Locating the first nodal set in higher dimensions

Transcription

1 Locating the first nodal set in higher dimensions Sunhi Choi, David Jerison and Inwon Kim March 12, 27 Abstract We extend the two dimensional results of Jerison [J1] on the location of the nodal set of the first Neumann eigenfunction of a convex domain to higher dimensions. If a convex domain in IR n is contained in a long and thin cylinder [, N] B ǫ () with nonempty intersections with {x 1 = } and {x 1 = N}, then the first nonzero eigenvalue is well approximated by the eigenvalue of an ordinary differential equation, by a bound proportional to ǫ, whose coefficients are expressed in terms of the volume of the cross sections of the domain. Also, the first nodal set is located within a distance comparable to ǫ near the zero of the corresponding ordinary differential equation. 1 Introduction Let be a bounded convex domain in IR n. Let u be an eigenfunction for associated with the smallest nonzero eigenvalue λ of the Neumann problem for, that is, u = λu in, u ν = on (1.1) where u ν = ν u and ν denotes the outer normal unit vector at each point on. The purpose of this paper is to locate the first nodal set Λ = {u = } and to estimate the first nonzero eigenvalue λ. We show that Λ is near the unique zero of the first nonconstant eigenfunction of a certain ordinary differential equation, and we estimate the difference between λ and the first nonzero eigenvalue of the corresponding ordinary differential equation. Assume {(x,y 1,...,y n 1 ) : x [,N],(y 1,...,y n 1 ) B ǫ ()} IR n (1.2) and suppose further that (s) := {x = s} is nonempty for < s < N. (1.3) 1

2 Let u be the first nonconstant eigenfunction for. Denote by φ 1 the first nonconstant eigenfunction with the smallest nonzero eigenvalue µ 1 for the Neumann problem on [,N] given by (wφ 1 ) = µ 1 wφ 1 on (,N); w(s)φ 1 (s) as s + or s N (1.4) where w(s) is the (n 1)-dimensional volume of (s). Let s 1 (,N) be the unique zero of φ 1, i.e., φ 1 (s 1 ) =. The main results of this paper are as follows. Theorem 1.1. If u is the first nonconstant Neumann eigenfunction of and satisfies (1.2) and (1.3), then there is a dimensional constant C such that (a) u(x,y 1,...,y n 1 ) = implies x s 1 < Cǫ (b) (1 Cǫ/N)µ 1 λ µ 1. where s 1 and µ 1 are given in (1.4). Theorem 1.1 was proven by Jerison ([J1]) for n = 2. By taking a new coordinate system, he bounds the first eigenvalue λ from below by a formula, whose coefficients are expressed in terms of the width of the cross section (x) = {y : (x,y) }. (Here y-axis is chosen so that the projection of onto y-axis has the shortest possible length.) Using ODE eigenvalue estimates, he first locate the nodal set, and then using the location of the first nodal set, he estimates the first eigenvalue. However for n > 2, a parallel approach leads to a weaker result mainly due to the fact that we do not have a proper coordinate system (s,t 1,...,t n 1 ) (x,y 1,...,y n 1 ) satisfying N s y j ds Cǫ, j = 1,...,n 1. (1.5) (see the Remarks at the end of this section.) In this paper we extend results in [J1] to higher dimensions, with a different approach to the problem. We first estimate the difference between the eigenvalues of the original PDE and the corresponding ODE, by taking an one-dimensional test function and using a sharp result of Kröger on the upper bound of a gradient of a Neumann eigenfunction, and also using a new coordinate system which will be constructed in section 4. (we will only need a weaker pointwise estimate (see (II ) in section 4) than (1.5).) Based on these estimates, it turns out that we can find a bound on the width of the first nodal set. Once we prove that the nodal set is thin, i.e., the diameter 2

3 of the nodal set is comparable to ǫ, then we are able to locate the nodal set near the zero of the eigenfunction of the corresponding ODE. First, we derive the second inequality of (b). Recall that u minimizes the Dirichlet integral J(v) = v 2 (1.6) among all functions v on satisfying v 2 = 1, v =. (1.7) The minimum value of J is the eigenvalue λ. If we consider functions of x alone, i.e., v(x,y 1,...,y n 1 ) = φ(x), then J(v) = I(φ) := and the constraints (1.7) become N φ (s) 2 w(s)ds (1.8) N φ(s) 2 w(s)ds = 1, N φ(s)w(s)ds =. (1.9) As in [J1] we observe that the minimizer of (1.8) under the constraints (1.9) is the first nonzero eigenfunction φ 1 given in (1.4) and I(φ 1 ) = µ 1. Hence λ µ 1. (1.1) Remark 1. If we normalize N = 1, then by [L] and [ZY], C 1 λ for some absolute constant C 1 >, and by plugging in the test function φ(x) = sin πx 1, we get λ C 2 for some dimensional constant C 2. Remark 2. In the case of Dirichlet problem on a planar convex domain, Jerison [J2] obtained results corresponding to Theorem 1.1. Later Grieser and Jerison ([GJ1], [GJ2]) showed that the nodal line is in an x-interval of much shorter length Cǫ/N (possibly at distance Cǫ from s 1 ). We expect that there is an analogous bound in the Neumann problem. Remark 3. An analogous approach to the method in [J1] for higher dimensions, by modifying the methods in [J1]-[J2], leads to a result weaker than Theorem 1.1, i.e., with ǫlog(n/ǫ) instead of ǫ. 3

4 2 Preliminary results and corollaries Throughout the paper we normalize such that N = 1. As mentioned in the introduction, a key ingredient in the proof of Theorem 1.1 is Kröger s comparison theorems on the gradient of eigenfunction u, which we state below. Theorem 2.1 (Kröger, [K1], [BQ] ). Let be a convex domain in IR n with smooth boundary. Let u be the first eigenfunction for the Laplace operator on with the associated eigenvalue λ >, under Neumann boundary conditions. Let v be a solution on some interval (a, b) of the differential equation v (x) + n 1 x v (x) = λv(x) on (a,b) v (a) = v (b) = (2.1) such that v on (a,b) and [min u,max u] [min v,max v]. Then (v 1 u) 1. Theorem 2.2 (Kröger, [K2]). Let, u and λ be given as in Theorem 2.1. Suppose maxu minu. Let b be a positive number such that λ is the first nonzero eigenvalue of ψ (x) + n 1 x ψ (x) = λψ(x) on [,b], ψ () = ψ (b) =. (2.2) If ψ is the corresponding eigenfunction with ψ() >, then maxu minu maxψ min ψ. Corollary 2.3. Let and u be given as in Theorem 2.1. If N = 1 and max u = 1, then u C for some dimensional constant C. Proof. First, we claim that if a < b then the solution v to (2.1) satisfies max v λ v(a). (2.3) To see this, multiply v to both sides of (2.1) to obtain for x >. Hence, v v + λv v = 1 2 ((v ) 2 + λv 2 ) = n 1 x (v ) 2 (2.4) (v ) 2 (x) + λv 2 (x) (v ) 2 (a) + λv 2 (a) = λv 2 (a) 4

