Harmonic maps on domains with piecewise Lipschitz continuous metrics

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1 Harmonic maps on domains with piecewise Lipschitz continuous metrics Haigang Li, Changyou Wang Abstract For a bounded domain Ω equipped with a piecewise Lipschitz continuous Riemannian metric g, we consider harmonic map from (Ω, g) to a compact Riemannian manifold (N, h) R k without boundary. We generalize the notion of stationary harmonic maps and prove their partial regularity. We also discuss the global Lipschitz and piecewise C 1,α -regularity of harmonic maps from (Ω, g) to manifolds that support convex distance functions. 1 Introduction Throughout this paper, we assume that Ω is a bounded domain in R n, separated by a C 1,1 -hypersurface Γ into two subdomains Ω + and Ω, namely, Ω = Ω + Ω Γ, and g is a piecewise Lipschitz metric on Ω that is g C 0,1 (Ω + ) C 0,1 (Ω ) but discontinuous at any x Γ. For example, Ω = R n is the unit ball, Γ = {x = (x, 0) R n }, and g 0 x B + 1 ḡ(x) = = {x n > 0}, kg 0 x B 1 = {x n < 0}, where g 0 (x) = dx is the Euclidean metric and 1 k is a positive constant. Let (N, h) R k be a l-dimensional, smooth compact Riemannian manifold without boundary, isometrically embedded in the Euclidean space R k. Motivated by the recent studies on elliptic systems in domains consisting of composite materials (see Li-Nirenberg [17]) and the homogenization theory in calculus of variations (see Avellaneda-Lin [1] and Lin-Yan [18]), we are interested in the regularity issue of stationary harmonic maps from (Ω, g) to (N, h). In order to describe the problem, let s first recall some notations. Throughout this paper, we use the Einstein convention for summation. For the metric g = g i j dx i dx j, let (g i j ) denote the inverse matrix of (g i j ), g = det (g i j ), and dv g = g dx denotes the volume form of g. For 1 < p < +, define the Sobolev space W 1,p (Ω, N) by { ( ) } W 1,p (Ω, N) = u : Ω R k u(x) N a.e. x Ω, E p (u, g) = u p g dv g < +, School of Mathematical Sciences, & Laboratory of Mathematics and Complex Systems, Ministry of Education, Beijing Normal University, Beijing , P. R. China. Department of Mathematics, University of Kentucky, Lexington, KY 0506, USA. Ω 1

2 where u g g i j u x i, u x j is the L -energy density of u with respect to g, and, denotes the inner product in R k. Denote W 1, (Ω, N) by H 1 (Ω, N). Now let s recall the concept of stationary harmonic maps. Definition 1.1. A map u H 1 (Ω, N) is called a (weakly) harmonic map, if it is a critical point of E (, g), i.e., u satisfies g u + A(u)( u, u) g = 0 (1.1) in the sense of distributions. Here g = 1 ( ) gg i j g x i x j is the Laplace-Beltrami operator on (Ω, g), A( )(, ) is the second fundamental form of (N, h) R k, and ( u A(u)( u, u) g = g i j A(u), u ). x i x j Definition 1.. A (weakly) harmonic map u H 1 (Ω, N) is called a stationary harmonic map, if, in additions, it is a critical point of E (, g) with respect to suitable domain variations: d dt t=0 Ω u t dv g g = 0, with u t (x) = u(f t (x)), (1.) where F(t, x) := F t (x) C 1 ([ δ, δ], C 1 (Ω, Ω)) is a C 1 family of differmorphisms for some small δ > 0 satisfying F 0 (x) = x x Ω F t (x) = x (x, t) Ω [ δ, δ] ( F ) (1.3) t Ω ± Ω ± t [ δ, δ]. It is readily seen that any minimizing harmonic map from (Ω, g) to (N, h) is a stationary harmonic map. It is also easy to see from Definition 1. that a stationary harmonic map on (Ω, g) is a stationary harmonic map on (Ω ±, g) and hence satisfies an energy monotonicity inequality on Ω ±, since g C 0,1 (Ω ± ). We will show in that a stationary harmonic map on (Ω, g) also satisfies an energy monotonicity inequality in Ω under the condition (1.) below. The first result is concerned with both the (partial) Lipschitz regularity and (partial) piecewise C 1,α -regularity of stationary harmonic maps. In this context, we are able to extend the well-known partial regularity theorem of stationary harmonic maps on domains with smooth metrics, due to Hélein [1], Evans [5], Bethuel []. More precisely, we have

