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11 Applications to Boundary Value Problems
Applications t o Boundary Value Problems
In this chapter we present solutions to boundary value problems, arising in various disciplines of mathematical physics, for axially symmetric bodies, including dumbbells, elongated rods, and prolate bodies, of which spheres and spheroids are special cases. The method rests on exploring the fundamental solutions of partial differential equations, as presented in the previous chapters, and then taking a suitable axial distribution of the Dirac delta function and its derivatives on a segment of the axis of symmetry of the body. This idea is extended to include distribution of these functions on arbitrary straight lines, curves, and disks [31-391.
11 . I . POISSON'S EQUATION
Let us consider Poisson's equation in n dimensions,
where u ( x ) is the generalized n-dimensional potential, F ( x ) is some forcing function, and x = (xl, x 2 , . . . , x,) is the position vector. In Section 10.4 we
288
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
found that for the special case F(x) tion E(x):
=
6(x) we obtain the fundamental solu-
r = 1x1. (2) u(x) = E(x) = l / F 2 , This value of u(x) gives the generalized electrostatic potential due to an n-dimensional sphere of radius I . For a distribution F ( x ) of delta functions over a volume V, the corresponding value of u(x) is
where * denotes the convolution and the variable of integration is x’. Our aim is to use ( 3 ) to obtain the generalized electrostatic potential of a large class of axially symmetric bodies for which a sphere, a dumbbell, and a spheroid are special cases. For this purpose, we prescribe an axial distribution of sources along the axis of symmetry, which we take to be the x1 axis. Accordingly, F(x)
=
.f’( x1)6(x2)6(x,)...6(xn)
(-el
2
x1
I
c2),
(4)
wheref(x,) is the strength of the distribution and c 1 and c 2 are two positive constants. For bodies with fore-and-aft symmetry we have c 1 = c2 = c. In terms of cylindrical polar coordinates, the shape of an axially symmetric body can be written p
=
po(x),
--a
I x Ia,
po(+u)
=
0,
(5)
where we have let x denote x 1 and where p 2 = x: + x i + . . . + x i . In view of the symmetry, po(x) = po(-x). The geometry of this configuration is explained in Fig. 1 1.1. In the next stage, we substitute F given by (4)in ( 3 ) and obtain
where we have used the sifting property of the delta function. If the potential u is prescribed on the surface S to be g(x), then (6) becomes
This is a Fredholm integral equation of the first kind for,f(x). Using this technique we can solve the potential problems for a wide class of axially symmetric configurations and in various fields of mechanics. Furthermore, we can solve both the direct and inverse problems. For the direct problem, the body profile p = po(x), as well as g(x), is prescribed, and we evaluate the
11.1,
289
POISSON'S EQUATION x2
b
Fig. 11.1. Line distribution of delta functions along the .vl axis in c, 5 .vl 5 c2 o f a n axially symmetric body.
distribution f(x) and the parameters c I and c 2 from (7). For the inverse problem we are given ,/'(x), g(x), and the parameters c I and c2: then (7) provides a body profile. The value g(x) = 1 is of special interest bccause it leads to the evaluation of the n-dimensional capacity of S. This is defined as C(n) = (11 - 2 )
where r l x
= dx'
rlx' . . . tlx"
=
s,,
F(.Y) ( I s ,
(8)
t1V. This rclation, in view of (4), becomes (2
C(n) =
(I1
- 2)
(9)
/'(<) ti<.
Let us now consider an axial distribution of dipoles along the axis of symmetry such that F ( x ) = h(x)h'(V)h(x,). . . h(xn),
1'1
I s I
('2.
(10)
where x 1 = x, x2 = y , and h ( x ) is the strength of the distribution. Thus, the dipoles are directed along the y axis. Substituting (10) in ( 3 ) , wc obtain
290
11.
Let u(x, p )
= yl(x)
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
on the surface S. Then ( 1 1) becomes
which, like (7), is a Fredholm integral equation of the first kind for h(x), provided that the equation of the surface S , p = p,(x), is known. Consequently, if h(x), l(x), e l , and c2 are prescribed, then (12) yields us the body profile. In the sequel we take c 1 = c2 = c.
11.2. DUMBBELL-SHAPED BODIES
Let us take two isolated delta distributions of equal strength such that
+ c) + A,h(x
,f(x) = A,h(x
- c),
where A , is a constant. This gives a dumbbell-shaped body, as shown in Fig. 11.2. We substitute in ( 1 1.1.7), and find that for g(x) = 1 we have
1
=
Ao(l/R;-2
+ I/R;-2),
(1)
where Rl
= [(x
+ c)' + p2I1I2,
R2 =
+
(2)
[(x - c ) ~ p2]*I2.
To obtain the relationship between the geometric parameters c/a and d/a (see Fig. 11.2), we evaluate ( I ) at the terminal points x = f a , p = 0 and at the dumbbell neck x = 0, p = +d. The result is 1 A,
--
1
(a
1
+ cy-2 + (a - cy-2'
1
A, - (c2 -
2
(3)
+ d2)'"-2)/2'
Elimination of A, yields the desired relation:
Furthermore, from ( I ) and (3) we derive the equation for the n-dimensional dumbbell :
1
[(x
+ c)2 + p2]'"-2'/2
1
+
[(x - c)2
+
p2]'"-2"2 - (u
1 + (a + cy-2 1
Cy-2'
(5)
11.2.
291
DUMBBELL-SHAPED BODIES
P
-0
1
,
A0
x
-C
Fig. 11.2. Dumbbell-shaped bodies that can be represented by two delta functions
The capacity of this configuration follows from (1 1.1.9): C(n) = 201
-
2 ) A , = 2(n - 2)
1
(a
+ c ) ' - ~+ (a
-
cy-2
I-'
Electrostatics
This corresponds to the case n respectively,
C(3)
=
3. In this case ( 1 1.1.6) and (4)-(6) become,
= (a2 - c2>/a=
(c2
+ d2)1/2.
(10)
Thus, the capacity of a dumbbell in three dimensions is equal to that of a sphere of radius (c2 + d2)1/2. Axially symmetric configurations are presented in Fig. 11.3 for different values of d/u.
292
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11.
I
-1.0
I
-0.8
I
-0.0
I
I
-6.4
-0.2
V 0
I
0.2
I
1
1
0.4
0.6
0.0
1
1.0
x/o
Fig. 11.3. A class of axially symmetric bodies generated by a pair of delta functions. The numbers along the p/a axis designate values of d/a.
Perfect Fluids
The relation +(n - 2)
=
UpnP3(n- 3)-’[1 - ~ ( n ) ]
(1 1)
relates the stream function +(n - 2) for an ( n - 2)-dimensional body of revolution moving with a uniform speed U along its axis of symmetry in a perfect fluid (or, equivalently, the fluid streams past the body with speed U ) with potential u(n)in a space of n dimensions. This potential function assumes the value unity on the surface of the body and vanishes at infinity. Thus, for the flow of a perfect fluid in three dimensions, we need the value of 4 5 ) . Now, from relations (1 1.1.6), (1 1.2.3), and (1 1.2.4), for n = 5 it follows that
C(5) =
Then (1 1) yields
6(a’ - c ’ ) ~ a(a’
+ 3c2) = 3(c’ + d*)3”.
11.2.
293
DUMBBELL-SHAPED BODIES
This gives the value of the stream function for a wide class of dumbbellshaped bodies. Cavities in Elastic Medium
The solution of a torsion problem for a shaft with a cavity can be obtained from the solution of the electrostatic potential for a seven-dimensional body, with g( x ) = 1, provided the cavity and the body have the same meridian profile. This isoperimetric equality is (15) 11/(5) = ap4c1 - u(7)~, where 11/ now stands for the elastic stream function. The strain energy E is E
=
(2G/n)[(2~~/15)C(7) - V(5)],
(16)
where C(7) is the capacity, V(5) is the volume of the body, and G is the shear modulus. The present method is suitable for solving some of these problems in a very simple fashion. For instance, for a spherical cavity we takef‘(x) = a56(x). Then (1 1.1.6) and (1 1.1.9) yield u(7)
= a5/(X2
+ p2)5’2,
C(7)
= 5a5.
Substituting these values in (1 5) and (16) and using (3.3.3) for V ( n )for n we obtain 11/(5) = ip4[1 - a5/(x2
+
p2)5’23,
E
= ~ ~ 7 1 0 5 .
=
5,
(17)
Similarly, for a dumbbell-shaped cavity,
C(7)
=
5(a2 - c 2 ) 5
a(a4
+ 10a2c2 + 5c4)’
Accordingly,
and E
=
nG
4(a2 - c 2 ) 5
3a(a4
+ 1oa2c2+ 5c4)
(19)
294
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11.3. UNIFORM AXIAL DISTRIBUTIONS
Capacity (n = 3)
Let the uniform distribution be
f (x) = A, a const. Then for g(x) = 1, (1 1.1.7) yields 1
-
A
=
In
R2
- (X
R1 - (X
- C)
+ c)’
where R , and R2 are defined in (1 1.2.2). This equation is equivalent to (p2
(p’
+ (x - C)2)1” - (x - c ) + (x + c)2)1’2 - (x + c ) =
where A is a constant, or X2
[c2(4A3
+ 8A2 + 4A)]/4A( I - A)’
+
[c2(4A3
P’
+ 81’ + 4A)]/( 1 -
= ’’
which is the equation of a spheroid. The equation of a prolate spheroid is x2/u2
+ p2/b2 = 1,
b2 = ( I
-
e2)u2, c = ue,
(3)
where e is the eccentricity. Accordingly (2) becomes 1 l + e -=InA l - e
=
2 tanh- e.
(4)
The value of the potential function u(x, p), from ( 1 1.1.6), is u ( x , p ) = (2 tanh-’ e)-’ In
R2
- (X - C)
R,
-
-
(x
+ c>’
(5)
and applying (1 1.1.9), we obtain the value of the capacity: ue(tanh-’ e)-’. (6) An interesting limiting c a w i s that of a slender body, which is obtained by setting b/u 6 1 . In this case (6) reduces to C
where
E =
b/u
=
=
(1 - e2)lj2 is called the slenderness parameter. Equation
11.3.
295
UNIFORM AXIAL DISTRIBUTIONS
(5) can be put in familiar form by using prolate-spheroidal coordinates
t, 4 4 :
x
=
ctu,
+ iz = c[(t2
y
e
l)(I -
-
l/to. (7)
=
Then (5) takes the form
where Qo(<) is the Legendre function of the second kind. The corresponding results for an oblate spheroid can be derived by replacing c by ic and e by ie(1 - e 2 ) -112 in (5) and (6) and by replacing t by i t in (8). The corresponding formulas are C(3)
=
c/sin-' e
=
be/sin-' e,
(9)
and u(x, p )
=
'
Qo(it)/Qo(iSo) = cot - t/cot -
' to.
(10)
+ 1 or to-+ 0, the oblate spheroid reduces to a thin circular disk of radius b and (9) and (10) reduce to
As r
C(3)
=
2b/~,
U(X,p )
=
2/7~cot - ' 5.
The results for a sphere follows upon setting e
=
0.
Perfect Fluids (n = 5 )
In this case, if we set,f(x) = A and g(x) = 1 in (1 1.1.7), we obtain
where R and R 2 are given by ( 1 1.2.2).For the semimajor axis (x and the semiminor axis (x = 0, p = b), (1 1) yields the relations 2ac(a2 - c2)-2 = 1/A,
2cb-'(b2
+ c')''~ = 1/A.
=
a, p
=
0)
(12)
Hence (a2 - c ~ = )ab2(b2 ~ + c~)'/~,
(13)
which relates the parameters b/a and c/a. The shape function p = po(x; a; c) is now known from (1 I)-( 13). These bodies are prolate shaped. The value of the potential 4 5 ) follows from relations ( 1 1.1.6): (14)
296
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
Next, we use (11.2.1 1) to obtain the stream function for the motion of these configurations in a perfect fluid,
where A is given by (12). The capacity C(5) is C(5) = 6Ac =
3(a2 - c 2 ) 2 U
=
3b2(b2 + c2)’j2.
(16)
Two limiting cases are of special interest.
(I)
c + 0, A
+ cx) such
that c A
+
iu3.
In this limit we recover the results for a sphere. (2) b/a 6 1 leads to an elongated rod. In this case (13)-(16) reduce to C b 7 b 4 - - - 5 - 6 (;)3 o(;) ,
+
U
(20) respectively. The body shape in this limit, from ( I I), is b
x2(3c2 - x2) 4(c2 - x2)4
+
(c’ - x2) >> b2.
(21)
Cavities in an Elastic Medium ( n = 7)
The relation corresponding to (2) and (1 1) in this case is
while those corresponding to (12) are 2ac(a2 + c2) - 1 (a2 - c2)4 A’
2 4 2 2 + 3b2) - -1 3b4(c2 + b2)3/2- A .
(23)
11.3.
297
UNIFORM AXIAL DISTRIBUTIONS
Hence a(a2 (a2
+ c2)
-
2c2 + 3b2
-
- 3h4((.2 +
c2)4
(24)
h2)3/2'
which relates the parameters b/a and c/a. Furthermore, (22)-(24) provide the body shape p = p o ( x ; a ; c). The quantities u(7) and C(7) are u(7) =
["
+
3b4(c2 h2)3'2 _ +_c _ x_ ~ _ _ 2c(2c2 + 3b2)p4 R ,
C(7)
=
5(a2 - ~ ~ ) ~ / +a cz) ( n =~ 15h4(c2 + h2)3'2/2c2+ 3h2.
(26)
We can then appeal to relations ( 1 1.2.15) and (1 1.2.16) to obtain
[
$ ( 5 ) = ip4 1 x ~
3h4(cZ+ h 2 ) 3 / 2 2c(2c2 + 3h2)p4
+ c - .Y -
C'
R,
-
1 (x + c)3 3 R:
-
~
2c2 + 3h2
p-.
where we have used V ( 5 ) = in2 p:(x) dx. As c + 0, A + x' such that c A + 1/2a5, we recover the results for a spherical cavity in Section 11.2. For a cavity with the shape of an elongated rod, that is, for h/u 6 I , we have
C(7) =
[:' (a)" +
(c2 - x2) 2 h Z ,
+ O(&)'], E
=
[;
G7cab4 c ~
4
O(;)y,
-
8 - 3(!)4(;
+ i c y+ o + 2c4(a2 3(u2 -
(361
.
( 9 3
A class of configurations is sketched in Fig. 11.4 for various values of h/u.
298
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
x /a
Fig. 11.4. A class of axially symmetric cavities of uniform distribution of delta functions. The numbers along the p/a axis designate the values of h/a.
11.4. LINEAR AXIAL DISTRIBUTIONS
In this section we study the polarization potential, so that g(x) = -Eox, where E o is the uniform electrostatic field in which a conductor is lying. Furthermore, we take (1)
J(x) = Bx.
Then (1 1.1.7) yields -Eox
=
[
B xln
r2
- (x -
'I} - R ,
R, -(x+c)
+ R.].
Proceeding in the same manrier as we did following (1 1.3.2), we find that the right side of (2) is a multiple of x only if the shape of the body is a spheroid. For a prolate spheroid, as defined by (1 1.3.3), we find from (2) that B
=
[
E , 2e - In
(t ':)I-'
__
11.5.
299
PARABOLIC AXIAL DISTRIBUTIONS, n = 5
The potential 4 3 ) is obtained from (1 1.1.6): u(x, P ) =
2r - In[(l
Eo
+ t.)/(l
-
r)]
[ {”’ .Y
In
-
(x
-
‘)}-R1
R1 - ( x + c )
+ R,].
(3)
The polarization potential P ( 3 ) is
In terms of the prolate spheroidal coordinates defined by ( 1 1.3.7), (3) and (4) take the forms
4 5 , V ) = -cEo5VQi(O/Qi(5o) and
p(3)
-%5;c35i(5;
-
1)”’CQ~(0/Q1(5o)l7
respectively.
11.5. PARABOLIC AXIAL DISTRIBUTIONS, n = 5
Let us now take f(x) = A
Setting y(x)
=
+ B.u2,
Is1 5
(1)
C.
