H.K. Moffatt and B.R. Duffy Local similarity solutions and their limitations
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J . Fluid Meoh. (1980), vol. 96, part 2, p p . 299313Printed in Great Britain
299
Local similarity solutions and their limitationsBy H. K . MOFFATT A N D B. R. DUFFY
School of Mathematics, University of Bristol, England(Received 16 March 1979)
Two problems exhibiting breakdown in local similarity solutions are discussed, andthe appropriate asymptotic form of the exact solution is determined in each case. Thefirst problem is the very elementary problem of pressure driven flow along a ductwhose crosssection has a sharp corner of angle 201. When 2 a < +T, a local similaritysolution is valid, whereas, when 2a > &r, the solution near the corner depends on theglobal geometry of the crosssection. The transitional behaviour when 2 a =5 +T isdetermined.
The second problem concerns lowReynoldsnumber flow in the wedgeshapedregion 1/31 < a when either a normal velocity proportional to distance from the vertexis imposed on both boundaries, or a finite flux 2Q is introduced or extracted a t thevertex (the JefferyHamel problem). It is shown tha t the similarity solution in eithercase is valid only when 2 a < 201, z 257 .6"; a modified problem is solved exactlyrevealing the behaviour when a > a,, nd the transitional behaviour when E = a,(when the similarity solutions become infinite everywhere), An interesting conclusionfor the JefferyHemel problem is that when a > a,, nertia forces are of dominantimportance throughout the flow field no matter how small the source Reynoldsnumber 2 Q / v may be.
1. Introductionordinates Oxyzwith Oz parallel to the flow, the equation for the velocity w(~,) sConsider Poiseuille flow along a straight duct of crosssection 9. hoosing co
V2w =  G / p in 9 , (1.1)whereG s the applied pressure gradient, and ,U the viscosity of the fluid. The boundarycondition is
w = O on 8 9 . (1 .2)1f the boundary 8 9 has any sharp corners (as is the case if, for example, it has
square or triangular crosssection), then i t is very natural to enquire into the natureof the solution of the problem (1 l ) , 1 . 2 )near the corner. Choosing polar coordinates( r ,8) ith origin at the corner and with 8 = 0 along the bisector of the angle, so thatlocally the boundary consists of the two planes 8 = ? a (figure l ) ,one might reasonablyconjecture that w must have a local similarity form depending only on r , 8, G/,u anda (i.e. independent of the 'remote' geometry of the duct) . If this is so , then ondimensional grounds,
w (G IP ) "f (@ (1 .3)00221 120/80/43608100 $02.00 0 1980 Cambridge University Press
www.moffatt.tc

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300 H . K . Moffatt and B . R . Duffy
wV w =  GIP0=  (YFIGURE. Poiseuille flow along duct whose crosssection3 has a corner of angle 2a.(the dependence on a being implicit), and the only possible form of f(0) ompatible
with (1.1) and (1.2)is
This solution explodes when 201 = $ 7 ~ and also when 2a = $n), nd this leads oneto question the validity of the above conjecture.? It is physically clear that thesolution must remain finite when 2 a = in, nd that the similarity form (1.3) cannottherefore be correct for this angle. The difficulty is in fact quite an elementary one,and we shall resolve it in 9 2.A similar difficulty, which may be described as an inapplicability of a local similaritysolution, can however arise in more complex physical contexts, when the means ofresolution of the problem may not be so readily apparent. A second example, whichwe shall treat in $9 35, is the following. Suppose that viscous fluid is contained in thespace 101 < a between two planes 8 = fa which are hinged at their intersection (sothat the angle a may be varied), and suppose that the normal velocity is instantaneously prescribed on the planes, by rotating the planes about 0 as indicated infigure 2(a) , s
U , = T r w on 8 = +a. (1.5)t A similar breakdown was noted by Fraenkel (1961) in the context of inviscid corner flow
with constant vorticity; Fraenkel correctly resolved the behaviour when 2a = +T ,but, by limitingattention to acute angles, the inapplicability of the similarity solution for 2a z &I (see 5 2 below)was not explicitly appreciated.The problem has been touched on in a number of other studies, but the particular difficultydiscussed here does not appear to have been explicitly recognized. For example, Collins & Dennis(1976) study flow along a curved duct of triangular crosssection, and they consider particularlythe case of a ightangled triangle. They sta te that the solution of (1.1)is O(r2)near r = 0 regardless of the angle of the corner ,but, as suggested by (1.4) , and as demonstrated in $ 2 below, this isnot true when 2a > (although the subsequent argument of Collins & Dennis remains substantial ly unaffected).A second example is provided by the work of Allen & Biggin (1974) who studied the flow of aliquid filament down an inclined plane, the shape of the free surface being controlled by surfacetension and the contact angle a t the 3phase contact line. When th is contact angle is a, ocalanalysis using a zero stress condition on the free surface8 = 0 and a noslip condition on the solidboundary 0 = a again yields (1.4) , which breaks down when a = $77. Alleii & Biggin in fact chosea = in in their numerical computation of the velocity distribution over the crosssection of thefilament; a inite element procedure provided an apparently wellbehaved solution. It was in factthrough attempting to understand the structure of this solution th at we were led to investigatethe paradox presented by (1.4) when a = 4

