Chapter IX The Huygens Property

IX

Chapter

THE HUYGENS PROPERTY

1 INTRODUCTION

If the temperature of an infinite bar extended along the x-axis of an

x , t-plane is given by the function cp(x) at time t = 0, the temperature u ( x , t ) at later times t is usually given by the Poisson integral

=I W

u ( x , t ) = etD2cp(x)

+ 6 by u ( x , t + 6 ) = e('+6)D2cp(x)

or at time t

-m

k ( x - y , t)cp(y) dy,

=I

m

-w

k ( x - y, t

(1)

+ 6)cp(y) dy.

But if the integral (1) converges absolutely we have by substitution

J-

W

m

155

IX. THE HUYGENS PROPERTY

156

Here we have used Theorem 3, Chapter 111. Symbolically, equation (2) is

Physically, equation (2) means that the temperature of the bar, u ( x , 6), at time 6 may be used as the data for Cauchy’s problem to determine temperatures at later time. A similar phenomenon in optical theory was described by C. Huygens. The phenomenon does not obtain for all temperature functions, but when it does we shall describe it as the Huygens property. In Chapter 11, $6, we saw the existence of “null” solutions of the heat equation, solutions u ( x , t) for which u ( x , 0) = 0 but which do not vanish identically for t > 0. Obviously, the Poisson integral (1) cannot be used to represent such a function for t > 0 with q ( y ) = u(y, 0). Accordingly, it becomes important to distinguish between those functions which have the Huygens property and those which do not. Ddlnltlon 1.1 u ( x , t) E H” (Huygens property), @

A.

u ( x , t) E H ,

B.

u(x, t)

=I” -m

< t < b,

< t < b,

a

k(x

a

-

y , t - t ’ ) u ( y , t’) dy

(3)

for every t and t‘, a < t‘ < t < b, the integral converging absolutely. Let us introduce a further useful class of functions included in H’, using the notation H Aintroduced by Gehring [ 1960; 3371. Ddlnltlon 1.2

1. @

u(x, t ) E HA, 4 x 9

a

< t < b,

t ) = u , ( x , 1 ) - uz(x, t ) ,

where u,(x, t), u2(x,t ) E H ,

> 0,

a

< t < b.

The function k ( x , t ) clearly belongs to both these classes in 0 for it is positive there and from Theorem 3, Chapter 111, k(x,t)

=/-,kc. m

- y , t - t’)k(y,t’) dy,

In fact, we can show at once that H A c H”.

0<

1’


< t < 00.

00,

(4)

1. INTRODUCTION

157

Theorem 1

< t < b, < t < 6.

1. u ( x , t) E H ~ , a

*

u(x, t ) E H",

It will be sufficient to assume u ( x , t) Theorem 3, Chapter VIII, u(x, t ) =

where a ( y ) E t( - 00,

a

> 0 and to take a

J" k ( x - y , t ) da(y),

o < t < c,

- W

00).

= 0. Then by

To establish (3) we have

Here we have used Fubini's theorem to interchange the order of integration, applicable since k > 0 and a(.) E t. But the inner integral ( 5 ) is k ( x - z , t) by the addition formula, (4), and equation (3) is proved. We point out that the classes H" and H A are not identical. The function h ( x , t) provides an example of a temperatui;: function in H" but not in H A in the half-plane 0 < t < 00. For this function, (3) becomes h(x,r) = k ( x , t

-

r')

h ( x , t'),

0<

1'

< t < m.

(6)

To prove this we use the product theorem for bilateral Laplace transforms. Using Theorem 7.1 and Corollary 7.1, Chapter 111, for the transforms of k and h. we must show that - 2sers2 =

e(t-r')sz(-2se'sz).

But this is an obvious identity. That h ( x , t) 62 H A is not immediately evident. We defer the proof. Corollary l a

0
1. u ( x , 1 ) E H",

< c,

2. o < s < c

*

u(x, t ) E HA,

6

< t < c.

IX. THE HUYGENS PROPERTY

158

By Definition 1, u(x,t) =

Srnk ( x - y, --m

t

6

- 6)u(y, 6 ) dy,

< t < c,

the integral converging absolutely. Replace u(y, 6) by the difference of two nonnegative functions:

Each integral represents a function of H,

> 0, so that u

E

HA.

