ENTIRE DOWNWARD TRANSLATING SOLITONS TO THE MEAN CURVATURE FLOW IN MINKOWSKI SPACE

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1 ENTIRE DOWNWARD TRANSLATING SOLITONS TO THE MEAN CURVATURE FLOW IN MINKOWSKI SPACE JOEL SPRUCK AND LING XIAO Abstract. In this paper, e study entire translating solutions ux to a mean curvature flo equation in Minkoski space. We sho that if Σ = {x, ux x R n } is a strictly spacelike hypersurface, then Σ reduces to a strictly convex rank k soliton in R k, after splitting off trivial factors hose blodon converges to a multiple λ 0, of a positively homogeneous degree one convex function in R k. We also sho that there is nonuniqueness as the rotationally symmetric solution may be perturbed to a solution by an arbitrary smooth order one perturbation.. Introduction Let R n, be the Minkoski space ith Lorentz metric n ḡ = dx i dx n+. i= We ill say that a hypersurface Σ = {x, ux x Ω} R n, is strictly spacelike if u C Ω and Du c 0 < in Ω. Ecker and Huisken [5] studied the mean curvature flo ith forcing term in cosmological spacetimes V and constructed a spacelike hypersurface ith prescribed mean curvature in V. More specifically, they studied the parabolic evolution equation. F = H Hν, t here Fp, 0 : M n V is a compact n-dimensional manifold and H is the forcing term. Later, M. Aarons [] proved the folloing convergence result. Theorem.. [] Let M 0 be a smooth spacelike hypersurface ith bounded curvature. Suppose M 0 never intersects future null infinity I + or past null infinity I. Then M t converges under the flo. u t = Du [ div Du Du c Research of the first author is partially supported in part by the NSF. ],

2 JOEL SPRUCK AND LING XIAO to a convex donard translating soliton, that is, an entire solution of.3 H = c + a, u here c a < 0. It s easy to see that hen a = c, the only convex translating soliton is the trivial one. Therefore, in the folloing, e ill focus on the case here c < a < 0. After rescaling, e may assume a = and consider H = C Du, here C > is a constant. We obtain u.4 div = C Du Du or in nondivergence form.5 δ ij + u iu j u Du ij = C Du. Notice that any u = v x + b, v = C a maximal hypersurface is a solution of.5. The existence of a unique up to translation radially symmetric solution of.5 as shon by Ju, Lu and Jian [9]. Aarons [] in fact conjectured that any solution u of.3 is either rotationally symmetric about some point x 0 or is a hyperplane. Hoever, this conjecture is not correct hen c < a < 0. Let x = x,..., x k and set ux = n i=k+ a ix i + hx here h is strictly convex in x and a = a k+,..., a n is chosen such that C a >. Then u satisfies.5 if and only if h satisfies k h i h j.6 δ ij + a Dh h ij = C a Dh a. i,j= No let h = λ hλx, C = λc > ith λ = a. Then h satisfies.7 k δ ij + h i hj D h h ij = C D h. i,j= Thus h is a rank k solution of.5 in R k ith C replaced by C = λc >. In fact, e ill sho a splitting theorem analogous to hat Choi-Treibergs [4] proved for spacelike constant mean curvature hypersurfaces.

3 ENTIRE TRANSLATING SOLITONS IN MINKOWSKI SPACE 3 Theorem.. Let u be a strictly spacelike solution of.4 and let Σ = {x, ux x R n } be the graph of u. Then Σ is convex ith uniformly bounded second fundamental form. Moreover after an R n, rigid motion, R n, splits as a product R k, R n k such that Σ also splits as a product Σ k R n k here Σ k R k, is a strictly convex graphical solution in R k, Thus it is natural to ask if Aarons conjecture is correct for u, a strictly convex solution of.4 in R n? In other ords, is Σ = {x, ux x R n } rotationally symmetric? The anser is no. Theorem.3. Let f C S n C, C = C. Then there exists an entire strictly spacelike hypersurface u satisfying equation.4 such that ux C x n C log x + f Cx as x. As in the ork of Treibergs [] and Choi-Treibergs [4], the blo-don of a convex urx strictly spacelike solution V u = lim r converges uniformly on compact subsets r to the space CQ of convex homogeneous degree one convex functions hose gradient has magnitude C herever defined. It as shon in [4] that the space Q is in one to one correspondence ith the set of lightlike directions L u := {x S n : V u x = }. It may be possible that any cone in CQ arises as the blo-don of a solution to.4 but e have not shon this. An outline of the paper is as follos. In section e sho the strictly space like assumption implies that the graph Σ is mean convex. Then in section 3 e sho Σ is in fact convex and then prove the splitting Theorem.. In section 4 e study the blo-don V u and finally in section 5 folloing [], e construct counterexamples for the radial cone in CQ and prove Theorem.3. Strictly spacelike implies mean convex Let a ij = δ ij + u iu j, here = Du / ; then equation.5 becomes. a ij u ij = C

