DISCUSSION AND CONCLUSION

-68-

Since no solution of this integral equation has been found we can only speculate as to the outcome of the variational method. To sound a pessimistic note, it may turn out that the solution fails to have direct physical application; it might, for instance, fail to satisfy the pres - sure condition

r{g)

~ 0 or it could be that

r

(±1)

'f

0 in which case 13

l l

would become logarithmically infinite as

g -

± 1 and the flow would be many sheeted.

Nonetheless, such a solution (if it actually does minimize I) would provide an absolute lower bound for the drag which could then be used as a basis for comparison with other minimum drag profiles which are physically relevant (such as those found by the parameter method}.

Final judgment of the usefulness of the variational technique as a design tool should await a more thorough investigation (most likely numerical) of the equations of Ch. V, § 3.

>'c

In the parameter method, expressions for D ^{0 ,} s

0 , and y

0 ,

were found in terms of {N+2) parameters - - A, K, a , a ,

l z

- - by setting a = ; and expanding

r

1 and 13

1 in finite Fourier series with N terms. For a minimum I, we solve the (N+2) equations oI/ oA = 0,

oI/aK

= 0, and 81/oan = 0, n = 1, 2, . . . , N. These equations plus the equation s /y = k (k, a given number greater than unity) are all that are need-

o 0

ed to solve for the {N+3) unknowns - - A. , A. , K, a , • . . , a N- - since

1 2. l

A drops out of these equations. Finally, the drag coefficient is given

>'c

·by CD= D' /y

0•

This procedure was carried out for the case N

=

1 and found to give reasonable results, although it is difficult to tell just how good they are. To do this, one should solve the variational problem (which

essentially corresponds to N = oo) or do the parameter method for N

=

2,3, etc., as was done in the two examples considered in Ch. IV,

§ 1.

The parameter problem for N

>

l would most likely require use of the computer, although it may be possible to find analytic solu- tions by series expansion for K near unity.

Aside from investigation of the variational equations and exten- sion of the parameter method to N

>

1, several areas for further study include the following: (1) An extension of the variational calculus method of Ch. IV to handle inequality constraints and constraints on the values of f (or g) at the endpoints, the case f(±l) = 0 being particular- ly important; (2) Application of the methods of this paper to the mini- mum drag problem and the hydrofoil problem with the possible use of more complicated finite cavity flow models, and; (3) Application of the variational technique of Ch. IV to other fields. The linearity of the integral equations for the case of functionals of quadratic form makes this method particularly well suited for application to problems involv- ing energy constraints.

-70- REFERENCES

1. Gilbarg, D. (1960) Jets and Cavities. Handbuch der Physik, Vol.

IX, p. 311 - 438. Berlin: Springer-Verlag.

2. Milne-Thomson, L. M. (1960), Theoretical Hydrodynamics, 4th edition. Macmillan Company.

3, Wu, T. Y. (1962), A Wake Model for Free-Streamline Flow Theory.

J. Fluid Mech, .!2_, p. 161 - 181.

4., Muskhelishvili, N. I. (1946), Singular Integral Equations.

Groningen, Holland: P. Noordhoff Ltd.

5. Tricomi, F. G. (1957), Integral Equations. Interscience Publishers Inc.

6. Gradshteyn, I. S. and Ryshik, I. M. (1965 ), Table of Integrals, Series, and Products. p, 316, formula 3. 365. Academic Press.

7. Abramowitz, M. and Ste gun, I. A. (1964 ), Handbook of Mathematical Functions. National Bureau of Standards.

111

-71-

x

1-1

u

0

-h

y

-72-

Fig. ·2(a) - Lavrentieff¹s Solution

y

Fig. 2(b) - A Sequence of Plate Shapes with Zero Drag· in the Limit as h - O.

l

---

s .,,.,,,. , -

FREE STREAMLINE

0 x

FREE STREAMLINE

Fig. 3 - Example of Changing the Plate Shape without Changing the Flow,

I --J w

i

f - plane

"'

^.

S A

0

¢

s'

^B

(a )

I

C -

^plane -.] ^,.!::.

77

^I

f=~AU~

²

T

B .. s'

ol ^s

^A

^...

^~

I I

-c ^-I +I +c

( b )

Fig. 4 - Transfo'rmation from Complex Potential Plane to (.-plane.

,.=-J_(v+l..)

