On existence theorems of nonlinear partial differential systems of Poisson type in R n

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1 On existence theorems of nonlinear partial differential systems of Poisson type in R n Yifei Pan September 1, 212 Abstract We prove some general existence theorems for nonlinear partial differential systems of Poisson type. More importantly, the solutions obtained are all neither radially symmetric nor positive. Our method is only based on Newtonian potentials through the framework set forth in the case of dimension two from complex analysis point view in our recent work. 1 Introduction In this paper we consider solvability problem for nonlinear partial differential systems in R n. In dimension two, the same study was carried out in [P1] using complex analysis with high order Cauchy-Riemann operators. Here we will solely base on Newtonian potential and its Hölder estimate in R n. We obtain existence of classical solutions nonlinear systems of Poisson type under rather general conditions and further all these solutions are not radial. The following are the main results of this paper. Theorem 1.1. Let a(x, p, q, r) (a 1 (x, p, q, r),..., a N (x, p, q, r)) be of class Cloc k (2 k, < α < 1), where x R n, p R N, q R n R N, and r Sym(n) R N. Assume that a(), (1) r a(), (2) 2 ra(). (3) Then the following system: u(x) (u 1 (x),..., u N (x)) : { x R} R N, u(x) a(x, u(x), u(x), 2 u(x)) (4) has infinitely many solutions of C k+2+α ({ x R}) of vanishing order two at the origin for sufficiently small values of R. Consequently, all these solutions are neither radially symmetric nor positive. MSC 21: 35G2 (Primary); 32G5, 3G2 (Secondary) 1

2 Here we make comments on notations used above. First the notation r means that derivatives are taken with respect to the variables of r and Sym(n) denotes the set of n n symmetric matrices. A solution u is said of vanishing order two if u(), u(), but 2 u(). The vanishing order two ensures that the solutions obtained are non-trivial; namely they are neither constant nor linear ones. More importantly, the vanishing order two ensures that solutions of such are not radial, and this shows that the solutions could not come from solving an ODE if the equation could be ever possible. If a is independent of r, then the following existence theorem can be regarded as a variant of ODE with initial values. We point out, though, that there is no uniqueness result as in ODE; in fact instead there are infinitely many solutions by construction. Theorem 1.2. Let a(x, p, q) (a 1 (x, p, q),..., a N (x, p, q)) be of class C k+α loc (1 k, < α < 1), where x R n, p R N, and q R n R N. Then, for any given c R N, c 1 R n R N, the following system: u(x) (u 1 (x),..., u N (x)) : { x R} R N, u(x) a(x, u(x), u(x)) (5) u() c (6) u() c 1 (7) has infinitely many solutions of C k+2+α ({ x R}) for sufficiently small values of R. In particular, all solutions are neither radially symmetric nor positive if c 1. If a is independent of x, that is, the system is so-called autonomous, then we can solve non-trivial semi-global solutions, i.e., solutions that are defined in any given ball { x R}. Theorem 1.3. Let a(p, q, r) (a 1 (p, q, r),..., a N (p, q, r)) be of class Cloc k (2 k, < α < 1), where p R N, q R n R N, and r Sym(n) R N. Assume that a contain no linear term, i.e., a(), (8) a(). (9) Then the following system: u(x) (u 1 (x),..., u N (x)) : { x R} R N, u(x) a(u(x), u(x), 2 u(x)) (1) has infinitely many solutions in { x R x R n } of C k+2+α with vanishing order two at the origin for any given value of R. Consequently, all these solutions are neither radially symmetric nor positive. We remark that the conditions (8) and (9) imply that the nonlinearity of a does not contain linear terms and that u is a trivial solution of (1). These conditions are necessary due to a well-know result of Osserman [O]. In fact, let us solve the scalar equation u e u with the initial value u() a, which exists for a small R according to Theorem 1.2. On other hand, by Osserman s theorem applied to this case, we have R 1 e u() e a. 2

3 Letting a +, we see that R. Of course, a(p) e p does not satisfies the condition (8). This example also shows in Theorem 1.1 and 1.2, R needs to be small in general. Another example, we may consider, is the eigenvalue equation u λu, which, of course, has no non-zero solutions for most values of λ. A classical and well-known result of Gidas-Ni-Nirenberg [GNN] says that if u is a positive solution of the Poisson equation u f(u) in x < R with u { x R}, then u is radial. However the theorem does not prove the existence of such a solution. As an application of our results, we prove the following corollary for existence of non-radial solutions. Corollary 1.4. Let f(t) be a function of class C k (R) (k 2). Assume that f() f (). Then the equation u f(u) has, for any given R, infinitely many solutions of class C k+2+α ({ x R}) which are of vanishing order 2 at the origin and are neither radially symmetric nor positive. For example, much study has been conducted on seeking positive solutions of the equation u u n+2 n 2 in R n. Corollary 1.4 implies there are indeed infinitely many non-positive solutions for a ball of any radius. However it does not conclude the existence in the whole R n. We would like to point out that for solvability of linear partial differential equations, there is a well-known so-called Nirenberg-Treves conjecture. This conjecture was recently solved by Dencker [D], following previous important works in [L], [NT], [H], [BF], [LE]. Our consideration on nonlinear cases is very different from linear ones technically. The method of this paper and of [P1] can be applied to prove identical results for power of Laplace: m u(x) a(x, u, u,..., m u) which will be in a future paper [P2]. The paper is organized as follows. First we introduce a Banach space with vanishing order from which we are seeking possible solutions. Then, we study Newtonian potential as an operator on the Banach space, which seems to have been overlooked in the literature. Finally, we construct an operator map from which we will try to produce a fixed point on a closed subset of the Banach space. A large portion of the paper is on estimating Hölder norm of involved functions in order to apply Fixed point theorem. 2 Function spaces and their norms In this paper throughout, we let D denote the closed ball {x R n x R} and C its boundary {x R n x R}. Unless otherwise stated, all functions considered will be real-valued and integrable, with domain D. We will consider some classes of functions. 3

