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Multiplicity for semilinear elliptic equations involving singular nonlinearity
Nonlinear Analysis 65 (2006) 265 – 283 www.elsevier.com/locate/na
Multiplicity for semilinear elliptic equations involving singular nonlinearity夡 J. Hernándeza,∗ , J. Karátsonb , P.L. Simonb a Departemento de Matemáticas, Universidad Autónoma de Madrid, Spain b Department of Applied Analysis, Eötvös Loránd University, Budapest, Hungary
Received 1 January 2003; accepted 1 November 2004
1. Introduction We investigate the number of positive solutions of the semilinear boundary value problem u + f (u) = 0, u|jBR = 0,
u>0
in BR , (1)
where BR ⊂ Rn is the ball centered at the origin with radius R, and f : (0, ∞) → (0, ∞) is a locally Lipschitz continuous function satisfying the following conditions: (H1) f (u)m > 0 (u > 0); (H2) there exists lim
u→∞
f (u) ∈ (0, ∞) for some 1 < p < 2∗ , up
where 2∗ = (H3)
n+2 n−2
if n > 2 and 2∗ = ∞ if n 2.
lim sup u f (u) ∈ [0, ∞) for some 0 < < 1. u→0 夡 This research was supported by the Hungarian National Research Funds AMFK under Magyary Zoltán Scholarship and OTKA under grants no. F034840 and T031807.
∗ Corresponding author.
E-mail addresses: [email protected] (J. Hernández), [email protected] (J. Karátson), [email protected] (P.L. Simon). 0362-546X/$ - see front matter 䉷 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2004.11.024
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The motivating example, involving singular nonlinearity, is the problem u + + up = 0, u
u>0
in B1 ,
u|jB1 = 0
(2)
with some constants > 0,
0<<1
and
1 < p < 2∗ .
This problem can be reduced to (1) with f (s) = s − + s p using a suitable transformation. Problems with singular nonlinearities have been studied extensively recently involving more general nonlinearities which may also depend on x. For the case of singularities in x, see [10,13] and references therein. We are mainly interested here in singularities in u. For nonlinearities where the term up does not appear, existence and uniqueness has been proved in [3,5,9,11,12,17]. For equations of the form u + u − up , existence and uniqueness for p > 0 (and also for p = − < 0, < ) can be found in [12]. (For p = − < 0, < < 1 non-existence is expected [12].) In the presence of up the problem exhibits a different behavior according to the sign of . If < 0, then there may be no positive solution; existence and non-existence results for the concave case p ∈ [0, 1) can be found in [6,21,24] and generalized for x-dependent nonlinearities in [4,12]. In the latter case non-existence of positive solutions under the critical value ∗ still allows existence of non-negative solutions with a free boundary (see [4]). Related results can be found in [20] for the nonlinearity (up − u− ). When the parameter multiplies the term up , then existence and uniqueness holds for all if u− has a negative coefficient [12]. Here non-existence can also occur (namely for with some > 0) when u− has a changing-sign coefficient or when the term up is replaced by a bounded non-negative function f (x) (see [6,12]). Conversely, a unique positive solution exists exactly for with some > 0 in two related cases when the equation contains three terms: u + u + f (x) equals u− or −u− . For p = 0 there are also some multiplicity results in [1]. For > 0, existence is easier to prove on a ball and has also been obtained on smooth bounded domains. For p ∈ (0, 1) there is always a positive solution, which is also unique [12,21]. On the other hand, for p > 1 existence only holds for small [2,12]. Existence of a second positive solution has been proved for associated non-singular problems by using variational arguments. Some related results for different concave and/or convex nonlinearities can be found in [14–16,19]. For the singular case we prove here the existence of a second positive solution in the case of a ball and, moreover, that there are exactly two radial positive solutions for n = 1. At the best of our knowledge, the celebrated result by Gidas et al. [8] has not been extended to the case of singular nonlinearities. Hence we cannot be sure that our problems have only radial solutions in the case of a ball, and consequently our results for the case n > 1 refer only to positive radial solutions. The main results of our paper are formulated as follows.
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Theorem 1. Let f satisfy conditions (H1)–(H3). Then there exists R ∗ > 0 such that 1. (1) has at least two (radial) solutions for R < R ∗ ; 2. (1) has at least one (radial) solution for R = R ∗ ; 3. (1) has no radial solution for R > R ∗ . In order to have exact multiplicity for n = 1, f is assumed to satisfy a further condition: (H4) f is a strictly convex C 2 function (i.e. f 0 and f does not vanish identically on any interval). Theorem 2. Let n = 1 and let f satisfy conditions (H1)–(H4). Then there exists R ∗ > 0 such that 1. (1) has two solutions for R < R ∗ ; 2. (1) has one solution for R = R ∗ ; 3. (1) has no solution for R > R ∗ . Concerning the main example (2), Theorems 1 and 2 yield the following result using the transformation p−1
R = 2(p+)
and u( ˜ x) ˜ =
1 − p+
1−p
u( 2(p+) x). ˜
Corollary 1. There exists ∗ > 0 such that 1. (2) has at least two (radial) solutions for < ∗ ; 2. (2) has at least one (radial) solution for = ∗ ; 3. (2) has no radial solution for > ∗ . Further, if n = 1 then these multiplicity results are exact. The proof of our results relies on the shooting method. Theorem 1 is proved in Section 3 via characterizing the shape of the time-map. The exact multiplicity is dealt with in Section 4 in a little more general setting than in Theorem 2, namely, we assume a certain admissibility condition (H5), which is seen to be always satisfied in the case n = 1.
2. Preliminaries In the study of radial solutions we regard u as a function of r = |x| and use notation u = jr u. Then (1) is equivalent to (r n−1 u ) + r n−1 f (u) = 0,
u (0) = 0, u(R) = 0.
u>0
in [0, R), (3)
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We apply the shooting method, which consists of the investigation of the solutions u(·, c) to the initial value problems (r n−1 u ) + r n−1 f (u) = 0, u(0, c) = c, u (0, c) = 0.
(4)
C2
solution u(·, c) for any c > 0 [22], which is defined as long These IVPs have a unique as it is positive. In the sequel we omit c from the argument where it does not cause ambiguity and simply write u(r). The integral form of (4) will be frequently used r −r n−1 u (r) = s n−1 f (u(s)) ds. (5) 0
Together with (H1), this implies immediately that u (r) < 0
(r > 0).
(6)
The main tool of our investigation is the time-map, which associates the first zero of u(·, c) to c: Definition 1. The time-map corresponding to the above initial value problem is the following function T: T (c) = sup{r > 0: u(s, c) > 0 ∀s ∈ [0, r)}; D(T ) = c > 0: ∃R0 > 0, lim u(r, c) = 0 . r→R0
As mentioned above, u(r, c) is defined for r ∈ [0, T (c)) since f is allowed not to be defined at 0. It is straightforward to extend u(r, c) to [0, T (c)] by u(T (c)) = 0. We note that, owing to (6), c ∈ D(T ) whenever the sup in the definition of T (c) is finite. The proof of our results relies on the characterization of the shape of the time-map. Namely, the number of radial solutions of (3) on BR coincides with the number of solutions c > 0 of the equation T (c) = R. The latter is determined by the following three properties: • the domain of the time-map; • the limit of the time-map at the boundary points of its domain; • the monotonicity of the time-map on the maximal subintervals of its domain. This characterization uses the following variational equations w.r.t. parameters. Differentiating (4) with respect to c and introducing the notations h(r, c) = jc u(r, c)
and z(r, c) = j2c u(r, c),
we get (r n−1 h ) + r n−1 f (u)h = 0, h(0, c) = 1,
h (0, c) = 0
(7)
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269
(r n−1 z ) + r n−1 (f (u)z + f (u)h2 ) = 0, z(0, c) = 0,
z (0, c) = 0.
(8)
Differentiating (4) with respect to r and introducing the notation v(r, c) = jr u(r, c) we get n−1 n−1 n−1 v = 0, f (u) − 2 v ) +r (r r v(0, c) = 0,
v (0, c) = −
f (c) . n
(9)
3. Multiple solutions In this section Theorem 1 is proved. The theorem follows in an elementary way from the following properties of the time-map: • D(T ) = (0, ∞), • limc→0 T (c) = limc→∞ T (c) = 0, • T is continuous. These properties are proved in the following subsections. 3.1. The domain of the time-map Lemma 1. Let f satisfy condition (H1). Then D(T ) = (0, ∞) and limc→0 T (c) = 0. Proof. Dividing (5) by −r n−1 , using (H1) and integrating on [0, r] one obtains u(r, c)c −
m 2 r . 2n
(10)
√ This implies D(T ) = (0, ∞) and T (c) 2nc/m from which the second statement of the Lemma follows. 3.2. The limit of the time-map at infinity We will need the following propositions. Proposition 1. Let us assume that lim
u→∞
f (u) =∞ u
(11)
and that the solution of (4) is decreasing for any c > 0. Then for all > 0 there exists c > 0 such that for the solutions of (4) u(, c)c
for all c c .
(12)
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Proof. Let > 0 be an arbitrary positive number. Then there exists K > 0 such that for the solution of the linear initial value problem (r n−1 w ) + r n−1 K w = 0, w(0) > 0, w (0) = 0
(13)
we have w() = 0 and w(r) > 0 for r ∈ [0, ), i.e. the first root of w is . According to (11) there exists c > 0 such that f (u) > K u
for all u > c .
