FORMATION OF TRAPPED SURFACES II. 1. Introduction

FORMATION OF TRAPPED SURFACES II SERGIU KLAINERMAN, JONATHAN LUK, AND IGOR RODNIANSKI 1. Introduction. Geometry of a null hypersurface As in [?] we consider a region D = D(u, u ) of a vacuum spacetime (M, g) spanned by a double null foliation generated by the optical functions (u, u) increasing towards the future, 0 u u and 0 u u. We denote by H u the outgoing null hypersurfaces generated by the level surfaces of u and by H u the incoming null hypersurfaces generated level hypersurfaces of u. We write S u,u = H u H u and denote by H (u 1,u u ), and H (u 1,u ) u the regions of these null hypersurfaces defined by u 1 u u and respectively u 1 u u. Let L = g αβ α u β, L = g αβ α u β, L be the geodesic vectorfields associated to the two foliations and define, g(l, L) := Ω = g αβ α u β u (1) Observe that the flat value 1 of Ω is 1. As well known, our space-time slab D(u, u ) is completely determined (for small values of u, u ) by data along the null, characteristic, hypersurfaces H 0, H 0 corresponding to u = 0, respectively u = 0. Following [?] we assume that our data is trivial along H 0, i.e. assume that H 0 extends for u < 0 and the spacetime (M, g) is Minkowskian for u < 0 and all values of u 0. Moreover we can construct our double null foliation such that Ω = 1 along H 0, i.e., Ω(0, u) = 1, 0 u u. () We denote by r = r(u, u) the radius of the -surfaces S = S(u, u), i.e. S(u, u) = 4πr. We denote by r 0 the value of r for S(0, 0), i.e. r 0 = r(0, 0). For simplicity we assume r 0 = 1. Throughout this paper we work with the normalized null pair (e 3, e 4 ), e 3 = ΩL, e 4 = ΩL, g(e 3, e 4 ) =. 1991 Mathematics Subject Classification. 35J10. 1 Note that our normalization for Ω differ from that of [?] 1

SERGIU KLAINERMAN, JONATHAN LUK, AND IGOR RODNIANSKI Given a -surfaces S(u, u) and (e a ) a=1, an arbitrary frame tangent to it we define the Ricci coefficients, Γ (λ)(µ)(ν) = g(e (λ), D e(ν) e (µ) ), λ, µ, ν = 1,, 3, 4 (3) These coefficients are completely determined by the following components, χ ab = g(d a e 4, e b ), χ ab = g(d a e 3, e b ), η a = 1 g(d 3e a, e 4 ), η a = 1 g(d 4e a, e 3 ) ω = 1 4 g(d 4e 3, e 4 ), ω = 1 4 g(d 3e 4, e 3 ), (4) ζ a = 1 g(d ae 4, e 3 ) where D a = D e(a). We also introduce the null curvature components, α ab = R(e a, e 4, e b, e 4 ), α ab = R(e a, e 3, e b, e 3 ), β a = 1 R(e a, e 4, e 3, e 4 ), β a = 1 R(e a, e 3, e 3, e 4 ), (5) ρ = 1 4 R(Le 4, e 3, e 4, e 3 ), σ = 1 R(e 4, e 3, e 4, e 3 ) 4 Here R denotes the Hodge dual of R. We denote by the induced covariant derivative operator on S(u, u) and by 3, 4 the projections to S(u, u) of the covariant derivatives D 3, D 4. Observe that, ω = 1 4(log Ω), ω = 1 3(log Ω), η a = ζ a + a (log Ω), η a = ζ a + a (log Ω) (6) We recall the integral formulas for a scalar function f in D, d ( df f = du S(u,u) S(u,u) du + Ωtrχf) = d ( df f = du du + Ωtrχf) = In particular, S(u,u) S(u,u) dr du = 1 Ωtrχ, 8π S(u,u) S(u,u) S(u,u) Ω ( e 4 (f) + trχf ) Ω ( e 3 (f) + trχf ) (7) dr du = 1 Ωtrχ (8) 8π S(u,u) We also recall the following commutation formulas: We record below commutation formulae between and 4, 3 : see for example Lemma 3.1.3 in [?]