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Local similarity solutions 301
FIGURE. (a )The hinged plate problem, w r 2 / vQ 1.( b )The JefferyHamel problem, Q / v < 1.Sufficiently near 0, inertia is clearly negligible, a.nd the stream function $ ( r , 0 )satisfies the biharmonic equationand the boundary conditions V4$ = 0 (1.6)
Again it is tempting t o seek a similarity solution for 9 epending only on CL) , r , 0 anda ;on dimensional grounds this must have the form
$ = h r 2 f ( 0 ) , (1 .8)and the unique functionf 0 )compatible with (1.6) and (1.7) (Moffatt 10640) is
sin 20 20 cos 2af ) = sin 2a  2a cos 2a *
This solution explodes when tan 201 = 2 a , i.e. when 2a = 257.45". For this criticalangle,f 0) apparently becomes infinite for all (01 4 a, and the corresponding velocitycomponents derived from (1.8) become infinite also. The assertion tha t the Reynoldsnumber is low near the corner therefore requires reexamination, as does the relevanceof a similarity solution of the form (1.8).A closely related difficulty has been noted (Fraenkel 1062) in the contest of thelow Reynolds number limit of the JefferyHamel problem (Batchelor 1067, $5.6).Suppose that the planes 0 = f are fixed and t hat fluid is introduced at a volume rate2& (per unit length) at the intersection (figure 2 b ) ; in this case, the low Reynoldsnumber limit of the JefferyHamel stream function (which provides an exact solutionof the NavierStokes equations) is
sin 20  20 cos 2a'= Qsin2a2acos2a' (1.10)
with precisely the same angular dependence as (1.8) and the same pathologicalbehaviour a t 2a z 257'.

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302 H. . Hoffatt and B. R.DzGffv
= a
( U ) ( b )FIGURE. The sector geometry considered in 5 2.The same type of difficulty has been encountered and resolved in an entirelydifferent physical context by Sternberg & Koiter (1958). When an elastic wedge is
subjected to a couple that is concentrated a t the vertex, the dilsplacement field (whichmay be derived from a function satisfying the biharmonic equation) involves thesame factor (sin2a a cos2a)  l which blows up at 261 x 257". Sternberg & Koiterresolved this apparently paradoxical behaviour by spreading out the force fieldproviding the couple overa small neighbourhoodof the vertex and solving the resultingproblem exactly. We shall follow a similar procedure in the fluid context in $$35,and shall in this way providea simultaneous resolution of the hinged plate and JefferyHamel paradoxes.
2, The Poiseuille problem at a sharp cornerConsider the problem defined by ( 1 . 1 )and (1.2) when9 =Ba)s the domain bounded
by the lines f3= f and the circular arc T = a.We may consider the full range of angles0 < 261 < 2n, he limiting case 2 a = 2 n representing a circular pipe with a radial' lange (figure 3 ) .Suppose first that 2 a ?= an or #n.Then the general solution of (1.1) symmetricabout 8 = 0 and satisfyingw = 0 on 8 = f swhere
Gr2 cos28w = I) + 5 AnrhnC0sA,8,4p cos2a n 0A,a = 4 ( 2n+ l)n, n = 0 , 1 , 2 , ... .
The coefficientsA, are determined by the condition w = 0 on r = a (101 < a); singthe orthogonality
we find
conditioncosA, ecoaA, eae = ad,,,!

