An exterior boundary value problem in Minkowski space

Transcription

1 Math. Nah. 81, No. 8, (008) / DOI /mana An exteio bounday value poblem in Minkowski spae Rafael López 1 1 Depatamento de Geometía y Topología, Univesidad de Ganada, Ganada, Spain Reeived 0 July 005, evised 16 May 006, aepted 15 June 006 Published online 8 July 008 Key wods Young Laplae equation, Loentzian meti, stationay sufae MSC (000) 35Q35, 76B45, 35J65, 53A10 In thee-dimensional Loentz Minkowski spae L 3, we onside a spaelike plane Π and a ound dis Ω ove Π. In this atile we seek the shapes of unbounded sufaes whose bounday is Ω and its mean uvatue is a linea funtion of the distane to Π. These sufaes, alled stationay sufaes, ae solutions of a vaiational poblem and govened by the Young Laplae equation. In this sense, they genealize the sufaes with onstant mean uvatue in L 3. We shall desibe all axially symmeti unbounded stationay sufaes with speial attention in the ase that the sufae is asymptoti to Π at the infinity. 1 Fomulation of the poblem Let L 3 denote the 3-dimensional Loentz Minkowski spae, that is, the eal veto spae R 3 endowed with the meti, = dx 1 + dx dx 3,wheex =(x 1,x,x 3 ) ae the anonial oodinates in R 3. An immesion x : M L 3 of a smooth sufae M is alled spaelike if the indued meti on the sufae is positive definite. We identify S =: x(m) with M. Fo spaelike sufaes, the notions of the fist and seond fundamental fom, and the mean uvatue ae defined in the same way as in Eulidean spae. Constant mean uvatue (spaelike) sufaes ae obtained as solutions of a vaiational poblem, exatly, they ae itial points of the aea funtional fo vaiations whih peseve a suitable volume funtion. In geneal, onstant mean uvatue spaelike submanifolds of a Loentzian manifold ae inteesting in elativity theoy. In this setting, thee is inteest of finding eal-valued funtions on a given spaetime, all of whose level sets have onstant mean uvatue. Then the mean uvatue funtion may then be used as a global time oodinate whih has many appliations. See [4, 11]. In this wok, we genealize onstant mean uvatue sufaes in L 3 by sufaes whose mean uvatue is a linea funtion of the time oodinate and that follow being solutions of a moe geneal vaiational poblem. We expliit ou mise in sene. Conside the hoizontal plane Π={x 3 =0} and let us fix a ound dis Ω of adius R>0 at distane b fom Π, namely, Ω:=Ω R,b = { x L 3 ; x 3 = b, x 1 + x R }. We assume the existene of a timelike potential Y = κx 3 + λ, κ>0, λ R, whih measues at eah point, up onstants, the distane to Π. LetS be an unbounded spaelike sufae whose bounday is the ile Ω and onside all the petubations of S in suh way that S emains adheed to Ω along its bounday. Fo eah elatively ompat domain D in M suh that M D, we define the enegy funtional E = D osh β Ω + YdM, D whee D and Ω denote the aeas of D and Ω espetively. As in the ase of onstant mean uvatue sufae, we ae inteested in those onfiguations whee the enegy of the physial system is itial unde any petubation of the system that does not hange the volume of x(d) and the adheene of x(d) on Ω along the ile Ω. We amino@ug.es

2 1170 López: An exteio bounday value poblem in Minkowski spae say then that S is a stationay sufae. Aoding to the piniple of vitual wok, and when the equilibium of the system is ahieved, the possible onfiguations that adopts a stationay sufae ae haateized as follows: Poposition 1.1 With the above assumptions, the sufae S is stationay if and only if 1. The mean uvatue H of S satisfies the elation H(p) =κx 3 (p)+λ, p S, κ > 0, λ R. (1.1). The sufae S meets the suppot dis Ω in a onstant hypeboli angle β, that is, osh β = N,N Ω along S, whenn and N Ω denote futue-dieted oientations on S and Ω espetively. We addess the eade to [1,, 3, 10] fo a speifi bakgound of the poblem. As a patiula ase, if κ =0, the mean uvatue of S is onstant, namely, H = λ/. Although the study of the possible shapes of a stationay sufae pesents diffiulty of analysis in all its geneality, at least in the situation that S is an embeddedsufae, that is, without self-intesetions, one an expet that S inheits the symmeties of its bounday, that is, S is a sufae of evolution. Fo this eason, we shall fous in the ase of axially symmeti onfiguations of the poblem: S is otational symmeti with espet to the x 3 -axis. See Figue 1. x 3 x 1 x S Fig. 1 Desiption of the physial system Moeove, we ae inteested by those stationay sufaes, if thee