Lecturer: K.S. Narain , ICTP

This lesson starts with the study of the **hypergeometric differential equation**** **and its analytic solution** **the **hypergeometric function** F(a,b,c,z) , where a,b,c are free parameters and c **is not** an integer. The lesson focused first in the solution near of the **regular singular points** z_{0}=0, the general solution from previous lesson can be written as U(z)=z^{R}?C_{n}z^{n}. The analytic solution is when R=0 therefore F(a,b,c,z) = ?C_{n}z^{n}. By substituting this solution in the **hypergeometric differential equation **it can be obtain a recursive relation for the coefficients C_{n} and then to derive an equation for C_{n} as a function of C_{0 }and the **Gamma function **then a normalization is choused such that C_{0}=1 and the expression for F(a,b,c,z) is found. It was shown that this solution converge for |z|<1 and that F(a,b,c,z) =F(b,a,c,z) and that F(a,b,c,z) is analytically continuous in all the complex plane.

The second solution U_{2}(z) is supposed to be of the form U_{2}(z)=z^{(1-c)}g(z) substitute this guest in the **hypergeometric differential equation **and then obtain a new differential equation for g(z). After some transformations it was obtain that the differential equation for g(z) is also a **hypergeometric differential equation **with new parameters a',b',c' that are functions of the initial parameters a,b,c, therefore we have that g(z)= F(a',b',c',z) thus U_{2}(z)=z^{(1-c)}F(b-c+1,a-c+1,2-c,z).

The solution near the **regular singular points** z_{0}=1 is simpler to be found, actually by doing the change of variable z'=1-z we have that when z goes to one z' goes to zero. Doing this transformation we obtain again a **hypergeometric differential equation **and the solution** ** U_{1}(z)=F(a,b,a+b+1-c,1-z). The second solution can be also obtain by the same type of trick used before giving U_{2}(z)=z^{(c-a-b)}F(c-b,c-a,1+c-a-b,1-z).

Finally the solution near the **regular singular points** z_{0}=infinity was analyzed. Now you define z'=1/z and the solution U_{1}(z) is supposed to be of the form U_{1}(z)=z^{?}h(z) a quadratic equation is obtained for the possible ? and the roots are ?_{1}=a and ?_{2}=b. For ?_{1}=a the solution is U_{1}(z)=z^{a}F(a,a+1-c,a+1-b,1/z) and for ?_{2}=b it is obtained U_{2}(z)=z^{b}F(b,b+1-c,b+1-a,1/z). It was explained that even if there were obtained two solution for each **regular singular point** only two of them are linear independent the others 4 can be written as a linear combination of the two solution obtained for one of this tree **regular singular point.**

Written by Philippe Dennery,Andre Krzywicki, Chapter 1, 2, 3 and 4