4 2.1 Hölder space C α (D) is the set of all functions f on D for which { } f(x) f(x ) H α [f] sup x x α x, x D is finite. C k (D) is the set of all function f on D whose k th order partial derivatives exist and are continuous, k an integer, k. C k+α (D) is the set of all functions f on D whose k th order partial derivatives exist and belong to C α (D). The symbol f or f D denotes sup x D f(x). For f C α (D) we define the norm f f + (2R) α H α [f]. The set of N-tuples f (f 1,..., f N ) of functions (vector functions or maps) of C α (D) is denoted by [C α (D)] N, and H α [f] is defined as the maximum of H α [f i ](i 1,.., N). In a similar fashion we define f A sup x A f(x) for functions and vector functions, and write f when the domain is understood. Finally, in this paper throughout, the norm of R N is taken as v max 1 j N v j. The following lemma is well-known; see ([GT]). Lemma 2.1. The function defined on C α (D) is a norm, with respect to which C α (D) is a Banach algebra: fg f g. 2.2 Function spaces with vanishing order at the origin Our idea of solving differential equations or systems of order m is to look for solutions that vanish up to m 1 order at the origin; this way the norm estimate of the function space to be considered later is made possible using only mth order derivatives. This is rather different from classical norms used for higher order derivatives in partial differential equations. The results in this section have been given in [P1] in dimension two using all complex notations, and the method would work for any dimension. For convenience of readers, we produce all results in real variable notations in R n. We denote for k 1, C k+α (D) the set of all functions in C k+α (D) whose derivatives vanish up to order k 1 at the origin. Specifically C k+α (D) {f C k+α (D) β f(), β k 1}, where we have used β (β 1,..., β n ) and β β β n. One has the following obvious nesting C m+α (D) C m 1+α (D) C 1+α (D) C α (D). We now define a function (k) on C k+α (D): f (k) max β k { β f }. We point out that the function (k) on C k+α (D) is not a norm since f (k) if and only if f is a polynomial of degree at most k 1. However it becomes norm when restricted to the subspace C k+α (D), which is to be proved below and is one of important facts used in this paper. First we obtain some useful estimates. 4

5 Lemma 2.2. If f C k+α (D), then, for x, x D, f(x ) k l Proof. Expanding at x, we have the formula 1 β f(x)(x x) β 1 { } H α [ β f] x x k+α. l! k! β l β k f(x ) k 1 l tk 1 tk 1 1 β f(x)(x x) β l! β l t1 { } d k dt k f(tx + (1 t)x) dtdt 1 dt k 1 t1 { β f(tx + (1 t)x)(x x) }dtdt β 1 dt k 1. β k Hence, we have, by subtracting kth term, tk 1 Thus we have, This completes the proof. f(x ) Lemma 2.3. If f C k+α (D), then Proof. Let f C k+α (D), then k l 1 β f(x)(x x) β l! β l t1 { { β f(tx + (1 t)x) β f(x)}(x x) }dtdt β 1 dt k 1. β k f(x) k f(x 1 ) β f(x)(x x) β l! l β l 1 H α [ β f] x x k+α. k! tk 1 tk 1 { β k β k f (3n)k R k f (k). k! t1 t1 tk 1 { } d k dt f(tx) dtdt k 1 dt k 1 { β f(tx)x }dtdt β 1 dt k 1 β k t1 5 β f(tx)dtdt 1 dt k 1 }x β.

