Calc. Var. 3, 155-192 (1995) 9 Springer-Verlag 1995

The capillary problem for an infinite trough Henry C. Wente Department of Mathematics, University of Toledo, Toledo, OH 43606, USA Fax: (419)537-21 57; e-mail: fac3330 @ uoft01.utoledo.edu Mathematisches Institut, Universitfit Bonn, D-53115 Bonn, Germany Received December 22, 1993/Accepted January 21, 1994

Abstract. Consider an infinite trough (or wedge) with dihedral angle 2a, 0 < a < n and a quantity of fluid inside contacting the edge. In equilibrium the free interface of the fluid will be a surface of constant mean curvature meeting the planar walls at a constant angle 7 determined from physical considerations. One obvious configuration is for the free surface to be a section of a round circular cylinder parallel to the axis of the wedge whose position is determined by the angles ~ and 7. For ~ + 7 > n/2 the cylinder configuration is unstable and bifurcation occurs. We exhibit the full family of bifurcating solutions starting with the round cylinder solution and proceeding through a "beading up" process into a series of spherical sections suitably positioned. Furthermore, if the edge of the wedge is a re-entrant corner (~ > n/2) then there are further bifurcating families. One is a secondary bifurcation from the family initially constructed while the other is a primary bifurcation from the cylinder which are less symmetric than the initial families.

Mathematics Subject Classification: 53A10, 35B32, 76B45 I. Introduction Consider an infinite wedge Z consisting of two half planes Z 1 , 2 2 meeting at a c o m m o n edge l with dihedral angle 2 ~, 0 < ~ < re. Let T denote a quantity of fluid perhaps of infinite volume in contract with the wedge and also its edge. The configuration will be in equilibrium if its potential energy is an extremum relative to all compactly supported perturbations of T which pre* This paper was completed while the author was on a subbatical from the University of Toledo and was a visitor at Stanford University and also with SFB 256, University of Bonn.

156

H.C. Wente


J

Fig. 1

serve volume. If A denotes the free surface of T and if X' is the wetted part of 27 then for any large ball B the energy of T inside the ball is (1.1)

#/~ (T) = ~ r l A ~ B I - k c r l S ' ~ B I .

Here IA n B I and I S ' n B I are the areas of the respective surfaces inside B, ois the surface tension and k is the wetting energy. F r o m the Calculus of Variations one knows that (a) A has mean curvature H for some constant H and (b) the angle of contact y between the free surface and the walls of 27 is constant with cos 7 = k. For an infinite wedge one obvious equilibrium solution is for A to be a section of an infinite round cylinder correctly positioned with its axis parallel to the edge of the wedge. This configuration is uniquely determined by the angles ~ and 7 up to a scaling factor. The infinite wedge configuration has applications to the study of fluids in a container with sharp edges in conditions of micro-gravity. It is the model for a thin strip of fluid wetting an edge of such a container. One wishes to understand the stability, possible bifurcations, and shapes of equilibrium configurations. Two earlier papers on this problem were by D. Langbein [5] and T. Vogel [6]. Vogel showed that in the case ~ + 7 < re/2 the cylinder configuration is stable relative to all compactly supported volume preserving perturbations and that this solution is the unique one which is periodic in the direction of the edge. Note that if ~ + 7 < z/2 the cylindrical region is not convex while for ~ + 7 = ~z/2 the free surface is planar [see Fig. 1]. For c~+ 7 >re/2 Vogel showed that the cylinder configuration is no longer stable and that there is a bifurcation of the cylinder into a one-parameter family of equilibria. He considers only the case e < ~/2. Langbein somewhat earlier had considered the same problem and developed a Fourier series expansion for the bifurcation. He allowed 0 < c~< re. The bifurcation results of these papers are local and describe the bifurcation equilibria only in a neighborhood of the cylinder solution. We exhibit the complete bifurcating family. The following observation allows us to do this. The free interface A is topologically an infinite strip such that its two boundary components meet the walls of 27 at a constant angle 7. It follows from a theorem of Joachimsthal [3] that these boundary curves are curvature lines. They are planar curvature lines. We now make the ansatz that all curvature lines of this family on A are planar. A surface satisfying this condition is said to be of Joachimsthal type. If we normalize the surface by a homothety so that its mean curvature H = 1/2

The capillary problem for an infinite trough

157

it will follow that the entire family of such surfaces can be completely described, forming a two-parameter family with parameters (a, b). In this paper we will use the solutions as presented by U. Abre~ch [l]. His beautiful paper is based on results of classical Differential Geometry as it applies to the construction of surfaces of Gauss curvature K = - 1 (see G. Darboux [2] or L.P. Eisenhart [3]). R. Walter [7] also has carried out a similar program but in a slightly different manner. Every member of the family is represented by a conformal immersion of the plane into IR 3 such that the curvature lines correspond to straight lines parallel to the coordinate axes in the domain. The image of any horizontal curvature line will be a planar curvature line. The immersions initially constructed will have the property that the image of both axes will lie in symmetry planes for the surface. We restrict any such mapping to a horizontal strip ~ = {(u, v) I Iv l < v}. The resulting surface A (a, b,v) will have constant mean curvature H = 1/2 and its boundary curves will be planar curvature lines with contact angle 7 = 7(a,b,v). The two planes of the wedge 21, Zz will meet at an angle 2 e where c~= ~ (a, b, v). For any admissible (e, 7) we expect a smooth one-parameter family of equilibrium configurations and this is what we get. There is a single curve of solutions. One end starts from the round cylinder while at the other limit we find a series of spheres of radius two suitably positioned. These surfaces are all symmetric with respect to the bisecting plane of the wedge. There will also be a series of planes H,, perpendicular to the axis I of the wedge such that the surface is symmetric with respect to each Hr,. If zl is the common distance between two successive planes the surface will be periodic in the axial direction with periodic 2 zl. In Section II we describe the results of Abresch and derive the needed formulas for ~ and 7. In Section III we show that for admissible (c~,7) there is a curve of solutions as described above while in Section IV we analyse the beading-up process and show that it converges to a limit configuration which we shall describe shortly. The configurations constructed in Section III will all be symmetric about the bisecting plane of the wedge. In section V we exhibit a family of secondary bifurcations which break this symmetry. This will occur for any admissible (c~,7) where c~> re/2 and 0 < 7 < re, so that the wedge has a re-entrant corner. These new families continue to be of Joachimsthal type. In Section VI under the same conditions on (e, 7) we construct yet another family of primary bifurcations from the round cylinder which also break by symmetry with respect to the bisecting plane of the wedge. The set of allowable values for (c~,7) can be found by examining a cross-section of the round cylinder solution. The free surface is represented by a circle of unit radius while the wedge is represented by two rays symmetrically positioned. The possibilities are indicated in Fig. 2. There are no bifurcations if e + 7 < re/2. If e > re/2 them from geometrical considerations one must have ~ - re/2 < 7 < (3 re/2)- e. For c~< re/2 the value 7 = rt is allowed [see Fig. 3]. The case 7 = re/2 is easily discussed. The bifurcating family will be sections of the rotationally symmetric Delaunay unduloids. In this case the limit configuration is a series of round spheres of radius 2 whose centers lie on a common line and which are successively tangent to each other.

158

H.C. Wente T

= ~/2

/'~

"\\\\ I !

\

~

J ~ .

/1

/

T =~+n/2

~--..

\ Fig. 2

,,f

--

geometrically

=/2

-

-

unrealistic

--~---> /

o~

~/2

no bifurcation

\ \

-

)

g ~ or- ~z/2

o~- rd2 < 'y< 3rc/2- o~

y = 3rt/2 - o~

Fig. 3

The capillary problem for an infinite trough

159

-

Fig. 4. The limit configuration: y < c~+ 7t/2; dotted circle = K = + 1 case; cross-section in plane Qo; solid circle = H = 1/2 case

We now describe the limit configurations of the primary bifurcating families more generally. It is useful to recall the following. If A is a cmc surface with H = 1/2 then the parallel surfaces .,i obtained by moving a unit distance in the direction of the oriented unit normal vector gives a surface of constant Gauss curvature K = + 1 away from those points on A where one of the principal curvatures vanishes. For a given (e, 7) each surface in the family A (e, 7, e) which we construct is a section of a complete cmc immersed surface of Joachimsthal type with H = 1/2, At(e, 7, ~). There are three cases to consider. Case 1. 7 < ~ + n/2

Referring to Fig. 2 we notice that for the initial cylinder configuration the axis of the wedge lies inside the cylinder. The limit spheres will have a similar cross-section except that their radii will now be two. The complete configuration is an arrangement of such spheres. The limit K = + 1 configuration is a sequence of unit spheres arranged in succession and tangent to one another. The centers of these spheres are co-planar and the tangency points for successive unit spheres lie on a line 10, the axis for this configuration. For 7 = n/2 we have l0 containing the centers of all spheres while for n/2 < 7 < ~ + n/2 the axis is determined by observing a cross-section of one of these spheres by a plane perpendicular to lo. This is identical to the cylinder cross-section of Fig. 2. In fact, if2 is the distance from the center of the sphere to the axis 10 we have the formula (1.2)

a) 2 s i n ~ = s i n ( 7 - n / 2 )

of

7>n/2

b) 2 s i n , = s i n ( r e / 2 - 7 )

of

7
As 7 increases from n/2 to ~ + n/2 the axial line moves away from the center of the sphere to its edge. The H = 1/2 configuration is found by enlarging every other sphere to have radius two while the remaining spheres shrink to a point. The new H = 1/2 spheres have a similar cross-section when finding the wedge and the new axis I is shifted parallel to l0. This new axis passes through the centers of those K = + 1 spheres which were reduced to a point [see Fig. 4].

-

I0

160

H.C. Wente

//

\ \ K = +1 circle

/

9 "I / H = 1 / 2 circle

Fig. 5. The limit configuration: y = ~ + n/2

Observation. Suppose ~ < n/2 (? < ~ + n/2). Take that part of the H = 1/2 limit configuration which spans the wedge S. We obtain a sequence of sections of spheres which touch on the axis that do not otherwise intersect. This would seem to be a physically reasonable solution to the beading up process. Otherwise the limit sections of the H = 1/2 spheres will intersect. In other words at some prior point the bifurcating family had started to self-intersect. Our mathematical solutions do not feel this problem but the real world would. Case 2. ? = c~ + 7z/2

In this case the distance 2 from the axial line lo to the centers of the K = + 1 sphere now equals one. The line lo is a tangent line to the unit spheres, Furthermore the sequence of spheres has merged together leaving just two spheres touching on the c o m m o n tangent line l o . For the limit H = 1/2 configuration one sphere expands to have radius two while the other shrinks to a point. The new axis I is parallel to the axis lo passing through the center of the collapsed sphere [see Fig. 5]. Case 3. 7 > ~ + n/2

Now the distance 2 from the center of the K = + 1 spheres is greater than one and the axis l o lies outside the unit sphere. L e t / 7 o be that plane perpendicular to lo and bisecting this sphere. Each K = + 1 sphere of our sequence has its center on this plane. They are all equidistant from lo and are successively tangent to each other. The configuration resembles a beaded bracelet. The corresponding H = 1/2 configuration is obtained by enlarging every other unit sphere by a concentric sphere of radius two and letting the others shrink to a point. Each of these enlarged spheres has it own axis l determined by the angles (~, 7). This axis is parallel to l o and lies in the plane determined by lo and the center of the sphere under consideration. Its distance from the center of the H = 1/2 sphere is twice the distance of lo to the center o f the K = + I sphere [see Fig. 6].

The capillary problem for an infinite trough

161

Cross-section in plane rl o K = +1 circles only

Fig. 6. The limit configuration: y > c~+ =/2

II. Constant mean curvature immersions of Joachimsthal Type L e t x (u, v) be a p a r a m e t r i c surface of c o n s t a n t m e a n c u r v a t u r e H = 1/2 c o n f o r m a l l y i m m e r s e d in R 3 such t h a t the c u r v a t u r e lines c o r r e s p o n d to s t r a i g h t lines p a r a l l e l to the c o o r d i n a t e axes in the p a r a m e t e r d o m a i n . T h e first a n d s e c o n d f u n d a m e n t a l forms c a n be w r i t t e n ds 2 = e 2~ (du 2 + dv 2)

(2.1)

F / = - (dx" d~) = e" (sinh co du 2 + coshco dv 2) k 1 = e -" sinhco,

k2 = e - " c o s h c o .

