数理解析研究所講究録922 巻 1995 年 25-35

25$\mathrm{L}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{z}$橋本英典の定理について(法政大

工)Lorentz Theorem on the Stokes Equation
Hidenori HASIMOTO

College of Engineering, Hosei University,
Koganei, Tokyo 184, Japan
Abstract. A simple derivation of the Lorentz theorem is presented
which gives the perturbation pressure and velocity due to thepresence of a plane wall introduced into an unlimited viscous fluidof given pressure and velocity obeying tlle Stokes equation.
An extension to the case of spherical boundary is made in thesame manner, leading to the case of plane boundary as a limit oflarge radius. The sphere theorem is revised and applied. to threeelementary solutions by Lamb .

1.

Introduction
In 1896 Lorentz [1] developed a theorem yielding a mirror imageof the pressure and velocity of the Stokes flow due to a plane wall in
In a previous review [2] this theorem has
a unlimited viscous fluid
been shown to be derived by the use of the general solution in terms.
of three harmonics proposed by
As to the case of spherical boundary the present author [4]presented a sphere theorem for the perturbed stream function of theaxisymmetric flow due to a sphere introduced into an unlimited
viscous flow in axisymmetric motion obeying the Stokes equation.
This theorem has been presented by Collins in a more compact form$\mathrm{I}\mathrm{m}\mathrm{a}\mathrm{i}[3]$[5].

Recently Paraniappan et al $[6,7]$ have extended this theorem to a
general non-axisymmetric . flow represented by a biharmonic
function and a harmonic function, on the basis of an inversion
This formalism
due totheorem for thehas been applied by their group to several internal flows [9] and has
been extended to the case of spherical interface [10]In this paper will be presented a simple procedure giving thereflected pressure and velocity due to a plane or spherical boundary$[11,12]$ directly from the original flow without recourse to auxiliary
functions, in a similar way to the original Lorentz formula .
The sphereis given in an alternative form and is appliedelementary
types
ofsolutions given by Lamb [13]to three$\mathrm{p}\mathrm{o}.1\mathrm{y}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{m}\mathrm{o}\mathrm{n}\mathrm{i}_{\mathrm{C}}\mathrm{S}$$\mathrm{C}\mathrm{h}_{\mathrm{W}\mathrm{a}}\mathrm{n}\mathrm{g}[8]$.

$\iota \mathrm{h}\mathrm{e}\mathrm{o}\mathrm{r}\mathrm{e}\ln$.

26

2. Plane wall

2.1

Preliminaries

Let us derive the image field due to the presence of a plane wall$=0$ in the Cartesian system $(x,y,z)$ with the unit vector
normalto define the
For this purpose, it is convenientto the wall.
reflection of any function $f(X, y, z)$ by $J^{*}(x, y, z)=f$ ( y,- ) andstart from the following lemmas for any harmonic function $H(x,y,z)$
and biharmonic function $B(x,y,z)$ .
Lemma 1
$e$

$z$

$X,$

$\tilde{H}=-H*$$z$

is harmonic :(2.1)$\Delta\tilde{H}=0$

and

satisfies

at $z=0$ .

$\tilde{H}+H=0$

Lemma 2Let$\tilde{B}=-$$[B- \mathit{2}z B’+z^{2}\Delta B]^{*}$

the prime

where
Then

’

,

denotes the$z$derivsative$e\cdot \mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}$.

is biharmonic$\tilde{B}$(2.2)$\Delta^{2}\tilde{B}=0$and satisfies$\tilde{B}+B=0$

as well as $\tilde{B}’+B’=0$ at $z=0$ .