6 Applying norm inequality, we obtain f β k β k 1 k! β f x β 1 k! β f x k nk k! (3R)k f (k), where we have used x i 3R, which is easily verified. Lemma 2.4. If f C k+α (D), then, for β k, β f (3n)k β (k β )! Rk β f (k). Proof. Let f C k+α (D). If β k, then β f C k β +α (D). By Lemma 2.3, we have β f (3n)k β (k β )! Rk β β f (k β ) (3n)k β (k β )! Rk β f (k). An immediate corollary is the following: Lemma 2.5. If f C k+α (D), then, for l k, f (l) (3n)k l (m l)! Rk l f (k). In order to verify that C k+α (D) is a Banach space with norm (k), we need the following simple lemma similar to ([NW], 7.1.a). Lemma 2.6. Let {f m } m1 be a sequence in C α (D), with f m M; assume that {f m } converges to a function f at each point of D. Then f C α (D), f M. Proof. Since f m + (2R) α H α (f m ) f m M, we also have f m (x) f m (x ) (2R) α M x x α for all x, x D. Letting m in the above inequality, we conclude and Therefore f M. f(x) f(x ) (2R) α M x x α, f + (2R) α H α (f) M. Lemma 2.7. Let {f m } be a sequence in C k+α (D), with f m (k) M, and if { β f m }, for all β, β k, are Cauchy sequences in the norm, then there is a function f C k+α (D) such that β f m β f as m for all β, β k, and with f (k) M. 6

7 Proof. For β, β k, consider gm β β f m, then gm β C α (D) and gm β M. Applying to Lemma 2.6, we have functions g β C α (D) such that gm β g β as m, with g β M. Define f by tk 1 t1 { f(x) g β (tx)x }dtdt β 1 dt k 1. β k We have f m (x) { β k Therefore f m (x) f(x) { whence β k tk 1 tk 1 f m f Rk k! t1 } β f m (tx)dtdt 1 dt k 1 x β t1 { } β f m (tx) g β (tx) }dtdt 1 dt k 1 x β, β f m g β, β k which goes to as N, implying f N f in the norm. For γ l k 1 we want to prove that { γ f m (x)} are Cauchy sequences. Indeed, Since f m vanishes up to k 1 order at the origin, then for γ l k 1, γ f m vanishes to k 1 l order at the origin. Thus, we have the formula, Then γ f m (x) β k l tk 1 l tk 1 l γ f m (x) γ f m (x) β k l tk 1 l d k l dt k l γ f m (tx)dt dt k 1 l β γ f m (tx)x β dt dt k 1 l β k l tk 1 l β γ f m (tx)dt dt k 1 l x β, (11) β γ {f m (tx) f m (tx)}dt dt k 1 l x β, (12) Then γ f m (x) γ f m (x ) Rk l (k l)! β f m β f m (13) β k which proves that { γ f m (x)}( γ k 1) are Cauchy sequences since { β f m (x)}( β k) are. We assume that for β l k 1, α f m (x) converges to g β in norm. Thus, applying 7

8 Lemma 2.2, we have 1 k! f m(x ) k l 1 β f m (x)(x x) β l! β l H α [ β f m ] x x k+α β k nk k! (2R) α f m (k) x x k+α 2k k! (2R) α M x x k+α, which is independent of m. Letting m, we have f(x ) f(x) k l1 1 g β (x x) β 2k l! k! (2R) α M x x k+α, β l which implies, by definition of differentiability, g β β f. This implies f (k) M by taking limit from f N (k) M. The convergence for β k 1, follows from that of β k by (12). Lemma 2.8. The function space C k+α (D) equipped with the function (k) is a Banach space. Proof. Let {f m } be a Cauchy sequence in (k). Then for any ɛ >, there is m so that if m, m > m it holds which implies f m f m (k) < ɛ, (14) f m (k) f m (k) < ɛ, which implies { f m (k) } are Cauchy sequence and therefore bounded by say M. Also by definition of (k), (14) implies β f N for β k are Cauchy sequence in. By Lemma 2.7, there is a function f C k+α (D) such that β f m β f for β k as m, and f (k) M. Now the sequence {f m f m } m m+1 is bounded in (k) by ɛ, and converges to f m f, with α (f m f m ) α (f m f) in as m, β k. By Lemma 2.7 again, it holds f m f (k) < ɛ, which implies that f m f in (k). The proof is complete. 8

9 3 Newtonian potentials and Hölder estimate 3.1 Definitions and basic facts The fundamental solution of Laplace s equation is given by for n > 2, Γ(x y) 1 n(2 n)ω n x y 2 n Γ(x y) 1 ln x y 2π for n 2. For an integrable function f on D, the Newtonian potential of f is the function N (f)(x) defined on R n by N (f)(x) Γ(x y)f(y)dy. Lemma 3.1. Let f C α (D). Then it holds, for x Int(D), ij N (f)(x) ij Γ(x y)(f(y) f(x))dy δ ij n f(x) D D The proof of this lemma can be found in [F] (also [GT]) and is based on the following (not so obvious) integrals: N (1)(x) 1 1 2n x 2 + 2(n 2) R2 for n 3, x Int(D) and N (1)(x) 1 4 x R2 (log 1 R ) for n 2, x Int(D). Now we discuss a technical result in order to deal with Hölder estimate of functions. Let x be an interior point of D, and D the intersection of D with the open ball of radius δ and center x. The following is proved in [F]. Lemma 3.2. If i j, then D\D ij Γ(x y)dy, x Int(D), otherwise, there is a constant M only depending on n (independent of R) ii Γ(x y)dy M, x Int(D). D\D We note Lemma 4.4 in [GT] states that N maps C α ( 1 2 D) to Cα (D) continuously. In fact, more is true. The following is essentially proved in Lemma A.16 in [F], although not stated specifically; in particular, the second part of Theorem 3.3 below. This result is essential to our method and we give here a complete proof using lemmas above for convenience of readers and for the further reference in a sequential paper. 9

