ(a+bc)Γ(c)Γ(a)Γ(b)Γ(2c)Γ(cab)Γ(1a)Γ(1b)Γ(2c)Γ(a+bc)Γ(ac+1)Γ(bc+1))(y1[1]y2[1]).\begin{pmatrix} y^{[0]}_1\\ y^{[0]}_2 \end{pmatrix}= \begin{pmatrix} \frac{\Gamma (c) \Gamma (c-a-b)}{\Gamma (c-a) \Gamma (c-b)} & \frac{\Gamma (a+b-c) \Gamma (c)}{\Gamma (a) \Gamma (b)} \\ \frac{\Gamma (2-c) \Gamma (c-a-b)}{\Gamma (1-a) \Gamma (1-b)} & \frac{\Gamma (2-c) \Gamma (a+b-c)}{\Gamma (a-c+1) \Gamma (b-c+1)} \end{pmatrix} \begin{pmatrix} y^{[1]}_1\\ y^{[1]}_2 \end{pmatrix}.

Boundary value problems between z=0z=0 and z=1z=1 can be analytically solved by means of the connection formula.

The normal form of the hypergeometric equation is

(d2dz2+14θ02z2+14θ12(z1)2+θ02+θ12θ214z(z1))ψ(z)=0.\left(\frac{d^2}{dz^2}+\frac{\frac14-\theta_0^2}{z^2}+\frac{\frac14-\theta_1^2}{(z-1)^2}+\frac{\theta_0^2+\theta_1^2-\theta_\infty^2-\tfrac14}{z(z-1)}\right)\psi(z)=0\,.

which is related to the standard form by ψ(z)=z1/2θ0(1z)1/2θ1y(z)\psi(z)=z^{1/2-\theta_0}(1-z)^{1/2-\theta_1}y(z) and

θ0=12(1c),θ1=12(cab),θ=12(ba).\theta_0=\frac{1}{2}(1-c),\qquad \theta_1=\frac{1}{2}(c-a-b),\qquad \theta_\infty=\frac{1}{2}(b-a)\,.

The three regular singular points and the corresponding exponents are represented by the Riemann scheme as

ψ=P{0112θ012θ112θ;z12+θ012+θ112+θ}.\psi=P \begin{Bmatrix} 0 & 1 & \infty & {}\\ \frac{1}{2}-\theta_0 & \frac{1}{2}-\theta_1 & \frac{1}{2}-\theta_\infty & ;z\\ \frac{1}{2}+\theta_0 & \frac{1}{2}+\theta_1 & \frac{1}{2}+\theta_\infty & {} \end{Bmatrix}.

The two linearly independent solutions near z=0z=0 are

ψ±[0]=z1/2θ0(1z)12θ12F1(12θ0θ1θ,12θ0θ1+θ;12θ0;z).\psi^{[0]}_{\pm}=z^{1/2\mp\theta_0}(1-z)^{\frac12-\theta_1}\,{}_2F_1(\tfrac12\mp\theta_0-\theta_1-\theta_\infty,\tfrac12\mp\theta_0-\theta_1+\theta_\infty;1\mp 2\theta_0;z)\,.

The two linearly independent solutions near z=1z=1 are

ψ±[1]=(1z)12θ1z1/2θ02F1(12θ0θ1θ,12θ0θ1+θ;12θ1;z).\psi^{[1]}_{\pm}=(1-z)^{\frac12\mp\theta_1}z^{1/2-\theta_0}\,{}_2F_1(\tfrac12-\theta_0\mp\theta_1-\theta_\infty,\tfrac12-\theta_0\mp\theta_1+\theta_\infty;1\mp 2\theta_1;z)\,.

The connection formula can be written as

ψϵ[0]=ϵChyp(ϵθ0,ϵθ1,θ)ψϵ[1],ϵ,ϵ=±,\psi^{[0]}_{\epsilon}=\sum_{\epsilon'}\mathsf{C}_{\text{hyp}}(\epsilon\theta_0,\epsilon'\theta_1,\theta_\infty)\psi^{[1]}_{\epsilon'},\qquad \epsilon,\epsilon'=\pm\,,

where

Chyp(θ0,θ1,θ)=Γ(12θ0)Γ(2θ1)Γ(12θ0+θ1+θ)Γ(12θ0+θ1θ).\mathsf{C}_{\text{hyp}}(\theta_0,\theta_1,\theta_\infty)=\frac{\Gamma(1-2\theta_0)\Gamma(2\theta_1)}{\Gamma(\frac{1}{2}-\theta_0+\theta_1+\theta_\infty)\Gamma(\frac{1}{2}-\theta_0+\theta_1-\theta_\infty)}\,.

[1] J. Ren and Z. Yu, Holographic thermal correlators from recursions, JHEP 06 (2025) 183 [arXiv:2412.02608].