Lemma 3
(2.3)$\Delta\tilde{B}=[4B’’- 3\Delta B- \mathit{2}z\Delta B’]^{*}$

2.2

Lorentz

Theoremof theand the pressure
Let us start from the velocity
viscous flow satisfying the continuity and the Stokes equation:
$u$

(2.4)

$\mathrm{d}\mathrm{i}\mathrm{v}u=0$a$p$$\mathrm{n}\mathrm{d}$$\Delta u=\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}p/\mu$

(2.5)

or
rot

where

$\omega$$\omega=-\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}p/\mu$(2.6)

is the vorticity$\omega=$

rotu

(2.7)

27
Then it is easily seen thati) $p,$ $p$$’=e\cdot \mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}p,$ii) $u,w=u\cdot e$ ,$e\mathrm{x}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}p$and$e\mathrm{x}u$,$z\omega$

as

well

$\omega$andare all harmonic.

$\omega_{\mathrm{z}}=e\cdot\omega$are all biharmonic, satisfying (2.2) as(2.8)$\Delta(e.u-)=p’/\mu$a$\mathrm{n}\mathrm{d}$$\Delta(z\omega)=2\omega$

(2.9)

’

It should be noted that theseare valid also for the
perturbed quantities denoted by tilde
Applying the Lemma 2 to the biharmonic function $w=e\cdot u$
satisfying $\tilde{w}+w=0$ and $(\tilde{w}+w )$ $’=0$ at $z=0$ , we have$\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{p}\mathrm{e}\mathrm{r}\sim \mathrm{t}\mathrm{i}\mathrm{e}\mathrm{s}$$\tilde{w}=- e\cdot q[u , p]*$$\mathrm{w}$here(2.10),$q[u , p]=u-2zu’+z$ 2grad $p/\mu$ ,(2.11)

use of (2.8).
and we have
The application of Lemma 3 to (2.10) yields$\mathrm{m}\mathrm{a}\dot{\mathrm{d}}\mathrm{e}$$\tilde{p}’=\mu\Delta\tilde{w}=[4\mu w1’$

-3p

\dag er(2.12)$- 2_{Zp]}"*$

Integrating (2.12) with respect to $z=-$

we have$z^{*}$(2.13)$\tilde{p}=[p+\mathit{2}_{Zp-}’ 4\mu w\uparrow]^{*}=[(2zp-4\mu w)’ - p ]^{*}$

Making use of the relation(2.14)$e\mathrm{x}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}p/\mu=(e\cdot \mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d})\omega-\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}(e\cdot\omega )$derived from (2.6), we have for the perturbed vorticity$\tilde{\omega}=\int$

where

the

component$e\cdot\tilde{\omega}$(2.15),$[e\mathrm{x}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}\tilde{p}/\mu+\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}(e\cdot\tilde{\omega})]\mathrm{d}z$$\mathrm{z}$$\tilde{\omega}$in the integrand is found to be(2.16}$e\cdot\tilde{\omega}=- e\cdot\omega^{*}$

from Lemma 1$e\cdot\tilde{\omega}$applied to the harmonic function satisfying$=$-e$\cdot\omega$

at

.

$z$

$=0$

.

(2.17)

28
Introducing (2.13) and (2.16) into (2.15), the tangential component
is obtained to be$\omega_{t}=\omega-(e\cdot\omega)e$(2.18)$\tilde{\omega}_{t}=[\omega_{t}+e\mathrm{x}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d} (4w - \mathit{2}zp/\mu)]*$where we have made use of (2.13).
The tangential velocity can be obtained by applying Lemma 2 tothe biharmonic function:$U$$U=e\mathrm{x}u$-z(2.19)$\omega$satisfying the boundary condition$\tilde{\mathrm{U}}+U=0$

Making use ofand$(\tilde{\mathrm{U}}+U)$(2.8) ,$(\mathit{2}.9)$$’=0$ atand $(2.11)$$\mathrm{z}=0$.