10 Theorem 3.3. If f C α (D), then N (f) C 2+α (D). Moreover, there is a constant C(n, α), independent of R, such that N (f) (2) C(n, α) f. Proof. First let us recall the norm Let N (f) (2) max 1 i,j n { ijn (f) }. φ(x) Then, ij N (f) φ(x) δ ij f(x), and therefore n D ij Γ(x y)(f(y) f(x))dy. ij N (f) φ + f. Now we proceed to estimate φ. First, for x Int(D), φ(x) ij Γ(x y) f(y) f(x) dy D y x α CH α [f] D x y dy n 2R r α r n 1 CH α [f] drds C(n, α)h r n α [f]r α. (15) S n 1 To compute the Hölder constant of φ, let x, x be two (distinct) points of D. Let δ be the ball of radius ρ 2 x x with center at x. Let D D δ. We consider φ(x) φ(x ) ij Γ(x y)(f(y) f(x))dy ij Γ(x y)(f(y) f(x ))dy D D ij Γ(x y)(f(y) f(x))dy + ij Γ(x y)(f(y) f(x))dy D\D D ij Γ(x y)(f(y) f(x ))dy ij Γ(x y)(f(y) f(x ))dy D\D D ( ij Γ(x y) ij Γ(x y))(f(y) f(x))dy D\D + (f(x ) f(x)) ij Γ(x y)dy D\D + ij Γ(x y)(f(y) f(x))dy ij Γ(x y)(f(y) f(x ))dy D D I 1 + I 2 + I 3 + I 4. (16) 1

11 We are ready to estimate each of I j : I 1 ( ij Γ(x y) ij Γ(x y))(f(y) f(x))dy D\D H α [f] x x ij Γ(ˆx y) x x α dy D\D CH α [f] x x x x α ˆx y n 1 dy D\D CH α [f] x x 2R ρ r α r n 1 r n 1 dr CH α [f]ρ α. (17) where ˆx is a point on the line segment between x, x, and a polar coordinate is used at ˆx, for which we have for y D \ D, ρ y ˆx 2R. By Lemma 3.2, which is a crucial difference from [GT. Lemma 4.4], we have I 2 (f(x ) f(x)) ij Γ(x y)dy D\D CH α [f] x x α (18) Similarly, I 3 ij Γ(x y)(f(y) f(x))dy D H α [f] ij Γ(x y) x y α dy D CH α [f] x y n x y α dy D CH α [f] x x α. (19) For I 4, the estimate is identical to I 3. Combining (15)-(19), we complete the proof. 4 An integral system We consider the integral system Namely, u h + N (a). (2) u 1 h 1 + N (a 1 ) u 2 h 2 + N (a 2 ). u N h N + N (a N ) 11

12 where h (h 1 (x),..., h N (x)) with h j (x) being any harmonic function, and N (a) is the Newtonian potential of a, namely N (a i )(x) Γ(x y)a i (y, u(y), u(y), 2 u(y))dy. D It is clear that any solution of (2) is a solution to (4). We further modify the equation (2) to fit our Banach space C 2+α (D). To do so, we let, for any f C 2+α (D), i 1,..., N, and define Θ i (f)(x) ω i (f)(x) ω i (f)() ω i (f)(x) N (a i (y, f(y), f(y), 2 f(y))(x) (21) n j (ω i (f))()x j 1 2 j1 n k l;k,l1 k l (ω i (f))()x k x l. (22) We note that by Theorem 3.3, if f C 2+α (D), then ω i (f) C 2+α (D), and Θ i (f) C 2+α (D) by construction (22). More importantly, we have k l (Θ i (f))() (23) if k l. Now we introduce the Banach space from which we will seek our solutions. Define B(R) C 2+α (D) C 2+α (D) which consists of N-copies of C 2+α (D), and define the function on B(R) as f (2) max 1 j N f j (2). By Lemma 2.8, (B(R), (2) ) is a Banach space. Now we are ready to consider a map as follows Θ : B(R) B(R) Θ(f) (Θ 1 (f),..., Θ N (f)). In order to apply a fixed point theorem on the Banach space B(R), which has R as a parameter, we will have to estimate Θ(f) Θ(g) (2), and Θ(f) (2) for any f, g B(R). This is to be done in the next subsections. 4.1 Θ(f) Θ(g) (2) estimates It suffices to estimate Θ i (f) Θ i (g) (2). To this end, we see from (22) Θ i (f) Θ(g) (2) ω i (f) ω i (g) (2) + n k l (ω i (f) ω i (g))(). (24) First we have ω i (f) ω i (g) (2) N (a i (, f, f, 2 f) a i (, g, g, 2 g)) (2) by (21). By Theorem 3.3, we have N (φ) (2) C φ for any φ C α (D), where C is a constant dependent only on n, in particular, independent of the radius R. Therefore we have k,l1 ω i (f) ω i (g) (2) C a i (, f, f, 2 f) a i (, g, g, 2 g)). 12