exist, that ae almost flat at the infinity of Π. This means that we seek unbounded stationay sufaes S suh that lim x 3 (p) =0, p S π(p) + whee π : R 3 Π is the othogonal pojetion onto Π. In the pesent wok, we will show that suh sufae exist. Besides the esults that we shall state in the next setion, it is wothwhile to announe hee the following two onlusions. The fist one shows that pat of the spae L 3 an be paametized by slies that ae stationay sufaes. Exatly [C. 1] Given R, κ > 0 and b R, it is possible to foliate the set of spaelike dietions that go out fom Ω R,b by a unipaameti family of unbounded stationay sufaes that ae solutions of Equation (1.1). Eah leaf of the foliation is otational symmeti with espet to the x 3 -axis. The seond onsequene infoms us that one slie of the above family is lose to the level efeene Π at inflinity. [C. ] Let us fix R, κ > 0. Foeahb R, thee exists an unbounded stationay otational symmeti sufae whose bounday is Ω R,b and asymptoti to Π at infinity. Peliminaies and statement of esults Fo a spaelike sufae M in L 3, the mean uvatue H of M is given by H = tae ( I 1 II ).WhenM is the gaph of a smooth funtion u = u(x 1,x ), the spaelike ondition is equivalent to Du < 1 and H wites as div(tu)=h, T u = Du 1 Du, (.1)

3 Math. Nah. 81, No. 8 (008) 1171 whee the oientation N on M is N = (Du, 1)/ 1 Du. Aoding to (.1), the Eule equation (1.1) onvets into div(tu)=κu + λ, κ > 0, λ R. (.) We will oient all the sufaes by the Gauss map N aoding to N,(0, 0, 1) < 0, thatis,n points upwads, that is, N is futue-dieted. In patiula, if M is the gaph of a smooth funtion u defined in R \ Ω R,0,then u satisfies (.). Let us emak that Equation (.) is a quasilinea ellipti equation of divegene-type due to the spaelike ondition on the intefae S. This allows us, fo example, to use lassial maximum and ompaison piniples. On the othe hand, the ondition on the onstany of the angle of ontat between the dis Ω and S along S wites now as N,N Ω = 1 1 Du =oshβ on Ω R,0. (.3) In this atile, we disuss the ase that S is a gaph on the plane Π and otational symmeti with espet to a staight-line L othogonal to Π. This hypothesis on axial symmety omes fom some evidenes in othe simila onfiguations. Fo example, we have: Theoem ([9]) We onside a spaelike plane Π in the Minkowski spae L 3. Then any bounded stationay sufae whose bounday lies in Π is otational with espet to a staight-line othogonal to Π and the sufae is a gaph ove Π. Moeove, any (nonempty) intesetion with a paallel plane to Π is a ound ile. A simila statement is obtained fo bounded stationay sufaes tapped in paallel (spaelike) planes. Unde these assumptions, S is detemined by the otation of the pofile of a funtion u : [R, ) R with espet to L = { x R 3 ; x 1 = x = 0 } and S an be epesented as S = {( os θ, sin θ, u()); [R, m), θ R}, fosomem. Equation (.) fo the pofile uve that defines the intefae S beome an odinay diffeential equation given by d d and (.3) is now ( u () 1 u () ) = κ u()+λ, R < < m, (.4) u (R + )=tanhβ. (.5) Without loss of geneality, and afte a vetial tanslation, we assume in the pesent wok that λ =0.Givena positive numbe κ, we begin in Setion 3 with the study of the exteio bounday value poblem P: ( u () 1 u () ) = κ u(), > R, (.6) u(r) =b, u (R + )=. (.7) Hee R > 0, b R and ( 1, 1). The fist esult assues existene and uniqueness of the initial value poblem P. Theoem.1 Fo eah (R, b, ), thee exists a unique solution of P. The maximal domain is [R, ). We denote by u = u(; b, ) the dependene on the initial onditions. Next, we ompae the solutions with espet to the onstant κ and the initial values. Theoem. Let κ 1 and κ be two positive onstants. Denote u i = u i () two solutions of (.6) (.7) fo κ = κ i, i =1,. Ifκ 1 <κ,then fo >R. u 1 (; b, ) <u (; b, ) and u 1(; b, ) <u (; b, )

