An L Bound for the Neumann Problem of the Poisson Equations

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1 An L Bound for the Neumann Problem of the Poisson Equations Xiaojing Ye and Shulin Zhou Abstract In this paper we establish an L -bound for the Neumann problem of the Poisson equations. We first develop some estimates for the bounds of solutions in several spaces using Poincarés inequality, Trace theorem and Sobolev s embedding theorem, and then prove our main theorem utilizing the De Giorgi-Nash estimates. Keywords. L -bound, Poincaré s inequality, Sobolev s theorem, De Giorgi-Nash estimates. Introduction Suppose is a bounded and connected subset of R n. Furthermore, satisfies consistent cone condition, and its boundary with external normal vector n satisfies local Lipschitiz condition. A-priori maximum estimate for the Neumann problem of the Poisson equations u(x = f(x in u(x n = g(x on ( has not been established yet. Our purpose in this paper is to provide an L - bound for the solutions of the Neumann problem. In this paper we assume that f(x and g(x in ( meet f(xdx + g(xds(x = 0. (2 This is a reasonable assumption, since if there is a solution u to (, then we should have u(x f(xdx = u(xdx = n ds(x = gds(x. (3 Furthermore, we define two constants F and G as follows, F := f L ( and G := g L ( (4 and let u := udx be the average of u in. We want to show that u u has a similar a-priori maximum estimate as the third problem of the Poisson equation: Department of Mathematics, LMAM, School of Mathematical Sciences, Peking University, Beijing, 0087, P.R. China. xyex9@gmail.com, szhou@math.pku.edu.cn.

2 Theorem.. There exists a constant C depending only on and its dimension n, such that u u L ( C(F + G (5 for all u H 2 ( H ( that solves (. We first quote several important theorems and corollaries that will be used in this paper. Theorem.2 (Poincaré s inequality. Let be a bounded, connected subset of R n with satisfying local Lipschitz condition. Assume p <. Then there exists a constatnt C depending only on n, p and such that for all u W,p (. Proof. See, e.g. [2, 6, 5]. u u L p ( C Du L p ( (6 Definition.3. For constant p, define p by np if p < n, n p p = q if p = n, p q <, if p > n. (7 Theorem.4 (Sobolev s Embedding Theorem. Let be a bounded, connected subset of R n and satisfies consistent cone condition. Assume that p n, then there exists a constant C depending only on n, p and (if p = n there is also an q, such that u L p ( C u W,p ( (8 for all u W,p (. Proof. See []. Theorem.5 (Trace Theorem. Let be a bounded set of R n. Then there exists a linear operator T : W,p ( L p ( (9 and a constant C depending only on n, p and, such that for all u W,p (. Proof. See [3, 4]. T u Lp ( C u W,p ( (0 Corollary.6. Let be a bounded set of R n. Then there exists a constant C such that u L p ( C u W,p (. ( for all u W,p (. 2

3 Proof. Let ũ W,p ( satisfy ( ũ = u in \, and (2 T ũ = u on. So there is u Lp ( = T ũ Lp ( C ũ W,p (. (2 Note that has measure 0 in R n, we obtain and the conclusion follows. u W,p ( = ũ W,p (, (3 Proposition.7. Let be a bounded, connected subset of R n and satisfy consistent cone condition, with satisfying local Lipschitz condition. Assume that Neumman problem u(x = f(x in u(x n = 0 on (4 and u(x = 0 in u(x n = g(x on (5 both have a-priori maximum estimate, that is, there exist two constants C and C 2 depending only on n and, such that and v v L ( C F (6 w w L ( C 2G (7 for all v H 2 ( H ( that solves (4 and w H 2 ( H ( that solves (5. Proof. Assume that u, v solve ( and (4, respectively. In addition, u = 0, v = 0. Since ( is linear, we know that u v solves (5, and (u v = 0. So we obtain u L ( u v L ( + v L ( C 2 G + C F C(F + G. (8 Remark. In the following part of this paper, we will prove Theorem. in g(x = 0 and f(x = 0 cases separately, and combine the results with proposition.7 to prove theorem.. For notation simplicity, we assume that u = 0 and let C be a constant depending on n and unless otherwise noted. Note that the exact value may vary in different situation. 2 Case that g(x = 0 In this section we prove Theorem. in the case that g(x = 0. Also, we consider the n > 2 and n = 2 cases separately. 3

4 2. The dimension n > 2 Let p R and p 2, from u = f, we know u p 2 u udx = u p 2 ufdx. (9 In addition, there is u p 2 4(p u udx = p 2 4(p = p 2 Du p/2 2 dx u p 2 u u n ds(x Du p/2 2 dx. (20 From the two equations above we can obtain the following inequality Du p/2 2 dx p2 F u p dx (2 4(p Using Hölder s inequality, we can obtain Du p/2 2 dx p2 /p F 4(p From Theorem.4 we know there is ( 2/2 ( Du p/2 2 dx C u p dx + Du p/2 2 dx C ( Let p = np/(n 2, then there is ( u p 2 dx / p C /p ( ( Define a sequence { } k=0 as follows, Then we know that ( (p /p u dx p. (22 u p dx + p2 /p F 4(p /p u p dx + p2 /p F 4(p ( u p dx /p ( (p /p u p dx. (23 (p /p 2 (24 ( k n := 2, k = 0,, 2,. (25 n 2 ( k n S(n := = 2 = n n 2 4. (26 k=0 k=0 For notation simplicity, we use u pk estimate based on (24 as follows, to denote u L (. We can obtain the u pk+ C / ( u pk + α F / u ( /, (27 4