Corollary Ib 1. u ( x , t ) =

1

00

--m

k(x - y, t ) da(y)

w

u(x, I ) E

HA,

converges absolutely 0 0


< t < c.

If we express a ( y ) as the difference of two nondecreasing functions [Widder, 1946; 61, it becomes clear that u ( x , t) is the difference of two functions of H, > 0, and the sufficiency of condition 1 is proved. The necessity follows in an obvious way from Theorem 3, Chapter VIII. Corollary l c 1. u ( x , t) =

O0

-m

h(x

- y, t ) da(y)

u ( x , t ) E H",

=$

converges absolutely 0 0


< t < c.

Recall that h ( x , t ) is an odd function of x so that we no longer have a positive kernel. By Definition 1.1 we must prove for 0 < t' < t < c that

100

h(x - y, t ) da(y) =

00

lW k ( x - y, - t ' ) dy t

--m

2. BLACKMAN'S EXAMPLE

159

By Fubini's Theorem, applicable in the presence of hypothesis 1, the iterated integral (7) is equal to

-

J_, 00

h ( x - z, t ) da(z).

Here we have evaluated the inner integral on the left by equation (6). Equation (7) is thus established. 2 BLACKMAN'S EXAMPLE

It is important to note that if u ( x , t) E H" in two adjoining or overlapping horizontal strips of the x, t-plane it need not do so in their union. The function

is a case in point. The example was introduced by Blackman [1952; 6781. Hitherto we have dealt chiefly with real functions, but solutions of the heat equation may well be complex. If a real illustration is desired, the real part of B may be substituted for B in what follows. Explicitly,

where that branch of is chosen which is positive when b = 0, t > 0. The formal differentiation used to prove k ( x , t) E H for t > 0 now shows that B ( x , t) E H for - 00 < t < 00. We prove the following result: Theorem 2 1.

o < u < a

A.

B(x,t) E Ho,

B.

B ( x , t ) E H",

C.

B(x, t) E

H O ,


IX. THE HUYGENS PROPERTY

160

It will be sufficient to prove the first result since the other two are limiting cases, a + 0 and a + m. In Definition 1.1, choose t’ = - 6, 0 < 6 < a. Then we must show that

-

s_,

00

k ( x - y, I

+ 6)k(y, i - 6 ) dy

(1)

for - 6 < t < 1/ a , the integral converging absolutely. Formally, this is just the addition formula for the source solution, Theorem 3, Chapter 111, but we must reexamine its validity for complex values of the time variable. We prove the absolute convergence of the integral (1) first. The first factor of the integrand is real and positive. For the second,

The first factor predominates when 1 t + 6

>-y

6 a2+1

or when I < 1/6, and hence when t < l / a . That is, the integral ( 1 ) converges absolutely for - 6 < t < l / a , as desired. To prove the extended addition formula we use analytic continuation. Set F(z) =

-

6

-m

k ( x - y, t

+ 6)e-y’(42)

dy,

z=a

+ ib,

(3)

so that the integral ( 1 ) is F(i - 6). If y2 is replaced by r and z by l/s, the integral (3) becomes a Laplace integral, so that F(l/s) is an analytic function of s in the half-plane of convergence of the Laplace integral [Widder, 1946; 571. Hence, F(z) is analytic for 1 Rez

>-t

-1 + 6 ’

a

a2+b2

+ -t + 6 > o .

This is a region D of the complex z-plane exterior to a disk with center at z = -(r + 6)/2 and of radius ( t + 6)/2 (Figure 10). The positive real

3. CONDITIONALLY CONVERGENT POISSON INTEGRALS

161

b

Flgun 10

axis as well as the point in question, positive real axis, F(a) = k ( x , a

t =

-6

+ t + 6),

+ i , are in D. But on the a

> 0,

by Theorem 3, Chapter 111. But we see by inspection that k ( x , t + t + 6) is analytic over the whole t-plane cut along the real axis from - 00 to - t - 6, and hence in D less the negative real axis. Since F ( t ) and k ( x , t + t + 6) coincide all along the whole positive axis of reals, they must coincide where both are analytic. Hence F(-6