TRAPPED SURFACES 3 Lemma.1. For a scalar function f: [ 4, ]f = 1 (η + η)d 4f χ f (9) [ 3, ]f = 1 (η + η)d 3f χ f, (10) For a 1-form tangent to S: [ 4, a ]U b = χ ac c U b + ac β b U c + 1 (η a + η a )D 4 U b χ ac η b U c + χ ab η U [ 3, a ]U b = χ ac c U b + ac β b U c + 1 (η a + η a )D 3 U b χ ac η b U c + χ ab η U In particular, [ 4, div ]U = 1 trχ div U ˆχ U β U + 1 (η + η) 4U η ˆχ U [ 3, div ]U = 1 trχ div U ˆχ U + β U + 1 (η + η) 3U η ˆχ U.. Christodoulou s heuristic argument. We recall here the assumptions needed in Christodoulou s heuristic argument for the formation of a trapped surface as described in [?]. As mentioned above we assume that our data is trivial along H 0, i.e. assume that H 0 extends for u < 0 and the spacetime (M, g) is Minkowskian for u < 0 and all values of u 0. We introduce a small parameter δ > 0 and restrict the values of u to 0 u δ, i.e. u = δ. Main Assumptions. We assume that throughout D = D(u, u ) we have the following estimates: MA1. For small δ, Ω is comparable with its standard value in flat space, i.e. MA. The Ricci coefficients χ, ω, η, χ, ω verify Ω = 1 + O(δ 1/ ). MA3. Also for some c > 0, ˆχ, ω = O(δ 1/ ), trχ, η = O(1), ˆχ, trχ + r, ω = O(δ1/ ). η = O(δ 1/+c ).

4 SERGIU KLAINERMAN, JONATHAN LUK, AND IGOR RODNIANSKI Note that in view of (8) we also have, dr du = 1 + O(rδ1/ ), dr du = O(r) (11) Thus, for δ sufficiently small, we infer that r is decreasing along the incoming null hypersurfaces and remains bounded, 0 r r 0 + 1 =, in D. Christodoulou s argument for the formation of trapped surfaces in [?] rests on the equations, 4 trχ + 1 (trχ) = ˆχ 1 (trχ) ωtrχ 3 ˆχ + 1 trχˆχ = η + ω ˆχ 1 trχˆχ + η η In view of our Ricci coefficients assumptions we can rewrite, Multiplying the second equation by ˆχ, 4 trχ = ˆχ + O(δ 1/ ) 3 ˆχ + 1 trχˆχ = O(δ 1/+c ) Using also our assumptions for u, u, Ω we deduce, Integrating (1) we obtain, 4 ˆχ + trχ ˆχ = O(δ 1+c ) d du trχ = ˆχ + O(δ 1/ ) (1) d du ˆχ + trχ ˆχ = O(δ 1+c ) (13) trχ(u, u) = In view of our assumptions for trχ and dr du Therefore, r(u, 0) u 0 ˆχ (u, u ) du + O(δ 1/ ) (14) d du (r ˆχ ) = r d du ˆχ + r dr du ˆχ = r [ trχ ˆχ + O(δ 1+c ) ] + r [ 1 + O(rδ c ) ] ˆχ = r O(δ 1+c ). r ˆχ (u, u) = r (0, u) ˆχ (0, u) + r O(δ 1+c ) As in [] we freely prescribe ˆχ along the initial hypersurface H (0,δ) 0, i.e. ˆχ(0, u) = ˆχ 0 (u) = O(δ 1/ ) (15)

for some traceless tensor ˆχ 0. We deduce, (need 0 < c 1 ), or, since u δ and r(u, u) = r 0 + u u + O(δ c ), Thus, returning to (14), TRAPPED SURFACES 5 ˆχ (u, u) = r (0, u) r (u, u) ˆχ 0 (u) + O(δ 1+c ) ˆχ (u, u) = r 0 r (u, 0) ˆχ 0 (u) + O(δ 1+c ) trχ(u, δ) = = We have thus proved the following. r(u, 0) r 0 r (u, 0) r(u, 0) r 0 r (u, 0) u 0 u 0 ˆχ 0 (u )du + O(δ c ) ˆχ 0 (u )du + O(δ c ) Proposition.3. Under the assumptions MA1- MA3 we have, for sufficiently small δ > 0 and fixed c > 0, trχ(u, δ) = δ r(u, 0) r 0 ˆχ r 0 (u )du + O(δ c ) (16) (u, 0) 0 Since r(u, u) = r 0 u + u + O(δ c ) formula (16) can also be written in the form, trχ(u, δ) = r(u, δ) r 0 r (u, δ) Corollary.4. The necessary condition to have trχ(u, u = δ) < 0 δ δ 0 ˆχ 0 (u )du + O(δ c ) (17) r(u, 0) < ˆχ r0 0 + O(δ c ) (18) 0 for sufficiently small δ > 0. Since r(u, 0) = r 0 u + O(δ c ), condition (18) can also be written in the form, (r 0 u) r 0 < δ 0 ˆχ 0 + O(δ c ) (19) 3. Change of foliation 3.1. Main transformation formula. To improve on (18) we plan to change the u foliation along u = δ and compute the corresponding incoming expansion trχ. More precisely, given the foliation induced by u, we look for a new foliation v = v(u, ω) defined by the equations u v = e f, v S0 = u S0 = 0 u f = 0, f S = f 0 (0)