~ 2 11

OI

I I I I I

: / L _j/

_E

I I I I I

I

E

'-/

11- plane

_1,s -! ^~I '. _______ +: ^!1

^er

Fig. 5 - The Path of Integration in the v-plane.

I ---!

U"1 I

A

B

( i )

a

₁

=

-·7T/2

K

=

f.0

0 0

A

B (ii)

a₁.=

-7T/4

K = 0.2

s

0

s'

A

8 ( iii ) a _I

= 0

K

=

^0.3

0

Fig. 6 - Various Plate Shapes for the Case N

=

1.

A

B

(iv)

0 I

= 1.0

K

= 0.4

I ..._.]

O' I

0.10

0.08

a1

0.06.

0.04:

0.02 I

0 0.1

6<0

0.2 0.3 0.4

...

0.5 K

0.6

6>0

0.7

Fig. 7 - The Curve a = f(K) on which b:(a ,K) = 0.

1 1

I I

-.J -.J I

0.8 0.9

1.0

Co Coo

1.2

1.0

0.8

0.6

_o

1 =f(K)

0.4

0.2

K

= I,

_{0 1}

< O ro,=-7T

!_

²

4 +7T

Co /Coo,...., ^1T

+

8k

OS k __,.... 00

OL....---1---"---'---...J..---'---

_1.0

_I.

_I

_1.2

_1.3

_1.4

k =

S₀

/y

₀

Fig, 8 .., CD vs k Plo~s for the Two Cases for which b.= 0, CDo = 2;r/ (4+n) is the Drag Coefficient of a Flat Plate.

I -J cc

-79-

A

1.0

k =

1.03

k =

1.05

k =

1.10

k =

l.25 0.5

k

= ^1.50

0

x

s'

8

Fig. 9 - Optimum Plate Profiles for the Case N = 1.

-80-

0.1

-1.0 -0.5 ₀ 0.5 1.0

c =I.

I

-0.1

c = 1.4

c =

^1.8

-0.4

c = 2.0

-0.5 ^c = 3.0

c =

⁰⁰

-0.6

Fig. 10 - Linearized Solution.

APPENDIX A

CALCULATION OF PLATE LENGTH AND CHORD From Eq. (3. 8),

2i 1 ) vz

+ 7

^dv

4l A (I

+

2iI - 2I

1 z 3 2iI

+

I )

4 5

(A. 1) N

where O(v) =

l

^a n ^{v 2n-l} This expression is evaluated by taking the n=l

path of integration along Le defined as follows: The imaginary axis, from v = i to v = ie; the arc

l

^v

l

= e, from v = ie to v = E; and the real axis, from v = e to v =K (see Fig. 5). We then let the radius of the circular arc go to zero. Integrals I and I are

4 5

integrated by parts before integrating along LE.

The following notation for

n (

v} on v = i 7 is used:

N

-iO(i7}:: l\;(7)

= l

(-l}n+ian7zn- ^l n=l

Integrals I and I can be evaluated directly ori L (i.e. ,

1 ? 0

E = O}.

(A. 2}

• (S

0

1

• --v<r>tdt

+ s:

co• n(t)tdt)

+

i

S

0

1( sinl"l(t)tdt

-82-

I

=

\Kein(v)dv

=

ro e-lj;(T\dT+

SK

ei11(a)d0"

z

Ji

^{j 1 0}

S

^{K ( r• K}

s

^{1 )}

=

cos n(t)dt + i

j

sin n(t)dt - e -lj;{t)dt

0 0 0

I is evaluated on L :

3 E

I

= ~

^K ein( v)

~ = s

^e e -4; (T) d T +

3

Ji

^{V l T}

S

^K i11(a) da

+

e - _O"

E

In the limit ^E

-o,

S

_1T ^{o el·H• ee . ...,( HJ).}

^ide

2

{A. 3)

{A. 4)

The asterisks on the integral signs mean that the singular parts of the integrands should be combined; i.e. ,

>!pK

j

cos n(t) .t dt ):Iol e-4;{t) dtt -

SOK{

cos n(tt)-e -lj;{t) }dt

0

on L .