13 Now we are estimating a i (, f,...) a i (, g,...). First we use coordinates for x (x k ), p (p j ), q (q j k ), and r (rj kl ). From this point on, we will use constant C depending only on n, N, α, which varies from line to line. We begin with a i (x, f, f, 2 f) a i (x, g, g, 2 g) d dt ai (x, tf + (1 t)g, t f + (1 t) g, t 2 f + (1 t) 2 g)dt N n N N n A j (f j g j ) + B j k k(f j g j ) + C j kl k l (f j g j ) (25) j1 k1 j1 j1 k,l1 where A j B j k C j kl p j a i (x, tf + (1 t)g, t f + (1 t) g, t 2 f + (1 t) 2 g)dt (26) q j k r j kl a i (x, tf + (1 t)g, t f + (1 t) g, t 2 f + (1 t) 2 g)dt (27) a i (x, tf + (1 t)g, t f + (1 t) g, t 2 f + (1 t) 2 g)dt. (28) Taking norm on (25), and using algebraic property of the norm, we obtain a i (x, f, f, 2 f) a i (x, g, g, 2 g) N n N A j f j g j + B j k k(f j g j ) + j1 f g k1 j1 N A j + f g (1) j1 N C(R 2 A j + R j1 n k1 j1 n N k1 j1 N N B j k + N j1 k,l1 N B j k + f g (2) j1 k,l1 where we have used, according to Lemma 2.5, that n C j kl k l (f j g j ) j1 k,l1 n C j kl n C j kl ) f g (2) (29) f g CR 2 f g (2), and f g (1) CR f g (2). Since we will apply Fixed point theorem for a closed subset of B(R), we consider the following closed subset A(R, γ) {f B(R) f (2) γ}. Let W k t k f(x) + (1 t) k g(x), for k, 1, 2. We need to study the range of W k for f, g A(R, γ). The following is needed. 13

14 Lemma 4.1. If f, g A(R, γ), then W k CR 2 k γ for k, 1, 2. Proof. This follows from Lemma 2.4 and 2.5 since W k k f + k g. We now consider the compact set E(R, γ) D {p R N p CR 2 γ} {q R N R n q CRγ} {r Sym(n) R n r Cγ} In order to carry out estimates of (29), we need to introduce constants that would measure solvability of partial differential equations we consider in this paper. A[R, γ] max{ a i p j i 1,..., N; j 1,..., n} (3) E(R,γ) [ ] a Hα A i [R, γ] max{h α i 1,..., N; j 1,..., n} (31) p j E(R,γ) B[R, γ] max{ a i q j i 1,..., N; k, j 1,..., n} (32) k E(R,γ) [ ] a Hα B i [R, γ] max{h α q j i 1,..., N; k, j 1,..., n} (33) k E(R,γ) C[R, γ] max{ a i r j i 1,..., N; j 1,..., n} (34) kl E(R,γ) [ ] a Hα C i [R, γ] max{h α i 1,..., N; j 1,..., n} (35) E(R,γ) r j kl Now we need more important constants on Lipschitz in variables of r. Lipschitz constant in r as follows Here we denote the H 1 [f] sup{ f(, r) f(, r ) }. r r Now we define: [ ] a H1 A i [R, γ] max{h 1 i 1,..., N; j 1,..., n} (36) p j E(R,γ) [ ] a H1 B i [R, γ] max{h 1 q j i 1,..., N; k, j 1,..., n} (37) k E(R,γ) [ ] a H1 C i [R, γ] max{h 1 i 1,..., N; j 1,..., n} (38) E(R,γ) r j kl 14

15 4.1.1 Estimate of A j It is obvious that A j A[R, γ]. Now we are estimating Hα A [A j ]. To do so, we begin with and it follows A j (x) A j (x ) + + a i p j (x, W (x), W 1 (x), W 2 (x))dt a i p j (x, W (x ), W 1 (x ), W 2 (x ))dt a i p j (x, W (x), W 1 (x), W 2 (x))dt a i p j (x, W (x), W 1 (x), W 2 (x))dt a i p j (x, W (x ), W 1 (x), W 2 (x))dt a i p j (x, W (x ), W 1 (x ), W 2 (x))dt a i p j (x, W (x ), W 1 (x ), W 2 (x))dt a i p j (x, W (x ), W 1 (x ), W 2 (x ))dt (39) A j (x) A j (x ) H A α [R, γ] x x α + H A α (R, γ) + H A α [R, γ] + H A 1 [R, γ] N ( f j (x) f j (x ) + g j (x) g j (x ) ) α j1 N ( f j (x) Df j (x ) + g j (x) Dg j (x ) ) α j1 N ( 2 f j (x) 2 f j (x ) + 2 g j (x) 2 g j (x ) ) (4) j1 Now we need a lemma on Lipschitz property of C m+α (D). Lemma 4.2. Let f C k+α (D). For x, x D, and β k 1, we have β f(x) β f(x ) f ( β +1) x x. 15

