Mathematical methods Lecture 31 of 34

November 14, 2012 by Multimedia Publications and Printing Services

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 z0=0, the general solution from previous lesson can be written as U(z)=zR?Cnzn. The analytic solution is when R=0 therefore F(a,b,c,z) = ?Cnzn. By substituting this solution in the hypergeometric differential equation it can be obtain a recursive relation for the coefficients Cn and then to derive an equation for Cn as a function of Cand the Gamma function then a normalization is choused such that C0=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 U2(z) is supposed to be of the form  U2(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  U2(z)=z(1-c)F(b-c+1,a-c+1,2-c,z).

The solution near the regular singular points z0=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  U1(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 U2(z)=z(c-a-b)F(c-b,c-a,1+c-a-b,1-z).

Finally the solution near the regular singular points z0=infinity was analyzed. Now you define z'=1/z and the solution U1(z) is supposed to be of the form U1(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 U1(z)=zaF(a,a+1-c,a+1-b,1/z) and for ?2=b it is obtained U2(z)=zbF(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. 

Mathematics for Physicists (Text Book from Google e-books preview)

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

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