(14)
We will argue by contradiction. Let us assume that there exists c c , for which u(, c) > c . For short, let us denote u(r, c) by u(r). Since u is decreasing, therefore u(r) > c for all r ∈ [0, ]. Let us denote by w the solution of (13) satisfying the initial condition w(0) = c. It is easy to see from (13), (4) and (14) that w (0) = −
f (c) K c >− = u (0), n n
that is for small r > 0 we have w(r) > u(r). However, w() = 0 and u() > c imply that there exists r1 ∈ (0, ) such that c < u(r1 ) = w(r1 ),
u (r1 ) > w (r1 ),
u(r) < w(r) for all r ∈ (0, r1 ).
(15)
Let us introduce the function A(r) = r n−1 (u (r)w(r) − u(r)w (r)).
(16)
Then using (4), (13) and (14) A (r) = r n−1 w(r)(K u(r) − f (u(r))) < 0
for all r ∈ (0, ).
(17)
From (15) and (16) we get A(r1 ) > 0, further, from A(0)=0 and (17) there follows A(r1 ) < 0, which is a contradiction. Proposition 2. Let n2 and f satisfy (H1)–(H3). Then 1. there exist u1 > 0 and K > 0 such that 2nF (u) − (n − 2)uf (u)Kup+1
for all u > u1 ,
2. there exists a > 0 such that 2nF (u) − (n − 2)uf (u) − a
for all u 0.
Proof. According to (H2) there exist L > 0 and p ∈ (1, 2∗ ) such that L = lim
u→∞
f (u) . up
J. Hernández et al. / Nonlinear Analysis 65 (2006) 265 – 283
Let
n + 2 − (n − 2)p ∈ 0, L min 1, 5n − 2 + (n − 2)p
271
.
(18)
Then there exists u0 > 0 such that f (u) ∈ (L − , L + ) for all u > u0 . up Let u1 =
1/(p+1)
L −1
u0
Then, using the formulas above, we get for u > u1 u u up+1 . F (u) f (t) dt (L − ) t p dt (L − 2) p+1 u0 u0 Hence for u > u1 2n (L − 2)up+1 − (n − 2)(L + )up+1 p+1 2n 4n = − (n − 2) L − + n − 2 up+1 , p+1 p+1
2nF (u) − (n − 2)uf (u)
from which, using (18), statement 1 follows. The second statement follows from statement 1 and the continuity of 2nF (u) − (n − 2)uf (u) on [0, ∞) (which is a consequence of (H3)). Proposition 3. Let us assume that f satisfies (H2). For a given c > 0 let us define rc by the equation c u(rc , c) = . 2
(19)
Then there exist M > 0 and u2 > 0 such that n 1−p c rc2 M
for all c > 2u2 .
Proof. Assumption (H2) implies that there exist M > 0 and u2 > 0 such that f (u)Mup
for all u u2 .
From (5) and (6) we get r rn s n−1 Mup (s) ds Mcp −r n−1 u (r) n 0
(20)
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if u(r)u2 . Dividing this inequality by r n−1 and integrating on [0, t] one obtains c − u(t)
M p 2 c t 2n
(21)
if u(t)u2 . If c > 2u2 , then applying (21) for t = rc we get the desired inequality (20).
Lemma 2. Let f satisfy conditions (H1)–(H3). Then limc→∞ T (c) = 0. Proof. For n = 1 the lemma was proved in [15] (in a little more general case). Now we assume n 2. Step 1: Let > 0 be an arbitrary positive number. Let us choose c according to Proposition 1. We will prove that there exists d > c such that |u (r, c)| >
c
for all r ∈ (, 2), c > d .
(22)
This inequality implies T (c) < 2 for all c > d , since u(, c) c and u(r, c) decreases at least by c on the interval (, 2). Step 2: To prove (22) we will apply the following general form of Pohozhaev identity [7]. r 2 s n−1 [2nF (u(s)) − (n − 2)u(s)f (u(s))] ds r n u (r) + 2r n F (u(r)) = 0
− (n − 2)r n−1 u(r)u (r). According to (6) 2
u (r) > r
−n
r
s n−1 [2nF (u(s)) − (n − 2)u(s)f (u(s))] ds − 2F (u(r)).
(23)
0
Let us choose K and u1 according to Proposition 2, and M and u2 according to Proposition 3. Let c = max{2c , 2u1 , 2u2 }. Let c > c and introduce rc according to (19). For short, we use the notation u(r) instead of u(r, c). From (23) one obtains rc 2 s n−1 [2nF (u(s)) − (n − 2)u(s)f (u(s))] ds u (r) > r −n 0
+ r −n
r
s n−1 [2nF (u(s)) − (n − 2)u(s)f (u(s))] ds − 2F (u(r)).
(24)
rc
Step 3: Now, we will estimate each term in the r.h.s. of (24). For the first term s < rc implies u(s) > u(rc ) =
c > u1 . 2
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273
Hence from Proposition 2 2nF (u(s)) − (n − 2)u(s)f (u(s)) Kup+1 (s) > K
c p+1 2
for all s < rc ,
yielding for r ∈ (, 2) rc c p+1 r n c r −n . s n−1 [2nF (u(s)) − (n − 2)u(s)f (u(s))] ds (2)−n K 2 n 0 Using c > c 2u2 Proposition 3 implies rc r −n s n−1 [2nF (u(s)) − (n − 2)u(s)f (u(s))] ds 0 c p+1 K n n/2 c(1−p)n/2 . (2)−n 2 n M For the second term in (24) we use statement 2 of Proposition 2: r a a rn −n r =− . s n−1 [2nF (u(s)) − (n − 2)u(s)f (u(s))] ds − n r n n rc
(25)
(26)
For the third term r ∈ (, 2) and c > c imply u(r) c , hence (using F = f > 0) F (u(r))F (c ).
(27)
Step 4: Now let us substitute the estimates (25)–(27) into (24). Then for all r ∈ (, 2) and c > c K n n/2 p+1+(1−p)n/2 a 2 u (r, c) > (2)−n p+1 c − − 2F (c ). (28) n2 M n Since p < 2∗ , therefore p + 1 + (1 − p)n/2 > 0, hence the r.h.s. of (28) tends to infinity as c → ∞, which proves the desired inequality (22). 3.3. The continuity of the time-map For any > 0 let us introduce the function T as follows: T (c) = min{r ∈ (0, T (c)) : u(r, c) = }.
(29)
Obviously, D(T ) = (, ∞) and T is continuous. Lemma 3. Let f satisfy (H1). Then T tends uniformly to T on any compact subinterval of (0, ∞) as → 0. Proof. Let [a, b] ⊂ (0, ∞) be a compact subinterval. Let 0 < a, then for all 0 D(T ) ⊃ [a, b]. Let R0 = min T0 , [a,b]
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J. Hernández et al. / Nonlinear Analysis 65 (2006) 265 – 283
then (since < 0 implies T > T0 ) T (c)R0
for all c ∈ [a, b], ∈ (0, 0 ].
(30)
Let c ∈ [a, b], ∈ (0, 0 ] and r ∈ (T (c), T (c)], then using (H1) and (5) −u (r, c)
m m m r T (c) R0 , n n n
integrating this inequality on [T (c), T (c)]
m R0 (T (c) − T (c)), n
thus |T (c) − T (c)|
n , mR 0
from which the statement follows.
Corollary 2. Let f satisfy (H1). Then T is continuous. Remark 1. In fact, the condition (H1) for Lemma 3 (and hence that of Corollary 2) is too restrictive. For example, the condition lim inf u→0 f (u) > 0 is also sufficient. 4. Exact multiplicity in one dimension In the sequel we study the exact number of radial positive solutions. When n = 1, it can be seen similarly to the non-singular case that all positive solutions are radial, that is symmetric to 0 in (−R, R). The crucial part of Theorem 2 is the following property (H5) concerning the function h = jc u (see (7)). It often arises in several investigations with different nonlinearities, e.g. [7,18,19,23]. Such problems are sometimes called disconjugate [19] and the corresponding values of c admissible [18]. (H5) For all c > 0, the function h(·, c) has at most one root in [0, T (c)]. Remark 2. Condition (H5) is equivalent to the following [23]: for any radial solution u of (1) the Morse index of u is at most 1, and if it equals 1 then u is non-degenerate. The multiplicity result of Theorem 2 in fact exploits (H5) instead of the restriction n = 1. Hence in this section we will verify Theorem 3. Let f satisfy conditions (H1)–(H5). Then there exists R ∗ > 0 such that 1. (1) has two radial solutions for R < R ∗ ; 2. (1) has one radial solution for R = R ∗ ; 3. (1) has no radial solution for R > R ∗ .
