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Multiplicity of positive solutions for p-Laplace equation with superlinear-type nonlinearity
Nonlinear Analysis 56 (2004) 867 – 878
www.elsevier.com/locate/na
Multiplicity of positive solutions for p-Laplace equation with superlinear-type nonlinearity S. Prashanth∗ , K. Sreenadh TIFR Centre, Post Box No. 1234, IISc Campus, Bangalore 560012, India Received 5 December 2002; accepted 23 October 2003
Abstract In this work, we consider the problem ∗ −2p u = up + uq in BR (0); (PR ) u ¿ 0 u=0 on @BR (0); where p ¿ 2; p∗ = Np=(N − p) and BR (0) is the open ball of radius R centered about the origin in RN , N ¿ 3. Under the condition that p − 1 ¡ q ¡ p2 =(N − p) − 1, we show that there exists R∗ ¿ 0 such that (PR ) admits atleast two solutions for all R ∈ (R∗ ; ∞) and no solution if R ¡ R∗ . This extends the results of (Nonlinear Anal. TMA 10(8) (1986) 755) to the case of p-Laplace equation. ? 2003 Elsevier Ltd. All rights reserved. MSC: 35J65; 34B15; 35J60 Keywords: Multiplicity; Critical exponent; p-Laplace equation; Emden–Fowler equation
1. Introduction Let BR (0) ⊂ RN denote the open ball of radius R centered about the origin, N ¿ 3. Let p ∈ (1; ∞) and de@ne p∗ = Np=(N − p) if p ¡ N and p∗ = ∞ if p = N . In this work we are interested in studying the existence of multiple solutions to the following
∗
Corresponding author. E-mail addresses: [email protected] (S. Prashanth), [email protected] (K. Sreenadh).
0362-546X/$ - see front matter ? 2003 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2003.10.026
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quasilinear equation: −2p u = u + uq u ¿0 (PR ) u=0
in BR (0); on @BR (0);
where 2p u=div(|∇u|p−2 ∇u) is the usual p-Laplace operator, 0 ¡ q ¡ 6 p∗ −1. We brieGy recall the many multiplicity results known for problems of the type (PR ). A more easily understood situation is when we study (PR ) with 0 ¡ q ¡ p − 1 ¡ 6 p∗ − 1. This is the so called “sublinear” situation. In this case, due to the “sublinear” nature of the nonlinearity u +uq as u → 0, we can use the sub–super solution approach to obtain one solution with small L∞ norm to (PR ) and the other solution can be obtained using a variety of methods like degree theory [2], ODE shooting methods [9], or variational methods [3,5,6]. We summarize all these “sublinear” multiplicity results as below: Theorem 1. Let 0 ¡ q ¡ p − 1 ¡ 6 p∗ − 1, if p ¡ N and 0 ¡ q ¡ p − 1 ¡ ¡ ∞ if p = N . Then, there exists R∗ = R∗ (q; p; ; N ) ¿ 0 such that problem (PR ) admits atleast two solutions for all 0 ¡ R ¡ R∗ and no solution for R ¿ R∗ . Therefore it is surprising to @nd that even in the case q ¿ p − 1, when there is no “sublinear” eMect to take advantage of, [4] have shown that when p = 2; = 2∗ − 1 and N = 3 we may obtain multiple solutions to (PR ) for all large R. More precisely, they showed the following result. Theorem 2. Let N =3; p=2; =2∗ −1 and 1 ¡ q ¡ 3. Then there exists R∗ ¿ 0 such that problem (PR ) admits atleast two solutions for all R ∈ (R∗ ; ∞) and no solution for all R ∈ (0; R∗ ). We extend the above result to the case of the p-Laplace equation using the same methods to prove the following: Theorem 3. Let 2 ¡ p ¡ N; p − 1 ¡ q ¡ p2 =(N − p) − 1, = p∗ − 1. Then there exists R∗ = R∗ (p; q; N ) ¿ 0 such that problem (PR ) admits atleast two solutions for all R ∈ (R∗ ; ∞) and no solution for all R ∈ (0; R∗ ). Remark 1.1. The restriction p−1 ¡ p2 =(N −p)−1 implies the dimensional restriction N ¡ 2p. The above result also shows the limitations in the exponent q and the dimension N for the uniqueness results for (PR ) proved in [1,8,10]. In particular, the following uniqueness result can be inferred from [8]: The following two results (Corollaries 4 and 5) follow from Theorem 3. Corollary 4. Let = p∗ − 1 and p∗ − 2 6 q. Then problem (PR ) admits a unique solution for all R ¿ 0.
