AN ABSTRACT OF THE THESIS OF
George Earl Barr
for the
Ph. D.
in
Physics
(Major)
Name)
Date thesis is presented
Title ON THE DIFFRACTION OF A CYLINDRICAL PULSE BY A
HALF PLANE
Abstract approved
Redacted for privacy
(Major professor)
The effect of a pulse produced by an acoustic or electro-
magnetic line source oriented parallel to the edge of a perfectly conducting half plane is considered. The Green's functions for the modified Helmholtz equation are found by solving for the Green's functions
for a cylinder of sectorial cross-section, then allowing the radius of
the sector to go to infinity to form a wedge, and finally opening the
wedge to form a half plane. The Green's function for the sectorial
domain itself, is determined by first solving the eigenvalue problem
for the sectorial domain. The solution for the half plane is repre-
sented in the form of a Laplace transform integral. Then the solu-
tion for a "Dirac" pulse, which is the inverse Laplace transform of
the Green's function for the modified Helmholtz equation, is directly
available and the result of an arbitrary pulse can be synthesized from
it by integration. The results for several pulses are produced by this
method and explicitly show the "transient" terms.
ON THE DIFFRACTION OF A CYLINDRICAL
PULSE BY A HALF PLANE
by
GEORGE EARL BARR
A THESIS
submitted to
OREGON STATE UNIVERSITY
in partial fulfillment of
the requirements for the
degree of
DOCTOR OF PHILOSOPHY
June 1 965
APPROVED:
Redacted for privacy
Professor of Mathematics
In Charge of Major
Redacted for privacy
Ww
Chairman ofgepartment of Physics
Redacted for privacy
Dean of Graduate School
Date thesis is presented
Typed by Carol Baker
TABLE OF CONTENTS
Page
Chapter
1
INTRODUCTION
2
GREEN' S FUNCTIONS
3
REPRESENTATION OF GREEN'S FUNCTIONS
16
AS LAPLACE TRANSFORM INTEGRALS
Time Harmonic Case
5
26
4
TRANSIENT SOLUTIONS
28
5
SOME SPECIAL CASES
32
BIBLIOGRAPHY
38
APPENDIX I
40
APPENDIX II
44
LIST OF FIGURES
Figure
Page
1
6
2
7
3
20
4
21
23
6
30
ON THE DIFFRACTION OF A CYLINDRICAL
PULSE BY A HALF PLANE
CHAPTER 1
INTRODUCTION
The diffraction of electromagnetic or acoustic pulses by
a wedge (and by its special case, a half plane) has been treated in a
number of recent contributions. The majority of these are based
upon a direct attack on the wave equation of arbitrary (not necessarily
time harmonic) time dependency. By this method the following cases
have been treated: Normal incidence of a plane pulse on a half plane
[3] , [41; two dimensional pulse incident on a half plane [4]; plane unit
step function pulse incident on a wedge [9], [11], [12]; arbitrary pulse
incident on a wedge [8].
Another approach centers around integral
transform methods. For instance, the case of a spherical pulse
(unit step function) incident on a half plane [18] uses the Fourier
transform method and the case of a cylindrical "Dirac" pulse incident
on a half plane [17] uses the Kontorovich-Lebedev transform method
[10]
.
The solution of the latter problem is given in the form of an
infinite series involving Legendre functions of such extremely slow
convergence as to make the result practically useless. It will be
shown here that the solution of this problem can not only be given in
elementary form but that it involves functions of algebraic character
2
only.
In addition the method employed here is much more straight-
forward and not just restricted to this particular case.
The procedure is as follows: First we determine Green's
functions
-a-1
for the so-called "modified" Helmholtz equation
2
- y2u
= 0 (exponential decay case), which is obtained from
Helmholtz's equation
there k = -iy.
Au + k u = 0 (time harmonic case) putting
Then the "Dirac" pulse solution is simply given by
the inverse Laplace transform of Gi
with respect to y (14).
2
Having obtained the "Dirac" pulse solution a further integration
yields the arbitrary pulse case [14]. Although only pulse diffraction
phenomena with a half plane are considered here, we shall deal
however in the first stage of our investigation with the somewhat
more general configuration of a sectorial domain, as indicated in
Figure 2, bounded by
0 = a,
0 = 0,
p = a.
In Chapter 2 we represent G1
in the form of an infinite
2
(double) series involving eigenfunctions for this sectorial domain.
One summation can be carried out using a special Fourier-Bessel
expansion formula in which -di
is expressed in the form of a
2
single series. Upon letting a (the radius of the sector) tend to
infinity, both Green's functions for a wedge of angle a are obtained
and then finally the special case
a = 2.7 leads to the two dimen-
sional Green's functions for the half plane.
In Chapter 3 we express
G1
in the form of a Laplace
2
transform integral with respect to the parameter y for the special
case of a half plane, i. e.
G
1
2
where
F(t)
F (te)
dt
0
does not depend on y
.
It should be pointed out that
F(t) is found to be a very simple elementary function, while the
usual integral or series expressions for Gi
involve higher tran-
2
scendental functions (see for instance [1] , [61, [7], [13] [17].
