Isoperimetric regions in H^2 between parallel horocycles
aa r X i v : . [ m a t h . DG ] O c t Isoperimetric regions in H between parallel horocycles M ´ARCIO F ABIANO DA S ILVA
Abstract.
In this work we investigate the following isoperimetric problemin the hyperbolic plane: to find the regions of prescribed area with minimalperimeter between two parallel horocycles. We give an explicit and detaileddescription of all such regions.
1. Introduction
For a Riemannian manifold M , the classical isoperimetric problem con-sists in classifying, up to congruency by the isometry group of M , the (com-pact) regions Ω ⊆ M enclosing a fixed volume that have minimal boundaryvolume. The existence and regularity of solutions for a large number of casesmay be guaranteed by adapting some results from the Geometric MeasureTheory (cf. [6]).For example, when M is the Euclidean plane R , the classical isoperi-metric problem has the disk as the unique solution. If M is a hyperbolicsurface, the least-perimeter enclosures of prescribed area are described in [1]and [7]. An interesting version of the isoperimetric problem is to study it ina slab. Physically, it corresponds to determine the shape of a drop trappedbetween two parallel planes, which was solved by Vogel in [8]. Independently,Athanassenas studied the isoperimetric problem between parallel planes of1 in [2]. If M is a slab between two parallel horospheres in the 3-dimensionalhyperbolic space H ( − R . The parallelhorocycles are represented by the horizontal straight lines of R . We willpresent in this paper a detailed and complete classification of the isoperimet-ric solutions.In Section 2 we give some basic definitions in the model R for the hyper-bolic plane like geodesics and curves of constant geodesic curvature. We alsopresent a more precise formulation for the considered isoperimetric problemand get some preliminary characterizations by adapting the results from [3].We will see that the possible isoperimetric regions must be delimited by suchcurves and meet the horocycles perpendicularly when this intersection is non-empty. We introduce the so-called geodesic halfdisk, horocycle halfdisk andequidistant halfdisk.In Section 3 we read off the expressions for perimeter and area for theregions obtained in Section 2 as the possible isoperimetric solutions.In Section 4 we compare the perimeter of the possible isoperimetric re-gions with prescribed area. In fact, we see that this is equivalent to investi-gating the regions of maximal area with prescribed perimeter.In Section 5 we give the isoperimetric profile for the region between twoparallel horocycles in R and prove the following result:Let c be a positive real constant and F c = { ( x, y ) ∈ R : 1 ≤ y ≤ c } . Let A > C c,A be the set of all Ω ⊂ F c with area | Ω | = A and perimeter | ∂ (Ω ∩ F ◦ c ) | < ∞ , where we suppose Ω to be connected, compact, 2-rectifiable2n F c , having as boundary (between the horocycles) a simple rectifiable curve. Theorem 1.1.
Let L c,A = inf {| ∂ (Ω ∩ F ◦ c ) | : Ω ∈ C c,A } . Then1. there exists Ω ∈ C c,A such that | ∂ (Ω ∩ F ◦ c ) | = L c,A ;2. if Ω ⊂ F c has minimal perimeter, the boundary of Ω has a single con-nected component made up with either(a) a halfdisk (geodesic, horocycle, equidistant) above { y = 1 } ;(b) a section of F c , namely S [ x ,x ] = [ x , x ] × [1 , c ] . More precisely, if d is the hyperbolic distance between the horocycles, we have: if d < , there exists A ( c ) such that • if A < A ( c ) then Ω is a geodesic halfdisk; • if A = A ( c ) then Ω is a geodesic halfdisk or a section; • if A > A ( c ) then Ω is a section; if d = 1 , there exists A ( c ) such that • if A < A ( c ) then Ω is a geodesic halfdisk; • if A = A ( c ) then Ω is a horocycle halfdisk or a section; • if A > A ( c ) then Ω is a section; if d > , there exist two constants A ( c ) < A ( c ) such that • if A < A ( c ) then Ω is a geodesic halfdisk; if A = A ( c ) then Ω is a horocycle halfdisk; • if A ( c ) < A < A ( c ) then Ω is an equidistant halfdisk; • if A = A ( c ) then Ω is an equidistant halfdisk or a section; • if A > A ( c ) then Ω is a section. Acknowledgments . The present work was supported by CAPES and CNPq.The author is specially grateful to Professor Doctor Luiz Amancio Machadode Sousa Junior, from Universidade do Rio de Janeiro, for his helpful sug-gestions.