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Remark 1.2. It is easy to check that the above restriction on q for the uniqueness always holds if 1 ¡ p ¡ 2; q ¿ p − 1 and N ¿ 6. Hence the hypothesis that p ¿ 2 in Theorem 3 is restrictive only when N = 3; 4; 5. We can also interpret the above multiplicity results in the language of bifurcation theory. For this, we set = Rp( −q)=( −p+1) and u(x) = 1=( −q) w(( −p+1)=p( −q) x). It can be easily checked that u solves problem (PR ) if and only if w solves the following problem: −2p w = w + wq in B1 (0); w ¿0 (P ) w=0 on @B1 (0): We may rewrite the multiplicity results for problem (PR ) given in Theorems 1–3 in terms of problem (P ) as below Corollary 5. (i) The sublinear case: Let 0 ¡ q ¡ p − 1 ¡ 6 p∗ − 1 if p ¡ N and 0 ¡ q ¡ p − 1 ¡ ¡ ∞ if p = N . Then, there exists ∗ = ∗ (q; p; ; N ) ¿ 0 such that problem (P ) admits atleast two solutions for all ∈ (0; ∗ ) and no solution for ¿ ∗ . (ii) Superlinear case: Let 2 6 p ¡ N; = p∗ − 1 and p − 1 ¡ q ¡ p2 =(N − p) − 1. Then there exists ∗ = ∗ (p; q; N ) ¿ 0 such that for all ¿ ∗ the problem (P ) has atleast two solutions and no solutions if ¡ ∗ . Therefore, the bifurcation diagrams in these two cases look like
|| u ||oo
|| u ||oo . . . .
. . . . . *
*
2. Preliminary lemmas By Gidas–Ni–Nirenberg type symmetry results (shown in [7] for example), we have that any solution of (PR ) is radial and radially decreasing about the origin. Since any solution of (PR ) is radial, it solves the following ODE (which we again denote by (PR )): −(r N −1 |u |p−2 u ) = r N −1 (u + uq ) in (0; R); (PR ) u ¿0 u (0) = 0; u(R) = 0:
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We prove Theorem 3 by using asymptotic analysis to analyse the ODE associated to problem (PR ). This was the approach used by Atkinson–Peletier in their elegant work [4] to prove Theorem 2. We will extend their results to the case of p-Laplace equation to prove Theorem 3. Because of the quasilinear nature of the problem, obtaining a lower bound for y in terms of z (see (2.3) and (2.4) below) is more delicate than in the semilinear case treated by Azorero and Peral Alonso [5]. We obtain such lower bound in Lemma 2.5. Similar delicate estimates are required also in Lemma 2.7. For simplicity of notations, we de@ne the following quantities: N −1 k −1 N −p ; k =p ; l= : (2.1) = p−1 N −p p−1 In what follows we @x = p∗ − 1; p − 1 ¡ q ¡ . Let u be a solution of (PR ). We now make the following transformation known as the Emden–Fowler transformation (below, r denotes the radial variable): t= ; y(t) = u(r): (2.2) r Then, it is easy to check that y solves the following Emden–Fowler equation: −(|y |p−2 y ) = t −k (y + yq ) in t ∈ (T; ∞); y ¿0 (PT ) y(T ) = 0; y (∞) = 0; T= : R To study the multiplicity of solutions to (PR ) it is natural to consider the following version of (PT ) which is parametrised by a parameter ¿ 0: −(|y |p−2 y ) = t −k (y + yq ); (P ) lim y(t) = ; y (∞) = 0: t→∞
The above “initial value” problem with the initial data prescribed at in@nity admits a unique solution (see the appendix). Let y(t; ) denote any solution of (P ). De@ne T () = inf {t ¿ 0; y(:; ) ¿ 0 in (t; ∞)}. Then it is easy to show that T () is the @rst zero of y(t; ) as t is decreased from in@nity. From continuous dependence on initial values (see Appendix) we note that T () is a continuous function on (0; ∞). It is also easy to see by integration that y(:; ) is a strictly increasing concave function in (T (); ∞). ∗ Let f(s) = sp −1 + sq , and de@ne for ¿ 0,
1=(p−1) 1=(1−l) f() l−1 −1 z(t; ) = t t + ; t ¿ 0: (2.3) k −1 It is easy to check that z solves the equation ∗
− (|z |p−2 z ) = t −k 1−p f()z p
∗
−1
;
lim z(t; ) = :
t→∞
We now prove several lemmas required for the proof of Theorem 3.