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Theorem 3 will be proved in a sequence of lemmas in the subsections below: • First we study the approximate time-maps T in Section 4.1, and prove that T has a unique maximum at some c0 , it is strictly increasing before c0 and strictly decreasing after c0 . (Together with the limits of T at 0 and ∞, this already proves Theorem 3 when limu→0 f (u) is finite, hence the sequel is devoted to the singular case.) • The main ingredient of the proof is provided by Section 4.2, stating that the time-map cannot be constant on any subinterval of (0, ∞). • Passing to the limit → 0 and using the obtained non-degeneracy of the time-map, in Section 4.3 we derive easily that T inherits the shape of T . Theorem 2 can be verified from Theorem 3 by checking condition (H5) in the following way. Proof of Theorem 2. In the case n = 1 the functions h and v in (7) and (9) satisfy the same linear equation, further, we have v < 0 in (0, T (c)] from (6). Hence Sturm comparison shows that h may have at most one root in [0, T (c)], i.e. condition (H5) is satisfied and thus the assertions of Theorem 3 hold. 4.1. The shape of the approximate time-maps For all > 0 the approximate time-maps T have been introduced in (29). The function T is determined by the implicit equation u(T (c), c) = .
(31)
Differentiating (31), one gets the following equations for the derivatives of T , using the notations h = jc u, z = j2c u, v = u = jr u (introduced in Section 2): v(T (c), c) T (c) + h(T (c), c) ≡ 0,
(32)
v (T (c), c) T (c)2 + 2h (T (c), c) T (c)
+ v(T (c), c) T (c) + z(T (c), c) ≡ 0.
Expressing T (c) from (32) and substituting in (33), T (c) is given as 1 v h2 h h T (c) = − +z , −2 v v2 v
(33)
(34)
where all functions on the r.h.s. have the argument (T (c), c). Lemma 4. Let conditions (H1)–(H5) hold, > 0. Then 1. limc→ T (c) = limc→∞ T (c) = 0. 2. Let c > 0. If T (c) = 0 then T (c) < 0. √ Proof. 1. From (10) follows T (c) 2n(c − )/m, implying lim T = 0. From Lemma 2 and T < T follows lim∞ T = 0.
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2. Eqs. (32) and (33) imply h(T (c), c) = 0 and T (c) = −
z(T (c), c) . v(T (c), c)
Since v = u < 0, therefore it is enough to prove that h(r ∗ ) = 0 implies z(r ∗ ) < 0. Let us introduce the function A(r) = r n−1 (h(r)z (r) − h (r)z(r)).
(35)
Then A(0) = 0 and using (7)–(8), (H4) and (H5) A (r) = −r n−1 f (u(r))h3 (r) < 0
for all r ∈ (0, r ∗ ),
(36)
since h > 0 in (0, r ∗ ) by (H5). Hence A(r ∗ ) < 0, which yields z(r ∗ ) < 0, because h(r ∗ ) = 0 and h (r ∗ ) < 0. Corollary 3. This characterization of the approximate time-maps implies that T has a unique maximum at some c0 > , it is strictly increasing before c0 and strictly decreasing after c0 . Hence, together with limc→0 T (c) = limc→∞ T (c) = 0, the multiplicity result of Theorem 3 holds with the boundary condition u|jBR = in (1). If limu→0 f (u) < ∞ (i.e. there is no singularity), then the differentiability properties of T remain valid for T (essentially for the case = 0). Hence Lemma 4 can be repeated with T, which means that Theorem 3 is proved for this case: Corollary 4. If limu→0 f (u) < ∞, then Theorem 3 holds. 4.2. The non-degeneracy of the time-map Since the non-singular case (limu→0 f (u) < ∞) has been handled in Corollary 4, in this subsection we restrict ourselves to the singular case, i.e. we assume (H6) lim f (u) = ∞. u→0
The aim of this subsection is to prove that the time-map cannot be constant on any nondegenerate subinterval of (0, ∞), which will be stated in Lemma 5. The difficulty in proving this plausible property lies in the singularity (H6). Hence Lemma 5 is preceded by careful estimates, formulated in propositions using the indirect degeneracy assumption (D) T (c) ≡ R
(c ∈ J )
for some non-degenerate compact interval J ⊂ (0, ∞) and number R > 0.
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Proposition 4. Let conditions (H1)–(H4) and (H6) hold. If T has property (D), then limR h = 0 uniformly, i.e. for any > 0 there exists r > 0 such that |h(r, c)| (r ∈ [r , R), c ∈ J ). Proof. Step 1: There exists r0 > 0 such that f (u(r, c)) < 0
(r ∈ (r0 , R), c ∈ J ).
Namely, from (H4) and (H6) there exists > 0 such that f (s) < 0
(s ∈ (0, ]).
(37)
For any c ∈ J let rc > 0 such that u(r, c) < (r > rc ). The continuous dependence of u on c implies that there exists r0 := max{rc : c ∈ J } < R. Step 2: For any r ∈ [r0 , R) and c ∈ J the following holds: if h(r, c) > 0 then h (r, c) < 0, and (conversely) if h(r, c) < 0 then h (r, c) > 0. To see this, we use the integral of (7): s n−1 f (u())h(, c) d s n−1 h (s, c) − s0n−1 h (s0 , c) = − s0
(s0 < s < R, c ∈ J ).
(38)
Namely, let first h(r, c) > 0 and assume indirectly that h (r, c)0. Setting s0 = r, for s a bit larger than r the right side of (38) is positive, hence there exists r˜ > r such that h(˜r , c) > 0 and h (˜r , c) > 0. Using the continuous dependence of h on c, there exists a > 0 and J˜ ⊂ J such that h(˜r , c) ˜ a and h (˜r , c) ˜ > 0 for all c˜ ∈ J˜. Now setting s0 = r˜ , there n−1 n−1 holds s h (s, c) ˜ > (˜r ) h (˜r , c) ˜ > 0 for all c˜ ∈ J˜ and s > r˜ as long as h(., c) ˜ > 0. This implies that h increases after r˜ , hence h(s, c)a ˜
(s ∈ [˜r , R], c˜ ∈ J˜).
(39)
However, this yields u(s, c2 ) − u(s, c1 ) =
c2 c1
jc u(s, c) dc =
c2
h(s, c) dc a(c2 − c1 ) > 0
(40)
c1
for all s ∈ [˜r , R] and c1 , c2 ∈ J˜. Letting s → R with fixed c1 , c2 , the left side tends to 0 and the right side is fixed, a contradiction. The counterpart with h(r, c) < 0 is proved similarly. Step 3: For any c ∈ J , h(., c) has constant sign on [r0 , R). Namely, assume that h(r1 , c)=0 for some r1 ∈ [r0 , R). Then h (r1 , c) = 0 from uniqueness. If e.g. h (r1 , c) > 0, then h(r2 , c) > 0 and h (r2 , c) > 0 for some r2 > r1 in contradiction to Step 2. In the same way h (r1 , c) < 0 is impossible. Step 4: There exists B1 > 0 such that |h (r, c)| B1
(r ∈ [r0 , R), c ∈ J ).
(41)
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Namely, let first h(r0 , c) > 0 for some c ∈ J . Then h(r, c) > 0 and hence h (r, c) < 0 for all r ∈ [r0 , R) (from Steps 3 and 2). Here (7) yields that (r n−1 h ) > 0 on [r0 , R), hence r n−1 h (r, c) > r0n−1 h (r0 , c) for any r ∈ [r0 , R). Since r > r0 , we obtain |h (r, c)| = −h (r, c) < − h (r0 , c) = |h (r0 , c)|
(r ∈ [r0 , R)).
(42)
The same result follows for the case h(r0 , c) < 0 as well. Hence for all c ∈ J , |h (r, c)| |h (r0 , c)| (r ∈ [r0 , R)). Introducing B1 := max{h (r0 , c) : c ∈ J }, we obtain the required estimate. Step 5: For all c ∈ J , lim h(r, c) = 0.
(43)
r→R
Namely, Steps 3 and 2 imply that h(., c) is (strictly) monotone on [r0 , R), hence the limit exists. Assume indirectly that e.g. limr→R h(r, c0 ) =: a > 0 for some c0 > 0. In this case, h(., c0 ) decreases on [r0 , R), hence h(r, c0 ) > a for r ∈ [r0 , R). Let r1 ∈ [r0 , R) such that R − r1 a/2B1 (with B1 from Step 4). Then, from the continuous dependence of h on c, h(r1 , c) > a (c ∈ J ) for some J ⊂ J . Hence for any r ∈ [r1 , R) and c ∈ J , r a a h (s, c) ds > a − B1 (R − r1 ) a − B1 · = > 0. h(r, c) = h(r1 , c) + 2B1 2 r1 This is impossible (see (39) and afterwards). Step 6: Let > 0 be given. Introducing r = R − (/B1 ), we obtain from Steps 5 and 4 that R |h (s, c)| ds B1 (R − r)B1 (R − r ) = |h(r, c)| r
(r ∈ [r , R), c ∈ J ).
The fact limR h = 0 itself and condition (H5) imply. Corollary 5. Let conditions (H1)–(H6) hold. If T has property (D), then for all c ∈ J h(r, c) > 0
(r ∈ [0, R)).
Proposition 5. Let conditions (H1)–(H6) hold. If T has property (D), then there exist m1 > 0 and r1 ∈ (0, R) such that z(r, c) − m1
(r ∈ [r1 , R), c ∈ J ).
