the zeros of the hankel function

Numerische Mathematik 2, 228--244 (1960)

The Zeros o f the Hankel Funct ion as a Funct ion o f its Order*

By

WILHELM MAGNUS and LEON KOTIN

1. Introduction

The theory of diffraction of electromagnetic waves by a sphere requires a knowledge of the zeros of certain transcendental functions.

In the simplest case of a perfectly conducting sphere of radius a, the equation in question is H l(ka) = 0, where H 1 is the Hankel function of the first kind of order v, and k is the wave number of the incident wave. The quant i ty to be determined is the order v, with ka being regarded as a parameter. Special attent ion is paid to the case in which z=ka is real.

We evaluate v only for sufficiently large values. For other values, we in- vest igate the behavior of v as a function of the argument. Specifically, we derive certain inequalit ies satisf ied by the real and imaginary parts of v and the argument z; we investigate the behavior of v as z--~0; and we obtain an expression

for g~. dz

Although many of the results can be extended, the argument z of H i will

be restr icted for the most part to the right half plane; i.e., ]argz]

The Zeros of the Hankel Function 229

order of an entire function to prove that there exists an infinite number of zeros, and also to obtain an infinite product representation.

From HEINE'S formula (see [1], p. 26), for 0< arg z< z~,

oo

g(v) ~ ~2 e("+l)~ii2 Hl'(z) = f e~:c~176 0

I f we let Ivl =~, oo

I g (~')1

230 WILHELM MAGNUS and LEON I~OTIN:

We can extend this result to the real axis by applying the same technique to the function h(v) defined below. When x>0, from NICHOLSON'S formula (U], P. 31),

(2.~) h(,) = v =' ~ (~) m (~) = f ~o(2z sinh t) cosh 2v t dt o

where K o is the modified Hankel function of order zero. Then if we put Iv] = ~, OO

lh(v)[ < f [Ko(2zsinht)I e2O' dt. o

Now for any v and z for which x>0 ([3], p. 181), OO

(2.2) K~ (z) = f e -~ ~o~h, cosh v s ds. 0

Therefore,

Thus

O0 CO

IKo(2z sinh t)] ~ f e--2*~h~~ ds


for x>O, where the vk are now the zeros of 1 H.H,. Moreover, equating coefficients of v 2, we have

oo

2f t 2 K 0 (2z sinh t) dt I _ h" (o ) _ 0

oo

Z v~ 2h(0) f Ko(2zsinht ) dt 0

For z real and positive, if v (i) is a zero of H{, we have @)=v~ 2). Hence

and

Theorem 2.2.

k

oo

f t2Ko(2xsinht) dt o

oo

f Ko(2xsinht ) dt 0

To show that H~ has only simple zeros, we require the following well-known

Lemma 2.1. Let F(t) be positive, continuous, and decreasing for t>0. Then oo

f F(t) sin t dt > O. 0

Pro@

f F(t) sin t dt = sin t dt + f F(t) sin t dt , 0 n=0L 2n~ (2n+l)r~

By the first mean value theorem,

(2 n+l ) ~ (2 n+ 2)

f F(t) s intdt+ f F(t)sintdt-~2F(t')--2F(t"), 2n~ (2n+l )~

for some t', t" such that 2n;z=O, p(t )>0, p ' ( t )o, t-~q-l(x)=t(x). Then OO OO

fp( t ) sinq(t ) dt = /" pit(x)] sin xdx. J q'[t(x)]

0 0

Applying Lemma 2.t to the function F(x)= P- completes the proof. q' Now we can prove the simplicity of the zeros. For if we let H~ (z)= 0 where

x>0, from a formula of WATSON (see [1], p. 30)

(20

H~(z) oI4~(z)ov 8j~ f Ko(2zsinht)e_~ dt. 0

17"

232 WILHELM MAGNUS add LEON KOTIN:

Since H~ (z) =~0, using (2.2) it suffices to show that

0 (see Theorems 3.3 and 4A), the integrand, as a function of t, satisfies the requirements

of Lemma 2.2 and is therefore positive for each s. Consequently OHm(z) ~0 Ov and we have

Theorem 2.3. The zeros of H}(z) are simple when x>0 and y>--0. We note that the proof is valid for the general cylindrical function of the

form aJ~ (z)+ b Y~ (z) in which the coefficients are independent of v.