3 Theorem 1.1. Let u H 1 (Ω, N) be a stationary harmonic map on (Ω, g). If, in additions, g satisfies the following jump condition on Γ for n 3 1 : for any x Γ, there exists a positive constant k(x) 1 such that lim y Ω +,y x g(y) = k(x) lim y Ω,y x g(y), (1.) then there exists a closed set Σ Ω, with H n (Σ) = 0, such that for some 0 < α < 1, (i) u Lip loc (Ω \ Σ, N), (ii) u C 1,α loc ((Ω+ Γ) \ Σ, N) C 1,α loc ((Ω Γ) \ Σ, N). We would like to remark that when the dimension n =, since the energy monotonicity inequality automatically holds for H 1 -maps, Theorem 1.1 holds for any weakly harmonic map from domains of piecewise C 0,1 -metrics, i.e., any weakly harmonic map on domains with piecewise Lipschitz continuous metrics is both Lipschitz continuous and piecewise C 1,α for some 0 < α < 1. Through the example constructed by Rivière [19], we know that weakly harmonic maps on domains with smooth metrics may not enjoy partial regularity properties in dimensions n 3. Here we consider weakly harmonic maps on domains with piecewise Lipschitz continuous metrics into any Riemannian manifold (N, h), on which dn (, p) is convex. Such Riemannian manifolds N include those with non-positive sectional curvature K N, and geodesic convex ball in any Riemannian manifold. In particular, we extend the classical regularity theorems on harmonic maps on domains with smooth metrics, due to Eells-Sampson [8] and Hildebrandt-Kaul-Widman [13], and prove Theorem 1.. Let g be the same as in Theorem 1.1. Assume that on the universal cover (Ñ, h) of (N, h), the square of distance function d (, p) is convex for any p Ñ. If Ñ u H 1 (Ω, N) is a weakly harmonic map, then for some 0 < α < 1, (i) u Lip loc (Ω, N), (ii) u C 1,α loc (Ω+ Γ, N) C 1,α loc (Ω Γ, N). The idea to prove Theorem 1.1 is motivated by Evans [5] and Bethuel []. However, there are several new difficulties that we have to overcome. The first difficulty is to establish an almost energy monotonicity inequality for stationary harmonic maps in Ω, which is achieved by observing that an exact monotonicity inequality holds at any x Γ, see below. The second one is to establish a Hodge decomposition in L p (B, R n ), for any 1 < p < +, on a ball B(= ) equipped with certain piecewise continuous metrics g, in order to adapt the argument by Bethuel []. More precisely, we will show that the following elliptic equation on B: v x i (a i j x j ) = div( f ) in B, v = 0 on B 1 this condition is needed for both energy monotonicity inequalities for u in dimensions n 3 and the piecewise C 1,α -regularity of u. Here the covering map Π : Ñ N is a Riemannian submersion from (Ñ, h) to (N, h). 3

4 enjoys the W 1,p -estimate: for any 1 < p < +, v C f L p (B) L p (B) provided that (a i j ) C ( B ±) C ( B δ) for some δ > 0 is uniformly elliptic, and is discontinuous on B + \ B δ, where B δ = { x B : dist(x, B) δ }. This fact follows from a recent theorem by Byun-Wang [3], see 3 below. The third one is to employ the moving frame method to establish a decay estimate in suitable Morrey spaces under a smallness condition, which is similar to [1]. To obtain Lipschitz and piecewise C 1,α -regularity, we compare the harmonic map system with an elliptic system with piecewise constant coefficients and extend the hole-filling argument by Giaquinta-Hildebrandt [10]. The paper is organized as follows. In, we derive an almost monotonicity inequality for the renormalized energy. In 3, we show the global W 1,p (1 < p < ) estimate for elliptic systems with certain piecewise continuous coefficients, and a Hodge decomposition theorem. In, we adapt the moving frame method, due to Hélein [1] and Bethuel [], to establish an ɛ-hölder continuity. In 5, we establish both Lipschitz and piecewise C 1,α regularity for Hölder continuous harmonic maps. In 6, we consider harmonic maps into manifolds supporting convex distance square functions and prove Theorem 1.. Acknowledgement. Part of this work was completed while the first author visited University of Kentucky. He would like to thank the Department of Mathematics for its hospitality. The first author was partially supported by SRFDPHE ( ) and NNSF in China ( ) and Program for Changjiang Scholars and Innovative Research Team in University in China. The second author is partially supported by NSF grant Energy monotonicity inequality This section is devoted to the derivation of energy monotonicity inequalities for stationary harmonic maps from (Ω, g) to (N, h). More precisely, we have Theorem.1. Under the same assumption as in Theorem 1.1, there exist C > 0 and r 0 > 0 depending only on Γ and g such that if u W 1, (Ω, N) is a stationary harmonic map on (Ω, g), then for any x 0 Ω, there holds s n u dv g g e Cr r n u dv g g (.1) B s (x 0 ) for all 0 < s r min{r 0, dist(x 0, Ω)}. Since the metric g C 0,1 (Ω ± ), it is well-known that there are K > 0 and r 0 > 0 such that (.1) holds for any x 0 Ω ± and 0 < s r min{r 0, dist(x 0, Ω ± )}. In particular, (.1) holds for any x 0 Ω\Γ r 0 and 0 < s r min{r 0, dist(x 0, Ω)}, where B r (x 0 )

5 Γ r 0 = {x Ω : dist(x, Γ) r 0 }. We will see that to show (.1) for x 0 Γ r 0, it suffices to consider the case x 0 Γ. It follows from the assumption on Γ and g, there exists r 0 > 0 such that for any x 0 Γ there exists a C 1,1 -differmorphism Φ 0 : B r1 (x 0 ), where r 1 = min{r 0, dist(x 0, Ω)}, such that Φ 0 (B ± 1 ) = Ω± B r1 (x 0 ) Φ 0 (Γ 1 ) = Γ B r1 (x 0 ), where Γ 1 = {x : x n = 0}. Define ũ(x) = u(φ 0 (x)) and g(x) = (Φ 0 ) (g)(x), x. Then it is readily seen that (i) g is piecewise C 0,1, with the discontinuous set Γ 1, and satisfies (1.) on Γ 1 3, (ii) ũ : (, g) (N, h) is a stationary harmonic map, if u : (B r1 (x 0 ), g) (N, h) is a stationary harmonic map. Thus we may assume that Ω =, g is a piecewise C 0,1 -metric which satisfies (1.) on the set of discontinuity Γ 1, and u : (, g) (N, h) is a stationary harmonic map. It suffices to establish (.1) in. We first derive a stationarity identity for u. Proposition.. Let u W 1, (, N) be a stationary harmonic map on (, g). Then ( g i j u, u ) g Yi k u ( gg x k x gdivy ) dx = i j Y k u, u dx (.) j x k x i x j holds for all Y = (Y 1,, Y n 1, Y n ) C 1 0 (, R n ) satisfying 0 for x n > 0 Y n (x) = 0 for x n = 0 0 for x n < 0, (.3) where Y k i = Yk x i and div Y = n i=1 Y i x i. Proof. Let Y satisfy (.3), it is easy to see that there exists δ > 0 such that F t (x) = x + ty(x), t [ δ, δ], is a family of differmorphisms from to satisfying the condition (1.3). Hence 0 = d (u(f t (x)) g dv g = d ( ) dt t=0 (u(f t (x)) g dv g + (u(f t (x)) g dv g. dt t=0 B + 1 B 1 3 In fact, since (Φ 0 ) (g) i j (x) = g kl (Φ 0 (x)) Φk 0 x i (x) Φl 0 x j (x), (1.) implies that for any x Γ 1 lim (Φ 0) g(y) = k(φ 0 (x)) lim (Φ 0) g(y). y Ω +,y x y Ω,y x 5