Proof. Step 1: Let A(r) = A(r, c) ≡ r n−1 (z (r)h(r) − h (r)z(r))
(r ∈ (0, R), c ∈ J )
as in (35). We have seen that A(r) is negative and decreasing, see (36) and afterwards. Step 2: There holds z(r, c) < 0 for all r ∈ (0, R), c ∈ J . Namely, let first f (c) > 0. Then z(0) = 0, and (r n−1 z ) (r) < 0 for small r from (8). Hence z(r) < 0 for small r. If z had a
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root in (0, R), then its first root r1 would satisfy z (r1 ) 0, hence A(r1 ) 0 in contradiction to Step 1. Now, let f (c) = 0. Since z(0) = 0, therefore z(r, c) must have constant sign for r ∈ (0, ) with some > 0. Since the values of c with f (c) > 0 are dense in J owing to (H4), it follows by continuity from the previous case that this initial sign of z can only be negative. Then z(r) < 0 extends to r ∈ (0, R) similarly as above using the function A. Step 3: Let r0 ∈ (0, R) be as in Steps 1–2 of Proposition 4 (i.e. h(., c) decreases on [r0 , R) for all c ∈ J ). Then there exists > 0 such that z (r)h(r) − h (r)z(r) −
(r ∈ [r0 , R), c ∈ J ).
(44)
Namely, by Step 1, A(r) is negative and decreasing, hence R n−1 (z (r)h(r) − h (r)z(r)) < A(r)A(r0 ) < 0. Therefore = −R −(n−1) min{A(r0 , c) : c ∈ J } is suitable. Step 4: Let r 1 G(r, c) = h(r, c) ds. 2 (s, c) h r0 Then there holds z(r, c) − G(r, c)
(r ∈ (r0 , R), c ∈ J ).
(45)
Namely, (44) implies ( hz ) − h12 on [r0 , R), hence z(r, c) z(r0 , c) − h(r, c) h(r0 , c)
r
r0
1 h2 (s, c)
ds.
Multiplying this inequality by h(r, c) and using h > 0 and z < 0, we obtain (45). Step 5: For all c ∈ J , G(., c) is convex on [r0 , R). Namely, r r 1 1 1 ds + , G (r) = h (r) ds 0 G (r) = h (r) 2 (s) 2 (s) h h(r) h r0 r0
(46)
since h(., c) is convex on [r0 , R) from (42). Step 6: For any r1 ∈ (r0 , R) there exists a1 > 0 such that G(r, c)a1
(r ∈ [r1 , R), c ∈ J ).
(47)
Namely, (46) yields G (r0 , c) = 1/ h(r0 , c)d, where d = min{1/ h(r0 , c) : c ∈ J } > 0 from the continuous dependence of h on c. The convexity of G yields G(r, c)G (r0 , c)(r − r0 ) d(r − r0 ) d(r1 − r0 ) =: a1 . Step 7: Fixing arbitrary r1 ∈ (r0 , R), the proposition follows from (45) and (47).
Proposition 6. Let conditions (H1)–(H6) hold. If T has property (D), then there exist M > 0 and 0 > 0 such that T (c) − M
(c ∈ J, ∈ (0, 0 ]).
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Proof. Step 1: T (c) can be expressed by rearranging (34) as h v − v h h 1 T (c) = h − z , + v v2 v where all functions on the r.h.s. have the argument (T (c), c). In virtue of Proposition 5, now our aim is to prove that v is bounded and the term containing h tends to 0, both uniformly for c ∈ J . Step 2: There exist r2 ∈ (0, R) and 2 1 > 0 such that (r ∈ [r2 , R), c ∈ J ).
− 2 v(r, c) − 1
Namely, let r˜ ∈ (0, R) be arbitrary. Then r n−1 v = r n−1 u is decreasing owing to (4), hence for any r ∈ [˜r , R) v(r, c) (˜r /r)n−1 v(˜r , c)(˜r /R)n−1 max{v(˜r , c) : c ∈ J } =: − 1 ,
(48)
where 1 ∈ (0, ∞) exists by continuous dependence. On the other hand, integrating (48) on [r, R], we obtain u(r, c) 1 (R − r)
(r ∈ [˜r , R)).
Let r2 = max{˜r , r0 }, where r0 is from Step 1 of Proposition 4. Then u(r, c) <
(r ∈ [r0 , R), c ∈ J ),
where f is decreasing on [0, ], hence f (u(r, c))f ( 1 (R − r))
(r ∈ [r2 , R)).
Integrating (4), we obtain n−1
r
v(r, c)
r2n−1 v(r2 , c) −
r
s n−1 f (u(s, c)) ds r n−1 R v(r2 , c) − f (u(s, c)) ds . r2
r2
Here
r
r2
f (u(s, c)) ds
R
f ( 1 (R − s)) ds =
r2
1
1
1 (R−r2 )
f (t) dt =: ,
0
where < ∞ owing to assumption (H3) and is independent of c. Hence n−1 R · (min{v(r2 , c) : c ∈ J } − ) =: − 2 (r ∈ [r2 , R)). v(r, c) r2 Step 3: There exists B2 > 0 such that |h (r, c)v(r, c) − v (r, c)h(r, c)|B2
(r ∈ [r2 , R), c ∈ J ),
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where r2 ∈ (0, R) is from Step 2. Namely, let A(r) = A(r, c) ≡ r n−1 (h (r)v(r) − v (r)h(r))
(r ∈ (0, R), c ∈ J ).
(49)
Then A (r)=(r n−1 h ) v −(r n−1 v ) h=−(n−1)r n−1 hv from Eqs. (7) and (9). Here r2 r0 by its definition, hence (41) and (43) imply |h(r, c)|(R − r2 )B1
(r ∈ [r2 , R), c ∈ J ).
Together with Step 2, we obtain for all r ∈ [r2 , R), c ∈ J |A (r, c)| (n − 1)R n−1 (R − r2 )B1 2 , hence |h (r)v(r) − v (r)h(r)| =
1
r
|A(r, c)| n−1 1
r2n−1
1 r2n−1
|A(r2 , c)| +
R
|A (r, c)|
r2
(max{|A(r2 , c)| : c ∈ J }
+(n − 1)R n−1 (R − r2 )2 B1 2
=: B2 .
Step 4: Using Steps 1–3, further, from (41) and Propositions 4 and 5, we obtain the estimate
1 B2 B1 T (c) |h(T (c), c)| − m1 (50) +
2
1
21 whenever T (c) ∈ [r2 , R) and c ∈ J . From Proposition 4 there exists r3 ∈ [r2 , R) such that m1 B2 B1 |h(T (c), c)| + 2
1 2
1 whenever T (c) ∈ [r3 , R) and c ∈ J . Since by Lemma 3 and property (D), T → R
uniformly on J ,
(51)
therefore there exists 0 > 0 such that T (c) ∈ [r3 , R) for all ∈ (0, 0 ], c ∈ J . For such the expression in large brackets in (50) is at most −m1 /2, hence, letting M := m1 /2 2 , the required estimate follows. We note that the uniform estimates of Propositions 4–5 were required to ensure that M in Proposition 6 is uniform in c and . Now we are in the position to exclude the indirect assumption (D). Lemma 5. Let conditions (H1)–(H6) hold. Then the time-map cannot be constant on any non-degenerate subinterval of (0, ∞).
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Proof. Assume for contradiction that property (D) holds. Using Taylor’s formula, Proposition 6 implies that there exist M > 0 and 0 > 0 such that T (c1 ) − T (c2 ) T (c2 )(c1 − c2 ) −
M (c1 − c2 )2 2
(c1 , c2 ∈ J, ∈ (0, 0 ]).
Let → 0. Then (51) implies that the left side tends to 0. Further, by (32) T (c2 ) = −
h(T (c2 ), c2 ) → 0, v(T (c2 ), c2 )
since T (c2 ) → R, and hence (43) yields that h(T (c2 ), c2 ) → 0 and (48) implies that v is bounded away from 0. From these we obtain the contradiction 0 −
M (c1 − c2 )2 2
(c1 , c2 ∈ J ).
4.3. Proof of Theorem 3 The results obtained in Sections 4.1–4.2 enable us to prove Theorem 3 immediately: namely, we can verify that T inherits the shape of the T . In virtue of Corollary 4, it suffices to do this in the singular case limu→0 f (u) = ∞. The three properties of T, formulated at the very beginning of Section 3, imply that T has a maximum at some c0 > 0. First we prove that T increases on [0, c0 ] and decreases on [c0 , ∞). Namely, let first 0 c1 < c2 c0 and assume for contradiction that T (c1 ) > T (c2 ). Since T → T uniformly on [c1 , c2 ] by Lemma 3, hence for some 0 > 0 we have T (c1 ) > T (c2 ) for all ∈ (0, 0 ]. In virtue of Corollary 3, T is decreasing after c1 , hence T (c2 ) T (c0 ). Letting → 0, we obtain T (c2 ) T (c0 ), which gives T (c1 ) > T (c0 ) in contradiction to the maximum assumption on T (c0 ). The other part on [c0 , ∞] is proved similarly. Using Lemma 5, we obtain that T is strictly increasing on [0, c0 ] and strictly decreasing on [c0 , ∞]. Let R ∗ = T (c0 ). Together with limc→0 T (c) = limc→∞ T (c) = 0 and the continuity of T, we conclude that T assumes each value R twice if 0 < R < R ∗ , once if R = R ∗ and does not assume R if R > R ∗ .