(2.4)
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Lemma 2.1. Let y solve (P ); ¿ 0. Then y(t; ) ¡ z(t; ) for all t ∈ (T (); ∞). Proof. See [4] for the case p = 2 and [9] for general p. Lemma 2.2. Let y solve (P ); ¿ 0. Then, the 7rst zero T () of y is positive. Proof. See again [5,9]. Lemma 2.3. Let y solve (P ). Then, T () = O((q−1+p)=(k−p) )
as → 0:
Proof. Integrating the ODE in (P ) twice, we obtain, y(t) = −
∞
t
s
∞
1=(p−1)
−k f(y( )) d
ds:
(2.5)
We estimate the integral on the right of Eq. (2.5) as below: t
∞
s
∞
−k
1=(p−1) f(y( )) d
6 [f()] =
1=(p−1)
∞
t
∞
s
(p − 1)(f())1=(p−1) (k − 1)1=(p−1) (p − k)
−k
1=(p−1) d ds
t (p−k)=(p−1)
= K(k; p)(f())1=(p−1) t (p−k)=(p−1) :
(2.6)
Hence, using (2.6) in (2.5) we obtain y(t) ¿ − K(k; p)(f())1=(p−1) t (p−k)=(p−1) ;
t ¿ T ():
(2.7)
Hence, letting t = T () in (2.7) we obtain the estimate 0 ¿ − K(k; p)(f())1=(p−1) (T ())(p−k)=(p−1) :
(2.8)
Noting that f()1=(p−1) ∼ q=(p−1) as → 0, we obtain from (2.8) that T () = O((q−p+1)=(k−p) ) as → 0. This proves the lemma.
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Lemma 2.4. Let ! = p=(p − 1). We have the following estimates for the solution y of (P ) and the function z given in (2.3): (i) There exists k1 ¿ 0 such that y(t; ) ¡ k1 1−! t for t ¿ T (). (ii) For every a ¿ 0, we have z(a! ; ) ¿ for all ¿ 0. (iii) z(t; ) ¿ 1−! t for T () 6 t 6 ! . Proof. (i) By an easy estimate we obtain k1 ¿ 0 such that z(t; ) 6 k1 1−! t for all t ¿ 0. Now (i) follows by using Lemmas 2.1 and 2.2. (ii) Let a ¿ 0. Then, we compute
1=(p−1) 1=(1−l) f() −1 ! !+1 l−1 !(l−1) a + z(a ; ) = a k −1 ¿ a!+1 [al−1 !(l−1) ]1=(1−l) = : This proves (ii). (iii) Since t 6 ! , we obtain by a simple estimate,
1=(p−1) 1=(1−l) f() !(l−1) −1 + z(t; ) ¿ t k −1 T () 6 t 6 ! :
¿ 1−! t; This proves (iii).