,
A
number of summation formulae which are necessary to arrive at
such a simple representation are derived in Appendix I.
In Chapter 4 we derive the "Dirac" pulse solution as an
inverse Laplace transform of Gi with respect to y . But this
2
property as expressed in Chapter 3, itself occurs in the form of a
Laplace transform, hence the "Dirac'? pulse solution (as the Laplace
inverse of a Laplace transform) is simply the elementary function
F(t) itself. The solution for an arbitrary pulse excitation
g(t)
is
then given in the form of an integral involving the pulse function g(t)
and the "Dirac" pulse solution.
In Chapter 5 we give a few special cases. They include
pulse functions
a)
g(t) of the following character:
Unit step function [instantaneous line source, g(t)= 1]
time harmonic pulse [g(t) = exp
rectangular time harmonic pulse of duration T,
[g(t) = exp it, t < T, g(t) = 0. t> T. ]
In all three cases
g(t) = 0 for t <
0,
1. e. the pulse excitation be-
gins at t = 0.
Finally it is shown that the two Green's functions for the
Helmholtz equation (time harmonic case) can be obtained from the
case
b)
before, as the limiting case for t = 00.
For this purpose
the pulse solution corresponding to the case b) is represented as
the sum of two contributions, the "transient" part and the "quasi-
stationary" part. The latter part is precisely Green's function for
the Helmholtz equation while the "transient" part becomes negligible
when
t
becomes large enough.
5
CHAPTER 2
GREEN'S FUNCTIONS
As mentioned in the introduction we set, in Helmholtz's
equation
Au + k2u
(v
=
k=
0,
(1)
is the velocity of propagation, w is the frequency and
X.
the
wave length)
(2)
and obtain
y2u
=
(3)
0
which we shall call the "modified" Helmholtz equation. The transi-
tion from k to y amounts to the transition from a wave problem
to an exponential decay problem. At first the exponential decay
problem is solved and the wave problem solution is established by
returning to k. Often this simplifies questions with respect to
convergence considerably, since in this case we are dealing with
monotonic functions instead of with oscillatory functions.
(See for
instance [13] , [14] , [19]). Consequently the definition of the two
Green's functions
G1
of
(3) for a domain D bounded by a curve
2
C,
if we regard u to be independent of
z,
is (See Figure 1):
6
Figure 1
a)
A-G-1
y-di
-
2
b) G1+
0
everywhere in D save for
PQ)
is regular inside
2
,17..
x = x' and y = y'
C,
2
c) When P is on C then:
for the first Green s function,
0
or
aG2
for the second Green's function.
an
and
aG
an
is the differential quotient
of G, with respect to the direction normal to
C.
Ko(z) is the
7
modified Hankel function (see Appendix II). The distance between P
and Q is denoted by PQ.
Having obtained Gi for (3) one can obtain G1 for
(1)
2
2
by replacing y by ik. Since the main interest here centers around
pulse problems and since the solution of those requires GI rather
2
than G1 the latter will receive only casual treatment.
2
We now consider the sectorial domain D as indicated in
Figure 2 i. e. our configuration is a cylinder whose cross-section
is a circular sector of radius a and angle
Figure 2
Since we shall give an expansion of Gi in terms of the normalized
2
eigenfunctions of Helmholtz's equation (1), we observe that a solution
of (1) in cylindrical coordinates can be given in the form (see for
instance [16], Chapter V)
8
U(p, 0) =
(4)
(P)f2(0)
where
f
(
)=
J (kp) +
A1 v
Y (kp)
(5)
B1 v
f2(0) = A2cos(4) + B2sin(vsb)
(6)
[Jv(z) is the Bessel function and Yv(z) is the Neumann function, both
of order v and argument z, (see Appendix II)]
In order to establish the eigenfunctions for D we have to find solutions of (1) which are finite in D and which satisfy certain boundary
We establish here the results for the first boundary value
conditions.
problem, only ( = 0 on the boundary of
second boundary value problem
(
8u
The analysis for the
D).
on the boundary of D)
On
follows the same line of reasoning.
The condition u = 0 for 0 = 0 and 0 = a gives, by (6):
A
The function f
f
= 0,
(p)
sin(va) = 0 or
v=
nr
a
y
n = 1, 2, 3,
in (5) then reduces to
1(p) = Al Jnr (kp)
.
a
The corresponding Neumann function has to be rejected since
to be finite for
have the form:
p=
U
is
0; hence B1 = 0. Therefore the eigenfunctions
9
U=
nil
a
JnTr
95 )
a
They vanish for çb = 0 and tp = a. In order that U vanishes for
p = a also, we have
J nTr (ka) = 0
a
This equation determines the eigenvalues.
We denote by x = T v, m the mth positive root of J(x)
= 0. Then
v
ka = T
nTr
a
n, m = 1, 2, 3, ... ,
m
and the eigenvalues are
= kn, m
=TriTr
n, m = 1, 2, 3,
.
(7)
a
The eigenfunctions U
U
= Un, m
for D are then
=
J
n-rr
a
T
nit
a m
pia sin (3.1r±)f
a
In order to obtain the normalized system
U1
of eigenfunctions we
form
a
,c
p= 0
Hence
0= 0
U2 do- ,
do- = pdpd0
.