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Local similarity solutions 303Now A , = nr/2a, so that A, >< 2 according as 2 a 5 in.Hence the particular integral.
is the dominant part of the solution (2 .1) near r = 0 only i f 2 a < )R. If 2 a > in, hesolution is dominated near r = 0 by the leading term of the series of eigenfunctions,namely
The particular value of the coefficient A , depends on the 'global' geometry of 9,ndso the asymptotic form ( 2 . 6 )cannot be obtained from purely local considerations.When 2 a > &r, A, and A, are both less than 2 , and so the first two terms of theseries in ( 2 . 1 ) both dominate over w , ( r , 0 ) for small r . In the extreme case 201 = 2 n(figure3 b ) , we find
The cases 2 a = in and $R require special attention. If we let 2a = +n+c in theexpression ( 2 , 1 ) ,we find
so that, in the limit E + 0 (i.e. 261 = tn),
2u Leading terms in expansion of 'U: near r = 0r a ,thenr2 log r , then r2 , then T srn lza , then r2 , hen r3nlzad, hen r2 log r , then ra, then rL8rn12a, then r3n/2a,then r2 , hen r6n12a
TABLE1
A similar treatment of the neighbourhoodof 201 = $71 shows that, when 2 a = $n,
1= E "(log:) cos20+0sin203np 4In this case, the singularity in wo(r,O) is compensated by an equal and oppositesingularity in the second term (n= 1)of the series in ( 2 . 1 ) .The above results concerning the leading terms in the asymptotic expansion ofw ( r ,0) near T = 0 are summarized in table 1 .
As noted in the footnote on p. 300, the same type of mathematical problem has

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304 H . K . Moffatt and B . R.Du#y
(C )FIGURE. Flow gmerated in domain ga y rotation of bounding cylinder keeping 6 = f ixed:( a ) assuming no separation; ( b )and ( c ) , likely streamline pa tte rns allowing for boundarylayerseparation.been considered by Fraenkel (1961) n a different physical context. An example ofthe sort of situation envisaged by Fraenkel (who limited attention to acute angles201 < &r) is indicated in figure 4( a ) ,which shows a method of generating a flow withclosed streamlines in the domain ga , y rotation of the cylinder r = a with angularvelocity Q , keeping the radial boundaries 8 = 1 fixed. If R = Qa2/v B 1 , and if theflow does not separate from the boundary a t any point, then the vorticity w is uniformthroughout the core region (Batchelor 1956) w = wo say  and the stream function$ ( r , 8) is determined by
V2$ = wo in ga , $ = 0 on 8ga. (2.10)The solution for $ is then precisely analogous to the solution for w( r ,O)describedabove. The above proviso concerning nonseparation is however physically unrealistic. I f 2a < T , the flow will separate at some point on 8 = a where the pressuregradient impressed on the boundary layer is adverse (figure 4 b ) i ; and if 2a > T , theflow will separate at the sharp vertex (figure4c) . In either case, the vorticity will takedifferent constant values in different subdomains of $Ba separated by thin layers inwhich viscous forces are nonnegligible; the specification (2.10) is then clearly inapplicable.
t The seoondary eddy so formed will then also presumably separate when R is sufficientlylarge; and indeed, when R +CO, the limiting steady solution presumably involves an infinitesequence of eddies, within each of which the vorticity takes a different constant value.

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Local sim ilarit y solutions 305
FIGURE.Modified JefferyHamel problem treated in $ 5 35. The hinged plate problem is obtainedin the limit a + 00, and the conventional JefferyHamel problem in the limit a+ 0, w +CO ,Q = +was = constant.
3. A modified JefferyHamel problemConsider now the following modification of the JefferyHamel problem (figure 5):
suppose that viscous fluid is contained in the space 181 c a between the planes 8 = k 01on which the radial velocity is zero and the normal velocityuo s specified as
The total flux introduced into the region 181 c a s then
The hinged plate problem described in 3 1 is obtained in the limit a+m, and theJefferyHamel problem is obtained by the limiting process a+O, w+m, with Qconstant.
The stream function $ ( r , 8)satisfies the biharmonic equationV4$ = 0
and the boundary conditions (3.3)
This problem may be solved by means of the Mellin transform (cf. Tranter 1948,Sternberg & Koiter 1958, who considered problems involving elastic wedges, andMoffatt 19643 in which a similar problem with specified tangential velocity on 0 = fwas treated). This is defined by
with inverse relation