Let S() be de@ned by 1=(p−1) f() l−1 −1 : S () = k −1 Lemma 2.5. For each 7xed a ¿ 0 we may 7nd a positive function da such that da () → 0 as → ∞ and the following inequality is true: y(t; ) ¿ (1 − da ()) z(t; )
for t ¿ aS():
Proof. Integrating the ODE in (P ) we obtain, ∞ ∞ −k p∗ −1 p−1 s y (s) ds + s−k yq (s) ds: (y ) (t) = t
t
Since by Lemma 2.1, y(t) ¡ z(t) for t ¿ T (), we get, ∞ ∞ ∗ s−k z p −1 (s) ds + s−k z q (s) ds (y )p−1 (t) 6 t
= I1 (t) + I2 (t):
t
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Now a simple computation gives that, z(aS(); )=%(a) where %(a)=a(1+al−1 )−1=(l−1) . Since t ¿ aS() and z is increasing, we get ∞ p∗ −q−1 I1 (t) ¿ (%(a)) s−k z q (s) ds t
= (%(a))p
∗
−q−1
I2 (t):
Therefore, (y (t))p−1 6 (1 + (%(a))q−p
∗
+1
)
∞
0
s−k z p
∗
−1
(s) ds:
(2.9)
We note that z satis@es the diMerential equation −((z )p−1 (t)) = t −k −p
∗
+1
f()z p
∗
−1
(t):
Integrating the above diMerential equation from t to ∞, we obtain ∞ ∗ ∗ (z )p−1 (t) = (1 + q−p +1 ) s−k z p −1 (s) ds: t
(2.10)
Therefore, letting &(c; ) =
1 ; (1 + (c)q−p∗ +1 )
we obtain from (2.9) and (2.10) that &(%(a); )(y )p−1 − &(1; )(z )p−1 6 0. Therefore, 1=(p−1) &(1; ) y (t) 6 z (t); ∀t ¿ max{aS(); T ()}: &(%(a); ) Integrating the above inequality once more between t and ∞, 1=(p−1) &(1; ) − y(t) 6 ( − z(t)): &(%(a); )
(2.11)
Now we note that [&(1; )=&(%(a); )]1=(p−1) ¡ 1 for a small enough. Let da () = 1 − [&(1; )=&(%(a); )]1=(p−1) . Then (2.11) implies y(t) ¿ (1 − da ()) z(t) + da () ¿ (1 − da ()) z(t). It is now easy to see that da () → 0 as → ∞ for @xed a ¿ 0 which proves the lemma. Lemma 2.6. T () = o(! ) as → ∞. Proof. Clearly, it is suPcient to show that for any a ¿ 0; y(t) ¿ 0 for all t ¿ a! . But this follows from Lemma 2.5 by noting that S() ¿ C! for some C ¿ 0. Lemma 2.7. Let p ¿ 2. Then as → ∞, the following hold: (i) If 0 ¡ q ¡ k − p; T () = O((p−q−1)=(k−q−1) ). 2 (ii) If q = k − p; T () = O((p−1−q)=(p−1) log ). (iii) If k − p ¡ q ¡ p∗ − 1; T () = O(1=(p−1)(!p−!k+q+1) ).
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Proof. (i) q ¡ k − p: Integrating the ODE satis@ed by y from t to ∞ and using Lemma 2.1, we get ∞ ∗ (y )p−1 (t) 6 s−k (z p + z q )(s) ds: t
Now using the fact that z solves the ODE given in (2.4), from the above inequality we obtain p∗ −1 ∞ p−1 p−1 (z ) (t) + s−k z q (s) ds: (y ) (t) 6 f() t We note that 1=(p − 1) ¡ 1 since we assumed p ¿ 2. We now apply the elementary inequality (a + b)1=(p−1) 6 a1=(p−1) + b1=(p−1) to the last inequality to obtain p∗ −1 1=(p−1) ∞ 1=(p−1) y (t) 6 z (t) + s−k z q (s) ds : f() t Integrating the above inequality once more between t and ∞, we obtain as → ∞, 1=(p−1) ∞ ∞ − y(t) 6 − z(t) + −k z q ( ) d ds: (2.12) t
s
We now note the following simple estimates for the function z: ∃k1 ¿ 0
such that z(t) ¿ k1 1−! t for t ∈ [T (); ! ];
(2.13)
∃k2 ¿ 0
such that z(t) 6 k2 1−! t for all t ¿ T ():
(2.14)
Using (2.13) and (2.14) in (2.12) and using the fact that q ¡ k − p, we obtain for all t ∈ [T (); ! ], as → ∞, − y(t) 6 − k1 1−! t + C(1−!)q=(p−1) t (p+q−k)=(p−1) ;
C ¿ 0:
Rearranging the terms, we get as → ∞, k1 1−! t − C(1−!)q=(p−1) t (p+q−k)=(p−1) 6 y(t):
(2.15)
It can be checked easily that if we choose t = (p−1−q)=(p−1)(k−q−1) with large enough but @xed, then LHS in (2.15) becomes positive. Hence, T () = O((p−1−q)=(p−1)(k−q−1) ). This proves (i). (ii) q = k − p: We proceed as in case (i) to derive inequality (2.12). But we evaluate the integral term in this inequality diMerently as below: 1=(p−1) 1=(p−1) ∞ ∞ ! ∞ −k q −k q z ( ) d ds 6 z ( ) d ds t
s
s
t
+
∞
!