+ i ) = k ( x , t + i),

and this is equivalent to equation (I), which we wished to prove. Thus conclusion A is established. Note that if I > 1/S the reverse of inequality (2) holds and

so that the integral (1) fails to converge absolutely. That is, B ( x , t ) B H" in - 6 < t < (1/6) + E, E > 0. We can now substantiate the first statement in this section. For example, B ( x , t) E H" in (- 1, 1) and in (- 4 , 2) but not in ( - 1, 2). 3 CONDITIONALLY CONVERGENT POISSON INTEGRALS

We give first a simple example to show that Poisson integrals exist which converge but not absolutely in a strip of the x , t-plane. We then prove that an integration by parts replaces the given integral by an absolutely

IX. THE HUYGENS PROPERTY

162

convergent one. Finally, we show that any convergent Poisson integral defines a function of H". The example is

where q ( y ) = y e 2 Y * sin e y ' .

Then

=

iw

sin r dr, 1 yp 4 6

eY1

= r;

p = - 1- 1 . 4r Since the familiar integral (2) is known to converge absolutely for p > 1 but only conditionally for 0 < p < 1, we see that the Poisson integral (1) converges absolutely in the strip 0 < t < i but only conditionally in i < t < The question now arises: Does such an integral define a function having the Huygens property throughout its strip of convergence or only where it converges absolutely?

a.

Theorem 3 1.

*

U(X,

t) =

/"-"k ( x - y , t ) d a ( y ) u ( x , t ) E H",

0< t

converges at ( x o , to)

< to.

Set P ( x ) = p x o - Y , to)

so that, by hypothesis 1, p ( x ) is bounded on (-

00, 00)

and

4. H o #

H"

163

In the quotient of the integrand the dominant factor near y = 2 00 is exp((y2/4to) - ( y 2 / 4 t ) )and this + 0 as IyI + 00 for 0 < t < t,,. Integrating by parts and using the fact that k' = - h / 2 , we obtain

Each integrand is

~ ( l ~ l ~ ~ ~ 1,( f - to)/(%)

IYI+*,

so that each integral converges absolutely for 0 < t < to. Hence Corollaries l b and lc are applicable, and the proof is complete. 4 H"

# HA

In $ 1 we saw that class H A is included in class H". We show now that the two classes are not identical. To do so we need only exhibit a function in H" which is not in H A . As indicated earlier, one such function is h ( x , t). In fact, any derivative of k ( x , t) is also such a function.

Theorem 4 1. n = l , 2 , 3 , . . .

*

an

t) E

A.

7k ( x ,

B.

-k ( x , t )

ax

an

ax,

H O ,

q! H A ,

O
0

< t < 00.

It is easy to show by induction that

where P , ( x , t) is a polynomial of degree n for which

P"( - x , t )

=

( - l)"P"(X, 1).

164

IX. THE HUYGENS PROPERTY

For n = 1, X

k'(x, t ) = - - k ( x , t ) 2t

so that P , ( x , t) = - x / 2 . Similarly, P2(x, t ) = ( x 2 - 2 t ) / 4 . Since k ( x , t ) > 0 for t > 0, k ( x , t ) E H" and

J-m

By Chapter IV, 53, we may differentiate under the integral sign to obtain t)k'"'(x - y , 6 ) dy

Thus, k(")(x,t ) satisfies equation l(3). To complete the proof of A we need only point out that the integral (3) converges absolutely. This is clear from equation (I). An alternative proof of (3) is provided by the product theorem for bilateral Laplace transforms, for,

and the image of equation (3) is the obvious identity

Equation (4) follows easily from

by successive integrations by parts. To prove B we proceed by contradiction. Suppose k(")(x,t) E HA, 0 < t < 00. Then by Theorem 3, Chapter VIII,

5. HEAT POLYNOMIALS AND ASSOCIATED FUNCTIONS

165

the integral converging absolutely. We may assume that a ( y ) is normalized,

"(V + ) + .(y

.(Y)

=

2

-)

Then by Theorem 6, Chapter IV, a ( x ) - a(0) = lim r+O+

r 0

k(")(y,t ) dy = k(n-I)(x, 0

+ ) - k ( " - ' ) ( ~o, + ).