5 and the claim follows. Now, for sufficiently large M >, consider an interval (a,b) (M, ) such that λ is the first nonzero eigenvalue of the Neumann problem (2.1). Let v be the corresponding eigenfunction. If M is large enough, then v is close to a constant multiple of cos λx, and thus we can normalize v so that 1 min v 2 and 1 max v 2. Then by Theorem 2.1, we get sup u sup v C where the second inequality follows from Remark 1 and the above claim. Corollary 2.4. Let and u be given as in Theorem 2.1. Suppose that N = 1 and 1 = maxu minu. Let k(ǫ) be the smallest integer such that 2 k(ǫ) ǫ. For integers 1 k < k(ǫ), let I k = [2 k 1,2 k ] [1 2 k,1 2 k 1 ] and let I k(ǫ) = [,2 k(ǫ) ] [1 2 k(ǫ),1]. Then there exists a dimensional constant C > such that sup u C2 k for 1 k k(ǫ). (2.5) x I k Proof. Suppose u < on () and u > on (1). Denote m = min u. We claim that max u m + 3M ǫ. (2.6) x () where M is the upper bound for u. To see this, let s = min{x : min (x) u = m}. By Corollary 2.3, u M (2.7) with a dimensional constant M, and thus max (s ) u m+2m ǫ. Assume that max () u > m + 3M ǫ, then {u > m + 2M ǫ} has a component A such that A > and A {x s }. Let { u for x A ũ = m + 2M ǫ for x A 5

6 then u 2 u 2 > ũ 2 ũ 2 and we get a contradiction. A parallel argument yields min u 1 3M ǫ. (2.8) x (1) Now we show (2.5) using Theorems 2.1 and 2.2. First, consider the case k = k(ǫ), on the interval [,2 k(ǫ) ] [,ǫ]. Recall that m := minu max u = 1 and u m + 3M ǫ on (). Let ψ be the first eigenfunction of (2.2) with ψ on (,b), ψ() > and min ψ = m. Observe that ψ is decreasing and thus ψ(b) = m. By Theorem 2.2, maxψ maxu and thus [min u,max u] [min ψ,max ψ]. Also, since ψ satisfies (2.4) with v replaced by ψ, ψ (x) 2 = b x 2(n 1) ψ (t) 2 + λ(ψ(t) 2 ) dt t = b x Hence for x J := {ψ m + Cǫ} ψ (x) 2 b x 2(n 1) ψ (t) 2 dt + λ(ψ(b) 2 ψ(x) 2 ) t 2(n 1) ψ (t) 2 dt + 2Cmλǫ t Assume that ψ (x) 2 = Mǫ for the first time in [x,b] J. Then by above inequality Mǫ Mǫ 2(n 1)log b x + 2Cmλǫ, which yields log b x 1 M 1 C(2mλ). Therefore if we let M := 4Cmλ, 2(n 1) then we obtain x e 1/4(n 1) b. Choose a sufficiently small dimensional constant c > such that if ǫ c then J = {ψ m + Cǫ} [e 1/4(n 1) b,b]. It follows that if ǫ c, then ψ C 1 ǫ, C1 = 4Cmλ on J. 6

7 By Theorem 2.1 and above argument with C = 3M, we obtain that if ǫ c then Du M 1 ǫ in {u m + 3M ǫ} (2.9) where M 1 = 2 3M mλ. Using the improved gradient bound (2.9) instead of (2.7), (2.6) improves to u m + 3M 1 ǫ 3/2 on { x ǫ}. (2.1) Next we repeat the argument with the improved bound (2.1), i.e., with {ψ m+cǫ} replaced by {ψ m+3m 1 ǫ 3/2 }. It follows that for ǫ c where Du M 2 ǫ 1/2+1/4, u m + 3M 2 ǫ 1+1/2+1/4 on { x ǫ} 3M 2 := 6(6(3M mλ) 1/2 mλ) 1/2 = 6 1+1/2 3M 1/4 (mλ) 1/2+1/4. Iteration of above argument will yield that, if ǫ < c then Du 36mλǫ C ǫ on { x ǫ} where C is a dimensional constant. For other intervals [2 k 1,2 k ] (k < k(ǫ)), a similar iteration can be applied with the first step max u m + 4M 2 k, (2.11) x [,2 k ] which follows from (2.6) and (2.7). Then a parallel argument with ǫ replaced by 2 k proves (2.5) on [2 k 1,2 k ] for k such that 2 k c. Note that other than to derive (2.6), we do not use the fact that ǫ is the specific constant depending on. For intervals near x = 1, (i.e., for [1 2 k,1 2 k 1 ]) the proof is divided into two cases. First if maxu = 1 = min u, then for w := u(1 x,y 1,...,y n 1 ), min w = max w = 1, w < on {x = }, and w > on {x = 1}. Therefore the argument for w on intervals near x = gives the result for u on intervals near x = 1. Secondly if maxu = 1 > min u, then as in the proof of Corollary 2.3, choose sufficiently large constants a and b such that λ is the first nonzero eigenvalue of (2.1) and the first eigenfunction v satisfies maxv = 1 and min v minu. 7

8 Since [min u,maxu] [min v,max v] and min u 1 3M ǫ = maxv 3M ǫ, (1) a similar reasoning as in the interval [2 k 1,2 k ], yields the result for the interval [1 2 k,1 2 k 1 ] near x = 1. From Lemma 2.3, we obtain Corollary 2.5, which states that the first nodal set is located in the middle part of. Later in section 4, Corollary 2.5 and Theorem 2.1 will be used along with a new coordinate system to estimate the first nonzero eigenvalue λ. Based on the bound on λ, the width and location of the nodal set are derived, again by using Theorem 2.1. Corollary 2.5. Let and u be given as in Theorem 2.1. Suppose N = 1 and sup u = 1. Then there exist dimensional constants c 1 > and c 2 > such that c 2 u 1 for x [,c 1 ] [1 c 1,1]. Proof. Without loss of generality, we may assume max u = u( x,ỹ) = 1 (ỹ IR n 1 ) and u > on {x = 1}. By Corollary 2.3, there exists a dimensional constant c 1 > such that u(x,y) > 1/2 if x x < c 1. Since the Courant nodal domain theorem [CH, p.452] implies that + and are connected, u > 1/2 in {x > x c 1 }. On the other hand, since u = u +, Lemma 2.3 implies min u + c 2 for some dimensional constant c 2 >. Hence we obtain Corollary 2.5 by a similar reasoning for u as in u +. 3 ODE eigenvalue estimates In this section we prove several lemmas on ODE eigenvalue estimates, which will be applied to the one-dimensional eigenfunction φ 1 in section 4. In particular Lemma 3.5 will be used to locate the nodal set in section 4, and Lemma 3.6 yields the bound on the width of nodal set. The proof for Lemmas 3.1, 3.2, 3.3 and 3.4 are parallel to those of the corresponding lemmas in [J1]. The only difference in the proof is that, instead of the concavity of the height of the cross-section h(x) for n = 2, we have the concavity of w 1/n 1 (x) by the Brunn-Minkowski inequality for the volume of cross-section w(x). 8

