# Appell hypergeometric identity

In this paper http://www.sciencedirect.com/science/article/pii/S0022247X09000997 the authors conclude that the Appell hypergeometric function $$F_4\left(\begin{matrix}a; b \\ c, a+b-c+1\end{matrix} \bigg| \, \, t^2, (1-t)^2 \right)$$ and the hypergeometric function $${}_2 F_1\left(\begin{matrix}a; b \\ c \end{matrix}\bigg| \, \, t \right)^2$$ satisfy the same differential equation. Does this mean they are equal? What good is this information for otherwise?

• Broadly speaking, two solutions of a same differential equation (and sharing the same definition domain) are equal if they verify a same set of initial conditions. – Jean Marie May 29 '17 at 22:36
• Related : (math.stackexchange.com/q/812748) – Jean Marie May 29 '17 at 22:38
• Thank you very much for your answer! So, just to see if I understood correctly: in my case, the solution of my differential equation gave me the Appell hypergeometric function, with $t$ between 0 and 1. I would very much like it to be equal to the Gauss hypergeometric squared, so in order to check if it is I must see if the values for t=0 and t=1 are the same? – GKiu May 30 '17 at 10:01
• The relevant differential equation is given in Theorem 6.1. As its order is 3, it seems that one should check 3 initial conditions to show the equality. – Start wearing purple May 30 '17 at 10:14
• No, what you are mentionning are called Neumann conditions (en.wikipedia.org/wiki/Neumann_boundary_condition). It's not the same. Same initial conditions means 3 constraints: $f(t)=g(t)$, $f'(t)=g'(t)$,$f''(t)=g''(t)$, say, in $t=0$ (for a second order differential equation). But it may be slightly more intricated than that. – Jean Marie May 30 '17 at 10:48