3. Zeros of H~(z) for a fixed, complex

In this section we shall derive several inequalities relating the zeros v and the complex arguments z. For the most part, z will be restricted to the upper half plane. If we include the positive real axis this corresponds to the actual physical situation, since the wave number, which is essentially z, lies in the first quadrant.

Theorem 3.1. Given f l~0 , there exists a positive y such that H~(iy)~O. Proo/. For b real, we have (see Eli, p. 34)

oo

f e-t cosh b dt 2i e ~ ~I2 sin (fl b) HIS (i t) sinh fl ~sinh b "

0

Then the integral vanishes. Since Choose/3 = b"

it is pure imaginary, and must vanish for some t> O.

In this case we can show that ]fi] >y by means of the following classical argument. The function H~ (ze t) =u (t) +iv (t) satisfies the equation

a2 H~ (~ ~') + (~ ~' -- ~) H~ (~ ~') = 0 dl2

If we multiply by H~ (ze% separate the real and imaginary parts, and integrate, we obtain for y>O

oo

(3.1) f (~ y ~ ' - ~t~) (~+ ~) dt =o to

where t o is a root of u(t)+iv(t) and OO OO

(3.2) f[(x2--y2) e~t--(o~--fln)l(u2+v2)dt=f(u'2+v'e)dt>O. to to


We now let z=v. Then equation (3.t) becomes 0o

/3f _ t) (u2 + dt = 0. to

If to_>0, then we must have ef t=0. But then ~=0 s ince /3=y>0, and (3.2) becomes

oo

f (1 - (,3 + d, > o. to

This, however, is impossible if t0>0. Therefore, t o must be negative. We have thus obtained the following result.

Theorem 3.2. If H ) (av)=0 for a>0 and /3>0, then ao, then I/3I >y . We can exploit equation (3.t) somewhat further. For instance let x>0,

and recall that y>0. If t is sufficiently large, xye2t--o~/3>O. Then for the integral in (3.1) to vanish, since the above function is monotonic in t, we must have xye~to--~/30 are pure imaginary. If we now consider equation (3.2) and assume x 2 -- y=< 0, the previous reason-

ing indicates that (x 2- y2)e2t.--((X2--/32)>0. This is the content of Theorem 3.4. If H}(z)=0 with x2O, then o:2--/32

234 WILHELM MAGNUS and LEON KOTIN:

The derivation of this formula for real values of v is reproduced in WATSON (E3~, pp. 21t--212). This can be extended to complex values by only slight changes. Without going through this lengthy analysis, we merely state the following inequality for R. If 21zl=2r>=~-- 89 , fl~O, and x>0, then

= 2r - t 2 r ] / ' (v+ 89 "

But it can easily be shown that when (~+ t) (~ -- 89 89 fl=>0, and x>0, then

2lz [

The Zeros of the Hankel Function 2}5

or eo

x

from the behavior of the Hankel functions at infinity, and the fact that

H~ (0 =H~ (0. Since the last integral converges, we have

Theorem 4.1. If H~(x)=0 for x>0, then ~f l>0. As far as ~ is concerned, this tells us only that l~l> 0. We can improve this

inequality, but first we require the following well-known extension of Lemma 2.]. co

Lemma 4.2. Let f / ( t ) cos bt dt exist for b~0, where/(t ) is twice differenri- 0 co

able for t>0 with ['(t)


Then

(4A)

oo

cos g 7~ K~:(2xs inht) cos2f l td t=~ 0

oo

f ~' (4.2) sinc~z K~(2xs inht )s in2 f l td t=~6 (IH~l~e~=--IHil2e-~" ) . 0

After subtracting we obtain

co

=' f K2~ (2 x sinh t) cos (2fl t + ~ 7c) dr. 8 IHi(x)l~e-a~ = 0

After dividing both sides by cos ~, integrating by parts, and applying equation (2.2) for K2~ we find that when x>f l tan~+~, Lemma2A is applicable.

We can close the ~ interval under consideration in the following manner. If we let v = 89 + ifl, then

4 9 1 2 - H=_~ H~) = i(Hi H~ -- H*~ H~) = IHil ~ e-as + IH~I ~ ~ - - * (H~ H~- I 1 2 ~t

where the argument of both Hankel functions is t. This is as easily proved for ~----- 89 and incidentally provides an extension of (4.1). Now if we divide both sides by t and integrate from x to o% applying Lemma 4.1 we find

co

x

whence

1 Then ]f l )> x . Theorem 4.4. Let Hi (x )=0 with x>O, a= ~: ~ . 2

If we apply Lemma 2A to equation (4.2), we obtain

Theorem 4.5. I f x>0 and ]=[< 89 then

sgn [[H~ (x) I e ~= -- ]H 1 (x)]] = sgn ~ ft.