6 For t [ δ, δ], set G t = Ft 1. Direct calculations yield d (u(f t (x)) g dv g t=0 dt B ± 1 = d dt t=0 B ± 1 = gg i j u, u (δ ki Y l j B ± x 1 k x + δ l jyi k ) dx l d ( + g i j (G t (x)) g(g t (x))jg t (x) ) u, u dx B ± dt t=0 x 1 i x j ( = g i j u, u Y l j x i x gi j u, u divy ) g dx l x i x j B ± 1 B ± 1 g(x)g i j (x) u y k, u y l (x + ty(x))(δ ki + ty k i )(δ l j + ty l j ) dx x k ( gg i j ) Y k u x i, u x j dx, where we have used d dt JG t (x) = divy, t=0 d dt G t (x) = Y(x), t=0 d ( dt g i j (G t (x)) g(g t (x)) ) ( gg = ) i j t=0 x k Y k. This completes the proof. Proposition.3. Let u W 1, (, N) be a stationary harmonic map on (, g). Then there exists C > 0 such that (i) for any x 0 = (x 0, xn 0 ) \ Γ 1, there exists 0 < R 0 min{ 1, xn 0 }, such that r n u gdv g e CR R n u gdv g, 0 < r R < R 0. (.) B r (x 0 ) (ii) for any x 0 Γ 1, there holds r n B r (x 0 ) u gdv g e CR R n In particular, for any x 0, there holds r n B r (x 0 ) u gdv g e CR R n B R (x 0 ) B R (x 0 ) B R (x 0 ) u gdv g, 0 < r R 1. (.5) u gdv g, 0 < r R 1. (.6) Proof. (i) By choosing Y Cc (B + 1, Rn ) or Y Cc (B 1, Rn ), we have that u is a stationary harmonic map on ( B + 1, g) and ( B 1, g). Thus the monotonicity inequality (.) is standard. 6

7 (ii) For simplicity, consider x 0 = (0, 0). For ɛ > 0 and 0 < r 1, let Y ɛ(x) = xη ɛ (x), where η ɛ (x) = η ɛ ( x ) C 0 () satisfies Then 0 η ɛ 1; η ɛ (s) 1 for 0 s r ɛ; η ɛ (s) 0 for s r; η ɛ 0; η ɛ ɛ. (Y ɛ ) j i = δ i j η ɛ ( x ) + η ɛ( x ) xi x j. (.7) x Substituting Y ɛ into the right hand side of (.), and using we have ( gg ) i j C, x k ( gg ) i j Yɛ k u, u dx Cr u dx Cr u g dv g. (.8) x k x i x j B r B r Substituting (.7) into the left hand side of (.), we obtain ( g i j u, u ) g (Y ɛ ) k i u x j x gdivy ɛ dx k = ( n) u gη ɛ (x) g dx u g x η ɛ(x) g dx + g i j u x i, u Set the piecewise constant metric g by Then we have xk x j η x k x ɛ(x) g dx. (.9) lim g(y) if g(x, x n y 0, y ) = n 0 xn 0 lim g(y) if y 0, y n <0 xn < 0. It follows from (1.) that we can assume g 0 if x n 0 g(x) = kg 0 if x n < 0, g(x) g(x) C x, x. (.10) for some positive constant k 1. Thus we can estimate g i j u, u xk x j η x i x k x ɛ(x) g dx = g i j u, u xk x j η x i x k x ɛ(x) g dx + (g i j g i j ) u, u xk x j η x i x k x ɛ(x) g dx = I ɛ + II ɛ. (.11) 7

8 Since g i j u, u xk x j x i x k x x u r h(x) := if x n 0 1 u x k r if x n < 0, and h(x) 0 for x, we have I ɛ = h(x)η ɛ( x ) g dx 0. (.1) For II ɛ, by (.10) we have II ɛ Cr u dv g Cr u g dv g. (.13) B r B r First substituting (.1) and (.13) into (.11), and then plugging the resulting (.11) into (.9), and finally combining (.9) and (.8) with (.), we obtain, after sending ɛ to zero, ( n) u gdv g + r u g g dh n 1 Cr u gdv g. B r B r B r This implies ) d (e Cr r n u dr gdv g 0, B r which clearly yields (.5). To show (.6), it suffices to consider the case x 0 / \ Γ 1, B R (x 0 ) B + 1 > 0 and B R(x 0 ) B 1 > 0. For simplicity, assume x 0 B 1. We divide it into two cases: (i) d(x 0, Γ 1 ) = xn 0 1R: If R r 1 R, then it is easy to see r n u gdv g n R n B r (x 0 ) If 0 < r < 1 R( d(x0, Γ 1 )), we have B R (x 0 ) B 1 r n B r (x 0 ) u gdv g e CR ( R ) n B R (x 0 ) B R (x 0 ) u gdv g. so that (.) implies u gdv g e CR n R n B R (x 0 ) u gdv g. (ii) d(x 0, Γ 1 ) = x 0 n < 1 R: If R r 1 R, then r n B r (x 0 ) u gdv g n R n 8 B R (x 0 ) u gdv g.

