POSITIVE NON-SYMMETRIC SOLUTIONS OF A NON-LINEAR BOUNDARY VALUE PROBLEM

This paper deals with a non-linear second order ordinary differential equation with symmetric non-linear boundary conditions, where both of the non-linearities are of power type. It provides results concerning the existence and multiplicity of positive non-symmetric solutions for values of parameters not considered before. The main tool is the shooting method.


Introduction
In this paper we study positive non-symmetric (i.e.non-even) solutions of the problem u (x) = au p (x), x ∈ (−l, l), u (±l) = ±u q (±l) (1) for p ∈ (−1, 1), q > p+1 2 , a, l > 0. (The choice of these conditions will soon be clarified.) The first systematic study of positive solutions of (1) was done by M. Chipot, M. Fila and P. Quittner in [5].They also studied the N -dimensional version of (1), but they were interested mainly in global existence and boundedness or blow-up of positive solutions of the corresponding N -dimensional parabolic problem where Ω ⊂ R N is a bounded domain, n is the unit outer normal vector to ∂Ω, u 0 : Ω → [0, ∞), p, q > 1 and a > 0. The same questions were independently studied in [13] for N = 1.
The reader can find the complete answer to the question of the existence and multiplicity of positive symmetric solutions of (1) for p, q > 1 in [5].It was also proved there that (1) can possess positive non-symmetric solutions only for EJQTDE, 2013 No. 69, p. 1 q > p+1 2 , but their existence and multiplicity was determined only under some additional condition.The solvability of (1) in the class of positive symmetric solutions was examined in [15] for all p > −1, q ≥ 0 and p = −1, q = 0.The results of [5] concerning positive non-symmetric solutions were extended in [16] to all p ≥ 1, q > p+1 2 .In view of the cited studies it is natural to ask the question of the existence and number of positive non-symmetric solutions of (1) for p ∈ (−1, 1), q > p+1 2 .This is what we investigate in this article.
It is known from [16] that given any p ≥ 1, q > p+1 2 and a > 0, (1) has either two or no positive non-symmetric solutions, depending on the value of l > 0.Here we prove that (1) possesses at least four positive non-symmetric solutions for certain p ∈ (−1, 1), q > p+1 2 and a, l > 0, and even infinitely many for some special choices of p, q, a and l.Moreover, the sets of (p, q) for which (1) has different multiplicity of solutions are separated by line segments and also some implicitly given curves.See Theorems 3.2 and 3.10 for the exact formulations.
Some further extensions and generalisations of the results from [5] can be found in the following studies: In [17], the behaviour of positive solutions of (2) was determined for all p, q > 1. Sign-changing solutions of the parabolic problem were considered in [6] for p ≥ 1, q > 1-in that case, u p and u q are replaced by |u| p−1 u and |u| q−1 u respectively.The results from [6] regarding sign-changing stationary solutions for N = 1 were completed in [16].Positive solutions of the elliptic problem with −λu + u p on the right-hand side of the equation were dealt with in [14] for λ ∈ R, p, q > 1, and later in [11] for λ ∈ R, p, q > 0, (p, q) / ∈ (0, 1) 2 .In [12] and [18], positive and sign-changing solutions of the parabolic problem with more general non-linearities f (u), g(u) instead of au p , u q were studied, while f (x, u), g(x, u) were considered in [2].Many results concerning elliptic problems with non-linear boundary conditions were summarised in [19].See also [1,3,4,7,8,9,10].