16 Proof. We have It follows from (41) β f(x) β f(x ) d dt β f(tx + (1 t)x )dt β f(tx + (1 t)x ) (x x )dt (41) β f(x) β f(x ) f ( β +1) x x. Applying Lemma 4.2 and 2.5, we have f(x) f(x ) f (1) x x 3nR f (2) x x, It follows from (4) j βf(x) j f(x ) f (2) x x. Now we have by (42) A j (x) A j (x ) Hα A [R, γ] x x α + Hα A [R, γ]n(3nr) α ( f (2) + g (2) ) α x x α + Hα A [R, γ]n( f (2) + g (2) ) x x α N + H1 A [R, γ] (H α [ 2 f j ] + H α [ 2 g j ]) x x α. (42) j1 A j (x) A j (x ) (H A α [R, γ]+2n(3nr) α γ α H A α [R, γ]+2nγh A α [R, γ]+2nγh A 1 [R, γ](2r) α ) x x α which implies and H α [A j ] H A α [R, γ] + 2N(3nR) α γ α H A α [R, γ] + 2NγH A α [R, γ] + 2NγH A 1 [R, γ](2r) α (43) A j A j + (2R) α H α [A j ] A[R, γ] + (2R) α (1 + 2N(3nR) α γ α + 2Nγ)H A α [R, γ] + 2NγH A 1 [R, γ]. (44) Similarly, we have estimates and B j k B j + (2R) α H α [B j k ] B[R, γ] + (2R) α (1 + 2N(3nR) α γ α + 2Nγ)H B α [R, γ] + 2NγH B 1 [R, γ], (45) C j kl Cj kl + (2R)α H α [C j kl ] C[R, γ] + (2R) α (1 + 2N(3nR) α γ α + 2Nγ)H C α [R, γ] + 2NγH C 1 [R, γ], (46) 16

17 To simplify the notation, we denote the right side of (44), (45), and (46) respectively by δ A (R, γ), δ B (R, γ) and δ C (R, γ). Then by (29) we, have a i (x, f, f, 2 f) a i (x, g, g, 2 g) C(n, N, α)(r 2 δ A (R, γ) + Rδ B (R, γ) + δ C (R, γ)) f g (2) (47) Lemma 4.3. Let f C α (D). Then it holds Proof. By Theorem 3.3, we have k l N (f)() C(n, α) f. k l N (f)() k l N (f) N (f) (2) C(n, α) f. where Finally, we combine (24),and Lemma 4.3 to conclude that Θ i (f) Θ i (g) (2) δ(r, γ) f g (2) (48) δ(r, γ) C(n, N, α)(r 2 δ A (R, γ) + Rδ B (R, γ) + δ C (R, γ)) (49) δ A (R, γ) A[R, γ] + (2R) α (1 + 2N(3nR) α γ α + 2Nγ)Hα A [R, γ] + 2NγH1 A [R, γ] (5) δ B (R, γ) B[R, γ] + (2R) α (1 + 2N(3nR) α γ α + 2Nγ)Hα B [R, γ] + 2NγH1 B [R, γ] (51) δ C (R, γ) C[R, γ] + (2R) α (1 + 2N(3nR) α γ α + 2Nγ)Hα C [R, γ] + 2NγH1 C [R, γ] (52) 4.2 Θ(f) (2) estimates Our proof here is mostly similar to that of (48). We will point out minor differences without repeating the arguments. It suffices to estimate Θ i (f) (2). To this end, we see from (22) Θ i (f) (2) ω i (f) (2) + n k l (ω i (f))(). (53) First we have ω i (f) (2) N (a i ) (2). By Theorem 3.3, we have N (a i ) (2) C a i, where C is a constant dependent on only on n, in particular, independent of the radius R. Now we are estimating a i. First we use coordinates for x (x j ), p (p j ), q (q j k ), and r (r j kl ). We begin with k,l1 a i (x, f, f(x), 2 f(x)) a i (,..., ) d dt ai (tx, tf, t f(x), t 2 f(x))dt n N n N N n D j x j + A j (f j ) + B j k k(f j ) + C j kl k l (f j ) (54) j1 j1 k1 j1 j1 k,l1 17