9 If 0 < r d(x 0, Γ 1 ) = xn 0 < 1R, then by setting x0 = (x 0 1,, x0 n 1, 0) we have so that (.5) yields r n B r (x 0 ) B x 0 n (x 0 ) B x 0 n (x 0 ) B R (x 0 ) B R (x 0 ) B r (x 0 ) u gdv g x 0 n n B x 0 n (x0 ) n ( x 0 n ) n ( R n e CR e CR R n If d(x 0, Γ 1 )(= xn ) 0 r < 1 R, then we have so that (.5) yields r n ) n B R (x 0 ) u gdv g B x 0 n (x0 ) B R (x 0 ) u gdv g. B r (x 0 ) B r (x 0 ) B R (x 0 ) B R (x 0 ), B r (x 0 ) u gdv g n (r) n ( R n e CR e CR R n B r (x 0 ) ) n B R (x 0 ) u gdv g u gdv g u gdv g B R (x 0 ) u gdv g. u gdv g Therefore (.6) is proven. 3 W 1,p -estimate for elliptic equations with certain piecewise continuous coefficients In this section, we will show the global W 1,p -estimate for elliptic equations with certain piecewise continuous coefficients, for 1 < p < +. As a corollary, we will establish the Hodge decomposition Theorem 3. for certain piecewise continuous metrics g, which is a key ingredient to prove Theorem 1.1 and may also have its own interest. For a ball B = R n, denote B ɛ = {x B : dist(x, B) ɛ} for ɛ > 0. Let (a i j (x)) 1 i, j n be bounded measurable, uniformly elliptic on B, i.e., there exists 0 < λ Λ < + such that λ ξ a i j (x)ξ α i ξ j β Λ ξ, a.e. x B, ξ R n. (3.1) 9

10 Theorem 3.1. Assume (a i j ) satisfies (3.1), and there exists ɛ > 0 such that (a i j ) C ( B ±) C (B ɛ ) and is discontinuous on B + \ B ɛ. For 1 < p < +, let f L p (B, R n ). Then there exists a unique weak solution v W 1,p 0 (B, Rn ) to ( ) v ai x i, j i j x j = f i x i i in B, (3.) u = 0 on B, and for some C > 0 depending only on p and (a i j ). v L p (B) C f L p (B) (3.3) Proof. By our assumption, it is easy to verify that for any δ > 0, there exists R = R(δ) > 0 such that the coefficient function (a i j ) satisfies the (δ, R)-vanishing of codimension 1 conditions (.5) and (.6) of Byun-Wang [3] page 65. In fact, we have a stronger property: lim r 0 max a i j (x, x n ) a i j (x x 0 =(x 0,xn 0 ) B 0, xn ) L (B r ((x 0,xn 0 ))) = 0. Thus the conclusion of Theorem 3.1 follows by direct application of [3] Theorem., page 653. As an immediate consequence of Theorem 3.1, we have the following Hodge decomposition on B equipped with suitable piecewise continuous metrics g. Theorem 3.. Let ḡ be a piecewise continuous metric on B such that ḡ C ( B ±) C ( B δ) for some δ > 0, and is discontinuous on B + \ B δ. Then for any 1 < p < +, F = (F 1,, F n ) L p (B, R n ), there exist G W 1,p 0 (B) and H Lp (B, R n ) such that F = G + H, 0 = divḡh (:= and there exists C = C(p, n, ḡ) > 0 such that ḡ 1 ( ḡḡ i j H j )) in B, (3.) x i G L p (B) + H L p (B) C F L p (B). (3.5) Proof. Set a i j = ḡḡ i j on B for 1 i, j n. It is easy to verify that (a i j ) satisfies the conditions of Theorem 3.1. Thus Theorem 3.1 yields that there exists a unique solution G W 1,p 0 (B) to ( ḡḡ ) ( i j G x i x j = ḡḡ ) i j x i F j, in B (3.6) G = 0 on B, and G L p (B) C ḡḡ i j L F j C F p L (B) p (B). 10

11 Set H = F G. Then we have divḡh = ḡ 1 x i ( ( ḡḡ i j F j G )) = 0 on B, x j and H L p ( ) F L p ( ) + G L p (B) C F L p (B). This completes the proof. Hölder continuity In this section, we will prove that any stationary harmonic maps on (, g), with a piecewise Lipschitz continuous metric g C 0,1 (B ± 1 Γ 1), is Hölder continuous under a smallness condition of u g dv g. The idea is based on suitable modifications of the original argument by Bethuel [] (see also Ishizuka-Wang [1]), thanks to the energy monotonicity inequality and the Hodge decomposition theorem established in previous sections. More precisely, we have Theorem.1. There exist ɛ 0 > 0 and α 0 (0, 1) depending only on n, g such that if the metric g C 0,1 (B ± 1 Γ 1) satisfies the condition (1.) on Γ 1, and u W 1, (, N) is a stationary harmonic map on (, g) satisfying r n 0 B r0 (x 0 ) u g dv g ɛ 0 (.1) for some x 0 and 0 < r 0 1, then u Cα 0 (B r 0 (x 0 ), N) and [ u ] C α 0(B r 0 (x 0 )) C(r 0, ɛ 0 ). (.) Proof of Theorem.1. The proof is based on suitable modifications of [] and [1]. First, observe that if x 0 = (x 0, xn 0 ) B±, it follows from the monotonicity inequality (.6) that we may assume (.1) holds for some 0 < r 0 < x n 0. Then the ɛ 0-regularity theorem by Bethuel [] (see [1] for domains with C 0,1 metrics) implies that for some 0 < α 0 < 1, u C α 0 (B r 0 (x 0 )) and (.) holds. Hence it suffices to consider the case x 0 = (x 0, 0) Γ 1. By translation and scaling, we may assume x 0 = (0, 0) and proceed as follows. Step 1. As in [] [1] [1], assume that there exists an orthonormal frame on u T N. B1 For 0 < θ < 1 to be determined later, let {e α} l α=1 W1, (B θ, R k ) be a Coulomb gauge orthonormal frame of u T N : Bθ div g ( e α, e β ) = 0 in B θ (1 α, β l), l e B α θ gdv g C u B θ gdv g. α=1 11 (.3)