Preliminaries
In this section we recall the shooting method from [5] and [15].Let p, q ∈ R, a, l > 0. If u is a positive solution of (1), then u (−l) < 0 < u (l), therefore u has a stationary point x 0 ∈ (−l, l).So the function u(• + x 0 ) solves for some m > 0. Since u → au p is locally Lipschitz continuous on (0, ∞), (3) has a unique maximal solution, which is apparently even and strictly convex.We will denote it by u m,p,a and its domain by (−Λ m,p,a , Λ m,p,a ).
Let p, q ∈ R, a > 0. Then the following statements are equivalent for arbitrary m, l > 0: (i) l < Λ m,p,a and u m,p,a (l) = u q m,p,a (l), (ii) the equation ) with the unknown x > 0 has some solution R > m, and where for y ≥ 1.
From now on we will consider only However, the definition and the properties of I −1 will be needed for the proofs of Lemmata 3.8 and 3.9-that is the reason why we formulated Lemma 2.1 for p ∈ R.
Using Lemmata 2.1 and 2.2, we can describe N (l) by means of the time maps: 2.4 Lemma.If (6) holds, then for all l > 0.
Thus, to determine the number of positive non-symmetric solutions of (1) for given p, q, a, l, one needs to calculate the limits of L 1 + L 2 at 0 and M , to examine its monotonicity and to estimate its possible relative extrema.In doing so, we will use (( 8) can be obtained by substituting √ V − 1 in (5)) and other properties of I p from [15], as well as the following theorem.
Proof: Firstly, we prove that p → I p (y)/ √ p + 1 is continuously differentiable on (−1, ∞) for any y > −1, and fulfils (9).So chose arbitrary y > 1 and p 0 > −1.We have (Taylor polynomials can be used).Consequently, p → I p (y)/ √ p + 1 is differentiable on (p 0 , ∞), and (9) holds.Moreover, p → J p (y) is continuous on (p 0 , ∞) due to the continuity of ∂µ ∂p (V, •) for all V ∈ (1, y).In order to obtain the continuous differentiability of (y, p) → I p (y)/ √ p + 1 (or equivalently of (y, p) → I p (y)), we have to validate the continuity of its partial derivatives: Since J p (y) is continuous in p, and is apparently continuous and decreasing in y, it is indeed continuous.And the continuity of
The situation is much more complicated for p < 1, and we have succeeded only in describing the behaviour of L 1 + L 2 near 0 and M , except two special cases dealt with in the following theorem.
Proof: We have to calculate L 1 + L 2 , and the statement of the theorem will follow from Lemma 2.4.
In the rest of this article we determine the values of (p, q) for which L 1 + L 2 is greater than lim m→0 (L 1 + L 2 )(m) near 0 and for which it is less.The same will be done for the neighbourhood of M .
Standard asymptotic notations will be used: If f , g are functions defined in some punctured neighbourhood of a point a ∈ R ∪ {±∞}, then 3.4 Lemma.Assume that (6) holds with p < 1.Then See Figure 1 showing these two sets in the (p, q)-plane.
EJQTDE, 2013 No. 69, p. 6 The statement of the lemma for the remaining pairs (p, q) can be verified finding the second term of the asymptotic expansion of (L 1 + L 2 )(m) for m → 0 and determining its sign.For this purpose, we will join 3] (its proof was done only for p > 1, but it holds for all p > −1) with several equalities from the proof of [15,Lemma 8.5].
All the asymptotic expansions will concern m → 0.
To determine the behaviour of L 1 + L 2 near M is much more difficult.For this purpose, the second term of the corresponding asymptotic expansion will be investigated, the finding of which requires the following lemma: Proof: Assume (6).From [15, Lemma 8.1] we already know the first term of the asymptotic expansion of R 1,2 (m)/R(M ) for m → M −.Before calculating the next two terms, let us notice that (4), as an equation in m, has the explicit solution which determines the inverse functions of R 1 and R 2 , and will be an important tool of this proof.All the asymptotic expansions appearing below will concern m → M − or z → 0.
and decreasing for i = 2, which explains the choice of the sign of c i .)Using the substitution one obtains where z → 0∓ means z → 0− for i = 1 and z → 0+ for i = 2.This limit (which should be finite and non-negative, determining the value of c i ) will be calculated using the asymptotic expansion of the denominator of the last fraction.Therefore, it is convenient to derive the equality . EJQTDE, 2013 No. 69, p. 8 Approximating (1 + z) 2q−p−1 with its 2nd order Maclaurin polynomial, one obtains h(z) = qz 2 + o(z 2 ), which results in 2. Now we seek c i = 0 and So we have to calculate the corresponding limit 12) was used again), which requires the knowledge of one more term of the asymptotic expansion of h(z).Therefore, we derive that which yields The next step is to calculate the expansion of L 1 + L 2 .

M
(and follows from the definition of the Taylor polynomial), we obtain It can be inserted in (13), resulting in which confirms the conclusion of the lemma.
In addition, for all p ∈ [− 1 7 , 1), q = q(p) is given as the only solution of .
In the sequel we EJQTDE, 2013 No. 69, p. 10 Figure 2: The graphs of q, q and the two sets from Lemma 3.7 (i), (ii).

3.
In this step we prove that f (p, 1) < 0 for all p ∈ (−1, 0), or equivalently, Our method is to gradually derive simpler and simpler sufficient conditions for (16), the last of which will be proved directly.
The next lemma describes the basic properties of q.
(a) The first sufficient condition for (21) is (again, ω(x) = x 1/(x−1) ), which can be derived in a way completely analogous to the corresponding part of the proof of Lemma 3.7.
EJQTDE, 2013 No. 69, p. 15 (b) The opposite inequality of (18) does not hold for all x > 2. Instead, will be used, which is equivalent to and the validity of which follows from the facts that κ(2) = 0 and Due to (23), 1 can be replaced with ω(x)/2 on the right-hand side of ( 22), yielding a sufficient condition for ( 22), which can be rewritten as (c) The final simplification will be done by virtue of the inequality which holds since Q(2) = 0 and Q (x) > 9 > 0 for x > 2. So now the only assertion to prove is And to do so, we just have to recall part (c) of step 3. of the proof of Lemma 3.7, and to realise that η(x) > 0 for x > 2 because η (2) = As suggested by numerical calculations, q(−1) ≈ 2.151, and q seems to be convex, having min q ≈ 0.822 ≈ q(0.495).It can be proved that setting q(1) := 1, q (1) = 1  2 holds.Recall that the line q = −p forms the border between those sets of (p, q) where L 1 + L 2 < L 2 (0) and L 1 + L 2 > L 2 (0) near 0 (see Lemma 3.4).According to Lemma 3.7, the graph of q plays a similar role in the behaviour of L 1 + L 2 near M .Therefore, if we are interested in the behaviour of L 1 + L 2 on (0, M ), we have to know the mutual position of these to curves.
3 ).We have Thanks to Theorem 2.5, Φ is differentiable, and Numerical calculations indicate that Φ is decreasing.If we could prove it, the proof would be complete (since we know that Φ(− 1 2 ) = 0).The nonpositivity of H is a sufficient condition for it.

Figure 3 :
Figure 3: The behaviour ofL 1 + L 2 for p > −1, q > p+12 , a > 0 according to Theorem 3.2 and Lemmata 3.1, 3.3, 3.4, 3.7, 3.8 and 3.9.The dashed graphs mean that for those values of p and q the behaviour of L 1 + L 2 has been examined only near 0 and M , and the graph has been plotted assuming that L 1 + L 2 has at most one stationary point.(This assumption is consistent with numerical calculations.) ).