10
a
7
sin2(_nArtp )
j 2 /Tn-rr
do
00= 0
p/a\ pdp .
0 nn
a
a\
P-
The integration over
m
yields a.12
95
.
In order to obtain the integral
over p we apply a well-known formula from the theory of Bessel
functions [2, vol. 2 p.
901:
2
xJ2 (a x)dx =
2
v
(ax) - J v+.1 (a x) J
v-1
(a x)]
with this formula we obtain
c a J2
r,
v
) pdp
T
mr
a
a.
=
12-a2
m
r2
/
JniT (iTnTr
a\a
_nn-
a
Now,
\
(T
\
a
/
(Tmr,
a.
a
by definition and the first term within the
=0
J Tin /Tmi.
,...-,..
m
CL
rn
a
bracket on the right-hand side vanishes. And, we use the recurrence
relation [ 2, vol. 2, p. 121
2v
Jv -1(x) =
X
V
(X)
Jv+1
(X)
hence
/yr
a
-1
Tmr
a
m.
=
+1k a,
-j nrr
a
nrr
11
[since again Jnir 7TnTr
a
=
01
.
M
Therefore
ra
0
7
nTr
nir
a.
a.
a) pdp
1
=
a
2
2
J nir
a
(7nil
1-
M.)
and hence, finally
N2 = 12
aa
4
a
a
The system of normalized eigenfunctions of
(1)
for a domain as
sketched in Figure 2 is then
1
1
U =U n, m =2 aa7 J
(.1"
+1
a
a
rn)
(8)
fir
sin ( a
3nTr
nTr
a
a
,
n, m = 1, 2, 3, ...
)p/a
We now make use of the well-known expansion for Green's function
G(PQ) of Helmholtz's equation
(1)
into eigenfunctions of (1)
[1 6, p. 183]
G(PQ) =
tp)ui (Q)/,..2..,_ 2,
:IC
3
... 1
'
(9)
where ki are the eigenvalues and U(P) and U(Q) are the
normalized eigenfunctions of
(1)
for the domain under consideration,
12
with reference to the point of observation P = P(p,sb) and the source
point
Q = Q(p', 0'1 respectively.
Upon putting k = -iy according to (2) and using (7) and
(8)
for the eigenvalues and the normalized eigenfunctions respectively,
we obtain finally, for the first Green's function of
for the sector-
(3)
ial domain:
nircp
sin
2
n-rr
.
a
-2
(Tnir m/a)
iTnTr
a
(9)
)
a
(TnTr
+1
a
a
nir
Oa)
3n Tr
a
,
.
a
We will now show that the summation over m in the above double
series can be carried out and thus reduce
(9)
to a single series.
For this purpose we use the following Fourier -Bessel series expansion [2, vol. 2, p. 1043
00
Y-J
(
T 2 )v, m
Iv(x
v+ 1
(T
v, m
)]
v
v, m
93 v(Tv, m X)
(10)
()K ( x)
[1
v
v
-
K (0I
v
v
,x)]Ivq)
13
valid for
[Again,
0_ x< X < 1 and
arbitrary complex.
t
is the mth positive root of J (x) = 0, while Iv and
Tv
Kv are the modified Bessel and Hankel functions, respectively].
This summation formula is applied to the case
v= ,ax = p/a, X =
ya ,
and then G1 becomes
G1
=
[sin
-2a
t
n1
(nin1))
a
sin
(n")')
a
(11)
InTr. (yp)
7i-
a_
In'rr (ya)
under the condition that
p<
Inrr -ya)KnTr(yp
p'
a
a
)
_
- Imr(yp')Kmr(ya)
a
a
.
Since the right-hand side of (10) is symmetric in x and X it
follows that then p and
p'
have to be interchanged in (11) when
p> p'. The same analysis applies to the second Green's function
G2 .
In this case, for the system of normalized eigenfunctions, instead of
(8) we have
14
U
H4
=
Un,
1
1
T
a
cos
+1
(nTr95
)
a
JnTr (Tnir m
a
a
\
nTr
a
m;
p/ a)
(12)
n= 0,1, 2,
m = 1, 2, 3, ...
In this case the eigenvalues are
k=
kn, m
,
Tnir
n= 0, 1, 2, .
(13)
a
m = 1, 2, 3,
If we observe that
sin
sin
[ 22Z (p]
cos
cos
[ "95']=4
{cos{(-)]11-1-1a
a 00,
T cosr-LIT(0+990
a
we can represent both Green' s functions of the two dimensional
modified Helmholtz equation for the cylindrical sector of cross section
D in the form
; cos 11(0-95')]/
Gl= -(2a )-11 E { cos[122-1.(95-091
a
2
n=0
(14)
I
nTr
(yp)
a
I
nTr
a
(ya)
I
nTr
a
ya)KnTr (yp' ) -InTr (yp' )K nTr (ya.)
a
a
a
15
inter-
valid when p < p' and the same expression with
p and
changed when p> p'
are defined as
The Neumann numbers E
.