1 and substituting this value of,/’(.v)in ( I 1.1.7) yields X+(’
xz x + c
x+c +
p’
(T-
-)
x -c RZ
This relation is again valid only for a spheroid. For a prolato spheroid as defined by ( I 1.3.3), we find that
300
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
which determines the values of both A and B. Appealing to (11.1.6), we obtain
This in turn enables us to use ( 1 1.2.11) to obtain the stream function for perfect fluids,
'+I,
In prolate spheroidal coordinates (1 1.3.7), (4) and (5) become u(x, p ) =
(5)
- In c2 A [ t225 - 1 5-1
-
~
respectively, furthermore, from (1) and (1 1.1.9) we obtain C(5) =
4a3e3 2e/( 1 - e2) - In[( 1
+ e)/( 1 - e)]
'
The values of these quantities for an oblate spheroid can be obtained by using the limiting process, as explained for the n = 3 case in Section 11.3. They are
. [ 2Ai
5
u(x, p ) = 7 - cotc 5 +1
$(3)
=
+
' 51.
fUcZ(1 t2)(1 - 1 2 ) - iUA(1 x [-cot-' 5 5/(1 593,
+
+
+ 52)(1 - $)
C(5) = -2b3e3[e(1 - e2)1'2 - sin-' e l- ' ,
'
'.
where in this case A = +b2e2i[e(1 - e2)1'2 - sin- e l - As e -+ 1 we derive the corresponding results for a thin circular disk of radius b : 4x9
P)
=
(2/.)C5/(t2
+ 1)
-
'
cot- 51.
$(3) = + U P ( 1 + 52)( 1 - q 2 ) - ( U P / n ) (1 x C-ccot-l t 5/(1 t2)3,
+
+
+ t 2 ) ( 1 - $)
C(5) = 4b3/n. Similarly, if we set e
=
0, we obtain the corresponding formulas for a sphere.
11.6.
FOURTH-ORDER POLYNOMIAL DISTRIBUTION
301
11.6. THE FOURTH-ORDER POLYNOMIAL DISTRIBUTION, n = 7; SPHEROIDAL CAVITIES
Let us take
With this value of,f'(x)and y(x)
=
1, (1 1.1.7) yields
where
n = o , 1 , 2, . . . , m = - l , l , 3 , 5 , . . . . These satisfy the recurrence relation
Thus
302
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11,
Ill., = In
R , - (X - C) R1 - (X c)'
+
By evaluating these quantities on the surface of the prolate spheroid (1 1.3.3) and with a slight algebraic manipulation, it follows that c4BS., - 2c2B5.,
l + e _ _ _2e + In ____ + €5, = 3(1 4e3 . - e2), 1 e2 I-e -
(8)
Comparing (2) and (8) we obtain A,
=
A4c4, A , = -2c2A,,
A,
=
[3(1
y:2)2
-__ 2e_ 1 - e2
+ In Eel. 1-e
Next, we make use of (1 1.1.6) and (1 1.1.9) and obtain
4 7 ) = A4[c4B5. 0 - 2c2B5,2 C(7)
=
+ B5,
(10)
41,
yu5e5A4.
( 1 1)
We substitute these values in (1 1.2.15) and ( 1 1.2.16) and derive
11/(5) = aP4[1 E
=
- A4(cL4Bs.o-
2c'Bs.z
(16Gnu5/45)[4e5A, - 3(1
-
+ BS.,)],
e2),],
where we have used V ( 5 ) = (8n2a5/15)( 1 - e2),,
The corresponding relations for an oblate spheroid are C(7) = -%b5es[(2e3 + e)(l - e 2 ) ' I 2 - sin-' E = -(16.rrG/45)[8h5e5~(2e3+ e)(l + 3b5(1 - e 2 ) ' I 2 ] .
el-'
e2)l', -
sin-' e}-'
When e -+ 1, these relations reduce to their corresponding values for a penny-shaped crack of radius b, C(7)
=
16b5/3n,
E
=
EGh5.
11.7.
303
THE POLARIZATION TENSOR FOR A SPHEROID
11.7. THE POLARIZATION TENSOR FOR A SPHEROID
The polarization potentials ui,i = 1, 2, 3, created by uniform electric fields in the x i (x = x I , y = x 2 , z = x3) directions, respectively, are obtained by solving the boundary value problems V2Ui =
-47cF,(x), on S ,
ui = - x i
+
ui
=
o ( l / r 2 ) as r
--+
x.
(1)
where r2 = x2 y’ + z 2 and S is the surface of the spheroid, as defined in ( I 1.3.3). The boundary value problem (4) for i = 1 can be solved by taking the line distribution of delta functions along the axis of symmetry (x axis), that is, F , ( x ) = .1;(x)h(y)h(z).Then, following the analysis of Section 11.1, we find that
Letting (x, p ) approach S and using the boundary condition, we have
which is a Fredholm integral equation of the first kind. This equation has been solved in Section 11.4; the solution is
Then (2) becomes ul(x, p )
=
[
2e - In
~
‘3”.
I-e +
hiR2 - (x R1 - (X
+ ‘I}
-
C)
-
R,
+ R2],
(5)
where R and R 2 are defined by (1 1.2.2). In prolate spheroidal coordinates, (5) takes the form ~ 1 ( 5 9
YI)= -doYIQi(O/Qi(to).
(6)
Solutions of the boundary value problems (1) for i = 2 , 3 are obtained by taking the line distributions of dipoles, that is, F 2 ( x ) = i2(x)h’(y)b(z)and F , ( x ) = j;(x)h(y)d’(z)along the axis of symmetry. The values of ui(x, p), i = 2, 3, follow by appeal to ( 1 1.1.1 1). Indeed,
304
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
In view of the boundary conditions, we find that (7) reduces to the Fredholm integral equation
The solution of (8) is achieved by taking
Then (8) yields .u+c
x - c
Rz - (X - C ) Rl
- (X
+
C)
11
,
1x1 I n .
(10)
This relation is valid for the prolate spheroid ( 1 1.3.3) if
Combining (7), (9), and ( I 1) we obtain the values of
Iti,
i
=
2, 3,
In terms of prolate spheroidal coordinates, these relations take the forms ~ z ( 5 Vt ,
CP)=
-C(COS
~p)C(sIi-
-
V ~ ) ] ” ~ Q ~ ( O / Q (13) ~(~~)~
and 143(5.
YI.cp) = -+in
q>C(5;- 1)(1 - Y I ~ ) I ” ~ Q ! ( ~ ) / Q : ((14) ~~).
The formula for the components Qij
= -
s
ui
Qij
of the polarization tensor is
du . rlS dn
From the symmetry of the spheroid it is clear that Qij = 0,
i # .j.
(16)
11.7.
305
THE POLARIZATION TENSOR FOR A SPHEROID
In terms of the eccentricity c Qii
=
I/C0, these take the forms
=
+
4ntr3(1 - e2)[ln---1 e 3 I - r
Q 2 2 = Q33 =
-
x
4ntr3(1 - e 2 ) 2e--- 2 e 3 3 I - e2 2t?
[I
1
= -
4nh3(1 3
+ "I-' .
In
LEI I-e
In (20) I-r e2 in (19) and (20), we derive the corresponding ~~~~
-
Replacing e by ie( I - e 2 ) values for an oblate spheroid: QI
fi
2e ][ln - 2 ~ ] l ' , (19) 1-r2 I-e
("
[sin-' e
-
e2)Il2] sin- r x [r( I
- e2)Ij2 -
sin-
I
03- I .
(22)
-+ 1, the oblate spheroid reduces to a thin circular disk of radius h, and (21) and (22) become
As
Q11
=
0,
Q22
= Q33 =
16b3/3,
respectively. The limiting case of a sphere follows by letting r have
(23) -+
0. We then
(24) Q I 1 = Q 2 2 = Q 3 3 = 8xu3/3, as is well known. The limiting case of a slender body is obtained by taking h/u = ( 1 - L ) ~6 )1. ~Then ~ ~(19) and (20) reduce, respectively, to
306
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11.8. THE VIRTUAL MASS TENSOR FOR A SPHEROID
In this section we consider the potentials ui(x), i boundary value problems
=
1, 2, 3, that satisfy the
V2ui = -4nFi(x),
i = 1, 2, 3.
i l . Vui = - A . i i on S,
ui = O( l/r2)
(1)
at
E,
(2)
where A is the unit vector normal to the surface S and the i i ,i = 1,2,3, are the unit vectors in the x, y, z, directions, respectively. Then, the u i , i = 1,2,3, are the velocity potentials of a body with surface S and moving with unit velocity in a perfect fluid in the x, y, z directions, respectively. For a prolate spheroid, fi is
A
=
( I :+,
( a 2 - e2x2)-'I2 - x,tl
CI
-
yi2
+
(3)
We solve these boundary value problems by the same distribution of singularities as in Section 11.7. That is, we set F,(x) = ,fl(x)S(y)G(z),F2(x) = .f2(X)b'(Y)8(Z), and F3W = .f2(X)&Y)WZ). The integral representation formulas for u l , u 2 , u3 are given by (11.7.2) and (1 1.7.7). Applying the boundary conditions (2) to these formulas, we obtain the integral equations
The solution of (4) is obtained by settingfl(x) =
A[a2B3. 1 - xB3.
=
Ax, 1x1 I c, so that
21,
where the quantities B,,,n are defined by (11.6.3). On the surface of the spheroid (1 1.3.3) this relation becomes x
=
2e
A [ -1 - e
-
In I+e] 1 - e X,
307
11.8. THE VIRTUAL MASS TENSOR FOR A SPHEROID
which yields the value of A . Thus,
Equations ( 5 ) are solved by setting
,fi(x) = A.
-
A2x2,
i
=
(7)
2, 3, 1x1 I c.
Substituting (7) in (9,we obtain, after a slight algebraic manipulation,
Hence,
Next, we substitute these values of ,fi(x) in the integral representation formulas ( I 1.7.2) and ( 1 1.7.7) and obtain
U,(.X,
p) =
-
u3(x,
p) =
-
2e(l - 2eZ) -
[
2e(l - 2 2 ) 1-e2 ~~
The virtual mass tensor
In Ice]-1y(czB3.0 1-e
-
1-e
B,.,),
(10)
Z(C~B, . ~B,.,).
(11)
In%]-', I-e
(13)
-
w.jis defined as
Substituting (9)-( 11) in (12), we obtain
w,,
=
W,,
=
4nu (1 - e2) 3 ~
W,,
=
4TU3 3 x
and
2e
--
I +e In-1-e
[
wj
=
0
I-e
I'
2e(I - 2eZ) 1 - LJ2 for i # ,j.
-'
308
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
By letting e -,0 in these relations, we recover the corresponding values for a sphere of radius a : W,,
=
W,,
=
W,, = 2na3/3,
which are well known. For a slender body, these relations reduce to
w,,= w,,=
~
w.j
The value of for an oblate spheroid follows from ( I 3)-( 15) on replacing e by ie(1 - e2)-'I2: W,,
=
W2,
= W33 =
(4nb3/3)[(1- e2)'12sin-' e - e][e(l
- e2)'12 -
(4nb3/3)(I - e')[e(I - e2)lI2 - sinx [(I - e2)lI2sin-' e - e - e 3 ] - ' ,
' el
sin-'
el-',
(19) (20)
and Rj = 0 for i # j . In the limiting case of a thin circular disk of radius b, (19) and (20) reduce to W1, = $b3,
W,,
=
W,, = 0.
(21)
11.9. THE ELECTRIC A N D MAGNETIC POLARlZABlLlTY TENSORS
The electric and magnetic polarizability tensors P i j and M i j are defined in terms of the tensor Qij and as follows:
wj
where Iij is the identity tensor and V is the volume of the body. Accordingly, we can use the results of Sections 11.7 and 11.8 to find the values of P i j and Mij.
11.9.
THE ELECTRIC AND MAGNETIC POLARlZABlLlTY TENSORS
For a prolate spheroid, 1+e 8na3e3 PI,= pn- 2e] 3 I-e
-1
(3)
,
~
309
In
I f e l ' , (4) I-e
~ , , 8na3e3 = ~ [ ~2e - l n ! + e , ] I-e - 1
(5)
(6)
1 - e2
P.. = M . . = 0,
i # j.
For an oblate spheroid,
P II
=
-(4nh3e3/3)(1 - e2)"'[(1
P,,
=
P 3 3 = -(8nb3e3/3)[e(1 - e2)li2- sin-' el-',
-
e2)'l2 sin-' e - e l - ' ,
(7)
(8)
M I , = -(4nh3e3/3)[e(l - e2)'/, - sin-' e l - ' , M,,
=
M,,
-(8nb3e3/3)(1 - e2)'/,[(1 - e 2 ) ' / 2 s i n - L e - e - e 3 ] - I ,
=
P,,
=
P2? =
MI,
=
M,,
P33 =
=
M,,
P . . = M.. = 0,
For a slender body, 4na (a2 - h2)(ln: 3
P,,
=
P,,
= P 3 3 = __
~
2nab2 (41n:
-
l)-'[l
Mll =
2na(a23- 3b2) (a'p
M,,
M,,
2na -(2a2 3
-
- jI 3b2)
In
4nu3,
=
ha3,
i # j.
,I):(.
(1 1) (12) (13)
+
- l ) ( l n T 2a
3
=
(10)
P I.J . = M IJ. . = 0, i # ,j.
For a sphere,
=
(9)
;)-'[
+
-
1
+
"')-'[I
+)'I,
+ (I(:)(15) ,],
310
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
For a thin circular disk,
PI,
P22
0,
=
M I , = $b3,
=
P33
=
yb3,
M22 = M 3 3 = 0.
(18) (19)
P I.J . = MlJ. . = 0,i # j . The values of component M, of the magnetic polarizability tensor and component PZ2of the electric polarizability tensor for various dumbbelland prolate-shaped profiles can be found using the isoperimetric equalities
,
M11(3) + V(3) = (2743)C(5), PA3)
+ V(3) = 2CMI1(3)+ W I ,
(20) (21)
where C(5) for these profiles is given in Sections 11.2 and 11.3. Indeed, using (1 1.2.6) with n = 5 and (1 1.3.16),we derive the following results. For dumbbell-shaped bodies, defined by ( 1 1.2.5),
+ d 2 ) 3 / 2- J : o p i ( ~ )dx],
(22)
where p = p,(x), is the equation of the profile and V(3) = ny-, pi(x) dx. Finally, for prolate-shaped bodies, defined by (1 1.3.1 l),
[
M, ,(3) = n 2b2(b2 + c2)l/' -
[:
p i ( x ) dx],
a
4b2(b2 + c2)II2 - /:ap;(~) dx].
11.lo. THE DISTRIBUTIONAL APPROACH TO SCATTERING THEORY
In this section we derive an approximation for the scattered field by distributing delta functions along the axis of symmetry. We discuss only acoustic scattering and start with the inhomogeneous Helmholtz equation
(V'
+
k2)U"X)
=
-4nF(x).
(1)
11.10. THE DISTRIBUTIONAL APPROACH TO SCATTERING THEORY
For the special case F ( x )
=
31 1
6(x), (1) has the solution (see Section 10.9)
(2)
q X ) = eiklxl l l x l .
Let the scatterer S be defined as
Then for F ( x )
= ,f(x)h(y)6(z)
we find that
where R = [ ( x - < ) 2 + p 2 ] 1 / 2and p2 = y 2 + z2. We can now discuss the scattering of an acoustic wave by a soft spheroid (the Dirichlet problem) and that by a rigid spheroid (the Neumann problem).
The Dirichlet Problem
In the case of a soft spheroid the boundary condition is
where we have taken the incidence to be along the axis of symmetry. From
(4)and ( 5 ) we derive
J-c c
eikR
- f ( t )d t =
R
-eikx,
(x,p ) E S.
(6)
Now for small values of k we expand the functions eikx,eikR/R,andf(x) as follows : (ik)" n=O
R
and
n=O
(7)
312
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
Substituting these expressions in (6), we obtain the following simple set of integral equations:
where (x, p ) E S. Proceeding as in earlier sections, we find that the solutions of (lo)-( 14) are
+ e)/(I - e)}]-' = a - ' , = -2ueaP2 + ( a - 2e)-'x,
.fo(x) = [In{(] .f;(x)
a(3 - e 2 ) - 2e X2, 2a[(3 - e2)a - 6el
-
,f4(~)
+
+
2u2e3[-6e 2e3 a(3 - 2e2)] 2e)'[ - 30e 8e3 3(5 - 3e2)a]
+ ae [- 10e + 4e3 + ( 5 - 3e2)aI - _ x2 + x3, 3a2 2(a - 2e)[-30e + 8e3 + 3(5 - 3e2)a] = Do + Dlx + D2x2+ D4x4, (a
-
+
11 .lo. THE DISTRIBUTIONAL APPROACH TO SCATTERING THEORY
31 3
where
- 2e}{2e + 2e3 - (1 + ( 4 3 - e 2 )32a{(3 e2)a - 6e}
- e’)’.}
- 2e2(3 - 2e2) 9a2
-
1 + 192a {10e - 6e3 + 3(1 ~
+ 2e
-
(1 - e2)a D, 2u2
-
e2)2a)
+ $ { - 6 e + 10e3 + 3(1
D,
=
2u3e3/9(a - 2 c ~ ) ~ ,
D2
=
-2a2[(3 - e2)a - 6el-l
-
i . ’ ) ’ a } ~ ~ ] (21a) .