9 Lemma 3.1. For a 1/2, inf {φ:φ(a)=} a φ (x) 2 w(x)dx a φ(x)2 w(x)dx 1 2 n 2 a 2. Proof. The proof is the same as that of Lemma 4.2 of [J1], using the fact that w(x) 2 n 1 w(t) for x t a. Lemma 3.2. There exists a constant C > depending on n such that b a φ (x) 2 w(x)dx b a φ(x)2 w(x)dx inf {φ:φ(a)=} C (b a) 2. Proof. Take a test function φ(x) = x. We need to show that w(x)dx C x 2 w(x)dx, where C depends on n. Multiply w by a constant so that w1/(n 1) (t)dt = 1. Due to the normalization and the concavity of w 1/(n 1), the arguments in the proof of Lemma 4.3 in [J1] yield that By Hölder inequality, ( x x w 1/(n 1) (t)dt (1 x) 2 for x 1. w(t)dt) 1/(n 1) (1 x) (n 2)/(n 1) x w 1/(n 1) (t)dt (1 x) 2, and thus W(x) := x w(t)dt (1 x)n. Therefore by integration by parts, x 2 w(x)dx = x 2 W (x)dx = 2xW(x)dx 2x(1 x) n dx = 2/(n + 1)(n + 2), where the second inequality holds because x 2 W(x) = for x =,1. On the other hand, since w 1/(n 1) (t) is concave with w 1/n 1 (),w 1/n 1 (1), the graph of w 1/(n 1) (t), t 1 is above the triangle generated by (,),(1,) and (t,w 1/n 1 (t )) where w(t ) = max w. It follows that w(t)dt w(t ) (2 9 w 1/(n 1) (t)dt) n 1 = 2 n 1

10 and our lemma holds with C = C(n) = 2 n (n + 1)(n + 2). Lemma 3.3. Let s 1 be the zero point of φ 1 given in (1.4). Then there exist constants c 1 > and c 2 > depending on n such that c 1 < s 1 < c 2. Proof. The lemma follows from Lemmas 3.1, 3.2 and the proof of Lemma 4.4 in [J1]. For a [,1], define E[a,1] = inf {φ:φ(a)=} a φ (x) 2 w(x)dx a φ(x)2 w(x)dx. Lemma 3.4. Suppose that c a 1 c for some < c < 1. Then there exists a constant C > depending on n and c such that Proof. Normalize w so that ( / a)e[a,1] C. max w(x) = 1. (3.1) x N By concavity of w 1/(n 1) and ( 3.1), w 1/(n 1) (c ) c. Let φ be the unique nonnegative minimizer for E[a,1] with the normalization a φ(x) 2 w(x)dx = 1. (3.2) Following the proof of Lemma 3.4 in [J1], we only need to prove that φ(x) CE n for a x 1, (3.3) where E = E[a,1] and C is a constant depending on n and c. Observe that, since φ satisfies (wφ ) = Ewφ and φ (1) =, wφ (x) E( x x Ew(t)φ(t)dt (3.4) φ 2 (t)w(t)dt) 1/2 ( 1 x w(t)dt) 1/2

11 In particular, (3.1) and (3.2) imply that wφ E for a x < 1. Since φ(a) =, we have t ds φ(t) E a w(s). (3.5) On the other hand, by concavity of w 1/n 1 (t), for a s t 1, w 1/n 1 (s) is above the line l(s) = αs+β connecting (c,c ) and (t,w 1/n 1 (t)). Without loss of generality, we may assume that c > w 1/n 1 (t) and α < (Other cases are better.) For n 3 t a ds w(s) t c ds (αs + β) n 1 Hence by (3.4), (3.5) and (3.6) φ (x) E w(x) E2 w(x) CE2 w(x) C (αt + β) n 2 = x x x w(t)φ(t)dt t w(t)( a w 1/(n 1) (t)dt CE 2 (1 x) w (n 2)/(n 1) (x) CE 2 1 w (n 3)/(n 1) (x) C w (n 2)/n 1 (t). (3.6) 1 w(s) ds)dt where the fourth and fifth inequalities follow respectively from and (3.7) w 1/(n 1) (t) C(n)w 1/(n 1) (x) for a x t 1 (3.8) w 1/(n 1) (x) min{1 x,c } for a x 1. (3.9) (3.8) and (3.9) are due to the concavity of w 1/(n 1) and the normalization (3.1). (For (3.8), see Remark 4.1 (b) in [J1].) (3.7) and (3.8) yield that φ(x) x a φ (t) dt C 1 E 2 w (n 3)/(n 1) (x) for a x 1, (3.1) where C 1 is a constant depending on n and c. We go back to the first inequality of (3.7) and apply (3.1) and then (3.8) and (3.9) to obtain φ (x) C 1E 2 w(x) x w(t) 2/(n 1) dt 11 C 2 E 3 w(x) (n 4)/(n 1)

12 where C 2 is a dimensional constant. Now by similar reasoning as in (3.1), the improved estimate on φ holds: φ(x) C 3 E n w(x) (n 4)/(n 1). We repeat the above process (n 4) more times to obtain φ(x) C(n)E n. Lemma 3.5. Let φ 1 and s 1 be given as in (1.4) and suppose N = 1. If φ is a function on (,1) such that φ(s 1 ) = and s 1(φ ) 2 wds 1 s φ 2 wds 1 (1 + Mǫ) s 1 (φ 1 )2 wds s 1 φ 2 1 wds, then s 1 s 1 + Cǫ for some constant C depending on n and M. Proof. The lemma follows from Lemmas 3.3, 3.4 and from the fact E[s 1,1] = µ 1 = s 1 (φ 1 )2 wds s 1 φ 2 1 wds. Next we show that if the energy associated with φ is bounded by (1 + Mǫ)µ 1, then sup φ is bigger than ǫ on any interval of length Cǫ. Lemma 3.6. Let N = 1 and µ 1 be given in (1.4). Suppose φ(s) is a function on (,1) such that φwds =, sup φ = 1, sup φ C 1 and µ 1 (φ ) 2 wds φ2 wds (1 + Mǫ)µ 1. (3.11) If for some < a < b < 1 and C 2 > sup φ 2C 1 ǫ and [a,b] a φwds C 2 wds, then b a < Cǫ for some C > depending on C 1, C 2 and M. 12