E0 = 1, E
The transition to a
angle a.
p'
2 for n=1, 2,3, ...
co leads to the domain of a wedge of opening
Since I (x)-4- co as x-.00 and Kv(x) 0 as x -.00 we
obtain, from (14), for the two Green' s functions for a wedge-shaped
domain
oo
G1 =
2
2a )1:En { cos[B-Lra (04'
)] ; cos[(+
(0+ 0' )] } Inff(yp)K
_a
0
(yp' )
n,n.
a
P < 13'
which are the well-known expressions for the open wedge [13] , [14].
Since our main results will be concerned with half plane diffraction
problems we specialize (15) for the case a
2 Tr
to obtain
00
G1=
2
En cos[ i-1(4)-01)]
m=0
cos[
(yp)Kn (ypi ),
(0-0. )] }
2
7
P<
and if p> p'
,
the same formula with p and
p'
interchanged.
This last expression will be the starting point of our investigation.
16
CHAPTER 3
REPRESENTATION OF GREEN'S FUNCTIONS AS LAPLACE
TRANSFORM INTEGRALS
Our next goal is to express (15) as a Laplace transform
integral with y as transformation parameter, i.e. G1 should be of
2
the form
op
F )e
dt
(17)
0
2
where F(t) contains, besides the integration variable t, the proper-
ties a,
,
p, p'
as parameters but independent of y. Of
,
course F(t) can vanish over one part of the interval of integration
(0,00).
For this purpose it is desirable to represent the term
tnrr
(yp)
a
Kmr (yp' )
a
occurring in (16) in such a form. We use the representation
IV(bs)Kv(as) =
1
2
+
a+b
(ab)
2
,
21a-b Pv-1( a 2ab°
cos (Trv)
TT
for a > b, Re s > 0
g °3
a+b
.
Q v-
3.
7
(
2_ 2
-st at +
t2-a2-b2 -stdt}
2ab
17
P and Q are the Legendre functions of the first and second kinds,
respectively ( see Appendix II).
The formula (18) seems to be new and is derived in Appendix I.
now replace
I
tyrr
(yp)KnTr
We
(yp' ) in (16) by (18) while putting there
a
v= a
s= y, b= p, a = p'.
,
Upon interchanging the order of summation and integration [ which is
easily seen to be permissible because of the absolute convergence of
both sum (16) and integral (18)] we encounter terms of the form:
oo
EP
n=0
2
n a.
a2
1
2
2
-t
2pp
(P
,
)
v)
cos (n"
a
(1 9)
and
Ec
n0
s(.12-:-") Q
a
222
nT
1
(t -P2pp-'13
2
cos (n"v)
a
(20)
a
The series (19) can be summed for an arbitrary opening angle a of
the wedge. However, it does not seem possible to do the same for
(20) if a is arbitrary. But for the case a = ZiT, i. e. if the wedge
becomes a half plane, (20) can also be summed. Then Green' s
functions --G-1
2
for the two dimensional modified Helmholtz equation
18
for the half plane can be given in the desired form of a Laplace
transform integral of the form (16). In order not to overburden
the text we will present this summation in Appendix I. From now on
we will consider only the half plane problem.
Then by (16), (18)
oo
EnIn
( yp) K (yp' )cos
=
oo
1
1
2
171
2
P
Encos(111)
Pn
2
(PP' )2 {
+---2 2
P -P n=0
r
+
(P
1
co
,
2
2
-u )1e-Ytdt +
-'
2pp'
2
2
P
-1 flE co s (nv )Q nt -2pp
2
)1e-Ytdt}
2
JP1+
The sums under the integral signs are those given in (2), (3), Appendix I. We consider at first the second term on the right-hand side
of (21) and we obtain, by (3), Appendix I:
2
2
-1)nEcos(nv)Q
n
n--z1
--
2 2
,t
-0
t
2 271
('t
p
'
2PP
jP+PI
= 71p'it 1
op
r
P+ (3'
2
12
-Papp'
-P
e-Ytdt
=
2
+ cos v)
(22)
e- Ytdt
1
2
(p + p'
2
- 2pp' cosv)j
yt
e-dt
,
19
2
Since
z =
2
t -P
-P
2ppt
2
> 1 along the whole interval of integration
from t = p + p' to t = co the condition for the validity of (3),
Appendix I, is met. As for the first term on the right hand side of
(21) we have at first, by (2), Appendix I
op
2
E cos(2--.v)P
n
2
1
2
3
2
)= 22[cosv-P +
n (P +PI
2pp'
n= 0
2PPI
--1
2
2
2
2
-t
(23)
1
= 4(pp9
2
(t -1)
, 2 + 2ppl cosy)
subject to the conditions in (2), Appendix I.
Put
22
2
p
+p' -t
cos U =
2PP`
then, since in the integration of (21) cos U varies from 1 to
when t varies from p' -p to
p' + p,
U varies from 0 to
-1
Tr.
But
the sum (2, Appendix I) vanishes when U <I vl [Here we write Iv I
since v =
± 01]
.
Hence, if (23) is inserted into (21) the integra1
tion does not start with p'-p but with (p2+ p' 2-2pp' cosv)2
and
ends with p' + p (corresponding to U = TT). If moreover v > Tr
then the sum vanishes altogether in 0 < U < Tr and henceforth the
integral. Since v =
figuration 0 ± 0' = Tr.