4 117 López: An exteio bounday value poblem in Minkowski spae Theoem.3 Let R > 0. Conside u = u(; b, ) be the solution of P. If η > 0,then u(; b + η, ) η>u(; b, ), fo >R. It follows that if the uve u(; b + η, ) is moved igidly downwad a distane η, it will lie ompletely above the uve u(; b, ) exept at the single point (R, b) of ontat. In the last Setion 4, we etun ou inteest in those solutions that oespond with unbounded stationay sufaes that ae almost hoizontal at infinity. Fo this eason, we impose an aditional ondition given by lim u() =0. (.8) Fo a fixed numbe b, thee exists a (unique) solution asymptoti to the plane at infinity. Theoem.4 Let R>0and let b R. Thee exists a unique value δ, δ ( 1, 1), suh that the solution u(; b, δ) of the initial value poblem P satisfies lim u() =0. Moeove, if b>0,then b(1 + 1+κR ) R + b (1 + 1+κR ) <δ<0. As a onsequene of the lassial maximum piniple fo ellipti equations, we onlude: Theoem.5 Let κ>0. Conside a solution v of (.) (.3) on Ω R,0 suh that v(x) =b fo all x Ω R,0 and lim v(x) =0. x Then v(x) depends only on x and v( x ) =u(; b, δ), = x,wheeu is the funtion obtained in Theoem.4. Remak.6 The bounday value poblem (.) with v = b along Ω R,0 is equivalent to the minimum poblem ( κ F (v) = v ) 1 Dv dx min, v C R \Ω R,0 whee { } ( C = v; v + Dv ) dx <, v ΩR,0 = b; Dv 1 in R \ Ω R,0. R \Ω R,0 The stong onvexity of the funtional F guaantees an unique solution v suh that v(x) 0 fo x. Thus, v must agee with the solution u found in Theoem.4. We ontinue in Setion 4 obtaining esults of monotony and ontinuity, as well as, estimates of the height of an unbounded stationay sufae in tems of the angle of ontat with the ound dis. We hope that otational symmeti stationay sufaes shall allow in the futue the study fo othe moe geneal onfiguations of the domain Ω. We show an example of this giving an estimate of the ontat angle in the ase that Ω satisfies an inteio sphee ondition. In Eulidean spae, analogous poblems have been studied by a numbe of authos: see [6] and efeenes theein. While the analysis to follow is aied out using methods appopiate to the Eulidean ase, the theoems yield new esults when we onside Loentzian setting. We point out that some diffeenes appea in the Loentzian setting, mainly due to the spaelike haate of ou sufaes. It is wothwhile to bing out two of them:

5 Math. Nah. 81, No. 8 (008) We pove that the maximal inteval of a solution of (.4) (.5) is [R, ). On the ontay, the solutions of the (analogous) exteio bounday value poblem ae not defined in all eal line, and we stop at points 1 < suh that u ( 1 )=.. Ou sufaes do not pesent vetial points. In patiula, eah solution goes out fom the ight solid ylinde C =Ω R defined by Ω. In ontast to this, in Eulidean setting, thee exist stationay sufaes that eente in C and next, they go out fom C [1, 13, 14]. In these sufaes appea vetial points. 3 Existene of solutions of the exteio bounday value poblem P o o f o f T h e o e m.1. Equation (.6) also wites as u = κu ( 1 u ) 3/ u ( 1 u ). (3.1) As usual, set x = u, y = u and f() =(x(),y()). Then a solution of the value poblem P is equivalent to ( f () =F (, f ()), F(, x, y) = y, κx(1 y ) 3/ y(1 ) y ), f(r) =(b, ) whee F is C 1 in (R, ) R ( 1, 1). It follows fom usual existene theoems fo O.D.E. that we have loal existene and uniqueness of solutions fo eah (b, ) R ( 1, 1). We pove that the maximal inteval of existene is [R, ). Integation of (.4) fom R to leads to u () 1 u () = R + κ Denote v() the ight side of this expession, that is, v() = R fo eah >R.Then u () = + κ R R tu(t) dt. tu(t) dt, (3.) v() 1+v(). (3.3) Conside u a solution of (.4) (.5) on the maximal inteval [R, m). Ifm<, and ombining (3.) and (3.3), we would have v() α<1 as m. Sineu() =b + R u (t) dt, it follows that u() b + α R 1+α and so, u() M < as m. Standad agument says us then that we an extend the solution u beyond of the point = m. This ontaditionimplies that m =. Moeove, we have ontinuity on the initial paametes. On the othe hand, it is immediate that u(; b, ) = u(; b, ). Without loss of geneality, thoughout the emainde of the pape we shall assume that b 0 is fulfilled. When b =0, we also suppose 0. Notie that u(;0, 0) = 0. Next lemma will be useful fo futhe esults. The statement is the same that in Eulidean ambient [8]. We do the poof by ompleteness. Lemma 3.1 Let u 1 and u be two solutions of the O.D.E. (.6) with u 1 ( 0 ) u ( 0 ), u 1( 0 ) u ( 0 ) and u 1 ( 0 )+u 1 ( 0) <u ( 0 )+u ( 0) fo some 0 R. Thenu 1 <u and u 1 <u fo > 0. P o o f. We have two possibilities. If u 1 ( 0) <u ( 0), thenu 1 <u and u 1 <u in some inteval ( 0, 1 ). If u 1( 0 )=u ( 0 ),thenu 1 ( 0 ) <u ( 0 ). Taking into aount (3.1), u 1( 0 ) <u ( 0 ) and, again, u 1 <u and u 1 <u in some inteval ( 0, 1 ). Anyway, we onside the maximal domain ( 0, 1 ) suh that u 1 <u and u 1 <u. We pove that 1 =. On the ontay ase, the funtion u = u u 1 satisfies u( 0 )=0and u > 0 in the inteval ( 0, 1 ). Thus, at the point = 1, u ( 1 )=0,thatis,u 1( 1 )=u ( 1 ). Then (3.1) implies u 1 ( 1) <u ( 1), whih means u 1 ( 1) <u ( 1): a ontadition.