12 For 1 α l, consider ((u u θ )η), e α, where u θ = u is the average of u on B θ B θ, and η C 0 () satisfies 0 η 1; η = 1 in B θ ; η = 0 outside B 7 θ; η θ. Let g 0 be the standard metric on R n. We define a new metric g on B θ by letting Then it is easy to see that g(x) = η(x)g(x) + (1 η(x))g 0 (x), x B θ. g g on B θ, g g 0 outside B 7 θ, and g C(B ± θ ) C(B θ \ B 7 θ). In particular, g satisfies the condition of Theorem 3.. Hence, by Theorem 3., we have that for 1 < p < n n 1, there exist φ α W 1,p 0 (B θ) and ψ α L p (B θ ) such that ((u u θ )η), e α = φ α + ψ α, div g (ψ α ) = 0 in B θ, (.) φ α L p (B θ ) + ψ α L p (B θ ) ((u u θ )η) L p (B θ ) u L p (B θ ). Since u satisfies the harmonic map equation (1.1), we have Thus we obtain div g ( u, e α ) = g i j i u j e α, e β e β in B θ. (.5) g φ α = g i j i u j e α, e β e β in B θ. (.6) Set φ α = φ (1) α + φ () α, where φ (1) α solves g φ (1) α = 0, in B θ, φ (1) α = φ α, on B θ, and φ () α solves g φ () α = g i j i u j e α, e β e β, in B θ, φ () α = 0, on B θ. (.7) (.8) Step. Estimation of φ (1) α : It is well-known (cf. [11]) that φ (1) α C α 0 (B θ ) for some α 0 (0, 1), and for any 0 < r θ [ ] φ (1) p α C α 0(B r ) θp n φ (1) α p dx Cθ p n u B θ B p dx, (.9) θ and (τθ) p n B τθ φ (1) α p Cτ pα u where M p,p ( ) denotes the Morrey space: M p,p (E) := { f : E R : f p M p,p (E) = sup B r (x) R n 1 M p,p ( ) {r p n, 0 < τ < 1, (.10) B r (x) E } f p dx < + }, E R n.

13 Step 3. Estimation of φ () α : First, denote by H 1 (R n ) the Hardy space on R n and BMO(E) the BMO space on E for any open set E R n. By (.13) of [1] page 35, for p = p p 1 > n, there exists h W1,p 0 (B θ ), with h L p (B θ ) = 1, such that φ () α C L φ p (B θ ) B () α, h g dv g. θ Hence by the equation (.8), (.), and the duality between H 1 and BMO, we have φ () α C L p (B θ ) gg i B j i u j e α, e β (e β h) dx θ = C gg i j j e α, e β i (e β h)u dx B θ C gg i j [ ] j e α, e β i (e β h) u H 1 (R n ) BMO(B θ ) gg i j j e α, e β L (B θ ) (e β h) L (B θ ) [u] BMO(Bθ ) u L (B θ ) u M p,p ( ) θ n p n, (.11) where we have used: (i) Since div g ( e α, e β ) = 0 in B θ and h W 1,p 0 (B θ ), we have gg i j j e α, e β i (e β h) H 1 (R n ) and gg i j j e α, e β i (e β h) C gg i j j e α, e β L H 1 (R n ) (B θ ) i (e β h) L, (B θ ) (ii) Since p > n, the Sobolev embedding implies h C 1 n p (B θ ) and so that h L (B θ ) Cθ 1 n p, (e β h) L (B θ ) e β L (B θ ) h L (B θ ) + h L p (B θ )θ n p n Cθ n p n, (iii) By Poincaré inequality, it holds [u] BMO(Bθ ) C u M p,p ( ). Putting the estimates of φ (1) α and φ () α together, we obtain ) 1 p [ ] ((τθ) p n φ α p dx C τ α 0 + τ 1 n p ɛ0 u M p,p ( ), 0 < τ < 1. (.1) B τθ 13

14 Step. Estimation of ψ α : Since div g (ψ α ) = 0 on B θ, we have ψ α g dv g = (ψ α + φ α ), ψ α g dv g B θ B θ = ((u u θ )η), e α, ψ α g dv g B θ = (u u θ )η e α, ψ α g dv g B θ g g i j i e α ψα j [ H 1 (u uθ )η ] BMO [ ψ α L (B θ ) e α L (B θ ) (u uθ )η ] BMO u L (B θ ) ψ α L (B θ ) u M p,p ( ), where we have used the fact [ (u uθ )η ] BMO C [u] BMO(B θ ) C u M p,p ( ). This, combined with Hölder s inequality, implies ) 1 p (θ p n ψ α p Cɛ0 u M p,p(b1). (.13) B θ Step 5. Decay estimation of u: Putting (.1) and (.13) together, we have that for some 0 < α 0 < 1, ) 1 p ( ) ((τθ) p n u p C ɛ0 + τ α 0 + τ 1 n p ɛ0 u B τθ M p,p ( ) (.1) holds for any 0 < τ < 1 and 0 < θ < 1. Now we claim that for some α 0 (0, 1), it holds u M p,p (B τ ) C ( ɛ 0 + τ α 0 + τ 1 n p ɛ0 ) u M p,n p ( ), 0 < τ < 1. (.15) To show (.15), let B s (y) B τ. We divide it into three cases: (a) y B τ B ± and s < y n. As remarked in the begin of proof, we have that for some 0 < α 0 < 1, (s p n ) 1 p u p B s (y) ( ) α0 s C y n p n y n ( ) α0 s C ( y n ) p n y n ( τ p n C u ) p B τ (y 0) u p B yn (y) u p B yn (y,0) 1 p 1 p (since y n τ ) C(ɛ 0 + τ α 0 + τ 1 n p ɛ0 ) u M p,p ( ) (by (.1)). 1 p 1