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Multiplicity for semilinear elliptic equations involving singular nonlinearity夡 J. Hernándeza,∗ , J. Karátsonb , P.L. Simonb a Departemento de Matemáticas, Universidad Autónoma de Madrid, Spain b Department of Applied Analysis, Eötvös Loránd University, Budapest, Hungary
Received 1 January 2003; accepted 1 November 2004
1. Introduction We investigate the number of positive solutions of the semilinear boundary value problem u + f (u) = 0, u|jBR = 0,
u>0
in BR , (1)
where BR ⊂ Rn is the ball centered at the origin with radius R, and f : (0, ∞) → (0, ∞) is a locally Lipschitz continuous function satisfying the following conditions: (H1) f (u)m > 0 (u > 0); (H2) there exists lim
u→∞
f (u) ∈ (0, ∞) for some 1 < p < 2∗ , up
where 2∗ = (H3)
n+2 n−2
if n > 2 and 2∗ = ∞ if n 2.
lim sup u f (u) ∈ [0, ∞) for some 0 < < 1. u→0 夡 This research was supported by the Hungarian National Research Funds AMFK under Magyary Zoltán Scholarship and OTKA under grants no. F034840 and T031807.
∗ Corresponding author.
E-mail addresses: [email protected] (J. Hernández), [email protected] (J. Karátson), [email protected] (P.L. Simon). 0362-546X/$ - see front matter 䉷 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2004.11.024
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The motivating example, involving singular nonlinearity, is the problem u + + up = 0, u
u>0
in B1 ,
u|jB1 = 0
(2)
with some constants > 0,
0<<1
and
1 < p < 2∗ .
This problem can be reduced to (1) with f (s) = s − + s p using a suitable transformation. Problems with singular nonlinearities have been studied extensively recently involving more general nonlinearities which may also depend on x. For the case of singularities in x, see [10,13] and references therein. We are mainly interested here in singularities in u. For nonlinearities where the term up does not appear, existence and uniqueness has been proved in [3,5,9,11,12,17]. For equations of the form u + u − up , existence and uniqueness for p > 0 (and also for p = − < 0, < ) can be found in [12]. (For p = − < 0, < < 1 non-existence is expected [12].) In the presence of up the problem exhibits a different behavior according to the sign of . If < 0, then there may be no positive solution; existence and non-existence results for the concave case p ∈ [0, 1) can be found in [6,21,24] and generalized for x-dependent nonlinearities in [4,12]. In the latter case non-existence of positive solutions under the critical value ∗ still allows existence of non-negative solutions with a free boundary (see [4]). Related results can be found in [20] for the nonlinearity (up − u− ). When the parameter multiplies the term up , then existence and uniqueness holds for all if u− has a negative coefficient [12]. Here non-existence can also occur (namely for with some > 0) when u− has a changing-sign coefficient or when the term up is replaced by a bounded non-negative function f (x) (see [6,12]). Conversely, a unique positive solution exists exactly for with some > 0 in two related cases when the equation contains three terms: u + u + f (x) equals u− or −u− . For p = 0 there are also some multiplicity results in [1]. For > 0, existence is easier to prove on a ball and has also been obtained on smooth bounded domains. For p ∈ (0, 1) there is always a positive solution, which is also unique [12,21]. On the other hand, for p > 1 existence only holds for small [2,12]. Existence of a second positive solution has been proved for associated non-singular problems by using variational arguments. Some related results for different concave and/or convex nonlinearities can be found in [14–16,19]. For the singular case we prove here the existence of a second positive solution in the case of a ball and, moreover, that there are exactly two radial positive solutions for n = 1. At the best of our knowledge, the celebrated result by Gidas et al. [8] has not been extended to the case of singular nonlinearities. Hence we cannot be sure that our problems have only radial solutions in the case of a ball, and consequently our results for the case n > 1 refer only to positive radial solutions. The main results of our paper are formulated as follows.
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Theorem 1. Let f satisfy conditions (H1)–(H3). Then there exists R ∗ > 0 such that 1. (1) has at least two (radial) solutions for R < R ∗ ; 2. (1) has at least one (radial) solution for R = R ∗ ; 3. (1) has no radial solution for R > R ∗ . In order to have exact multiplicity for n = 1, f is assumed to satisfy a further condition: (H4) f is a strictly convex C 2 function (i.e. f 0 and f does not vanish identically on any interval). Theorem 2. Let n = 1 and let f satisfy conditions (H1)–(H4). Then there exists R ∗ > 0 such that 1. (1) has two solutions for R < R ∗ ; 2. (1) has one solution for R = R ∗ ; 3. (1) has no solution for R > R ∗ . Concerning the main example (2), Theorems 1 and 2 yield the following result using the transformation p−1
R = 2(p+)
and u( ˜ x) ˜ =
1 − p+
1−p
u( 2(p+) x). ˜
Corollary 1. There exists ∗ > 0 such that 1. (2) has at least two (radial) solutions for < ∗ ; 2. (2) has at least one (radial) solution for = ∗ ; 3. (2) has no radial solution for > ∗ . Further, if n = 1 then these multiplicity results are exact. The proof of our results relies on the shooting method. Theorem 1 is proved in Section 3 via characterizing the shape of the time-map. The exact multiplicity is dealt with in Section 4 in a little more general setting than in Theorem 2, namely, we assume a certain admissibility condition (H5), which is seen to be always satisfied in the case n = 1.
2. Preliminaries In the study of radial solutions we regard u as a function of r = |x| and use notation u = jr u. Then (1) is equivalent to (r n−1 u ) + r n−1 f (u) = 0,
u (0) = 0, u(R) = 0.
u>0
in [0, R), (3)
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We apply the shooting method, which consists of the investigation of the solutions u(·, c) to the initial value problems (r n−1 u ) + r n−1 f (u) = 0, u(0, c) = c, u (0, c) = 0.
(4)
C2
solution u(·, c) for any c > 0 [22], which is defined as long These IVPs have a unique as it is positive. In the sequel we omit c from the argument where it does not cause ambiguity and simply write u(r). The integral form of (4) will be frequently used r −r n−1 u (r) = s n−1 f (u(s)) ds. (5) 0
Together with (H1), this implies immediately that u (r) < 0
(r > 0).
(6)
The main tool of our investigation is the time-map, which associates the first zero of u(·, c) to c: Definition 1. The time-map corresponding to the above initial value problem is the following function T: T (c) = sup{r > 0: u(s, c) > 0 ∀s ∈ [0, r)}; D(T ) = c > 0: ∃R0 > 0, lim u(r, c) = 0 . r→R0
As mentioned above, u(r, c) is defined for r ∈ [0, T (c)) since f is allowed not to be defined at 0. It is straightforward to extend u(r, c) to [0, T (c)] by u(T (c)) = 0. We note that, owing to (6), c ∈ D(T ) whenever the sup in the definition of T (c) is finite. The proof of our results relies on the characterization of the shape of the time-map. Namely, the number of radial solutions of (3) on BR coincides with the number of solutions c > 0 of the equation T (c) = R. The latter is determined by the following three properties: • the domain of the time-map; • the limit of the time-map at the boundary points of its domain; • the monotonicity of the time-map on the maximal subintervals of its domain. This characterization uses the following variational equations w.r.t. parameters. Differentiating (4) with respect to c and introducing the notations h(r, c) = jc u(r, c)
and z(r, c) = j2c u(r, c),
we get (r n−1 h ) + r n−1 f (u)h = 0, h(0, c) = 1,
h (0, c) = 0
(7)
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(r n−1 z ) + r n−1 (f (u)z + f (u)h2 ) = 0, z(0, c) = 0,
z (0, c) = 0.
(8)
Differentiating (4) with respect to r and introducing the notation v(r, c) = jr u(r, c) we get n−1 n−1 n−1 v = 0, f (u) − 2 v ) +r (r r v(0, c) = 0,
v (0, c) = −
f (c) . n
(9)
3. Multiple solutions In this section Theorem 1 is proved. The theorem follows in an elementary way from the following properties of the time-map: • D(T ) = (0, ∞), • limc→0 T (c) = limc→∞ T (c) = 0, • T is continuous. These properties are proved in the following subsections. 3.1. The domain of the time-map Lemma 1. Let f satisfy condition (H1). Then D(T ) = (0, ∞) and limc→0 T (c) = 0. Proof. Dividing (5) by −r n−1 , using (H1) and integrating on [0, r] one obtains u(r, c)c −
m 2 r . 2n
(10)
√ This implies D(T ) = (0, ∞) and T (c) 2nc/m from which the second statement of the Lemma follows. 3.2. The limit of the time-map at infinity We will need the following propositions. Proposition 1. Let us assume that lim
u→∞
f (u) =∞ u
(11)
and that the solution of (4) is decreasing for any c > 0. Then for all > 0 there exists c > 0 such that for the solutions of (4) u(, c)c
for all c c .
(12)
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J. Hernández et al. / Nonlinear Analysis 65 (2006) 265 – 283
Proof. Let > 0 be an arbitrary positive number. Then there exists K > 0 such that for the solution of the linear initial value problem (r n−1 w ) + r n−1 K w = 0, w(0) > 0, w (0) = 0
(13)
we have w() = 0 and w(r) > 0 for r ∈ [0, ), i.e. the first root of w is . According to (11) there exists c > 0 such that f (u) > K u
for all u > c .