6 SERGIU KLAINERMAN, JONATHAN LUK, AND IGOR RODNIANSKI with f 0 a function on S 0 = S(0, δ) to be carefully chosen later. NOTE CHANGE: BEFORE WE HAD 3 v = e f WHICH LEADS TO THE UNDESIRED TERM e f log Ω IN THE EQUATION FOR G. We introduce the new null frame adapted to the v-foliation, e 3 = e 3, e a = e a e f Ωe a (v)e 3, e 4 = e 4 e f Ωe a (v)e a + e f Ω v e 3 (1) Indeed since u = Ω 3 we have, e a(v) = e a (v) e f Ωe a (v)e 3 (v) = e a (v) e f e a (v) u (v) = 0. Also, ince e 3 is orthogonal to any vector tangent to H we easily check that We prove the following. g(e a, e b) = g(e a, e b ) = δ ab, g(e 4, e a) = g(e 4, e 4) = 0, g(e 3, e 4) =. Lemma 3.. The new incoming expansion trχ verifies the transformation formula, trχ = trχ e f div (e f F ) trχ F 4ˆχ bc F b F c (η + ζ) F () where F a = e f Ω a v and trχ, ζ, trχ, ˆχ, ω are connection coefficients for the given double null foliation (u, u). Proof. We have, χ (e a, e b) := g(d ae 4, e b) = g(d a e 4, e b) e f Ω 1 e a (v)g(d 3 e 4, e b) Now, writing e 4 = e 4 F + F e 3 with F = F c e c and e b = e b F b e 3, g(d a e 4, e b) = g ( D a (e 4 F + F e 3 ), e b F b e 3 ) = χ(e a, e b ) F b ζ a a F b + F b g(d a F, e 3 ) + F g(d a e 3, e b F b e 3 ) = χ ab ζ a F b a F b F b χ(f, e a ) + F χ ab Also, = χ ab ζ a F b a F b F b F c χ ac + F χ ab g(d 3 e 4, e b) = g ( D 3 (e 4 F + F e 3 ), e b F b e 3 ) = g(d 3 e 4, e b ) F b g(d 3 e 4, e 3 ) 3 F b Hence, = η b + 4F b ω 3 F b χ ab = χ ab ζ b F a a F b F b χ(f, e b ) + F χ ab F a ( ηb + 4F b ω 3 F b ) = χ ab a F b + F a 3 F b ζ b F a F a η b + ( ) F χ ab F b F c χ ac 4ωFa F b By symmetry in a, b we deduce the formula, χ ab = χ ab ( a F b + b F a ) + 3 (F a F b ) (ζ b + η b )F a + (ζ a + η a )F b (3) + ( ) F χ ab F b F c χ ac F a F c χ bc 4ωFa F b