13 Proof. Let φ 1 and s 1 be as given in (1.4). Suppose that sup φ 2C 1 ǫ and b = a + Cǫ, [a,b] for some a and sufficiently large C >. Without loss of generality we may assume a s 1 + Cǫ/4 or b s 1 Cǫ/4. (3.12) ( If a < s 1 < b, then s 1 a Cǫ/2 or b s 1 Cǫ/2. If b s 1 Cǫ/2, replace a with (s 1 +b)/2. If s 1 a Cǫ/2, replace b with (s 1 +a)/2. Lastly if a < b < s 1 or s 1 < a < b, by replacing a or b with (a+b)/2, we get (3.12).) Changing the sign of φ if needed, we also set a φwds = a φwds. Define a A φ = φ2 wds φ2 wds, B a φ = φ2 wds φ2 wds, A φ + B φ = 1. (3.13) By our hypothesis, (φ ) 2 a wds φ2 wds = A (φ ) 2 wds φ a φ2 wds + B a (φ ) 2 wds φ a φ2 wds (1 + Mǫ)µ 1. If C is sufficiently large, then Lemma 3.4 and above inequality imply that a (φ ) 2 wds a φ2 wds µ a 1 (φ ) 2 wds a φ2 wds. (3.14) We will construct a test function ψ such that ψwds = and (1 + Mǫ) (ψ ) 2 wds ψ2 wds < (φ ) 2 wds φ2 wds, (3.15) which contradicts our hypothesis. First, construct a continuous function ψ such that ψ = φ on the left interval [,a] and ψ = φ + C 2 (b a)/1 on the left interval [b,1]. For α = C 2 /1, let φ(s) for s a ψ(s) = φ(s) + α(s a) for a s b = a + Cǫ φ(s) + α(b a) for b s 1. 13

14 From a straightforward calculation, it follows that the energy of ψ on [a,1] gets smaller than the energy of φ on [a,1] by some amount. More precisely ψ a 2 wds ψ a 2 wds (1 C a Cǫ) φ 2 wds a φ2 wds (3.16) for some C > depending on C 1 and C 2. Next we perturb ψ and get ψ such that ψwds = and will show that ψ satisfies (3.15). Set ψ(s) = φ(s) for s a ψ(s) = β( ψ(s) φ(a)) + φ(a) for a s 1 where β > is chosen to satisfy ψwds =, i.e., a ψwds = a φwds. Then a β = 1 α b a wds 1 a φwds + O(ǫ2 ). Since it follows that a a φwds φ 2 wds wds, (3.17) ψ 2 a wds (1 + 2α b a wds 1 a φwds + C 3ǫ 2 ) a a a φ 2 wds. Therefore ψ 2 wds (1 + C 3 ǫ 2 ) a a φ 2 wds and B ψ (1 + C 3 ǫ 2 )B φ (3.18) where B ψ is defined as in (3.13) (Recall that A ψ = 1 B ψ ). Therefore, 14

15 (ψ ) 2 wds ψ2 wds a = A (ψ ) 2 wds ψ a ψ2 wds + B a (ψ ) 2 wds ψ a ψ2 wds a = A (φ ) 2 wds ψ a φ2 wds + B a (ψ ) 2 wds ψ a ψ2 wds a (1 + C 3 ǫ 2 )(A (φ ) 2 wds φ a φ2 wds + B a (ψ ) 2 wds φ a ψ2 wds ) (1 + C 3 ǫ 2 )(A φ a (φ ) 2 wds a A φ a (φ ) 2 wds a φ2 wds + B φ(1 C Cǫ) φ2 wds + B φ(1 C Cǫ) (φ ) 2 wds φ2 wds B φc Cǫ a (φ ) 2 wds a φ2 wds ) a (φ ) 2 wds a φ2 wds + C 3ǫ 2 (1 + Mǫ)µ 1 a (φ ) 2 wds a φ2 wds + C 3ǫ 2 (1 + Mǫ)µ 1 where we obtain the first inequality from A φ +B φ = A ψ +B ψ = 1, (3.18) and (3.14), the second inequality from (3.16), the third inequality from (3.11). From the hypothesis a φwds C 2 wds and (3.17), one can observe that B φ is bounded below by a constant depending on C 2. Hence if we choose a sufficiently large C depending on C 1, C 2 and M, and if ǫ is sufficiently small compared to C, then (3.15) holds. In the next lemma, we show the nondegeneracy of φ 1 near the zero s 1. Lemma 3.7. Let φ 1 be the first nonzero eigenfunction of (1.4) with N = 1 and let s 1 be the zero of φ 1 - Note that c 1 s 1 c 2 by Lemma 3.3. Normalize φ 1 such that φ 1 > on (s 1,1] and maxφ 1 = 1. Then φ 1 on [,1]. Moreover there exists a dimensional constant c = c(n) such that φ 1 c(n) on [s 1 c 1 /2,s 1 + (1 c 2 )/2]. Proof. By the normalization φ 1 > on (s 1,1] and φ 1 < on [,s 1 ). Hence from (1.4) we observe that wφ 1 has its maximum at x = s 1 and wφ 1 is increasing on [,s 1 ] and decreasing on [s 1,1]. From the boundary condition it follows that φ 1 on [,1] and thus φ 1(1) = 1. We will only prove that φ 1 c(n) on [s 1,s 1 + (1 c 2 )/2]: a parallel argument leads to the rest of the claim. 15

16 Observe that from (1.4) we have φ 1(x) = x µ 1w(t)φ 1 (t)dt. (3.19) w(x) We will use (3.19) to find lower bounds on φ respectively near t = 1 and then near t = s 1, using that wφ has its maximum at s 1. Note that for c 1 x t 1, the concavity of w 1/(n 1) (x) implies (1 t)w 1/(n 1) (x) w 1/(n 1) (t). (3.2) (See Remark 4.1 (a) in [J1].) Thus (3.8) and (3.2) imply that C 1 (1 x) n w(x) x w(t)dt C 2 (1 x)w(x) (3.21) where C 1 and C 2 are positive dimensional constants. Let φ 1 (s 2 ) = 1/2 for s 2 [s 1,1]. Then by (3.19) and (3.21), it follows that for x [s 2,1], C 1 µ 1 (1 x) n+1 φ 1(x) C 2 µ 1 (1 x). (3.22) Since φ 1 (1) = 1, (3.22) implies that φ 1 (x) 1 C 3 µ 1 (1 x) 1/2 if x [A,1] where C 3 and A are dimensional constants. Therefore s 2 A and φ 1(s 2 ) C 1 µ 1 (1 s 2 ) n+1 C 4, (3.23) where C 4 is a dimensional constant. Now we have, for s 1 t s 2, φ 1 (t) w(s 2) w(t) φ 1 (s 2) C 5 φ 1 (s 2) C 6 where the second inequality is due to t s 2 < A and the last from (3.23). For t [s 1,s 1 + (1 c 2 )/2], (3.22) implies φ 1 (t) C 7 and our claim is proved. 4 A new coordinate system In this section, we define a new coordinate system for satisfying certain properties, which will enable us to use Fubini s Theorem when we construct a one-dimensional test function with small energy. For the proof of the main lemma (Lemma 4.1), first we will prove Lemma 4.2 for general dimensions. Then we construct in detail coordinate systems for n = 3 and n = 4 satisfying the properties given in the main lemma. We will then discuss the general dimension based on the three and four-dimensional cases. 16