18 where A j B j k C j kl D j p j a i (x, tf, t f, t 2 f)dt (55) q j k r j kl a i (x, tf, t f, t 2 f)dt (56) a i (x, tf, t f, t 2 f)dt (57) x j a i (x, tf, t f, t 2 f)dt. (58) We notice here we have kept notations similar to (26)-(28), although the integrand is different. By virtue of the same argument as for (24), we can arrive to the following estimate a i (x, f, f, 2 f) a i () + 3Rn D j N n N N + C(R 2 A j + R B j k + n C j kl ) f (2). (59) j1 k1 j1 j1 k,l1 Similarly, we need to define constants D[R, γ)], H D α [R, γ], and H D 1 [R, γ] as in (3)-(38). Without repeating, we will come out estimate where Θ i (f) (2) η(r, γ) (6) η(r, γ) C(n, N, α)( a() + Rδ D (R, γ) + γδ(r, γ)) (61) δ D (R, γ) D[R, γ] + (2R) α (1 + 2N(3nR) α γ α + 2Nγ)Hα D [R, γ] + 2NγH1 D [R, γ]. (62) 4.3 A general estimate Here we collect the estimates together for a later quick reference. Theorem 4.4. Let Θ : B(R) B(R) be defined as in (22). If f, g A(R, γ), then Θ(f) Θ(g) (2) δ(r, γ) f g (2) (63) Θ(f) (2) η(r, γ) (64) where δ(r, γ) and η(r, γ) are defined by (49), (61) respectively. 5 Proof of theorems In this section, we will give proofs of all results presented in the introduction. It suffices to work with a of class C 2 or C 1+α since the regularity of insures higher regularity. 18

19 5.1 Proof of Theorem 1.1 Here we will prove a result slightly more general than Theorem 1.1. Theorem 5.1. Let a(x, p, q, r) (a 1 (x, p, q, r),..., a N (x, p, q, r)) be of class Cloc k (2 k, < α < 1), where x R n, p R N, q R n R N, and r Sym(n) R N. There are a (small) constant δ depending on n, N, α such that if a(), (65) r a() + 2 ra() < δ, (66) then the following system: u(x) (u 1 (x),..., u N (x)) : { x R} R N, u(x) a(x, u(x), u(x), 2 u(x)) (67) has solutions of C k+2+α (D) of vanishing order two at the origin for sufficiently small values of R. In the proof, our goal is to find R, γ sufficiently small so that we have δ(r, γ) < 1, η(r, γ) < γ 2. Once we have these, we can consider an operator T defined by T (f) h + Θ(f) where f A(R, γ) and h is a harmonic vector-function such that h() h(), and h (2) γ. Therefore T is a contration operator from A(R, γ) to A(R, γ), for which there is a 2 fixed point that becomes a solution of (2). To this end, we first simply δ(r, γ), η(r, γ), and we can replace them by δ(r, γ) C(C[R, γ] + γh1 C [R, γ]) + ε(r, γ) (68) η(r, γ) C( a() + γc[r, γ] + γ 2 H1 C [R, γ]) + ε(r, γ). (69) where C is a constant only depending on n, N, α, and ε(r, γ) is such that lim R ε(r, γ) for each γ >. We now give estimates of C[R, γ], H C 1 [R, γ] in terms of conditions (1)-(3). Let σ r j kl be a component variable of r. We want to estimate the Lipschitz constant of σa. In fact, we have σ a i (x, p, q, r) σ a i (x, p, q, r ) d dt ai (x, p, q, tr + (1 t)r )dt r j σ a i ( )(r j kl r j kl )dt. (7) kl Here a i ( ) is defined naturally as given in the integrand. It follows from (7) that H 1 [ σ a i ] E(R,γ) 2 ra i E(R,γ) 2 ra i () + o(r + γ), (71) 19

20 where o(r+γ) as R, γ by continuity of C 2 smoothness of a and lim R,γ E(R, γ) {}. From (71) and definition (38), we have On the other hand, we have H C 1 [R, γ] 2 ra() + o(r + γ). σ a i (x, p, q, r) σ a i () d dt ai (tx, tp, tq, tr)dt { xj σa i ( )x j + pj σa i ( )p j + q j σ a i ( )q j k + k r j kl σ a i ( )r j kl } dt. (72) Notice that for (x, p, q, r) E(R, γ), we have x R, p CR 2 γ, q CRγ, and r Cγ. Hence by (72), putting terms with R, γ together, we have σ a i E(R,γ) C( r a i () + 2 ra i () ) + ε(r, γ) where ε(r, γ) as R for each given γ. This implies Letting we can have C[R, γ] C( r a() + 2 ra() ) + ε(r, γ) τ r a() + 2 ra(), δ(r, γ) C(τ + γ(τ + o(r + γ)) + ε(r, γ) η(r, γ) C(γτ) + γ 2 (τ + o(r + γ)) + ε(r, γ) By o(r + γ), there exist R, γ (< 1) such that for R R, γ γ. Hence we have Now choose δ 3. If τ < δ, we have 1C o(r + γ) Cτ δ(r, γ ) Cτ + γ 2Cτ + ε(r, γ ) η(r, γ ) γ 3Cτ + ε(r, γ ) δ(r, γ ) γ 1 + ε(r, γ ) η(r, γ ) 3γ 1 + ε(r, γ ). 2