H e r e s t r a i g h t lines p a r a l l e l to the u-axis c o r r e s p o n d to k l - c u r v a t u r e lines while the k / - c u r v a t u r e lines are in the d i r e c t i o n of the v-axis (see E i s e n h a r t [3] for example). T h e G a u s s e q u a t i o n is (2.2)

Aco + sinhco cosh co = 0 ,

a n d the choice of f u n d a m e n t a l forms g u a r a n t e e s t h a t the C o d a z z i e q u a t i o n s

are satisfied. By a theorem of Bonnet, for any solution of (2.2) there is a conformally immersed H = 1/2 surface x (u, v) with fundamental forms (2.1) u n i q u e u p to a n E u c l i d e a n m o t i o n in IR 3. T h e e q u a t i o n s to be i n t e g r a t e d are x , . = cou x, - coy xv + e ~ sinh co Xuv = (Do Xu + cou Xv

(2.3)

xw = -- cox xu + co~ x~ + e ~ c o s h co

~u= - k a x . ,

~v=-k2xv.

Let X = [e~, e2, e3] r be the m a t r i x r e p r e s e n t i n g the D a r b o u x frame for the i m m e r s i o n so t h a t el = e -~ x , , e2 = e - ~ x~ a n d e3 = r is the o r i e n t e d unit

162

H.C. Wente

n o r m a l vector. T h e equations determining the frame are simply Xv = d X , Xv = N X where (2.4)

d=

(0 .sir/ wv -sinhw

0 0

0

,

~=

/

(0

--w, 0

.

0 -coshw

0)

coshw 0

.

The G a u s s and Codazzi equations are the integrabitity conditions for this system. If x (u, v) is our conformally i m m e r s e d cmc surface with H = 1/2 then the parallel surface y (u, v) = x (u, v) + ~ (u, v) is an i m m e r s e d K = + 1 surface except where the conformal function co (u, v) = 0. Its f u n d a m e n t a l forms are ds 2 = cosh2w du 2 q- sinh2 w dv 2

(2.5)

/7 = (sinhco coshco)(du 2 + dv 2) ~l = k j k 2 = s i n h w / c o s h c o ,

~2 = kz/kl .

N o t e that a l t h o u g h y (u, v) fails to be an i m m e r s i o n when 09 = 0 it shares the same s m o o t h frame as the H = 1/2 surface. We n o w impose the condition that the k , - c u r v a t u r e lines be p l a n a r so that x (u, v) is a surface of J o a c h i m s t h a l type, following the d e v e l o p m e n t as presented by Abresch [1]. Suppose that x(u, ~) lies in the plane fl(~). The vectors e 1 and (el), b o t h lie in that plane. This fact along with (2.4) allows us to c o m p u t e the unit n o r m a l vector N (~) to f2(~). One obtains (2.6)

N(~) = ( - w v ~ - sinh o) e2)/x/w 2 + sinh2 w .

By J o a c h i m s t h a l ' s t h e o r e m the angle r (~) between the surface and the plane f2(~) is constant along the curve x(u,~). We have that cosq~(~) = N . 4. The derivative with respect to u must vanish giving us a) (wv/sinhw), = (w,/sinhw)~ = 0 (2.7)

b) w, = - f ( u )

sinhw,

c% = - g(v) s i n h w

c) cot q~(v) = 9 (v) for suitable functions f (u) and g (v). Following Abresch one finds f ' ( u ) + 0 (v) c o s h w = 1 + f 2 ( u ) + gZ(v)

(2.8)

where f (u) and g (v) are solutions to the following differential equations which were put into a convenient form by Abresch. a) f ' ( u ) z = f 4 + (1 + a 2 - b 2 ) f 2 + a 2 = V l ( f ) , f(O) = O, f'(O) = a (2.9)

b)

0(v) 2 = g 4 + ( l - a

2 + b 2) g 2 + b 2 = v 2 ( g ) ,

g(0)=0,

0(0)=b.

If we assume that a and b are non-negative then it follows from (2.8) that a + b >_ 1 and if a + b = 1 then w (u, v) = 0. In this case the c o r r e s p o n d i n g H = 1/2 surface x (u, v) is the r o u n d cylinder.

The capillary problem for an infinite trough

163

A n o t h e r w a y of writing (2.7) is to set W = cosh co. Abresch then derives the identities

I a2

D

a) Wu2 = (W 2 - 1) 1 + g2 (2.10)

b)

W~2 = (W 2 - 1) 1

c) Wu = - f ( u ) ( W

+ f2

(l+f2)(W-l+f2}

2-1),

w, = -

2

s' VIj

a(v) (w ~ -

1).

It will be useful for us to use (2.10a) when v = 0 (so that g = 0 and 0 = b) and (2.10b) when u = 0 (so that f = 0 and f ' = a), giving us (2.11)

a)

W,2 = ( W 2 - 1 ) [ a 2 - ( W - b )

2]

when v = 0

b)

Wo2 = ( W 2 - 1 ) [ b 2 - ( W - a )

2]

when u = 0 .

Consider the e q u a t i o n (2.9a) for f(u). If b = a + 1 one finds that V1 ( f ) = is non-negative but vanishes at f = xfa. We have

(a -- f2)2

f (u) = x / ~ tanh (u,~/a),

b = a + 1.

F o r b < a + 1, V1 ( f ) is always positive and u p o n solving (2.9 a) one finds that f (u) diverges to ___c~ at a value u = A where (2.12)

a = 7 0 x/f 4+(l+a

af 2 - b 2) f 2 + a

2"

F r o m (2.8) we conclude that coshco = 1 when u = A and so co(u,A) = 0 (i.e. the line u = A is a vertical nodal line for co (u, v)). At the origin cosh co = a + b > 1. We select co to be positive here. Across a nodal line co (u, v) can be extended by o d d reflection. Thus if b < a + I the solution co (u, v) will have the lines u = (2m + 1)A as vertical nodal lines and co (u, v) will be an o d d function a b o u t these lines. It also follows from (2.8) that co (u, v) will be s y m m e t r i c a b o u t the lines u = 2 m A.

(2.13)

a) co (2mA + u, v)

= co (2mA - u, v)

b) c o ( [ 2 m + 1]A + u , v ) = - c o ( [ 2 m +

1]A--u,v).

F o r b = a + 1 there is no vertical nodal line but the function co (u, v) is symmetric a b o u t u = 0. I f b > a + 1 then 1/1 ( f ) will have a first zero at a value f,, > 0. In this case f ( u ) can be extended to be a s m o o t h periodic function defined for all u. It is an o d d function of u which is increasing on an interval [-.4,,4] where f (-~) = fro. T h e function f (u) is s y m m e t r i c a b o u t u = ,4. In this case there are no vertical nodal lines. Since f(u) vanishes at u = 2 m.4 it will follow that co (u, v) is symmetric a b o u t these lines, satisfying a condition (2.13a). F r o m (2.10c) we see that W(u,O) is decreasing from W(0,0) = a + b to W(2_~, 0) = - a + b. We can n o w use (2.11a) to find ,4. We find (2.14)

2,4=

a+b dW ! b a~q/~-lx/a2-(W-b)2

b>a+l

164

H.C. Wente

A similar discussion applies to the function g (v). F o r b > a - 1 the function 9 (v) becomes infinite at a value B where (2.15)

B = S dg o ~/g4 + (1 - a 2 + b 2) g2 + b E

T h e function co (u, v) will have the lines v = (2 n + 1) B as horizontal nodal lines and co(u, v) is an o d d function a b o u t these lines. It will also be symmetric a b o u t the lines v = 2 n B. (2.16)

a) c o ( u , 2 n B + v )

=co(u,2nB-v)

b) c o ( u , [ 2 n + 1 ] B + v ) = - - c o ( u , [ 2 n + l ] B - v ) .

F o r b < a - 1 there are no horizontal nodal lines. In this case g(v) is a periodic function increasing on an interval [ - / ~ , / 3 ] with g ( / ~ ) = gm and g (2n/~) = 0. We can c o m p u t e / 3 using (2.11 b) obtaining (2.17)

2/3=

,+b S

o-b

dW

l ,/b2 -1W-

a) 2

The function co (u, v) is s y m m e t r i c a b o u t the lines v = 2 n/3. The lines b = a + I and b = a - 1 divide the p a r a m e t e r d o m a i n ~ into three zones 9 = {(a,b)la>_ O,b >_ O,a + b >_ 1}. In each of the three zones the functions co (u, v) are doubly periodic but the nodal line structure differs. Zone L' a - 1 < b < a + 1. Here there are b o t h horizontal and vertical nodal lines u = (2m + I ) A and v = (2n + 1)B. T h e function co(u,v) is an o d d function a b o u t these lines. T h e dual lattice u = 2 m A and v = 2n B are s y m m e t r y lines for co (u, v) [see Fig. 7]. At b = a + 1 the period A b e c o m e s infinite. The function co(u, v) remains periodic in the v direction with nodal lines v = (2 n + 1)B and s y m m e t r y lines v = 2 n B. T h e line u = 0 is the only vertical s y m m e t r y line and there are no vertical nodal lines. F u r t h e r m o r e , co(u,v) converges to 0 uniformly as ]ul --* ___oe. O n e has cosh [co (u, v)] a s e c h Z ( u x / a ) + bseca(v'c/b) b = a + 1. i + a tanh 2 (u x/~) + b tan 2 (v x / b ) ' Zone II." a + 1 < b [see Fig. 7]. The solutions co (u, v) are again doubly periodic. The nodal lines are the lines v = (2n + 1)B. T h e s y m m e t r y lines are u = 2 r n ~ and v = 2 n B . Zone III." b < a - 1. T h e situation is analogous to Z o n e II with the values of u, v being reversed. The nodal lines are u - - ( 2 m + 1)A while the s y m m e t r y lines are u = 2 m A and v = 2 n B . We n o w consider the i m m e r s e d H = 1/2 surface x(u, v) and the K = + 1 surface y(u,v). We have x(u,~) c O(~) with n o r m a l vector (2.6) and angle of intersection cot q) (9) = g (~). The parallel surface y (u, ~) also is a p l a n a r curvature line with y (u, ~) c ~2(9) with the same angle of intersection. The two planes are parallel and the distance between t h e m is cos q)(~)= 9 / x / 1 + 92 (Fig. 8).

The capillary problem for an infinite trough

165

Zone I

V

b=a+l

b=a-1

4-

+ __4v

----4-----

= 2B I

(o,1)

+ I v=B +

>a

(1,o)

+

I I I

-

u

- - - - 4 - - - - - -----4------->

----'Jr-----

u=A u=2A v

Zone II

4` --

--

--4- .... I I I I _ _ _L . . . . . I I

Iv=B I I

I I

_.

u

.i.___> I

+

+

_

I

I

I -

I

u = 2.,~ Fig. 7. Nodal regions for various zones

2

~

+

+I

I I

I v=2g

I

I I

+

I

]v--4~

I

[

Iv=2B T

+

Zone III

V

4, r

I I I

-

_

I "F . . . . .

HI

+

+ I J. I I +

II

---->

+

U

-

I T -- --

u=A u=2A

H = 1/2 n@)

e2#

" X ~ f

K=+I fi@)

/

e2

Fig. 8. The curves s (0, v) and ~ (0, v)

It is well k n o w n t h a t the lines of c u r v a t u r e in the v-direction lie o n spheres. To see this we set (2.18)

p = x + ~ + ( l / f ) e 1 = y + ( 1 / / ) el

a n d check t h a t p ~ = 0 so t h a t p = p ( u ~ . T h u s for u = ~ we h a v e x (fi, v) = S [p(~), R 1 (fi)] w h e r e R1 (~) = [ f I/%/14-f2 while the c o r r e s p o n d i n g

166

H.C. Wente

K = + 1 surface satisfies y ( a , v ) c S [p(~),R2(fi)] where Rz(ft ) = 1/I f[. F r o m (2.18) one sees that p(a) c ~(v) for all v. This means that all points p(u) lie on a line lo where (2.19)

lo = ~ ~03).

The K = + 1 immersion is a Joachimsthal surface (in the traditional sense, see Eisenhart [3]) while the H = 1/2 immersion is only of Joachimsthal type. Furthermore if we let ~u(a) be the angle of intersection of these surfaces with their respective curvature spheres one finds using (2.18) that (2.20)

a) c o s ~ P ( f i ) = f / x / 1 + f 2 b) cos~k(fi) = 0

or cot ~u(O)=f

( H = 1/2),

(K = + 1).