(14)
We will argue by contradiction. Let us assume that there exists c c , for which u(, c) > c . For short, let us denote u(r, c) by u(r). Since u is decreasing, therefore u(r) > c for all r ∈ [0, ]. Let us denote by w the solution of (13) satisfying the initial condition w(0) = c. It is easy to see from (13), (4) and (14) that w (0) = −
f (c) K c >− = u (0), n n
that is for small r > 0 we have w(r) > u(r). However, w() = 0 and u() > c imply that there exists r1 ∈ (0, ) such that c < u(r1 ) = w(r1 ),
u (r1 ) > w (r1 ),
u(r) < w(r) for all r ∈ (0, r1 ).
(15)
Let us introduce the function A(r) = r n−1 (u (r)w(r) − u(r)w (r)).
(16)
Then using (4), (13) and (14) A (r) = r n−1 w(r)(K u(r) − f (u(r))) < 0
for all r ∈ (0, ).
(17)
From (15) and (16) we get A(r1 ) > 0, further, from A(0)=0 and (17) there follows A(r1 ) < 0, which is a contradiction. Proposition 2. Let n2 and f satisfy (H1)–(H3). Then 1. there exist u1 > 0 and K > 0 such that 2nF (u) − (n − 2)uf (u)Kup+1
for all u > u1 ,
2. there exists a > 0 such that 2nF (u) − (n − 2)uf (u) − a
for all u 0.
Proof. According to (H2) there exist L > 0 and p ∈ (1, 2∗ ) such that L = lim
u→∞
f (u) . up
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Let
n + 2 − (n − 2)p ∈ 0, L min 1, 5n − 2 + (n − 2)p
271
.
(18)
Then there exists u0 > 0 such that f (u) ∈ (L − , L + ) for all u > u0 . up Let u1 =
1/(p+1)
L −1
u0
Then, using the formulas above, we get for u > u1 u u up+1 . F (u) f (t) dt (L − ) t p dt (L − 2) p+1 u0 u0 Hence for u > u1 2n (L − 2)up+1 − (n − 2)(L + )up+1 p+1 2n 4n = − (n − 2) L − + n − 2 up+1 , p+1 p+1
2nF (u) − (n − 2)uf (u)
from which, using (18), statement 1 follows. The second statement follows from statement 1 and the continuity of 2nF (u) − (n − 2)uf (u) on [0, ∞) (which is a consequence of (H3)). Proposition 3. Let us assume that f satisfies (H2). For a given c > 0 let us define rc by the equation c u(rc , c) = . 2
(19)
Then there exist M > 0 and u2 > 0 such that n 1−p c rc2 M
for all c > 2u2 .
Proof. Assumption (H2) implies that there exist M > 0 and u2 > 0 such that f (u)Mup
for all u u2 .
From (5) and (6) we get r rn s n−1 Mup (s) ds Mcp −r n−1 u (r) n 0
(20)
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if u(r)u2 . Dividing this inequality by r n−1 and integrating on [0, t] one obtains c − u(t)
M p 2 c t 2n
(21)
if u(t)u2 . If c > 2u2 , then applying (21) for t = rc we get the desired inequality (20).
Lemma 2. Let f satisfy conditions (H1)–(H3). Then limc→∞ T (c) = 0. Proof. For n = 1 the lemma was proved in [15] (in a little more general case). Now we assume n 2. Step 1: Let > 0 be an arbitrary positive number. Let us choose c according to Proposition 1. We will prove that there exists d > c such that |u (r, c)| >
c
for all r ∈ (, 2), c > d .
(22)
This inequality implies T (c) < 2 for all c > d , since u(, c) c and u(r, c) decreases at least by c on the interval (, 2). Step 2: To prove (22) we will apply the following general form of Pohozhaev identity [7]. r 2 s n−1 [2nF (u(s)) − (n − 2)u(s)f (u(s))] ds r n u (r) + 2r n F (u(r)) = 0
− (n − 2)r n−1 u(r)u (r). According to (6) 2
u (r) > r
−n
r
s n−1 [2nF (u(s)) − (n − 2)u(s)f (u(s))] ds − 2F (u(r)).
(23)
0
Let us choose K and u1 according to Proposition 2, and M and u2 according to Proposition 3. Let c = max{2c , 2u1 , 2u2 }. Let c > c and introduce rc according to (19). For short, we use the notation u(r) instead of u(r, c). From (23) one obtains rc 2 s n−1 [2nF (u(s)) − (n − 2)u(s)f (u(s))] ds u (r) > r −n 0
+ r −n
r
s n−1 [2nF (u(s)) − (n − 2)u(s)f (u(s))] ds − 2F (u(r)).
(24)
rc
Step 3: Now, we will estimate each term in the r.h.s. of (24). For the first term s < rc implies u(s) > u(rc ) =
c > u1 . 2
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Hence from Proposition 2 2nF (u(s)) − (n − 2)u(s)f (u(s)) Kup+1 (s) > K
c p+1 2
for all s < rc ,
yielding for r ∈ (, 2) rc c p+1 r n c r −n . s n−1 [2nF (u(s)) − (n − 2)u(s)f (u(s))] ds (2)−n K 2 n 0 Using c > c 2u2 Proposition 3 implies rc r −n s n−1 [2nF (u(s)) − (n − 2)u(s)f (u(s))] ds 0 c p+1 K n n/2 c(1−p)n/2 . (2)−n 2 n M For the second term in (24) we use statement 2 of Proposition 2: r a a rn −n r =− . s n−1 [2nF (u(s)) − (n − 2)u(s)f (u(s))] ds − n r n n rc
(25)
(26)
For the third term r ∈ (, 2) and c > c imply u(r) c , hence (using F = f > 0) F (u(r))F (c ).
(27)
Step 4: Now let us substitute the estimates (25)–(27) into (24). Then for all r ∈ (, 2) and c > c K n n/2 p+1+(1−p)n/2 a 2 u (r, c) > (2)−n p+1 c − − 2F (c ). (28) n2 M n Since p < 2∗ , therefore p + 1 + (1 − p)n/2 > 0, hence the r.h.s. of (28) tends to infinity as c → ∞, which proves the desired inequality (22). 3.3. The continuity of the time-map For any > 0 let us introduce the function T as follows: T (c) = min{r ∈ (0, T (c)) : u(r, c) = }.
(29)
Obviously, D(T ) = (, ∞) and T is continuous. Lemma 3. Let f satisfy (H1). Then T tends uniformly to T on any compact subinterval of (0, ∞) as → 0. Proof. Let [a, b] ⊂ (0, ∞) be a compact subinterval. Let 0 < a, then for all 0 D(T ) ⊃ [a, b]. Let R0 = min T0 , [a,b]
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then (since < 0 implies T > T0 ) T (c)R0
for all c ∈ [a, b], ∈ (0, 0 ].
(30)
Let c ∈ [a, b], ∈ (0, 0 ] and r ∈ (T (c), T (c)], then using (H1) and (5) −u (r, c)
m m m r T (c) R0 , n n n
integrating this inequality on [T (c), T (c)]
m R0 (T (c) − T (c)), n
thus |T (c) − T (c)|
n , mR 0
from which the statement follows.
Corollary 2. Let f satisfy (H1). Then T is continuous. Remark 1. In fact, the condition (H1) for Lemma 3 (and hence that of Corollary 2) is too restrictive. For example, the condition lim inf u→0 f (u) > 0 is also sufficient. 4. Exact multiplicity in one dimension In the sequel we study the exact number of radial positive solutions. When n = 1, it can be seen similarly to the non-singular case that all positive solutions are radial, that is symmetric to 0 in (−R, R). The crucial part of Theorem 2 is the following property (H5) concerning the function h = jc u (see (7)). It often arises in several investigations with different nonlinearities, e.g. [7,18,19,23]. Such problems are sometimes called disconjugate [19] and the corresponding values of c admissible [18]. (H5) For all c > 0, the function h(·, c) has at most one root in [0, T (c)]. Remark 2. Condition (H5) is equivalent to the following [23]: for any radial solution u of (1) the Morse index of u is at most 1, and if it equals 1 then u is non-degenerate. The multiplicity result of Theorem 2 in fact exploits (H5) instead of the restriction n = 1. Hence in this section we will verify Theorem 3. Let f satisfy conditions (H1)–(H5). Then there exists R ∗ > 0 such that 1. (1) has two radial solutions for R < R ∗ ; 2. (1) has one radial solution for R = R ∗ ; 3. (1) has no radial solution for R > R ∗ .
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275
Theorem 3 will be proved in a sequence of lemmas in the subsections below: • First we study the approximate time-maps T in Section 4.1, and prove that T has a unique maximum at some c0 , it is strictly increasing before c0 and strictly decreasing after c0 . (Together with the limits of T at 0 and ∞, this already proves Theorem 3 when limu→0 f (u) is finite, hence the sequel is devoted to the singular case.) • The main ingredient of the proof is provided by Section 4.2, stating that the time-map cannot be constant on any subinterval of (0, ∞). • Passing to the limit → 0 and using the obtained non-degeneracy of the time-map, in Section 4.3 we derive easily that T inherits the shape of T . Theorem 2 can be verified from Theorem 3 by checking condition (H5) in the following way. Proof of Theorem 2. In the case n = 1 the functions h and v in (7) and (9) satisfy the same linear equation, further, we have v < 0 in (0, T (c)] from (6). Hence Sturm comparison shows that h may have at most one root in [0, T (c)], i.e. condition (H5) is satisfied and thus the assertions of Theorem 3 hold. 4.1. The shape of the approximate time-maps For all > 0 the approximate time-maps T have been introduced in (29). The function T is determined by the implicit equation u(T (c), c) = .