(2.20)

we have(2.21)$\tilde{\mathrm{U}}=-[e\mathrm{x}q(u, p)+z\omega]*$by noting (2. 19)

which yields$e\mathrm{x}\tilde{\mathrm{u}}=$

$=$

-e

$\tilde{\mathrm{U}}+z\tilde{\omega}$$\mathrm{x}[u- 2zu’+z^{2}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}p/\mu]^{*}+z(\tilde{\omega}+\omega*)$, (2.22)is given by (2.16) and (2.18)
whereCombining (2.10) and (2.22) we have$\tilde{\omega}$$\tilde{\mathrm{u}}=-[u- 2zu ’ +z^{2}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}p/\mu]*$-z$e\mathrm{x}(\tilde{\omega}+\omega^{*})$(2.23)

or$\tilde{w}=-[w- 2zw|+z^{2/*}p’\mu]$ ,an$(2,24)$$\mathrm{d}$$\tilde{\mathrm{u}}_{t}=-[u_{t}+2z \mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}_{t}w - z 2 \mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}_{t}p/\mu]^{*}$

where we have made use of (2.18) and (2.14).
The expressions (2.13), (2.24) and (2.25) for
essentially the formulae given by Lorentz. [1]$\tilde{p},\tilde{w}$(2.25),

and$\tilde{\mathrm{u}}_{t}$are3. Spherical boundary
3.1

Preliminaries
theorem for the harmonics
It is convenient to rewrite$H$ $(X)$and Chwang’s theorem for the biharmonics $B(x )$ into thefollowing form:$\mathrm{K}\mathrm{e}\mathrm{l}\mathrm{V}\mathrm{i}\mathrm{n}’ \mathrm{s}$Theorem 1(Kelvin’s

exterior theorem)

29

Let $H(x)$ be the harmonic function of $x=re$ and be regular in
the domain $r<a$ . Then$\tilde{H}(x)=-$is regular harmonic$\tilde{H}+H=\mathit{0}$a $H(x^{*})/r$(3.1)

in $r>a$ and satisfies the boundary condition(3.2)

at $r=a$is the radius from the center and is the unit radial vector.
wheredenotes the inversion
Here and hereafter the asterisk *
$e$

$r$

$[f(X)]*=f(X^{*})$$wi$th$x^{*}=a$ $x/r^{2}=r*e$and$r^{*}=a^{2}/r$

.

(3.3)

Theorem 2 (Chwang’s exterior theorem)

Let $B(x)$ be the biharmonic function satisfying(3.4)$\Delta^{2}B=0$

and regular in the domain $r<a$ . Then$\tilde{B}$given by$\tilde{B}=-[(r/a+a/\Gamma)B/\mathit{2}+a(\mathit{1}- r2/a^{2})B’+a^{\mathit{3}}(l- r^{2}/a^{2}).

\Delta B/(\mathit{4}r)]*$

.

(3.5)

and satisfies the following boundary,is regular biharmonic in $r>$
conditions on the sphere $r=a$ :
$a$

and

$\tilde{B}+B=\mathit{0}$where the prime’$(\tilde{B}+B)’=\mathit{0}$denotes the derivative(3.6)

at $r=a$($\partial/\partial r=(l/r)$ $x$. grad)$=e\cdot \mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}$The theorem 2 is a generalization of the sphere theorem
for the axisymmetric stream function in the Stokes flow derived by
the author [4] and Collins [5].
:We may note the following formula for$\Delta\tilde{B}$$\Delta\tilde{B}=-$$B$)$/a^{3}$[( B-4r $2’-4\Gamma^{3}B’’$$r$$+(r/a+5r^{3}/a^{3})\Delta B/\mathit{2}- a(r2/a2_{-}r^{4}/a4)(\Delta B)’]^{*}$(3.7)

which is easilv derived from (3.4) and (3.3) by use of equalities

.

30

(3.8)

$\Delta(f^{*}/_{\Gamma})=a(4\Delta f)^{*/}r^{5}$a$\mathrm{n}\mathrm{d}$

.

(3.9)

$\Delta(rf’)=\Delta(x\cdot \mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d})f]=(X , \mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}+\mathit{2})\Delta f$

Interior theorems
1’ and 2’
In the interior problem where we interchange $r>a$ and $r<$in theabove Theorems we have to restrict $H$ andto be respectively
$a$

$B$

$(l/r)$

and3.2$O(l/r^{2})$Sphereas $r->$$O$.