5. An application of HURWITZ'S Theorem

We can extend the resu]ts of the preceding section by using the following theorem due to HURWlTZ (I2], p. tt9).

Theorem 5.1. Let / , (z ) be a ~equence of analytic functions which converge uniformly to /(z) ~ 0. Then z o is a zero of /(z) if and only if it is a limit point of the set of zeros of ]n(z).

In particular, if ],(z) is the partial sum of a power series, we can use the following result due to CAUCHY to get a bound on the zeros of the limit.

Lemma 5.1. Let /~(z )~ao+alz+a2z2+. . .+a ,z"=O with ao~0. Then

[ ~ iI -~ I~ l~ ~+~maxl~; 9

The Zeros of the Hankel Function 23 7

Now for x > 0,

where

oo

-g-=~ H ~, ( x) tt~ ( x) = f Ko ( 2 x sinh t) cosh 2 v t d t 0

~, an~ 2n n=O

oo

' f< ~ a, (2 n) ! 2 t) (2 x sinh l) d t o

after reversing the order of summation and integration. Applying HURWITZ'S Theorem, we obtain

Lemma 5.2. If H~(x)=0 and x>0, then

and

We note that

1 max a,,] -~ . [~1 > t+ ~o

oo

a,~ < f K o (2 x sinh t) cosh 2 t dt 0

__ 2~ 2

s [HI(*)I~

yg$

ao = ~ l~0 ~ (x)l ~

This results in the following simpler inequality.

Theorem 5.2. If H~(x)=0 and x>O, then

[ tti (x) 3]-~ i~l >t l+ Ha*i- ] '

which implies

I~1 > 1+ ~ and, if x > {,

[ i~ ' l>t t+ s . -~ ] "

The latter inequalities are obtained from an application of the asymptotic formula (3.3) and (3.4).

These inequalities can be improved at the expense of the range of x.

Theorem 5.3. Let H~(x)=0. If x=>0.6312 then

2 --,}

xiHo /x>l ] .Proo]. I t can be shown that arc sinh s=0. Then

oo

~"" f s~""Ko(2Xs)ds. 0


According to [3], p. 388, this is

__ 2 2n-2 /,2 (k n + fl_) = bn" (2n) l x 2k~+l

Forming b~+l/b n we find from the fact that F(x) increases for x~t .5 that b,+:0.5. For n=0, b:>b o if xe~=>0.606 . . . . 2F~(k+{)jm Thus a~O, co

igl(x)[~ = s f Ko(2xsinht ) cosh 2tdt 0

co

> -:~ fK~ c~ 0

2 - - :7"~X

6. Behav ior of the zeros as x -~ 0

The inequalities [ e [ > x -V~//2 and x


Since~--~0, IF(v)l ,~ [F(-v)I and we find that lim ~log ~- + =0. It remains

to show that fl-+0, for then F ( f )~ 1 and the proof would be complete. This is shown in *'

Theorem 6.2. Let H~(x)=0 with x---~0+. Then v--~0. Moreover for some integer k, fl log x + k ~ = o (v x).

Proo/. We know that ~--~0. Suppose fl-r Then since fl decreases with x (Corollary 7A), fl approaches a limit flo which we may assume positive. Applying

(~)~ ,-~e -~12, from the proof of the preceding theorem, to (6.2), we obtain

ea'~/~I~(ifl~ [c~ flol~ 2) - is in( fl~176 2)] +

+e~~176 [c~ flol~ 2) + i sin (fl~ l~ =o(x) .

Choose a sequence of x's converging to zero such that sin (rio log -2-) = 0 for each x. k

Then ~(iflo ) +F( - - i f l o ) =0. Similarly choose another sequence of x's such that

c~176176 =0. Then F( i f lo ) - -F ( - - i f l o )=O. These results are incompatible; t

thus fl0=O.

(2) Therefore from Theorem6.1 we have Also I"(v) N v . from (6.2)

= :

= -- ~2: sin (fl log 2)

$'

for some integer k, depending on which zero v we take. Then/~ log x + k 7r--~ o (vk).