TRAPPED SURFACES 7 and, taking the trace, trχ = trχ div F + 3 F (η + ζ) F + ( F trχ χ bc F b F c ) 4ω F = trχ div F + 3 F (η + ζ) F ˆχ bc F b F c 4ω F We next calculate 3 F using (0) and the commutation formula [ 3, ]h = ( log Ω) 3 h χ h or, [ u, ]h = Ωχ h Since u f = 0 and F = Ω 1 e f v we deduce, or, i.e., from which, Therefore, as desired. u F a = u (Ωe f v) = Ωe f u v + u Ωe f v = Ωe f u v Ωe f Ωχ v + u Ωe f v = Ω f Ω e f χ v + u Ωe f v = Ω f Ωχ F Ω 1 u ΩF 3 F = F χ F Ω 1 3 ΩF = F χ F + ωf 3 F + 1 trχf = f ˆχ F + ωf (4) 3 F = trχ F + F f ˆχ bc F b F c + 4ω F trχ = trχ div F (η + ζ) F ˆχ bc F b F c 4ω F trχ F + F f ˆχ bc F b F c + 4ω F = trχ div F + F f trχ F 4ˆχ bc F b F c (η + ζ) F = trχ e f div (e f F ) + F f trχ F 4ˆχ bc F b F c (η + ζ) F Remark. Note that we can eliminate ζ from the formula () by writing the term (η + ζ) F = 4η F Ω 1 Ω F Thus, trχ = trχ e f Ωdiv (Ω 1 e f F ) trχ F 4ˆχ bc F b F c 4η F (5)

8 SERGIU KLAINERMAN, JONATHAN LUK, AND IGOR RODNIANSKI To understand how trχ differs from trχ it only remains to derive a transport equation for div G with G = e f F. 3.3. Transport equation for div G. In view of (4) and e 3 (f) = 0 we have for G := e f F. 3 G + 1 trχg = e f f ˆχ G + ωg (6) To derive a transport equation for div G we make us of the following Lemma 3.4. Assume that the S-tangent vectorfield V verifies an equation of the form, Then, 3 V + 1 trχv = ˆχ V + W 3 (div V ) + 1 trχdiv V = div W + W (log Ω) ˆχ V trχ V + ( trχζ ˆχ ζ ˆχ (log Ω) ) V Proof. 3 (div V ) + 1 trχdiv V = div ( ˆχ V + W ) 1 trχ V + [ 3, div ]V We make use of the commutation formula, see lemma.1, Therefore, [ 3, div ]V = 1 trχdiv V ˆχ V + ( β η ˆχ ) V + (log Ω) 3 V 3 (div V ) + trχdiv V = div ( ˆχ V + W ) ˆχ V + ( β 1 trχ η ˆχ ) V + (log Ω) 3 V = div W ˆχ V + ( div ˆχ + β 1 trχ η ˆχ ) V + (log Ω) ( 1 ) trχv ˆχ V + W = div W + W (log Ω)) ˆχ V + ( div ˆχ 1 trχ + β η ˆχ 1 trχ (log Ω) ˆχ (log Ω)) V

TRAPPED SURFACES 9 Using the Codazzi equation, div ˆχ = 1 trχ + β + ζ (ˆχ 1 trχ) as well as η = ζ + (log Ω) we derive, div ˆχ 1 trχ + β η ˆχ 1 trχ (log Ω) ˆχ (log Ω)) = trχ ζ (ˆχ 1 trχ η ˆχ 1 trχ (log Ω) ˆχ (log Ω) Hence, as desired. = trχ ˆχ (ζ + η + (log Ω)) + 1 trχ(ζ + (log Ω)) = trχ ˆχ (ζ + (log Ω)) + trχη 3 (div V ) + trχdiv V = div W + W (log Ω) ˆχ V trχ V + ( trχη ˆχ ζ ˆχ (log Ω) ) V Applying the lemma to equation (6) we derive, 3 (div G) + trχdiv G = div W + W (log Ω) ˆχ G trχ G with W = e f f + ωg. Thus, + ( trχη ˆχ ζ ˆχ (log Ω) ) G div W + W (log Ω) = div (e f f) + e f (log Ω) f + div (ωg) + (log Ω)ωG We deduce the following transport equation for div G, with error term, 3 (div G) + trχdiv G = div (e f f) + ωdiv G + Err 1 (7) Err 1 = e f (log Ω) f ˆχ G trχ G + ( trχη ˆχ ζ ˆχ (log Ω) + ω + ω log Ω ) G In the same manner we deduce a transport equation for the principal term div (e f f) on the right hand side of (7). Indeed, since 3 f = 0 we derive, Therefore, using lemma 3.4, 3 ( f) + 1 trχ f = ˆχ f 3 div (e f f) + trχdiv (e f f) = ˆχ (e f f) trχ e f f + ( trχη ˆχ ζ ˆχ (log Ω) ) e f f.