2. Preliminaries
In this section we will introduce some basic facts and notations that willbe used along the paper. There is a large literature about the subject (wesuggest beginning with [4]). We also adapt some important results of [3] toget the possible isoperimetric regions in the hyperbolic plane.Let L = ( R , g ) be the 3-dimensional Lorentz space endowed with themetric g ( x, y ) = x y + x y − x y and the hyperbolic plane H := { p = ( x , x , x ) ∈ L : g ( p, p ) = − , x > } . We use the upper halfplane model R := { ( x, y ) ∈ R ; y > } for H ,endowed with the metric <, > = ds = dx + dy y . The Euclidean straight line { y = 0 } is the infinity boundary of R , de-noted by ∂ ∞ R .The curves of constant geodesic curvature k ≥ R are described asfollows: 4. Geodesic: ( k = 0). Represented by vertical Euclidean straight linescontained in R and Euclidean semicircles perpendicular to ∂ ∞ R andcontained in R ;2. Geodesic circles: ( k > R ;3. Horocycles: ( k = 1). Represented by horizontal Euclidean straightlines of R and Euclidean circles of R tangent to ∂ ∞ R .4. Equidistant curves: (0 < k < R with the straight lines of R that are neither parallel nor perpendicularto { y = 0 } , and by the Euclidean circles not entirely contained in R and are neither tangent nor perpendicular to { y = 0 } .The isometries of R are the M¨obius transformations of b C that leave R invariant. For our purposes, we are interested in the following Euclideanapplications: horizontal translations, reflections with respect to a verticalgeodesic, homotheties and inversions (with respect to circles centered in { y =0 } ).For R the isoperimetric problem may be formulated as follows: “tominimize the perimeter of a region inside two parallel horocycles (representedby two horizontal Euclidean straight lines), with prescribed area, but notcounting its part of the boundary contained in the horocycles”. By the perimeter of a region we mean the length of its boundary.Since the Euclidean homothety is an isometry of R , we take the lowerhorocycle as { y = 1 } to study the isoperimetric problem, so that any solutionis obtained by homothety. 5y adapting the demonstration of Theorem 1.1 from [3] to our case,namely R , together with Lemma 2.1 of [1], we have that there exists regu-lar isoperimetric solutions and they are regions whose boundary consists ofcurves of constant geodesic curvature perpendicular to the horocycles (whenthe intersection is non-empty). Essentially, this proves the first item of ourTheorem 1.1 in this present paper, stated at the Introduction.Before we start to calculate the expressions for the perimeter and area ofthe regions delimited by curves of constant geodesic curvature, we presentthe polar coordinate system for R and conclude this section by giving amore precise formulation for the isoperimetric problem.If ( x, y ) are the cartesian coordinates in R and γ is the geodesic y > ρ, θ ) of a point p ∈ R as follows: ρ is thehyperbolic distance from p to the origin O = (0 ,
1) and θ is the angle betweena fixed geodesic radius γ + , given by { x = 0; y ≥ } , and the geodesic through O and p , measured counterclockwise.The relation between these systems of coordinates is:( x, y ) = 1cosh ρ − sinh ρ cos θ (sinh ρ sin θ, , (1)and the metric of R in polar coordinates is dσ = dρ + sinh ρ dθ .We now obtain the expression for the arclength of a geodesic circle andthe area of a sector as functions of the central angle β . For the sake ofsimplicity we take the circle of hyperbolic radius ρ centered in O .The geodesic circle can be parametrized by α ( θ ) = ( ρ, θ ), with constant ρ and 0 ≤ θ ≤ β . Then dσ ( α ′ ) = sinh ρ . Therefore, the arclength corre-6ponding to β in the hyperbolic metric is L ( α ) = Z β p dσ ( α ′ ) dθ = β sinh ρ, (2)and the area A of a sector of the disk corresponding to β is A = Z β Z ρ sinh ρ dρ dθ = β (cosh ρ − . (3)As we mentioned above, the isoperimetric solutions are regions delimitedby curves of constant geodesic curvature perpendicular to the horocycles(when the intersection is non-empty). So we have the following possibilitiesfor barriers: vertical geodesics, geodesic circles, horocycles represented byEuclidean circles of R tangent to ∂ ∞ R , and equidistant curves representedby Euclidean circles not entirely contained in R and neither tangent nor per-pendicular to { y = 0 } . The region in F c delimited two vertical geodesics willbe called a section . The region in F c delimited by geodesic circles perpendic-ular to { y = 1 } or { y = c } will be called geodesic halfdisk . The region in F c delimited by horocycles and equidistant curves perpendicular to { y = 1 } willbe called horocycle halfdisk and equidistant halfdisk , respectively. We meanby halfdisk above (respectively below ) { y = c } , the part of the Euclideanhalfdisk above (respectively below) the horocycle { y = c } (see Figure 4). Isoperimetric problem for F c : fix an area value and study the domains Ω ⊂ F c with the prescribed area which have minimal free boundary perimeter. Definition 2.1:
A (compact) minimizing region Ω for this problem willbe called an isoperimetric solution or region in F c .7 . Expression for perimeter and area In this section we get expressions for the perimeter and area of the possi-ble isoperimetric solutions Ω contained in F c . For our purposes we consideronly the regions that are 2(-dimensional)-rectifiable (with respect to Haus-dorff’s measure) with boundary 1(-dimensional)-rectifiable. We denote thismeasure by | · | , so that any Ω has area | Ω | and perimeter | ∂ Ω | , but it never counts ∂ Ω ∩ ∂ F c . For more details, see [6]. Let c > x < x be real constants. For the sake of simplicity, weconsider the vertical geodesics { x = x } and { x = x } , contained in R , andthe parallel horocycles { y = 1 } and { y = c } . Lemma 3.1.1.