$\infty[\mathit{6}]$theoremIf we replace $ze$ in the section 2 by . the procedure
proceeds analogously to the case of plane boundary.
We havei) $p,$
andare all harmonic.
$x$$rp’arrow-x\cdot \mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}p,$$x$ii) $u,$ $ru_{r}=xu,$ $x\mathrm{x}u$

as well as$x\cdot\omega=r\omega_{r}$$\mathrm{x}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}p$and$r^{2}\omega$are all biharmonic, satisfying(3.10)$\Delta(xu)=rp’/\mu$ ,$\Delta(x\mathrm{x}u)=2\omega+x$

an

(3.4)

(3.11)$\mathrm{x}\mathrm{g}.\mathrm{r}\mathrm{a}\mathrm{d}p/\mu$$\mathrm{d}$(3.12)$\Delta(r^{2}\omega)=4r\omega’+6\omega$

It should be noted that these properties are valid also for the
perturbed quantities denoted by tilde
Applying the theorem 2 to the biharmonic function $x.u$ satisfying
andat $r=a$ , we have$x\cdot(\tilde{\mathrm{u}}+u)=0$$x\cdot(\tilde{\mathrm{u}}+u)^{\mathfrak{l}}=0$(3.13)$\tilde{u}_{r}=x\cdot\tilde{\mathrm{u}}=-\{x\cdot q(u, p)\}^{*},$$\mathrm{w}$here$rq(u, \mathrm{p})=a\{(3 - r^{2}/a^{2})u/2+(1-\gamma^{2}/a^{2})ru$’$+a2(1 - r 2/2a)^{2}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}p/(4\mu)\}$(3.14)

and we have made use of (3.10).
The application of (3.10) and (3.7) yields$\tilde{p}^{\dagger}=\mu\Delta(_{X}\cdot\tilde{\mathrm{u}})/r$$=\{\mu a(4r^{2}u\gamma\dagger’+1\mathit{2}ru_{\gamma’}+^{3u_{r}})+a^{3}r(1- r^{2}/a^{2})p’’$$+a3(1- 7r^{2}/a2)p’/2\}^{*/3}r\cdot(3.15)$

The perturbed pressure
with respect to $r=a$:$2/\gamma*$$\tilde{p}$is obtained by the integration of (3.15)

31$\tilde{p}=-\{(4\gamma^{3}u_{\Gamma}’+3\int ru_{r}dr)/a^{3}+a(r^{2}/a^{2}-r^{4}/a^{4})p’-I/(2a)\}^{*}$

with(3.16)$I= \int(3-r^{2}/a^{2})rp^{\mathrm{t}}dr=(3-r^{2}/a^{2})rp-3\int(1-r^{2}/a^{2})pdr$,

where we assume that the integrals are convergent and take thegauge pressure corresponding to. the lower limit ( $r=0$ or infinity
according as outer or inner problem) to be zero.
Making use of the relation’-$x\mathrm{x}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}p/\mu=(\gamma\omega)$(3.17)$\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}(x\cdot\omega)$

derived from (2.6), we have for the perturbed vorticity :$r \tilde{\omega}=\int-$$[X\mathrm{x} \mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}\tilde{p}/\mu - \mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}\{a(X\cdot\omega)^{*}/r \}]$$\mathrm{d}r$(3.18),

sinceis harmonic and satisfy the condition (3.2)
theorem 1 on the sphere $r=a$ . We have$x\cdot\omega$$=$$r\omega_{r}$$r\tilde{\omega}=x\cdot\tilde{\omega}=- a(X\cdot\omega)^{*}/r$from theorem 1
The tangential velocity
2 to the biharmonic function(3.19),

can be obtained by applying$u_{t}$

oftheorem$V$(3.20)$\mathrm{V}=\mathrm{x}\cross \mathrm{u}-(r-2a)2\omega/2$

satisfying the boundary conditions of the theorem.

We get,$\tilde{\mathrm{V}}=-\mathrm{t}\mathrm{x}\cross \mathrm{q}(\mathrm{u},p)-a(\gamma/a-\gamma 3/a^{3})\omega/2\}^{*}$(3.21)

where $q(u,p)$ is given by (3.14) and we have made use of (3.11) and(3.12).