5 where α = (/2 max{, }. First, since there is u = f, we can obtain that u udx = Du 2 dx F u dx. (28 From Theorem.2, we know u 2 dx C Using Hölder s inequality, we get From (27 we can obtain that Du 2 dx. (29 u p0 CF. (30 u p CF and u p2 CF. (3 Suppose that there is an estimate as follows, k j=0 u pk αβc p j (k pk F, (32 such that it is true for k = 2, and β is a constant depending only on n and. Thus, for all k N (32 is true: if (32 is true for some k, then we can show that ( C / k j=0 αβc p j (k pk F + α F u pk+ k j=0 αβc p j kpk+ F (33 So by induction, (32 is true for all k. Note that here we assumed that C >. Otherwise the estimate is much simpler and the results can follow directly. Plugging (32 into (27, we obtain that ( u pk+ C / /pk k j=0 αβc p j (k pk F + α F u ( / (αβcs(nk F / u ( / (Ck F / u ( / (34 Let k =, 2,,, we can obtain the relation between u p and u pk as follows, u pk Here { A k = C λ p 2 k p [p p 2 (2p 2 A k F µ k u λ k p, k. (35 p 2 ] p 3 p 3 } p [(k ] k, λ k = p p 2 pk, p p 2 {( p 2 µ k = + p3 + p p 2 p 2 p 3 ( p 2 = pk p p 2 = λ k. + + } pk + (36 5

6 Note that 0 λ k < and λ k is nonincreasing as k, we know the limit exists and depends only on n and. Thus also exists. In terms of A, we know and it is easy to prove that exists. Thus we obtain that λ := lim k λ k (37 µ := lim k µ k (38 k A k C λ k (jp j /pj, (39 lim k k j= j= A := lim A k C λ exp lim k log(jp j p j (40 k k j= log(jp j (4 p j exists and depends only on n and. Take the results back to (35 and let k, we can get u p AF µ u λ p. (42 Since u p CF, we now can conclude that 2.2 The dimension n = 2 u L ( CF. (43 In this case we can also obtain (2. With 2 in Theorem.4 being set to 4, we can get a similar inequality as (24 where p = 2p. So let we still can get 3 Case that f(x = 0 u L ( CF (44 In this case we prove Theorem. in the case that g(x = 0. Again, we consider two cases n > 2 and n = 2 separately. 3. The dimension n > 2 Since there is u = 0, we have u p 2 4(p u udx = p 2 Du p/2 2 dx u p 2 ugds(x = 0 (45 6

7 Thus there is Du p/2 2 dx From corollary.6, we know that ( u p uds(x u p dx + (p Using Hölder s inequality, we have p2 G u p uds(x (46 4(p u p 2 Du dx. (47 ( /2 ( /2 u p 2 Du dx u p 2 dx u p 2 Du 2 dx ( /2 ( /2 (48 G u p 2 dx u p ds(x p Plugging (47 into (48, there is u p ds(x ( C u p dx + ( /2 ( /2 G(p u p 2 dx u p ds(x. (49 Solving this inequality, we can obtain u p ds(x 4C u p dx + CG 2 (p Form Theorem.4, we can get ( u p dx / p C /p ( C /p ( u p dx + u p dx + /p Du p/2 2 dx p2 G 4(p u p 2 dx (50 /p (5 u p dx here p = np/(n 2, and from (49 we can obtain ( u p dx ( C /p / p u p dx + CG p2 p u p dx + C2 p 2 G 2 4 u p 2 dx /p. (52 Using Hölder s inequality, we have u p dx /p ( u p 2 dx 2/p ( (p /p u p dx (p 2/p (53 u p dx 7

8 Plugging them into (49, we obtain that ( u p C /p u 2 p + CG p2 p /p u p + C2 p 2 G 2 2/p /p u (p 2/p p 4 C /p ( u p + CGp /p 2/p u (p 2/p p (54 We still let Then we can get u pk+ α = max{, } and = 2 ( k n, k = 0,, 2,. (55 n 2 C /p ( u pk + αcg 2/ u ( 2/, k = 0,, 2,. (56 Since there is u = 0, we can obtain u udx = Du 2 dx Therefore, we have By Theorem.5, we know that ( u ds(x C u dx + Theorem.2 indicates that Thus we get ugds(x = 0. (57 Du 2 dx G u ds(x. (58 u dx C u 2 dx C Du dx Du 2 dx Du dx. (59 (60 u 2 dx CG. (6 Using the same method as in section 2., form (56 we can prove that 3.2 The dimension n = 2 Similar to Section 2.2 we can readily show that in the case that n = 2. u L ( CG. (62 u L ( C G. (63 8

9 4 Conclusion Combine the results in Sections 2 and 3, and apply Proposition.7, we can prove Theorem.. References [] R. A. Adams, Sobolev Spaces, Pure and Applied Mathematics, Vol. 65. Academic Press, New York-London, 975. [2] Yazhe Chen, Lancheng Wu, Elliptic Equation of Order 2. Science Press, 99. [3] Lawrence C. Evans, Partial Differential Equations, Berkeley Mathematics Lecture Notes, University of California, Berkeley, 994. [4] David Gilbarg, Neil S. Trudinger, Elliptic Partial Differential Equations of Second Order. Springer-Verlag, 997. [5] Lishang Jiang, Yazhe Chen, Method of Mathematical Physics. Higher Education Press, 986. [6] Shulin Zhou, Partial Differential Equations. Peking University Press,

Sobolevology. 1. Definitions and Notation. When α = 1 this seminorm is the same as the Lipschitz constant of the function f. 2.

Sobolevology. 1. Definitions and Notation. When α = 1 this seminorm is the same as the Lipschitz constant of the function f. 2. Sobolevology 1. Definitions and Notation 1.1. The domain. is an open subset of R n. 1.2. Hölder seminorm. For α (, 1] the Hölder seminorm of exponent α of a function is given by f(x) f(y) [f] α = sup x