Under the notations above, if T is a section then | ∂T | = 2 ln c and | T | = ( x − x )( − /c + 1) . Proof : Since the length of a vertical geodesic segment 1 < y < c isln( c/
1) = ln c , then | ∂T | = 2 ln c . And | T | = Z x x Z c y dy dx = ( x − x )( − /c + 1) . q.e.d.8 .2. Perimeter and area for a geodesic halfdisk and a horocyclehalfdisk Consider c ∈ R ∗ + and { y = c } a horocycle in R . We take the Euclideancircle S centered in (0 , c ) with radius r < c . The circle S can be viewed as ageodesic circle S H with hyperbolic center C H = (0 , h ) and hyperbolic radius ρ . We want to relate the centers and the radii of S and S H .In the hyperbolic metric, since C H equidists from both (0 , c − r ) and(0 , c + r ), it is easy to see that h = √ c − r , C H = (0 , √ c − r ) and ρ = Z hc − r t dt = ln hc − r = 12 ln (cid:16) c + rc − r (cid:17) . (4)From (4) we have that rc = e ρ − e ρ + 1 = tanh ρ .Later we will use the relation between | S + | and | S − | , where S + and S − are halfdisks above and below { y = c } , respectively. They are given by | S + | = 2 Z r Z c + √ r − x c y dy dx = 2 c Z r √ r − x c + √ r − x dx, and | S − | = 2 Z r Z cc −√ r − x y dy dx = 2 c Z r √ r − x c − √ r − x dx. Notice that | S − | > | S + | . Similarly, one has | ∂S − | > | ∂S + | . (5)In Figure 1, θ denotes the angle between { x = 0 , y ≥ h } and the geodesic˜ S through C H and ( r, c ). Thus, θ measures the half of the central anglecorresponding to the arc of the geodesic semicircle above { y = c } , and ˜ S hascenter ( r,
0) and radius c . 9ince the Euclidean and hyperbolic metrics are conformal, in order tomeasure θ we parametrize ˜ S as α ( t ) = ( c sin t + r, c cos t ), with − π/ 2. Then C H = (0 , √ c − r ) = α ( t ) = ( c sin t + r, c cos t ) so thatsin t = − r/c and cos θ = − sin t = r/c. (6)If ¯ S is the region delimited by ˜ S , axis y and { y = c } then | ¯ S | = Z ch r − p c − y y dy = − r/c + π/ − arcsin( h/c ) . Suppose c > { y = 1 } and { y = c } .Let S be the circle centered at (0 , 1) with radius r < 1, and S the circlecentered at (0 , c ) with radius r < c − S can beviewed as a geodesic circle S H with hyperbolic center (0 , h ) = (0 , p − r )and hyperbolic radius ρ = 12 ln (cid:16) r − r (cid:17) , and S as a geodesic circle S H with hyperbolic center (0 , h ) = (0 , p c − r ) and hyperbolic radius ρ =12 ln (cid:16) c + r c − r (cid:17) .Let β be the central angle of S H corresponding to the arc above { y = 1 } ,and β the central angle of S H corresponding to the arc below { y = c } . By(6) we have β = 2 arccos( r ) and β = 2 π − r /c ) . For geodesic halfdisks it holds the following result (see Figure 2): Lemma 3.2.1. Under the notations above, let ˜ S be the geodesic through C H = (0 , h ) and ( r , , and ˜ S the geodesic through C H = (0 , h ) and ( r , c ) .Let θ = β / and θ = π − β / , < θ , θ < π/ . Let S +1 be the geodesichalfdisk delimited by S H and above { y = 1 } , and S − the halfdisk delimited by S H and below { y = c } . Then | ∂S +1 | = 2 θ cot θ , | ∂S − | = 2( π − θ ) cot θ , (7)10 nd | S +1 | = 2 θ sin θ − π + 2 cos θ , | S − | = 2( π − θ )sin θ − π − θ . (8) Proof : By (2), the arclengths determined by β and β are | ∂S +1 | = β sinh ρ and | ∂S − | = β sinh ρ . We have sinh ρ = sinh (cid:16) 12 ln (cid:16) r − r (cid:17)(cid:17) = r p − r , sinh ρ = sinh (cid:16) 12 ln (cid:16) c + r c − r (cid:17)(cid:17) = r p c − r . Since cot θ = r p − r and cot θ = r p c − r , the first part of thelemma is proved.Now observe that | S +1 | / | ˜ S | − | ¯ S | , where ˜ S is the sector correspond-ing to θ and ¯ S is the region delimited by ˜ S , axis y and the horocycle { y = 1 } . In the same way, | S − | / | ˜ S | + | ¯ S | , where ˜ S is the sector corre-sponding to β / π − θ and ¯ S is the region delimited by ˜ S , axis y andthe horocycle { y = c } .Therefore, by (3) | S +1 | = 2 θ (cosh ρ − − − r + π/ − arcsin( h )) , | S − | = 2( π − θ ) (cosh ρ − 1) + 2( − r /c + π/ − arcsin( h /c )) . (9)But cosh ρ = cosh (cid:16) 12 ln (cid:16) r − r (cid:17)(cid:17) = 1 p − r , cosh ρ = cosh (cid:16) 12 ln (cid:16) c + r c − r (cid:17)(cid:17) = c p c − r . (10)11ince r + (cid:16)p − r (cid:17) = 1 and ( r /c ) + ( p c − r /c ) = 1, we havearccos( r ) = arcsin( p − r ) = arcsin( h ) , arccos( r /c ) = arcsin (cid:16) p c − r c (cid:17) = arcsin( h /c ) . (11)Furthermore, by (6) it follows that cos θ = r and cos θ = r /c .Therefore, sin θ = q − r and sin θ = p c − r c . (12)By (9), (10), (11) and (12), the proof of (8) is complete q.e.d.We observe that a horocycle H can be viewed as a limit geodesic circlewith hyperbolic center in ∂ ∞ R . By (6), we have cos θ = r and the horocy-cle is obtained when r converges to 1, that is, θ converges to 0. Hence weget the expressions for the perimeter and the area of the horocycle halfdiskas the following consequence of Lemma 3.2.1: Corollary 3.2.2. Let H be the horocycle halfdisk above { y = 1 } repre-sented by a Euclidean semicircle with center (0 , and radius . Then | ∂H | = 2 and | H | = 4 − π. (13) Proof : It is enough to calculate | ∂S +1 | and | S +1 | from (7) and (8) for thelimit case when θ → Let ¯ E be the equidistant curve represented by a Euclidean circle withcenter (0 , 1) and radius r > 1. The Euclidean equation of ¯ E is given by12 + ( y − = r . Then ¯ E ∩ ∂ ∞ R = { ( −√ r − , , ( √ r − , } . Thecurve ¯ E is equidistant from the geodesic η with equation x + y = r − 1. If ρ denotes the hyperbolic distance between ¯ E and η , then ρ is the hyperbolicdistance between (0 , r ) and (0 , √ r − ρ = ln (cid:16) r + 1 r − (cid:17) , whence r = coth ρ. (14)If α is the non-oriented angle between ¯ E and η , 0 < α < π/ 2, then (forinstance, see Proposition 3 in Chapter 5 of [4])tanh ρ = sin α. (15) Lemma 3.3.1. Under the notations above, let E be the equidistanthalfdisk above { y = 1 } . Then | ∂E | = 2cos α ln (cid:16) α + cot α (cid:17) , | E | = 2sin α − π + 2cot α ln (cid:16) α + cot α (cid:17) . (16) Proof: In order to calculate | ∂E | , we parametrize E by β ( t ) = ( r cos t, r sin t ) , ≤ t ≤ π. Then | ∂E | = 2 Z π/ r r sin t dt = 2 r √ r − (cid:16) r + tan( t/ − √ r − r + tan( t/ 2) + √ r − (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) π/ = 2 r √ r − r + √ r − . By (14) and (15) we have r = 1 / sin α , whence √ r − α , because0 < α < π/ 2. Therefore, | ∂E | = 2cos α ln (cid:16) α + cot α (cid:17) , and the first part13f (16) is proved. Now, | E | = 2 Z r Z √ r − x y dy dx = 2 r − π + 1 √ r − (cid:12)(cid:12)(cid:12) r √ r − r − r √ r − − ( r − (cid:12)(cid:12)(cid:12) . By (14) and (15), it follows that | E | , as function of the equidistance angle α , is given by | E | = 2sin α − π + 2cot α ln (cid:16) α + cot α (cid:17) , which proves thesecond part of (16) q.e.d. 4. Comparison of perimeters of regions with prescribed area In this section we analyze the expressions of the perimeter and area ofthe regions delimited by curves of constant geodesic curvature. Their isoperi-metric profiles in F c will be obtained in the next section as functions of itshyperbolic width d . Since we have been taking the horocycles { y = 1 } and { y = c } , the constant c must satisfy the following condition: if H is a horo-cycle halfdisk above { y = c } and T is a section in F c , then | ∂H | = | ∂T | . By(13), this means 2 = 2 ln c , whence c = e and d = 1. This is why we compare d with 1 in Theorem 1.1.Let S be a Euclidean circle with radius r and center (0 , 1) above { y = 1 } .When 0 < r < S delimits a geodesic halfdisk. Consider the limit cases θ → θ → π/ 2, which correspond to a horocycle halfdisk ( r → r → θ → | ∂S +1 | = 2 , lim θ → π/ | ∂S +1 | = 0 , (17)and from (1), lim θ → | S +1 | = 4 − π, lim θ → π/ | S +1 | = 0 . (18)14herefore, | ∂S +1 | increases from 0 to 2 when r varies from 0 to 1, while | S +1 | increases from 0 to 4 − π .If r > α → π/ r → 1) and α → r → ∞ ).By (16), lim α → π/ | ∂E | = 2 , lim α → | ∂E | = ∞ , (19)and lim α → π/ | E | = 4 − π, lim α → | E | = ∞ . (20)Therefore, both | ∂E | and | E | increase infinitely while r increases.Since c > 1, let S be a Euclidean semicircle with radius r and center(0 , c ) below { y = c } . When 0 < r < c , S delimits a geodesic halfdisk. By(6), the limit cases θ → θ → π/ r → c and r → θ → | ∂S − | = ∞ , lim θ → π/ | ∂S − | = 0 , (21)and by (8), lim θ → | S − | = ∞ , lim θ → π/ | S − | = 0 . (22)Then we see that | ∂S − | and | S − | increase infinitely while r → c . If r ≥ c , S delimits a horocycle halfdisk or an equidistant halfdisk. By (21)and (22), if r ≥ c , then both | ∂S − | and | S − | diverge to infinity.From the analysis we have just done for the perimeter and area of thepossible isoperimetric solutions, there are only