(21b)
e(2e - (1 - r2)a){2ea(4- e 2 ) - 12r2 - m’} 2a3{(3 - r 2 ) @- 6 e ) ~~
+ (5
a{
-
6eZ + e4)a - 2e 32a
+ 6e3
2e4 3a2
--
1
+ 26e3 + 3(5 - e2)(1 - e2)a}D4 , D4 = (1/8a)[6ea(42 36e2 + 6e4) (3 - e2)(35 - 3 0 2 + 3e4)a2 84e2 + 44e4][210e 1 10e3 - 3(35 - 30e3 + 3e4)a]-’ x [(3 - e2)a - 6 ~ 1 ’. -
-
30e
-
-
-
(21c)
-
(214
We have obtained the value o f f ’ ( x )up to 0(k4). To find the far-field behavior of the scattered field u ( x ) we appeal to (4). In terms of spherical polar coordinates, x = r cos 0, y = r sin 8, we have for r > > 1, R
= J(x
-
o2+ p 2
2:
I / R = I / J ( x - t>2+ /T”
r -
< cos O,
= l/r,
and eikR/~
(eikr/,.)e-ik
Substituting this value in (4) we obtain iis(r,
0)
N
(eikr/r)jl(8),
(22a) (22b)
314
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11,
where j l ( U ) is the far-field amplitude, given by .jr(8)=
J
e-ikCCOSef(5) dt.
~
(24)
-C
We next substitute in (24) the value of,f’(x)we have obtained. Then (23) yields the far-field value of the scattered fields:
+--3a
u3e3
-
-
2u3e3 cos u 3(cr - 2e)
ik3{7 2u4e3
(- 8’
+? !
cr2
3r
3
D2
u3e3 + __ cos2 8 3cr
2u4e4
5e) cos’ 8 9 3cr2
-
2a5e5
+ ___ 5 D4
}
2a5e5[30e2 - 14e4 - 5(25 - 14e’)ecr + 3(5 - 3e2)a2] 15(cr - 2e)’[ -30e 8e3 3(5 - 3e2)a]
+
+
COS
40e - 2cr(10 + 3e2) + 3ecr2(3 - eZ) a5e4jcos2 t) +{F+ 30a2[(3 - e2)a - 6e] use5 cos3 o + uses C O S ~8 . 15(ln(l + e)/(l - e ) 2e) 60a 4a5e6
~
-
The value of the scattering cross-section u , is given by u1 = 2n: Joljl(0)12 sin 5
16n:a2[ ( : ) 2
+ 2)4{:r
1
o (10 -
(6)’
+ 3 c(5) r - +e2}
1
40e2 - 2ecr(10 + 3e2) + 3e2cr2(3- e2) 30a2((3 - e2)a - 6e)
(94[-q2 + 4 ():
+3cr
-
%]}I.
O
315
11.lo. THE DISTRIBUTIONAL APPROACH TO SCATTERING THEORY
Neumann Problem for a Prolate Spheroid
In this case the boundary condition is
. VL,i
fj .VUS(X) = - f j
= -fj
.v p ,
(27)
where Applying (27) to (4), we obtain
I-, ',ikR
c
1 - ikR)(a2 - x ( ) , f ( ( )d t , (x, p ) E S. (29) R3 Next we use the same expansions as in the case of Dirichlet problem except that now $..p -
-(
2 ,fn(x)(iky+l. m
.f(x) =
(30)
n=O
Proceeding as in the previous case, wc obtain the following system of simple integral equations:
where (x. p ) E S. The values of,fi(x),i = 0, I , 2,3, are derived by a method similar to that of earlier sections. The result is
(Iz
.f;(-x) = - - ( a
2P
+ [I-
2r
=
2[4e(l -
2e)[2r
+ 2e(3 - 2r2) - 3(1 - OZ)a]-l(l
I-'
+ 4e - 3a x z , 6e~+ (3 - 2)a][22e + 9e2a -~22e/( I ~ - r 2 ) + 26r - 3(5 - ez)a][2e/( 1 - I?)
- ez
uz[.f2(x)
-
-
2)'
- eZ)] -
_
( 5 - @')a - 6e - 4r/( 1 - C J ~ ) 2(2~/(1- r z ) - a)[4e/(l - P') + 26c - 3(5 - (.')a] ~
~-
_
.Y33
(37)
31 6
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11.
and .f;(x) = Go
+ Glx + G2x’ + G4x4,
(38)
where G0
+ (a
G,
I - e’ 2e a - e)(( 2 I - e’
2e
-
=
(-
2e)G2
-
2e
1 - e’
+ 9a2[6e
-
+ 4e - 3a
4e3 - 3(1 - e 2 ) a 2 ] G 4 }
+ 4e - 3a) -’a’{? 1 [a’(1 2e- e’) + a - 4 e
2e
+ 4e - 3a 3e’)a
-
I-’
-
[8e3 - 30e
+ye3
66e
-
~
+ 3(5 - eZ)a]G4
1 - e2
+ (5 - 3e’)a +
’),
+ 4e - 3a]-
2e
and a is defined by (20). Putting this value off’(x)in (4) and using (22)-(24) withj, replaced by j 2 , we obtain the far-field behavior: u’(x)
N
-
eikr [ p 3 ( i k’ r
r(
e’)
-
- k’
2aeG0 + %a3e3G2+ 3a5e5G4
_ _a5e5 _ 30
+ 6(50
a)-’[ I
2e
~-
1 - e’
1392e’
-
48e’ 16e’)a’
-
1+
+ 5 a 5 e1 5- (e 4
ye2+ { + ’} [- + 26e
808
~
-
4e3] cos’ 8
a)-’
(3083 O}].
-
I-’
3(5 - e2)a
512 1-e
use2 2e cos 0 - __ 30 1 - e
x [5(l - e2)a - 1Oe -
2a3e3
I-’
4e - 3u
(39)
11.11.
31 7
STOKES FLOW
The scattering cross-section 0 , is given by 6,
5
47t -((ku)4u2 3 -kio2[ -
2e
(1 - e 2 ) 2
1
2e(I
-
u4
-
c2
c2)
e2(l - e 2 ) [5(l - e2)a - 10e 90
2e + 4e + 4e3][I- e2
-
I-'
+ 2 6 -~ 3(5 - e2)ci
+ 6(50
-
+
~6e')o2]
I-'
3ci
&?(A - a)-']}.
I-e
11.11. STOKES FLOW
Let us now study the flow of a viscous fluid in which a solitary body or a large number of bodies of microscopic scale are moving. They may be carried about passively by the flow, like solid particles in sedimentation, or moving actively, as in the locomotion of microorganisms. The linear differential equations (the inertia forces have a negligible effect on the flow) that describe their motion are the Stokes equations
v . u(x) = 0, V p ( x ) = pV2u(x)
(1)
+ F(x),
(2)
where u is the velocity vector, p is the pressure, p is the constant viscosity, and F is the external forcing function. We shall study only rotary motion. For this purpose we assume F to be solenoidal, that is, V . F = 0, so that (3)
F ( x ) = 47tpv x Q(x),
where Q is the vector potential. The far-field behavior of u and p is u
-+
0, p
--t
p m , a constant,
as r
=
1x1 -+ co.
Taking the divergence of both sides of (2), we find that V2p = p V 2 ( V . U )
+ V . F = 0,
(4)
318
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
which means that p is a harmonic function having a constant value pm at infinity. Accordingly, p = pa, in the entire space, and the differential equation (2) reduces to V2l1 = - ( l / p ) F ( x ) = -471v x 51(x). (5) The fundamental solution of (5) is obtained by taking 51(x) = $(x) where
y is a constant vector. It is readily verified that
(6)
is then the solution of (5). This solution is called a rotlet or couplet. Physically, this solution provides the velocity generated by a sphere of radius a rotating about the y axis with angular velocity oo= 1 y !/a3. For a volume distribution 51(x) that is a superposition of delta functions the solution is
Following the method of this chapter, let us take an axial distribution of rotlets of the form
(8) wheref(x) is the line distribution and 2 is the unit vector along the x axis. Thus 51 = &j(x)6(y)6(z)
(-c1 I x I CZ),
For a body with fore-and-aft symmetry, this equation yields
where p2 = y 2 + z2. Thus
In terms of cylindrical polar coordinates (p, 8, z), (1 1) is equivalent to u = (0, 0, u~),and
11.12.
DISPLACEMENT-TYPE BOUNDARY VALUE PROBLEMS
31 9
The boundary condition is u d x , P ) = W,P, on P = pO(x>, (13) where coo is the angular velocity of the body. From (12) and (13) we get the Fredholm integral equation,
This equation is a special case of (1 1.1.7). Accordingly, we can obtain the solution of the present problem by takingj'(5) given in Sections 11.2-11.6. For various examples in fluid mechanics solved by this method, the reader is referred to the work of Chwang and Wu [35]. The analysis for the rectilinear motion of bodies in Stokes flow can be obtained as an appropriate limit of their displacement field in elastostatics as given in the next section.
11.12. DISPLACEMENT-TYPE BOUNDARY VALUE PROBLEMS IN ELASTOSTATICS
For a homogeneous, linearly elastic, and isotropic medium, the displacement equations of equilibrium are [32]
1+
1
vI v . un + v2u " [
.f' = 0,
where u = ( u l , u 2 , u 3 ) denotes the displacement field, p the shear modulus, v the Poisson ratio of the material, and,fthe body force per unit of volume in a three-dimensional Euclidean space. In Exercise 13 of Chapter 10 we found that by taking f to be the Kelvin force ,fk
=
16np(1
-
v)b(x)a,
(2)
where x is the position vector of the field point and t~ is a constant vector that characterizes the direction and magnitude of the force, we can obtain the fundamental solution Uk(x; a ) =
(3 - 4v)a r
+-( a rx ) x , '
r
=
1x1.
(3)
This solution expresses the displacement field corresponding to a concentrated force located at the origin of the coordinates. Let us note in passing that U k ( x ;a ) = O(l/r)
as r
-+
m or r + O ,
(44
320
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
and, for any constant A , (4b)
Uk(x;Aa) = ALlk(x; a).
The net force F experienced by the medium bounded by a control surface S enclosing the singular point is given by F
=
s,
fi. T dS =
Jv
V . T dV =
i
-fk
dV = - 1 6 ~ ~ / ~ (V1) U ,
(5)
where ii is the unit outward normal at S, T is the stress tensor, and V is the control volume enclosed by S . The net moment M is obviously zero. In the absence of boundaries, a derivative of (3) of any order in an arbitrary fixcd direction is also a solution of (l), the corresponding forcing function being the derivative offk of the same order and in the same arbitrary direction. These derivatives are readily obtained by considering the displacement field corresponding to a concentrated force located at a point p # 0. The formal multipole expansion (similar to that in classical potential theory) in a Taylor series about x yields
Uk(x - p ; X )
=
U k ( x ;a) - ( p . V ) U k ( x a) ; + + ( / ~ * V ) ' U ' ( Xa)+ ; ... .
(6)
The interpretation of various terms on the right of (6) is obvious: The first term has already been explained in (3), the second term represents a Kelvin dipole characterized by the vectors a and /3, the third term represents a Kelvin quadrupole, etc. We write a Kelvin dipole
+ 3(a . x)(r5p . x ) x
(74
the corresponding forcing function being
j k d= - I ~ T C / L ( I - v ) ( p . V ) [ G ( x ) ~ r=] -16np(l
-
v ) [ ~ . V G ( X ) ] ~(7b) C,
where the superscript kd stands for Kelvin dipole. For fixed x , a Kelvin dipole (7a) is, in general, not symmetric with respect to an interchange of a and p. It is more convenient to handle a Kelvin dipole in terms of its symmetric and antisymmetric parts. The antisymmetric part is itself a physical entity of special significance:
3 [ U k d ( x a, ; p) - U k d ( x /), ; a ) ] = -2(1 - v)(a x
P)
x x/r3.
(8a)
This singularity is called a center ofrotation, and we denote it Ur.Next, set y = -2(1 - v)(a x
8).
(8b)
11.12.
DISPLACEMENT-TYPE BOUNDARY VALUE PROBLEMS
321
Then (8a), in view of (8b), can be rewritten
where the superscript r indicates the center of rotation. The corresponding forcing function is ,fr =
f [ , f k d ( xa, ; p) - ,fkd(x;p, a)]
=
4 7 V~ x~ [ h ( x ) ~ ] .
(8d)
The displacement field (8c) has a vector potential y/r and no scalar potential. The net force F and the torque M experienced by a control volume V containing the singular point and bounded by a surface S are given by
and
= JV(fr
x X) dV = -8xy,
respectively. The symmetric part of a Kelvin dipole yields another fundamental singularity, called a strrsslet. Its displacement field and the forcing function are
+ 3(a . x)(r5
x)x
and ,fk"
= -8
7 ~I ~-( v ) [ p . VG(X)C~ + ~ 1 VS(x)fl], .
(lob)
respectively. Physically, a stresslet represents a self-equilibrating system and hence contributes no net force or moment to the medium. It is of the nature of two double forces with moments in the plane determined by the vectors cy and p, the moment of one double force being equal in magnitude but opposite in direction to that of the other double force.
322
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
Yet another type of singularity is useful in the present investigation. It is called a center qfdilatarion, and it is characterized by a scalar. In this case the displacement field and the forcing function are
respectively. Thus, the displacement field is derivable from a scalar potential l/r and, like a stresslet, is a self-equilibrating singularity. A dipole formed by two centers of dilatation yields U dd (x;a)
=
-V(a.x/r3)
=
-(a.V)(-l/r)
=
-a/r3
+ 3(a.x)x/rS,
(13)
whereas a quadrupole formed by two dipoles or by four centers of dilatation leads Ud4(x; a, p) =
-(p
*
V)Lldd(x; a)
=
( f l . V)(a. V)Ud(X) = ( a . V)(fl. V)Ud(X)
= Ud4(x; /J, a ) = - V(fl . V)(a . V)( l/r).
(l4a)
This shows that the quadrupole Ud4 is symmetric with respect to an interchange of a'and p. In particular,
where C,, C,,, 2z are the base vectors in the x, y, z directions, respectively. These singular solutions are useful in constructing solutions to several boundary value problems involving different modes of displacement of rigid prolate and oblate spheroids, including their limiting configurationsthe sphere, slender body, and thin circular disk. We now proceed to demonstrate this by an example. Various other examples are given in reference [32]. Translation of a Prolate Spheroid
Consider a rigid prolate spheroid, as defined by (1 1.3.3), embedded in an unbounded elastic medium. It is given the displacement U
=
Uxex
+ U,e,.
(15)
There is no loss of generality in assuming the displacement to be in the form of (15) because the y and z axes can be so chosen that the z component of the specified displacement is zero. The displacement field in the elastic medium must (a) satisfy the homogeneous equations of equilibrium (1) (with.f = 0), (b) vanish at infinity, and
11.12.
DISPLACEMENT-TYPE BOUNDARY VALUE PROBLEMS
323
(c) meet the boundary condition (15) on the surface S of the spheroid. We attempt to construct the appropriate solution by employing a line distribution of Kelvin solutions (3) and dipoles formed by centers of dilatation (13) between the foci x = - c and x = c ; i.e., [a,Uk(x -
5 ; b x ) + a2 U k ( X - 5 ; by)] (15
J --c
where 4 = 52,. The first integral in (16) represents a line distribution of Kelvin solutions of constant strength a , and a2 oriented in the x and y directions, respectively. The second integral represents a line distribution of dipoles formed by centers of dilatation,each of parabolic density and pointing in the x and y directions. It is thus obvious that (16) satisfies the homogeneous (withf = 0) equations (1) of equilibrium in the elastic medium and vanishes as ( xl -+ x).To satisfy the boundary condition (2) on the surface S of the spheroid, we use the integrated form of (16), namely,
where b,, = ( J + ~+ _-bZ)/pis the unit radial vector in the y , z plane, R , and R 2 are defined in (1 1.2.2), and the quantities B,,,n are given by (1 1.6.3). The particular values B , . o , B , . , , I?,.,, B 3 . and B5.0have already been evaluated in (1 1.6.7); the remaining particular values needed are BL.1
=
R,
-
Rl
+ xBI,o,
B5.1 =
&l/R;
-
l/R:)
+ xBS.0.
(18)
Next we evaluate these quantities on the surface of the spheroid S. We find from (17) that the displacement on S is
+
b2xyb, u2yp?, b2(a2- e2x2) ’
324
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
where l + e L = ln-= 1-tJ
2tanhK'e.
Accordingly, the boundary condition (1 5) on the surface S is satisfied if we choose
+ 1 + 3e2 - 4ve2)L]-',
a, = -2e28,/(1 - e2) = U,e2[-2e
a2 = -2e2p2/(1
-
(20a)
e2) = 2U,e2[2e - (1 - 7e2 + 8ve2)L]-'.