15 (b) y B τ B ± and s y n. Then we have B s (y) B yn +s(y, 0) B s (y, 0). Hence (s p n ) 1 p u p B s (y) n p p ((s) p n ) 1 p u p B s (y,0) C ( ɛ 0 + τ α 0 + τ 1 n p ɛ0 ) u M p,p ( ) (by (.1)). (c) y B τ Γ 1, i.e. y n = 0. Then it follows directly from (.1) that (s p n ) 1 p ( ) u p C ɛ0 + τ α 0 + τ 1 n p ɛ0 u M p,p ( ). B s (y) Combining (a), (b) and (c) together and taking supremum over all B s (y) B τ, we obtain (.15). It is now clear that by first choosing sufficiently small τ and then sufficiently small ɛ 0, we have u M p,p (B τ ) 1 u M p,p ( ). Iterating this inequality finitely many time yields that there exists α 1 (0, 1) such that for any x and 0 < r 1, it holds r p n u p dx C r pα 1 u p M p,p ( ). B r (x) This immediately implies u C α 1 ( ). The proof is now completed. 5 Lipschitz and piecewise C 1,α -estimates In this section, we will first establish both Lipschitz and piecewise C 1,α -regularity for stationary harmonic maps on domains with piecewise C 0,1 -metrics, under the smallness condition of renormalized energies. Then we will sketch a proof of Theorem 1.1. Theorem 5.1. There exist ɛ 0 > 0 and β 0 (0, 1) depending only on n, g such that if the metric g C 0,1 (B ± 1 Γ 1) satisfies the condition (1.) on Γ 1, and u W 1, (, N) is a stationary harmonic map on (, g) satisfying r n 0 B r0 (x 0 ) u g dv g ɛ 0 (5.1) for some x 0 and 0 < r 0 1, then u C1,β 0( B r 0 (x 0) B ±, N ), and u C 0,1( B r 0 (x 0 ), N ). Proof. The proof is based on both the hole filling argument and freezing coefficient method. It is divided into two steps. 15

16 Step 1. u C α (B 3r 0 (x 0 ), N) for any 0 < α < 1. To see this, recall Theorem.1 implies that there exists 0 < α 0 < 1 such that u C α 0 (B 7r 0 (x 0 )) and for any y B 7r 0 (x 0 ), it 8 8 holds ( s ) α0 s n u dx C r n u dx, 0 < s r < r 0 r 8, (5.) B s (y) and osc Br (y)u Cr α 0, 0 < r < r 0 8. (5.3) For y B 7r 0 (x 0 ) and 0 < r < r 0, let v : B 8 8 r(y) R k solve g v = 0 in B r (y) (5.) v = u on B r (y). B r (y) Then by the maximum principle and (5.3), we have osc Br (y)v osc Br (y)u Cr α 0. Moreover, since g C 0,1 (B ± 1 Γ 1), it is well-known (cf. [17] Theorem 1.1) that v C 0,1 (B r (y), R k ) and v C 1,β (B r (y) B ±, R k ) for any 0 < β < 1. Now multiplying both the equations (1.1) and (5.) by (u v) and subtracting each other and then integrating over B r (y), we obtain (u v) dx u u v r n +3α 0. Since B r (y) B r (y) B r (y) v dx C v, L (B r (y) )rn we obtain that if 0 < α 0, then 3 ( r n u ) dx C ( v L (B r (y)) r + r 0) 3α Cr 3α 0. B r (y) This, combined with Morrey s decay lemma, yields u C 3α 0 (B 7r0 (x 0 )). Repeating this 8 argument, we can show that u C α (B 3r 0 (x 0 )) for any 0 < α < 1, and r n u dx Cr α, y B 3r 0 (x 0 ), 0 < r < r 0. (5.5) B r (y) Step. There exists 0 < β 0 < 1 such that u C 0( 1,β B r (x 0 0) B ±, N ). The proof is divided into two cases. Case I. x 0 = (x 0, xn 0 ) B± 1. We may assume 0 < r 0 < x n 0 so that B r 0 (x 0 ) B ±. For B r (x) B r0 (x 0 ), let v : B r (x) R k solve g v = 0 in B r (x), (5.6) v = u on B r (x). 16