17 Lemma 4.1. Let be a bounded convex domain in IR n satisfying (1.2) and (1.3). Suppose N = 1. Then there exists a coordinate system (s,t 1,...,t n 1 ) [,1] B 1 () to (x,y 1,...,y n 1 ) such that the following statements hold. (I) s =x (II) s y j Cǫmax{1/s,1/(1 s)} for a dimensional constant C. (III) the mapping f : (s,t 1,...,t n 1 ) (s,y 1,...,y n 1 ), has a constant Jacobian a n w(s), where w(s) is the volume of cross-section (s) and a n is a dimensional constant. Here we denote by s y j = y j / s, the partial derivative of y j with t 1,..., t n 1 held fixed. For the proof of Lemma 4.1, we need the following lemma. Lemma 4.2. Let D be a planar bounded convex domain. Suppose (,ǫ) D, (, C 1 ǫ) D (C 1 > ), and (±ǫ,) D. Further suppose that the length of projection of D on the y-axis is less than C 2 ǫ for some C 2 >. If (r cos θ,r sin θ) D and (s cos(θ θ),ssin(θ θ)) D, then there exists C depending on C 1 and C 2 such that for sufficiently small θ. r s Cr 2 θ/ǫ Proof. Suppose θ π/4. Consider a line l passing through (,ǫ) and (r cos θ,r sinθ). Let and ( s cos(θ θ), s sin(θ θ)) l (s cos(θ θ),ssin(θ θ)) D. Since is convex, s s. The equation for l is y = r sin θ ǫ x + ǫ and thus r cos θ If θ is sufficiently small, s sin(θ θ) = r sin θ ǫ s cos(θ θ) + ǫ. r cos θ s s r(1 + 2( r cos θ ǫ + r sin θ tan θ ǫ ) θ) r(1 + C r θ ). ǫ 17

18 On the other hand, if l is a line passing through (, C 1 ǫ) and (r cos θ,r sinθ), by a similar reasoning r s(1 + C s θ ǫ ) for C depending on C 1 and the proof is done for θ π/4. Next, suppose π/4 θ π/2. From the hypothesis on, the angle between the tangent line to at (r cos θ,r sinθ) and the line connecting the origin to (r cos θ,r sin θ) is bounded below by an angle depending on C 2. Hence, s r Cr θ C r2 θ ǫ for some C depending on C 2, where the second inequality follows from r Cǫ. Using Lemma 4.2, we prove Lemma 4.1. Proof of Lemma 4.1. Without loss of generality, we may assume that (,,...,),(1,,...,), since if l = {(x,l 1 (x),...,l n 1 (x)) : x 1} is a line segment connecting a point on {x = } to a point on {x = 1}, then 1 length(l) 1+Cǫ and l j Cǫ for 1 j n 1 and C depending on n. Denote a i = 2 i, b i = 1 2 i (a 1 = b 1 ). For i 1, set I i = (a i+1,a i ), I i = (b i,b i+1 ). With this notation, the condition (II) of Lemma 4.1 becomes equivalent to the following statement: (II ) s y j C2 i ǫ for s I ±i which will be verified in the proof instead of (II). Recall that (s) = {x = s}. Denote by h 1 ((x)), the minimum length of projection of (x) IR n 1 on a line l 1 IR n 1, and denote by h j ((x)),j = 2,...,n 1, the minimum length of projection of (x) on a line l j which is perpendicular to l k for 1 k j 1 (See Figure 1). Since is convex, each i := {x I i } has orthonormal basis e i1,e i2,...,e in 1 IR n 1 18

19 ( x) a ( i ) x = x= a i +1 a i ei 2 e i 2 ε i2 e i1 ε i2 e i 1 ε i1 ε i 1 Figure 1 such that for x I i, p ei1 ((x)) h 1 ((x)) h 1 ((a i )) p ein 1 ((x)) h n 1 ((x)) h n 1 ((a i )) (4.1) where p eij ((x)) denotes the length of projection of (x) on a line parallel to e ij. Denote ǫ i1 = h 1 ((a i )),..., ǫ in 1 = h n 1 ((a i )), so that (x) has dimensions comparable to ǫ i1,..., ǫ in 1 for x I i where ǫ i1 ǫ i2... ǫ in 1. For a domain D IR n, define the center of mass of D as 1 D D xdx, where D denotes the volume of D. Let (x,l(x)) be the curve in, that is linear on each interval I ±i and equal to the center of mass of the crosssection at each endpoints of I i. In other words, L(a i ) is the center of mass 19

20 of (a i ), and similarly for b i. On I i, denote L(x) = L 1 (x)e i L n 1 (x)e in 1, x I i. Since the line segment connecting (,,...,) and (1,,...,) is contained in, (a i,,...,) (a i ) and (a i+1,,...) (a i+1 ), which imply L j (a i ) L j (a i+1 ) L j (a i ) + L j (a i+1 ) Cǫ ij and hence L j(x) C2 i ǫ ij for x I i. (4.2) 4.1 n = 3 First, when n = 3, we define a new coordinate system (s,t 1,t 2 ) [,1] B 1 () to (x,y,z) as follows. (i) x = s, (y,z) = f(s,t 1,t 2 ) (ii) f(s,,) = L(s), f(s,1,) f(s,,) is parallel to e i1 for every s I i (iii) the mapping f : (t 1,t 2 ) (y,z), with s held fixed, is linear on every line segment from (,) to B 1 () (iv) the mapping f : (t 1,t 2 ) (y,z), with s held fixed, has a constant Jacobian w(s)/π, where w(s) is the area of (s). Note that with s held fixed, f(t 1,t 2 ) is defined so that B 1 () { θ a} is mapped onto a sector (s) { (e i1,(y 1,y 2 ) L(s)) b} with volume V satisfying a 2π = V (s). (Here, (v 1,v 2 ) denotes the angle between v 1 and v 2, that is (v 1,v 2 ) = arccos( (v 1,v 2 ) v 1 v 2 ).) Observe that if g 1 (s) is a function on I i such that then by the convexity of, {(s,g 1 (s)e i1 ) : s I i }, g 1 (s) C g 1(s) / s C2 i ǫ i1 for s I i. (4.3) 2

21 Similarly, if {(s,g 2 (s)e i2 ) : s I i } is a curve on, then g 2(s) C2 i ǫ i2 for s I i. (4.4) (4.3) and (4.4) imply that if x I i and h > is sufficiently small, where (x) (x + h) C2 i hǫ i1 ǫ i2 (4.5) (x) (x + h) := ((x) (x + h)) ((x) (x + h)). Now to prove (II ) for x I i, fix (t 1,t 2 ) B 1 (). Since L(x) satisfies (4.2) and ǫ ij ǫ, we may assume L(x) = L(x+h) without loss of generality. Denote θ x = (e i1,f(x,t 1,t 2 ) L(x)) and r x = f(x,t 1,t 2 ) L(x). Change the coordinates in IR 2 so that L(x) = L(x + h) = and f(x,t 1,t 2 ) = (r x cos θ x,r x sin θ x ). Fix a sufficiently small h >, and denote f(x + h,t 1,t 2 ) = (r x+h cos θ x+h,r x+h sin θ x+h ). Since f has a constant Jacobian, (4.5) implies since r 2 x θ x+h θ x C2 i hǫ i1 ǫ i2 (4.6) (x) { θ θ x } (x + h) { θ θ x } (x) (x + h) C2 i hǫ i1 ǫ i2. Let s > be a number such that (s cos θ x+h,ssin θ x+h ) (x). Then Lemma 4.2 implies which yields s r x Cr 2 x θ x+h θ x /ǫ i1, (4.7) (r x cos θ x,r x sin θ x ) (s cos θ x+h,ssin θ x+h ) Cr x θ x+h θ x + s r x C2 i hǫ i2 21