4 4 JOEL SPRUCK AND LING XIAO Then i = u ku ki. and ij = u kiu kj = u ku kij u ku kij + u ku ki j u ki u kj + u ku ki u l u lj = u ku kij akl u ki u lj. Lemma.. Suppose Σ = {x, ux x R n } is a strictly spacelike hypersurface, and ux satisfies equation.5. Then Σ is mean convex, that is H 0. Proof. We differentiate equation. ith respect to x k to obtain.3 a ij k u ij + a ij u ijk = C k. Since.4 and u ijk = u kij, this gives a ij uik u j k u ij = + u iu jk u iu j k u 3 ij = uik u j + u iu j u l u lk u 3 ij = i u ik u iu l u lk i = δ ij + u iu j i u kj = aij i u kj,.5 a ij u kij aij i u kj = C k. Multiplying.5 by u k and using u ku kij = ij akl u ki u lj, e obtain.6 a ij ij aij i j + C u k k = aij a kl u ki u lj. We no observe that since A = a ij a kl u ki u lj e can rerite.6 as.7 a ij ij + C u k k = A. The Omori-Yau maximum principle see for example [3], [] implies that achieves its maximum at infinity and moreover, there exists a sequence {P N } such that P N sup, P N < /N, and ijp N /Nδ ij. Therefore,.8 n H A C N at P N.

5 ENTIRE TRANSLATING SOLITONS IN MINKOWSKI SPACE 5 Thus HP N 0 at infinity. Since H = C e obtain inf H = Mean convexity implies convexity and constant rank In this section, e ill use ideas due to Hamilton [7] to prove that under the strictly spacelike assumption, Σ is in fact convex. We use the folloing approximation of Heidusch[8]. Definition 3.. The δ-approximation to the function minx, x is given by µ x, x = x x + x x + δ for any δ > 0. The δ-approximation to the function minx, x,, x n is defined recursively by µ n = n µ x i, µ n x,, x i, x i+,, x n n i= The folloing lemma is elementary see []. Lemma 3.. For every δ > 0 and n e have:. µ n is smooth, symmetric, monotonically increasing and concave.. µn x i. 3. minx,, x n nδ µ n minx,, x n. 4. For x R n e have n µ n µ n x i µ n + nδ, x i and n µ n i= x i x i µ n nδ nδ 4 i= i<j n x i + x j. Lemma 3.3. Assume Σ = {x, ux x R n } is mean convex, and u satisfies equation.5. Then the principal curvatures of A are nonnegative, i.e. Σ is convex. Proof. Let p be a fixed point in Σ e may assume p = 0, 0 and let r be the distance function from p restricted to the geodesic ball B Σ p, a of radius a centered at p in the induced metric on Σ. Let fx = A = i,j h ij. By a ell-knon calculation see equation.4 of [3] 3. h ij = = h ijk + i,j i,j,k i,j h ij H ij + hij h ij h jm h mi H i,j i,j,k

6 6 JOEL SPRUCK AND LING XIAO 3. H ij = i j H = i j ν, e n+ In the folloing, e ill denote ν n+ by V. 3.3 = i h jk τ k, e n+ = h ijk u k h ik h jk ν n+. f = h ijk h ij h ijk u k h ij h ik h jk V + f i,j,k fv 4 + f Cf 3/ C 4 f + f Cf 3/ f C Let ηx = a r and set g = η f. Then in B Σ p, a, h ij h jm h mi C V 3.4 η g C + η g = C + η g η 3 < η, g > +g η. At the point x here g assumes its maximum, g = 0 and g 0. Since R ii H C, e have by Lemma of [3] that 4 4 r C 3 + r. Hence at x, 3.5 g C η 4 + gη 4 η = C η 4 gη η + 6g η C a 8 + ga r + r C 4 a 8 + a 4 g It follos that gx C 5 a 4. Therefore e can let a to conclude A C 5. Next e ill sho that the smallest principal curvature λ min of Σ is nonnegative. Let µ n λ,, λ n = F γ ik h kl γ lj, assume µ n achieves its minimum at an interior point x 0. Then at this point e have 3.6 Thus, µ n = F ij h ijkk + F rl,st h rlk h stk = F ij H ij Hh ij + h ij h kk + F rl,st h rlk h stk i,j,m F ij k h j i τ k, e n+ F ij h k i h jk ν n+ HF ij h ij + µ n + nδ A k µ n, e n+ µ n + nδ + nδ λ i + λ j 4 i<j n + H µ n + nδ + nδ λ i + λ j + µ n + nδ A. 4 i<j n µ n + µ n + nδ A + nδ + nδ 4 i<j n λ i + λ j.