6 1174 López: An exteio bounday value poblem in Minkowski spae By setting u 1 =0, we onlude fom Lemma 3.1 Coollay 3. Let u be a solution of the O.D.E. (.6) suh that fo some 0 R, u( 0 ) 0, u ( 0 ) 0 and u( 0 )+u ( 0 ) > 0. Thenu() > 0 and u () > 0 fo > 0. We study the pofiles of solutions of the initial value poblem P. Theoem 3.3 Let u(; b, ) be a solution of P. 1. If 0(with >0if b =0), then u(; b, ) is stitly ineasing.. If <0and fo some 0 >R, u ( 0 ) 0, thenu(; b, ) is positive and thee exists 1, R< 1 0, suh that u is stitly deeasing on (R, 1 ) and stitly ineasing in ( 1, ). At = 1, u pesents a minimum. In both ases, u is a onvex funtion and u() lim = lim u () =1. P o o f. The fist item is a onsequene of Coollay 3. by taking 0 = R. Fo the seond one, we know that u is stitly deeasing of some inteval on the ight of = R. By ontadition, we assume that u is not positive and set s>rthe fist zeo of u. By uniqueness of solutions, u (s) 0.Ifu (s) > 0, Coollay 3. implies that u is positive on (s, ). Thus s would be a minimum of u and so, u (s) 0: this is a ontadition to Equation (3.1). Consequently u (s) < 0. The same agument as in Lemma 3.1 and Coollay 3. fo the funtion u yields that the both u and u ae negative in (s, ). Thus 0 (R, s). In patiula, u( 0 ) > 0. By applying Coollay 3. again, u is positive in ( 0, ): a ontadition. As a onsequene, u is always positive in the inteval of definition. Theefoe, u( 0 ) > 0 and u is stitly ineasing in ( 0, ) by Coollay 3.. Sine u (R) < 0, thee exists 1, R< 1 0, suh that u ( 1 )=0.Let 1 be the fist itial point of u. Asu( 1 ) > 0, Coollay 3. yields u > 0 in ( 1, ), poving that 1 is the only zeo of u and so, the (unique) minimum of u. We show now that u > 0. We teat both ases, with 1 = R if 0. It is lea that in the inteval [R, 1 ], u 0 and (3.1) yields u > 0. Thus, it suffies to study the ase that > 1. This ous if 0 but an integation of (.6) between 1 and gives v() = κ tu(t) dt < κu() 1 1, (3.4) whee we have bounded the integand by u() sine u is monotone ineasing. On the othe hand, equation (.6) wites (v) = κu(). Then v ()+ v() = κu(). Fom this equation and the inequality (3.4) we obtain v () =κu() v() >κu() κu() 1 >κu() κu() > 0. Sine v = u / ( 1 u ) 3/, we onlude that u > 0. Fo the last statement, onside u in ( 1, ). Sine u is ineasing, we bound (3.4) fom below by tu( 1 ). This leads to v() >κu( 1 ) 1 as. Thus (3.3) implies u 1 as.nowl Hôpital theoem yields the othe limit. As onsequene, we have

7 Math. Nah. 81, No. 8 (008) 1175 Coollay 3.4 Let u(; b, ) be a solution of P. Assume b > 0. If thee exists some point 0 suh that u( 0 )=0,thenu is monotone deeasing in (R, ), u has a unique zeo and lim = u() = lim = u () = 1. P o o f. We know fom Theoem 3.3 that <0and u < 0 in [R, ). Ifu( 0 )=0, then the monotoniity of u yields that 0 is the unique zeo. Coollay 3. implies that u<0in ( 0, ). With simila aguments as in Theoem 3.3 fo the funtion u, we onlude the asymptoti behaviou. Remak 3.5 Some positive solutions of Theoem 3.3 show examples of otational symmeti ompat stationay sufaes M in L 3 whose bounday is fomed by two axial onenti iles in paallel planes but M does not lie ompletely in the slab domain detemined by these planes: it suffies to onside a positive solution u(; b, ) that initially deeases nea the value = R andtoestitu in the inteval [R, m], wheem is any value with u(m) >b. Let u = u(; b, ) be a positive