22 where the last inequality follows from (4.6), (4.7) and ǫ i1 Cr x. On the other hand, since A := (s cos θ x+h,ssin θ x+h ) and B := (r x+h cos θ x+h,r x+h sinθ x+h ) are points on (x) and (x + h) with the same angle from e i1, a similar argument as in (4.3 ) and (4.4) yields Hence we conclude f(x,t 1,t 2 ) f(x + h,t 1,t 2 ) A B C2 i hs C2 i hǫ i2. = (r x cos θ x,r x sin θ x ) (r x+h cos θ x+h,r x+h sin θ x+h ) C2 i hǫ i2 C2 i hǫ and this implies (II ) by sending h to. 4.2 n = 4 Next we consider n = 4. Recall that for x I i, p eij ((x)) ǫ ij with ǫ i1 ǫ i2 ǫ i3, and (x,l(x)) = (x,l 1 (x)e i1 + L 2 (x)e i2 + L 3 (x)e i3 ) is a line segment connecting the center of mass of (a i ) to that of (a i+1 ) on I i. To define a new coordinate system, we first divide (x) into two parts with the same volume as follows. Let L 3 (x) be a number such that {y 1 e i1 + y 2 e i2 + y 3 e i3 (x) : y 3 L 3 (x)} = (x) /2. Let (r,θ 1,θ 2 ) be a polar coordinate in (x) with r = at L(x) := L 1 (x)e i1 + L 2 (x)e i2 + L 3 (x)e i3. (4.8) Let g x be a density function on (x) with total mass 1, i.e., for any a 1,a 2,b 1,b 2 2π {(r,θ 1,θ 2 ) (x) : a 1 θ 1 b 1,a 2 θ 2 b 2 } = (x) g x dθ 1 dθ 2. (4.9) (x) {a j θ j b j, j=1,2} 22

23 (Note that g x (y) = C r(y)3 (x), r(y) = y L(x) where C > is a dimensional constant.) Now we construct a coordinate system (s,t 1,t 2,t 3 ) [,1] B 1 () to (x,y 1,y 2,y 3 ) as follows. (i) x = s, (y 1,y 2,y 3 ) = f(s,t 1,t 2,t 3 ) (ii) f(s,,,) = L(s), f(s,1,,) f(s,,,) is parallel to e i1 for every s I i (iii) the mapping f : (t 1,t 2,t 3 ) (y 1,y 2,y 3 ), with s held fixed, is linear on every line segment from (,,) to B 1 () (iv) the mapping f : (t 1,t 2,t 3 ) (y 1,y 2,y 3 ) satisfies the following properties (a) f maps the half sphere onto B right 1 () := {(t 1,t 2,t 3 ) B 1 () : t 3 } right (x) := {y 1 e i1 + y 2 e i2 + y 3 e i3 (x) : y 3 L 3 (x)}. (b) f maps the half cone onto B(θ) := {t B right 1 () : (t,e 1 ) θ} (x,φ(x,θ)) := {y right (x) : ( y,e i1 ) φ(x,θ)} where y is a vector from L(x) to y, and φ(x,θ) is the angle such that B(θ) B right 1 () = (x,φ(x,θ)) right. (x) (c) Let B(θ) = {t B right 1 () : (t,e 1 ) = θ} and (x,φ(x,θ)) = {y right (x) : ( y,e i1 ) = φ(x,θ)}. 23

24 Then the mapping f : B(θ) (x,φ(x,θ)), with s and θ held fixed, satisfies for any A B(θ). g x f(a) = g x (x,φ(x,θ)) A B(θ) (4.1) (d) f satisfies parallel properties for B left () := B 1 () B right 1 () and left (x) := (x) right (x). (f might be discontinuous on the intersection of B right () and B left ().) The conditions (i)-(iii) are parallel to those when n = 3, and the only difference between n = 3 and n = 4 is the condition (iv). In fact, when n = 3, there is a unique map f : (s,t 1,t 2 ) (y,z) with a constant Jacobian on each cross sections, if we fix one direction e i1 (See condition (ii)). However when n = 4, there are infinitely many maps f : (s,t 1,t 2,t 3 ) (y 1,y 2,y 3 ) with a constant Jacobian, even if we fix any two directions. In other words when n = 4, the properties stated for three dimensional case do not suffice to construct a function f which satisfies (II ). Hence when n = 4, we construct a (unique) map f with a constant Jacobian, under the constraint (iv) that f maps a two dimensional surface in B 1 () with a fixed angle θ from e 1, to a two dimensional surface in (x) with a fixed angle φ(x,θ) from e i1. Then the map f, with θ fixed, is two-dimensional and area-preserving with respect to the normalized density function g x / g x. Hence we can proceed as in the case n = 3. Here we divide into two parts right (x) and left (x), so that the shorter arcs of are mapped to the shorter arcs of {y (x) : (y,e i1 ) = φ(x,θ)} {y ( x) : (y,e i1 ) = φ( x,θ)} by f x f 1 x, where f x and f x denote the map f with s = x and s = x, respectively. Now to prove (II ), let x I i and fix a sufficiently small h >. Translate so that (x, L(x)) =. Let z 1 = a 1 e i1 + a 2 e i2 and z 2 = b 1 e i1 + b 2 e i3 be points on (x, φ(x, θ)) (See Figure 2). 24

25 e i 1 ( x, φ ( x, θ ) ) z 1 δ1 δ 2 z 2 left x ( ) r 1 r 2 right ( x) e i3 e i 2 Figure 2 In other words, z 1 (x) is a point on e i1 e i2 plane such that the angle between e i1 and z 1 equals φ(x,θ). Similarly, z 2 (x) is the point on e i1 e i3 plane such that the angle between e i1 and z 2 equals φ(x,θ). Denote by δ 1, the distance from z 1 to e i1 -axis and by δ 2, the distance from z 2 to e i1 - axis. Since ǫ i1 ǫ i2 ǫ i3, one can observe that δ 1 Cδ 2 for a dimensional constant C. Also denote r 1 = z 1 and r 2 = z 2, then r 1 Cr 2. Observe that since (x,φ(x,θ)) is convex, ǫ i1 δ 1 δ 2 /C (x,φ(x,θ)) Cǫ i1 δ 1 δ 2 (4.11) for a dimensional constant C. On the other hand, by similar arguments as in (4.3) and (4.4) for x I i and a small h >. This implies and thus (x) (x + h) C2 i h (x) (4.12) ǫ i1 ǫ i2 L 3 (x) L 3 (x + h) C2 i h (x) L 3 (x) C2i ǫ i3, L 1 (x) C2i ǫ i1, L 2 (x) C2i ǫ i2 (4.13) 25

