K.S. Narain , ICTP
This lesson continues with the study of the hypergeometric differential equation. First it is shown that dF(a,b,c,z)/dz proportional to F(a+1,b+1,c+1,z). First it is shown that dF(a,b,c,z)/dz proportional to F(a+1,b+1,c+1,z) and in general we have that dFn(a,b,c,z)/dnz is proportional to F(a+n,b+n,c+n,z).
Then the hypergeometric function F(a,b,c,z) was calculated for the case b=c and a=-m were m is a positive integer. This case is important because they Jacobi polynomials . In fact if in the Jacobi differential equation the change of variable z’=(1-z)/2 is done we obtain an hypergeometric differential equation for z’ with solution for the Jacobi polynomials via the hypergeometric function is F(-?,?+?+?+1,?+1,(1-z)/2). The Legendre polynomials are obtained for ?=?=0.
In the second part of the lesson the Confluent hypergeometric functions
were introduced it starts by the general definition of hypergeometric differential equation given in lesson 30 and select the regular singular points z1=0, z2 and z3 then b is selected to be equal to b=z2 and b=2z3 and then take the limit b goes to infinity, the resultin equation is the confluent hypergeometric differential equation. This equation will have only two parameters a and c and two singularities at z=0 and z=infinity. The singularity at z=0 is a regular singular point but z=infinity is an irregular singular point.
The solution for the case of z=0 can be found using the method explained in the previous lessons. Here is used the method of finding a solution is expressed in terms of an integral like in the case of lesson number 32. Here the solution is guest to be U1(z)=IntegralC [Exp[zt]f(t)dt], this solution is substituted in the confluent hypergeometric differential equation and then the expression for f(t) is found to be f(t)=ta-1(t-1)c-a-1. So one solution is obtained when Re[c]>Re[a]>0 for this case the contour C is chosen to be from 0 to 1. This solution that is the analytic solution near z=0 is called Confluent hypergeometric function ?(a,c,z). It was demonstrated that this is just the limit when b goes to infinity of the hypergeometric function F(a,b,c,z/b). The guest for the second solution is suposed to be of the form U2(z)=(1-z)1-cV(z) and then V(z) is found to be V(z)=?(a+1-c,2-c,z).
Finally the behavior near the irregular singular point z=infinity was studied for this case the same integral solution is proposed ?(z)=K IntegralC [Exp[zt]f(t)dt] but now the contour C is taken from minus infinity to zero and K is a constant and to the integrand be zero in the boundary we have to impose Re[a]>0 and Re[z]>0. Then the behavior of U1(z), U2(z) and ?(z) at z=infinity was studied and it was found that U1(z) and U2(z) grows exponentially while ?(z) does not, so as ?(z) can be written as linear combination of U1(z) and U2(z) this must be in such a way that the exponential in z cancels.
Mathematics for Physicists (Text Book from Google e-books preview)Written by<span class="addmd"> Philippe Dennery,Andre Krzywicki, Chapter 1, 2, 3 and 4</span>