Combining the expression for(3.13) we have$x\mathrm{x}u$obtained from (3.20), (3.21) and$\tilde{\mathrm{u}}=-\{(3\gamma/a-r^{3}/\mathit{0}^{3})\mathrm{u}/2+(_{\Gamma^{2}/a^{2}r/a}-44)a\mathrm{u}’+\gamma(a^{2}-\gamma^{2})^{2}gradp/(4\mu a^{3})\}*$$-[( \gamma/a-\gamma^{3}/a^{3})\{agr\mathit{0}d3\downarrow\int\tilde{p}*/(\mu\gamma^{2})dr+\mathrm{e}\mathrm{x}(\Gamma\omega+grad\int x\cdot\omega dr)\}]^{*}/2$(3.22)

where
denotes the tangential derivative
Separating the radial and the tangential component we have$\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}_{t}$$\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}-e(e\cdot \mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d})$$\tilde{u}_{r}=\tilde{u}\cdot e=-\mathrm{t}(3r/a-r^{3}/a^{3})\mathrm{u}_{r}/2+(r^{2}/a^{2}-_{\Gamma^{4}}/a^{4})au,+\gamma(|2-r^{2}a)^{2}p^{1}/(4\mu a^{3})\}^{*}$.

(3.23)

32$\tilde{\mathrm{u}}_{\iota}=\tilde{\mathrm{u}}-\tilde{u}_{r}e$$=-[r \mathrm{u}_{\mathrm{t}}/a-(1-\gamma/2a^{2})g\Gamma ad_{1}[(r^{2}u_{r}-3\int ru_{\Gamma}/2d\gamma)/a+a\mathrm{t}\Gamma(1-\gamma/2pa^{2})-3\int(1-r^{2}/a)2dp\Gamma\}/(4\mu_{)}^{\backslash }$(3.24)

where we have made use of the relation(3.25)$x\mathrm{x}\omega=-(\Gamma u)’+\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}(x.u)$

as well as (3.17) to eliminate the vorticity and $(3,16)$ to eliminate $p.$ ,and assume that the integral constand cam be taken to be zero at
the lower limit.
The expression (3.24)
previous section.

is new and corresponds directly to (2.25) of the
3.3 Limit of large radius

Let us put(3.26)$r=a$ $(1 +z/a+\mathrm{o}(z/a ))$in the several expressions and retain the lowest order in
For the pressure , it is convenient to adopt the last expression of
I in (3.16) and neglect the integral in comparison with the .rst term.
-derivative as z- derivative to the lowest
as $w,$
We may put.
approximation.obtaining (2.10) for(3.13) leads to thevelocity
theradialseen
that
It is easily. In the. same manner, (2.25) is derived fromexpression (2.24) for(3.24) by neglecting integral terms of O(z/a) in comparison withother terms.
$z/\mathrm{a}$$p$$\mathrm{f}\mathrm{i}$

$r$

$u$

$p$

$w$

3.4

ExampleAs an illustrating example, three components of Lamb’s generalsolution [13] are considered. For the sake of simplicity we may taketo be unity without loss of generality
and1) external flow
.

$a$

$\mu$

$\geq 1$$r$i) $u=\mathrm{r}\mathrm{o}\mathrm{t}(xH)=$

where

$H$

integer,sinceand$\mathrm{n}=0$$C$$x\mathrm{x}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}H=- C$-,

are solid harmonics of n-th degree( . is positive$\mathrm{n}$is trivial).