solution of P and assume that <0. We know fom Theoem 3.3 that eithe u < 0 in [R, 1 ) with u ( 1 )=0,ou () < 0 fo all R. Conside the inteval [R, 1 ), with 1 = eventually. As u is a onvex funtion, u is monotone ineasing, and it follows u () (1 u () ) = κu() u () 3/ 1 u () <κb R := M. Integating between R and 1, we obtain u () 1 u () < + M( R). The ight side in the above inequality is negative if < with = R M. (3.5) Hene R< 1. Beause 0 < 1 u < 1, in the inteval (R, ) we have u () < + M( R). We do a new integation in this inequality. Fo any, with R<, one obtains u() <b+ Setting =,weinfe b + ( R)+M ( R). ( R)+ M ( R) > 0. With the value of defined in (3.5), put δ (b) = α, α = b (1+ ) 1+κR 1+α R. Then a neessay ondition fo u() to be positive in its domain is that >δ (b). Assume now that u() is negative in some point. In this ase, we know that <0and u( 0 )=0,fosome 0 >R.Futhemoe, 0 is the unique zeo of u and u ( 0 ) < 0. Sineu is negative, one obtains fom equation (.6) that u () >κu(). (1 u () ) 3/

8 1176 López: An exteio bounday value poblem in Minkowski spae Multiplying by u () and integating between R and 0, we dedue 1 1 u ( 0 ) < 1 κb. Sine the left side is geate than 1, we onlude <δ + (b) with δ + (b) := b κb +κb +4κ. This gives a neessay ondition fo u to be not positive in the maximal inteval. Consequently, Poposition 3.6 Let b>0be a fixed numbe and let u = u(; b, ) be a solution of P. 1. If δ + (b), the solution u(; b, ) is positive in its domain.. If δ (b), the solution u(; b, ) is not positive in some point. We finish this setion poving Theoems. and.3. P o o f o f T h e o e m.. We onside again the funtion u = u u 1 defined in [R, ). Thenu satisfies u(r) =u (R) =0. Fom (3.1) and 0 <κ 1 <κ, we obtain u (R) > 0. This implies that u is positive on the left of = R. Thee exists a maximal 1 suh that u 1 <u and u 1 <u in the inteval (R, 1). If 1 <, and beause u is positive in = 1, we onlude that u 1 ( 1 ) <u ( 1 ) and u 1( 1 )=u ( 1 ).Byusing(3.1) again at = 1, one obtains u 1 ( 1) <u ( 1), thatis,u ( 1 ) > 0. Consequently, u is stitly ineasing aound the point = 1, in ontadition to the maximal popety of 1. This implies that 1 =. Poof of Theoem.3. Lemma3.1saysthatu(; b, ) <u(; b + η, ) and u (; b, ) <u (; b + η, ) fo >R. Define the funtion w() =u(; b + η, ) u(; b, ) η. Thenw(R) =w (R) =0. By using (3.1), we have u (R; b + η, ) =u (R; b, )+κη ( ) >u (R; b), and thus w (R) > 0. This implies that w is stitly ineasing on the ight of R. Let(R, 1 ) be the maximal inteval whee w is positive. If 1 <,thenw ( 1 )=0<w( 1 ), whih is false. This ontadition poves that 1 =, and onsequently, w>0, as was to be shown. 4 Unbounded stationay sufaes asymptoti to a spaelike plane As a onsequene of the above esults, it emains the study of ase <0and that u is positive and monotone deeasing. Theoem 4.1 Given R, b > 0 and <0suh that the solution u(; b, ) satisfies Then u() > 0, and u () < 0, fo R. (4.1) lim u() = lim u () =0. (4.) P o o f. We know that u < 0 and that u is monotone deeasing. By (3.1) and (4.1), u > 0, thatis,u is stitly ineasing on (R, ). Moeove we know that the numbe 0 is a lowe and uppe bound fo u and u espetively. Fom the theoy of odinay diffeential equations, the limits lim u() =l, lim u () =L exist and ae finite, with L 0 l. In patiula, u is a bounded funtion and so, L = 0. Fom (3.1), u ( ) =κl. Again,asu is bounded, lim u () =0and sine κ>0, l =0.