10 SERGIU KLAINERMAN, JONATHAN LUK, AND IGOR RODNIANSKI We summarize the results of this subsection in the following proposition. We summarize the results of this subsection in the following proposition. Proposition 3.5. Let v, f be defined according to (0), F = Ω 1 e f v and G = e f F. The trace of the null second fundamental form χ, relative to the new frame (1), is given by the formula (), i.e., F verifies the transport equation and div G verifies, where, Also, with error term, trχ = trχ e f div G trχ F 4ˆχ bc F b F c (η + ζ) F (8) 3 F + 1 trχf = f ˆχ F + ωf (9) 3 (div G) + trχdiv G = div (e f f) + Err 1 (30) Err 1 = e f (log Ω) f ˆχ G trχ G + ( trχζ ˆχ ζ ˆχ (log Ω) + ω + ω log Ω ) G 3 f = 0 (31) 3 ( f) + 1 trχ f = ˆχ f (3) 3 [e f div (e f f)] + trχ[e f div (e f f)] = Err (33) Err = ˆχ ( f f f ) trχ f + ( trχη ˆχ ζ ˆχ (log Ω) ) f 3.6. Additional assumptions. To proceed we need to make stronger assumptions than those of section.. More precisely, we need, in addition MA1 -MA3 the following, MA-S. The Ricci coefficients η, η, log Ω verify the stronger assumptions, η, η = O(δ c ) MA3-S For a fixed c > 0, η, η = O(δ 1/+c ), χ, β = O(δ c )

TRAPPED SURFACES 11 As a corollary of proposition 3.5 and these assumptions we deduce first, trχ = trχ e f div (G) + r F + F O(δ c ) (34) From the equation (3) 3 ( f) + 1 trχ f = ˆχ f we deduce, Therefore, u (r f ) = O(δ 1/ ) r f r f = r 0 f 0 (1 + O(δ 1/ ) ) (35) We can also deduce in the same manner an estimate for r f. Indeed, differentiating (3) and commuting with 3, according to lemma.1 we deduce, 3 ( f) + trχ f = ˆχ f + O(δ c )(1 + 1 r ) f Note that the 1 term is due to the contribution of the term trχη f which appear in the commutation r lemma. Hence, since r r 0 r we deduce using (35), and we infer that, u (r f) = O(δ 1/ )r f + O(δ c )(1 + 1 r )r f = O(δ 1/ )r f + O(rδ c )r 0 f 0 r f C ( r 0 f 0 + O(δ c )r 0 f 0 ) (36) Proceeding in the same manner with (33) we derive, for H = e f div (e f f) We deduce, Hence, or, Now, from equation (38), 3 H + trχh = O(δ 1/ ) ( f + f ) + O(δ c )(1 + 1 r ) f u (r H) = O(δ 1/ )r f + O(δ c )r f = O(δ c ) [ r 0 f 0 + r 0 f 0 ] r H = r 0H 0 + O(δ c ) [ r 0 f 0 + r 0 f 0 ] r div (e f ) f = r 0 (e f 0 ) + O(δ c ) [ r 0 f 0 + r 0 f 0 ] e f 0 (37) 3 F + 1 trχ F = f + O(δ1/ )F we deduce, u (r F ) = O(δ 1/ )r F + r f = O(δ 1/ )r F + r 0 f 0 ( 1 + O(δ c ) )

1 SERGIU KLAINERMAN, JONATHAN LUK, AND IGOR RODNIANSKI and therefore, since F 0 = e f 0 v 0 = 0, r F = ur 0 f 0 ( 1 + O(δ c ) ) (38) with C > 0 independent of δ or f 0. We next calculate F. Using the commutation lemma.1 we deduce, 3 F + trχ F f + O(δ 1/ ) F + O(δ c ) F Thus, according to (36) u (r F ) r f + O(δ 1/ )r F + O(δ c )r F O(δ 1/ )r F + C ( r0 f 0 + O(δ c )r 0 f 0 ) + O(δ c )rr 0 f 0 We deduce, r F C ( r0 f 0 + O(δ c )r 0 f 0 ) (39) Since G = e f F we also deduce, r G C ( r0 f 0 + O(δ c )r 0 f 0 ) e f 0 (40) Next we calculate div G from (30) which we write in the form, 3 (div G) + trχdiv G = div (e f f) + O(δ c )I 0 e f 0 I 0 := ( r0 f 0 + O(δ c )r 0 f 0 ). Hence, making use of (37) u (r div G) = r div (e f f) + O(δ c )r I 0 e f 0 = r0 (e f 0 ) + O(δ c )I 0 + O(δ c )r I 0 e f 0 Therefore, r div G = ur0 (e f 0 ) + O(δ c )I 0 (41) Finally, going back to (34), and formula (38) for F, trχ = trχ e f div (G) + r F + F O(δ c ) = trχ + ur r0e f 0 (e f 0 ) + O(r δ c )I 0 + r 3 u r0 f 0 ( 1 + O(δ c ) ) = trχ + ur r0( f0 + f 0 ) + r 3 u r0 f 0 ( 1 + O(δ c ) ) + O(r δ c )I 0 ( = trχ + ur 0 f r 0 + [ 1 + u/r ( 1 + O(δ c ) ] f 0 ]) + O(r δ c )I 0 We summarize the result in the following proposition.