We have also

$p=0$ andIntroducing these into the corresponding expressions in 3.2, we
obtain$u_{r}=0$$\tilde{p}=0$

$i.e$

. .

and

$\tilde{u}_{r}=0$$\tilde{\mathrm{u}}=-u*/r$.

so that$\tilde{\mathrm{u}}_{\mathrm{I}}=[rC]*$

,

33

ii)

$u=\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}H$

we have$u_{r}=nH/r$$u_{t}=\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}_{t}H=T/r$

with$T=r\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}{}_{t}H$

as well as $p=0$ , $ur\iota=n(n- 1)H/r^{2}$ , $u_{t}’=(n- 1)T/r^{2}$an$\mathrm{d}$$\int ru_{r}d\gamma=nrH/(n+1)$(3.24) give$n(4n^{2}- 1)(rH)^{*}/(n+1)$ ,Then (3.16), $(3,23)$ and$\tilde{p}=-$$\tilde{u}_{r}=- n[\{n+1/2-(n- 1/\mathit{2})_{\Gamma\}H}2]^{*}$a$\mathrm{n}\mathrm{d}$-$\tilde{\mathrm{u}}_{1}=$$[\{1 - n(1 - r 2)+3n(1- r^{2})/(2n+2)\}T]^{*}$$=[\{(n- 2)(\mathit{2}n+1)- n(2n- 1)r^{2}\}T ]^{*}/(\mathit{2}n+\mathit{2})$iii) $p=H$ ,

$u=$

[

$(n+3)r^{2}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}$.

H-2 $nxH$ ] $/(\mathit{2}n+\mathit{2})(2n+3)]$

We have$ll_{r}=nrH/(4n+6)$ ,$u_{t}=(n+3)rT/[(2n+2)(\mathit{2}n+3)]$$\int;\cdot u_{r}dr=n\prime^{3}.H/[(n+3)(4n+6)],$$. \int(1- r^{2})p\mathrm{d}r=[r/(\mathrm{n}+1)-\Gamma^{3}/(\mathrm{n}+3)]H$

Introducing these expressions into (3.19), (3.23) and (3.24) we obtain$\tilde{p}=- n$$(\mathit{2}n- 1)(rH)^{*}/(2n+2)$$\tilde{u}_{r}=- n[\{\mathit{2}n+3 - (2n+1)r^{2}\}H]^{*}/(8n+12)$a$\mathrm{n}\mathrm{d}$$\tilde{\mathrm{u}}_{1}=[\{(\mathrm{n}-\mathit{2})- \mathrm{n}(2\mathrm{n}+1)r^{2}/(2\mathrm{n}+3)\}T]*/(4\mathrm{n}+4)$2) Interior flow

$r$

$<1$

.

this case we have only to take solid harmonics ofor replace $H$ by $(rH)$ *etc. in the exterior problem$\ln$i) $u=\mathrm{r}\mathrm{o}\mathrm{t}(xH*/f)=$

where

$H$

and

$\tilde{p}=0$

,

$C$

-

$x\mathrm{x}$so thatdegree,$\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}H*/r=- C*/r$

are solid harmonics of$\tilde{u}_{r}=0$$-(\mathrm{n}+1)$$u=C=$-$\mathrm{n}$th degree.and we obtain$u*/r$.

34
ii)$u=\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}(H*/\mathrm{r})$. $p=0,$$\mathrm{n}>0$

we have$\tilde{p}=-(\mathrm{n}+1)(2\mathrm{n}+1)(2\mathrm{n}+3)H/\mathrm{n}$,$\tilde{u}_{r}=-(\mathrm{n}+1)\{(\mathrm{n}+1/2)\mathrm{r}-(\mathrm{n}+3/2)r\}H$a$\mathrm{n}\mathrm{d}$$\tilde{\mathrm{u}}_{\mathrm{t}}=-\{(\mathrm{n}+3)(\mathrm{n}+1/\mathit{2})r-(\mathrm{n}+1)(\mathrm{n}+3/2)/r\}T/\mathrm{n}$iii) $p=(rH)^{*},$$u=-[(\mathrm{n}- 2)^{r}2\mathrm{g}\Gamma \mathrm{a}\mathrm{d}(rH)^{*}- 2(\mathrm{n}+1)x(rH)^{*}]/(2\mathrm{n})(\mathit{2}\mathrm{n}- 1)]$

. $n>0$

we obtain$\tilde{p}=(\mathrm{n}+1)(\mathit{2}\mathrm{n}+3)H/(\mathit{2}\mathrm{n})$$\tilde{u}_{r}=(\mathrm{n}+1)\{\Gamma^{2} - (2\mathrm{n}+1)/(2\mathrm{n}- 1) \}^{H}/(4r)$a$\mathrm{n}\mathrm{d}$.