17 Then by Step 1, we have that for any < α < 1, 3 (u v) dx C u u v dx C r 3α+n. (5.7) B r (x) B r (x) Moreover, since g C 0,1 (B r0 (x 0 )), we have that for any 0 < β < 1, v C 1,β (B r (x)) and v ( v) Bs (x) dx C( s r )β u ( u) Br (x) dx, 0 < s r. (5.8) B s (x) Henceforth, we will denote E f = 1 E B r (x) E f dx. Combining (5.7) and (5.8), we obtain that for any 0 < θ < 1, u ( u) Bθr (x) dx [ u v dx + v ( v) Bθr (x) dx ] B θr (x) B θr(x) B θr (x) C [ θ β u ( u) Br (x) dx + θ n r 3α ]. For 3α B r (x) < β 0 < β, let 0 < θ 0 < 1 be such that Cθ β 0 = θ β 0 0. Then we have u ( u) Bθ0 r(x) dx θ β 0 u ( u) Br (x) dx + Cr 3α. (5.9) B θ0 r(x) 0 B r (x) Iterating (5.9) m-times, m 1, yields u ( u) Bθ0 r(x) dx ( θ m ) β0 0 B θ m 0 r(x) B r (x) +C(θ m 0 r)3α This clearly implies that u C 3α (B r0 (x 0 )). u ( u) Br (x) dx m j=1 θ j(β 0 (3α )) 0 (5.10) (θ m 0 )3α [ u ( u) Br (x) dx + Cr 3α ]. B r (x) Case II. x 0 = (x 0, 0) Γ 1. For simplicity, we assume x 0 = 0. Define the piecewise constant metric ḡ on by letting lim t 0 + g(0, t) x B + 1 ḡ(x) = lim t 0 g(0, t) x B 1. Then we have g(x) ḡ(x) C x, x. (5.11) Moreover, by suitable dilations and rotations of the coordinate system, (1.) implies that there exists a positive constant k 1 such that note that (5.8) trivially holds for r s r. ḡ(x) = (1 + (k 1)χ B 1 (x))g 0, x, 17

18 where χ B 1 is the characteristic function of B 1. For 0 < r < r 0, let v : B r(0) R k solve ḡv = 0 in, v = u on. (5.1) Then we have osc Br (0)v osc Br (0)u Cr α, v dx C u Cr n +α. Multiplying (1.1) and (5.1) by (u v) and integrating over, we obtain (u v) dx g i j (u v) i (u v) j g dx C u u v dx + gg i j ḡḡ i j v i (u v) j dx Cosc Br (0)v u dx + Cr v + 1 (u v) dx Cr n +3α + Cr n+α + 1 (u v) dx. This implies (u v) dx Cr n +3α. (5.13) It is well-known that v C ( B ± s (0) ) for any 0 < s < r. In fact, (5.1) is equivalent to: we conclude (i) v x n satisfies the jump property on Γ 1 : lim x n 0 + ( n (1 + (k 1)χB x 1 ) v ) = 0, in Br (0), (5.1) i x i v x n (x, x n ) = k n lim xn 0 v x n (x, x n ), (x, 0) Γ 1. (ii) α v C 0 () for any α = (α 1,, α n 1, 0) N n. (iii) v L (B s (0)) for any 0 < s < r, and v L () Cr n u. (5.15) For f : R k, set D f := ( f f,,, (1 + (k n 1)χB x 1 x 1 ) f ), (5.16) n 1 x n 18

19 and let ( D f ) s = B s (0) for any 0 < β < 1, ( Dv ( Dv) s ) β s dx C r B s (0) D f dx denote the average of D f over B s (0). Then we have that Du ( Du) r dx, 0 < s r. (5.17) Combining (5.13) with (5.17) yields that for any 0 < θ < 1, Du ( Du) θr dx Cθ β Du ( Du) r dx + Cθ n r 3α. (5.18) B θr (0) As in Case I, iteration of (5.18) yields that for any 0 < s r, it holds ( Du ( Du) s ) 3α s dx C Du ( Du) r dx + Cs 3α. (5.19) r B s (0) This, combined with Case I, further implies that for any B r (x) B r0 (x 0 ) and 0 < s r, ( Du ( Du) s ) 3α x,s dx C Du ( Du) x,r dx + Cs 3α, (5.0) r B s (x) where ( Du) x,s denotes the average of Du over B s (x). It is readily seen that (5.0) yields 3α 1, u C (B r 0 (x 0 ) B ± 1 ) and u C0,1 (B r 0 (x 0 )). This completes the proof. B r (x) Now we sketch the proof of Theorem 1.1. Proof of Theorem 1.1. Define the singular set { Σ = x Ω : lim r n u dx ɛ 0 r 0 B r (x) Then by a simple covering argument we have H n (Σ) = 0 (see, for example, Evans- Gariepy [7]). For any x 0 Ω \ Σ, there exists 0 < r 0 < dist(x 0, Ω) such that u dx ɛ 0. r n 0 B r0 (x) Hence by Theorem.1, Theorem.1, and Theorem 5.1, we have }. u C 1,α( B r 0 (x 0 ) Ω ±, N ) and u C 0,1( B r 0 (x 0 ), N ), for some 0 < α < 1. In particular, we have lim r n u r 0 B dx = 0, x B r 0 (x 0 ) r (x) so that B r 0 (x 0 ) Σ = and hence Σ is closed. This completes the proof. 19