9 Math. Nah. 81, No. 8 (008) 1177 We summaize the above esults as follows. Let us fix b>0 and we take values vaying fom =+1to = 1. Notie that the solution u(; b, ) is defined fo all R. Fo nonnegative values of, we know that u is monotone ineasing on and u is positive in all its domain. As we ae going to deease the value of, that is, the shooting slope, u deeases on the ight of = R until a minimum. Next, u ineases again towad infinity. The funtion emains positive. Howeve, it aives a itial slope δ suh that the funtion u deeases monotonially towads. See Figue. By using a shooting agument, we shall pove that fo the slope = δ, the funtion u(; b, δ) is positive but asymptoti to the axis aoding (4.). This is ontained in Theoem.4. If S denotes the sufae obtained by otating u(; b, ) with espet to the x 3 -axis, then the family S, ( 1, 1), is a foliation of the set of spaelike dietions fom Ω R,b,whee indiate the slope of eah spaelike dietion. Only the sufae S δ is asymptoti to Π at infinity. This was stated in [C. 1] and [C. ] of Setion =0.9 =-0.5 =-0.7 =-0.75 =-0.9 Fig. Fo {0.9, 0.6, 0.3, 0, 0.5, 0.7, 0.75, 0.9}, we show the family of solutions u = u(;1,), fo R =and κ =1/. The pofile uves oesponding to = 0.5 and = 0.7 have negative shooting slope but they lie ove the -axis Poof of Theoem.4. Ifb =0,thenδ =0. Conside b>0. Set A + = { ( 1, 1); the solution u(; b, ) is positive in [R, )}, A = { ( 1, 1); the solution u(; b, ) is negative in some point of [R, )}. Poposition 3.6 assues that both sets ae not empty, with δ + (b) A +, δ (b) A. Futhemoe, and a onsequene of the peeding esults, ( 1, 1) = A + A, A + A =. Also, we have [ ɛ, 1) A +,fo some ɛ>0. On the othe hand, Lemma 3.1 implies that if A and <,then A. In the same way, if A + and >,then A +. This shows that both A + and A ae intevals of the eal line. Set δ =supa. Claim. The numbe δ satisfies δ A +. On the ontay ase, assume δ A. Then thee exists 1 >Rsuh that u( 1 ; b, δ) < 0. By the ontinuity of paamete fo O.D.E., thee exists >δsuh that u( 1 ; b, ) is also negative, poving then that A,whihis a ontadition with the definition of δ. This poves the laim. Claim. The solution u(; b, δ) satisfies u < 0 fo any R. By ontadition, we suppose that thee exists 0 >Rsuh that u ( 0 ; b, δ) =0.Sineu( 0 ) > 0, Coollay 3. and Theoem 3.3 imply u > 0 fo any > 0, and a onsequene, the point 0 is the unique minimum of u. Theefoe, u() u( 0 ) > 0 fo any R. By using the ontinuity on the initial values, thee exists nea to δ, and<δsuh that u(; b, ) is positive in all its domain. Thus A + : this ontadits the fat that δ is the supemum of A. As onlusion of both laims, the solution u = u(; b, δ) satisfies (4.1) and onsequently, by Theoem 4.1, it satisfies (4.). This shows the existene.

10 1178 López: An exteio bounday value poblem in Minkowski spae We pove the uniqueness. Let u(; b, ) be othe solution of (.6) that satisfies (4.). Then A beause in suh ase, u would be stitly deeasing and goes to as : see the agument of Theoem 3.3 fo the funtion u. Theefoe A +.If>δ,thenu (R; b, ) >u (R; b, δ) and Lemma 3.1 implies that u(; b, ) >u(; b, δ), u (; b, ) >u (; b, δ), fo any >R. Then the funtion u(; b, ) u(; b, δ) vanishes at = R, it is positive fo >Rand stitly ineasing. Thus u(; b, ) is bounded away fom 0, in ontadition with (4.). Numeially (fo example with the Mathematia softwae), it is diffiult to obtain the exat value δ due to the disetization in the omputations. Fo fixed positive numbes R, b and κ, we an appoximate values of nea to δ = δ(r, b, κ), whee the solution u(; b, ) is vey lose to the -axis. See Figue 3. The value obtained fo is geate than δ. Fo the same values as above, in Figue 4 we show as u() omes bak to inease to as.howeve,u is vey nea to the -axis in the inteval (10, 4). In fat, it would be possible to efine the omputations to obtain the pofile as lose to 0 as we desie Fig. 3 [1, 10]. Hee R =1and κ =1/4. The solution is u(;5, ). We show the uve in the inteval Fig. 4 The same solution as Fig. 3 in the inteval [1, 9] P o o f o f T h e o e m.5. The poof is a onsequene of the maximum piniple (see fo example, [7] as geneal efeene, and [5] in the apillay ontext). We do it fom a geometi viewpoint. Let u = u(; b, δ) be the solution given by Theoem.4 and denote Σ u and Σ v the sufaes detemined by the funtions u and v espetively. In the same way, let us denote H u and H v thei mean uvatues. Reall that the spaelike sufaes ae futue-dieted oiented. By a easoning by the absud, we assume that Σ v Σ u. Without loss of geneality, we suppose that thee ae points of Σ v below Σ u. Fist, we laim that the sufae Σ v lies ompletely ove the plane Π. On the ontay ase, and beause Σ v is lose to Π at infinity, thee would be points p Σ v with hoizontal tangent plane and with non-positive thid