The immersions x (u, v) and y(u, v) have the following symmetry properties. Suppose ~o(u,v) is symmetric along the horizontal line v = 2 n B (or v = 2 n B), satisfying (2.16 a). The curvature planes f2 (2 n B) = ~ (2 n B) are in this case identical. Call it f2,. Both surfaces intersect this plane at right angles and the immersions are equivariant. We have (2.21)

x(u, 2 n B -- v) = ~ , ~ x(u, 2 n B + v)

where ~ , is the reflection map about f2,. Similarly the function oJ(u,v) is symmetric about the lines u = 2 m A (or u = 2 m 4). The corresponding curvature lines again lie in symmetry planes 1I m {the spheres have infinite radius} and the mappings are equivariant with respect to reflections about this plane. Each f2, contains the axial line l0 while e v e r y / / , , is perpendicular to lo. A succession of two reflections is either a translation of a rotation. It follows that a) x (u + 4 A, v) = x (u, v) + 2 rl m (2.22) b) x ( u , v + 4B) = ~ 2 ~ x(u,v) where ~1 is the directed distance between the parallel planes H0 a n d / I 1 , m is the unit vector parallel to lo (say e 1 (0, 0)) and ~2 is a rotation through an angle 2 0 2 where ~2 is the angle between f2o and f21 . If v = ( 2 n + 1)B is a horizontal nodal line it follows from (2.7c) that the curvature plane f2 [(2 n + 1) B] is also a tangent plane to the (H = 1/2) surface along the curve x (u, (2n + I)--B). The same property is true for the K = + 1 surface. The two curves x (u, (2 n + 1) B) and y (u, (2 n + 1) B) are congruent as ~ is a constant vector along this curve. The map y(u,v) fails to be an immersion when v = ( 2 n + l ) B . Its image is a cusp line on the K = + 1 surface. For a vertical nodal line u = (2m + 1)A of ~o(u, v) we see (2.18) that the curvature line x ([2m + 1] A, v) lies on a unit sphere and the H = 1/2 surface is tangent to this sphere along this curve. Under the K = + 1 map y (u, v) the nodal line is mapped to the center of this sphere. The different nodal line structure for co(u, v) in each of the three zones leads to qualitatively different immersions for x(u,v) and y(u,v) [see Fig. 9].

The capillary problem for an infinite trough v 4'

I

4l_

167

2 (u, v) and 37(u, v) in Zone I

54

I --/I + / /

61.___-.> -

-I

u

6 4

x(O,v) N,y/f,~

~

SectO i npl17 aoinen

\ \ \

2(u,v)and~(u,v)in zoneIII v

~ (O,v)

(P~) Po

\

P2

p~=+1 o Surface ) ~ ~ Kp

V .... / x(Po)

Slicein

X/YtPo)

piane I] 0

3)

, = 1/2

/

S u r f a c e ~

Fig. 9.

The Periods zl and 02 The angle O 2 between the successive s y m m e t r y planes #20 and 01 is calculated as follows. F o r b > a - 1 one has v = B a nodal line for o)(u,v) with x (u, 0) c #2o a n d x (u, 2 B) c f21 . The angle is m e a s u r e d by following the curve x (0, v) lying in the s y m m e t r y plane H e . We have (2.23)

0 2 = S k2 ds = S cosh o dr.

N o w use (2.8) when u = 0 so that f ( 0 ) = 0, f ' ( 0 ) = a and note that g(B) = + or. We find (2.24 a)

dv

0 2 (a, b) = rc + 2 a i0 1 -[- 9 2

(b > a - 1)

168

H.C. Wente

where g = g (v, a, b) solves (2.9 b). We can also write

dg (2.24b)

O 2 (a, b) = n + 2 a Jo (1 + g2)x/94 + (1 - a 2 + b 2) g2 d- b 2

b > a - 1.

F o r b = a - 1, 0 2 is infinite while for b < a - 1 one again uses (2.23) along with (2.11 b), giving (2.25a)

02( a,b)=

a+b S

W dW

o_b

_

,/b2

b < a-1

_ (W_

a)2

If we set W = a + b 4 , - l < ~ < l w e o b t a i n (2.25b)

1

(a + b 4) d~

Oa(a,b ) = ~ -1

-

r

b 0 2 -

b < a-

1.

1

I n particular we have the following information. = (2.26)

~

when

b=a+l

b) O2 = 7 ~ / ~

when

a + b= 1

"J- O0

when

b = a - 1.

C)

O 2 =

In order to calculate the directed distance zl between/7o a n d / / 1 one starts with the formula (2.18) for the centers p(u) of the curvature spheres. A direction vector for the axial line lo is found by c o m p u t i n g p' (u) using (2.18). O n e finds (2.27)

m = ( f 2 cosh co - f ' ) e 1 - co, f e 2 -F f sinh co e 3

where ] m ] = a is constant. F o r any t3, x(u,0) or y(u,~3) are planar curves c o n n e c t i n g / 7 0 to H1. The unit tangent vector to either curve is e I (U, 0). If a is the angle between the curve and axis then cos a = ( f ' - - f 2 c o s h c o ) / a . We shall p r o d u c e three different formulas for z~.

Casel: ZoneI, a-l
2 ~ [(1 - 2q t2)/(x/t 4 - 2q t 2 + I ) ( t 2 + x / t 4 - 2q t 2 + 1)] dt x/~ o

h-

2 --

F(q)

where (a + q)2 _ b 2 = q2 _ 1. This formula is similar to one given by Abresch. In particular one notes that "q = 0 when F(q) = 0 which occurs at a value q* ~ 0.67. In Z o n e I we have - 1 < q _< 1. The h y p e r b o l a (a + q,)2 _ b 2 = q,2 _ 1 divides the zone into two regions. A b o v e the h y p e r b o l a z~ is negative while below it is positive. a) If b = a + l

then

q=+l

and

zl=-oe

b) If b 2 = a 2 + 1

then

q=0

and

z1 > 0

c) If b = a - 1

then

q=-I

and

z1>0.

The capillary problem for an infinite trough

169

Case II: b < a + 1, Z o n e s II and III. We carry out the integration along the s y m m e t r y line v = 0 using (2.8) and (2.23) (2.29) z~(a,b)=ZS~'-~ f4+(1 +a2-2b2)f2+a2 df. a o (1 + ~ - ) S ~ 4 + ( 1 + ~ - b ~ + a2 Case III: b > a + 1, Z o n e I. Once again one integrates along the line v = B. Following Abresch one sets U = cos a and calculates

U 'a = 2 a (1 - U 2) (U + q), (a + q)2 _ b z = q 2 . N o w zl = ~ U ds but along v = B we have du = ds. 1

(2.30)

1

UdU

b>a+l.

I f b > a + 1 t h e n q > 1 a n d z t is negative while for b = a + 1 we have q = 1 and

z 1 = -oe.

In our discussion of the p r i m a r y and secondary bifurcations it will be helpful to describe the H = 1/2 and K = + 1 surfaces when we are in Z o n e III, 0 < b < a - 1, a > 1. In this case the function o)(u, v) has vertical nodal lines at u = (2m + 1)A and vertical s y m m e t r y lines at u = 2 m A where A is given by (2.12). There are no horizontal nodal lines but there are horizontal s y m m e try lines at v = 2n/3 w h e r e / 3 is given by (2.17). Proposition

2.1. The K = + 1 surface in Z o n e I l L

L e t G be the rectangle in the (u,v) plane bounded by the four vertices Po = (0, 0), Pl = (0, 2/~), P2 = (A, 0), P3 = (A, 2/3). The function co (u, v) is positive in G and vanishes on the segment P2 P3. W e consider the K = + 1 surface y(u,v) restricted to this rectangle. L e t c 2 be the image o f the segment Po P l, Co the image o f the segment Po P2, cl the image o f p~p 3, and c 3 the image o f pzP3 under the map y(u,v). The following are true.

1) L 2 = length (c2) = zc where c 2 is a geodesic curvature line lying in the symmetry plane 1I o perpendicular to the axis l o. 2) c o and cl are geodesic planar curvature lines lying in the symmetry planes 0 o and 01 respectively. I f L o = l e n g t h (Co) and L 1 = l e n g t h (c 0 then Lo + L1 = ~. 3) c 3 the image o f p2P3 is a point p* lying on the axis l o. Clearly c o and c 1 meet at p*. 4) The area o f y (G) is exactly n. 5) L e t P and Q be opposite points on the equator o f the unit sphere. L e t F 2 be the half great circle joining P to Q running over the north pole and let F 1 be half o f the equator. The region on S 2 o f area ~c bounded by F 1 w F2 is isometric to our K = + 1 surface y (G). The arc c 2 is mapped onto F2 while c o w c 1 is mapped onto F 1 . The image o f t * on F 1 is such that its geodesic distance f r o m P is L o .

170

H.C. Wente

Proof. We calculate L2 = length (c2) by ~ . 1 1 b) along with the observation that ds = sinhco dv and sinhco = ~ / W 2 - 1 as W = cosh~o, a+b

(2.31) In

dW

L2 = S a-b x/b 2 order

to

calculate

(W

a) 2

-

Lo=length(co)

-n. we

note

that

v=0

and

ds = cosh~o du. With the aid of (2.8) n

i

du

f=f(a,b,u).

L o =~-+ b o 1+/2'

Similarly in calculating L 1 = length(c 0 we have v = 2/~. Here we have 9 (2/3) = 0 and ~ (2/~) = - b giving us

n L l=~-b!

A

du l+f2"

Obviously, Lo + L 1 = n and L o can be more explicitly written using (2.9 a) as (2.32)

L o=~-+b

o (l+f2) x/f4+(l+a

2 - b 2) f a + a

2

or upon using (2.11 a) (2.33)

,+b

WdW

Lo = S

1

1 ,/a 2 -(W-

b) 2

Since y (G) has constant Gauss curvature K = + 1 it is isometric to some region on the unit sphere. Clearly c2 being a geodesic of length n is m a p p e d onto a half great circle. N o w Co and cl are both perpendicular to c2, they meet at one end and their total length is n. Their union is also a half great circle perpendicular to the image of c2. Our discussion allows us to conclude that the area of y (G) is n + 4 k n for some non-negative integer n. However, the area depends continuously on the parameters (a, b) in Zone III. F r o m (2.5) 21] A

[y (G)[ = S S sinh ~o cosh ~ du dv = constant. 0 0

We evaluate this when b = 0. In this case g = 0 and we have cosh~o = f ' / ( l + f 2) by (2.8). When b = 0 formula (2.9a) reduces to f , 2 = ( f 2 + 1) ( f 2 + aZ), giving sinheo = ~ / x / l + q2. It now follows that ly(G)l = n j

df

0 (1 + f 2 ) 3 / 2

- n.

Q.E.D.

The selection of a value for L o between ~ and n determines a one p a r a m eter family of isometric surfaces with K = + 1 known to the Classical Differential Geometers. In practice they are realized by taking a unit sphere and slicing it along a great circle to a length o f 2 L 1 = 2n -- 2Lo. Let lo be the line joining the end points of this slit. N o w pull these two points apart keeping the boundary curves planar and maintaining the symmetry plane H o perpendic-

The capillary problem for an infinite trough

171

kt

r

L2=~

t Po~....~

j..~

Pl

,o ~((3) is isometricto a quartersphere Fig. 10. K = + 1 surface in Zone III

ular to I. There is a limit configuration when the short arc becomes a straight line lying on lo. This corresponds to the curve L 1 (a, b) = constant hitting the curve b = a - 1. This terminal surface is known as Rembs' surface (see [4] for details) [see Fig. 10].

IlL The primary bifurcating family In this section we show how for a given choice of angle (cq ?) with 0 < c~< n, 0 < 7 < n and ~ + ? > n/2 there is a one-parameter family A (a, b, e) consisting of sections of complete H = 1/2 surfaces of Joachimsthal type whose boundary meets the walls of the wedge at an interior contact angle ? and where 2 c~is the opening angle of the wedge. The family is to bifurcate from the initial round cylinder configuration. As noted earlier the solution consists of sections of Delaunay unduloids when ? = n/2.

Case 1: ? > n/2. Here the symmetry plane bisecting the wedge will be the plane f20 containing the curvature line x (u, 0) and the immersion will be restricted to a domain = {(u, v) [ Iv[ _< v} for a suitable choice of v. For any (a, b) in Abresch's parameter domain our complete H = 1/2 immersion is a function x (a, b, u, v) -~ x (u, v) [we will often omit the (a, b) dependence unless greater clarity is needed]. The curvature line x ( u , - v ) lies in a plane Z1 = f 2 ( - v ) and x(u,v) c E z = (2 (v). The two planes Z1, Z2 intersect along a line l at an angle 2 ~. We obtain the needed formulas for (~, ?) by looking at the trace of x (0, v) or y (0, v) lying in the plane Ho. F r o m the discussion in Section II, the function g (v) -= 9 (a, b, v) defined by (2.9 b) satisfies: i) If b > a - 1 then 9 (v) increases from 0 to + oo on an interval [0, B] where B is given by (2.15). ii) If b = a - 1 then 9 (v) increases from 0 to x/~ on the interval [0, + oo) 9 (v) = v/b tanh (v ~/b).