± 0' we have to investigate for which con-
20
Let the source point Q be in the upper half plane (Because of
symmetry this is no restriction), i.e.
0<
< Tr and consider the
cases as indicated in the Figures (3) and (4).
+
= Tr
Q(131, 4:4)
II E
0.211/401/.0r.11%."....1:0"...0505/110,..40MIOrooralP1111%.411r4r4/...010/
II
(1) -(1)'= Tr
Figure 3
Q'(p', -40
21
We form the image
Q' with respect to the half plane or its extension
and draw the straight lines QE and Q' E where E is the edge of the
half plane. The domain 0 <
< 2Tr is now subdivided into three
domains I, II, III as indicated in the above figures.
Region I is bounded by the upper boundary of the obstacle
(the upper part of the half plane) and the boundary of geometric reflection.,
Region II is bounded by the boundary of geometric re-
flection and the boundary of geometric shadow.
Region III is bounded by the boundary of geometric
22
shadow and the boundary of the obstacle (lower part of the half plane)
(shadow region).
For a point of observation P(p, 0) lying in the respective regions:
Region I
+ 0' < Tr
,
Region II
0 +9' > Tr
,
Region III
0 + 0' > ii,
I
H
- 0'
- 0'
Tr
,
<
Hence, by (23)
2
E cos[(-12-0: 01 )]
n
2
ID1
--2
22
n( P +213'pp:t )1e-Ytdt=
2
(24)
1
=
4(pp' )
P+
[t2-[p2+p' 2-2pp' cos(0-F 0' )1} ae-Ytdt
1
[p2+ pi 2-2pp' cos(0-F
in the regions I and II for the upper sign and in region I for the
lower sign, while (24) is zero in region III for the upper sign and in
regions II and III for the lower sign.
Denote by R and R' the distance of the point of observation P
from the source point Q and its image point Q'
then:
,
respectively,
23
1
R = 15.0 = [p2
p' 2
- 2pp' cos(-' )}
1
PIT= [P2+p
2
- 2pp' cos(0+95'
(25)
di
P(P,14))
Figure 5
One then obtains, from (16), (21), (22), (24), for the two Green's
functions of the two dimensional modified Helmholtz equation;
for a point of observation P lying respectively in
Region I
G1 =(t P+ P
1
2'
1
2 -yt dt
P+PI, 2
)2-e -yt dt] +
R'
R
2
2
(26a)
2
1
- 4,7
P+
-2- -yt
e
-
2
oo
dt -F
(t--R' 2) 2e-Ytdt]
Pi
24
Region II
[ C 2-dt
1
GI =-
P+13t2
(t
2-71
2
-
1
-Nt
co
1
+1
)
2
2
(t
P-FPI
(26b)
Region III
1
G1
00
= - -477-1
2
oo
(t -R'
dt T
P+ PI
2
2
)
2e -yt dti ,t26c)
i3+
This is the desired form (17). Note that the integrand is elementary.
We write
1
1
P+PI
1
co
(
t
2
2e-Ytd
r- (t2 -R
t2-R
=
`JR
-Ytdt
p+pt
and observe that 2, vol. 2, p. 82].
1
(t2-
2
2e- yt dt = K(yR)
Then G1 can be written in the form
2
Region
1
1
1- -Tfr1
2
0
(yR) T Ko(yR')] +
4Tr
(c) (t2-R
P+PI
e
-
dt-F
(t -R'2 ) 2e-ytdtj
'N-P'
cc°
2yt
(27a)
25
Region II
1
sx (t2 -R2)
Gi=---K
(yR)-F[
0
2 27
p+ p,
LITr
1
1
e
-ytdt-
(t P+Pt
2
2e-Ytdt]
(27b)
Region III
1
GI= 2
r°° 2
4Tr
2
(t -R )
faf 170
1
e
_vt
00
7
-
dt]
(27c)
p+pI
These equations admit a physical interpretation.
Region I
The field consists of the result of two line sources, one
located at Q(original source) and one located at its image point Q'
plus an additional term which we shall call the "diffraction" term.
Region II
The field consists of the result of a single line source
located at Q(the original source) plus a " diffraction" term.
Region III
The field consists of a " diffraction" term only.
This " diffraction' term is equal to:
26
1
1
oo
DI
- 4-rr j (t2-R2)
(t2-R'
dt]
P+Pt
P+131
_1
1
oo
(t2-R2)2e-
I
2,
2
dt]
jP+
1
oo
III = -
D
(t2-R2)
4Tr
e
P+ PI
for the regions I, II, III, respectively.
Time Harmonic Case
In order to obtain the corresponding expressions for the
time harmonic case (Helmholtz's equation) we have, according to (2)
to replace
y
by ik.