(20b)
This choice of a,, p,, a 2 , and b2 in (17) furnishes the required solution. Next, we compute the net force F experienced by the spheroid by superposition of (9, F
= - 16np(l -
v)
s1,
(a,C,
+ a28,) d(
=
F,C,
+ F,&,,
(21a)
where the longitudinal and transverse components of the force are
+ (1 + 3e2 - 4ve2)L]-',
F,
=
-32np(l - v)U,ae3[-2e
F,
=
-64np(I - v)Uyae3[2e - (1
-
7e2
(21b)
+ 8ve2)L]-',
(21c)
respectively. By the limiting processes v + f and 2pvV , u/( 1 - 2v) -+ - p (where p is the pressure), the elastostatic equations (1) reduce to the equations of motion for Stokes flow as given in Section 11.11. In this case u denotes the velocity field. By setting v = 3 in (21b) and (21c), we obtain the corresponding drag formulas when a rigid spheroid moves in a viscous fluid with velocity U . In the limiting case of a sphere ( b + a, or e -+ 0), the displacement field (17) and the net force (21) reduce respectively to 3a
u ( x ) = 10 - 12v
and
F r 1(3 - 4v)U
+
( U .x)x r3
a3 u.x (22) 10 - 12v ( 7) 3
+
F,
=
-24np(l - v)aU,/(5 - 6v),
(234
F,
=
- 24np( 1 - v)a U,/(5
(23b)
-
6v).
These results can be summarized as follows: An embedded rigid sphere of radius a is given a displacement U . The field in the elastic medium is the same as would be produced by (i) a concentrated force - 24np( 1 - v)aU/(5 - 6v) and (ii) a dipole of strength -a3U(10 - 12v) formed by centers of dilatation, both located at the center of the sphere, namely,
325
11.13. THE EXTENSION TO ELASTODYNAMICS
For the other extreme case, a very slender spheroid,
F,
=
-
32r~p(1 - v)uU, (6 - 8 v ) ln(2/~)+ 1 [ I
+ O(E)1.
When the elastic medium is under torsion, the displacement field is quadratic even without any inclusion. The problem is therefore rather complicated when a rigid inclusion is embedded in it. However, by this method this problem can be solved effectively, as explained in reference [36].
11.13. THE EXTENSION TO ELASTODYNAMICS
The concepts in the previous section can be extended to cases of steady state dynamic singularities [37-391. The singularities can then be applied to obtain the solutions for the rectilinear oscillations of rigid inclusions in an infinite homogeneous isotropic medium. The dynamical equations of elasticity are
( A + p ) grad div ZI + pV2u - p
aZu
at
+ ,f = 0,
where u is the displacement vector, L and p are Lam& constants of the elastic medium, p is the density of the medium, andf is the body force per unit volume. Let us qssume that f = foeiw', u = uoei'". Substituting these values in (1) and dropping the zero subscript, we have
( A + p ) grad(div u)
+ p V2u + p o 2 u + .f = 0,
or 1
grad div u
1
+ V2u + m2u + f
=
0,
(2) (3)
where m2 = p o 2 / p , and v is the Poisson ratio. The fundamental solution Udkcorresponding to a force .fdk(X) = 4np6(x)u,
(4)
where x = ( x l , x 2 , x3)is the field point, u is a constant vector, and the superscript d and k stands for dynamical Kelvin, located at the origin of the coordinate system is (see Exercise 14 of Chapter 10) - imr
r
(5)
326
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11.
where r = 1x1 and T~ = (1 - 2v)/2(1 - v). This solution will be called the dynamical Kelvin solution. The net dynamical force F experienced by a control surface S enclosing the singular point is given by F =
I
Jv Jv
n.Tds= -
= 47rpa
+ m2p
V . T d V = /vfdkdV+m2pJv~dk(x;a)dV
Udk(x,a) d V ,
where A is the unit outward normal to S , T is the stress tensor, and V is the control volume enclosed by S. The net moment M is clearly zero. In free space a derivative of ( 5 ) of any order in arbitrary direction is also a solution of (3). The corresponding forcing function is the derivative off dk of the same order in the same direction. These derivatives can be obtained easily by expanding Ddk(x,p; a), where p is a given vector, in a Taylor series about x . This expansion is Udk(x,B; a)
N
udk(x,a) - (j?.V ) u d k ( x a) ;
+ $ p . V ) 2 U d k ( a)~ ;+ . . . .
(7)
The interpretation of various terms on the right-hand side of (7) is as follows: The first term is the dynamical Kelvin solution just discussed, the second term represents the dynamical Kelvin dipole characterized by the vectors a and p, the third term gives the corresponding quadruple, and so on. The dynamical Kelvin dipole can be written u d k d ( x a, ; p)
=
- ( ~ . V ) U ~ ' ( Xa);
=
-a(/?.V)- r
- imr
1
-m2
[T
v(p.V ) ( a .V )
-1.
- imrr
- imr
-
r
(8)
while the corresponding forcing function is f
dkd
-
- 4 7rpL(p* V)G(x)a = -4lrpcp * VG(x)]r,
(9)
where the superscript dkd stands for dynamical Kelvin dipole. From (8) it follows that a dynamical Kelvin dipole is not symmetric with respect to a and p. Its symmetric and antisymmetric parts yield other important fundamental solutions. The antisymmetric part is + [ u d k d ( xa,; p) - U d k d ( xp, ; a)]
= +[ a( p. V)e-imr/r- /?(a.V)e-imr/r] =
-V x +(a x p)eWimr/r.
We shall call it the dynamical center of rotation and denote it Udr.If we set y = -12(a x p), the foregoing relation becomes Udr(x,y)
=
(V x y)e-'"'/r.
(10)
11.13.
327
THE EXTENSION TO ELASTODYNAMICS
The corresponding forcing function is
Since Ud' has only a vector potential ye-'"'/r and no scalar potential, the net force F vanishes, while the torque M experienced by the control volume V containing the singular point and bounded by S is
M
=
=
Js
x x ( f i . T ) d S =-
Jv,fdr
J"
(V.T) x xdV
x x dl/ + rn2p J " ~ d ' ( x ;x,
0)
x x dl/
The symmetric part of the dynamical Kelvin dipole gives another fundamental solution, which will be called the dynamical stresslet (dks). The corresponding displacement field and the forcing function respectively are
and
This fundamental solution represents a self-equilibrating system and accordingly contributes neither a net force nor a moment to the medium. Another useful fundamental solution in the present investigation is the potential dipole, which is characterized by a scalar function and is also useful in discussing vibrations in Stokes flow. The displacement field due to this potential is
We now demonstrate that these fundamental solutions are very effective in solving boundary value problems in elastodynamics. To fix the ideas we consider the case of a spheroid.
328
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
Translation of a Prolate Spheroid Let the rigid spheroid S, x2/a2 + p2/b2 = 1,
p2 = y 2
+ z2,
c2 = ( a 2 - h2) = a2e2,
(16)
where e (0 Ie I1) is the eccentricity and 2c is the focal length, be embedded in an isotropic and homogeneous elastic medium. It is excited by a periodic force with period 2n/o acting in the direction of its axis of symmetry, so the displacement of the points on S is
u = u2,,
(17)
where 2, is the unit vector along the x axis and U is a constant. Our aim is to find the displacement field in the elastic medium so that (3) and the boundary condition (17) as well as the far-field radiation condition are satisfied. For this purpose we apply the technique of distributing the appropriate singularities. In the present situation we have the following line distributions between the foci x = - c and x = c: U(X)
=
s1,
F1(<)Udk(X- <,2,) dx
+ [ I C F 2 ( < ) ( c 2- t2)Udd(x- <,ex) d5,
x E S,
(18)
<
where 4 stands for 2,. The first integral in (18) represents the line distribution of dynamical Kelvin solutions (5) of variable strength F,(<), while the second integral is the line distribution of potential dipole (15) with strength (c2 - t2)F2(<).Clearly (18) satisfies the differential equation and vanishes as 1x1 + m. To find F,(<) and F,(<) we apply the boundary condition (17) on the surface S of the spheroid. We obtain Ug,
=
fcF1(C)Udk(x - t, 8,) d t
SIC
F2(<)(c2 - l2)Udd(x-
+
t, 2,)
d<,
x E S.
(19)
It is obvious that we cannot easily find a closed form solution to equation (19). Accordingly, we use a perturbation technique for small values of the parameter in. For this purpose, we expand all the functions in (19) in powers of in:
11.13.
329
THE EXTENSION TO ELASTODYNAMICS
where Uk(X,
a) =
(3
-
4v)a
r
+-(a .r3x)x ’
u, (x, a ) = dd
a
- -
r3
+ 3(a.x)x r5
________
are the Kelvin solution and potential dipole, respectively, for the elastostatic field, as derived in the previous section. Substituting the expansions (20) in (19) and equating equal powers of m, we get the following system of equations: 1
rc
330
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11.
The solutions of (21) and (22) are available in Section 1 1.12, and we thus have 2r2
1 1
4(1 - v ) '1'l 2
e2
2e2
-
-
1
.
2
f21 = Ue2[-2e+(1 +3r2-4ve2)L]-',
'
f2 =
2i(2
+ ?)
3
(24)
.fl ~4~
+
where L = In[( I e)/( 1 - e)]. Next we substitute (24) and (25) in (23) and obtain
where $0
+ T ~ +) (1 - e 2 ) L ] - u2J21(-2e + L ) + $iue(2 + = afll[e(l + r4) - ( I - e ' ) ~ ] - .f21[4f? - (2 - e2)L], 1 2 '
= zu
.lll[e(I
t3).f1 2,
$2
=
and C, = (yC, (26) we set
-[%I
+ zCJp
-
+ ,f21(-2e + L)I,
~~).f11e
(274 (27b) (27c)
is the unit radial vector in the y , z plane. To solve
Substitution of (28) in (26) and simplification yields
11.13.
THE EXTENSION TO ELASTODYNAMICS
where
4o = - 1/2(1 - v)e + L, 4l = a'[e - { ( I - e2)(3 - 2v)/4(1 4 2
=
-2u2[6e
Al
=
[(14 - 1 2 ~ -) (7 ~
A2
=
-36e
0,
=
-(2e
02 =
-
(3
-
-
6v
-
+ 2(9 - 3e2)L, + L)/2(1 -
+
2e2)L],
-
v)}L],
3e2 + 2ve2)L]/4(1 - v ) ,
\I),
1 6 ~ 8e/(l - r 2 ) - 12L.
Equation (29) is satisfied if we choose 1
4( 1
-
(A0
V)
+ c2A2) = -
This is a linear system of equations with the following solution:
The net force F experienced by the spheroid can be computed by superposition of (6);
J
-5
Jv
332
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
where the volume integral is taken over the spheroid. Substituting the values of .fl ‘,f1 ,f, 3 , and II in this formula, we have =
PO[{l
+
12npuu2,
+
+ A,e’ + f 2 1 [ 2- 4(1 -(e-
where Po
=
1 1+
’ + 4e - ~ v e ) (urn)’ ~ ]
32npa(l
-
v)Ue3[-2e
zX
-
v)e2
O(~rn)~,
(39)
+ ( 1 + 3e2 - 4ve’)L]-’.
Various other rotary and rectilinear oscillation problems can be solved by this method and are presented in references [38, 391.
11.14. DISTRIBUTIONS O N ARBITRARY LINES
The concepts of the distribution of the delta function and its derivatives can be extended to include any straight line in the plane. Indeed, in the theory of bending and buckling of elastic plates one has to apply concentrated loads on lines inside the plate, the so-called line loads. The classical approach is to take a strip of finite width, so that the problem is first solved for the case of loads on an area; then the width is made to approach zero. The problem can also be solved easily with the help of the theory of distributions. As an illustration, let us find the bending of a rectangular plate 0 I x I a, 0 Iy I b of small thickness h when a periodic concentrated load of uniform strength P is applied on its diagonal line x 2 = (b/u)x, (see Fig. 11.5). This load p ( x , , x 2 , t ) can therefore be written [34] p ( x l , x 2 , t ) = PeiwrS(x2- bx,/u)(l
+ b2/a2)li2,
where the factor ( 1 + b2/a2)’/’has been added for dimensional reasons. The equation of motion for the plate in the foregoing situation is 9 V4v
+ h d2v/dt’
=
P e ilDi 6 ( x 2 - b x , / a ) ( l
+ b’/uZ)”’,
(1)
(2)
where u denotes the deflection of the plate in the x 3 direction, 9 = Eh3/12(l - v’) is the flexural rigidity (also called bending stiffness), E is Young’s modulus, v is Poisson’s ratio, p is the density, and the differential operator V4 is v4 = a4/ax:
+ 2 a 4 / a x : ax: + a4/ax;.
11.14.
333
DISTRIBUTIONS O N ARBITRARY LINES
Fig. 11.5. x 2
= (h/a)x,.
Let us assume that the plate has simply supported edges so that the boundary conditions are u
=
d2v/8x:
=
0,
x1 = 0 and
x 1 = a;
u
=
J2v/dx:
=
0,
x2 = 0 and
x2
=
(3)
b.
The solution of the boundary value problem (2) and (3) is given by the double series
where A,, is the amplitude, +,
is the phase, and
Vmn(xl,x 2 ) = sin(mnx,/u) sin(nnx2/b), are eigenfunctions for this boundary. The corresponding eigenvalues A,,, are
A,,
= x2[mZ/o2
+ n2/b2](9/ph)1/2.
Substituting (4) in (2), we obtain the following values for the amplitude and phase, respectively:
where
Amn =
(Frnn/Ain){[l
+,
tan- '[2[rnndm,,(&
=
+ 41in((0/1rnJ2)-
- ((~/Arnn)~I~ -
to2)- '3,
1'2,
334
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
a
0 Fig. 11.6. x 2 = cx,
+ d.
XI
and l,, is the modal damping factor. Using the sifting property of the delta function, we have F,, = 2P(u2
+ b2)'l2phub(l + b2/~2)1/2&,,n,
where,a, is the Kronecker delta. We have thereby obtained all the values of the required solution (4) to the deflection v of the plate. By the same token the load can be distributed on any straight line x2 = cxI d that intersects the plate (see Fig. 11.6). In this case the corresponding load is Peiorb[x2- (cxl + d)](l + c2)'I2.
+
11.15. DISTRIBUTIONS ON PLANE CURVES
Let us now distribute the generalized functions on a plane curve x2 = f(xl) that intersects the elastic plate as in Fig. 11.7. Accordingly, the value of the load is d x , , x2, t ) = PeiW'G[x2 - f(x1)lCl
+ (f'(x1)) 21
112
.
We have already discussed the delta function 6[x2 - f(xl)] in Section 5.8. As an example, let us take the curve to be a parabola, x2 = b(xl/u)2. In this case the differential equation for the deflection v of the plate is
Proceeding as in Section 11.14 [see (1 1.14.4)], we again obtain the value of
11.16.
DISTRIBUTIONS ON A CIRCULAR DISK
335
Fig. 11.7. x 2 = f ( x , ) .
the deflection ti(xl, x2, t ) , where the quantities Fmn are now
The next natural extension is to distribute the generalized functions on laminae and surfaces and to solve the corresponding boundary value problems for configurations such as shells. This is easily accomplished with the help of the surface distributions explained in Chapter 5. We content ourselves with discussing in the next section the distribution on a circular disk.
11.16. DISTRIBUTIONS ON A CIRCULAR DISK
The ideas of the previous sections can be extended by distributing the singularities on a circular disk. We take F(x) = ,f(p)h(x),
0 I 0 I 2 n , 0 I p I b,
(1)
where (x, p, 0) are cylindrical polar coordinates andf(p) is defined over the disk 0 p I b, 0 5 8 27c. Then (1 1.1.3) reduces to
336
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11.
If we set cc
=
8,
is easy to see that
- 8, it
Jo2n
+ p2 + 5’
[x’ =
do, - 2 5 p cos(8 - 8 1 ) ] 1 / 2 dcc
Jo2’
[x’
+ p2 + 5 2 - 2 5 p cos cc]””
Thus (2) takes the form
Next, we use the expansion [x’
+ p 2 + t 2 - 25p cos a ] - ” 2
in (3), where So, is the Kronecker delta and we use the orthogonality of the cosine functions. The result is u(x, p ) = 271
Suppose that u(x, p ) becomes
J; J ~ . I ( t ) t J o ( P P ) J o ( p 6 ) e -
(4)
l x l p d 5 dP.
= A(x, p )
on the surface S , p
= po(x), so
that (4)
where the kernel K ( x , 5, p ) is
Equation ( 5 ) is a Fredholm integral equation ofthe first kind forf(5) provided the equation p = p o ( x ) of the surface S is known. The inverse problem pertains to finding the equation of the surface when A(x, p),f(5), and b are prescribed. For the special case ,f(5) = S(5 - b), (4) reduces to u(x, p ) = 2 n b
J;
J o ( p p ) J o ( p b ) e - l x l pd p ,
which is the electrostatic potential due to the point sources on a ring.