172

H. C. Wente

Fig. 11. ~z+ 7 = 02 + g/2

iii) If b < a - 1 then g (v) increases from 0 to gM (a, b) = gM on an interval [0,/~] where g ~ is the smaller positive root to V2 (g) = g4 + (I - a z + b 2) g2 + b E = 0 and B is given by (2.17). In this case g(v) then extends as a periodic function of period 4/~ and zeros when v = 2 n/~. In the ensuing discussion we shall restrict our cut-off level ]v[ < v so that

(3.1)

i) 0 < v _ < B

if b > a - I

where g ( B ) = + o o

ii) 0
if b = a - 1

iii) 0 < v < 2 / ~

if b < a - l , 9 ( 2 B ) = O .

L e m m a 3.1. Let (a, b) be given parameters and consider the corresponding H = 1/2 immersion x(u,v) restricted to the strip ]vl -< v where v satisfies the conditions (3.1). Let Z~ = f 2 ( - v ) and Z z = f2(v) be the curvature planes defining the wedge. I f y is the contact angle of the surface with the planes X~ and X; and 2~ is the angle between Z1, Z a then a) cot 7 =- - g (v) ----- g (3.2)

b) e + 7 = 0 2 (v) + n/2

where 0 2 (v) is the angle of rotation of the curve x (0, v) as it moves from the plane f2o to Z (v) = Z 2 in the symmetry plane 11o . Proof. F o r m u l a (3.2 a) is just (2.7 c), noting that if we are entering the wedge with increasing v then cot 7 = g (v) while if we are leaving the region then cot ? = - g (v). It follows that our configuration contacts b o t h walls Z1, X2 at the same angle. F o r m u l a (3.2b) is elementary geometry [see Fig. 11]. [] We now use (3.2 b) to obtain the needed formula for ~. L e m m a 3.2. Suppose (a, b) are given parameters and that v is chosen in accordance with conditions (3.1). We have 7 given by (3.2a) while cr is given by c ~ = a io 1 + evg2 - - F ( a , b , v )

(3.3)

where g (a, b, v) is the solution to (2.9 b). I f g (a, b, v) is increasing (in v) on the interval [0, v] as is always the case when b > a - I and true for O <_ v <_ B when b < a - I then one can replace v by g -=- g (a, b, v) as parameter and we have a) cot ~ = -- g (3.4)

b)

dg Jo(l+g2) x/q4+(1--a

a f

2 + b 2) q 2 + b 2

G (a, b, g).

The capillary problem for an infinite trough

173

I ~~~1 = 82(a,b)/2

/ Fig. 12

s

Finally if B < v < 2 B and if g (v) = g with cot 7 = - g then the analogous formula to (3.4b) will be

(3.5)

~ = 02 (a, b) - G (a, b, g)

b < a - 1.

Here 6) 2 (a, b) is the angle between the planes (2 o, f21 as given in formula (2.25 a). P r o o f We use (3.2b) along with the formula (2.23) for 6)2(v) and (2.8) for

coshc~ giving us (3.6)

~. dv 6)2 (v) = a ! ~

+ t a n - l(g),

g = g (v).

But t a n - 1 (y) = 7 - n/2 which upon using (3.2b) gives us (3.3). F o r m u l a (3.4) follows from a change of variables and (3.5_) follows by making use of the symmetry of the function g (a, b,v) about v = B. Q.E.D. Remark 3.1. I f b < a - 1 then for v = 2 b B where g(2/~) = 0 we have ~, = n/2 and ~ = 6) 2 (a, b) the angle between ~o and ~1. For v = B we have 7 = 7~,x where cotTm = - g(B) = -- gm and ~(/~) = ~ (2/~)/2 = 6)2 (1, b)/2 [see Fig. 12]. P r o o f The function g (a, b, v) is symmetric about v =/~. Therefore 21~ dv ~ ~ 1-~92 =2a!

~(a,b, 2 B ) = a

0

dv

1+

gz-

Z~(a,b,B).

Remark 3.2. We note the behaviour of the function 6)2 (a, b) in Zone I I I as given by (2.25) measuring the angle between Qo and ~1. One finds a

a) 6)2(a,0)

b)

+ 00 where b = a - 1

1 -- 1 d~ < 0 0-a- = -~1 x/1 - 42 [(a + b 4) 2 - 1] 3/2 06)2= 1 _ d) J d4>O Ob -1 x/1 - 4 2 [(a + b 4) 2 -- 1] 3/2 C)

(3.7)

6)2 =

~

06)2

e) l i m 6)2 (a, b) = n

f)

lim [92 (a, a a~oo

1 -

e) ~-- 7c

any e > O.

174

H.C. Wente

It follows that in Zone III, 0 2 ( a , b ) > ~ and f o r any e > O, 0 2 ( a , b ) will converge uniformly to ~r as a goes to infinity so long as 0 < b < a - 1 - e. R e m a r k 3.3. Let b >_ a - 1 and set cot 7 = - g. The angle ~ = G (a, b, g) given by (3.4 b) satisfies 1 (3.8 a) ~ = - ~ t a n - 1 ( g / K ) -- t a n - 1(g) /f a + b = 1 .

Thus as b decreases from one to zero the angle o~ increases f r o m 0 to infinity. (3.8b)

1

c~ = t a n - 1(g)

,~

t a n - t (g/x/b)

/f

b = a + 1.

As a increases from 0 to infinity the angle ~ increases from zero to t a n - 1 (g) = 7 - re~2. 4

(3.8c)

a=tan-l(g)+~-tanh-l(g/x/b

,/b

)

/f

b=a-1.

Thus as b increases from g2 to infinity the angle c~ decreases from infinity to 7 - re~2. T h e o r e m 3.1. L e t (~, 7) be given with ~/2 < 7 <- ~ and 0 < ~ < 7c. There is a

curve cg (~, ~) in the (a, b) plane given in the f o r m of a graph b = b (a) starting f r o m a point (ao, bo) in the segment a + b = 1 defined for all a >_ ao with positive slope b' (a). There is also a function v = v(a) where v satisfies the restriction (3.1) such that the corresponding H = 1/2 immersion x (u, v) =- x (a, b, u, v) restricted to the strip Iv[ _< v has its boundary lying on the two planes Z 1 = ~ ( - v ) , Z 2 = ~v). The surface intersects each plane at the prescribed angle 7 and the two planes intersect at an angle 2 ~ with ~2o a symmetry plane for the configuration. Each curve cg(~,7) has a limiting slope b' (a) ~ 2 and also b/a ~ 2 as a becomes infinite where 2 satisfies the condition (1.2a). I f 7 > ~ + re~2 this slope is greater than one and the curve ends up in Zone II. For y = ~ + ~/2 we have 2 = 1 and the curve c~ (~, 7) remains in Zone I while for 7 < ~ + ~/2 the limit slope is less than one and the curve ends up in Zone I I I [see Fig. 13]. P r o o f Let ~ be the set of points (a, b, v) where v > 0 satisfies the restrictions (3.1), and consider the m a p 7J: ~ . _ ~ , 2 given by 7J (a, b, v) = (~, 7) where cot~ - - - g(a,b,v) and the angle a = F ( a , b , v ) is given by (3.3). We want to show that ~P-1 (~, 7) is a s m o o t h curve parameterized by a. Consider the case where either b > a - 1 or b < a - I but 0 < v < / ~ (a, b), the subset of ~ for which 0v (a, b, v) is positive. Since cot7 = - 0 (v) = - g we can in this instance use g rather than v as a p a r a m e t e r and so ~ = G (a, b, g) from (3.4). F o r a given y > n/2 (i.e. given g) we m a y use (3.4b) to find b = b (a) for a given ~. We c o m p u t e a) G~

g (0 2 q- b 2) r > J0 [0 4 + (1 -- a 2 + b 2) 0 2 + b2] 3/2 d9 0

b) Gb

g a b dy < 0 Jo [0 4 q-(l - - a 2-t- b 2) 0 2 q-b2] 3/2

(3.9)

and so b' (a) = - G , / G b is always positive. By R e m a r k (3.3 a) along the initial curve a + b = 1 for any given g ( c o t 7 --- - g) we have c~ increasing from 0 to

The capillary problem for an infinite trough

b

C(oqT)when /

175

b=a+l C(~,y)when / ',f=cc+~/2

~

(0,1

h

e

n T< o~+~:/2 Fig. 13. The curvesCg(~,T) when 7 > ~/2. Case I: 7 < e + n/2; Case 2:7 = ct + ~/2; :~ a ' Case 3 : 7 > ~ + ~ / 2

(1,o)

infinity. T h u s in a n e i g h b o r h o o d of a + b = 1 and any pair (e, 7) where re/2 < and 0 < e < oo there is s curve ~f (~, 7) e m a n a t i n g from this segment into the interior of Z o n e I. (If 7 = rc we just set g = + oo.) Consider the function G(a, b, g ) = ~ in the region b o u n d e d by a = 0, a+b=l and b = a - 1 . A l o n g a = 0 we have G = 0 , along a + b = l G increases from 0 to 0% while along b = a - 1, G (a, b, g) decreases from + oo when g = ~ to e = Y - ~/2 when a is infinite (Remark 3.3). We also have lim G (a, b, g) = 0 by (3.4 b). It follows that the level curves to c~ = G (a, b, g)

b~+oo

all have positive slope which e m a n a t e from the segment a + b = t, 0 < a < 1. F o r 7 = e + re/2 this level curve is the g r a p h b = b (a) of a function starting from a point (ao, bo) in a + b = 1. It will be defined for all a > ao and by R e m a r k 3.3 it must remain in Z o n e I. If 7 > c~ + W2 the level curve ~ (~, 7) emanates f r o m a point (a 1 , b 0 in the segment a + b = 1, and will be defined by a g r a p h b = b (a) for a~ < a < oo. This follows from (3.8) and (3.9) as the function G (a, b, g) will decrease from an angle greater than e to zero as b increases to oo. I n either of these cases the ratio b/a has a limit. F o r let (a,, b,) be a sequence of points on ~ (e, 7) with b,/a, converging to 2. The integral (3.4b) for G (a, b, g) has a limit

(3.10)

g 0~ f

Jo

dg + g2)

/(22 _ 1)g 2 + 22

L(2,g).

[This is equivalent to (1.2 a).] N o w 2 >_ 1 and L (2, g) is m o n o t o n i c in 2 with L (1, g) = t a n - 1 (g) = y _ re/2 and L ( ~ , g ) = 0. It follows that the limit is independent of the sequence chosen and so b/a--*2 with 2 given by (3.10). If y = ~ + ~/2 we must have 2 = 1 while for ~ < 7 - re/2 the limit 2 will be greater than one and the curve (8 (~, 7) will end up in Z o n e II. Finally by use of (3.9) it will follow that b' (a) will have the same limit.

176

H.C. Wente

F o r y < ct + re/2 the curve cg (e, y) will start from some point (al, bl) on the initial segment a + b = 1 continuing as a g r a p h and leaving Z o n e I at a point (a2, b2) where b2 = a2 - 1 (see R e m a r k 3.3). Once this curve is in Z o n e I I I it will remain there. We are to show that it continues as a g r a p h b -- b (a) defined for cq < a < oe and has a limiting slope. We recall R e m a r k 3.1 which asserts that w h e n (a,b) is in Z o n e I I I and v = / ~ then 9v(a, b,B(a,b))= 0 and = F(a,b,B) = 02(a,b)/2 > 7z/2 as 02(a,b ) > ~z. First consider the case re/2 < y < e + re/2 and c~ < re/2. W h e n cg (e, y) enters Z o n e I I I we have # (a, b, v) < 9~,x and 9~ (a, b, v) > 0 for 0 < v < v. This situation persists until a = + oo. If not, there would exist a first point (a*, b*) with (a,b(a)) converging to (a*,b*) where 0 < b* < a* - 1. If b* = a* - 1 then by continuity there would be two points on the line b = a - 1 with e = G ( a 2 , b 2 , g ) = G ( a * , b * , g ) . But by (3.8c) the function G is strictly m o n o t o n i c on B = a - 1, so we have b* < a* - 1. Since F (a, b,/3) > re/2 it follows that there is a unique v, 0 < v < / ~ with g = 9 (a*, b*, v). Thus at (a*, b*) one can continue to use g as p a r a m e t e r and by the implicit function t h e o r e m the curve cg(ct, y) continues as a g r a p h b = b(a) for al < a < c~ where g m a y be used as p a r a m e t e r rather than v. T h e previous discussion still applies. There is a n u m b e r 2, 0 < 2 < 1 so that b o t h b/a and b' (a) a p p r o a c h 2 where 2 is determined by (3.10) which is just (1.2 a). Finally we consider the case re/2 < y < e + ~/2 and re/2 < ~ < re. Here the curve cg(e,7) will be given by a g r a p h b = b(a) starting from (al,bO on a+b=l with v = v ( a ) chosen so that cotT=-e[a,b(a),v(a)] with 9,(a,b(a),v) > 0 for 0 < v < v(a) until a point (a3,b3) in Z o n e I I I is reached where O (a3, b3,/3) = g and 9~ (a3, b3,/3) = 0. At this point we must use (a, b, v) as p a r a m e t e r s rather than (a, b, g). So c~ = F (a, b, v) and cot 7 = - 9 (v) using (3.3). Consider the m a p ~ (a, b, v) = (e, 7) at the point (a3, b3,/~ (a3, ba)). We show that D 7~ has r a n k two at this point. In fact, O(e'7~) = act ( Fb 8 (b, v) ,%

F:) r 0 .