26 where the second and third inequalities follow by the same argument as in (4.2). Recall that by definition of φ(x,θ) and φ(x + h,θ), (x, φ(x, θ)) (x) = B(θ) B 1 () (x + h,φ(x + h,θ)) =. (x + h) The above equality and (4.12) imply (x,φ(x,θ)) (x + h,φ(x + h,θ)) C2 i h (x,φ(x,θ)). (4.14) To simplify notations, denote = (x,φ(x,θ)), 2 = (x + h,φ(x + h,θ)) and let 1 be an intermediate region such that 1 = right (x) {y : ( y,e i1 ) φ(x + h,θ)}. (Note that 1 is an intermediate region in the sense that and 1 are contained in (x), but the angle from ǫ i1 to 1 is the same as the angle from ǫ i1 to 2.) Then by similar reasoning as in (4.3) and (4.4), and by (4.13) 1 2 C2 i h 1 C2 i h if h is sufficiently small so that φ(x,θ) φ(x+h,θ) is small enough. Hence 1 = C2 i h C2 i hǫ i1 δ 1 δ 2 where the first inequality is due to the fact that either 1 or 1, the third inequality follows from (4.14), and the last equality follows from (4.11). Denote φ = φ(x + h,θ) φ(x,θ) then by the above inequality, φ r 2 2 δ 1 1 C2 i hǫ i1 δ 1 δ 2 and which yields φ C2 i hǫ i1 δ 2 /r 2 2. (4.15) 26

27 Fix t = (t 1,t 2,t 3 ) B 1 (). Without loss of generality, we may assume t 2,t 3. Denote by z and w, the points on (x) and (x + h) such that z = f(x,t), w = f(x + h,t). (Note that (e i1, z ) = φ(x,θ) and (e i1, w) = φ(x + h,θ) for the angle θ between e 1 and t.) For the proof of (II ), it suffices to show z w C2 i hǫ i3 C2 i hǫ. (4.16) To prove (4.16), introduce intermediate points z 1 and z 2 on (x) such that (e i1, z 1 ) = (e i1, z ), (e i2, p(z 1 )) = (e i2, p(w)) and (e i2, p(z 2 )) = (e i2, p(z 1 )), (e i1, z 2 ) = (e i1, w) where p is a projection on the e i2 e i3 -plane. Using the bound (4.15) on φ, Lemma 4.2 implies z 1 z 2 C(r φ + r 2 φ/ǫ i1 ) C2 i hδ 2 r 2 /r2 2 C2 i hǫ i3 (4.17) where r = z 1 L(x), the second inequality follows from ǫ i1 Cr and the last inequality follows from r Cr 2 and δ 2 Cǫ i3. Moreover, since z 2 and w are points on (x) and (x + h) with the same angles from e i1 and e i2, (4.13) and a similar argument as in (4.3) yield that z 2 w C2 i hǫ i3. Hence it suffices to prove z z C2 i hǫ i3 for the proof of (4.16). To simplify notations, denote θ 2 (z) = (e i2, p(z)) where p is a projection on e i2 e i3 -plane. Also let S S 1 S h S1 h = {y (x) : (e i1,y) = φ(x,θ), θ 2 (y) π}, = {y (x) : (e i1,y) = φ(x,θ), θ 2 (y) θ 2 (z)}, = {y (x + h) : (e i1,y) = φ(x + h,θ), θ 2 (y) π}, = {y (x + h) : (e i1,y) = φ(x + h,θ), θ 2 (y) θ 2 (w)}. 27

28 Then (4.1) implies S 1 g x S g x = S h 1 S h g x+h g x+h. (4.18) We claim that the denominators in (4.18) are very close to each other, more precisely, g x g x+h C2 i h g x. S S h S h To prove the claim, recall that g x (y) = Cr(y) 3 / (x) where r(y) = y L(x) and C is a dimensional constant. If y S and y S h are points satisfy θ 2(y) = θ 2 (y ), then g x (y) g x+h (y ) Cr(y)3 (x) Cr(ỹ)3 (x) + C2i hg x (y) = ( 1 r(ỹ)3 r(y) 3 + C2i h)g x (y) ( C2i hδ 2 r(y) 2 r 2 2 r(y) C2 i hg x (y) + C2 i h)g x (y) where the first inequality follows from (4.12), the second inequality follows from a similar argument as in the second inequality of (4.17), and the last inequality follows from δ 2 r(y) r2 2. Thus g x g x+h C2 i h g x S and due to (4.18) g x S 1 S h 1 S h g x+h C2 i h g x S 1 S h = C2 i r(y) h S1 3 (x) C2 i r(y) hr(z) S1 2 (x) C2i hr(z) 2 r 1 (x) 28 = C2i hr 2 r 1 (x) (4.19)

29 where r = r(z) = z L(x). Here the second inequality is from the fact r(y) Cr(z) on S 1 (See Figure 2). Thus r 3 θ 2 (z) θ 2 (w) C g x g x+h C2i hr 2 r 1 (4.2) (x) S 1 (x) and which yields S h 1 θ 2 := θ 2 (z) θ 2 (w) C2 i hr 1 /r. (4.21) By applying Lemma 4.2 on (x,φ(x,θ)) with ǫ = r 1, we get z z 1 C(r θ 2 + r 2 θ 2 /r 1 ) C2 i hr C2 i hǫ i3. (4.22) Combining (4.22) with the bounds on z 1 z 2 and z 2 w, (4.16) is proved. Remark Observe that in the construction of f we divided (x) into right (x) and left (x) in a way that any point y on (x) right (x) left (x) has the shortest distance from the center L(x) among the points on {z (x) : (e i1, z ) = (e i1, y )}. In fact, if the shorter arcs were not fixed by f, then our bound in (4.21) would be C2 i h r/r for some r r 1, which is not strong enough to obtain the second inequality of (4.22). 4.3 n > 4 As in n = 4, in each I i we fix orthonormal basis e i1,...,e in 1 as before (with ǫ i1... ǫ in 1 ) and construct one-to-one map f(s,t 1,...,t n 1 ) with a constant Jacobian, under the constraint f(s, ) : B 1 () {t n 1 } (x) {y n 1 L n 1 (x)}, and f(s, ) maps the (n 2)-dimensional surface {t B 1 () : (t,e k ) = φ k (y,θ k )} to {y (x) : (x,e ik ) = φ k (x,θ k )}. 29