20 6 Harmonic maps to manifolds supporting convex distance functions In this section, we consider weakly harmonic maps u from (Ω, g), with g the piecewise Lipschitz continuous metric as in Theorem 1.1, to (N, h), whose universal cover (Ñ, h) supports a convex distance function square d (, p) for any p Ñ. We will establish Ñ both the global Lipschitz continuity and piecewise C 1,α -regularity for such harmonic maps u. This can be viewed as a generalization of the well-known regularity theorem by Eells-Sampson [8] and Hildebrandt-Kaul-Widman [13]. The crucial step is the following theorem on Hölder continuity. Theorem 6.1. Assume that the metric g is bounded measurable on Ω, i.e. there exist two constants 0 < λ < Λ < + such that λi n g(x) ΛI n for a.e. x Ω. Assume also that the universal cover (Ñ, h) of (N, h) supports a convex distance function square d Ñ (, p) for any p Ñ. If u H1 (Ω, N) is a weakly harmonic map, then there exists α (0, 1) such that u C α (Ω, N). Proof. Here we sketch a proof that is based on slight modifications of that by Lin [16]. Similar ideas have been used by Evans in his celebrated work [6] and Caffarelli [] for quasilinear systems under smallness conditions. First, by lifting u : Ω N to a harmonic map ũ : Ω Ñ, we may simply assume (N, h) = (Ñ, h) and dn (, p) is convex on N for any p N. We first claim that g d (u, p) 0. (6.1) In fact, by the chain rule of harmonic maps (cf. Jost [15]), we have g d (u, p) = u d (u, p)( g u) + ud (u, p)( u, u) g. Since g u T u N, u d (u, p) T u N, the first term in the right hand side vanishes. By the convexity of dn, the second term in the right hand side satisfies ud (u, p)( u, u) g 0. Since u H 1 (Ω, N), by suitably choosing p N and applying Poincaré inequality and Harnack s inequality, (6.1) implies u L loc (Ω, N). For a set E N, let diam N (E) denote the diameter of E with respect to the distance function d N (, ). For any ball B r (x) Ω, we want to show that u C α (B r (x)) for some 0 < α < 1. To do it, denote C r := diam N (u(b r (x))) < +. We may assume C r > 0 (otherwise, u is constant on B r (x) and we are done). Now we want to show that there exists 0 < δ 0 = δ 0 (N) 1 such that diam N ( u(bδ0 r(x)) ) 1 C r. (6.) 0

21 Since u r (y) = u(x + ry) : (0) N is a harmonic map ( (0), g r ), with g r (y) = g(x + ry), we may, for simplicity, assume x = 0 and r =. For any 0 < ɛ < 1, since u( ) N is a bounded set, there exists m = m(ɛ) 1 such that u( ) is covered by m balls,, B m of radius ɛc 1. Now we have Claim: There exists a sufficiently small ɛ > 0 such that u( ) can be covered by at most (m 1) balls among,, B m. To see this, let x i such that B i B ɛc1 (p i ), p i = u(x i ), for 1 i m. Let 1 m m be the maximum number of points in {p i } m i=1 such that the distance between 1 any two of them is at least C 3 1. Thus we may assume 16 C 1 (p i ), 1 i m, covers u( ). Then there exists i 0 {1,, m } such that and 1 C 1 sup x B d N(u(x), p i0 ) C 1, (6.3) H n (u 1 (B N (p i0, ) 1 16 C 1)) c 0, (6.) for some universal constant c 0 > 0, where B N (p i0, R) is the ball in N with center p i0 and radius R. In fact, since m ) u (B 1 N 1 (p i, 16 C 1), we have m i=1 i=1 H n (u 1 (B N (p i, ) 1 16 C 1)) H n ( ). Hence there exists i 0 {1,, m } such that ) H (u n 1 (B N 1 (p i0, 16 C 1)) c 0 := 1 m Hn ( ). This implies (6.). By the triangle inequality, (6.3) also holds. Define f (x) := sup z d N(u(z), p i0 ) d N(u(x), p i0 ), x. It is clear that f 0 in, and (6.1) implies By Moser s Harnack inequality, we have inf f C f C f C g f 0, in. u 1 (B N (p i0, 1 16 C 1)) ) C sup dn(u, p i0 ) sup dn(u, p i0 ) (B Hn 1 u 1 (B N (p i0, 16 1 u 1 (B N 1 (p i0, C 16 C 1)) 1)) ( 1 C C 1 1 ) 56 C 1 c 0 := θ 0 C 1 (6.5) 1 f

22 for some universal constant θ 0 > 0. This implies sup d N (u(z), p i0 ) sup d N (u(z), p i0 ) θ 0 C 1 = (1 θ 0 )C 1. (6.6) z z Now we argue that the claim follows from (6.6). For, otherwise, we would have that u( ) B ɛc1 (p j ) for all 1 j m. Let z 0 be such that ɛc 1 + d N (u(z 0 ), p i0 ) sup d N (u(z), p i0 ). Since u( ) m i= B ɛc 1 (p i ), there exists p i1 {p 1,, p m } such that u(z 0 ) B ɛc1 (p i1 ). Since u( ) B ɛc1 (p i1 ), there exists z 1 such that u(z 1 ) B ɛc1 (p i1 ). Therefore we have d N (u(z 1 ), u(z 0 )) ɛc 1. Therefore we have sup d N (u(z), p i0 ) sup d N (u(z), p i0 ) ɛc 1 + d N (u(z 0 ), p i0 ) d N (u(z 1 ), p i0 ) z z ɛc 1 + d N (u(z 0 ), u(z 1 )) 3ɛC 1, this contradicts (6.6) provide that ɛ > 0 is chosen to be sufficiently small. From this claim, we have either (i) diam N (u( )) 1 C 1. Then (6.) holds with δ 0 = 1, or (ii) diam N (u( )) > 1 C 1. Then we consider v(x) = u( 1 x) : N so that v is a harmonic map on (, g 1 ), with the metric g 1 (x) = g( 1 x). 1 C 1 < diam N (v( )) C 1. v( ) is covered by at most (m 1) balls,, B m 1 of radius ɛc 1. Thus the claim is applicable to v so that u( ) = v( ) can be covered by at most (m ) balls among,, B m 1. If diam N (v( )) 1 C 1, we are done. Otherwise, we can repeat the above argument. It is clear that the process can at most be repeated m-times, and the process will not be stopped at step k 0 m unless diam N u(b k 0 ) 1 C 1. Thus (6.) is proven. It is readily seen that iteration of (6.) implies Hölder continuity. Proof of Theorem 1.. First, by Theorem 6.1, and the argument from, we can show that for some 0 < α < 1, u dx Cr n +α, B r (x) Ω. B r (x) Then we can follow the same proof of Theorem 5.1 to show that u C 0,1 (Ω) and u C 1,α (Ω ± Γ, N).

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