the following cases to consider:1. to compare a geodesic halfdisk above { y = 1 } with a geodesic diskentirely contained in F c ; 15. to compare a geodesic halfdisk above { y = 1 } with a geodesic halfdiskbelow { y = c } ;3. to compare a horocycle halfdisk above { y = 1 } with a geodesic halfdiskbelow { y = c } ;4. to compare an equidistant halfdisk above { y = 1 } with a geodesichalfdisk below { y = c } .In order to prove the second part of Theorem 1.1, one must determine theleast-perimeter regions with prescribed area. For this purpose, we will usea strategy: we determine the regions with prescribed perimeter and biggestarea. In fact, it is enough to show that if a region has the maximum areaamong all regions with a prescribed perimeter, then it has the minimumperimeter among all regions with the same prescribed area (see Lemma 4.1below). Since we have just listed all possible isoperimetric solutions besidesthe section, Lemma 4.1 will then refer to the above case 2. The other cases areproved analogously. Without loss of generality we suppose that the geodesichalfdisk above { y = 1 } has maximum area when compared to any geodesichalfdisk below { y = c } with the same perimeter. Lemma 4.1. Let Ω be the geodesic halfdisk above { y = 1 } with | Ω | ≥| Ω | , whenever | ∂ Ω | = | ∂ Ω | , for any geodesic halfdisk Ω below { y = c } , c > .If Ω is a geodesic halfdisk below { y = c } with | Ω | = | Ω | , then | ∂ Ω | ≤ | ∂ Ω | . Proof: Suppose by contradiction that | ∂ Ω | > | ∂ Ω | . By (21) we canincrease the radius of the Euclidean circle that represents Ω till we get ageodesic halfdisk Ω ′ such that | ∂ Ω ′ | = | ∂ Ω | . This procedure could fail ifΩ ′ surpassed { y = 1 } , but then the section will prevail as the isoperimetric16olution. This fact will be proved later on in Section 5. By (22), the areaincreases with the radius. Therefore, | Ω ′ | > | Ω | = | Ω | and | ∂ Ω ′ | = | ∂ Ω | .This is a contradiction with the fact that Ω maximizes the area when com-pared to regions of the same perimeter, by hypothesis q.e.d.Till the end of this section we are going to compare the area of the possibleisoperimetric solutions for a prescribed perimeter.For case 1 described above, we compare the area of a geodesic halfdiskabove { y = 1 } with a geodesic disk entirely contained in F c , when theyhave the same perimeter. Let S be the Euclidean circle with radius r ,0 < r < y − 1, and center (0 , y ), 1 < y < c , which delimits the geodesichalfdisk (see Figure 4).Consider θ ∈ ]0 , π/ 2[ such that cos θ = r /c . By (2), (3) and (4), if S isthe geodesic disk corresponding to a central angle of 2 π then | ∂ S| = 2 π cot θ and |S| = 2 π sin θ − π .By (7), (8) and the information from the previous paragraph, we showthat | S +1 | > |S| when | ∂S +1 | = | ∂ S| in the next Lemma. Lemma 4.2. Let θ , θ ∈ ]0 , π/ such that θ cot θ = π cot θ . (23) Then θ sin θ + 2 cos θ − π > π sin θ − π. (24) Proof: For θ ∈ ]0 , π/ θ = 1 − sin θ , one has 1 / sin θ = p θ cot θ + π /π . Thus2 π sin θ − π = 2 r(cid:16) θ sin θ (cid:17) − θ + π − π. (25)17ow we replace the right-hand side of (24) by (25), and define A ( θ ) := 2 θ sin θ + 2 cos θ − π − r(cid:16) θ sin θ (cid:17) − (cid:16) θ (cid:17) + π + 2 π, so that (24) will hold if and only if A ( θ ) > θ → A ( θ ) = 4 + π − √ π + 1 > θ → π/ A ( θ ) = 0 . (26)Moreover, dA ( θ ) dθ = 2 cos θ (cid:16) sin θ cos θ − θ (cid:17) sin θ n − θ r θ + (cid:16) π − θ (cid:17) sin θ o < , because θ ∈ ]0 , π/ 2[ implies cos θ > 0, sin θ cos θ − θ < < θ r θ + (cid:16) π − θ (cid:17) sin θ < . Therefore, A ( θ ) decreases in ]0 , π/ A ( θ ) > , π/ { y = 1 } isthe isoperimetric solution, instead of the geodesic disk, which concludes case1. Now we study case 2. By (7) and (8), we will show in the next Lemmaand Corollary that | S +1 | > | S − | when | ∂S +1 | = | ∂S − | . In Figure 4, the dashedcircle was obtained from the lower by a Euclidean homothety so that thecorresponding geodesic halfdisks have the same perimeter. By (5), in orderto have | ∂S +1 | = | ∂S − | , it is necessary to decrease the radius of S − .18 emma 4.3. Let θ , θ ∈ ]0 , π/ such that θ cot θ = ( π − θ ) cot θ . (27) Then θ sin θ + cos θ ≥ π − θ sin θ − cos θ . (28) Proof: For θ , θ ∈ ]0 , π/ f ( θ ) = θ sin θ +cos θ and F ( θ , θ ) = f ( θ ) + f ( θ ) − π sin θ . We want to show that F ( θ , θ ) ≥ 0. By (27) we candefine θ implicitly as a function of θ . Namely, we get a function g suchthat θ = g ( θ ). Let h ( θ ) = F ( g ( θ ) , θ ) and h ( θ ) = h ( θ ) sin θ + π =sin θ f ( g ( θ )) + sin θ f ( θ ).The function h ( θ ) is C ∞ and h ′ ( θ ) = cos θ f ( θ ) + sin θ f ′ ( θ ) g ′ ( θ ) + cos θ f ( θ ) + sin θ f ′ ( θ ) . (29)Hence f ′ ( θ ) = sin θ − θ cos θ sin θ − sin θ = − cos θ (2 θ − sin 2 θ )2 sin θ . (30)From the Implicit Function Theorem we have g ′ ( θ ) = − cot θ − ( π − θ ) csc θ cot θ − θ csc θ . Now observe that cot θ − θ csc θ = sin 2 θ − θ θ . Therefore, g ′ ( θ ) = 2( π − θ ) + sin 2 θ θ θ θ − sin 2 θ . (31)By substituting (30) and (31) in (29), we obtain h ′ ( θ ) = 2 cos θ + θ cos θ sin θ − ( π − θ ) cos θ sin θ . (32)19ince h ′ ( θ ) = h ′ ( θ ) sin θ + h ( θ ) cos θ , it follows from (32) that h ′ ( θ ) sin θ = 2 cos θ + θ cos θ sin θ − ( π − θ ) cos θ sin θ − F ( θ , θ ) cos θ = cos θ − cos θ cos θ − ( π − θ )sin θ (cos θ − cos θ )= (cos θ − cos θ )(cos θ + π − θ sin θ ) . (33)For θ , θ ∈ ]0 , π/ θ ≤ π − θ , which by (27) impliescot θ = (cid:16) π − θ θ (cid:17) cot θ ≥ cot θ ⇒ θ ≤ θ ⇒ cos θ ≥ cos θ . (34)By (33) and (34) we have h ′ ( θ ) ≤ 0. Thus F ( θ , θ ) = F ( g ( θ ) , θ ) = h ( θ ) is a decreasing function for θ ∈ ]0 , π/ θ = π/ θ = g ( π/ g ( π/ g ( π/ 2) = π/ 2. Since h ( π/ 2) = F ( g ( π/ , π/ 2) = F ( π/ , π/ 2) = 0 , then F ( θ , θ ) ≥ F ( g ( π/ , π/ 2) = 0. Consequently, f ( θ ) + f ( θ ) ≥ π sin θ ,whence (28) is proved.The equality occurs if and only if θ = θ = π/ 2. q.e.d. Corollary 4.4. Let θ , θ ∈ ]0 , π/ such that θ cot θ = ( π − θ ) cot θ . (35) Then θ sin θ + cos θ > π − θ sin θ − cos θ . (36)20e finally conclude from Corollary 4.4 that the geodesic halfdisk above { y = 1 } is the isoperimetric solution for case 2.By (13), (7) and (8), we show in the next Lemma that | H | > | S − | when | ∂H | = | ∂S − | . Case 3 is illustrated in Figure 5. Lemma 4.5. Let θ ∈ ]0 , π/ such that π − θ ) cot θ . Then > π − θ sin θ − cos θ . Proof: Since the horocycle is obtained from the geodesic halfdisk above { y = 1 } when θ → 0, then it is enough to make θ → { y = 1 } is the isoperimetric solution, instead of the geodesic halfdisk below { y = c } .Now we analyze case 4. By (16), (7) and (8), we show in the next Lemmathat | E | > | S − | when | ∂E | = | ∂S − | . In Figure 5, the dashed circle wasobtained from the lower by a Euclidean homothety so that they have thesame perimeter. In order to have | ∂E | = | ∂S − | , it is necessary to decreasethe radius of S − . Lemma 4.6. Let α, θ ∈ ]0 , π/ such that α ln (cid:16) α + cot α (cid:17) = ( π − θ ) cot θ . (37) Then α + 1cot α ln (cid:16) α + cot α (cid:17) ≥ π − θ sin θ − cos θ . (38) Proof: For α, θ ∈ ]0 , π/ 2[ we define F ( α, θ ) = 1sin α + 1cot α ln (cid:16) α + cot α (cid:17) + θ sin θ + cos θ − π sin θ . By (37), we can implicitly define θ as a function of α . Namely, one gets afunction g such that θ = g ( α ). Consider the functions h ( α ) = F ( α, g ( α )),21 ( α ) = h ( α ) sin α . Then h ( α ) = (cid:16) α + 1cot α ln (cid:16) α +cot α (cid:17)(cid:17) sin α − (cid:16) π − g ( α )sin g ( α ) − cos g ( α ) (cid:17) sin α. The function h ( α ) is C ∞ and h ′ ( α ) = sin α cos α ln (cid:16) α + cot α (cid:17) − sin α cos α + sin α ln (cid:16) α + cot α (cid:17) ++ g ′ ( α ) cos θ (sin(2 θ ) + 2( π − θ ))2 sin θ sin α − ( π − θ )sin θ cos α + cos θ cos α. (39)From the Implicit Function Theorem we have g ′ ( α ) = − sin α cos α ln (cid:16) α + cot α (cid:17) − α cos α cot θ + ( π − θ ) csc θ . Since cot θ + ( π − θ ) csc θ = sin 2 θ + 2( π − θ )2 sin θ , then g ′ ( α ) = (cid:16) θ sin 2 θ + 2( π − θ ) (cid:17) (cid:16) α cos α − sin α cos α ln (cid:16) α + cot α (cid:17)(cid:17) . (40)By substituting (40) in (39), we obtain h ′ ( α ) = sin α cos α ln (cid:16) α + cot α (cid:17) − sin α cos α + sin α ln (cid:16) α + cot α (cid:17) ++ cos θ cos α − sin α cos α cos θ ln (cid:16) α + cot α (cid:17) − ( π − θ )sin θ cos α + cos θ cos α. (41)Since h ′ ( α ) = h ′ ( α ) sin α + h ( α ) cos α , it results from (41) that h ′ ( α ) sin α = 1 − sin α cos θ cos α (cid:16) tan α ln (cid:16) α + cot α (cid:17) − α (cid:17) . For α ∈ ]0 , π/ l ( α ) = tan α ln(1 / sin α +cot α ), then l ′ ( α ) = sec α k ( α ),where k ( α ) = ln(1 / sin α + cot α ) − cos α . Since k ′ ( α ) = − cos α/ sin α < k ( α ) is decreasing in ]0 , π/ 2[ and lim α → π/ k ( α ) = 0. So k ( α ) > , π/ l ( α ) is increasing in ]0 , π/ α → π/ l ( α ) = 1,then l ( α ) < α ln (cid:16) α + cot α (cid:17) − α < . Moreover, for α, θ ∈ ]0 , π/ 2[ we have 0 < sin α cos θ < 1. Thus, for α ∈ ]0 , π/ 2[ we conclude that h ′ ( α ) sin α < h ′ ( α ) < h is decreasing in ]0 , π/ h ( α ) ≥ lim α → π/ h ( α ) = lim α → π/ F ( α, g ( α )) = 2 − (cid:16) π − β sin β − cos β (cid:17) , where β = lim α → π/ g ( α ). From (13) it follows that | H | + π , where H is a horocycle halfdisk above { y = 1 } . By (8), | G | + π π − β sin β − cos β, where G is a geodesic halfdisk with the same perimeter as H (just take α → π/ θ = β ). But from Lemma 4.5 we conclude that h ( α ) = 2 − (cid:16) π − β sin β − cos β (cid:17) > , whence (38) is proved q.e.d.From Lemma 4.2, Corollary 4.4, Lemma 4.5 and Lemma 4.6, we concludethat the family of geodesic, horocycle and equidistant halfdisks above { y =1 } are the solutions to the isoperimetric problem, instead of the geodesichalfdisks below { y = c } , c > 1. 23 . Isoperimetric Profile in R In this section we study the isoperimetric profile for F c (see Figure 6).We adapt a well-known result from the Isoperimetric Problem Theory whichguarantees that the boundaries of the connected components of an isoperi-metric solution are curves with the same constant geodesic curvature (forinstance, see Lemma 2.1 of [1]). Before showing that a minimizing regionis made up with a single connected component, we prove that a connectedcomponent of an isoperimetric region must be either a section or a halfdiskabove the horocycle { y = 1 } . Here we need (17)-(22). The perimeter of thesection in F c is equal to 2 ln c . Now there are only three possibilities that weclassify according to the hyperbolic distance d = ln c . First Possibility: d < 11. Consider a horocycle { y = c } with 1 < c < e . Let A ( c ) be thearea of the geodesic halfdisk S above { y = 1 } , centered at (0 , 1) withEuclidean radius r ( c ) and | ∂S | = | ∂T | , where T is a section with | T | = A ( c ) (see Figure 7). Since c < e , then | ∂T | < { y = 1 } ).Consequently, • if A = A ( c ) then | ∂S | = | ∂T | and | S | = | T | = A . Therefore,the minimizing region Ω is a geodesic halfdisk or a section; • if A < A ( c ), let S be a geodesic halfdisk with area A , centeredat (0 , 1) and with Euclidean radius r . Since both | S | and | ∂S | r , we have r < r ( c ) and | ∂S | < | ∂S | . Let T be a section with | T | = A . Then | S | = | T | = A , but | ∂S | < | ∂T | = | ∂T | = | ∂S | . Therefore, the minimizing Ω is a geodesichalfdisk. In this case, we observe that | Ω | = A < | S | , so that Ωcan neither be a horocycle nor an equidistant halfdisk; • if A > A ( c ), let S be a geodesic halfdisk with | S | = A , centeredat (0 , 1) and with Euclidean radius r . Since both | S | and | ∂S | increase with r , then r > r ( c ) and | ∂S | > | ∂S | . Let T be asection with | T | = A . Then | S | = | T | = A , but | ∂S | > | ∂T | = | ∂T | = | ∂S | . Therefore, the minimizing Ω is a section. Second Possibility: d = 12. Suppose d = 1. Consider the horocycle { y = c } with c = e . Then A ( c ) = 4 − π is the area of the horocycle halfdisk S above { y = 1 } ,centered at (0 , 1) with Euclidean radius r ( c ) = 1 and | ∂S | = | ∂T | ,where T is a section with | T | = A ( c ) (see Figure 8). In this case, | ∂T | = 2.Consequently, • if A = A ( c ), then | S | = | T | = A . Therefore, the minimizing Ωis a horocycle halfdisk or a section; • if A < A ( c ), let S be a geodesic halfdisk with | S | = A , centeredat (0 , 1) and with Euclidean radius r . Since both | S | and | ∂S | increase with r till it becomes a horocycle disk, then r < | ∂S | < | ∂S | . Let T be a section with | T | = A . Then25 S | = | T | = A , but | ∂S | < | ∂T | = | ∂T | = | ∂S | . Therefore, theminimizing Ω is a geodesic halfdisk; • if A > A ( c ), let S be an equidistant halfdisk | S | = A , centeredat (0 , 1) and with Euclidean radius r . Since both | S | and | ∂S | increase infinitely with r , then r > | ∂S | > | ∂S | . Let T be a section with | T | = A . Then | S | = | T | = A , but | ∂S | > | ∂T | = | ∂T | = | ∂S | . Therefore, the minimizing Ω is a section. Third Possibility: d > 13. Suppose d > 1. Consider a horocycle { y = c } with c > e . Let A ( c ) =4 − π be the area of