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A.R.C.20,793

F.M.2786 !LUrD MOTION SUB-COMMITTEE

AERONAUTICAL RESEARCH COUNCIL

Proof of a Conjecture of Spenoe- By -

lVI. J. Lighthill

A.R.C.20,793

F.M.2786

i ST. NO·SR b~ ~f U.D.C.1!'r....-f·~f I J In.~,~ o,*"""w _,~

23rd Feb~uary, 1959

1By use of a physioal argument, Spence has conjectured that,if a function f satisfies

f(x) = ~! 00 (~ )t ~~~~~~~, f(O)71: t t-x

o

=fl (x)--1"-- dx =x2

- A, •.• (1)

then the value of A is ~7I:.

To prove the conjecture, we substitute t = xu in theintegro-differential equation, giving

f(x) = ••. (2)

a form which suggests the use of Mellin transforms. Suoh transformssatisfy

F(s) ! 00 1= xs- f(x)dx, f(x)

o

=1

!c+ioo

x -sF(s )ds,

c-ioo

for some real c.

Either of equations (3) can be used to determine the equationwhich the Mellin transform F(s) of the solution of (1) must satisfy.We use the latter because it gives a sufficient condition easily. Forit implies that

f

fl (x) __ __1_ J c+ico (-sx-s - 1 )F(s)ds =271:i .

C-J.c:o

1 ! c-1+ioo 1- --- sF(s)x-s - ds,.(4)

271:i l'c- -J..oo

provided that sF(s) is without singularities in the region c - 1 ~ Rs ~ c.In this case, substituting 8 = S - 1 and then dropping the bar,

ft (x) =1 J c+ioo

- --- (s -2?Ci .

C-J..oo

The right-hand side of equation (2) therefore becomes

~ J 00 -I-~:'--- (- -~-) Jc+ioo (8 - 1 )F(s - 1 )(xu)-Sds.71: u2 (u - 1) 271:i .

o c-J..oo

But/tA.3' REPORT

• •• (6)

-- 2 -

But if c = Rs satisfies Ie I < ~ we have

the integral bein? absolutely convergent at both limits. In this casethe expression (6) becomes

1 J c+ioo--- (tan ~s)(s - 1)F(s - 1)x-S ds, ••• (8)2~i .

C-J..oo

which ~nll be equal to the left-hand side of (2) if

F(s) = (tan ~s)(s - 1)F(s - 1).

The functional equation (9) is simplified by writing

• •• (9)

F(s)

when it becomes

(s - 1) w( s ) , ••• (10)

In terms of

G( s )

G(s), f(x) is

= (tan ~s)G(s - 1). • .. (11 )

f(x) =_1__ ! c+ioo

(s - 1)!G(s)x-sds.2'lCi .

c-J..oo

•.. (12)

To sum uPi (12) is a solution of (2) provided that (11) issatisfied, that - 2 < c < t and that sF(s) = s!G(s) has nosingularities for c - 1 ~ Rs ~ c. To secure also that f(O) = 1,we take 0 < c < t and

G( 0) = 1.

Then the integrand in (12) has residue 1 at its pole s = 0, andtherefore

• •• (13)

f(x) =1 ! c-1+ioo

1 + --- (s - 1)!G(s)x-s

ds2~i l'c- -J..oo

=

as x ~ O. Furthermore, from the result (5), that the Mellin transformof flex) is -(s - 1)F(s - 1), together with the reciprocal relationships(3) for Mellin transforms, we have

A = -! oox-~ f(x)dx =o

- [- (s - 1)F(s - 1)] 1S=-2

= - tF (-~) = G(-~) ~~ . • •• (15)

We now determine G(s) by deciding what poles and zeros itmust have, and hence evaluate (15). Equation (13), together with multipleapplication of (11), shows that as s ~ n (a positive or negative integer)

••• (16)

sol

-- 3 -

so that at s = n the functionn > 0, and a pole of order I n Ishows that as s ~ n - t

G(s) has a zero of order n ifif n < O. Similarly, equation (15)

G(s) ••• (17)

so that G(s)zero of order

has a pole of orderIn I if n < O.

n at s = n - t if n > 0 and a

These facts enable us to write down at once an Hadamardinfinite-product expression for G(s), namely

••• (18)G(s) =

I '""n

(-I t-~_~2~~_:_~~~~~2lj n-1 I i( s )( S) I

- : 1 + - 1 - ----~ It n n- 2 .J

where equation (13) has been used to exclude the need for a constantfactor outside the product. The product converges without the need foradding any exponential faotors, since it can be written

r fni S2 + ts I

00 '1 _ ------- I---I' 2 1

G( s ) = ~ ----~--~-:~ > 'I: S2 + ts I

n=1 11 - -;---~-Jl n-2TI

whence the nth term, as n ~ 00, is

= ••• (20)

It may be immediately verified that this G(s) does satisf~(11), since division of the infinite products for G(s) and G(s - 1)gives

••• (21 )tan 7t:s.G(s)

--------G(s - 1) =~ (:~:-:-=)~5(;-=-:-:~~ =

n=1

In addition, no other solution for G(s) appears to be possible,since it would have to be equal to this one times a periodic function ofperiod 1, whose singularities would be spaced periodically (unlike thoseof G(s), which have a gap from s = - 1 to s = + t) and thereforeprevent satisfaction of the condition that s~G(a) is regular forc - 1 < Rs ~ c.

We conclude,therefore, that the solution of Spencets problem canbe written in the form (12), with G(s) given by (19). But this gives atonce G(-t) = 1, from which, by (15), there follows immediately Spence'sconjecture A = ~7t:.

Reference!

No.

1

SB

Author

D. A. Spence

- 4- -

Reference

Title"etc.

A note on the lincarised theory ofthe two-dimensional jet flap.R.A.E. Technical Memorandum Acro.627.A.R.C.20,753 - F.M.2778.J'anuary, 1959.