(31)
Differentiating (31), one gets the following equations for the derivatives of T , using the notations h = jc u, z = j2c u, v = u = jr u (introduced in Section 2): v(T (c), c) T (c) + h(T (c), c) ≡ 0,
(32)
v (T (c), c) T (c)2 + 2h (T (c), c) T (c)
+ v(T (c), c) T (c) + z(T (c), c) ≡ 0.
Expressing T (c) from (32) and substituting in (33), T (c) is given as 1 v h2 h h T (c) = − +z , −2 v v2 v
(33)
(34)
where all functions on the r.h.s. have the argument (T (c), c). Lemma 4. Let conditions (H1)–(H5) hold, > 0. Then 1. limc→ T (c) = limc→∞ T (c) = 0. 2. Let c > 0. If T (c) = 0 then T (c) < 0. √ Proof. 1. From (10) follows T (c) 2n(c − )/m, implying lim T = 0. From Lemma 2 and T < T follows lim∞ T = 0.
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2. Eqs. (32) and (33) imply h(T (c), c) = 0 and T (c) = −
z(T (c), c) . v(T (c), c)
Since v = u < 0, therefore it is enough to prove that h(r ∗ ) = 0 implies z(r ∗ ) < 0. Let us introduce the function A(r) = r n−1 (h(r)z (r) − h (r)z(r)).
(35)
Then A(0) = 0 and using (7)–(8), (H4) and (H5) A (r) = −r n−1 f (u(r))h3 (r) < 0
for all r ∈ (0, r ∗ ),
(36)
since h > 0 in (0, r ∗ ) by (H5). Hence A(r ∗ ) < 0, which yields z(r ∗ ) < 0, because h(r ∗ ) = 0 and h (r ∗ ) < 0. Corollary 3. This characterization of the approximate time-maps implies that T has a unique maximum at some c0 > , it is strictly increasing before c0 and strictly decreasing after c0 . Hence, together with limc→0 T (c) = limc→∞ T (c) = 0, the multiplicity result of Theorem 3 holds with the boundary condition u|jBR = in (1). If limu→0 f (u) < ∞ (i.e. there is no singularity), then the differentiability properties of T remain valid for T (essentially for the case = 0). Hence Lemma 4 can be repeated with T, which means that Theorem 3 is proved for this case: Corollary 4. If limu→0 f (u) < ∞, then Theorem 3 holds. 4.2. The non-degeneracy of the time-map Since the non-singular case (limu→0 f (u) < ∞) has been handled in Corollary 4, in this subsection we restrict ourselves to the singular case, i.e. we assume (H6) lim f (u) = ∞. u→0
The aim of this subsection is to prove that the time-map cannot be constant on any nondegenerate subinterval of (0, ∞), which will be stated in Lemma 5. The difficulty in proving this plausible property lies in the singularity (H6). Hence Lemma 5 is preceded by careful estimates, formulated in propositions using the indirect degeneracy assumption (D) T (c) ≡ R
(c ∈ J )
for some non-degenerate compact interval J ⊂ (0, ∞) and number R > 0.
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277
Proposition 4. Let conditions (H1)–(H4) and (H6) hold. If T has property (D), then limR h = 0 uniformly, i.e. for any > 0 there exists r > 0 such that |h(r, c)| (r ∈ [r , R), c ∈ J ). Proof. Step 1: There exists r0 > 0 such that f (u(r, c)) < 0
(r ∈ (r0 , R), c ∈ J ).
Namely, from (H4) and (H6) there exists > 0 such that f (s) < 0
(s ∈ (0, ]).
(37)
For any c ∈ J let rc > 0 such that u(r, c) < (r > rc ). The continuous dependence of u on c implies that there exists r0 := max{rc : c ∈ J } < R. Step 2: For any r ∈ [r0 , R) and c ∈ J the following holds: if h(r, c) > 0 then h (r, c) < 0, and (conversely) if h(r, c) < 0 then h (r, c) > 0. To see this, we use the integral of (7): s n−1 f (u())h(, c) d s n−1 h (s, c) − s0n−1 h (s0 , c) = − s0
(s0 < s < R, c ∈ J ).
(38)
Namely, let first h(r, c) > 0 and assume indirectly that h (r, c)0. Setting s0 = r, for s a bit larger than r the right side of (38) is positive, hence there exists r˜ > r such that h(˜r , c) > 0 and h (˜r , c) > 0. Using the continuous dependence of h on c, there exists a > 0 and J˜ ⊂ J such that h(˜r , c) ˜ a and h (˜r , c) ˜ > 0 for all c˜ ∈ J˜. Now setting s0 = r˜ , there n−1 n−1 holds s h (s, c) ˜ > (˜r ) h (˜r , c) ˜ > 0 for all c˜ ∈ J˜ and s > r˜ as long as h(., c) ˜ > 0. This implies that h increases after r˜ , hence h(s, c)a ˜
(s ∈ [˜r , R], c˜ ∈ J˜).
(39)
However, this yields u(s, c2 ) − u(s, c1 ) =
c2 c1
jc u(s, c) dc =
c2
h(s, c) dc a(c2 − c1 ) > 0
(40)
c1
for all s ∈ [˜r , R] and c1 , c2 ∈ J˜. Letting s → R with fixed c1 , c2 , the left side tends to 0 and the right side is fixed, a contradiction. The counterpart with h(r, c) < 0 is proved similarly. Step 3: For any c ∈ J , h(., c) has constant sign on [r0 , R). Namely, assume that h(r1 , c)=0 for some r1 ∈ [r0 , R). Then h (r1 , c) = 0 from uniqueness. If e.g. h (r1 , c) > 0, then h(r2 , c) > 0 and h (r2 , c) > 0 for some r2 > r1 in contradiction to Step 2. In the same way h (r1 , c) < 0 is impossible. Step 4: There exists B1 > 0 such that |h (r, c)| B1
(r ∈ [r0 , R), c ∈ J ).
(41)
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Namely, let first h(r0 , c) > 0 for some c ∈ J . Then h(r, c) > 0 and hence h (r, c) < 0 for all r ∈ [r0 , R) (from Steps 3 and 2). Here (7) yields that (r n−1 h ) > 0 on [r0 , R), hence r n−1 h (r, c) > r0n−1 h (r0 , c) for any r ∈ [r0 , R). Since r > r0 , we obtain |h (r, c)| = −h (r, c) < − h (r0 , c) = |h (r0 , c)|
(r ∈ [r0 , R)).
(42)
The same result follows for the case h(r0 , c) < 0 as well. Hence for all c ∈ J , |h (r, c)| |h (r0 , c)| (r ∈ [r0 , R)). Introducing B1 := max{h (r0 , c) : c ∈ J }, we obtain the required estimate. Step 5: For all c ∈ J , lim h(r, c) = 0.
(43)
r→R
Namely, Steps 3 and 2 imply that h(., c) is (strictly) monotone on [r0 , R), hence the limit exists. Assume indirectly that e.g. limr→R h(r, c0 ) =: a > 0 for some c0 > 0. In this case, h(., c0 ) decreases on [r0 , R), hence h(r, c0 ) > a for r ∈ [r0 , R). Let r1 ∈ [r0 , R) such that R − r1 a/2B1 (with B1 from Step 4). Then, from the continuous dependence of h on c, h(r1 , c) > a (c ∈ J ) for some J ⊂ J . Hence for any r ∈ [r1 , R) and c ∈ J , r a a h (s, c) ds > a − B1 (R − r1 ) a − B1 · = > 0. h(r, c) = h(r1 , c) + 2B1 2 r1 This is impossible (see (39) and afterwards). Step 6: Let > 0 be given. Introducing r = R − (/B1 ), we obtain from Steps 5 and 4 that R |h (s, c)| ds B1 (R − r)B1 (R − r ) = |h(r, c)| r
(r ∈ [r , R), c ∈ J ).
The fact limR h = 0 itself and condition (H5) imply. Corollary 5. Let conditions (H1)–(H6) hold. If T has property (D), then for all c ∈ J h(r, c) > 0
(r ∈ [0, R)).
Proposition 5. Let conditions (H1)–(H6) hold. If T has property (D), then there exist m1 > 0 and r1 ∈ (0, R) such that z(r, c) − m1
(r ∈ [r1 , R), c ∈ J ).