7. The derivative of v In this section, we shall derive a formula which was first obtained by ~VATSON

(see [3J, p. 508), oo dc= f dv 2c Ko(2cs inht ) e-2~tdt ,

0

where v is real and c is a positive zero (in z) of the cylindrical function ~(z) coss+Y~(z)sins for constant s. This form immediately excludes H i. We shall generalize it to apply to complex values of z and v, as well as to any function of the form C, (z) = clJ~ (z) + c 2 Y, (z), where q and c 2 are independent of both z and r.

All the results of this section will be corollaries of Theorem 7.1. If C, (z) is of the above form, with x> 0, then any zero v satisfies

dv [ 'K (2zsinht) e-2~tdt = t (7A) 2z ~. ] o 9 0


Proo/. Let B, (z) = bl J~ (z) + b 2 Y~ (z) be a function of the same type as C, (z) and linearly independent of it. Then

6 = b 1 c 2 - - b2 c l =4= 0 .

The Wronskians of these functions with respect to v and z are respectively

B ec , C~ aB, _ ~ { , ~Y,, aj~ " av - - av \J" Tv-- - - Y" av ) oo

4af - - 7~ K 0 (2zsinh t) e-2~r 0

([1], p. 30), and

If C, (z) = 0, then

dC, __ C~ dBv 26 B~ dz dz -- :~z

(7.2)

and

(7.3)

B dC,. dv aC~ v dz +B~ - -0 , dz ~v

dd~ 4=0. Each term can now be replaced by one of the above Wronskians, whence

giving us the theorem.

We shall restrict C~ to H~, although many of the following results are valid for the general cylindrical function.

Corol lary 7.1. Let H~ (x )= 0 for a and x positive. Then ~'(x)> 0 when a~ or x>=~; and f l '(x)>O.

t~roo/. Solving (7.t) for dd~ and separating real and imaginary parts, we find

oo

dOt- = 22 -dv i2 f Ko(2xs inht ) e -2~'cos2t3tdt dx ax I d

0

oo

d3 dv 2 ~.. =2x ~ f Ko(2xsinht ) e-2~s in2f l td t . 0

I f we express the kernel as

/(t) =- K0(2x sinh t ) e -2~t oo

(7.4) = fexp I - - 2x sinh t cosh s -- 2~ tJ ds, 0

the corollary thus follows from an application of Lemmas 4.2 and 2.1 respectively.

Moreover, with ~>0, we have from (7.t) and (2.1)

oo

t

The Zeros of the Hankel Function 24t

Corollary 7.2. If H i (x) = 0 with x>O, then

which implies [/(x)[ > 2

We note from the first inequality that l im[/ (x) l =o0. X-+0

A similar inequality is obtained from (7.2) and (7.3). For from these,

[~'l = 2xlv'12 f Ko(2xt) e -~t dt 0

~-- -1 arccosh~x if e>x

o;~ ~-89 = lv ' l 2 t - x2} arc cOS-x, if ~ x.

Corol lary 7.3. If Hf f (x)=0 with x>0, then

dv _> sin(h),2y

If we consider the k-th zero ~, and let k--~ 0% since ~k-+ o0 (see section 8) we find that lim [v~[ = o~.

k--+ oo

These conditions on [v'l give us no picture of the trace of the zeros in the v-plane as x increases. We can remedy this by considering/5 as a function of

a~ and determining the behavior of d~ "

Corol lary 7.4. Let H~(x)=0 with x>0 and ~>{. Then

a~ < 2~

-as = Vso;- I

Proo/. From (7.2) to (7.4) ~'/(t) si~ 2~tdt dfl =

do: oo y /(t) cos 2fl t dt 0

dfl where ~>{. After integrating by parts, we find that do; V ~- 1"

We note that this value exceeds the preceding one, and applying Lemma 2A we d~ find that d~ < 2/3 ~. This implies the corollary.

242 WILHELM MAaNVS and LEON I{OTIN:

These results have so far been restricted to real values x. For complex values z, we get the following analogue of Corollary 7.t.

Corollary 7.5. Let H~(z)=0 with x, y, 0~, fl positive. Then

x~x -- y~x >0 when c~>~;

and am x~+y~>0.

Proo/. From (2.2) for x>0 oo oooo ooo0

f K 0 (2 z sinh t) e -~t dt = f f exp [2 z sinh t cosh s -- 2 v t_] ds dt = f f e- P - ' r d s dt 0 0 0 0 0

where p (t) = 2x sinh t cosh s + 2~ t

and q (t) = 2 y sinh t cosh s + 2fl t.