7 Letting δ 0 e find, ENTIRE TRANSLATING SOLITONS IN MINKOWSKI SPACE λ min λ min A, hich implies that λ min 0. Since e have already proven that A is bounded, e can again apply the Omori- Yau maximum principle this time on Σ and sho that, if µ n achieves its minimum at infinity then µ n 0. This completes the proof that mean convexity implies convexity. No that e have proved the convexity of Σ, e prove the splitting Theorem. of the introduction. Proof of Theorem.. Suppose that for some unit vector v and some x 0 R n, D v ux 0 = 0. Applying an isometry boost transformation of R n,, may assume x 0 = 0, v = e n, Du0 = 0, u nn 0 = 0 and u ij is nonnegative. Rerite.5 as 3.9 u = u iu j Du u ij + C Du Differentiating 3.9 tice in the x n direction, e can apply the argument of Korevaar a special case of [0] exactly as in Theorem 3. of Choi-Treibergs [4] to conclude u nn 0 and Σ is ruled by lines parallel to the x n axis. Therefore Σ = Σ n R and also R n = R n R. Since u as arranged to be nonnegative, u = hx, x = x,..., x n here 3.0 n δ ij + h ih j Dh h ij = C Dh. i,j= Proceeding inductively completes the proof of Theorem.. 4. The asymptotic cone at infinity. In this section, e ill study the asymptotic behavior of u at infinity. Proposition 4.. Let u be a convex space like solution of.5. Assume u0 = 0 and denote u h x = uhx. Define V h ux = lim u h x then V u x exists for all x and h is a positively homogeneous degree one convex function. Moreover for all x R n and δ > 0 there exists y R n so that y x = δ and V u x V u y = δ. In C particular DV u x = at every point of differentiability of V C u x.

8 8 JOEL SPRUCK AND LING XIAO Proof. Note that since u is convex, 0 = u0 uhx n i= hx iu xi hx e get d dh uh x 0. Then V u x, the projective boundary values blo-don of u at infinity in the terminology of Treibergs [] and Choi-Treibergs [4], is ell-defined, strictly spacelike, convex on R n and satisfies V u λx = λv u x, λ > 0, V u x V u y x y. C Claim: for all x R n and δ > 0 there exists y R n so that y x = δ and V u x V u y = and ε > 0 such that C δ. Suppose the claim is false. Then there exists x R n V u y V u x + ε δ y Bx, δ. C Since u h x V u x uniformly on compact subsets, e may choose h 0 large so that for all h > h 0, No u h y satisfies 4. H h := div u h y V u x + ε δ y Bx, δ. C Du h = hc 0 in Bx, δ Duh Duh We no make use of the radial solutions of the maximal surface equation H = 0 introduced by Bartnik and Simon []. Consider the barrier y = V u x+ ε y x C δ+ h t n + h dt δ h C t n + h dt. Note that on Bx, δ, y u h y 0 C δ 0 h t n + h dt > 0. Hence by the maximum principle, u h y < y in Bx, δ. In particular at y = x, u h x < V u x + ε C δ δ h C t n + h dt. No let h to conclude V u x V u x ε C claim is proven and the proposition is completed. 0 0 δ, a contradiction, so the

9 ENTIRE TRANSLATING SOLITONS IN MINKOWSKI SPACE 9 5. Construction of Counterexamples. We ill follo Treiberg s idea see [] to construct counterexamples, more precisely e ill construct solution to equation.4 such that ux C x log x + n C f Cx x, as x, here C = C and f C S n C. We extend the function f to R n \ {0} by defining f Cx = f Cx x e have for all x, y S n :. Since f C, 5. f Cx f Cy Df Cy Cx Cy M Cx Cy = CMy Cx Cy. Let p Cy = Df Cy + M Cy and p Cy = Df Cy M Cy, so that 5. p Cy Cx Cy f Cx f Cy p Cy Cx Cy. No let ψx denote the rotationally symmetric solution to.4 see [9]. We kno that ψx C x n log x + o as x. Let z C x; Cy = f Cy p Cy Cy + ψx + p Cy and z x; Cy = f Cy p Cy Cy + ψx + p Cy. Then by equation 5. e have 5.3 f Cx z rx; Cy Cr + n C log r as r, x, y S n, and 5.4 f Cx z rx; Cy Cr + n C log r as r, x, y S n. Therefore, 5.5 for x S n. Let q x = lim z rx; Cy Cr + n log r f r C Cx lim z rx; Cy Cr + n log r r C sup z x; Cy and q x = y S n q i x i=, tends to f Cx + Cr n log r as r. C inf z x; Cy. Then, q x q x and y S n Lemma 5.. There exists a smooth solution u to the Dirichlet problem { a ij u ij C + = 0 in G 5.6 u = 0 on G here G is a convex C,α domain in R n.