(6) Two cases arise according as n is even or odd. If n is even (2) shows that k("-')(O, t) = 0 and a(x)

- a(O) = lim k("-')(x, t) 1+0+

=

0,

x # 0.

Here we have used (1) and Lemma 1, Chapter V. This equation shows that

a ( x ) is constant and the integral ( 5 ) is identically zero, a contradiction. If n is odd k("-')(x,0 ) = 0, x # 0, and Jk("-')(O, 0 )I = 00. Here

+

+

equation (6) itself produces a contradiction, the left-hand side being finite, the right side infinite. Note that equation ( 5 ) is possible if n = 0. The inversion formula becomes a
ab

= 0,

> 0,

so that a ( t ) is the step-function with a single unit jump at the origin. Corollary 4 h ( x , 1) E H A ,

0


00.

This follows since h ( x , t ) = -2k'(x, t). 5 HEAT POLYNOMIALS AND ASSOCIATED FUNCTIONS

In Chapter I, $5, we defined the heat polynomials u,,(x, t ) by the equation

IX. THE HUYGENS PROPERTY

166

A function w,,(x, t) associated with u,,(x, t ) is defined as the Appell transform thereof: w,,(x, t ) = ~ p [ u , , ( x ,t ) ] = k ( x , t)U,,(

,-

f ),

o < t < 00.

(2)

We show here that u,,(x, t) E H" over the whole x , t-plane and that w,,(x, t) E H" for 0 < t < 00. Preliminary to the proof we need a crude upper bound for lu,(x, t)l, but one that applies to all three variables, n, x , t.

Lemma 5 1.

-w
*

n = 0 , 1 , 2) . . .

-oo
<

n! elXl+l'l.

By Cauchy's theorem applied to the power series (1)

where

r is the unit circle r = Iz {Z

On

r

= eie,

o < e<2

~).

so that Iu,(x, t)l

as we wished to prove.

<

n! elXl+lrl,

Theorem 5.1 1. n = 0 , 1 , 2 , . . .

*

U,(&

1) E

H",

--oo

< t < 00.

To prove this result, we must show for an arbitrary number a that u,,(x, t )

=I" --m

k(x

- y , t + a)u,,(y, - a ) dy,

-a

< t < w.

(3)

5. HEAT POLYNOMIALS AND ASSOCIATED FUNCTIONS

167

We prove first .that this equation holds if u,,(x, t) is replaced by its generating function (1)

But the right-hand side, by Theorem 7.1, Chapter 111, is equal to

so that (4) is established. Substituting series (1) in (4), we have

if term-by-term integration is valid. It is so if

By Lemma 5 , series (6) is dominated by

The integral converges for t > - a and the series converges for 121 < 1. Hence equation ( 5 ) is valid, at least for It\ < 1. Comparing coefficients of z " on two sides of the equation we obtain (3), and the theorem is proved.

Corollary 5.1 1.

O
n = 0 , 1 , 2 ,...

Since u,(y, 0) = y", this is equation (3) with

(I

= 0.

IX. THEHUYGENSPROPERTY

168

Theorem 5.2 1. n = 0 , 1 , 2 , . . .

*

0<

w,(x, t ) E H",

t

< 00.

If we replace z by Az in equation (l), we see at once that u,(x, homogeneous polynomial of degree n,

t2)

is a

u,,(Ax, h2t) = X"u,,(x, t ) .

Replacing t by - t and then setting h = l / t , we obtain from (2) w,(x,

1 ) = U"(X,

- t)t-"k(x,

t).

(7)

If we apply the Appell transform to both sides of equation (I), we obtain

Thus k ( x - 22, t) is the generating function for the associated functions, just as exp(xz t z 2 ) is for the heat polynomials. Comparing coefficients in series (8) with those of the Maclaurin expansion

+

k ( x - 22, t ) =

we obtain w,(x,

1) =

k'"'( x , t ) ( - 22)" n!

n=O

(-2)"k'"'(x,

t),

7

n = 0, 1 , 2 , . . .

.

(9)

And now Theorem 5.2 is an immediate consequence of Theorem 4. Incidentally, equations (7) and (9) give an explicit determination of the polynomial Pn(x,t ) in equation 4(1). We have

so that