First 7, = (sin27) 9~ = 0. Also F, = a/(1 + 92) ~" 0 by (3.3) so we need only to show that 7b = (sin27) 9b # 0. Call 9b -- w. U p o n differentiating (2.9 b) with respect to b we find 29~ w~ - [493 + 2 ( 1 - a 2 +b2)9] w + 2b(1 + 9 2 ) . Since this expression vanishes at (a3, b3,/3) we conclude that w (a3, b3,/3) ~ 0. The curve ~ (e, 7) continues as a g r a p h b = b (a), v = v(a) t h r o u g h (a3, b3,/3) by the Implicit F u n c t i o n T h e o r e m . At (a3,b3,/3) we have 0~ = 0 and ~ = F ( a 3 , b 3 , B ) = 02(a,b)/2 where 02 (a, b) is the angle between O o and t?~. In Section V we show that this can h a p p e n at m o s t once (i.e. the curves ~2 (a, b) ~-- 2c~ and 9,,(a, b) = g have just one intersection point). We also have F (a, b, 2/3) > ~ so that as a a p p r o a c h e s infinity the condition ~z/2 < c~ < rc guarantees that v (a) lies between/3 and 2/3. In this range we write the function for e as (3.11)

ct = O2(a,b ) - G(a,b,g),

cot7 = - g .

By R e m a r k 3.2 concerning [92 (a, b) and (3.9) one concludes again that e, < 0 and eb > 0 so that the curve cg (e, 7) remains the g r a p h of an increasing function b = b (a).

The capillary problem for an infinite trough

177

As a tends to ~ we again find that b/a and b' (a) have a c o m m o n limit 2 determined by the f o r m u l a (3.12)

a = 7r - L(2,g)

,

cot2 = - g.

Here re~2 < a < rr and re/2 < 7 < ~z and the e q u a t i o n (3.12) again yields the expression (1.2 a) 2 sin ~ = sin (? - zc/2). Q.E.D. We n o w treat the case 7 < n/2, c~ + 7 = re~2. These configurations are f o r m e d by restricting (a, b) to lie in Z o n e I I I and letting/21 rather t h a n t2 o be the plane of symmetry. T h e essential information is contained in the following Lemma. L e m m a 3.3. Let (a, b) be in Zone I I I so that 9 (a, b, v) is periodic in v with half period 2 B where B is given by (2.17) with 9 (a, b, v) positive for 0 < v < 2 B with 9v(a,b,B) = O. Let x(u,v) and y(u,v) be the corresponding H = 1/2 (K = + 1) surfaces where we have suppressed their dependence on (a,b). We have x (u,0) [resp. y(u,0)] c / 2 o and x(u, 2/3) [resp. y (u, 2/~)] c O 1 where f2o and [21 are symmetry planes for the immersion. When 0 o was the symmetry plane our surface was the restriction of x (u, v) to the strip Iv I_< v with boundary planes Z 1 = f2 ( - v ) and $2 = f2 (v). The angle of contact was ?o where cOt?o = - 9 (v) = - v and the aperature of the wedge was 2 a o where cto is given by either (3.3) or (3.4). I f f2 i is to be the symmetry plane then we restrict our immersion to the strip v < v < 4 B - v. The planes determining the wedge are now s = O(v) and Z~ = f2 (4/~ - v). We have a new pair (~1,7i) where 2 ~i is the angle between Z'l and S,'2 and 7~ is the angle of contact mensured interiors to the wedge. We see easily that a) ct0 + c t l = O 2(a,b) (3.13) b) ? o + h = r e .

Here 692 (a, b) is the angle between the planes f2 o and f21 . It follows that (3.14)

a) c o t h = #(v) = g, 0 < ? < re/2 2B dv b) a ~ = a ~v l + g 2 - 0 2 ( a ' b ) - F ( a ' b ' v ) = F ( a ' b ' 2 ~ - v )

where F (a, b, v) is given in (3.3). I f we use g = g (v) as parameter then

(3.15)

a) ~ l = G ( a , b , g )

if / 3 < v < 2 / ~ , 9 ( v ) = g

b) ~ = O l ( a , b ) - G ( a , b , g ) , O < v < B

where G(a,b,B) is given by (3.14). Proof The formulas (3.13) follow from our definition. Clearly 7o + ?~ = 7z and also ~o + ~ = 0 2 since 0 2 measures the angle between f2 o and 12~. T h e formulas (3.14) and (3.15) follow directly from (3.3) and (3.4). Q.E.D. L e m m a 3.4. Let the angle be (~1, ?a) with symmetry plane f2~ as in Lemma 3.3. We then have ~a + 71 > 7r/2 and 71 < re~2. In fact ~l + 7~ = ~2 (v) + re/2

where 7"2(v) = 0 2 (a, b) - (9 (v) is the bending of x (0, v), v <_ v <_ 2 B. In particular if v = B then ~l + 71 = [02(a,b)/2] + ~/2 > re.

178

H.C. Wente

Proof Follows directly from R e m a r k 3.1 that 0 2 (B)

= 0 2

fact that 0 2 (a, b) > n.

(a, b)/2 and the Q.E.D.

We n o w prove the p r i m a r y bifurcation t h e o r e m for the case 7 < n/2. Theorem 3.2. Let (e, 7) be chosen so that e + 7 > n/2, 0 < 7 < ~/2, and 0 < e < n. There is a curve c~ (e, 7) in the (a, b) plane given by a graph b = b (a). The curve has a left hand limit at a point (a I , bl) on the line b = a - t and is defined for all a >_ a 1 with positive slope. There is a function v = v(a) where 0 < v < 2/~ (a, b) so that the corresponding H = 1/2 surface x (u, v) restricted to the strip v <_ v <_ 4 B - v has its boundary lying on two planes 221 = t2(v) and X 2 = (2 (4/] - v). The surface intersects the two planes at an interior angle 7 and

the two planes meet at an angle 2 e with g21 as a symmetry plane. For given (e, 7) the corresponding curve cg (e, 7) has a limiting slope b' (a) ~ 2 and b/a ~ 2 as well. The limit 2 satisfies the equation (1.2b). It is the distance from the axis of the original round cylinder to the axis of the wedge. As the curve (( (e, 7) approaches the point (at, bl) on the line b = a - 1 the family of immersions restricted to their respective strips converges uniformly to the initial round cylinder configuration. Proof. Consider first the case el + 71 > n/2, 71 < n/2 but ch + 71 -< n. The conclusion of L e m m a 3.3 allows us to conclude that if(a, b, v) is a triple giving us the pair ( e l , h ) then v > B(a,b) and we m a y use g = g(v) as a parameter so that by (3.15) cot71 = g

e 1 = G(a,b,g). Along the line b = a - 1 there is no ~21 as the p e r i o d / ] becomes infinite. N o n e the less the function G (a, b, g) has a s m o o t h extension across the line and by R e m a r k 3.3 its values decrease from + oo to el = t a n - 1 (g) = n/2 - 7~ as the parameter b increases from g2 to infinity. The range of e 1 on this ray is [n/2 - 71, oo]. Thus the equation e 1 = G(a, b, g), cot71 = g has a unique solution on the line b = a - 1. As in T h e o r e m 3.2 this e q u a t i o n determines a curve cg (el, 71) given by a g r a p h b = b (a) with positive slope starting at (a 1 , bl). The curve will continue to exist so long as g = g (a, b, v) and gv (a, b, v) < 0 for then cq + 71 > n by L e m m a 3.4. It follows that Cg(e1,71 ) extends as a g r a p h b = b(a) with positive slope on the interval al < a < oo. O n e can take a limit as a approaches oo in the integral defining G (a, b, g) = e 1 . There is a value 2, 0 < 2 < I with b/a ~ 2 and b' (a) --+ 2 where e l=L(2,g),

cot71=g,

0<71
O
and

el+71-
In this case the integral condition reduces to the e q u a t i o n (1.2b), 2 sine = sin (n/2 - 7). N o w suppose el + 71 > n/2, 71 < n/2 and el + 71 > n. In this case we must expect that there is a point (a*,b*) o n the curve ~ ( e l , 7 1 ) where g ( a * , b * , B * ) - - g and g v ( a * , b * , B * ) = 0. We have a m a p ~P(a,b,v)= ( e l , h ) using (3.14), and we need to show that

o(el,71)

. . / ell b

8(b,v)

- oet ~t771/8 b

ellOv 871/t?v] # O.

The capillary problem for an infinite trough

179

Since c o t ? l = g (a, b, v) we have (72), = ( - sin2 72)" gv -- 0 at (a*, b*,/~*) while

t?72/Ob = ( - sin 2 71)" gb" As in the p r o o f of T h e o r e m 3.1 and using (3.14) we have 0%/0v = a/[l + g 2] < 0. We need only to show that w = gb # 0 when v = B. This condition was verified in the p r o o f of T h e o r e m 3.1. We still m a y use a as a p a r a m e t e r and there are functions b = b (a), v = v(a) solving the equations (3.14). As in the p r o o f of the T h e o r e m 3.1 it will be seen in Section V that this can h a p p e n just once. Eventually v(a) will remain in the interval 0 < v < / ~ and we can then use expression (3.15b) for a2, cq = O2(a,b ) - G(a,b,g). It will follow that cg (cq, 72) will continue as a curve b = b (a) with positive slope with b/a and b' (a) having limit 2 where -

~1 = n - L(2, g) giving us the relationship (1.2b) again. It remains to consider the b e h a v i o r of the functions as cg (a2,71) a p p r o a c h es (a2, b 0 on the line b = a - 1. F o r (a, b) near b = a - 1 the functions x (u, v) and co (u, v) are restricted to the strip v < v _< 4/~ - v where g(v) = g, cot71 = g and also B < v < 2/~. Let v 2 = 2 B - v so that g(a,b, v2) = g(a,b,v) = g and g~ (u, v, a, b) > 0. The m a x i m u m of the function co (u, v, a, b) on the strip occurs at (u, v) = (0, v). M a k i n g use of the symmetries we have by (2.8)

a - gv(a,b, v2) cosh c%t = 1 + g 2 (a, b, v2) " N o w g (a, b (a), v) converges uniformly to g ( a l , bl, v) --- g2 (v) on any interval 0 < v < V, as (a, b ) ~ (al,bl). There is a unique V2 with 91 (V2)= g and so v2 ~ V2 as well. But 01 = b2 - g2 and so c~

~

al -- (g2)~ (V2) _ al -- [b2 - g2 (V2)] 1. 1 +g2(V:) 1 +g2(V2) =

T h e function co (u, v) converges to zero uniformly on the respective strips. It follows that the limit configuration is a section of the r o u n d cylinder. Q.E.D.

IV. The limit configuration In Section I I I we showed that for each admissible (~, 7) there is a curve cg (ct, 7) in the p a r a m e t e r plane given by a g r a p h b = b (a) as well as a cut-off value v = v ( a ) such that the c o r r e s p o n d i n g H = 1/2 i m m e r s i o n x(a,b(a),u,v) restricted to the a p p r o p r i a t e strip gives a solution to the free b o u n d a r y problem. F u r t h e r m o r e b o t h b/a and b' (a) have a limiting value 2 determined by (1.2). F r o m Section III, T h e o r e m s 3.1 and 3.2 we have the following information. 1. 7 = z/2, 0 < ~ < n ( D e l a u n a y unduloids). Here b = b (a) = 0 and 2 = 0. 2. 7 > ~z/2, ~ + 7 > n/2. In this case the plane f20 containing x (u, 0) is the s y m m e t r y plane a) 7 < 7 + n/2. H e r e 0 < 2 < 1 and the curve cg (~, 7) ends up in Z o n e III. b) 7 = ~ + n/2. Here 2 = 1 and ~ (e, 7) stays in Z o n e I. c) 7 > e + n/2. H e r e 2 > 1 and cg (ct, 7) ends up in Z o n e II.