E

S

¹ -lj;(t) dt

+

e -

K t

The integral ^I is first integrated by parts and then evaluated

4

l

^K ein(v) dV

I ₄

= v

l

e -in(v)

- -

_v

K

i

+ SK ein(l'~n•(v)

^dv

. l

In the limit E - 0,

I ^{COS ~2(K)}

+

4

= ( -

K al. ; -

t

^K "'(t)sin "(t)

d" )

+

i ( - sin ~2(K)

K

^{- e -ljJ(l)}

>l

1 e -ljJ(t)ljJ.(t) dtt

0

+'\ ^*

^K

^{~2 1} (t)cos~~t)

^{dtt )}

.;o

The int.egral I must first be integrated by parts twice:

5

I =

SK eH~v)

^{dv = -}

_!_ ei~l(v) I

^K

^{+_!_SK ei~l(v)}

H2 •(v) dv

" . 3 2 2. • 2 . ²

" l v v ^{l l} v .

=(_!_

eHl(v) + 1

i~l(v)·

^{H2·1(v) )/ k:}

+

I +I

2 v 2

2

e v i 51 sz

where

I =

_!_SK

eiU(v) Li~l •(v)

J2. -

^dv

51 2 . v

l

I ₌

_!_SK

e Hl(v)Hl¹¹(v) dv

-

52. 2 . v

l

Now,

(A. 5)

-84-

- ^~ s

^K eH1(J)[n'(o-)]2

~

€

Letting ^{E -} 0, gives

. s

^{K dt}

+

i !. a² - ~ [n¹(t)]2 sinO(t) -

4 l 2 . t

0

I can be evaluated directly on L

0 since the integrand is

52.

regular at v = 0,

= -

s

¹ e -l}J(t)lf;"(t) dtt -

s

^K O"(t)sinO(t) dtt

0 0

+

i

s

^{K 011}(t)cos O(t) dtt

0

Combining. these results gives

I = -

.!..

e-lj;{l) - - 1- cosn(K)

+ .!..~

¹

(l)e...p(i)

s 2 2^Kz. 2

:I.I

^K

s

¹

- .!..

^[n1 (t)] z.cos qt)

~

^-

.!..

^e

-W (t)~

^{11 (t)}

~

2 t 2 t

0 0

- ..!_ r

^{K n 11} (t}sinn(t)

~ ⁺

^{i (- - 1_- sinn (K)}

2

j

⁰ t 2l'- .

(A. 6)

1 . (' K . dt )

+ ₂ j

0

n ¹¹ (t)cos n (t)

T

Finally, substituting Eqs. (A. 2) - (A. 6) in (A. 1) and separating real and imaginary parts; we have

>!< K [ ]

+ ~ S

0

cosn(t)

(

²

-n~(t)

^{)2 - 2t dt}

-86-

Y

= l Ail \

^K sinn (t) [2t - (2-n '(t) )²]dt

0 4 2.; _{0 .} t

( 2 Q ¹ (K ) ) 1 . }

+ - -

2 ^{cos Q} (K) - - - s1nQ (K)

K K 2K?.

APPENDIX B

THE SOLUTION OF l;:(a, K) = 0 . FOR K - 1

1

-

Let a

=

^f(K) be the curve on which A(a ,K)

=

^{0, where A is}

l l

given by (3.15) and (3. 16); that is,

:t;'.

^(i(K),K)

=

⁰

Differentiating this expression with respect to K gives

where the subscripts denote partial differentiation. Thus, df(K)

C1K =

b_K (f(K ), K)

'1a

^(£ (K ) , K )

1

(B. 1)

We now propose to solve this differential equation for K near 1 by using the Taylor series expansion

i(K)

=

^f(l)

⁺

(K-l) : ( l )

+ ~

^{(K-l )2 d2£(l)}

dK2

+ }

(K-1) 3 d3£(1)

+

O(K-l )4

dK3

We first show that £(1) = 0; i.e., ~(O, ^{1) = O.} From (3.16), the

-

elements of the determinant A are given by

t = 2 - a

ll 1

t

=

^-2

Zl

t

=

0

31

\z =(K.z. + :z +

6

+ 2'TT )-( 1; ⁺ 2'TT) al+(~ ⁺ ~'TT) alz + O(a12)

(B. 2)

(B. 3)

-88-

\z = -

l

l ; + 21T ) + (

1: ⁺ 3; )

^{a 1 - (}

i; ⁺ 3: )

^{a 1 z +}

⁰

^{(a 1 3 )}

1 t _32.