(7)
In this chapter we present solutions to boundary value problems, arising in various disciplines of mathematical physics, for axially symmetric bodies, including dumbbells, elongated rods, and prolate bodies, of which spheres and spheroids are special cases. The method rests on exploring the fundamental solutions of partial differential equations, as presented in the previous chapters, and then taking a suitable axial distribution of the Dirac delta function and its derivatives on a segment of the axis of symmetry of the body. This idea is extended to include distribution of these functions on arbitrary straight lines, curves, and disks [31-391.
11 . I . POISSON'S EQUATION
Let us consider Poisson's equation in n dimensions,
where u ( x ) is the generalized n-dimensional potential, F ( x ) is some forcing function, and x = (xl, x 2 , . . . , x,) is the position vector. In Section 10.4 we
288
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
found that for the special case F(x) tion E(x):
=
6(x) we obtain the fundamental solu-
r = 1x1. (2) u(x) = E(x) = l / F 2 , This value of u(x) gives the generalized electrostatic potential due to an n-dimensional sphere of radius I . For a distribution F ( x ) of delta functions over a volume V, the corresponding value of u(x) is
where * denotes the convolution and the variable of integration is x’. Our aim is to use ( 3 ) to obtain the generalized electrostatic potential of a large class of axially symmetric bodies for which a sphere, a dumbbell, and a spheroid are special cases. For this purpose, we prescribe an axial distribution of sources along the axis of symmetry, which we take to be the x1 axis. Accordingly, F(x)
=
.f’( x1)6(x2)6(x,)...6(xn)
(-el
2
x1
I
c2),
(4)
wheref(x,) is the strength of the distribution and c 1 and c 2 are two positive constants. For bodies with fore-and-aft symmetry we have c 1 = c2 = c. In terms of cylindrical polar coordinates, the shape of an axially symmetric body can be written p
=
po(x),
--a
I x Ia,
po(+u)
=
0,
(5)
where we have let x denote x 1 and where p 2 = x: + x i + . . . + x i . In view of the symmetry, po(x) = po(-x). The geometry of this configuration is explained in Fig. 1 1.1. In the next stage, we substitute F given by (4)in ( 3 ) and obtain
where we have used the sifting property of the delta function. If the potential u is prescribed on the surface S to be g(x), then (6) becomes
This is a Fredholm integral equation of the first kind for,f(x). Using this technique we can solve the potential problems for a wide class of axially symmetric configurations and in various fields of mechanics. Furthermore, we can solve both the direct and inverse problems. For the direct problem, the body profile p = po(x), as well as g(x), is prescribed, and we evaluate the
11.1,
289
POISSON'S EQUATION x2
b
Fig. 11.1. Line distribution of delta functions along the .vl axis in c, 5 .vl 5 c2 o f a n axially symmetric body.
distribution f(x) and the parameters c I and c 2 from (7). For the inverse problem we are given ,/'(x), g(x), and the parameters c I and c2: then (7) provides a body profile. The value g(x) = 1 is of special interest bccause it leads to the evaluation of the n-dimensional capacity of S. This is defined as C(n) = (11 - 2 )
where r l x
= dx'
rlx' . . . tlx"
=
s,,
F(.Y) ( I s ,
(8)
t1V. This rclation, in view of (4), becomes (2
C(n) =
(I1
- 2)
(9)
/'(<) ti<.
Let us now consider an axial distribution of dipoles along the axis of symmetry such that F ( x ) = h(x)h'(V)h(x,). . . h(xn),
1'1
I s I
('2.
(10)
where x 1 = x, x2 = y , and h ( x ) is the strength of the distribution. Thus, the dipoles are directed along the y axis. Substituting (10) in ( 3 ) , wc obtain
290
11.
Let u(x, p )
= yl(x)
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
on the surface S. Then ( 1 1) becomes
which, like (7), is a Fredholm integral equation of the first kind for h(x), provided that the equation of the surface S , p = p,(x), is known. Consequently, if h(x), l(x), e l , and c2 are prescribed, then (12) yields us the body profile. In the sequel we take c 1 = c2 = c.
11.2. DUMBBELL-SHAPED BODIES
Let us take two isolated delta distributions of equal strength such that
+ c) + A,h(x
,f(x) = A,h(x
- c),
where A , is a constant. This gives a dumbbell-shaped body, as shown in Fig. 11.2. We substitute in ( 1 1.1.7), and find that for g(x) = 1 we have
1
=
Ao(l/R;-2
+ I/R;-2),
(1)
where Rl
= [(x
+ c)' + p2I1I2,
R2 =
+
(2)
[(x - c ) ~ p2]*I2.
To obtain the relationship between the geometric parameters c/a and d/a (see Fig. 11.2), we evaluate ( I ) at the terminal points x = f a , p = 0 and at the dumbbell neck x = 0, p = +d. The result is 1 A,
--
1
(a
1
+ cy-2 + (a - cy-2'
1
A, - (c2 -
2
(3)
+ d2)'"-2)/2'
Elimination of A, yields the desired relation:
Furthermore, from ( I ) and (3) we derive the equation for the n-dimensional dumbbell :
1
[(x
+ c)2 + p2]'"-2'/2
1
+
[(x - c)2
+
p2]'"-2"2 - (u
1 + (a + cy-2 1
Cy-2'
(5)
11.2.
291
DUMBBELL-SHAPED BODIES
P
-0
1
,
A0
x
-C
Fig. 11.2. Dumbbell-shaped bodies that can be represented by two delta functions
The capacity of this configuration follows from (1 1.1.9): C(n) = 201
-
2 ) A , = 2(n - 2)
1
(a
+ c ) ' - ~+ (a
-
cy-2
I-'
Electrostatics
This corresponds to the case n respectively,
C(3)
=
3. In this case ( 1 1.1.6) and (4)-(6) become,
= (a2 - c2>/a=
(c2
+ d2)1/2.
(10)
Thus, the capacity of a dumbbell in three dimensions is equal to that of a sphere of radius (c2 + d2)1/2. Axially symmetric configurations are presented in Fig. 11.3 for different values of d/u.
292
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11.
I
-1.0
I
-0.8
I
-0.0
I
I
-6.4
-0.2
V 0
I
0.2
I
1
1
0.4
0.6
0.0
1
1.0
x/o
Fig. 11.3. A class of axially symmetric bodies generated by a pair of delta functions. The numbers along the p/a axis designate values of d/a.
Perfect Fluids
The relation +(n - 2)
=
UpnP3(n- 3)-’[1 - ~ ( n ) ]
(1 1)
relates the stream function +(n - 2) for an ( n - 2)-dimensional body of revolution moving with a uniform speed U along its axis of symmetry in a perfect fluid (or, equivalently, the fluid streams past the body with speed U ) with potential u(n)in a space of n dimensions. This potential function assumes the value unity on the surface of the body and vanishes at infinity. Thus, for the flow of a perfect fluid in three dimensions, we need the value of 4 5 ) . Now, from relations (1 1.1.6), (1 1.2.3), and (1 1.2.4), for n = 5 it follows that
C(5) =
Then (1 1) yields
6(a’ - c ’ ) ~ a(a’
+ 3c2) = 3(c’ + d*)3”.
11.2.
293
DUMBBELL-SHAPED BODIES
This gives the value of the stream function for a wide class of dumbbellshaped bodies. Cavities in Elastic Medium
The solution of a torsion problem for a shaft with a cavity can be obtained from the solution of the electrostatic potential for a seven-dimensional body, with g( x ) = 1, provided the cavity and the body have the same meridian profile. This isoperimetric equality is (15) 11/(5) = ap4c1 - u(7)~, where 11/ now stands for the elastic stream function. The strain energy E is E
=
(2G/n)[(2~~/15)C(7) - V(5)],
(16)
where C(7) is the capacity, V(5) is the volume of the body, and G is the shear modulus. The present method is suitable for solving some of these problems in a very simple fashion. For instance, for a spherical cavity we takef‘(x) = a56(x). Then (1 1.1.6) and (1 1.1.9) yield u(7)
= a5/(X2
+ p2)5’2,
C(7)
= 5a5.
Substituting these values in (1 5) and (16) and using (3.3.3) for V ( n )for n we obtain 11/(5) = ip4[1 - a5/(x2
+
p2)5’23,
E
= ~ ~ 7 1 0 5 .
=
5,
(17)
Similarly, for a dumbbell-shaped cavity,
C(7)
=
5(a2 - c 2 ) 5
a(a4
+ 10a2c2 + 5c4)’
Accordingly,
and E
=
nG
4(a2 - c 2 ) 5
3a(a4
+ 1oa2c2+ 5c4)
(19)
294
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11.3. UNIFORM AXIAL DISTRIBUTIONS
Capacity (n = 3)
Let the uniform distribution be
f (x) = A, a const. Then for g(x) = 1, (1 1.1.7) yields 1
-
A
=
In
R2
- (X
R1 - (X
- C)
+ c)’
where R , and R2 are defined in (1 1.2.2). This equation is equivalent to (p2
(p’
+ (x - C)2)1” - (x - c ) + (x + c)2)1’2 - (x + c ) =
where A is a constant, or X2
[c2(4A3
+ 8A2 + 4A)]/4A( I - A)’
+
[c2(4A3
P’
+ 81’ + 4A)]/( 1 -
= ’’
which is the equation of a spheroid. The equation of a prolate spheroid is x2/u2
+ p2/b2 = 1,
b2 = ( I
-
e2)u2, c = ue,
(3)
where e is the eccentricity. Accordingly (2) becomes 1 l + e -=InA l - e
=
2 tanh- e.
(4)
The value of the potential function u(x, p), from ( 1 1.1.6), is u ( x , p ) = (2 tanh-’ e)-’ In
R2
- (X - C)
R,
-
-
(x
+ c>’
(5)
and applying (1 1.1.9), we obtain the value of the capacity: ue(tanh-’ e)-’. (6) An interesting limiting c a w i s that of a slender body, which is obtained by setting b/u 6 1 . In this case (6) reduces to C
where
E =
b/u
=
=
(1 - e2)lj2 is called the slenderness parameter. Equation
11.3.
295
UNIFORM AXIAL DISTRIBUTIONS
(5) can be put in familiar form by using prolate-spheroidal coordinates
t, 4 4 :
x
=
ctu,
+ iz = c[(t2
y
e
l)(I -
-
l/to. (7)
=
Then (5) takes the form
where Qo(<) is the Legendre function of the second kind. The corresponding results for an oblate spheroid can be derived by replacing c by ic and e by ie(1 - e 2 ) -112 in (5) and (6) and by replacing t by i t in (8). The corresponding formulas are C(3)
=
c/sin-' e
=
be/sin-' e,
(9)
and u(x, p )
=
'
Qo(it)/Qo(iSo) = cot - t/cot -
' to.
(10)
+ 1 or to-+ 0, the oblate spheroid reduces to a thin circular disk of radius b and (9) and (10) reduce to
As r
C(3)
=
2b/~,
U(X,p )
=
2/7~cot - ' 5.
The results for a sphere follows upon setting e
=
0.
Perfect Fluids (n = 5 )
In this case, if we set,f(x) = A and g(x) = 1 in (1 1.1.7), we obtain
where R and R 2 are given by ( 1 1.2.2).For the semimajor axis (x and the semiminor axis (x = 0, p = b), (1 1) yields the relations 2ac(a2 - c2)-2 = 1/A,
2cb-'(b2
+ c')''~ = 1/A.
=
a, p
=
0)
(12)
Hence (a2 - c ~ = )ab2(b2 ~ + c~)'/~,
(13)
which relates the parameters b/a and c/a. The shape function p = po(x; a; c) is now known from (1 I)-( 13). These bodies are prolate shaped. The value of the potential 4 5 ) follows from relations ( 1 1.1.6): (14)
296
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
Next, we use (11.2.1 1) to obtain the stream function for the motion of these configurations in a perfect fluid,
where A is given by (12). The capacity C(5) is C(5) = 6Ac =
3(a2 - c 2 ) 2 U
=
3b2(b2 + c2)’j2.
(16)
Two limiting cases are of special interest.
(I)
c + 0, A
+ cx) such
that c A
+
iu3.
In this limit we recover the results for a sphere. (2) b/a 6 1 leads to an elongated rod. In this case (13)-(16) reduce to C b 7 b 4 - - - 5 - 6 (;)3 o(;) ,
+
U
(20) respectively. The body shape in this limit, from ( I I), is b
x2(3c2 - x2) 4(c2 - x2)4
+
(c’ - x2) >> b2.
(21)
Cavities in an Elastic Medium ( n = 7)
The relation corresponding to (2) and (1 1) in this case is
while those corresponding to (12) are 2ac(a2 + c2) - 1 (a2 - c2)4 A’
2 4 2 2 + 3b2) - -1 3b4(c2 + b2)3/2- A .
(23)
11.3.
297
UNIFORM AXIAL DISTRIBUTIONS
Hence a(a2 (a2
+ c2)
-
2c2 + 3b2
-
- 3h4((.2 +
c2)4
(24)
h2)3/2'
which relates the parameters b/a and c/a. Furthermore, (22)-(24) provide the body shape p = p o ( x ; a ; c). The quantities u(7) and C(7) are u(7) =
["
+
3b4(c2 h2)3'2 _ +_c _ x_ ~ _ _ 2c(2c2 + 3b2)p4 R ,
C(7)
=
5(a2 - ~ ~ ) ~ / +a cz) ( n =~ 15h4(c2 + h2)3'2/2c2+ 3h2.
(26)
We can then appeal to relations ( 1 1.2.15) and (1 1.2.16) to obtain
[
$ ( 5 ) = ip4 1 x ~
3h4(cZ+ h 2 ) 3 / 2 2c(2c2 + 3h2)p4
+ c - .Y -
C'
R,
-
1 (x + c)3 3 R:
-
~
2c2 + 3h2
p-.
where we have used V ( 5 ) = in2 p:(x) dx. As c + 0, A + x' such that c A + 1/2a5, we recover the results for a spherical cavity in Section 11.2. For a cavity with the shape of an elongated rod, that is, for h/u 6 I , we have
C(7) =
[:' (a)" +
(c2 - x2) 2 h Z ,
+ O(&)'], E
=
[;
G7cab4 c ~
4
O(;)y,
-
8 - 3(!)4(;
+ i c y+ o + 2c4(a2 3(u2 -
(361
.
( 9 3
A class of configurations is sketched in Fig. 11.4 for various values of h/u.
298
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
x /a
Fig. 11.4. A class of axially symmetric cavities of uniform distribution of delta functions. The numbers along the p/a axis designate the values of h/a.
11.4. LINEAR AXIAL DISTRIBUTIONS
In this section we study the polarization potential, so that g(x) = -Eox, where E o is the uniform electrostatic field in which a conductor is lying. Furthermore, we take (1)
J(x) = Bx.
Then (1 1.1.7) yields -Eox
=
[
B xln
r2
- (x -
'I} - R ,
R, -(x+c)
+ R.].
Proceeding in the same manrier as we did following (1 1.3.2), we find that the right side of (2) is a multiple of x only if the shape of the body is a spheroid. For a prolate spheroid, as defined by (1 1.3.3), we find from (2) that B
=
[
E , 2e - In
(t ':)I-'
__
11.5.
299
PARABOLIC AXIAL DISTRIBUTIONS, n = 5
The potential 4 3 ) is obtained from (1 1.1.6): u(x, P ) =
2r - In[(l
Eo
+ t.)/(l
-
r)]
[ {”’ .Y
In
-
(x
-
‘)}-R1
R1 - ( x + c )
+ R,].
(3)
The polarization potential P ( 3 ) is
In terms of the prolate spheroidal coordinates defined by ( 1 1.3.7), (3) and (4) take the forms
4 5 , V ) = -cEo5VQi(O/Qi(5o) and
p(3)
-%5;c35i(5;
-
1)”’CQ~(0/Q1(5o)l7
respectively.
11.5. PARABOLIC AXIAL DISTRIBUTIONS, n = 5
Let us now take f(x) = A
Setting y(x)
=
+ B.u2,
Is1 5
(1)
C.
1 and substituting this value of,/’(.v)in ( I 1.1.7) yields X+(’
xz x + c
x+c +
p’
(T-
-)
x -c RZ
This relation is again valid only for a spheroid. For a prolato spheroid as defined by ( I 1.3.3), we find that
300
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
which determines the values of both A and B. Appealing to (11.1.6), we obtain
This in turn enables us to use ( 1 1.2.11) to obtain the stream function for perfect fluids,
'+I,
In prolate spheroidal coordinates (1 1.3.7), (4) and (5) become u(x, p ) =
(5)
- In c2 A [ t225 - 1 5-1
-
~
respectively, furthermore, from (1) and (1 1.1.9) we obtain C(5) =
4a3e3 2e/( 1 - e2) - In[( 1
+ e)/( 1 - e)]
'
The values of these quantities for an oblate spheroid can be obtained by using the limiting process, as explained for the n = 3 case in Section 11.3. They are
. [ 2Ai
5
u(x, p ) = 7 - cotc 5 +1
$(3)
=
+
' 51.
fUcZ(1 t2)(1 - 1 2 ) - iUA(1 x [-cot-' 5 5/(1 593,
+
+
+ 52)(1 - $)
C(5) = -2b3e3[e(1 - e2)1'2 - sin-' e l- ' ,
'
'.
where in this case A = +b2e2i[e(1 - e2)1'2 - sin- e l - As e -+ 1 we derive the corresponding results for a thin circular disk of radius b : 4x9
P)
=
(2/.)C5/(t2
+ 1)
-
'
cot- 51.