180

H.C. Wente

3. 2 < n / 2 , ~ + 7 > n / 2 . The symmetry plane is now 01 where x ( u , 2 B ) c f21 . Here 7 < ct + n/2 so that 0 < 2 < 1 and ~(e,7) is in Zone III. As cg (e, 7) goes to oo the fundamental rectangles determining the period of the function co(u,v) shrink while the solution itself blows up (e.g. cosh co (0, 0) = a + b becomes infinite). We side step this problem by rescaling the immersions x (u, v) by s (u', v') = x (5 u', e v') in such a way that with the new representation the first fundamental form is just du '2 + dv '2 at the blow-up point. If f2 o is our symmetry plane this point will be (0, 0) while if f21 is the symmetry plane then the appropriate point will be (0, 2/~). With this rescaling we can take the limit as we go to infinity and a form of blow-up occurs. The limit is found by use of the formulas (2.8) and (2.9) for co (u, v). L e m m a 4 . 1 . Let x(u,v) and y ( u , v ) = x ( u , v ) + ~ ( u , v ) be the H = l / 2 (K = + 1) surfaces whose fundamental forms are given by (2.1) and (2.5) where co (u, v) solves the Gauss equation (2.2). Consider a conformaI reparameterization of these surfaces given by a homothety w = F (w') = e w' for some e > 0 (where we set w = u + i v). Let ~ (u', v') = x (5 u', e v') be the new representation. With these parameters the H = 1/2 surface has fundamental forms a) (4.1)

ds '2 =

e2a(du '2 4- dv '2)

b) H ' = ( e 2 ~ 2 e 2 ) d u ' 2 + ( e 2 ~ 2 e 2 ) d v ' Z

where e 2~ = ~z e2~, or a (u', v') = In ~ + co (e u', e v'). The function a (u', v') satisfies the Gauss equation (4.2)

1 e2 A'a + ~-

e2

a _

-~- e- 2~ = 0.

For the corresponding K = + 1 surface we have a) d s ' 2 = ( - e ~ (4.3) b) H' -- (e- 2" - 4 2 e- 2~) (du'2 + dv'2) . The parameterized surface ~ (u', v') will have ds 2 = du '2 + dv '2 at the point (0, O) if a ( 0 , 0 ) = 0 which means that ee "~ = 1 or coo = in(I/e) where coo = co(0,0). Proof Just a rewriting of the fundamental forms (2.1) and (2.5) and the Gauss equation (2.2) once we observe that e 2~~= e 2 e 2'~ Q.E.D. Theorem 4.1. For a given (~,7) consider the curve cg(~,7) given by a graph b = b (a) where limit b/a = limb' (a) = 2 a) ) ~ s i n ~ = s i n ( 7 - n / 2 )

if 7 > n / 2 , O < ~ < n

b) 2 s i n ~ = s i n ( ~ / 2 - 7 )

/f 7 < n / 2 , O < ~ < n

as determined in Section 111. Let co (a, b (a), u, v) be the family of solutions to the Gauss equation (2.2) which determine the H = 1/2 immersion x(u,v) and the K = + 1 immersion y(u, v). Suppose we rescale by a homothety factor e as described in Lemma 4.1 where e is chosen as follows.

T h e capillary p r o b l e m for a n infinite t r o u g h

181

a) If7 > rc/2 where f2o is the symmetry plane we rescale about (0, O) {w' = e w} so that the rescaled function a (u', v') vanishes at (0, O) G (u', v') = co [e u', e v'] ( , ( 0 , 0 ) with e chosen so that l n ( l / e ) = coo -- co(0,0) and coshco o = a + b. b) I f 7 < ~/2 with (2~ the symmetry plane so that x ( u , 2 B ) c f21 we must rescale about (0, 2/~) [u = e u', v - 2 / ~ = e v'] where coshco 1 = a - b. Along the curve cg (~, 7) the rescaled functions a (a, b (a), u', v') converge uniformly in any bounded domain to a limit function a~ (u', v'). This function is a solution to the Liouville D.E. -

1

(4.4)

Aa + ~-e

2o

= 0

and the limit functions are: 1) ~ = ~/2 Delaunay case ,~ = 0 % (u', v') = In [sech (u'/2)] I.

(4.5)

2) 7 > n/2, Do the symmetry plane (4.6)

1-2

e "~ =

cosh

(

21 -

cos

(l

51

b) 2 = 1, 7 = c~ + re/2, Zone L (4.7)

e"l

1 1 + (r/4) 2

r2

= u'2

+

/)t2 "

c) 2 > 1, ~ > ct + re/2, Zone II, (4.8)

2-1

e ''~ = 2c~176

"

3) 7 < re/2, 7 < 0~ + re/2, Zone I I I with symmetry plane f2 t , (4.9)

1+2

e"* =

cosh

?/+

cos

3/

As the curve cg (~, 7) goes to infinity the rescaling parameter ~ goes to zero and the fundamental forms for the immersions become a) ds 2 = e 2~ (du '2 + dv '2) (4.10)

and

17 = (e2~

(du '2 + dv '2)

b) ds 2 = (e2~/4) (du '2 + dv '2) H = (e2"/4) (du '2 + dv '2)

(H = 1/2 case) k 1 = k 2 = 1/2

(K = + 1 case) kl = k2 = 1.

where cr(u', v') is a solution to (4.4). The surfaces represented by these limit functions are conformal immersions onto the sphere of radius two (H = 1/2) or the unit sphere (K = + 1). The immersions determined by these fundamental forms have a simple description. We consider the mapping for the unit sphere (K = + 1). Let the unit sphere be centered at the origin o f ~ 3 and suppose y(u, v) is normlized so that y (u, O)c s o is the equatorial plane z = 0 with y(0, 0 ) = ( 0 , - 1, 0) and

182

H.C. Wente

el (0, O) = (1, O, 0). It follows that y (0, v) c 1I 0 is the vertical plane x = 0 and we choose orientation so that e2 (0,0) = (0,0, - 1 ) and so 4(0,0) = (0, 1,0). In the Delaunay case one see that the image of the strip Iv t < 2re is a covering of the sphere with the points PI : (1,0,0) and Pz: (1,0,0) excluded. In

fact

lim y (u, v) = I"1

and

U ---~ + ~

lirn y (u, v) = P2 U~--~3

so that y (u, O) maps the real line onto an equatorial semi-circle. When u = O, ds = (1/2)dv so that y ( O , v ) : - 2 z _< v < 2~ traces out an entire great circle with y (0, ___2 7z) = (0, i, 0). Finally we have y (u, ++_2 re) lies in f2~ = 0 o and traces out the rest of the equatorial great circle. The map is just the conformal map v = eW/2followed by a stereographic projection. This is the Mercator projection of a cylinder onto the sphere. For the case (2 a) the omitted points are no longer at opposite poles but are located at Pz : (~/1 - 22, 2, 0) and P2: ( - N~ 1 -- 22, 2, O) with t i m y (u, O) = PI

and

lira y (u, 0) = P2.

Otherwise the description is similar to the Delaunay case. The actual map now includes a linear fractional transformation as well as the exponential map. Notice that the fundamental strip is now the set {(u, v) [I v l _< 2 ~ x/(1 - 2)/(1 + 2)}. As 2 ~ 1 we see that P~ approaches P2 with 1~ = P2 = (0, 1, 0). The map y (u, v) in case (2 b) is the classical stereographic projection with the point at o~ being mapped onto PI = P2. The mapping is no longer periodic. In case (2c) 2 > 1 we see that the map is similar to the case (2a) except that the roles of u and v are switched. In particular, the omitted points now lie on the vertical plane 1-10 with el: ( 0 , 1 / 2 , - 2 ~ / 2 ) and P2 = (0, 1 / 2 , , ~ 5 - ~ I - / 2 ) with lim y (u, v) = P1 and lira y (u, v) = 1"2. v --+ + oO

v --* - - oo

Finally for case (3) the map is similar to case (2a) except that the omitted points are now PI: (x/1 - 22, - 2, 0) and P2: ( - x/1 - 22, - 2, 0). Proof. We shall derive f o r m u l a (4.6) of case (2 a). T h e others follow in similar fashion. T h e derivation involves a direct calculation using the formulas (2.8) for coshco in terms of f ( u ) and 9(v) and (2.9). Start with (2.8). We rescale setting u = e u', v = e v' and 09 [e u', e v'] = coo + a (u', v') where coo = co (0, 0). We have coshcoo = a + b with COo > 0 and e -'~~ = e. This allows us to solve for getting e = [F + Fx/F5 - - 1]- 1 where F = a + b. N o w b/a -~ 2 where 0 _< 2 _< 1. Therefore (4.11) ,lim ~ a . e = 1/2(1 + 2). In particular e ~ 0 . N o w divide (2.8) by a, set u = au', v = ev' and take the limit as a becomes infinite. O b s e r v i n g that coshco = cosh~o o s i n h o - + sinh co cosh o one finds cosh co lim - (1 + 2) e ~ (''' v') a~o~

a

while the limit of the right side of (2.8) becomes lim R Ha S = 2 ( 1 + 2 ) ! i m

a ~

where we have m a d e use of (4.11).

e f ' ( e u ' ) + ~ O(ev') l+fZ(eu')+92(ev')'

The capillary problem for an infinite trough

183

We next show (4.12)

!imof(a,b(a),eu' ) = F ( u ' ) = x / l _ 2 z sinh

i +2

"

To see this we use (2.9 a) so that ,

df

/3Ur ~ ~

o ~ / f 4 + (1 + a 2 - b 2) f 2 + a 2"

Divide this expression by e and let e ~ 0 keeping in mind (4.11). We find f

u'=2(1 +2)!x /

df (1 - 2) 2

fz

+ 1

"

This is equivalent to (4.12). Similarly using (2.9b) one finds (4.13)

~-,olimg(a,b(a),ev') = G(v')

x/1 _ 2~ s i n

+2

"

Combining (4.12) (4.13) with our calculations involving (2.8) gives us (4.14)

e~

= 2 [_F] _F' (u') + G(v') ,_] + VZ(u ') + G2(v)J"

This equation along with (4.12) and (4.13) gives us (4.6). The formulas for the omitted points P1 and P2 under the immersion y (u, v) determined by (4.6) can be computed by direct calculation. It will also follow from the next theorem. Q.E.D. It is clear from the proof of Theorem 4.1 that in limit f (a, b(a), ~ u') = F (u') and limit 9 (a, b (a), e v') = G (v') the convergence is uniform on any bounded interval in u or v. It follows from (2.8) that o-(a, b (a), u', v') converges uniformly to ~rz(u', v') on any bounded domain of •. Since f (u) and 9 (v) satisfy explicit differential equations the same will be true for the derivatives of f and 9. Upon integration of the fundamental forms we conclude that x (a, b(a), e u', e v') and y (a, b(a), e u', e v') converges uniformly along with their partial directives on any bounded domain in ]R 2. Suppose cg (~, 7) determines a curve b = b (a) where b/a ~ 2 and 0 < 2 < 1. For sufficiently large a we are in Zone III and in this case the nodal line set for ~o(u, v) are the vertical lines u = (2m + 1)A. On any vertical strip we have seen in Theorem 4.1 how upon rescaling the map restricted to this vertical strip converges to a conformal representation of the unit sphere (K = + 1 case) omitting just two points. We now claim that successive spheres fit together in the manner described in the introduction. Theorem 4.2. Let (c~,7) be given so that we are in case (2a) with ~ > re~2, + 7 > re/2, ~ < ~ + re/2. Let cg (~, 7) be the corresponding curve in the parameter plane given by a graph b = b (a) where lim b/a = lira b' (a) = 2, 0 < 2 < 1 satisfying identity (1.2a). Let y(a, b(a), u, v) be the corresponding K = + 1 immersion. The following is true a) lim A/e = + ~ a~ot)

where

e -~~ = ~ .

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A' =- A/e determines the location of the vertical nodal lines upon rescaling. 7he nodal lines diverge to oo upon reparametrization b)

lim (2/~)/e = 2 n x / ( l + )0/(1 - 2) = lim 2/~' . a --* o o

a-*oo

This is consistent with the period of the function az (u', v') given by (4.6). c) l i m z l = 2 ~ / 1 - 2 2

.

This is in agreement with the location of Pa (1 - 2 2 , 2, 0) as determined in Theorem 4.1. d)

lim O~ (a, b (a)) = n . a --* oD

The angle between the symmetry planes (2o and t21 approaches n. e) Let Lo(a) be the length of the curve y(a,b(a),u,O)'O <_u <_A(a,b). We have n aim Lo (a) = y + sin-1 and if Ll (a) is the length of y(a,b(a);u,2B), 0 <_u < A then 7Z

lira L1 (a) = ~- - sin- 1 (2). a --+ o o

Thus the singular point of the mapping y(a,b(a),A,v) converges to 1"1

-

o).