11 Math. Nah. 81, No. 8 (008) 1179 oodinate x 3. We ompae then Σ v with the plane Π o = {x 3 = x 3 (p)} in a neighbohood of p. Ifx 3 (p) < 0, then H v (p) =κv(p) < 0: this is impossible by ompaing with Π o.ifx 3 (p) =0, the Hopf maximum piniple implies that an open set of Σ v aound p is inluded in Π o =Π,thatis,v =0in some open dis: ontadition. One poved that Σ v {x 3 > 0}, we move down Σ u until Σ u lies in the halfspae {x 3 < 0}. Thenwe move up Σ u until to aive the oiginal position. Sine we have assumed the existene of points of Σ v with lowe height than points of Σ u (with the same othogonal pojetion on Π), we infe that befoe Σ u aives to the height x 3 = b, Σ u intesets Σ v at a fist time. Let p be an intesetion point between both sufaes (p Ω). Then aound p, Σ v would lie stitly ove Σ u but both sufaes have the same mean uvatue at p: the maximum piniple togethe a ontinuation agument poves that both sufaes agee, whih is false. This ontadition poves the theoem. The same aguments used in this poof allows to estimate the hypeboli angle of ontat between a stationay sufae and a domain Ω moe geneal than a ound dis. Coollay 4. Let κ>0. Letb and σ be positive numbes. Then thee exists a onstant β 0 = β 0 (κ, b, σ) suh that the following holds. Let D be a bounded smooth domain inluded in the plane P = {x 3 = b} that satisfies an inteio ile ondition σ>0. Then any unbounded stationay sufae between D and Π obtained as the gaph of a solution v of (.) (.3) makes an ontat angle β with D along its bounday that satisfies β β 0. A simila lowe bound fo β is obtained if D satisfies an exteio ile ondition. P o o f. The inteio ile popety means that given p D, thee exists a ound dis Ω σ D of adius σ with p Ω σ. Conside a ound dis of adius σ in the plane P whose ente lies at the line L. Denote by Σ u the oesponding unbounded sufae obtained by Theoem.4. This gives an angle β 0 of ontat, with tanh β 0 = δ, aoding to the notation in Theoem.4. Given p D, and by a hoizontal tanslation, we put Σ u so Σ u = Ω σ. An agument as in Theoem.5 poves that Σ u lies below Σ v, and this yields the desied estimate fo the angle of ontat at the point p. We povethe monotonyof solutions of the poblemp and the ondition (4.) with espet to the initial values: Coollay 4.3 Conside κ, R > 0. Letb 1 and b be two positive numbes and u(; b 1,δ 1 ) and u(; b,δ ) the oesponding solutions of P with the ondition (4.). Ifb 1 <b,then fo >R. u(; b 1,δ 1 ) <u(; b,δ ), u (; b 1,δ 1 ) >u (; b,δ ), Poof. Denoteu i = u(; b i,δ i ), i =1,. We laim that thee is not 0 R suh that u 1 ( 0 ) u ( 0 ) and u 1( 0 ) u ( 0 ). On the ontay ase, some one of both inequalities is stit by the uniqueness of solutions of P. In suh ase, Lemma 3.1 assues u 1 <u and u 1 <u in ( 0, ). Thus u is bounded away fom 0, in ontadition with (4.). Sine b 1 <b, we onlude u 1 <u and u 1 >u fo any >R. If we denote δ = δ(r, b, κ) the dependene of the shooting angle on the initial value, we pove the ontinuity on thei vaiables. Sine we have dependene of the solutions of (.6) with espet to, we only onside the dependene of δ with espet to (b, κ). With the above notations, we have Theoem 4.4 The funtion δ = δ(b, κ) is ontinuous on (b, κ). P o o f. It suffies to onside b 0. Let b n b and κ n κ. If b =0,thenδ + (b n ),δ (b n ) 0. By the poof of Theoem.4, δ(b n,κ n ) 0 and 0=δ(0,κ). Thus, we assume that b>0. Again, the same poof yields δ (b n,κ n ) <δ(b n,κ n ) δ + (b n,κ n ). In patiula, the sequene δ(b n,κ n ) is bounded. Let λ be a limit point, that is, assume thee exists a subsequene (labeling in the same way) suh that δ(b n,κ n ) λ. Assume by ontadition that λ δ(b, κ) and onside the solution u(; b, λ, κ), assued by Theoem.1. By Theoems 3.3 and.4, thee exists o suh that u( o ; b, λ, κ) < 0 o u ( o ; b, λ, κ) > 0. In the fist ase, and by the ontinuity with espet to the initial onditions, thee exists (b n,δ(b n,κ n ),κ n ) lose to (b, λ, κ) suh that u( o ; b n,δ(b n,κ n ),κ n ) < 0, whih ontadits the popeties of the solutions of Theoem.4. With a simila agument, it is impossible the seond possibility. This ontadition shows the theoem.

