Paper 1, Section II, E

Further Complex Methods | Part II, 2020

Use the change of variable z=sin2xz=\sin ^{2} x, to rewrite the equation

d2ydx2+k2y=0\frac{d^{2} y}{d x^{2}}+k^{2} y=0

where kk is a real non-zero number, as the hypergeometric equation

d2wdz2+(Cz+1+A+BCz1)dwdz+ABz(z1)w=0\frac{d^{2} w}{d z^{2}}+\left(\frac{C}{z}+\frac{1+A+B-C}{z-1}\right) \frac{d w}{d z}+\frac{A B}{z(z-1)} w=0

where y(x)=w(z)y(x)=w(z), and A,BA, B and CC should be determined explicitly.

(i) Show that ( $)\$) is a Papperitz equation, with 0,1 and \infty as its regular singular points. Hence, write the corresponding Papperitz symbol,

P{0100A1CCABB}P\left\{\begin{array}{cccc} 0 & 1 & \infty \\ 0 & 0 & A \\ 1-C & C-A-B & B \end{array}\right\}

in terms of kk.

(ii) By solving ( \dagger ) directly or otherwise, find the hypergeometric function F(A,B;C;z)F(A, B ; C ; z) that is the solution to ()(\ddagger) and is analytic at z=0z=0 corresponding to the exponent 0 at z=0z=0, and satisfies F(A,B;C;0)=1F(A, B ; C ; 0)=1; moreover, write it in terms of kk and

(iii) By performing a suitable exponential shifting find the second solution, independent of F(A,B;C;z)F(A, B ; C ; z), which corresponds to the exponent 1C1-C, and hence write F(1+k2,1k2;32;z)F\left(\frac{1+k}{2}, \frac{1-k}{2} ; \frac{3}{2} ; z\right) in terms of kk and xx.

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