=

^K

+

^K

( 4 ) ( 2K³ t = 21T + 4K + - + - -

13 K 3

{ 2K³ 1T ) + 2K - -

3- + - a z + O(a 3 2 ^{l l} )

\3

=(2~3

^-

~

·_ 4K - 21T) +(4K -

4~3

^+1T)al

t =2 -(1 -K 2)a -KZa. 2 +O(a3 )

3 3 l 1 1

Thus,

2

a(O,l)= -2 -( 136 _{+ 21T} ) _-(16

3

+ 21T )

0 2 2

=

⁰

Using the well-known rule for the differentiation of determinants,

at at at

__.u__ t t t ^l 2

t t t 13

aa ₁ 12 1 3 1 1

aa-

l 3 11 12.

aa

1 1

at at

at

,._ _l:;,. 2.1 t t + t zz 23

=

a-a- aa

^t

⁺

^{t t}

aa-

a 2. 2. 23 2.1 _2. 3 ^Z.l 22

l l l 1

at

_at

0 t t 0 ³² t 0 t 33

3Z. 3 3

a-a

^{33 3 z.}

a-a

1 1

These partial derivatives can be read directly from Eqs. (B. 3). Set- ting a = 0, K = 1, we have

l

-1 'Xa(0,1)= 0

l

0

8 + 21T 8 + 21T -{1; +z'TT) -(1; +z'TT}

2 2

2

-{ 1

; +z'TT) s+2-rr 2 8 + 21T

₍ ¹

6 . }

- 3

+ 21T

+

-2

1; + 21T -(1; +.21T} + -2 -(136 + 2ir) ~+'TT

0 0 2 0 2 0

21T + 32

= 3

...

The evaluation of AK is done similarly:

at at

I 1 t t t ^{I Z}

d/C l z. ^{I 3 l l}

8K

at at

...

~

^{= 8 i { 2. I t} z. z. ^t z. 3

⁺

^t Zl ^{&K z z}

at

0 t t 0 ^3Z

3 z ^{3 3 ~}

at

t t ¹³

l l 1 z

1fK

at

+

^t _z. ^{t Z.3}

l zz ^~

at

0 t ^{3 3}

32 ~

Using (B. 3), it is easily shown that at a

l

at ..

lJ 0

~

=

for all i and j. Therefore,

and from (B. 1),

t

1 3

t Z.3

t 33

=0,K;::l,

(B. 4)

(B. 5)

df(l)

dK=

-90- ...,

!:::,. ( 0' 1) K

'l.a(O,l}

1

=

0

The second derivative of f is found from (B. 1 ):

...,

. ..., _21:::,.

dz.f !:::,.KK Ka

=

---

^...,

-

^...,

dKZ. !:::,. !:::,.

a a

l l

Since f' (K) vanishes at K = 1,

l f ¹ (K) -

'A

^(0,1)

..., KK

!:::,. (0' l) a

l

...,

!:::,.

aa

l I

(f' (K) )²

(X

)z.

a,

1

(B. 6}

(B. 7)

From Eqs. (B.4) and (B. 5), we see that the only terms contributing to AKK(O, 1) are those that involve the second derivatives of tij with

with respect to K; i.e.,

az.t

_ _ 11_

t t

oKz. ^{lZ. l3}

az.t ...,

.6.KK(O, 1) = ^{_ _ z.1_ t t}

oKz. z.z. Z.3

0 t t

3Z. 33

o

^{2t az.t}

t ^{_ _ l_Z} t t t __ n _

11 oKz. ^{13 11 12} oKz.

azt azt.·

+ t ^___£ t + t t ^~

2.1 oKl. ^2.3 z.r Z.l. oKl.

azt az.t

0 ^32. t 0 t ^3-l

oKz. ^{33 32.} oKz.

0 8 + 2'IT 8 + 2'IT

=

⁰

-l

l; + 2'IT)

-l

1; + 2'IT)

0 2 2

2 8 8 + 21T 2 8 + 21T 8

+ -2 0

-(1:

+ 21T) + -2

-( i;

+ 21T) 0 32

- - ₃

0 2 2 0 2 0

Therefore, from (B. 7),

dz.£( 1 ) ( - 3;) 16

(B. 8)

= =

_16+31T

dKZ. 32

21T +

3

Similar calculations for the third derivative of f give

""

...,

3.6. ( 0' 1)

d³f (1) ^.6.KKK(O, l) aK

1 f 11 ( 1 )

dK3.

=

..., ,..,.. .

.6. (0,1) ~ (0, 1)

a a

1 1