Substituting this in (7.1) and separating real and imaginary parts, we get

oooo

xo~ x - yfl, = 2lzv'12 f f e -Pcosqdtds 0 0

and oooo

xfl. + y~. = 2lzv'12 f f e-Psinqdtds 0 0

where we changed the order of integration. An application of Lemmas 4.2 and 2.t to the integrands then yields the result.

These inequalities can be expressed more simply in terms of the partial derivatives with respect to the polar coordinates r and ~0 of the z-plane, if we make use of the Cauchy-Riemann equations for the analytic function v (z).

Corollary 7.6. Let H i (z) = 0 with x, y, a, fl positive. Then ~cc < 0 and off > O. ~ ~r

t then ~ ~fl If in addition ~ >-- -- 2 ' -br > 0 and ~ > 0.

8. Behavior of the large zeros

In this section we shall approximate the large zeros by considering the behavior of H i (z) for ]vl large compared to I and ]z].

The first Hankel function may be defined by

Z e - i~

Hi (z) ---- ]-~ ( ) - J~ (z) i sin v It follows that

where

D(v) = v~ (--t)kz*kk! k : l

and

-- ivs inv~H~(2z) = A(v) + D(v)

Z--v e -- i ~ r~ zv

A(v)-- F(--v) + F(v)

Z -v e-- i v~ Z v ] F( - -v+k+l ) + F~k+l ) =--Dl(v) + D~(v)"


Using STIRLING'S approximat ion for the gamma function, we can approximate the large zeros of A(v). Then we can show that IA(v)l >[D(v)] on a circle of

radius [lo ~ ]vl ]-1 centered at each zero. By RoucHs Theorem, the zeros of

A(v) approximate those of H i(2z). These results are summarized in

Theorem 8.1. Let z=re ~ and 0 t), the approximate expression

for the n-th zero v,~ of He (x), where for large n

and where

and so on.

z o = t .856 e i'fl3

T 1 = 3.245 ei'~l a

9. Survey of results on the zeros o f / /~(x ) , x> 0

Let x be a f ixed posit ive number. Let v=m+i f l be a complex number and let H i (x) be the Hankel function of the first k ind and of order v with argument x. We have :

(t) The function H i (x) is an entire function of v which vanishes for inf initely many values of v= 4-v k, k = t, 2, 3 . . . .

(2) ~,v=/~ H i (x) = H0 ~ ( , ) ~ - - ~ ,

where the v~ depend on x.

(3) For values Ivl which are sufficiently large compared to I and to x, the zeros of Hi (x ) are given by v~= -4-(~+ifln) where

,o o ,,

and where n is a (large) posit ive integer.

244 WILHELM MAGNUS and LEON KOTIN : The Zeros of the Hankel Function

(4) For the zeros v~ of H i(x), with k : t, 2, 3 . . . . . we have

oo

oo " f t 2 K 0 (2x sinh t) dt Re~, I _ o

fK 0 (2x sinh t) dt 0

where K o denotes the modified Hankel function of order zero.

(5) The following inequalities hold: If v=o~+ifl and H~l)(ar)=O, where a is real and positive and f l>0, then a0, then

a) ~f l>0,

b) >x-Vx , c) x~f l tan~+ 88 when loci< 89

d) Ifl] > x when ]~ l - - 1 2 2 '

H~(x) ] '

f) l im v : 0. X---> 0

(6) If v=c~+if l and H~l)(x)=0, then ~, fl may be considered as functions of the positive real variable x. We have

d~ >A I d~ >0 whenever ~= 8 or x>- - .

= 4 d~ dx >0 whenever c r and x>0.

lim d~ +idxf l =oo . x--~O dx

dfl < 2/~ when ~ > t . = Vs -I s

References [1] MAGNUS, W., and F. OBERHETTINGER: Formulas and Theorems for the Functions

of Mathematical Physics. New York: Chelsea Publ. Co. 1949. ~2] TITCHMARSH, E. C.: The Theory of Functions, 2nd Ed. New York: Oxford Univ.

Press 1939. [3] WATSON, G. N. : A Treatise on the Theory of Bessel Functions, 2nd Ed. New York :

Cambridge Univ. Press 1944.

New York University 25 Waverly Place

New York 3, New York and

U. S. Army Signal Research and Development Laboratory Fort Monmouth, New Jersey

(Received November 9, 1959)

the zeros of the hankel function

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