Proof (a) F o r b < a + 1 we have A given by (2.12). Divide the expression by e, make use of (4.11) and take the limit. We find

df

lira A' = lim A/e -- 2 (1 + 2) . . . . . . o ~/1 + (1 - 2) f 2 -- + 0(3. (b) F o r b < a - 1 use formula (2.17) for/~. Again divide by e, use (4.11) and take the limit.

,/2 l i m Z B = l i ~ 2 B / e = 2(l + 2) ~ .

.

.

.

.

dO

~/2 1 + 2 sin0

fl + 2 2n ~ - . 2

(c) We use formula (2.29) for zl. We let a go to infinity with b/a ~ 2 and find 1 +(1- 222)f 2 lim ~l(a,b(a))= 2 i (1 + f 2 ) ~ ~ 2 ) f = d f = 2V/1 - ,U. (d) The angle between ~o and f21 is 0 2 (a, b). It follows from R e m a r k 3.2 and formulas (3.7) that limit Oz(a, b(a))= n. (e) We use (2.33) to c o m p u t e L1. M a k e the change of variable W = b + a { and we get , (b + a 4) d4 1 d4 + sin-l(2). L ~ b~f .x/(b + a 4) 2 _ l x / 1 _ 4 2 ~ -aJ"x / ~ -- 4~ _ 2 0

Since L o + L 1 = n we have limit L 1 = ~- -- sin- 1 (2).

Q.E.D.

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185

R e m a r k 4.1. From Theorem 4.1, 4.2 we see that if b/a ~ 2 where 0 < 2 < 1 then the image of any vertical strip determined by the nodal lines converges to a multiple cover of a unit sphere with two points P1 ( 2 ) = (x/1 - 2 2 , 2, 0) and P2 (2)= ( - x / 1 - 22,2, O) omitted. For any ~, the curves y(u, ~ are smooth curves connecting one nodal region to another. The point of connection approaches the points P1 (2), P2 (2). It follows that the limit configuration for the K = + 1 surfaces is as described in the introduction. The centers of the spheres are all coplanar lying in the plane t2 o = ~21. The axis lo is that line connecting P1 (2) and P2 (2). The image of a vertical region where ca (u, v) is negative is still a K = + 1 sphere but with the orientation reversed. For this reason in the H = I/2 map the image of a positive vertical strip for co (u, v) becomes a sphere of radius two concentric with the corresponding K = + 1 sphere while the image of a negative vertical strip collapses to a point. The discussion of the other cases is similar. In case (2b) where ~ = o~+ ~/2 and 2 = 1 we know that the curve ~(~,7) stays in Zone I (Theorem 3.1). As in Theorem 4.2, upon reparametrization the nodal lines diverge to infinity, and our limit map is the stereographic projection with P1 = P2 = (0,1, 0). The sequence of K = + 1 spheres now all collapse to two such spheres with the axis 1o being a common tangent line lying in s o . One of the corresponding H = 1/2 surfaces collapses while the other goes into a sphere of radius 2. I f 2 > 1 (~ > ~ + n/2) we are in Zone I I and the nodal region for co (u, v) are now horizontal strips. The result is one dual to Theorem 4.2. T h e o r e m 4.3. Suppose ~ > c~+ re~2 where we are in case (2 c) so that 2 > 1 and

we are in Zone II. a) lim B/e = lim B' = + ~ . a --+ oo

a --+ oo

The horizontal nodal lines diverge to infinity. b) lim 2.4' = lim 2.4/e = 2 zcx/(2 + 1)/(2 - 1). c) l i m L o = ~ where L o is the length of y(u, O): 0 < u < 2.~. d) limL~ = ~/2 + sin- 1(1/2) where L I is the length of y (O, v): 0 < v <_B. I f L2 length of y(2.4,v): 0 < v <_ B then limL~ + L 2 = ~z. e) l i m vl = 0.

Proof The proofs of (a), (b) are identical to the corresponding statements of T h e o r e m 4.2. One computes Lo by using (2.11 a) obtaining Lo

~+b

WdW

2+b

1,/a

- (W-

M a k e the change of variable W = b + a ~ and let a ~ Go, b/a ~ 2 and we find lim Lo = n. To c o m p u t e L~ use (2.11 b) to get ,+b

dW

U p o n taking a limit we find L~ ~ ~ + sin- ~ (1/2). The c o m p u t a t i o n of L 2 is similar. To show that z~ o 0 it suffices to show that L 3 ~ 0 where L 3 = length of y (u, B): 0 _< u ~ 2 ~, where v = B is a nodal line for co (u, v). We find A

L 3 = ~ coshco du = A ~ 0 by (a). 0

Q.E.D.

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H.C. Wente

H = 1/2

The non-symmetric secondary bifurcation (section V)

Fig. 14. Zone III non-symmetric /

configuration. Cross-Section in Plane Ho; lo = axis of immersion; l = axis of wedge

The complete configuration when 2 > 1, 7 > e + n/2 may be deduced from Theorems 4.1 and 4.3. Each horizontal nodal region determined by the nodal lines v = (2 n + 1) B has an image under the map y(u,v) (K = + i) which converges to a multiple cover of the unit sphere. The spheres will be successively tangent to one another touching at points P1(2)=(O, 1 / 2 , x / ~ - l / 2 ) and P2(2)= (0,1/2,-x/~-l/2) lying in the vertical symmetry plane F/o, forming a bracelet-like configuration. F o r the H = 1/2 mapping the positive nodal regions are mapped into spheres of radius two while the negative nodal regions collapse to a point.

V. The secondary bifurcations All of the equilibrium configurations described so far have the feature that they are symmetric about the bisecting plane of the wedge. However, for every > n/2 and 0 < 7 < n there is a secondary bifurcation away from the primary family. This secondary family breaks the symmetry about the bisecting plane of the wedge yet remains in the family of surfaces of Joachimsthal type and we can follow its course through the entire beading up process. It is also easy to describe [see Fig. 14]. Lemma 5.1. Let (a, b) be a point in Zone I I I and g (a, b, v) the solution to (2.9b) so that g is periodic in v with period 4/] where g (a, b,/]) = g~ the smaller positive root of V2 (g) = g 4 + (1 - a 2 + b 2) g 2 + b z = O. Let x (u, v) be the corresponding H = 1/2 immersion of J oachimsthal type and consider the equilibrium configuration whose free interface is given by x(u, v) restricted to the strip {(u, v)[v < v < v + 2/]} so that the boundary curvature lines lie in the planes X1 = (2(v) and Z 2 = (2(v + 2B). The following are true. a) The angle 2~ between X 1 and S 2 is independent of v and equals 0 2 (a, b) where 0 2 (a, b) is the angle between the symmetry planes f2 o and f21. b) The angle of contact 7 measured interior to the fluid is the same on both walls, with cot 7 = g (v). Proof F o r the initial wall Nt = ~2(v) we have cotT~ = g(v) while for the second wall S 2 = O(v + 21]) we have cot72 = - g(v + 2/]) = g(v) and so 71 =72"

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187

F r o m L e m m a 3.2 F o r m u l a (3.3) we have that 2~, the angle between S 1 and Z 2 is given by v+2~ dv (5.1) 2e=a ~, 1 + 9 2 ' 9=g(a'b'v)" But 92 (a, b, v) has period 2/~ in v and the f o r m u l a for 2 a is independent of v and so 2ct = 02(a,b ). Q.E.D. F o r any point (a,b) in Z o n e I I I there is a o n e - p a r a m e t e r family of n o n - s y m m e t r i c equilibrium configurations where the angle of the wedge 2~ = 02(a,b). If 9m = 9(a,b,.B) is the m a x i m u m value for O and if we set cot 7* = - g~ so that 7* > n/2 and cot 7* = 9,, with 7* + 7* = n then for any 7, 7* < 7 < Y* there is exactly one value v E [ - / 3 , B] with cot 7 = 9 (v). Therefore the surface x (u, v) restricted to the strip v _< v < v + 2 B spans a wedge with b o u n d a r y plane Z 1 = Q (% Z 2 = O (v + 2/3). The angle between these planes is 2~ = O 2 (a, b) and the c o m m o n contact angle is 7 where cot 7 = 9 (v). Since 02 (a,b) is always greater than n by R e m a r k 2.3 [or (2.25a)], we have that ~ > n/2. L e m m a 5.2. Let ~ > n/2 be given and consider the curve F (~) lying in Zone III consisting of those points (a, b) for which 0 2 (a, b) = 2 ~, F (o0 is a smooth curve

of slope greater than one which leaves the a-axis with infinite slope where O 2 ( a , 0 ) = rc [ a / ~ ] . It is given by a graph b = b(a) with b' (a) > 1 and is asymptotic to the line b = a - 1. Furthermore the function gm (a, b), the maximum of g (a, b, v), is an increasing function on F (~) with gm= 0 along the a-axis and limit gm= + 0o as a becomes infinite. Proof In Z o n e I I I [9 2 (a, b) is given by formulas (2.25 a, b) and its properties are discussed in R e m a r k 3.2. We have 02 (a, 0) as in the statement of this L e m m a and so it has a range of 0z, + ~). T h e partial derivatives of O2 (a, b) are calculated in R e m a r k 3.2. U p o n inspection we see that 6~02/t~a< 0, O02/t3b > 0, and [~02/~bl < I002/~a [. It follows that the curve F (a) is given by a g r a p h b = b(a) with b'(a) > I with infinite slope on the a-axis. We also claim that the curve F (ct) is a s y m p t o t i c to b = a - 1. Since F (~) has slope greater than one it follows that if F (a) were not a s y m p t o t i c to b = a - 1 then there would be an e > 0 so that 7 (~) lies below b = a - 1 - e. But then limit O 2 (a, b) -- rc by R e m a r k 3.2, a contradiction. We n o w consider the function g,, (a, b). F o r a given value of gin, the set of prints (a, b) for which g,, (a, b) = 9,, is given by solving (5.2)

V: (g,n) = g4 d- (1 -- a 2 -I- b 2) g2 + b 2 = O.

This m a y be rewritten in the form (5.3)

a2 1 + 92

b2 q2 - 1

9

This is a h y p e r b o l a with a s y m p t o t e b = (g,,,/x/1 + g 2 ) a , and vertex at (x/l+g2m,0). The envelope of the family of h y p e r b o l a s (5.3) is the line b = a - 1. E a c h point in Z o n e I I I meets just two of these hyperbolas. The actual gm is the smaller of the two positive roots of (5.2). This corresponds to

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H.C. Wente

that branch whose vertex is closer to the origin. Thus the smaller root of (5.2) will correspond to those arcs of hyperbolas (5.3) which lie to the right of their tangency point with b = a - 1. The slopes of these arcs are always positive but less than one. It follows that the map ( a , b ) ~ ( O 2 , g m ) is a local diffeomorphism and that along any curve F (a) the function g,, is strictly increasing from zero to infinity. Q.E.D. Theorem 5.1. Let ~ > re~2 and 0 < 7 < ~z be given. Let F (~) be that curve in Zone I I I for which the angle between fl o and t? 1 , 0 2 (a, b) = 2c~, as described in Lemma (5.2) which emanates from the point (a, O) where 2 ~ = ( a / ~ ) ~. Along F (ct) the function 9m (a, b) is increasing with range [0, + oo). Let (a l, b l) be that point on F (~) for which a) cot7

=-9,,

b) cot7

=

/f7>rc/2

gm i f 7 < ~ / 2

c) (al,bl) = (a,O) if 7 = n / 2 . The secondary bifurcation is given by that portion o f f (~) starting at (al , b ~) and continuing to oo. For each (a, b) e F (a) with a > a 1 there is a unique v e [-/3,/]] where B = B(a,b) [see (2.17)] with cot 7 = 9(v). The immersed surface x(u,v) restricted to the strip {(u, v) l v < v _< v + 2/~} gives us an admissible solution with S 1 = t2(v) and S 2 = fl(v + 2/~). The angle between Sz and Z 2 is 2~ and the contact angle of x(u, v) with S I and z~ 2 is 7" Proof The fact that the branch described in the statement of the theorem gives solutions to the free boundary problem follows from Lemmas 5.1 and 5.2. At the point (al,bz) where the bifurcation occurs we observe that if ? > n/2 then the branch starts by setting cot? = - gm= g ( - / 3 ) so that the initial strip is {(u, v) ] Iv] _ ~/2 we noted in Theorem 3.1 that the primary bifurcation branch had at least one point where cot 7 = - gin. This is that point. F r o m Lemma 4.2 we observe that this can happen just once. Similarly of ? < re/2 we choose our initial strip by setting cot ? =gm = g (/3) at (al, bl) and ~ = {(u, v) l/~ < v < 3/3}. Here the initial configuration has t21 as the symmetry plane which is broken by the secondary bifurcation. Finally for ? = re/2 we use the entire curve F (~) with (a, 0) in the bifurcation point. For this family the boundary planes are t? o and t2t. Q.E.D. Remark 5.1. We observe that if ~ > 7c/2, 7 = re~2 and ~ - re~2 is small then the secondary bifurcation from Delaunay unduloids will occur when the configuration is almost a seo~ence of spheres. We also note that ~ = rc when a / x f a 2 ---1 = 2 or a = 2,,/3/3. Remark 5.2. The curves F (~) defining the secondary bifurcations are asymptotic to b = a - 1, so that b/a and b' (a) converge to one. One may apply the analysis of Section I V to visualize the limit of this family.