Since
K0
(iz) = -
1
2
im-F1
(2)
0
(z)
it follows from (27a, b, c) that
Region I
1
(2)
(
G1 --7.0(kR)-f.1-1 0
2
1
)(kR)] + ---[
4Tr
c
(t -
* OP'
2
1
2-ild ST2 2,-2
e d -t t -R1 ) e
P+Pi
t i
dtj
27
Region II
G=
1
4
H
(2)
0
-1
TT
)+ -4
1
1
___
00
(t2 -R2) 2e-iktdt
2
P+ PI
(t2 -
2
e
-ikt dt.1,(28b)
P+ Pt
Region III
1
1
C 7t2..R
S12_012-e-iktdt
P+P'
(3+
13'
e
-ikt dti
(28c)
28
CHAPTER 4
TRANSIENT SOLUTIONS
We will now consider the same configuration as before but
under the assumption that a " Dirac" pulse is applied, starting at
t = 0, to the line source. The field produced by such a " Dirac" pulse
will be labeled
Then
D(t).
. D(t) =" -1 {Gi}
(29)
2
i. e.D(t) is the inverse Laplace transform of the Green's function
for the modified Helmholtz equation with respect to y as inversion
parameter [14].
One first has:
1
t>R
-1
y
,
{K0(yR)} =
0
,
t<R
The remaining terms in (27a) to (27c) are expressed as Laplace
transforms whose inversions are the function F(t) itself.
Then one immediately finds for a point of observation P lying in the
respective regions:
Region I
1
(t) = -[(t
ZIT
1
.
-
2
1
-f(t2 -R' 2)2]Li-4Tr1 [(t 2 -
7.
1
..._
2
1
2
-T. (t -R'
2 -21
)
1(30a)
29
Region II
1
1
(t) = --(t
D
2
-
2
-2-
+
1
4Tr
2
2
(t -R )
1
(t2 -R'
2
)
]
(30b)
Region III
1
(t2-R' 2)
2]
(30c)
The last two terms in (30a) to (30c) are equal to zero when
t< p+ p' while the others are zero for t<R and t<R'
, respectively.
Similarly as before one can interpret (30a) to (30c) as:
Region I:
The field consists of the result of a " Dirac" pulse due to
two line sources, one located at Q and the other at its image point
plus a ' diffraction" term.
Region II:
The field consists of the result of a " Dirac" pulse due to
line source at Q plus a "diffraction" term.
Region III:
The field consists of a "diffraction" term only and moreover this term is zero when t< p + p'
30
Figure 6
The formula (29) is valid under the assumption that the velocity of
propogation is unity. Hence it stands to reason that
t<R in the regions I, II and /.D(t) = 0 and
slD(t) = 0
D(t) = 0
when
when t < p- F
for region III(geometric shadow). We are now in a position to deter-
mine the field
1(t)
due to an arbitrary time dependent excitation
function g(t) of the line source at Q [14] , under the assumption that
g(t) = 0 for t< 0
.
(t-T)g(T)dT =
(t) =
0
where
(x)
g(t-
(x)dx
(31)
-oo
is the "Dirac' pulse solution. Then from (30a) to (30c)
and (31) we obtain:
31
for a point P lying in
Region I
1t
1
2
-
-R2)
2
g(t-
1
txx)dx
2-RT
2
--2
,
gtt-x)dx +
R,'
(32a)
1
g(t-x)dx 4:
-
4Tr
(x2-R' 2)g(t-x)clx]
P+ P'
P+13'
Region II
1
-
=
t
2
2g(t-x)dx
(x2-R2) 2g(t-x)dx
+
(32b)
(A (x2p+pI
,
2
j
)g(t- dx
Region III
1
1,(t) = -
4.Tr
[C
j (x2P+
2
2g(t-
1
(ix
5t(x2-R' 2)-70-x)dx]
P+13'
Integrals whose upper limit t is smaller than the lower limit
R, R'
,
p+p'
, respectively have to be set equal to zero.
(32c
32
CHAPTER 5
SOME SPECIAL CASES
As an example for (32a) to (32c) we choose
g(t)
ei(A)t
,
0< t< T
0
,
t< 0, t>T
=
This is a time harmonic cylindrical pulse starting at t
duration T. The special cases
0 and of
(,) = 0 (rectangular pulse) and
T = 00 (time harmonic pulse of infinite duration starting at t = 0)
are treated separately.
The integrals occurring in (32a) to (32c) are of the
form
1
g(t-x)dx
Since
g(t-x) = e
e
=0
for t-x> T two cases have to be
distinguished:
t-c<T
t-c>T
The lower limit of integration is therefore t-T in case a and c in
33
case b).
We then obtain
Region I
_1
(t)et=
C (x
_
1
2
2e -i(4 x dX
2Tr
-R'
2
2 -icox
e
)
dx
1
(33a)
1 ryt 2 2 --2ei
(x -R ) -o.)xdx
1
471.
1
c(x2t3
t3
2)2 e-iwx dxi
Region II
1
Vt)e-icI3t = -
Ct 2_ 2
Trj
2e
-icax
+
clx--
(33b)
1
1
2e -icoxdx
C (x2-R, 2)
[st(3(
3
e
-icox
dxj
3
Region III
1
.1)(t)e -iwt=
1 ryt
4Tr 1
t3
2
-R2 ) Ze1Xdx4--
1
t
s
(x-2 _R,
3
2) 2e-1(4xdx
(33c)
34
The limits are:
t1 =
R if t-R<T and
t1 = t-T when t-R>T,
t2 = R' if t-Rf<T and t2 =
t-T when t-Rf>T,
t3 = p+ p' when t-(p+ pf ) <T and t = t-T when t- (0-11)>T.