10 0 JOEL SPRUCK AND LING XIAO Proof. Let d = diamg be the diameter of G. For any y G, e can choose coordinates such that y = y, 0,, 0 and G {x x y }, here 0 < y d/. Let ux = Cx Cy, ū 0. Then u ū in G and u satisfies 5.7 a ij u ij C u + = 0 By the maximum principle, any solution u to the Dirichlet problem 5.6 satisfies 5.8 u u ū on G. so Dux C on Ḡ. Combined ith equation.7 e conclude that 5.9 Dux C on Ḡ. No it is standard see [6] to prove that a smooth solution u C,α G exists. Finally, e ill find a saniched solution u such that q u < q. Let φ be a strictly spacelike hypersurface q φ < q so that φ0 = q 0 and G m = φ, m is a convex domain ith C,α boundary. By lemma 5. e kno there is an analytic solution u m to the Dirichlet problem 5.0 a ij u ij C + = 0 on G m u = m on G m. Therefore, e find a sequence of finite solutions u m ith q u m < q defined on convex domains G m hich exhaust R n. Next, let K be a compact subset of R n. Then, by equation 5.9 there are constants r < r so that for sufficiently large m e have dist m 0, x < r, for all x K, dist m 0, x < r, for all x G m, here dist m 0, x is the intrinsic distance beteen the points 0, u m 0 and x, u m x on Σ m = {x, u m x x G m }. At last, folloing the proof of Lemma 3.3, e find u m has uniform C 3 bounds on compact subsets. Hence, a subsequence can be extracted that converges to a global solution of equation.4. Moreover, lim u m j = u satisfies ux C x m j n C log x + f Cx. Thus e have proved

11 ENTIRE TRANSLATING SOLITONS IN MINKOWSKI SPACE Theorem 5.. Let f C S n. Then there exists an entire strictly spacelike hypersurface u satisfying equation.4 such C that ux C x n C log x + f Cx as x. Acknoledgement The second author ould like to thank Cornell University for their hospitality during her visit in the spring 04 hile this project as initiated. In particular, she ould like to thank Professor Xiaodong Cao and Professor Laurent Saloff-Coste for their help and support. References [] Mark A. S. Aarons, Mean curvature flo ith a forcing term in Minkoski space. Calc. Var. Partial Differential Equations 5 005, [] R. Bartnik and L. Simon, Spacelike hyper surfaces ith prescribed boundary values and mean curvature, Comm. Math. Physics 87 98, 3 5. [3] S. Y. Cheng, S. T. Yau, Maximal space-like hypersuface in the Lorentz-Minkoski spaces. Ann. of. math , [4] H. I. Choi and A. E. Treibergs, Gauss maps of space like constant mean curvature hypersurfaces of Minkoski space, J. Diff. Geom , [5] K. Ecker and G. Huisken, Parabolic methods for the construction of spacelike slices of prescribed mean curvature in cosmological spacetimes, Commun. Math. Phys , [6] D. Gilbarg and N. Trudinger, Elliptic partial differential equations of second order, Springer Verlag Ne York 998. [7] Hamilton, R., Four manifolds ith positive curvature operator. J. Diff. Geom , [8] M. Heidusch, Zur Regularita?t des Inversen Mittleren Kru?mmungsflusses, Ph.d thesis, Eberhard-Karls-Universita?t Tu?bingen 00 [9] H. Ju, J. Lu and H. Jian, Translating solutions to the mean curvature flo ith a forcing term in Minkoski space, Comm. Pure Appl. Anal. 9 00, [0] N. Korevaar and J. Leis, Convex solutions to certain elliptic equations have constant rank Hessians, Arch. Rational Mech. Anal [] S. Pigola, M. Rigoli and G. Setti, Maximum principles on Riemannian manifolds and applications, Memoirs of the American Mathematical Society 74 Number 8, 005. [] A. E. Treibergs, Entire spacelike hypersurfaces of constant mean curvature in Minkoski space. Inventiones Math , [3] S. T. Yau, Harmonic functions on complete Riemannian manifolds. Comm. Pure Appl. Math , 0 8.

12 JOEL SPRUCK AND LING XIAO Department of Mathematics, Johns Hopkins University, Baltimore, MD 8 address: js@math.jhu.edu Department of Mathematics, Rutgers University, Piscataay, NJ address: lx70@math.rutgers.edu