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306 H . K . Moffatt and B . R. DuffyIn (3 .5 ) , p may be complex, its real part being such tha t the integral exists; clearly,provided
where e > 0, the integral (3 .5 ) will certainly exist in a neighbourhood of the lineRep =  1 in the complex p planet. We shall find that c =  is an appropriatechoice for the real constant c in ( 3 . 6 ) .The various operations tha t follow may all bejustified under the conditions (3 .7 ) ,which may be verified retrospectively.
Under these conditions, equation (3 .3 ) transforms to
m d the boundary conditions ( 3 . 4 )transform toi n 0 = +a. (3 .9 )The (unique) solution of (3 .8 )and (3 .9 )is
(3 .10)W(p) = (p+l)sin2asin2(p++)a. (3.11)
(3 .12)(3 .13)
 wa*+2 COS ( p+ 2 ) a sinp8 cospa sin ( p+ 2) 0$(P,O) =(P) [ P P+2whereNote thatand tha tIt follows from (3 .6 ) (with c =  1) hat
W (P) =  W ( P )p( p , e ) = al)iJ(p 2, e ) .
@( 8) = (r/a)2lW/,, 8). (3 .14)It will therefore be sufficient in what follows to restrict attention to the outer regionr > a; the stream function in the inner region r < a may then immediately be inferredfrom (3 .14) .
Substitution of (3.10)in (3 .6 ) now gives] d p . (3 .15)P 1 cos(p+2)asinpB ospasin(p+2)8()( r ,e) =512m 1t W(P) P P + 2
When r > a , this integral may be evaluated by closing the contour by a large semicircle in the righthand halfplane R e p >  1. The integrand has a removable singularity a tp =  1, and poles at the other zeros of the function W ( p ) .We therefore haveto locate these zeros, and calculate the corresponding residues. Further progress
t The condition of zero forcing a t infinity implicit in (3.1) may be relaxed somewhat withoutaffecting the structure of the solution. In particular, if (3.1) is replaced by uB= 7P(r)whereP(r)= O (T  8 ) a t 03, o ( r 8 ) as + O , then the same method of solution is applicable with very similarresults.

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LoCal similarity solzctiom 307
2 02 0I
FIGURE(a ) .For legend 8ee next page.requires rather careful consideration of the function W ( p ) ,which is analysed in thefollowing section.
4. Zeros of the function W(p),Re p > 1pConsider first the real zeros of W ( p ) .Let p + 1 = [ / 2 a ; thenand the real zeros of W ( p ) are given by the intersections of the straight lineLa: = m(a)[ and the sine curve S : = sin

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308 H . K . Moffatt and B . R.Duffy
FIGURE. (a)The intersections of the straight line y = m(a) and the sine curve y = sin5 givethe real zeros of W ( p ) . b) (e) Solid curves are the family defined by (4.4) and dashed curves arethe family defined by (4.5). The intersections marked 0 give the complex zeros of W ( p )withi+iv = 2 a ( p + 1 ) . Double zeros are indicated by 0 . 6) 2 a = 135'; (c) 2 a = 159'; ( d ) 2 a = 168';( e ) 2 a = 257'.
(iv) When 2 a = 2ac w 257.45" there is a double root of W ( p )at p = 0 (6 = 2 a c ) .(v) For 2 a c 2ac, the smallest root of W @ ) is p = 0; for 201 > 2ac, however, thesmallest root p , satisfies(with equality only at 201 = 2 n ) .
 1 2 < p , < o ( 4 . 2 )Consider now the complex zeros of W @ ) .Let
P + 1 = ( E+ 7 ) / 2 a , (4.3)and suppose that 7 + 0. Since W @ * )= ( W @ ) ) * ,he complex zeros occur in complex

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Local similarity solutions 309
5
4
7
i
1
0
 4
1
FIGURE. Sketch of variation of the first few zeros of W(p ) s functions of 2a.Where the curvesare solid, p , is real; where the curves are dashed, p , is complex, and only the real part of pn srepresented.conjugate pairs, and i t will be sufficient to consider those in the first quadrant 5 > 0,7 > 0. The real and imaginary parts of the equation W ( p )= 0 may be written
sin5 sin2a  6 2acoshq (4.4)
For given a, quations (4.4) and (4.5) represent two families of curves in the (5 , 7 )plane, and the intersections give the complex roots of W ( p ) . The intersections of(4.4) with the real axis 7 = 0 give the real roots treated above.] These curves aresketched in figures 6 ( b  e ) for the four particular choices of the angle 2a represented