the horocycle halfdisk S above { y = 1 } , centeredat (0 , 1) with Euclidean radius r ( c ) = 1 and | ∂S | = 2. Let T bea section with | T | = A ( c ) and A ( c ) be the area of an equidistanthalfdisk S above { y = 1 } , centered at (0 , 1) with Euclidean radius r ( c ) and | ∂S | = | ∂T | , where T is a section with | T | = A ( c ) (seeFigure 9). In this case, we observe that | ∂T | > • if A = A ( c ) = 4 − π then | S | = | T | = A , but | ∂T | > | ∂S | .Therefore, the minimizing Ω is a horocycle halfdisk; • if A = A ( c ) then | S | = | T | = A and | ∂S | = | ∂T | . Therefore,the minimizing Ω is an equidistant halfdisk or a section; • if A < A ( c ), let S be a geodesic halfdisk with | S | = A , cen-tered at (0 , 1) and with Euclidean radius r . Then r < r ( c )and | ∂S | < | ∂S | . Let T be a section with | T | = A . Then26 S | = | T | = A , but | ∂S | < | ∂T | = | ∂T | = | ∂S | . Therefore, theminimizing Ω is a geodesic halfdisk; • if A ( c ) < A < A ( c ), let S be an equidistant halfdisk with | S | = A , centered at (0 , 1) and with Euclidean radius r . Then r ( c ) < r < r ( c ) and | ∂S | < | ∂S | . Let T be a section with | T | = A . Then | S | = | T | = A , but | ∂S | < | ∂T | = | ∂T | = | ∂S | . Therefore, the minimizing Ω is an equidistant halfdisk; • if A > A ( c ), let S be an equidistant halfdisk with | S | = A ,centered at (0 , 1) and with Euclidean radius r . Then r > r ( c )and | ∂S | > | ∂S | . Let T be a section with | T | = A . Then | S | = | T | = A , but | ∂S | > | ∂T | = | ∂T | = | ∂S | . Therefore, theminimizing Ω is a section.REMARK 5.1: A minimizing region consists of only one connected com-ponent, and in fact it is enough to show that it can not have two. If this werethe case, their geodesic curvatures would agree. Consider A > ′ aregion with area A and two disjoint sections. Their “gluing” would result inanother section with area A but with smaller perimeter, because two verticalgeodesics would not count anymore. Then Ω ′ is not minimizing.The other case to consider is two connected components consisting of twogeodesic halfdisks above { y = 1 } . In this case, we use the fact that a non-regular region is not minimizing: let A > ′ be a region with area A and two geodesic halfdisks above { y = 1 } with the same Euclidean radius,hence the same geodesic curvature. By sliding one of them over { y = 1 } till it touches the other, since horizontal translations are isometries of thehyperbolic plane, we get a non-regular region Ω ′′ with area A . Then Ω ′′ does27ot have the least-perimeter among all regions with prescribed area A . Since | Ω ′ | = | Ω ′′ | , Ω ′ is not minimizing.Therefore, a minimizing region must consist of a single connected com-ponent.Now we prove Theorem 1.1. Proof: The first part of Theorem 1.1 was already discussed in the Prelim-inaries. The existence of such an isoperimetric region follows from adaptionsof some results from [5] and [6]: the group G of isometries of R that leave F c invariant consists of horizontal Euclidean translations and Euclidean re-flections with respect to a vertical geodesic, so that F c /G is homeomorphicto the interval [0 , References [1] C. Adams and F. Morgan, Isoperimetric curves on hyperbolic surfaces.Proc. Am. Math. Soc. 127 (1986), 1347–1356.[2] M. Athanassenas, A variational problem for constant mean curvaturesurfaces with free boundary. J. Reine Angew. Math. 377 (1986), 97–107.[3] R.M.B. Chaves, M.F. da Silva and R.H.L. Pedrosa, A free boundaryisoperimetric problem in the hyperbolic space between parallel horo-spheres. Pre-print at http://arxiv.org/abs/0811.1046v1 ( ,0) r c ( , ) θ S ~ Cy rc ( + ,0) y=c xS HH Figure 1: Arc of geodesic circle corresponding to a central angle θ . y= H H y S S ββ xy=c Figure 2: Perimeter and area for geodesic halfdisks.30 r r y= _ r r _( r (,1) η α (0,1) ,1) r (0, +1) y E x Figure 3: Perimeter and area for an equidistant disk. y=y xy=cy=cy=yy=x y Figure 4: Cases 1 (left) and 2 (right).31 y= y=x y xy=cy=c=ey Figure 5: Cases 3 (left) and 4 (right). section section section L e=c A equidistant halfdiskhorocycle halfdiskgeodesic halfdisk L Figure 6: Isoperimetric profile for the region between the parallel horocycles.32 y= A c ( ) y=cy=e y= y Figure 7: Case c < e . y= y=y=e x A c ( ) y Figure 8: Case c = e . y=y= y=ey=c A c ( ) A c ( ) xy Figure 9: Case c > ec > e