260 PROBLEMS AND SOLUTIONS

GANAPOL (Centre d’Etudes Nucleaires de Saclay, France), M. L. GLASSER(Battelle Institute, Ohio), R. N. HILL (University of Delaware), A. A. JAGERS(Technische Hogeschool Twente, Enschede, the Netherlands), W. B. JORDAN(Scotia, N.Y.), A. N. OTHER (Middlesex, England), O. G. RUEHR (MichiganTechnological University), F. W. STEUTEL (Technological University, Eindhoven,the Netherlands), A. J. STRECOK (Argonne National Laboratory) and W. E.WILLIAMS (University of Surrey, Surrey, England).

Problem 73-10", A Summation Identity, by H. M. SRIVASTAVA (University ofVictoria, Victoria, British Columbia, Canada).

Let

F(t) 3 -+- at2 + bt + c

and

G(p,q)=(2p+ q+a- 1)+(2p + q + a + 1)F(p)

F(-p q- a)

(2p + q + a + 3)V(p)F(p + 1)4- 4-...,F(-p q- a)F(-p q a- 1)

where a, b, c, p and q are constants. Prove or disprove that

G(p, q) G(q, p).

Solution by Oxxo G. RUEHR (Michigan Technological University).

Denote the zeros of F by x, y and z. In hypergeometric notation, we have

p--x

G(p, q) A

Ap-y, p-z, +-;,

z

r+x, r+y, r+ z,

1;

where A=2p+ q+ a- and r=p+ q+ a. In order to simplify the ex-pression for G and display the symmetry, we employ the following result[1, p. 28]:

AA, 1+-, B, C, D, E;

6F5A-, +A-B, +A-C, +A-D, +A-E;

F(1 + A D)F(1 + A E)r(1 + A)F(1 + A D 3F2

D, E;

I+A-B, I+A-C;

Making thesubstitutionsA 2p + q + a 1,B p x,C p y,D p z

Dow

nloa

ded

11/1

9/14

to 1

29.4

9.23

.145

. Red

istr

ibut

ion

subj

ect t

o SI

AM

lice

nse

or c

opyr

ight

; see

http

://w

ww

.sia

m.o

rg/jo

urna

ls/o

jsa.

php

PROBLEMS AND SOLUTIONS 261

and E 1, we obtain

G(p, q) (p + q + a + z- 1) 3F2[q z,

from which the symmetry is obvious.

p-z 1"

p+q+a+x, p+q+a+y;

REFERENCE

[1] W. N. BAILEY, Generalized Hypergeometric Series, Stechert-Hafner, New York and London, 1964.

Also solved by O. P. LOSSERS (Technological University, Eindhoven, theNetherlands).

Problem 73-12, A Nonlinear Differential Equation, by OTTO G. RUEHR (MichiganTechnological University).

Determine the general solution of

x2 + 2 f= -x + 2x -d-x + x

The problem arose in modeling the Helmholtz equation in two dimensions.

Solution by the proposer.

Assuming that f’ + 2x 0, we have

f"+2f’+2x

Integration yields

and hence

f’ X

log If’ + 2xl log Ifl + log Icl + dx,

f’+ =cexp dx

Letting x/f= u"/u’ and integrating, we obtain

f(u’)z c2 exp {cu}.Substitutingf= xu’/u" and noting the identity

(d2u/ dx2)/(du/dx)3 d2x/du2,

we obtain the linear differential equation

d2x exp {-cu}du2 - c 2

x--O.

Let

=t+c,Dow

nloa

ded

11/1

9/14

to 1

29.4

9.23

.145

. Red

istr

ibut

ion

subj

ect t

o SI

AM

lice

nse

or c

opyr

ight

; see

http

://w

ww

.sia

m.o

rg/jo

urna

ls/o

jsa.

php