21 Finally, we choose R small so that ε(r, γ ) min{ 4, 2γ 1 1 }. It follows that δ(r, γ ) 9 1 η(r, γ ) γ 2. This is equivalent to that for f, g A(R, γ ) Θ(f) Θ(g) (2) 9 1 f g (2) (73) Θ(f) (2) γ 2. (74) Now we are ready to apply Fixed point theorem for R, γ. Let h be any harmonic homogenous polynomial of degree 2 so that h (2) γ. Here let us be more specific. Let h n 2 k,l1 a klx k x l where a kl R N. It is easy to see that h is harmonic if and only if the trace of a is zero, i.e., n k1 a kk {}. We also see that h (2) max 1 k,l n a kl. So we simply let γ 2 be max 1 k,l n a kl with a kl {}. Then we consider the map T (u) h + Θ(u) which maps A(R, γ ) to A(R, γ ) as contraction map by (73),(74). So T has a fixed point u in A(R, γ ). We claim the vanishing order of u at the origin is 2. In fact, by (23), k l u() k l h() a kl {} for k l, and also u() and k u() by the construction. This completes the proof. Remark 5.2. Given γ as in the proof above, we actually prove that the solution space can be parameterized by at (n 2 1)N parameters from the coefficients of the harmonic polynomials. In fact, the different choice of a kl produces different u which is defined in the same domain of radius R. Therefore there exist infinitely many solutions of vanishing order two at the origin. This remark also applies to other theorems 5.2 Proof of Theorem 1.2 First we consider the case where u, u 1 as in (6), (7). Since a is independent of r, we have C[R, γ] H C 1 [R, γ] H C α [R, γ]. Here we note that C 1,α regularity of a is only needed with a careful inspection of the proof of Theorem 4.4. Therefore we can replace constants as follows δ(r, γ) ε(r, γ) where η(r, γ) C a() + ɛ(r, γ) lim ε(r, γ) lim ɛ(r, γ) R R 21

22 for each γ >. Now we choose γ large enough that γ > C a(). Then we choose R sufficiently 4 small that ε(r, γ ) < 1 and ɛ(r, γ 2 ) < γ. As a result, we have 4 δ(r, γ ) < 1 2 η(r, γ ) < γ 2. As in Theorem 1.1, we find a solution in A(R, γ ) which vanishes up to order 1 at the origin. To get general case, we consider a new system ũ a(x, ũ + c + c 1 x, (ũ + c + c 1 x)) ã(x, ũ, ũ) We can solve this system for ũ. Then u ũ + c + c 1 x is the solution we are seeking for. 5.3 Proof of Theorem 1.3 Here we fix R, and choose γ small to prove the existence of semi-global solutions. Since a is independent of x, so we have D[R, γ] H D α [R, γ] H D 1 [R, γ]. Therefore we can replace δ(r, γ), η(r, γ), using (8), (61), by η(r, γ) Cγδ(R, γ) (75) where δ(r, γ) is still given by (49). It suffices to prove that for each given R, we have If this is proved, then we can choose γ so that lim δ(r, γ). (76) γ δ(r, γ ) < 1 2 η(r, γ ) < γ 2. Our next goal is to show (76). Indeed, since a is independent of x, we will take E(R, γ) as E(R, γ) {p R N p CR 2 γ} {q R N R n q CRγ} {r Sym(n) R n r Cγ}. We notice that as set, lim E(R, γ) {}. γ This is important for what follows in proving (76). Let σ be one of component variables of {p, q, r}. We have by condition (9) σ a i (p, q, r) d dt ai (tp, tq, tr)dt pj σ a i ( )p j + q l σa i ( )q l k k + r j σ a i ( )r j kldt (77) kl 22

23 Since p j CR 2 γ, q j k CRγ, and rj kl Cγ, we have from (77) which implies, by definitions (3),(32),(34), σ a i E(R,γ) C a C 2 (E(R,γ))γ, A[R, γ], B[R, γ], C[R, γ] C a C 2 (E(R,γ))γ. (78) Here, of course, C 2 (E(R, γ)) denotes the maximum norm of 2 a on the set E(R, γ). On the other hand, we have σ a i (p, q, r) σ a i (p, q, r ) d dt ai (tp + (1 t)p, tq + (1 t)q, tr + (1 t)r )dt pj σ a i ( )(p j p j) + q l σa i ( )(q l k k q j k) + r j σ a i ( )(r j kl r j kl)dt. (79) kl From (79), we have, by definition (36), (37), (38), Similarly, from (79), we have H A α [R, γ], H B α [R, γ], H C α [R, γ] C a C 2 (E(R,γ))γ 1 α. (8) H A 1 [R, γ], H B 1 [R, γ], H C 1 [R, γ] C a C 2 (E(R,γ)). (81) Finally, we substitute (78), (8), and (81) into (5),(51), (52) and (49), we see that δ(r, γ) is a function in γ and γ 1 α, which proves (76). The rest of proof is similar and we omit it. 5.4 Proof of Corollary 1.4 In order to prove it, we consider the following equation u f(u) u() u() b where b R n \ {}. The C 2+α solutions exist in any given D by Theorem 1.3. Let u be one of the solutions. We show that u is not radial function. If so, we have u(x) v(r) where x rω and ω S n 1. Since u is C 2, then near, u(x) bx + O( x 2 ), and we have v v(r) v() () lim r r r lim r u(x) r Since ω is arbitrary, we conclude that b, a contradiction. lim r brω + O(r 2 ) r bω. 23

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