TRAPPED SURFACES 13 Proposition 3.7. Assume that MA1-MA3 and MA-S, MA3-S are verified in the space-time region D(u, δ) and f, v defined according to (0) The the expansion trχ of the v foliation verifies, for all 0 u u and 0 u δ, with I 0 = ( r0 f 0 + O(δ c )r 0 f 0 ) verifies, ( trχ = trχ + ur 0 f r 0 + [ 1 + u/r ( 1 + O(δ c ) ] f 0 ]) + O(r δ c )I 0 (4) In particular, if δ > 0 is sufficiently small, trχ trχ + ur [ 0 f0 + ( 1 + u )] f0 + O(r δ c )I r 0 (43) r 3.8. Main equation. We now combine the results of propositions.3 and 3.7. For simplicity we shall also assume that r 0 = 1. According to proposition.3 we have, Thus, inserting in (4) trχ (u, δ) r + u r where r = r(u, δ) and trχ(u, δ) = r(u, δ) 1 r (u, δ) δ 0 ˆχ 0 (u )du + O(δ c ) ( f 0 + [ 1 + u/r ( 1 + O(δ c ) ] f 0 ]) 1 r M 0 + O(r δ c )I 0 M 0 = δ 0 ˆχ 0 (u )du. Now, along a level surface 3 S 1 := {v = 1} H u we can express both u and r as functions along S 0 which we denote by U = U(f 0 ) and R = R(f 0 ). In fact, since v = ue f 0 we deduce U = e f 0. Moreover since according to (11) dr = 1 + du O(δ1/ r) = 1 + O(δ 1/ ) we can write, R = 1 U + O(δ 1/ ) U To have trχ non-positive along S v0 we need, R + U ( f R 0 + [ 1 + U/R ( 1 + O(δ c ) ] f 0 ]) 1 ( ) M0 O(δ c )I r 0 We deduce the following. Corollary 3.9. A necessary condition for S 1 to be a trapped surface, is that, f 0 + [ 1 + U/R ( 1 + O(δ c ) ] f 0 ] + R U 1 ( ) M0 O(δ c )I 0 U where, 3 with v the deformation function defined by (0) U = e f 0, R = 1 e f 0 + O(δ 1/ ) e f 0 (44)

14 SERGIU KLAINERMAN, JONATHAN LUK, AND IGOR RODNIANSKI Note that the inequality is only meaningful in the domain D, i.e. for U = e f 0 u, or for δ sufficiently small, R > 1 u (45) We now re-express (44) with respect to R = R(f 0 ). We have, On the other hand, In view of formula (7) and the equation, we deduce, Hence, from which we deduce, u (r u r) = 1 8π u = 1 16π R = R df (f 0) f 0 R = R df (f 0) f 0 + d R d f (f 0) f 0 dr df (f 0) = dr du du df = ur e f 0 d R d f (f 0) = ur e f 0 + u re f 0 3 trχ + 1 (trχ) = ωtrχ ˆχ S(u,u) S(u,u) Ωtrχ = 1 8π Ω trχ 1 8π S(u,u) S(u,u) r ur + ( u r) = Ω + ro(δ 1/ ) = 1 + O(δ 1/ ) Ω ( e 3 (Ωtrχ) + Ωtrχtrχ ) Ω ˆχ = 1 + ro(δ 1/ ) r ur = O(δ 1/ ). (46) Hence, R = f 0 e f 0 u r = f 0 e ( f 0 1 + O(δ 1/ ) ) R = f 0 e f 0 u r + ( ur e f 0 + u re ) f 0 f 0 = e f 0 ( f ) 0 1 + O(δ 1/ + ( 1 + O(δ 1/ )e f 0 f 0 + R 1 e f 0 f 0 O(δ 1/ ) = e ( f 0 f 0 f 0 ) + O(δ 1/ ) ( f 0 + (1 + R 1 ) f 0 )