Proof. Step 1: Let A(r) = A(r, c) ≡ r n−1 (z (r)h(r) − h (r)z(r))
(r ∈ (0, R), c ∈ J )
as in (35). We have seen that A(r) is negative and decreasing, see (36) and afterwards. Step 2: There holds z(r, c) < 0 for all r ∈ (0, R), c ∈ J . Namely, let first f (c) > 0. Then z(0) = 0, and (r n−1 z ) (r) < 0 for small r from (8). Hence z(r) < 0 for small r. If z had a
J. Hernández et al. / Nonlinear Analysis 65 (2006) 265 – 283
279
root in (0, R), then its first root r1 would satisfy z (r1 ) 0, hence A(r1 ) 0 in contradiction to Step 1. Now, let f (c) = 0. Since z(0) = 0, therefore z(r, c) must have constant sign for r ∈ (0, ) with some > 0. Since the values of c with f (c) > 0 are dense in J owing to (H4), it follows by continuity from the previous case that this initial sign of z can only be negative. Then z(r) < 0 extends to r ∈ (0, R) similarly as above using the function A. Step 3: Let r0 ∈ (0, R) be as in Steps 1–2 of Proposition 4 (i.e. h(., c) decreases on [r0 , R) for all c ∈ J ). Then there exists > 0 such that z (r)h(r) − h (r)z(r) −
(r ∈ [r0 , R), c ∈ J ).
(44)
Namely, by Step 1, A(r) is negative and decreasing, hence R n−1 (z (r)h(r) − h (r)z(r)) < A(r)A(r0 ) < 0. Therefore = −R −(n−1) min{A(r0 , c) : c ∈ J } is suitable. Step 4: Let r 1 G(r, c) = h(r, c) ds. 2 (s, c) h r0 Then there holds z(r, c) − G(r, c)
(r ∈ (r0 , R), c ∈ J ).
(45)
Namely, (44) implies ( hz ) − h12 on [r0 , R), hence z(r, c) z(r0 , c) − h(r, c) h(r0 , c)
r
r0
1 h2 (s, c)
ds.
Multiplying this inequality by h(r, c) and using h > 0 and z < 0, we obtain (45). Step 5: For all c ∈ J , G(., c) is convex on [r0 , R). Namely, r r 1 1 1 ds + , G (r) = h (r) ds 0 G (r) = h (r) 2 (s) 2 (s) h h(r) h r0 r0
(46)
since h(., c) is convex on [r0 , R) from (42). Step 6: For any r1 ∈ (r0 , R) there exists a1 > 0 such that G(r, c)a1
(r ∈ [r1 , R), c ∈ J ).
(47)
Namely, (46) yields G (r0 , c) = 1/ h(r0 , c)d, where d = min{1/ h(r0 , c) : c ∈ J } > 0 from the continuous dependence of h on c. The convexity of G yields G(r, c)G (r0 , c)(r − r0 ) d(r − r0 ) d(r1 − r0 ) =: a1 . Step 7: Fixing arbitrary r1 ∈ (r0 , R), the proposition follows from (45) and (47).
Proposition 6. Let conditions (H1)–(H6) hold. If T has property (D), then there exist M > 0 and 0 > 0 such that T (c) − M
(c ∈ J, ∈ (0, 0 ]).
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Proof. Step 1: T (c) can be expressed by rearranging (34) as h v − v h h 1 T (c) = h − z , + v v2 v where all functions on the r.h.s. have the argument (T (c), c). In virtue of Proposition 5, now our aim is to prove that v is bounded and the term containing h tends to 0, both uniformly for c ∈ J . Step 2: There exist r2 ∈ (0, R) and 2 1 > 0 such that (r ∈ [r2 , R), c ∈ J ).
− 2 v(r, c) − 1
Namely, let r˜ ∈ (0, R) be arbitrary. Then r n−1 v = r n−1 u is decreasing owing to (4), hence for any r ∈ [˜r , R) v(r, c) (˜r /r)n−1 v(˜r , c)(˜r /R)n−1 max{v(˜r , c) : c ∈ J } =: − 1 ,
(48)
where 1 ∈ (0, ∞) exists by continuous dependence. On the other hand, integrating (48) on [r, R], we obtain u(r, c) 1 (R − r)
(r ∈ [˜r , R)).
Let r2 = max{˜r , r0 }, where r0 is from Step 1 of Proposition 4. Then u(r, c) <
(r ∈ [r0 , R), c ∈ J ),
where f is decreasing on [0, ], hence f (u(r, c))f ( 1 (R − r))
(r ∈ [r2 , R)).
Integrating (4), we obtain n−1
r
v(r, c)
r2n−1 v(r2 , c) −
r
s n−1 f (u(s, c)) ds r n−1 R v(r2 , c) − f (u(s, c)) ds . r2
r2
Here
r
r2
f (u(s, c)) ds
R
f ( 1 (R − s)) ds =
r2
1
1
1 (R−r2 )
f (t) dt =: ,
0
where < ∞ owing to assumption (H3) and is independent of c. Hence n−1 R · (min{v(r2 , c) : c ∈ J } − ) =: − 2 (r ∈ [r2 , R)). v(r, c) r2 Step 3: There exists B2 > 0 such that |h (r, c)v(r, c) − v (r, c)h(r, c)|B2
(r ∈ [r2 , R), c ∈ J ),
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281
where r2 ∈ (0, R) is from Step 2. Namely, let A(r) = A(r, c) ≡ r n−1 (h (r)v(r) − v (r)h(r))
(r ∈ (0, R), c ∈ J ).
(49)
Then A (r)=(r n−1 h ) v −(r n−1 v ) h=−(n−1)r n−1 hv from Eqs. (7) and (9). Here r2 r0 by its definition, hence (41) and (43) imply |h(r, c)|(R − r2 )B1
(r ∈ [r2 , R), c ∈ J ).
Together with Step 2, we obtain for all r ∈ [r2 , R), c ∈ J |A (r, c)| (n − 1)R n−1 (R − r2 )B1 2 , hence |h (r)v(r) − v (r)h(r)| =
1
r
|A(r, c)| n−1 1
r2n−1
1 r2n−1
|A(r2 , c)| +
R
|A (r, c)|
r2
(max{|A(r2 , c)| : c ∈ J }
+(n − 1)R n−1 (R − r2 )2 B1 2
=: B2 .
Step 4: Using Steps 1–3, further, from (41) and Propositions 4 and 5, we obtain the estimate
1 B2 B1 T (c) |h(T (c), c)| − m1 (50) +
2
1
21 whenever T (c) ∈ [r2 , R) and c ∈ J . From Proposition 4 there exists r3 ∈ [r2 , R) such that m1 B2 B1 |h(T (c), c)| + 2
1 2
1 whenever T (c) ∈ [r3 , R) and c ∈ J . Since by Lemma 3 and property (D), T → R
uniformly on J ,
(51)
therefore there exists 0 > 0 such that T (c) ∈ [r3 , R) for all ∈ (0, 0 ], c ∈ J . For such the expression in large brackets in (50) is at most −m1 /2, hence, letting M := m1 /2 2 , the required estimate follows. We note that the uniform estimates of Propositions 4–5 were required to ensure that M in Proposition 6 is uniform in c and . Now we are in the position to exclude the indirect assumption (D). Lemma 5. Let conditions (H1)–(H6) hold. Then the time-map cannot be constant on any non-degenerate subinterval of (0, ∞).
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J. Hernández et al. / Nonlinear Analysis 65 (2006) 265 – 283
Proof. Assume for contradiction that property (D) holds. Using Taylor’s formula, Proposition 6 implies that there exist M > 0 and 0 > 0 such that T (c1 ) − T (c2 ) T (c2 )(c1 − c2 ) −
M (c1 − c2 )2 2
(c1 , c2 ∈ J, ∈ (0, 0 ]).
Let → 0. Then (51) implies that the left side tends to 0. Further, by (32) T (c2 ) = −
h(T (c2 ), c2 ) → 0, v(T (c2 ), c2 )
since T (c2 ) → R, and hence (43) yields that h(T (c2 ), c2 ) → 0 and (48) implies that v is bounded away from 0. From these we obtain the contradiction 0 −
M (c1 − c2 )2 2
(c1 , c2 ∈ J ).
4.3. Proof of Theorem 3 The results obtained in Sections 4.1–4.2 enable us to prove Theorem 3 immediately: namely, we can verify that T inherits the shape of the T . In virtue of Corollary 4, it suffices to do this in the singular case limu→0 f (u) = ∞. The three properties of T, formulated at the very beginning of Section 3, imply that T has a maximum at some c0 > 0. First we prove that T increases on [0, c0 ] and decreases on [c0 , ∞). Namely, let first 0 c1 < c2 c0 and assume for contradiction that T (c1 ) > T (c2 ). Since T → T uniformly on [c1 , c2 ] by Lemma 3, hence for some 0 > 0 we have T (c1 ) > T (c2 ) for all ∈ (0, 0 ]. In virtue of Corollary 3, T is decreasing after c1 , hence T (c2 ) T (c0 ). Letting → 0, we obtain T (c2 ) T (c0 ), which gives T (c1 ) > T (c0 ) in contradiction to the maximum assumption on T (c0 ). The other part on [c0 , ∞] is proved similarly. Using Lemma 5, we obtain that T is strictly increasing on [0, c0 ] and strictly decreasing on [c0 , ∞]. Let R ∗ = T (c0 ). Together with limc→0 T (c) = limc→∞ T (c) = 0 and the continuity of T, we conclude that T assumes each value R twice if 0 < R < R ∗ , once if R = R ∗ and does not assume R if R > R ∗ .
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