$(3) = + U P ( 1 + 52)( 1 - q 2 ) - ( U P / n ) (1 x C-ccot-l t 5/(1 t2)3,
+
+
+ t 2 ) ( 1 - $)
C(5) = 4b3/n. Similarly, if we set e
=
0, we obtain the corresponding formulas for a sphere.
11.6.
FOURTH-ORDER POLYNOMIAL DISTRIBUTION
301
11.6. THE FOURTH-ORDER POLYNOMIAL DISTRIBUTION, n = 7; SPHEROIDAL CAVITIES
Let us take
With this value of,f'(x)and y(x)
=
1, (1 1.1.7) yields
where
n = o , 1 , 2, . . . , m = - l , l , 3 , 5 , . . . . These satisfy the recurrence relation
Thus
302
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11,
Ill., = In
R , - (X - C) R1 - (X c)'
+
By evaluating these quantities on the surface of the prolate spheroid (1 1.3.3) and with a slight algebraic manipulation, it follows that c4BS., - 2c2B5.,
l + e _ _ _2e + In ____ + €5, = 3(1 4e3 . - e2), 1 e2 I-e -
(8)
Comparing (2) and (8) we obtain A,
=
A4c4, A , = -2c2A,,
A,
=
[3(1
y:2)2
-__ 2e_ 1 - e2
+ In Eel. 1-e
Next, we make use of (1 1.1.6) and (1 1.1.9) and obtain
4 7 ) = A4[c4B5. 0 - 2c2B5,2 C(7)
=
+ B5,
(10)
41,
yu5e5A4.
( 1 1)
We substitute these values in (1 1.2.15) and ( 1 1.2.16) and derive
11/(5) = aP4[1 E
=
- A4(cL4Bs.o-
2c'Bs.z
(16Gnu5/45)[4e5A, - 3(1
-
+ BS.,)],
e2),],
where we have used V ( 5 ) = (8n2a5/15)( 1 - e2),,
The corresponding relations for an oblate spheroid are C(7) = -%b5es[(2e3 + e)(l - e 2 ) ' I 2 - sin-' E = -(16.rrG/45)[8h5e5~(2e3+ e)(l + 3b5(1 - e 2 ) ' I 2 ] .
el-'
e2)l', -
sin-' e}-'
When e -+ 1, these relations reduce to their corresponding values for a penny-shaped crack of radius b, C(7)
=
16b5/3n,
E
=
EGh5.
11.7.
303
THE POLARIZATION TENSOR FOR A SPHEROID
11.7. THE POLARIZATION TENSOR FOR A SPHEROID
The polarization potentials ui,i = 1, 2, 3, created by uniform electric fields in the x i (x = x I , y = x 2 , z = x3) directions, respectively, are obtained by solving the boundary value problems V2Ui =
-47cF,(x), on S ,
ui = - x i
+
ui
=
o ( l / r 2 ) as r
--+
x.
(1)
where r2 = x2 y’ + z 2 and S is the surface of the spheroid, as defined in ( I 1.3.3). The boundary value problem (4) for i = 1 can be solved by taking the line distribution of delta functions along the axis of symmetry (x axis), that is, F , ( x ) = .1;(x)h(y)h(z).Then, following the analysis of Section 11.1, we find that
Letting (x, p ) approach S and using the boundary condition, we have
which is a Fredholm integral equation of the first kind. This equation has been solved in Section 11.4; the solution is
Then (2) becomes ul(x, p )
=
[
2e - In
~
‘3”.
I-e +
hiR2 - (x R1 - (X
+ ‘I}
-
C)
-
R,
+ R2],
(5)
where R and R 2 are defined by (1 1.2.2). In prolate spheroidal coordinates, (5) takes the form ~ 1 ( 5 9
YI)= -doYIQi(O/Qi(to).
(6)
Solutions of the boundary value problems (1) for i = 2 , 3 are obtained by taking the line distributions of dipoles, that is, F 2 ( x ) = i2(x)h’(y)b(z)and F , ( x ) = j;(x)h(y)d’(z)along the axis of symmetry. The values of ui(x, p), i = 2, 3, follow by appeal to ( 1 1.1.1 1). Indeed,
304
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
In view of the boundary conditions, we find that (7) reduces to the Fredholm integral equation
The solution of (8) is achieved by taking
Then (8) yields .u+c
x - c
Rz - (X - C ) Rl
- (X
+
C)
11
,
1x1 I n .
(10)
This relation is valid for the prolate spheroid ( 1 1.3.3) if
Combining (7), (9), and ( I 1) we obtain the values of
Iti,
i
=
2, 3,
In terms of prolate spheroidal coordinates, these relations take the forms ~ z ( 5 Vt ,
CP)=
-C(COS
~p)C(sIi-
-
V ~ ) ] ” ~ Q ~ ( O / Q (13) ~(~~)~
and 143(5.
YI.cp) = -+in
q>C(5;- 1)(1 - Y I ~ ) I ” ~ Q ! ( ~ ) / Q : ((14) ~~).
The formula for the components Qij
= -
s
ui
Qij
of the polarization tensor is
du . rlS dn
From the symmetry of the spheroid it is clear that Qij = 0,
i # .j.
(16)
11.7.
305
THE POLARIZATION TENSOR FOR A SPHEROID
In terms of the eccentricity c Qii
=
I/C0, these take the forms
=
+
4ntr3(1 - e2)[ln---1 e 3 I - r
Q 2 2 = Q33 =
-
x
4ntr3(1 - e 2 ) 2e--- 2 e 3 3 I - e2 2t?
[I
1
= -
4nh3(1 3
+ "I-' .
In
LEI I-e
In (20) I-r e2 in (19) and (20), we derive the corresponding ~~~~
-
Replacing e by ie( I - e 2 ) values for an oblate spheroid: QI
fi
2e ][ln - 2 ~ ] l ' , (19) 1-r2 I-e
("
[sin-' e
-
e2)Il2] sin- r x [r( I
- e2)Ij2 -
sin-
I
03- I .
(22)
-+ 1, the oblate spheroid reduces to a thin circular disk of radius h, and (21) and (22) become
As
Q11
=
0,
Q22
= Q33 =
16b3/3,
respectively. The limiting case of a sphere follows by letting r have
(23) -+
0. We then
(24) Q I 1 = Q 2 2 = Q 3 3 = 8xu3/3, as is well known. The limiting case of a slender body is obtained by taking h/u = ( 1 - L ) ~6 )1. ~Then ~ ~(19) and (20) reduce, respectively, to
306
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11.8. THE VIRTUAL MASS TENSOR FOR A SPHEROID
In this section we consider the potentials ui(x), i boundary value problems
=
1, 2, 3, that satisfy the
V2ui = -4nFi(x),
i = 1, 2, 3.
i l . Vui = - A . i i on S,
ui = O( l/r2)
(1)
at
E,
(2)
where A is the unit vector normal to the surface S and the i i ,i = 1,2,3, are the unit vectors in the x, y, z, directions, respectively. Then, the u i , i = 1,2,3, are the velocity potentials of a body with surface S and moving with unit velocity in a perfect fluid in the x, y, z directions, respectively. For a prolate spheroid, fi is
A
=
( I :+,
( a 2 - e2x2)-'I2 - x,tl
CI
-
yi2
+
(3)
We solve these boundary value problems by the same distribution of singularities as in Section 11.7. That is, we set F,(x) = ,fl(x)S(y)G(z),F2(x) = .f2(X)b'(Y)8(Z), and F3W = .f2(X)&Y)WZ). The integral representation formulas for u l , u 2 , u3 are given by (11.7.2) and (1 1.7.7). Applying the boundary conditions (2) to these formulas, we obtain the integral equations
The solution of (4) is obtained by settingfl(x) =
A[a2B3. 1 - xB3.
=
Ax, 1x1 I c, so that
21,
where the quantities B,,,n are defined by (11.6.3). On the surface of the spheroid (1 1.3.3) this relation becomes x
=
2e
A [ -1 - e
-
In I+e] 1 - e X,
307
11.8. THE VIRTUAL MASS TENSOR FOR A SPHEROID
which yields the value of A . Thus,
Equations ( 5 ) are solved by setting
,fi(x) = A.
-
A2x2,
i
=
(7)
2, 3, 1x1 I c.
Substituting (7) in (9,we obtain, after a slight algebraic manipulation,
Hence,
Next, we substitute these values of ,fi(x) in the integral representation formulas ( I 1.7.2) and ( 1 1.7.7) and obtain
U,(.X,
p) =
-
u3(x,
p) =
-
2e(l - 2eZ) -
[
2e(l - 2 2 ) 1-e2 ~~
The virtual mass tensor
In Ice]-1y(czB3.0 1-e
-
1-e
B,.,),
(10)
Z(C~B, . ~B,.,).
(11)
In%]-', I-e
(13)
-
w.jis defined as
Substituting (9)-( 11) in (12), we obtain
w,,
=
W,,
=
4nu (1 - e2) 3 ~
W,,
=
4TU3 3 x
and
2e
--
I +e In-1-e
[
wj
=
0
I-e
I'
2e(I - 2eZ) 1 - LJ2 for i # ,j.
-'
308
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
By letting e -,0 in these relations, we recover the corresponding values for a sphere of radius a : W,,
=
W,,
=
W,, = 2na3/3,
which are well known. For a slender body, these relations reduce to
w,,= w,,=
~
w.j
The value of for an oblate spheroid follows from ( I 3)-( 15) on replacing e by ie(1 - e2)-'I2: W,,
=
W2,
= W33 =
(4nb3/3)[(1- e2)'12sin-' e - e][e(l
- e2)'12 -
(4nb3/3)(I - e')[e(I - e2)lI2 - sinx [(I - e2)lI2sin-' e - e - e 3 ] - ' ,
' el
sin-'
el-',
(19) (20)
and Rj = 0 for i # j . In the limiting case of a thin circular disk of radius b, (19) and (20) reduce to W1, = $b3,
W,,
=
W,, = 0.
(21)
11.9. THE ELECTRIC A N D MAGNETIC POLARlZABlLlTY TENSORS
The electric and magnetic polarizability tensors P i j and M i j are defined in terms of the tensor Qij and as follows:
wj
where Iij is the identity tensor and V is the volume of the body. Accordingly, we can use the results of Sections 11.7 and 11.8 to find the values of P i j and Mij.
11.9.
THE ELECTRIC AND MAGNETIC POLARlZABlLlTY TENSORS
For a prolate spheroid, 1+e 8na3e3 PI,= pn- 2e] 3 I-e
-1
(3)
,
~
309
In
I f e l ' , (4) I-e
~ , , 8na3e3 = ~ [ ~2e - l n ! + e , ] I-e - 1
(5)
(6)
1 - e2
P.. = M . . = 0,
i # j.
For an oblate spheroid,
P II
=
-(4nh3e3/3)(1 - e2)"'[(1
P,,
=
P 3 3 = -(8nb3e3/3)[e(1 - e2)li2- sin-' el-',
-
e2)'l2 sin-' e - e l - ' ,
(7)
(8)
M I , = -(4nh3e3/3)[e(l - e2)'/, - sin-' e l - ' , M,,
=
M,,
-(8nb3e3/3)(1 - e2)'/,[(1 - e 2 ) ' / 2 s i n - L e - e - e 3 ] - I ,
=
P,,
=
P2? =
MI,
=
M,,
P33 =
=
M,,
P . . = M.. = 0,
For a slender body, 4na (a2 - h2)(ln: 3
P,,
=
P,,
= P 3 3 = __
~
2nab2 (41n:
-
l)-'[l
Mll =
2na(a23- 3b2) (a'p
M,,
M,,
2na -(2a2 3
-
- jI 3b2)
In
4nu3,
=
ha3,
i # j.
,I):(.
(1 1) (12) (13)
+
- l ) ( l n T 2a
3
=
(10)
P I.J . = M IJ. . = 0, i # ,j.
For a sphere,
=
(9)
;)-'[
+
-
1
+
"')-'[I
+)'I,
+ (I(:)(15) ,],
310
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
For a thin circular disk,
PI,
P22
0,
=
M I , = $b3,
=
P33
=
yb3,
M22 = M 3 3 = 0.
(18) (19)
P I.J . = MlJ. . = 0,i # j . The values of component M, of the magnetic polarizability tensor and component PZ2of the electric polarizability tensor for various dumbbelland prolate-shaped profiles can be found using the isoperimetric equalities
,
M11(3) + V(3) = (2743)C(5), PA3)
+ V(3) = 2CMI1(3)+ W I ,
(20) (21)
where C(5) for these profiles is given in Sections 11.2 and 11.3. Indeed, using (1 1.2.6) with n = 5 and (1 1.3.16),we derive the following results. For dumbbell-shaped bodies, defined by ( 1 1.2.5),
+ d 2 ) 3 / 2- J : o p i ( ~ )dx],
(22)
where p = p,(x), is the equation of the profile and V(3) = ny-, pi(x) dx. Finally, for prolate-shaped bodies, defined by (1 1.3.1 l),
[
M, ,(3) = n 2b2(b2 + c2)l/' -
[:
p i ( x ) dx],
a
4b2(b2 + c2)II2 - /:ap;(~) dx].
11.lo. THE DISTRIBUTIONAL APPROACH TO SCATTERING THEORY
In this section we derive an approximation for the scattered field by distributing delta functions along the axis of symmetry. We discuss only acoustic scattering and start with the inhomogeneous Helmholtz equation
(V'
+
k2)U"X)
=
-4nF(x).
(1)
11.10. THE DISTRIBUTIONAL APPROACH TO SCATTERING THEORY
For the special case F ( x )
=
31 1
6(x), (1) has the solution (see Section 10.9)
(2)
q X ) = eiklxl l l x l .
Let the scatterer S be defined as
Then for F ( x )
= ,f(x)h(y)6(z)
we find that
where R = [ ( x - < ) 2 + p 2 ] 1 / 2and p2 = y 2 + z2. We can now discuss the scattering of an acoustic wave by a soft spheroid (the Dirichlet problem) and that by a rigid spheroid (the Neumann problem).
The Dirichlet Problem
In the case of a soft spheroid the boundary condition is
where we have taken the incidence to be along the axis of symmetry. From
(4)and ( 5 ) we derive
J-c c
eikR
- f ( t )d t =
R
-eikx,
(x,p ) E S.
(6)
Now for small values of k we expand the functions eikx,eikR/R,andf(x) as follows : (ik)" n=O
R
and
n=O
(7)
312
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
Substituting these expressions in (6), we obtain the following simple set of integral equations:
where (x, p ) E S. Proceeding as in earlier sections, we find that the solutions of (lo)-( 14) are
+ e)/(I - e)}]-' = a - ' , = -2ueaP2 + ( a - 2e)-'x,
.fo(x) = [In{(] .f;(x)
a(3 - e 2 ) - 2e X2, 2a[(3 - e2)a - 6el
-
,f4(~)
+
+
2u2e3[-6e 2e3 a(3 - 2e2)] 2e)'[ - 30e 8e3 3(5 - 3e2)a]
+ ae [- 10e + 4e3 + ( 5 - 3e2)aI - _ x2 + x3, 3a2 2(a - 2e)[-30e + 8e3 + 3(5 - 3e2)a] = Do + Dlx + D2x2+ D4x4, (a
-
+
11 .lo. THE DISTRIBUTIONAL APPROACH TO SCATTERING THEORY
31 3
where
- 2e}{2e + 2e3 - (1 + ( 4 3 - e 2 )32a{(3 e2)a - 6e}
- e’)’.}
- 2e2(3 - 2e2) 9a2
-
1 + 192a {10e - 6e3 + 3(1 ~
+ 2e
-
(1 - e2)a D, 2u2
-
e2)2a)
+ $ { - 6 e + 10e3 + 3(1
D,
=
2u3e3/9(a - 2 c ~ ) ~ ,
D2
=
-2a2[(3 - e2)a - 6el-l
-
i . ’ ) ’ a } ~ ~ ] (21a) .