The following terms have to be omitted:
The first term in (33a), (33b) when t <R, the second
term in (33a) when t<Rf and the last two throughout when t<p+pl
.
We specialize (33a) to (33c) now to the case co = 0. This
corresponds to a rectangular pulse of strength unity and duration T.
In this case the above integrals become elementary since
1
g(x2-a")
--2dx
= in
x
+4x2-a2
a
and we obtain
Region I
(t) = -
in.rt+qtz- Rz
Lti+Niti
1
in
-R2
t2
t+ Nit2-R2
t +Nit 2-R2
3
t+Nit2-R 2
3
+Nit2
2
- R'
35
Region II
-
1
2Tr
t+4t2-R
in
t3
+Nit 2-R2'
3
in t + 4t2-R' 2
2-R, 2
3
Region III
41,(t)
t+4t2-R2
1
4Tr
t +Nit 2-R2
3
with
in t +4t2-RI 2
1
t3+Nit3 2-R' 2
3
t1' t'2t3' as given before. Certain terms have to be replaced
by zero under the following conditions:
The first term in (34a) and (34b), when t<R, the
second term in (34a) when t<R'
when t< p-f pl
,
and the last two terms throughout
.
We conclude our investigation with the case T = oo,
the incident cylindrical pulse is a time harmonic one, starting at
t = 0. Here only the case b) applies, with the limits t1, t2, t3,
in (33a) to (33c) being
t1 = R' t2= R'' t3 = p+ p' .
36
Of particular interest is the case t> p+p' i.e. none of the terms in
(33a) to (33c) vanishes. We then write
.5' (x2 -R
2 -iwx2
dx= tx -R2 -iwxc°2
dx = (x -R )
2
2
e
e
2
e
dx
-
t1
(35)
1
-
1
2
(
S oox
-R 2) 2 e-iwx dx =
--rr
i
2
(2
)(
0
1
t
-
s
oo
2
e -iwxdx
t
using for the first integral the well-known integral expression for the
second Hankel function [2, vol. 2, p. 21]. The same expression holds
for R and R' interchanged.
Similarly
t
1
2
.11(x -R
t
e -iwxdx =
t3
-1
2
(x P+Pt
2
2e
-I
wx
--2 -ix
e
dx 1
oo .,
.
=S(x"-R
P-t-Pi
oo
56(x2-R
2e-iwx
dx
.
P+13'
Again the same expression holds when R and R' are interchanged.
We insert these results into (33a) to (33c) and group time independent terms together and time dependent terms together. The field
appears then in the form:
1)(t) -ickt
-- G1 + Y (t).
2
.I(t)
37
Here G1 represents Green's function for the time harmonic case
(28a) to (28c) (Note, that since the velocity of propogation is assumed
to be unity,
o.)
and k are equal). Obviously YT(t) represents the
transient field, i. e. the deviation of the field produced by a time
harmonic excitation starting at t = 0 from the quasistationary case
e.
t>
when the time harmonic excitation started at t = oc)) provided
p'
.
We find for the transient field
,c-icox
1
oo
Y
(t) = -14Tr
.
oo
2
e-1()x2
x -11 )2 dx c Se
(38)
t
for all three regions I, II, III.
Obviously (38) tends to zero as t
oo, i.e. the transient
field becomes, of course, more and more negligible with increasing
time.
38
BIBLIOGRAPHY
1 .
Clemmow, P. C. A note of the diffraction of a cylindrical wave
by a perfectly conducting half plane. Quarterly Journal
of Mechanics and Applied Mathematics 3:377-384. 1950.
Erdelyi, A. Higher transcendental functions. Vol. 2. New York,
McGraw-Hill, 1953. 396 p.
Friedlander, F. G. The diffraction of sound pulses. Proceedings
of the Royal Society, London A186:322-351.
.
1946.
On the half plane diffraction problem. Quar-
terly Journal of Mechanics and Applied Mathematics 4:344357.
1951.
.
Sound pulses.
sity Press, 1958. 202 p.
Cambridge, Cambridge Univer-
Gast, R. C. Half plane diffraction with line source excitation.
Pittsburgh, Carnegie Institute of Technology Department
of Mathematics, 1956. 49 p. (Technical report No. 24)
Harrington, R. F. Current element near the edge of a conducting
half plane. Journal of Applied Physics 24:547-550. 1953.
Kay, I. The diffraction of an arbitrary pulse by a wedge. Communications on Pure and Applied Mathematics 6: 41 9-434.
1 953.
Keller, J. B. and A. Blank. Diffraction and reflection of pulses
by wedges and corners. Communications on Pure and
Applied Mathematics 4:75-94. 1951.
Kontorovich, M. J and N. N. Lebedev. On a method of solution
of some problems in diffraction theory. Journal of
Physics, Moscow 1:229-241. 1939.
Miles, J. W. On the diffraction of an acoustic pulse by a wedge.
Proceedings of the Royal Society, London A212:543-547.
1 952.