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310 H. . Boffatt and B . R.Duffyby the linesLa in figure 6 (a),From such curves, we may infer the following propertiesof the zeros of W @ ) .(i)For 0 < 2a c 1, n eaoh stripSan,where8, is defined by
8, 3 ,g;mn< ,g < (m + l )n , rn = 1,2, ...},there are either two real zeros of W ( p ) as discussed above)or two complex conjugatezeros. We shall denote the two zeros in the strip S,, by pznand pZn f l ;f they are realthen pznQ pZn+,and if they are complex pZn+,=p & (and we may adopt the convention Imp,, > 0,mp,,+, c 0).
(ii) For 7r < 2a c 271, in each strip S2,+, (n= 1,2,3, ...) there are either two realzeros of W ( p )(232n+l, ,,+~) or two complex conjugate zeros (p2n+aP&+~).n thestrip S,, there are always two real zeros ( p l , p 2 ; p l p , ) which coalesce to form adouble zero only for the critical anglea = a,.
The behaviour of the roots p,, p, , p 3 , . (or their real parts where they are complex)is indicated in figure 7, over the full range 0 < 2a Q 2n. Note the position of thecritical angle 2a, and the denumerable sets {2a,}, (2a3 , ( 2 a 3 (with limit points 71,n,27r respectively) on this figure.
5. Asymptotic behaviour of the stream function (3.15)W e are now in a position to evaluate the integral (3.15). When r > a, the contour,
closed by a large semicircle in the halfplane Rep >  1, encloses all the poles pndescribed in the previous section. Suppose first that a s such that all of these poles aresimple (i.e.a {a,}U {a;} U {a:} U {ac})then standard evaluation of (3.15) gives, forr > a,
and in this ormula, since the polesere simple,W(p,) = sin 2a a cos 2(pn+ 1 )a + 0. (6.2)
Note that the complex terms of the series (5.1)occur in complex conjugate pairs. Notefurther that, from (5.1),for r > a,
since W ( p n )= 0 and limpl W ( p )= W(0).This provides a check for consistency with(3.2), since the total flux introduced through the boundaries is
(5.4)P+O
$(r , a) (r , a) = waa = 2Q.Equation (5.1)has the form
CO+(r,e) = c (:)pnjn(e), r > a,n= 1 (5.5)
wheref,(6) depends implicitly on a.The result (3.14) then gives the correspondingexpansion

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Local similarity solutionsFor 2 a c ac = 267.46', p1= 0 and
where ( 0 ) s as given by ( 1 . 9 ) ,and so (5.5) ives$(r , e ) N imaf(e) Q f ( @ for r B a,
i.e. the JefferyHamel solution ( 1 . 1 0 ) .Similarly (5 .6 )gives$ (T , e ) * J
which is just the similarity solution given by (1 .8) .For 2 a > 2ac,however, as noted in $4 , the smallest root p1 ies in the range
where nowfl(e)= (Wl(pl))l[pi1 os(pl 2 )a inpl 8 (pl 2)l cospl a sin (pl+ 2 )81,
311
( 5 . 7 )
(5 .8 )
( 5 . 9 )
( 5 . 1 0 )( 6 . 1 1 )
( 5 . 1 2 )and clearly we do not recover the similarity solutions in this situation. The forms(5.10)and (5.11)do however satisfy the basic requirements ( 3 . 7 ) , incep, 2 4.
In the JefferyHamel limit (a+O, w+m, Q constant), ( 5 . 1 0 )gives, for any fixed T ,$(r, 0 ) 2 & ( r / 4  ~ ~ f l ( e ) . ( 5 .1 3 )
From this, we may construct a local Reynolds numberR  Q / v ) T/u)% ( 5 . 1 4 )
Since p , < 0 when U > a,, t would therefore appear that R > 1 (in the limit a+ 0, rfixed) no matter what the magnitude of the source Reynolds number 2 Q l v may be.Inertia forces are then of dominant importance for all T , and a lowReynoldsnumbertreatment of the JefferyHamel problem for a > a, would appear to be quite untenable, no matter how small Q / v may be.As mentioned in the introduction, the breakdown of the JefferyHamel solutiona t a = ac (when R = 2 Q / v+ 0 ) was noted by Fraenkel ( 1 9 6 2 ) who also described asimilar behaviour at finite R (see also Fraenkel 1973) . From Fraenkel's discussion, i temerges that there exists a domain 9 : c a , ( R ) in the a  R plane (with ac(0) 129"),within which the JefferyHamel stream function ?,hJ(8;, ) s a continuous functionof a, nd tha t ?,hJ t 00 (for all 8 )as a f a , ( R ) . t may be conjectured from the discussionof the present paper that when a > a , ( R ) , the JefferyHamel solution is irrelevant,and that the flow for such angles must depend on the manner in which the sourceQ is distributed in the neighbourhood of the vertex, as well as on its net strength.Returning now to the lowReynoldsnumber limit, when 201 = 2ac, as noted earlier,W has a double zero at p = 0, i.e.
W'(0)= sin 2 e a cos 2 a = 0, (5.15)and so the integrand in ( 3 . 1 5 )has a double pole at p = 0. The leading term of the