30 Then with θ 1,...,θ n 3 fixed, f is a one-to-one map between two-dimensional surfaces, with shorter arcs (of length r 1 ) being fixed, and area-preserving g x with respect to the normalized density function g x, where g x is a density function on (x) R n 1 defined similarly as in (53). Then parallel arguments as in n = 4 would yield the corresponding inequality to (61): φ k C2 i hǫ i1 δ n 2 /rn 2 2 where φ k = φ k (x,θ k ) φ(x + h,θ k ) and 1 k n 3. Thus we get (4.17) with z replaced by z k and ǫ i3 replaced by ǫ in 1, where ( z,e ik ) = φ k (x,θ k ), ( z k,e ik ) = φ k (x + h,θ k ) and 1 k n 3. Similar arguments as in (4.2) would yield that r n 1 θ n 2 (z) θ n 2 (w) C g x g x+h C2i hr n 2 r 1, (x) S 1 (x) which would yield (4.21) and thus (4.22) with ǫ i3 replaced by ǫ in 1. Combining these inequalities as in n = 4, we obtain (II ). 5 Proof of Theorem 1.1 Throughout the proof, all dimensional constants will be denoted by C. We start with assuming that has a smooth boundary - we will consider the general case at the end of the proof. Normalize u and so that < minu max u = 1 and N = 1. In the first part of the proof, the difference between the first eigenvalues λ and µ of (1.1) and (1.4) will be estimated by a bound Cǫ. From this bound, the nodal set will be located in an x- interval of length Cǫ, which is also near the zero s 1 of the first eigenfunction φ 1 of the corresponding ordinary differential equation (1.4). Let (s,t 1,...,t n 1 ) be the new coordinate system constructed in Lemma 4.1. Using Fubini s Theorem with this new coordinate system, we can choose ( t 1,..., t n 1 ) such that u 2 = u 2 a n w(s)ds...dt n 1 [,1] B 1 () S h 1 u(s,f(s, t 1,..., t n 1 )) 2 w(s)ds. (5.1) 3

31 Based on (5.1), we will construct a one-dimensional test function ψ to compare with φ 1 as follows. Let k be the smallest integer such that 2 k ǫ and let J i = [2 i 1,2 i ] [1 2 i,1 2 i 1 ] for 1 i k. Define a function φ in such that (i) φ(z) = φ(w) for z, w (x), x 1 (ii) φ = u on {(s,f(s, t 1,..., t n 1 )) : 2 k s 1 2 k } (iii) φ = φ(2 k,f(2 k, t 1,..., t n 1 )) for x 2 k (iv) φ = φ(1 2 k,f(1 2 k, t 1,..., t n 1 )) for 1 2 k x 1. Observe that φ is continuous on J i and may have a jump discontinuity at endpoints of J i. But since u C2 i on J i due to Corollary 2.4, and since (x) has a diameter less than 2ǫ, u(z) u(w) C2 i ǫ for z,w (x). (5.2) (5.2) implies that there exists a continuous function φ such that for 1 i k φ = φ + d i on [2 i 1,2 i ], d i is a constant with d i Cǫ and similarly on [1 2 i,1 2 i 1 ]. In other words, φ is constructed so that φ is continuous, φ = φ on J i and φ 2 1 Cǫ 1 + Cǫ. (5.3) φ 2 Now we will obtain an estimate on the first eigenvalue λ ((b) of Theorem 1.1) using a function perturbed from φ. On J i, φ = φ = φ/ x (1 + s y j ) u(s,f(s, t 1,..., t n 1 )) (1 + C2 i ǫ)v(s) 31

32 where v(s) is a one-dimensional function in such that v(s) = u(s,f(s, t 1,..., t n 1 ). Hence φ 2 = ( 1 i<k J i ) φ 2 k 1 v 2 + Cǫ2 i v 2 + C (1 + Cǫ) (1 + Cǫ) i=1 k 1 k=1 ǫ2 i v 2 (J i ) (J i ) 1 u 2 + Cǫ (5.4) u 2 where (J i ) := {x J i }, the second inequality follows from Corollary 2.4 and the third inequality follows from (5.1). On the other hand, by (5.2), Corollary 2.5 and (5.3) 1 Cǫ Therefore (5.4) and (5.5) imply φ 2 φ 2 φ 2 u Cǫ. (5.5) u 2 (1 + Cǫ) (5.6) u 2 Since φ does not satisfy the constraint φ =, we construct ψ by perturbing φ, which satisfies ψ = as well as (5.6). From (5.2) and u =, φ Cǫ. (5.7) 32

33 Let z = (z 1,...,z n ) be a point on {(s,f(s,f(s, t 1,..., t n 1 )) : s 1} such that u(z) =. (Note that z = (z 1,...,z n ) Λ.) Then by Corollary 2.5, c 1 z 1 1 c 1 for a dimensional constant c 1 >. This bound on z 1 and (5.7) imply that there exists c such that ψ(x) := φ(x) + c ǫ(x z 1 ) + satisfies ψ = with c < C for a dimensional constant C. Observe that (5.6) also holds for ψ, and thus ψ 2 u 2 µ 1 (1 + Cǫ) = (1 + Cǫ)λ (5.8) u 2 ψ 2 which proves part (b) of Theorem 1.1. For the proof of part (a), we first show that the projection of the nodal set Λ of u onto x axis is contained in an interval of length Cǫ and then we locate Λ near the zero of φ 1. Let p be a projection on the x-axis. Then p( + ) and p( ) are intervals because the Courant nodal domain theorem [CH, p.452] implies + and are connected. Hence p(λ) = [a,b] for some a b. By (5.2), max u < Cǫ [a,b] and since max ψ u Cǫ, By Corollary 2.5, a φwds > C sup ψ < Cǫ. (5.9) [a,b] wds for a dimensional constant C >. Hence Lemma 3.6 with (5.8) and (5.9) implies that b a < Cǫ for a dimensional constant C. In other words, the nodal set Λ is contained in an x-interval of length Cǫ. 33

34 Now, observe z 1 ψ 2 wds z 1 ψ 2 wds φ 2 wds u 2 z (1 + Cǫ) 1 1 (1 + Cǫ) φ 2 wds u 2 z 1 u 2 (1 + Cǫ) + = (1 + Cǫ)λ (1 + Cǫ)µ 1 u 2 + where = {z 1 x 1}, ψ(z 1 ) = φ(z 1 ) =, the second inequality follows from a similar argument as in (5.6) and the third inequality follows from b a < Cǫ, u 1 and u C. By Lemma 3.5, z 1 s 1 + Cǫ. By a similar argument on the interval [,z 1 ], we obtain s 1 Cǫ z 1. Since the length of the projection of the nodal set on the x-axis is less than Cǫ, part (a) is proved. Lastly we discuss the general case. For a general domain, let { k } k be an increasing sequence of smooth domains which converges to uniformly on each cross sections k (x 1 ). Let u k be the corresponding first nonzero eigenfunctions of k with sup u k = 1. Then by Kröger s theorem (Theorem 2.1), sup k u k is uniformly bounded. Hence there exists a subsequence {u kj } j which converges uniformly to u and {λ kj } j converges to λ as j. On the other hand, since the volume w k (x) of k (x) uniformly converges to w(x), we may assume that {φ kj } j converges uniformly to φ 1 and {µ kj } j converge to µ as j. Now by the nondegeneracy of φ kj (Lemma 3.7) and the nondegeneracy of u kj in the scale ǫ (Lemma 3.6), we obtain Theorem 1.1 for u and λ. References [BQ] D. Bakry, Z. Qian, Some new results on eigenvectors via dimension, diameter, and Ricci curvature Adv. Math., 155(2), [CH] R. Courant, D. Hilbert Methods of Mathematical Physics vol. I, Interscience Publishers, New York (1953). [GJ1] D. Grieser, D. Jerison, Asymptotics of the first nodal line of a convex domain Inventiones Math., 125(1996),

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