TRAPPED SURFACES 15 Thus, Note that, f 0 = e f 0 R + O(δ 1/ ) f 0 f 0 = e f 0 R + f 0 + O(δ 1/ ) ( f 0 + (1 + R 1 ) f 0 ) = e f 0 R + e f 0 R + O(δ 1/ ) ( f 0 + (1 + R 1 ) f 0 ) The left hand side of (44) becomes, L = f 0 + [ 1 + U/R ( 1 + O(δ c ) )] f 0 ] + R U Hence, L = e f 0 = e f 0 R e f 0 R + (1 + e f 0 R 1 )e f 0 R + Re f 0 + O(δ c )R 1 e f 0 R + O(δ 1/ ) ( e f 0 R + e f 0 R ) [ R + R 1 R ( 1 + O(δ c ) ) + R + O(δ 1/ ) ( ] R + e f 0 R ) The inequality (44) becomes, or, R + R 1 R ( 1 + O(δ c ) ) + R + O(δ 1/ ) ( R + e f 0 R 1 (M 0 O(δ c )I 0 ) R + R 1 R ( 1 + O(δ c ) ) + R 1 M 0 + O(δ c )J where J is a fixed smooth function depending only on R and R. We deduce the following. Proposition 3.10. A necessary condition such that S 1 is a trapped surface is that, for δ > sufficiently small there exists a smooth function R on 0 verifying R > 1 u and the differential inequality, R + ( 1 + O(δ c ) ) R 1 R + R 1 M 0 + O(δ c )J (47) To proceed we need to make the assumption R δ c/. Hence, it suffices to prove the inequality: or, for δ sufficiently small, R + R 1 R + R 1 M 0 + O(δ c/ )J R + R 1 R + R < 1 M 0. (48) In order that the initial surface, corresponding to R = 1, is not trapped we need 1 M 0 < 1. We also note, by the maximum principle that, max R < 1 max M 0

16 SERGIU KLAINERMAN, JONATHAN LUK, AND IGOR RODNIANSKI Thus, 1 u < 1 max S 0 M 0. (49) is a necessary condition for the formation of a trapped surface. Recall that u is the maximum of u for which our assumptions are satisfied in D(u, δ). Is it sufficient? Performing the transformation R = e φ we derive, φ + 1 < 1 M 0 e φ 4. Solutions to the deformation equation on a fixed sphere (S, γ). In this section we provide examples of solutions to our main deformation equation, φ + 1 < Me φ (50) on a smooth, compact, dimensional Riemnannian manifold S, diffeomorphic to the standard sphere, with strictly positive Gaussian curvature K. We define r = r(s) such that S = 4πr and define, k m = min r K, k M = max r K S S We consider geodesic balls B(p, ɛ) = B(p, ɛ) for sufficiently small ɛ > 0. We start by proving the following lemma. Lemma 4.1. Given a ball B(p, ɛ) S, there exists a function w ɛ, smooth outside the point p, such that w ɛ + K = 4πδ p (51) where δ p is the Dirac measure at p. Moreover, if λ denotes the distance function from p, w ɛ = χ ɛ log λ + v (5) with v C 1.1 (S), v(p) = 0, smooth in S \ {p} and χ ɛ a smooth cutoff function, { χ ɛ = 1 on B(p, ɛ) χ ɛ = 0 on B(p, ɛ) Assuming the lemma true we consider the cut-off function { ϕ ɛ = 0 on B(p, ɛ/) ϕ ɛ = 1 on B ɛ (p)

