### Mathematical Methods Lecture 32 of 34

December 12, 2011 by K.S. Narain

K.S. Narain , ICTP

This lesson starts recalling the results obtained in the previous lesson for the hypergeometric differential equation. <span class="toctext"><span class="toctext">In particular it was emphasized that only two solutions are linear independent and that the others 4 can be written as a linear combination of the two solution obtained for one of this tree regular singular point. But near the <span class="toctext"><span class="toctext">regular singular point</span></span> is more convenient to use the solution obtained for it. </span></span>

<span class="toctext"><span class="toctext"> In this lesson it is shown that the previous statement is actually true and how one can obtain the behavior of the solution near the <span class="toctext"><span class="toctext">regular singular point </span></span></span></span>z0=1 by using a linear combination of the solutions obtained for the <span class="toctext"><span class="toctext"> <span class="toctext"><span class="toctext">regular singular point </span></span></span></span>z0=0 this exercise is a good example of the application of the Stirling's approximation because the large n behavior is studied for the coefficients Cn of the solution near <span class="toctext"><span class="toctext"></span></span>z0=0, as well as the Taylor expansion of the term (1-z)-? present in the solution for z0=1. It was demonstrated that for large n Cn behave like (1-z)-? for ?=a+b-c.

Some time for practical applications it is useful to have the solutions U(z) of the hypergeometric differential equation in form of integral functions. The guest solution is of the form U(z)=IntgralC[(z-t)?f(t)dt] where C is the contour of integration, this equation was substituted into the hypergeometric differential equation by doing this and imposing a condition we have that ? can take just two values ?=a and ?=b. Then the idea is to write the integrand as a total derivative on dt this implies that if an appropiete contour is chousen the integral could vanish. By doing this and assuming Re[c]>Re[b]>0 it is found that the solution can be written as U(z)=IntgralC[(t-z)-af(t)dt] where f(t)=ta-c(1-t)c-b-1. Thus it is obtained that it is possible with the same integrand to obtain all the solutions obtained in the previous lesson the different contour of integrations has to be chosen. As an example of how this result can be applied it was calculated the value of F(a,b,c,z=1). This integral representation of the solution is also useful to calculate the linear combination of one solution in terms of the odder two solutions selected as a base<span class="toctext"><span class="toctext"></span></span>.

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>

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