(21b)
e(2e - (1 - r2)a){2ea(4- e 2 ) - 12r2 - m’} 2a3{(3 - r 2 ) @- 6 e ) ~~
+ (5
a{
-
6eZ + e4)a - 2e 32a
+ 6e3
2e4 3a2
--
1
+ 26e3 + 3(5 - e2)(1 - e2)a}D4 , D4 = (1/8a)[6ea(42 36e2 + 6e4) (3 - e2)(35 - 3 0 2 + 3e4)a2 84e2 + 44e4][210e 1 10e3 - 3(35 - 30e3 + 3e4)a]-’ x [(3 - e2)a - 6 ~ 1 ’. -
-
30e
-
-
-
(21c)
-
(214
We have obtained the value o f f ’ ( x )up to 0(k4). To find the far-field behavior of the scattered field u ( x ) we appeal to (4). In terms of spherical polar coordinates, x = r cos 0, y = r sin 8, we have for r > > 1, R
= J(x
-
o2+ p 2
2:
I / R = I / J ( x - t>2+ /T”
r -
< cos O,
= l/r,
and eikR/~
(eikr/,.)e-ik
Substituting this value in (4) we obtain iis(r,
0)
N
(eikr/r)jl(8),
(22a) (22b)
314
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11,
where j l ( U ) is the far-field amplitude, given by .jr(8)=
J
e-ikCCOSef(5) dt.
~
(24)
-C
We next substitute in (24) the value of,f’(x)we have obtained. Then (23) yields the far-field value of the scattered fields:
+--3a
u3e3
-
-
2u3e3 cos u 3(cr - 2e)
ik3{7 2u4e3
(- 8’
+? !
cr2
3r
3
D2
u3e3 + __ cos2 8 3cr
2u4e4
5e) cos’ 8 9 3cr2
-
2a5e5
+ ___ 5 D4
}
2a5e5[30e2 - 14e4 - 5(25 - 14e’)ecr + 3(5 - 3e2)a2] 15(cr - 2e)’[ -30e 8e3 3(5 - 3e2)a]
+
+
COS
40e - 2cr(10 + 3e2) + 3ecr2(3 - eZ) a5e4jcos2 t) +{F+ 30a2[(3 - e2)a - 6e] use5 cos3 o + uses C O S ~8 . 15(ln(l + e)/(l - e ) 2e) 60a 4a5e6
~
-
The value of the scattering cross-section u , is given by u1 = 2n: Joljl(0)12 sin 5
16n:a2[ ( : ) 2
+ 2)4{:r
1
o (10 -
(6)’
+ 3 c(5) r - +e2}
1
40e2 - 2ecr(10 + 3e2) + 3e2cr2(3- e2) 30a2((3 - e2)a - 6e)
(94[-q2 + 4 ():
+3cr
-
%]}I.
O
315
11.lo. THE DISTRIBUTIONAL APPROACH TO SCATTERING THEORY
Neumann Problem for a Prolate Spheroid
In this case the boundary condition is
. VL,i
fj .VUS(X) = - f j
= -fj
.v p ,
(27)
where Applying (27) to (4), we obtain
I-, ',ikR
c
1 - ikR)(a2 - x ( ) , f ( ( )d t , (x, p ) E S. (29) R3 Next we use the same expansions as in the case of Dirichlet problem except that now $..p -
-(
2 ,fn(x)(iky+l. m
.f(x) =
(30)
n=O
Proceeding as in the previous case, wc obtain the following system of simple integral equations:
where (x. p ) E S. The values of,fi(x),i = 0, I , 2,3, are derived by a method similar to that of earlier sections. The result is
(Iz
.f;(-x) = - - ( a
2P
+ [I-
2r
=
2[4e(l -
2e)[2r
+ 2e(3 - 2r2) - 3(1 - OZ)a]-l(l
I-'
+ 4e - 3a x z , 6e~+ (3 - 2)a][22e + 9e2a -~22e/( I ~ - r 2 ) + 26r - 3(5 - ez)a][2e/( 1 - I?)
- ez
uz[.f2(x)
-
-
2)'
- eZ)] -
_
( 5 - @')a - 6e - 4r/( 1 - C J ~ ) 2(2~/(1- r z ) - a)[4e/(l - P') + 26c - 3(5 - (.')a] ~
~-
_
.Y33
(37)
31 6
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11.
and .f;(x) = Go
+ Glx + G2x’ + G4x4,
(38)
where G0
+ (a
G,
I - e’ 2e a - e)(( 2 I - e’
2e
-
=
(-
2e)G2
-
2e
1 - e’
+ 9a2[6e
-
+ 4e - 3a
4e3 - 3(1 - e 2 ) a 2 ] G 4 }
+ 4e - 3a) -’a’{? 1 [a’(1 2e- e’) + a - 4 e
2e
+ 4e - 3a 3e’)a
-
I-’
-
[8e3 - 30e
+ye3
66e
-
~
+ 3(5 - eZ)a]G4
1 - e2
+ (5 - 3e’)a +
’),
+ 4e - 3a]-
2e
and a is defined by (20). Putting this value off’(x)in (4) and using (22)-(24) withj, replaced by j 2 , we obtain the far-field behavior: u’(x)
N
-
eikr [ p 3 ( i k’ r
r(
e’)
-
- k’
2aeG0 + %a3e3G2+ 3a5e5G4
_ _a5e5 _ 30
+ 6(50
a)-’[ I
2e
~-
1 - e’
1392e’
-
48e’ 16e’)a’
-
1+
+ 5 a 5 e1 5- (e 4
ye2+ { + ’} [- + 26e
808
~
-
4e3] cos’ 8
a)-’
(3083 O}].
-
I-’
3(5 - e2)a
512 1-e
use2 2e cos 0 - __ 30 1 - e
x [5(l - e2)a - 1Oe -
2a3e3
I-’
4e - 3u
(39)
11.11.
31 7
STOKES FLOW
The scattering cross-section 0 , is given by 6,
5
47t -((ku)4u2 3 -kio2[ -
2e
(1 - e 2 ) 2
1
2e(I
-
u4
-
c2
c2)
e2(l - e 2 ) [5(l - e2)a - 10e 90
2e + 4e + 4e3][I- e2
-
I-'
+ 2 6 -~ 3(5 - e2)ci
+ 6(50
-
+
~6e')o2]
I-'
3ci
&?(A - a)-']}.
I-e
11.11. STOKES FLOW
Let us now study the flow of a viscous fluid in which a solitary body or a large number of bodies of microscopic scale are moving. They may be carried about passively by the flow, like solid particles in sedimentation, or moving actively, as in the locomotion of microorganisms. The linear differential equations (the inertia forces have a negligible effect on the flow) that describe their motion are the Stokes equations
v . u(x) = 0, V p ( x ) = pV2u(x)
(1)
+ F(x),
(2)
where u is the velocity vector, p is the pressure, p is the constant viscosity, and F is the external forcing function. We shall study only rotary motion. For this purpose we assume F to be solenoidal, that is, V . F = 0, so that (3)
F ( x ) = 47tpv x Q(x),
where Q is the vector potential. The far-field behavior of u and p is u
-+
0, p
--t
p m , a constant,
as r
=
1x1 -+ co.
Taking the divergence of both sides of (2), we find that V2p = p V 2 ( V . U )
+ V . F = 0,
(4)
318
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
which means that p is a harmonic function having a constant value pm at infinity. Accordingly, p = pa, in the entire space, and the differential equation (2) reduces to V2l1 = - ( l / p ) F ( x ) = -471v x 51(x). (5) The fundamental solution of (5) is obtained by taking 51(x) = $(x) where
y is a constant vector. It is readily verified that
(6)
is then the solution of (5). This solution is called a rotlet or couplet. Physically, this solution provides the velocity generated by a sphere of radius a rotating about the y axis with angular velocity oo= 1 y !/a3. For a volume distribution 51(x) that is a superposition of delta functions the solution is
Following the method of this chapter, let us take an axial distribution of rotlets of the form
(8) wheref(x) is the line distribution and 2 is the unit vector along the x axis. Thus 51 = &j(x)6(y)6(z)
(-c1 I x I CZ),
For a body with fore-and-aft symmetry, this equation yields
where p2 = y 2 + z2. Thus
In terms of cylindrical polar coordinates (p, 8, z), (1 1) is equivalent to u = (0, 0, u~),and
11.12.
DISPLACEMENT-TYPE BOUNDARY VALUE PROBLEMS
31 9
The boundary condition is u d x , P ) = W,P, on P = pO(x>, (13) where coo is the angular velocity of the body. From (12) and (13) we get the Fredholm integral equation,
This equation is a special case of (1 1.1.7). Accordingly, we can obtain the solution of the present problem by takingj'(5) given in Sections 11.2-11.6. For various examples in fluid mechanics solved by this method, the reader is referred to the work of Chwang and Wu [35]. The analysis for the rectilinear motion of bodies in Stokes flow can be obtained as an appropriate limit of their displacement field in elastostatics as given in the next section.
11.12. DISPLACEMENT-TYPE BOUNDARY VALUE PROBLEMS IN ELASTOSTATICS
For a homogeneous, linearly elastic, and isotropic medium, the displacement equations of equilibrium are [32]
1+
1
vI v . un + v2u " [
.f' = 0,
where u = ( u l , u 2 , u 3 ) denotes the displacement field, p the shear modulus, v the Poisson ratio of the material, and,fthe body force per unit of volume in a three-dimensional Euclidean space. In Exercise 13 of Chapter 10 we found that by taking f to be the Kelvin force ,fk
=
16np(1
-
v)b(x)a,
(2)
where x is the position vector of the field point and t~ is a constant vector that characterizes the direction and magnitude of the force, we can obtain the fundamental solution Uk(x; a ) =
(3 - 4v)a r
+-( a rx ) x , '
r
=
1x1.
(3)
This solution expresses the displacement field corresponding to a concentrated force located at the origin of the coordinates. Let us note in passing that U k ( x ;a ) = O(l/r)
as r
-+
m or r + O ,
(44
320
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
and, for any constant A , (4b)
Uk(x;Aa) = ALlk(x; a).
The net force F experienced by the medium bounded by a control surface S enclosing the singular point is given by F
=
s,
fi. T dS =
Jv
V . T dV =
i
-fk
dV = - 1 6 ~ ~ / ~ (V1) U ,
(5)
where ii is the unit outward normal at S, T is the stress tensor, and V is the control volume enclosed by S . The net moment M is obviously zero. In the absence of boundaries, a derivative of (3) of any order in an arbitrary fixcd direction is also a solution of (l), the corresponding forcing function being the derivative offk of the same order and in the same arbitrary direction. These derivatives are readily obtained by considering the displacement field corresponding to a concentrated force located at a point p # 0. The formal multipole expansion (similar to that in classical potential theory) in a Taylor series about x yields
Uk(x - p ; X )
=
U k ( x ;a) - ( p . V ) U k ( x a) ; + + ( / ~ * V ) ' U ' ( Xa)+ ; ... .
(6)
The interpretation of various terms on the right of (6) is obvious: The first term has already been explained in (3), the second term represents a Kelvin dipole characterized by the vectors a and /3, the third term represents a Kelvin quadrupole, etc. We write a Kelvin dipole
+ 3(a . x)(r5p . x ) x
(74
the corresponding forcing function being
j k d= - I ~ T C / L ( I - v ) ( p . V ) [ G ( x ) ~ r=] -16np(l
-
v ) [ ~ . V G ( X ) ] ~(7b) C,
where the superscript kd stands for Kelvin dipole. For fixed x , a Kelvin dipole (7a) is, in general, not symmetric with respect to an interchange of a and p. It is more convenient to handle a Kelvin dipole in terms of its symmetric and antisymmetric parts. The antisymmetric part is itself a physical entity of special significance:
3 [ U k d ( x a, ; p) - U k d ( x /), ; a ) ] = -2(1 - v)(a x
P)
x x/r3.
(8a)
This singularity is called a center ofrotation, and we denote it Ur.Next, set y = -2(1 - v)(a x
8).
(8b)
11.12.
DISPLACEMENT-TYPE BOUNDARY VALUE PROBLEMS
321
Then (8a), in view of (8b), can be rewritten
where the superscript r indicates the center of rotation. The corresponding forcing function is ,fr =
f [ , f k d ( xa, ; p) - ,fkd(x;p, a)]
=
4 7 V~ x~ [ h ( x ) ~ ] .
(8d)
The displacement field (8c) has a vector potential y/r and no scalar potential. The net force F and the torque M experienced by a control volume V containing the singular point and bounded by a surface S are given by
and
= JV(fr
x X) dV = -8xy,
respectively. The symmetric part of a Kelvin dipole yields another fundamental singularity, called a strrsslet. Its displacement field and the forcing function are
+ 3(a . x)(r5
x)x
and ,fk"
= -8
7 ~I ~-( v ) [ p . VG(X)C~ + ~ 1 VS(x)fl], .
(lob)
respectively. Physically, a stresslet represents a self-equilibrating system and hence contributes no net force or moment to the medium. It is of the nature of two double forces with moments in the plane determined by the vectors cy and p, the moment of one double force being equal in magnitude but opposite in direction to that of the other double force.
322
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
Yet another type of singularity is useful in the present investigation. It is called a center qfdilatarion, and it is characterized by a scalar. In this case the displacement field and the forcing function are
respectively. Thus, the displacement field is derivable from a scalar potential l/r and, like a stresslet, is a self-equilibrating singularity. A dipole formed by two centers of dilatation yields U dd (x;a)
=
-V(a.x/r3)
=
-(a.V)(-l/r)
=
-a/r3
+ 3(a.x)x/rS,
(13)
whereas a quadrupole formed by two dipoles or by four centers of dilatation leads Ud4(x; a, p) =
-(p
*
V)Lldd(x; a)
=
( f l . V)(a. V)Ud(X) = ( a . V)(fl. V)Ud(X)
= Ud4(x; /J, a ) = - V(fl . V)(a . V)( l/r).
(l4a)
This shows that the quadrupole Ud4 is symmetric with respect to an interchange of a'and p. In particular,
where C,, C,,, 2z are the base vectors in the x, y, z directions, respectively. These singular solutions are useful in constructing solutions to several boundary value problems involving different modes of displacement of rigid prolate and oblate spheroids, including their limiting configurationsthe sphere, slender body, and thin circular disk. We now proceed to demonstrate this by an example. Various other examples are given in reference [32]. Translation of a Prolate Spheroid
Consider a rigid prolate spheroid, as defined by (1 1.3.3), embedded in an unbounded elastic medium. It is given the displacement U
=
Uxex
+ U,e,.
(15)
There is no loss of generality in assuming the displacement to be in the form of (15) because the y and z axes can be so chosen that the z component of the specified displacement is zero. The displacement field in the elastic medium must (a) satisfy the homogeneous equations of equilibrium (1) (with.f = 0), (b) vanish at infinity, and
11.12.
DISPLACEMENT-TYPE BOUNDARY VALUE PROBLEMS
323
(c) meet the boundary condition (15) on the surface S of the spheroid. We attempt to construct the appropriate solution by employing a line distribution of Kelvin solutions (3) and dipoles formed by centers of dilatation (13) between the foci x = - c and x = c ; i.e., [a,Uk(x -
5 ; b x ) + a2 U k ( X - 5 ; by)] (15
J --c
where 4 = 52,. The first integral in (16) represents a line distribution of Kelvin solutions of constant strength a , and a2 oriented in the x and y directions, respectively. The second integral represents a line distribution of dipoles formed by centers of dilatation,each of parabolic density and pointing in the x and y directions. It is thus obvious that (16) satisfies the homogeneous (withf = 0) equations (1) of equilibrium in the elastic medium and vanishes as ( xl -+ x).To satisfy the boundary condition (2) on the surface S of the spheroid, we use the integrated form of (16), namely,
where b,, = ( J + ~+ _-bZ)/pis the unit radial vector in the y , z plane, R , and R 2 are defined in (1 1.2.2), and the quantities B,,,n are given by (1 1.6.3). The particular values B , . o , B , . , , I?,.,, B 3 . and B5.0have already been evaluated in (1 1.6.7); the remaining particular values needed are BL.1
=
R,
-
Rl
+ xBI,o,
B5.1 =
&l/R;
-
l/R:)
+ xBS.0.
(18)
Next we evaluate these quantities on the surface of the spheroid S. We find from (17) that the displacement on S is
+
b2xyb, u2yp?, b2(a2- e2x2) ’
324
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
where l + e L = ln-= 1-tJ
2tanhK'e.
Accordingly, the boundary condition (1 5) on the surface S is satisfied if we choose
+ 1 + 3e2 - 4ve2)L]-',
a, = -2e28,/(1 - e2) = U,e2[-2e
a2 = -2e2p2/(1
-
(20a)
e2) = 2U,e2[2e - (1 - 7e2 + 8ve2)L]-'.