TRAPPED SURFACES 17 and define w ɛ = ϕ ɛ w ɛ. Note that w ɛ verifies the following properties: w ɛ = 0, on B(p, ɛ/) w ɛ = log ɛ + O(1), on S \ B(p, ɛ/) w ɛ = O(ɛ log ɛ) on S \ B(p, ɛ/) w ɛ + K = 0 on S \ B(p, ɛ) Consider now the function w ɛ = Λw ɛ, for a fixed constant Λ and observe that, on S \ B(p, ɛ), we must have, w ɛ + 1 = ΛK + 1 < 0 provided that Λ > km 1. Thus, with a fixed choice of Λ > km 1 we have, w ɛ = 0, on B(p, ɛ/) w ɛ = Λ log ɛ + O(1), on S \ B(p, ɛ/) (54) w ɛ = O(ɛ log ɛ) on S \ B(p, ɛ/) w ɛ + 1 < 0 on S \ B(p, ɛ) It remains to check under what conditions for M, the function w ɛ verifies (50). Clearly, on S \B(p, ɛ), (50) is trivially verified in view of the fact that M 0. Now let M ɛ = inf B(p,ɛ) M. Thus, for some constant C, Me wɛ M ɛ e wɛ Cɛ Λ M ɛ, w ɛ + 1 = O(ɛ log ɛ) Hence, to have (50) verified in B(p, ɛ) we need, This proves the following. O(ɛ log ɛ) < M ɛ ɛ Λ Proposition 4.. Let M ɛ = min B(p,ɛ) M and let Λ > (min S K) 1. Assume that, for some universal constant C > 0, (53) M ɛ > Cɛ Λ log ɛ (55) Then, for sufficiently small ɛ > 0, there exists a function φ ɛ verifying the inequality (50) and such that min φ ɛ > log ɛ + O(1) (56) φ ɛ = O(ɛ 1 log ɛ), φ ɛ = O(ɛ log ɛ). (57) In remains to prove lemma 4.1. This is a standard argument, see for example chapter in [?], which we sketch below. Let λ be the geodesic distance function from p. In a neighborhood of p we can write, ds = dλ + a (λ, θ)dθ, a(0) = 0, da (0) = 1. dλ

18 SERGIU KLAINERMAN, JONATHAN LUK, AND IGOR RODNIANSKI Let h be the geodesic curvature of the level curves of λ, i.e. denoting, e = γ( θ, θ ) 1/ θ the unit tangent vector to these curves, h verifies the second variation formula, or, h = γ( e λ, e) = a 1 λ a λ h = h K λa + Ka = 0. (58) Since a(0) = 0, da(0) = 1 we deduce from (58) that dλ λa(0) = 0. Consequently, Now, a(λ) = λ + O(λ 3 ), λ a(λ) = 1 + O(λ ), h(λ) = λ 1 + O(λ) Thus, for δ < ɛ converging to 0, Hence, passing to the limit, Note also that, S\B(p,δ) log λ = λ 1 h λ = O(1) S (χ ɛ log λ)dv γ = π + O(δ) S (χ ɛ log λ)dv γ = π for any smooth test function χ supported in B(p, ɛ). We now solve the equation, where, f is the bounded function { f = K + (χ ɛ log λ) f = 0 χ (χ ɛ log λ)dv γ = πχ(p) (59) S v = f. (60) on S \ {p} at p Note that (60) admits a C 1,1 solution in view of the fact that f L (S) and, fdv γ = Kdv γ + (χ ɛ log λ)dv γ = 4π 4π = 0. S We can also normalized v such v(p) = 0. We now define, S S w ɛ = χ ɛ log λ v

and note that, Moreover, in view of (59), as desired. TRAPPED SURFACES 19 w ɛ + K = 0, on S \ {p} w ɛ + K = 4πδ Remark 4.3. Note that the proof of the proposition only requires the existence of a smooth function w on S which verifies w + 1 < 0 in the complement of a closed domain D for which inf D M : M D > 0. Indeed if such a function exists we can produce a solution to our inequality simply by taking φ = log s + w for a sufficiently small constant s > 0. Indeed, with such a choice (50) is automatically satisfied in the complement of D. In D, φ = w, e φ = s 1 e w and therefore we need max D [ e w ( w + 1) ] < s 1 M D. Finally we note that such solutions can easily be constructed for balls B(p, δ) with δ < i p, the radius of injectivity of S at p. Department of Mathematics, Princeton University, Princeton NJ 08544 E-mail address: seri@@math.princeton.edu Department of Mathematics, Princeton University, Princeton NJ 08544 E-mail address: irod@@math.princeton.edu