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189

1/2 The non-symmetric primary bifurcation (section Vl)

Fig. 15. Zones I, II non-symmetric configuration Cross-Section in Plane/7 0 ; t = axis of immersion; 1 = axis of wedge

VI. Non-symmetric primary bifurcations when 9 > x/2 The families of admissible H = 1/2 configurations bifurcating from the round cylinder solutions as presented in Section II all possess symmetry with respect to the bisecting plane of the wedge. If 7 > hi2 this plane was ~2o while for 7 < ~/2 this plane was (21. When ~ < n/2 these seem to be the only primary bifurcations lying in the Joachimsthal family. When ~ = n/2 (i.e. the wedge is a plane) this uniqueness breaks down and for ~ > ~c/2there is another primary bifurcating family of Joachimsthal type which is not symmetric about the bisecting plane of the wedge. The construction of this family is based on the following lemma.

Lemma 6.1. Let (a, b) with b > a - I be a point in Zones I or II. In this case the function co (a, b, u, v) solving the Gauss equation (2.2) possesses horizontal nodal lines at v = ( 2 n + 1)B where B is given by (2.15). Let 7 be any angle 0 < 7 < n. Select v ~ ( - - B , B ) with g(a,b,v) = g = cot7 where g(a,b,v) is the solution to (2.9 b) and consider the H = 1/2 surface x (a, b, u, v) restricted to the strip ~ = { ( u , v ) [ v < v < 2 B - v } . This surface is bounded by the planes ~1 = (2 (v) and ~'~2 = ~'~(2B --v). The interior contact angle of the surface with this wedge is 7 on both planes and the angle 2~ between s and Z 2 is given by a) 2~ = O2(a,b ) - 2G(a,b,g)

(6.1)

see (3.4b) dg b) 2c~=g+2a~g (l+g2)x/g4+(l_a2+b2)g2+

b2

In particular 2~ > ~. Note that if 7 = ~/2 then Z 1 = (20 and Z 2 = (21 (see Fig. 15). Proof Since x (u, v) enters the wedge at s with increasing v at v = v we have cot71 = g(v) = g, while the surface leaves the wedge at 22 when v = 2B -- v so that cot72 = - g ( 2 B - v ) = - - g ( - v ) = g ( v ) = g as g(a,b,v) has a periodic extension of period 2B. This gives us 71 = 72 = 7. To prove (6.1) we observe first that for 7 = g / 2 we have v = 0 and 2c~ = 02(a,b), the angle between the planes (20 and (21. Now suppose 7 < ~ / 2 with c o t y = g ( v ) > 0 so that 0 < v < B . In this case 2 ~ = 0 2 (a, b) - 2 angle [~o, (2 (v)] = 02 (a, b) - 2 G (a, b, g) giving us (6.1 a). We get

190

H.C.

Wente

(6.1 b) by using formula (3.4b) for the angle between f2o and ~2(v). If 7 > n/2 with cot 7 = g < 0 then 2~ = O 2 (a, b) + 2 9 angle [Oo, g2(v)] which again leads to (6.1). Q.E.D. We have a mapping ~: (a, b, v) ~ (a, 7) where b > a - 1 and - B < v < B such that the preimage of each point is a non-symmetric equilibrium configuration for the given parameters (a, 7). Since O(a, b, v) is an increasing function on ( - B , B) we m a y replace v by g = g (a, b, v) as a parameter in the construction, setting 7/: (a, b, g ) ~ (0~,7) given by cot y = g and 9 given by (6.1). The non-symmetric families constructed in Section V produced secondary bifurcations. We now show that the families described in this section bifurcate from the round cylinder solutions. L e m m a 6.2. Let cg(~,7) be the set of points in the (a,b) plane so that 7J (a, b, g) = (~, 7) where cot 7 = g = g (v). This curve is a smooth curve with positive slope given by a graph b = b (a). Proof. Using (6.1 b) for the angle 2e, one finds ( b 2 _[_ g2)

dg [V2 (g)]3/2 > 0

~a

(6.2)

~b=--S

a b dg

[v2(g)] 3/2

<0,

(see (3.9)) V2(g)

= g4

+(1

_

a z + b 2)

g2

+

b2

Q.E.D.

We now state the main result of this section. Theorem 6.1. For every choice o f (~, 7) with ~ > n/2 and 0 < 7 < n there is a curve cg (~, 7) in the (a, b) plane with b > a - 1 defined by a graph b = b (a) for a >_ al with positive slope and a function v = v (a) where v ~ ( - B, B) with B (a, b) is given by (2.15) so that the corresponding H = 1/2 surface x (a, b, u, v) restricted to the strip v <_ v <_ 2 B - v and with boundary curves lying on the planes X1 = (2 (v) and X 2 = (2 (2 B - v) is an equilibrium configuration with the surface meeting the walls S 1 and ~2 at constant angle 7 and the angle between XI and ~2 being 2~. As a-~ a 1 the point (a, b(a)) approaches a boundary point (al,b O. Either a 1 + b~ = 1 or b~ = a~ - 1. In either case the limiting configuration converges to the round cylinder solution. For n/2 < ~ < (3 n/2) - 7 the curve ~ (~, 7) will cross the line b = a + 1 and there is a value 2 > 1 so that both b' (a) and b/a converge to 2 where )o is determined by taking a limit in formula (6.1 b) (6.3)

2~=~+Sg

dg (l+92) Xf(~L1)92+2

2'c~

Otherwise (g (o~,y) is trapped in Zone I and b/a and b' (a) have a limit 2 = 1. I f > (3 n/2) - 7 then cg (~, 7) is asymptotic to the line b = a - 1. P r o o f F r o m L e m m a 6.2 we know that if cg (~, 7) is not empty then it is given by a curve of positive slope b = b(a) and by (6.1 b) we know that ~ > n/2. Let 0 < 7 < n. We will show that for any ~ > n/2 there is a single curve cg (~, 7) determined by (6.1 b) lying in the region ~ = {(a, b) [ a > 0, a + b > 1, b > a - 1} and that this curve will terminate on its left at a point (a,, b,) where either al + b~ = 1 o r b I = a 1 - 1.

The capillary problem for an infinite trough

191

To see this observe that the function for c~ given by (3.1 b) extends to the b o u n d a r y of N. We find a) ~ = z ~ / 2

of

b) e - ~ (6.4)

a=O t a n - I (g/v/-b) _ t a n - l ( g

(+oo

if g < 0 ( 7 > ~ / 2 )

]

= ) + oo if g2 < b (g > 0) (7 < ~/2)

C)

~

,a + b= l

2t when b = a - 1.

4

I~tanh-l(x/b/g)+tan-ig

if 0 < b < g

',,/b

We need to check that the range of values for ~ given by (6.4) is (re/2, + oo) and that each value is taken on once. This is true. F o r 7 > ~z/2 the function (6.4) for a will increase from re/2 to + oo as we m o v e on a + b = 1 from (0, 1) to (1, 0). It remains at + oo on b = a - 1. F o r 7 < n/2 the function for ~ will increase from 0 to e * = r~/2 + ( t a n 7 - 7) = t a n - 1 g + (l/g) as we move along a + b = 1 from (0, 1) to (1, 0). Then by (c) c~will continue to increase along b = a - 1 from c~* to + oo as b increases from 0 to b = gZ. It follows that for any choice of (a, 7) with e > ~/2 and 0 < 7 < rc there is exactly one curve cg (e, 7) given by graph b = b (a) terminating on its left e i t h e r a + b = l orb=a-1. If the curve terminates on the segment a + b = I then it is clear that the Gauss function co(a,b,u,v) converges uniformly to c o - 0 and the H = 1/2 surface becomes a r o u n d cylinder. N o w suppose cg (~, 7) approaches the curve b = a - 1 at a point (al,bl) where cot7 = g and ~ satisfies (6.4c) so that ba < g 2 . Let (a,b)ECg(a,j be a point near (al,bj. The function co(a,b,u,v) is determined from (2.8) using the solutions f(a, b, u) and g (a, b, v) of (2.9). We are interested in co (a, b, u, v) on the strip v _< v < 2 B - v with v e ( - B, B) and g (v) = g = cot y. The m a x i m u m of co (a, b, u, v) will occur at (0, v) so it suffices to show that co (a, b, 0, v) converges to zero. As (a, b) approaches (al, b 0 the function f(a, b, u) converges to f(a, b, u) = ~ tan (u ~ ) . At (al, b~) the right side of (2.9 b) becomes V2 (a, b, g) = (g 2 - b j 2 with solution g (a, b, v) = ~ t a n h ( v x / ~ 0 which is asymptotic to v = x / ~ . However, as (a,b) approaches (a~,bt) b o t h v and B become infinite. The function g(a,b,v) will resemble x/b1 t a n h ( v ~ / ~ 0 for a while but will blow up to + o o at v = B (a, b). O u r strip is centered about the line v = B (a, b). If we translate our d o m a i n so that the function g(a, b, v) has a vertical asymptotic at v = 0 then the limit function will be a solution to (2.9 b) with positive slope with 9 z > b. In the limit we have g(a, b, v ) = - ~ 1 coth(v,~b-1). We use this function g (a, b, u) to calculate co (a, b, u, v) using formula (2.8). We find cosh co _-- I so that co (a, b, u, v) converges uniformly to co - 0 on the horizontal strip. Finally consider the behavior of cs (a, ~) for large a. O n the line b = a + 1 the function (6.1 b) for ~ takes the form (6.5)

~

=

(2

--

~b)rC/2+[tan-1

g

l~-tan-~(g/x/b)],cot7

,/b

J

~

g

.

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This is an increasing function of b on the interval (1, + oo) with range (n/2, (3 n/2) - 7). Therefore, if n/2 < e < (3 n/2) - 7 the curve cg(e, 7) will cross the line b = a + 1. If we have a sequence (a,,b,) with b , / a , converging to 2 with b, > a, - 1 + e for some e > 0 then one can take a limit in the integral (6.1 b) and obtain the formula (6.3). The formula (6.3) is a decreasing function F of 2 with F(1) = (3 n/2) - 7 and F(oe) = n/2. It follows that for n/2 < ~ < (3 n/2) - 7 the curve cg (e, 7) will cross the line b = a + 1 and have limiting slope 2 determined by (6.3). F o r e > (3 n/2) - 7 the curve cg (e, 7) is t r a p p e d in Z o n e I. It is given by a g r a p h b = b (a) with b/a ~ 1. W h e n e > (3 hi2) - 7 one sees that this curve must be asymptotic to the line b = a - 1. Q.E.D. Remark. F o r (e, 7) with e > n/2, 0 < 7 < rc there are two p r i m a r y bifurcations from the r o u n d cylinder configuration, the symmetric one constructed in Section I I I and the n o n - s y m m e t r i c one constructed here. The bifurcation in either case will occur from co (u, v) - 0 in a horizontal strip of vertical width L = 2c~. F r o m the nodal line structure of the linearized e q u a t i o n for co (a, b, u, v) at the initial bifurcation one finds that the horizontal period for the n o n - s y m m e t r i c bifurcation is greater. This suggests that the symmetric bifurcation occurs first.

Bibliography 1. Abresch, U.: Constant mean curvature tori in terms of elliptic functions. J. Reine Angew. Math. 374, 169-192 (1987) 2. Darboux, G.: Lemons sur la Th6orie G6n~rale des Surfaces. Vol. III, Gantier-Villars, Paris 1915 (Chelsea Reprint) 3. Eisenhart, L.P.: Treatise on the Differential Geometry of Curves and Surfaces, Dover, Reprint 1960 4. Fischer, G.: Mathematical Models. Vieweg 1986. Commentary on Differential Geometry by M.P. do Carmo, G. Fischer, U. Pinkall, H. Reckziegel 5. Langbein, D.: The shape and stability of liquid menisci at solid edges. J. Fluid Mech. 213, 251-265 (1990) 6. Vogel, T.: Stability and bifurcation of a surface of constant mean curvature in a wedge. Indiana Univ. Journal 41, 625-648 (1992) 7. Walter, R.: Explicit examples to the H-problem of Heinz Hopf. Geom. Dedicata 23, 187-213 (1987)