12 1180 López: An exteio bounday value poblem in Minkowski spae Now, we shall give an estimate of the ise fo an unbounded stationay sufae. The next alulations ae simila as in Eulidean spae [6, 1, 13]. Conside u = u(; b, ) a solution of (.6) (.7) with the ondition (4.). Denote by ψ() the hypeboli angle that makes u with the -axis at eah point u(). Setting sinh ψ() = v() = u () 1 u (), Equation (.6) wites as sinh ψ +(sinhψ) = κu. As u < 0, we an take u anewvaiable,ψ = ψ(u) and so, sinh ψ +(oshψ) u = κu. Let us integate between u() and u(r) =b: κ b u() = b u() sinh ψ Letting, and by (4.), we onlude: κb = b 0 sinh ψ du +oshβ osh ψ(). du +oshβ 1. (4.3) We bound the integand in two ways. Beause u < 0, the integand is negative and so, we have κb < osh β 1. Hene that b< (osh β 1). κ On the othe hand, (sinh ψ) > 0 sine u > 0. Thus sinh β<sinh ψ() and beause sinh ψ<0, we onlude sinh β < sinh ψ R < sinh ψ. By intoduing this inequality into (4.3), we aive κb > sinh β b +oshβ 1. R Hene we an obtain a lowe bound fo b = u(r). As onlusion, Theoem 4.5 Let κ>0and let Ω be a ound dis of adius R. Conside an unbounded stationay sufae obtained by moving up Ω a distane h>0fom the plane Π. Assume that β is the hypeboli angle of ontat with Ω along its bounday. Then the height h of the dis satisfies the inequalities: ( ) sinh β κr + sinh β (osh β 1) + <h< (osh β 1). κ κr κ As onsequene, b = ( ) 1 (osh β 1) + O κ R as R. In Table 1, we pesent some numeial omputations of the ontat angle β.

13 Math. Nah. 81, No. 8 (008) 1181 R b δ =tanhβ lowe estimate height h uppe estimate Table 1 Values of the angle of ontat β fo an unbounded stationay sufae and ompaison of the height estimates obtained in Theoem 4.5. Hee κ =1. Aknowledgements This wok has been patially suppoted by a MEC-FEDER gant No. MTM Refeenes [1] L. Alías and J. Pasto, Spaelike sufaes of onstant mean uvatue with fee bounday in the Minkowski spae, Classial Quantum Gavity 16, (1999). [] J. L. Babosa and V. Olike, Spaelike hypesufaes with onstant mean uvatue in Loentz spae, Mat. Contemp. 4, 7 44 (1993). [3] D. Bill and F. Flahety, Isolated maximal sufaes in spaetime, Comm. Math. Phys. 50, (1976). [4] Y. Choquet Buhat and J. Yok, The Cauhy poblem. Geneal Relativity and Gavitation, edited by A. Held (Plenum Pess, New Yok, 1980). [5] P. Conus and R. Finn, On apillay fee sufaes in a gavitational field, Ata Math. 13, (1974). [6] R. Finn, Equilibium Capillay Sufaes (Spinge-Velag, Belin, 1986). [7] D. Gilbag and N. S. Tudinge, Ellipti Patial Diffeential Equations of Seond Ode (Spinge-Velag, Belin 1983). [8] W. E. Johnson and L. M. Peko, Inteio and exteio bounday value poblems fom the theoy of the apillay tube, Ah. Ration. Meh. Anal. 9, (1968). [9] R. López, Stationay liquid dops in Loentz Minkowski spae, Po. R. So. Edinb., Set. A, Math., to appea. [10] R. López, Spaelike hypesufaes with fee bounday in the Minkowski spae unde the effet of a timelike potential, Comm. Math. Phys. 66, (006). [11] J. E. Masden and F. J. Tiple, Maximal hypesufaes and foliations of onstant mean uvatue in geneal elativity, Phys. Rep. 66, (1980). [1] D. Siegel, Height estimates fo apillay sufaes, Paifi J. Math. 88, (1980). [13] B. Tukington, Height estimates fo exteio poblems of apillaity type, Paifi J. Math. 88, (1980). [14] T. Vogel, Symmeti unbounded liquid bidges, Paifi J. Math. 103, (198).