(20b)
This choice of a,, p,, a 2 , and b2 in (17) furnishes the required solution. Next, we compute the net force F experienced by the spheroid by superposition of (9, F
= - 16np(l -
v)
s1,
(a,C,
+ a28,) d(
=
F,C,
+ F,&,,
(21a)
where the longitudinal and transverse components of the force are
+ (1 + 3e2 - 4ve2)L]-',
F,
=
-32np(l - v)U,ae3[-2e
F,
=
-64np(I - v)Uyae3[2e - (1
-
7e2
(21b)
+ 8ve2)L]-',
(21c)
respectively. By the limiting processes v + f and 2pvV , u/( 1 - 2v) -+ - p (where p is the pressure), the elastostatic equations (1) reduce to the equations of motion for Stokes flow as given in Section 11.11. In this case u denotes the velocity field. By setting v = 3 in (21b) and (21c), we obtain the corresponding drag formulas when a rigid spheroid moves in a viscous fluid with velocity U . In the limiting case of a sphere ( b + a, or e -+ 0), the displacement field (17) and the net force (21) reduce respectively to 3a
u ( x ) = 10 - 12v
and
F r 1(3 - 4v)U
+
( U .x)x r3
a3 u.x (22) 10 - 12v ( 7) 3
+
F,
=
-24np(l - v)aU,/(5 - 6v),
(234
F,
=
- 24np( 1 - v)a U,/(5
(23b)
-
6v).
These results can be summarized as follows: An embedded rigid sphere of radius a is given a displacement U . The field in the elastic medium is the same as would be produced by (i) a concentrated force - 24np( 1 - v)aU/(5 - 6v) and (ii) a dipole of strength -a3U(10 - 12v) formed by centers of dilatation, both located at the center of the sphere, namely,
325
11.13. THE EXTENSION TO ELASTODYNAMICS
For the other extreme case, a very slender spheroid,
F,
=
-
32r~p(1 - v)uU, (6 - 8 v ) ln(2/~)+ 1 [ I
+ O(E)1.
When the elastic medium is under torsion, the displacement field is quadratic even without any inclusion. The problem is therefore rather complicated when a rigid inclusion is embedded in it. However, by this method this problem can be solved effectively, as explained in reference [36].
11.13. THE EXTENSION TO ELASTODYNAMICS
The concepts in the previous section can be extended to cases of steady state dynamic singularities [37-391. The singularities can then be applied to obtain the solutions for the rectilinear oscillations of rigid inclusions in an infinite homogeneous isotropic medium. The dynamical equations of elasticity are
( A + p ) grad div ZI + pV2u - p
aZu
at
+ ,f = 0,
where u is the displacement vector, L and p are Lam& constants of the elastic medium, p is the density of the medium, andf is the body force per unit volume. Let us qssume that f = foeiw', u = uoei'". Substituting these values in (1) and dropping the zero subscript, we have
( A + p ) grad(div u)
+ p V2u + p o 2 u + .f = 0,
or 1
grad div u
1
+ V2u + m2u + f
=
0,
(2) (3)
where m2 = p o 2 / p , and v is the Poisson ratio. The fundamental solution Udkcorresponding to a force .fdk(X) = 4np6(x)u,
(4)
where x = ( x l , x 2 , x3)is the field point, u is a constant vector, and the superscript d and k stands for dynamical Kelvin, located at the origin of the coordinate system is (see Exercise 14 of Chapter 10) - imr
r
(5)
326
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11.
where r = 1x1 and T~ = (1 - 2v)/2(1 - v). This solution will be called the dynamical Kelvin solution. The net dynamical force F experienced by a control surface S enclosing the singular point is given by F =
I
Jv Jv
n.Tds= -
= 47rpa
+ m2p
V . T d V = /vfdkdV+m2pJv~dk(x;a)dV
Udk(x,a) d V ,
where A is the unit outward normal to S , T is the stress tensor, and V is the control volume enclosed by S. The net moment M is clearly zero. In free space a derivative of ( 5 ) of any order in arbitrary direction is also a solution of (3). The corresponding forcing function is the derivative off dk of the same order in the same direction. These derivatives can be obtained easily by expanding Ddk(x,p; a), where p is a given vector, in a Taylor series about x . This expansion is Udk(x,B; a)
N
udk(x,a) - (j?.V ) u d k ( x a) ;
+ $ p . V ) 2 U d k ( a)~ ;+ . . . .
(7)
The interpretation of various terms on the right-hand side of (7) is as follows: The first term is the dynamical Kelvin solution just discussed, the second term represents the dynamical Kelvin dipole characterized by the vectors a and p, the third term gives the corresponding quadruple, and so on. The dynamical Kelvin dipole can be written u d k d ( x a, ; p)
=
- ( ~ . V ) U ~ ' ( Xa);
=
-a(/?.V)- r
- imr
1
-m2
[T
v(p.V ) ( a .V )
-1.
- imrr
- imr
-
r
(8)
while the corresponding forcing function is f
dkd
-
- 4 7rpL(p* V)G(x)a = -4lrpcp * VG(x)]r,
(9)
where the superscript dkd stands for dynamical Kelvin dipole. From (8) it follows that a dynamical Kelvin dipole is not symmetric with respect to a and p. Its symmetric and antisymmetric parts yield other important fundamental solutions. The antisymmetric part is + [ u d k d ( xa,; p) - U d k d ( xp, ; a)]
= +[ a( p. V)e-imr/r- /?(a.V)e-imr/r] =
-V x +(a x p)eWimr/r.
We shall call it the dynamical center of rotation and denote it Udr.If we set y = -12(a x p), the foregoing relation becomes Udr(x,y)
=
(V x y)e-'"'/r.
(10)
11.13.
327
THE EXTENSION TO ELASTODYNAMICS
The corresponding forcing function is
Since Ud' has only a vector potential ye-'"'/r and no scalar potential, the net force F vanishes, while the torque M experienced by the control volume V containing the singular point and bounded by S is
M
=
=
Js
x x ( f i . T ) d S =-
Jv,fdr
J"
(V.T) x xdV
x x dl/ + rn2p J " ~ d ' ( x ;x,
0)
x x dl/
The symmetric part of the dynamical Kelvin dipole gives another fundamental solution, which will be called the dynamical stresslet (dks). The corresponding displacement field and the forcing function respectively are
and
This fundamental solution represents a self-equilibrating system and accordingly contributes neither a net force nor a moment to the medium. Another useful fundamental solution in the present investigation is the potential dipole, which is characterized by a scalar function and is also useful in discussing vibrations in Stokes flow. The displacement field due to this potential is
We now demonstrate that these fundamental solutions are very effective in solving boundary value problems in elastodynamics. To fix the ideas we consider the case of a spheroid.
328
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
Translation of a Prolate Spheroid Let the rigid spheroid S, x2/a2 + p2/b2 = 1,
p2 = y 2
+ z2,
c2 = ( a 2 - h2) = a2e2,
(16)
where e (0 Ie I1) is the eccentricity and 2c is the focal length, be embedded in an isotropic and homogeneous elastic medium. It is excited by a periodic force with period 2n/o acting in the direction of its axis of symmetry, so the displacement of the points on S is
u = u2,,
(17)
where 2, is the unit vector along the x axis and U is a constant. Our aim is to find the displacement field in the elastic medium so that (3) and the boundary condition (17) as well as the far-field radiation condition are satisfied. For this purpose we apply the technique of distributing the appropriate singularities. In the present situation we have the following line distributions between the foci x = - c and x = c: U(X)
=
s1,
F1(<)Udk(X- <,2,) dx
+ [ I C F 2 ( < ) ( c 2- t2)Udd(x- <,ex) d5,
x E S,
(18)
<
where 4 stands for 2,. The first integral in (18) represents the line distribution of dynamical Kelvin solutions (5) of variable strength F,(<), while the second integral is the line distribution of potential dipole (15) with strength (c2 - t2)F2(<).Clearly (18) satisfies the differential equation and vanishes as 1x1 + m. To find F,(<) and F,(<) we apply the boundary condition (17) on the surface S of the spheroid. We obtain Ug,
=
fcF1(C)Udk(x - t, 8,) d t
SIC
F2(<)(c2 - l2)Udd(x-
+
t, 2,)
d<,
x E S.
(19)
It is obvious that we cannot easily find a closed form solution to equation (19). Accordingly, we use a perturbation technique for small values of the parameter in. For this purpose, we expand all the functions in (19) in powers of in:
11.13.
329
THE EXTENSION TO ELASTODYNAMICS
where Uk(X,
a) =
(3
-
4v)a
r
+-(a .r3x)x ’
u, (x, a ) = dd
a
- -
r3
+ 3(a.x)x r5
________
are the Kelvin solution and potential dipole, respectively, for the elastostatic field, as derived in the previous section. Substituting the expansions (20) in (19) and equating equal powers of m, we get the following system of equations: 1
rc
330
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11.
The solutions of (21) and (22) are available in Section 1 1.12, and we thus have 2r2
1 1
4(1 - v ) '1'l 2
e2
2e2
-
-
1
.
2
f21 = Ue2[-2e+(1 +3r2-4ve2)L]-',
'
f2 =
2i(2
+ ?)
3
(24)
.fl ~4~
+
where L = In[( I e)/( 1 - e)]. Next we substitute (24) and (25) in (23) and obtain
where $0
+ T ~ +) (1 - e 2 ) L ] - u2J21(-2e + L ) + $iue(2 + = afll[e(l + r4) - ( I - e ' ) ~ ] - .f21[4f? - (2 - e2)L], 1 2 '
= zu
.lll[e(I
t3).f1 2,
$2
=
and C, = (yC, (26) we set
-[%I
+ zCJp
-
+ ,f21(-2e + L)I,
~~).f11e
(274 (27b) (27c)
is the unit radial vector in the y , z plane. To solve
Substitution of (28) in (26) and simplification yields
11.13.
THE EXTENSION TO ELASTODYNAMICS
where
4o = - 1/2(1 - v)e + L, 4l = a'[e - { ( I - e2)(3 - 2v)/4(1 4 2
=
-2u2[6e
Al
=
[(14 - 1 2 ~ -) (7 ~
A2
=
-36e
0,
=
-(2e
02 =
-
(3
-
-
6v
-
+ 2(9 - 3e2)L, + L)/2(1 -
+
2e2)L],
-
v)}L],
3e2 + 2ve2)L]/4(1 - v ) ,
\I),
1 6 ~ 8e/(l - r 2 ) - 12L.
Equation (29) is satisfied if we choose 1
4( 1
-
(A0
V)
+ c2A2) = -
This is a linear system of equations with the following solution:
The net force F experienced by the spheroid can be computed by superposition of (6);
J
-5
Jv
332
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
where the volume integral is taken over the spheroid. Substituting the values of .fl ‘,f1 ,f, 3 , and II in this formula, we have =
PO[{l
+
12npuu2,
+
+ A,e’ + f 2 1 [ 2- 4(1 -(e-
where Po
=
1 1+
’ + 4e - ~ v e ) (urn)’ ~ ]
32npa(l
-
v)Ue3[-2e
zX
-
v)e2
O(~rn)~,
(39)
+ ( 1 + 3e2 - 4ve’)L]-’.
Various other rotary and rectilinear oscillation problems can be solved by this method and are presented in references [38, 391.
11.14. DISTRIBUTIONS O N ARBITRARY LINES
The concepts of the distribution of the delta function and its derivatives can be extended to include any straight line in the plane. Indeed, in the theory of bending and buckling of elastic plates one has to apply concentrated loads on lines inside the plate, the so-called line loads. The classical approach is to take a strip of finite width, so that the problem is first solved for the case of loads on an area; then the width is made to approach zero. The problem can also be solved easily with the help of the theory of distributions. As an illustration, let us find the bending of a rectangular plate 0 I x I a, 0 Iy I b of small thickness h when a periodic concentrated load of uniform strength P is applied on its diagonal line x 2 = (b/u)x, (see Fig. 11.5). This load p ( x , , x 2 , t ) can therefore be written [34] p ( x l , x 2 , t ) = PeiwrS(x2- bx,/u)(l
+ b2/a2)li2,
where the factor ( 1 + b2/a2)’/’has been added for dimensional reasons. The equation of motion for the plate in the foregoing situation is 9 V4v
+ h d2v/dt’
=
P e ilDi 6 ( x 2 - b x , / a ) ( l
+ b’/uZ)”’,
(1)
(2)
where u denotes the deflection of the plate in the x 3 direction, 9 = Eh3/12(l - v’) is the flexural rigidity (also called bending stiffness), E is Young’s modulus, v is Poisson’s ratio, p is the density, and the differential operator V4 is v4 = a4/ax:
+ 2 a 4 / a x : ax: + a4/ax;.
11.14.
333
DISTRIBUTIONS O N ARBITRARY LINES
Fig. 11.5. x 2
= (h/a)x,.
Let us assume that the plate has simply supported edges so that the boundary conditions are u
=
d2v/8x:
=
0,
x1 = 0 and
x 1 = a;
u
=
J2v/dx:
=
0,
x2 = 0 and
x2
=
(3)
b.
The solution of the boundary value problem (2) and (3) is given by the double series
where A,, is the amplitude, +,
is the phase, and
Vmn(xl,x 2 ) = sin(mnx,/u) sin(nnx2/b), are eigenfunctions for this boundary. The corresponding eigenvalues A,,, are
A,,
= x2[mZ/o2
+ n2/b2](9/ph)1/2.
Substituting (4) in (2), we obtain the following values for the amplitude and phase, respectively:
where
Amn =
(Frnn/Ain){[l
+,
tan- '[2[rnndm,,(&
=
+ 41in((0/1rnJ2)-
- ((~/Arnn)~I~ -
to2)- '3,
1'2,
334
11.
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
a
0 Fig. 11.6. x 2 = cx,
+ d.
XI
and l,, is the modal damping factor. Using the sifting property of the delta function, we have F,, = 2P(u2
+ b2)'l2phub(l + b2/~2)1/2&,,n,
where,a, is the Kronecker delta. We have thereby obtained all the values of the required solution (4) to the deflection v of the plate. By the same token the load can be distributed on any straight line x2 = cxI d that intersects the plate (see Fig. 11.6). In this case the corresponding load is Peiorb[x2- (cxl + d)](l + c2)'I2.
+
11.15. DISTRIBUTIONS ON PLANE CURVES
Let us now distribute the generalized functions on a plane curve x2 = f(xl) that intersects the elastic plate as in Fig. 11.7. Accordingly, the value of the load is d x , , x2, t ) = PeiW'G[x2 - f(x1)lCl
+ (f'(x1)) 21
112
.
We have already discussed the delta function 6[x2 - f(xl)] in Section 5.8. As an example, let us take the curve to be a parabola, x2 = b(xl/u)2. In this case the differential equation for the deflection v of the plate is
Proceeding as in Section 11.14 [see (1 1.14.4)], we again obtain the value of
11.16.
DISTRIBUTIONS ON A CIRCULAR DISK
335
Fig. 11.7. x 2 = f ( x , ) .
the deflection ti(xl, x2, t ) , where the quantities Fmn are now
The next natural extension is to distribute the generalized functions on laminae and surfaces and to solve the corresponding boundary value problems for configurations such as shells. This is easily accomplished with the help of the surface distributions explained in Chapter 5. We content ourselves with discussing in the next section the distribution on a circular disk.
11.16. DISTRIBUTIONS ON A CIRCULAR DISK
The ideas of the previous sections can be extended by distributing the singularities on a circular disk. We take F(x) = ,f(p)h(x),
0 I 0 I 2 n , 0 I p I b,
(1)
where (x, p, 0) are cylindrical polar coordinates andf(p) is defined over the disk 0 p I b, 0 5 8 27c. Then (1 1.1.3) reduces to
336
APPLICATIONS TO BOUNDARY VALUE PROBLEMS
11.
If we set cc
=
8,
is easy to see that
- 8, it
Jo2n
+ p2 + 5’
[x’ =
do, - 2 5 p cos(8 - 8 1 ) ] 1 / 2 dcc
Jo2’
[x’
+ p2 + 5 2 - 2 5 p cos cc]””
Thus (2) takes the form
Next, we use the expansion [x’
+ p 2 + t 2 - 25p cos a ] - ” 2
in (3), where So, is the Kronecker delta and we use the orthogonality of the cosine functions. The result is u(x, p ) = 271
Suppose that u(x, p ) becomes
J; J ~ . I ( t ) t J o ( P P ) J o ( p 6 ) e -
(4)
l x l p d 5 dP.
= A(x, p )
on the surface S , p
= po(x), so
that (4)
where the kernel K ( x , 5, p ) is
Equation ( 5 ) is a Fredholm integral equation ofthe first kind forf(5) provided the equation p = p o ( x ) of the surface S is known. The inverse problem pertains to finding the equation of the surface when A(x, p),f(5), and b are prescribed. For the special case ,f(5) = S(5 - b), (4) reduces to u(x, p ) = 2 n b
J;
J o ( p p ) J o ( p b ) e - l x l pd p ,
which is the electrostatic potential due to the point sources on a ring.
(7)