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312 H . K . Hoffatt and B . R. Du&asymptotic expansion of $ for r 9 a comes from the residue at this double pole and,from (3.15), s given by
a inpe cospa sin ( p+ 2)8P + 2
1w22a,2sin 2a,{logk) e cos2a,  sin 281+ [ge cos2e sin 20+ac sin 2a,l(5.16)[which again satisfies $ ( ~ , a ) &m2,consistent with (3.2)]. Once again, as in thesimpler problem treated in 3 2, a term logarithmic in r appears when the angle 2a iscritical. Here, the situation is closely analogous to that resolved in the elasticitycontext by Sternberg & Koiter (1958). For r < a , the correspondence (3.14) showsthat$ N r210grasr+Owhena = a,.
We now have the additional interesting complication that W(p) has a double zero[and so the integrand in (3.15)has a double pole] whenever a akes one of the specialvalues (of which there is a countable infinity) for which La s tangent to S (see thediscussion of $3). For example, when 2a = 201, z 159", there is a double pole atp, = p 3 z 1.78 and the second and third terms of the expansion (4.1) have to be replaced by
(5.17)
giving a contribution proportional to ( a / r ) p 2 og ( a / r )for r > a . This contribution ishowever dominated (when r 9 a ) by the leading term of ( 5 . 1 ) , which is (5.8) (with2a = 159").Similarly when a akes any of the special values a,, a;, a: for which Latouches S, evaluation of the residue at the corresponding double pole, p , say, givesrise to a contribution to $ proportional to ( a / r ) p mlog ( a / r ) ; his is, however, alwaysdominated by the terms of the series (5.1)with n < m.
 wa2 d a p cos ( p+ 2)a inp8 cospa sin ( p+2) 0+k)"(P2) dP [ P P + 2 l l p = p 2 ,
2 a < 2 a , z 257'r2 , hen r p * +2ro, hen rnr
2 a = 2 a , 2 a > 2 a ,rp1 f 2 , then r2, hen r ) r + 2r  p l , then To , then r va
Hinged plate limitJefferyHamel limit
TABLE . Asymptotic dependence of @(r ,0) on r as r l a + 0 and r l a+ 0 for the problem defined by (3 .3) and (3.4). The exponents p , , p , and p 3 are the roots of W ( p ) = 0 ordered so that < p 1 < Rep, < Rep, (figure 7 ) .For 2 a > 2 a , ,  < p , < 0.
rla + or / a + 00
r2 og (al.), then r2 , hen rps+2log ( r / a ) , hen ro , then r*a
The conclusions of the above analysis are summarized in table 2, which shows theradial dependence of the leading terms in the asymptotic expansion of $ ( r , 6 ) forr/a+O and r la tco for the problem defined by (3.3)and (3.4). t should perhaps beemphasized that we have assumed throughout that the boundary conditions andresulting flow fields are symmetric about 0 = 0. If antisymmetric ingredients areincluded, then the conclusions of table 2 require modification (see concluding footnoteof Sternberg & Koiter 1958). The antisymmetric ingredients can be important in

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Local similarity solutions 313certain circumstances, and in particular in someof the situations envisaged by Jeffrey& Sherwood ( 1 980).
The 'explosion' in the solution (1.8) at 2a ;=: 257' was originally drawn to ourattention by Dr T. J. Pedley, whose provocative observation is gratefully acknowledged, as is a stimulating discussion of the phenomenon with Dr J.M. Rallison. Thework is supported by S.R.C.Research Grant no. GR/A/5993.4.
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