39
.
On the diffraction of an electromagnetic
pulse by a wedge. Proceedings of the Royal Society,
London A212:547-551. 1952.
Oberhettinger, F. Diffraction of waves by a wedge. Communications on Pure and Applied Mathematics 7:551-563.
1 95 4.
On the diffraction of waves and pulses by
wedges and corners. Journal of Research of the National
Bureau of Standards 61:343-365. 1958.
.
Tabellen zur Fourier Transformation. Berlin,
Springer, 1957. 213 p.
.
Sommerfeld, A. Partial differential equations in physics. New
York, Academic Press, 1964. 335 p.
Turner, R. D. The diffraction of a cylindrical pulse by a half
plane. Quarterly of Applied Mathematics 14:63-75. 1956.
Wait, J. Diffraction of a spherical wave pulse by a half screen.
Canadian Journal of Physics 35:693-696. 1957.
Wells, C. P. and A. Leitner. A Lebedev transform and the
"baffle" problem. Quarterly of Applied Mathematics
15: 430- 434.
1958.
APPENDI C ES
40
APPENDIX I
Some Integral and Sum Formulae
Summation of series involving Legendre functions
00
EnP
n=0
1
niT
2
a
,
(cos u)cos(---v)
=
a
1
2
1
Tt
a cosv-cosu
-u<v<u,
u<v<2a -u
0 <u< Tr
(E
is Neu.mann's number, E =1
E =2 n> 1).
' n ' _
The formula (1)
is a Fourier series with respect to v with period
v = 2a. To prove (1) we expand the function
I
f(v) =
J
1
(cosv-cosu)-1-
,
-u<v<u
,
otherwise
into a Fourier series of period 2a. Since this is an even function of
v we have, by Fourier's theorem:
Ancos( nTrv
a
f(v) =
n=0
41
with
An =
En
a dv
2a,j -af(v)cos (117v)
therefore for a > u
1
n C cosv-cosu)
j
2
v
cos()dv
a
This integral is known [2, vol. 1, p. 159. formula 271.
1
An =Ena -1
271.
mr
--+
2
a
1
(cosu)
This proves (1).
The special case a = 27r yields
oo
EP
-1 n
n=0
1
--+
2
n
(co su) co s,
nv t
2
=
2
3
1
2,
lcosv - cosu)
2
(2)
u
--u<v
v< 47r -u
0<u<7r
nE2. ,
nQ
1
--2+
(z)cos(nv)= 2
n
[2, vol.1, p. 166]
.
1
,
ir tz+cosv)
2
,
z> 1
(3)
42
Representation of the Product of Two Modified Bessel Functions as a
Laplace Transform
We prove the formulae:
I(bs)K(as)
=
V
v
ya+b
a-b
(ab)
2
1
x-
(
a2+b2-t2 )e -st dt +
2ab
v- -2-
(4)
'0°
+27 -1cos(Tr
a+b
0
(
1
t2-a2-b2 ) e -St dt
2ab
v---2-
a> b.
--2 yco
(ab)
1
KV(as)K v(bs) =
If (4) is established,
K (X)
V
-
(5)
1
2
p72
a+b
(t
1
v--2-
2
a2-b 2 -stdt
2ab
.
(5)
follows by the relation
II
2sin(Tr
-v
(X)
- I (x)1
(Note that K -v(x) = K (x) )
In order to prove (4) we start with the well-known formula [2, vol. 2,
p. 96]
I (bs)K as) =
v(v2 +s2
0
a.b, Re v > -1
J V(av)J V(bv)dv
43
In the integrand above we substitute
v(v2-Fs2)-1
Soo-st
sin.vtdt
e
=
0
and obtain, upon interchanging the order of integration
I V(bs)K v(as)
=st
-
0
0
J v(av)J v(bv)sin(vt)dv] dt.
The inner integral is known [15, p. 169]
icc°
0
J v(av)J v(bv)sin(vt)dv =
a-b
0,
1
V-7
a2-Fb2-t2
2ab
,
a-b (*. t
1
2co s (Try)Q
This proves
(4).
it
1
2-a2_
2ab
a-Fb ,
2
t> a.
44
APPENDIX II
List of Notations
00_on
J (z)
=
(i)v+2n
n! r(v+n+1)
n=
Y (z) = (sinIT v)-1[J (z)cos(Tr v) -
(z)]
-v
H(z)
= J v(z)i Y (z)
v
Be ssel function, Neumann function and Hankel function, respectively.
iTr
iTT v
I( )=e
2
J ze 2
v+ 2n
oo
)
n! r(v+n+1)
n=0
[sin ( IT 0].1-
v(
)
(
modified Bessel function and modified Hankel function, respectively.
iTr v
iTr
K (ze
)
=
-
1
2
K (z) = K-v(z)
(see[2, vol. 2, Ch. 7] )
iTr e
2
(2)(
45
P
(z)
crP v(z)
Q (z)
(see [2, vol. 1, Ch. 3]
Legendre function of the first kind
-1<x< 1 .
Legendre function of the first kind,
z not on the real axis between 1 and
- oo
Legendre function of the second kind,
z not on the real axis between 1 and -oo
).
.