$\tilde{\mathrm{u}}_{\mathrm{t}}=\{(\mathrm{n}+3)\gamma-2(\mathrm{n}+1)(\mathit{2}\mathrm{n}+1)/(2\mathrm{n}- 1)\}T/(4\mathrm{n}r)$

is to be excluded according to the interior
It is seen thatin ii)
Theorems 1’\dagger and 2\dagger in 3.1. For example the caseto the presence of a single source in the cavity
contradicting the conservation of mass. On the other hand, the caseof iii) becomes possible by the cancellation of formidable
terms. Summarising, our formulae ls valid iffor
as long as the total flux of singularities in $r<a$ iszero.and there are no singularities in the outside of the sphere.
$\mathrm{n}=0$$\mathrm{n}=0$$\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{a}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{d}_{\mathrm{S}}$$\mathrm{n}=1,\mathit{2}$$p$$(1/\mathrm{r})$4.

$=\mathrm{O}(1/_{\Gamma}2),u=\mathrm{O}$$\mathrm{r}->\infty\infty$

Summary

A simple derivation of the Lorentz theorem is presented which
gives the perturbation pressureand velocity
due to thepresence of a plane wall introduced into an unlimited viscous fluidand velocity
of given pressureobeying the Stokes equation.
An extension to the case of spherical boundary is made in thesame manner, leading to the case of plane boundary as a limit oftheorem is given in the form corresponding tolarge radius. The
the original formula of Lorentz for the plane boundary.
As an illustrating example three elementary solutions by Lamb[13] are chosen. Application to any case will be easily done except
some restrictions in the interior problem $i.e$ .
and zero total flux of singularities in $r<a.\cdot$
as$\tilde{\mathrm{u}}$$\tilde{p}$

$p$

$u$

$p$

$(1/\mathrm{r})$$\mathrm{r}->\infty\infty$$=\mathrm{O}(1/\gamma 2),u=\mathrm{O}$35

References
1. H. A. Lorentz, A general theorem concerning the motion of aviscous fluid and a few consequences derived from it. Zittingsverlag.
Akad. Wet. Amsterdam 5 (1896) 168-175;
see also Abhandl.Theor.Phys. Leibzig 1 (1907) 23-42.
and O. Sano, Stokeslet and eddies in creeping flow.
Ann. Rev. Fluid Mech. 12 (1980) 335-363.

2) H. Hasimoto

applications of function theory to fluid dynamics, 2ndInt. JSME.Symp., Fluid Machinery and Fluidics,(Tokyo) (1972) 15-23.

3) I. Imai, Somesphere theorem on the Stokes equation foraxisymmetric viscous flow, J. Phys. Soc. Jpn. 11 (1956) 793-797.

4) H. Hasimoto, A

D. Collins, Note on a sphere theorem for the axisymmetrictokes flow of a viscous fluid , Mathematika 5 (1958) 118-121.

5) W.
$\mathrm{S}$S. D. Nigam and T. Amaranath and R. Usha,
tokes flow, Mech. Res. Comm. 17 (1990) 173.

6) D. Palaniappan,

Theorem for$\mathrm{S}$SphereS. D. Nigam and T. Amaranath and R. Usha,solution of Stokes’s equation: A sphere theorem, Q. J. Mech.
Appl .Math. 45 (1992) 47-56.

7) D. Palaniappan,$\mathrm{L}\mathrm{a}\mathrm{m}\mathrm{b}^{\dagger}\mathrm{s}$$\cdot$A.T . Chwang, On spherical inversions of polyharmonic functions,
Quart. Appl. Math. 44 (1987) 793-799.
8)9) R. Usha and S. D. Nigam,

stokeslet,

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