= I1 + I2 :
s
∞
−k q
z ( ) d
1=(p−1) ds (2.16)
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875
It is easy to see (using (2.14)) that I1 6 C1
−q=(p−1)2
!
t
(q−k+1)=(p−1)
d = C1
(p−k)=(p−1)2
! log t
:
Now we use the fact that z(t) 6 ; t ¿ T () to obtain, I2 6 C2
q=(p−1)
∞
!
∞
s
−k
1=(p−1) d
2
ds = C2 (p−k)=(p−1) :
Therefore, substituting these estimates for I1 and I2 in (2.12) and rearranging the terms, we obtain for all t ∈ [T (); ! ] as → ∞, ! 2 1−! (p−k)=(p−1)2 k1 t − C1 − C2 (p−k)=(p−1) 6 y(t): log t 2
It is easy to check that if we choose t = (p−q−1)=(p−1) log for large enough, then 2 the LHS in the above inequality is positive. Hence T () = O((p−q−1)=(p−1) log ). (iii) q ¿ k − p: Again, in this case we proceed as in case (i) to derive inequality (2.12). We split the integral term in this inequality as in (2.16) and evaluate I1 and I2 : I1 6 C1
−q=(p−1)2
I2 6 C2
q=(p−1)
t
∞
!
!
(q−k+1)=(p−1) d = 6 C1 1=(p−1)(!p−!k+q) :
s
∞
−k
1=(p−1) d
ds 6 C2 1=(p−1)(!p−!k+q) :
Therefore, we obtain that for all t ∈ [T (); ! ] as → ∞, k1 1−! t − (C1 + C2 )[1=(p−1)](!p−!k+q) 6 y(t): In the above inequality, it is easy to check that if we choose t= [1=(p−1)](!p−!k+q)+!−1 , with large enough, then the LHS in the inequality is positive. Therefore, we get T () = O([1=(p−1)(!p−!k+q)+!−1 ) in this case. This proves the lemma.
3. Proof of Theorem 3 If we assume q ¡ p2 =(N − p) − 1, then thanks to Lemma 2.7 we obtain that lim→0 T () = lim→∞ T () = 0. Let T ∗ = sup¿0 T (). Then, for any T ¡ T ∗ we may @nd atleast two values 1 ; 2 ¿ 0 such that T (1 ) = T (2 ) = T . Also, if T ¿ T ∗ there does not exist any ¿ 0 such that T () = T . This proves the theorem.
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Appendix Cauchy problem Let p ¿ 2 and N ¿ 1 be an integer. Consider the Cauchy problem (|u |p−2 u ) + N − 1 |u |p−2 u + f(u) = 0; r ¿ 0; r (P) u(0) ¿ 0; u (0) = 0: Theorem A.1. Let f : [0; ∞) → [0; ∞) be a locally Lipschitz function on (0; ∞) with f(0) = 0 and f(s) ¿ 0 for all s ¿ 0. Then there exists r0 ¿ 0 and a unique solution denoted by u(:; +) of (P) on (0; r0 ) which depends continuously on the initial data +. Proof. Let h = 0 be small so that + + h ¿ 0. Let u and v be solutions of (P) with initial conditions u(0) = +, v(0) = + + h. De@ne R(%) = |%|p−2 % and w = R(u ) − R(v ). Then w satis@es w + where
N −1 w = (r); r
(r) = f(v(r)) − f(u(r)). Then we have r t N −1 w(r) = (t) dt; r ¿ 0; r 0
and hence, |w(r)| 6
r sup | (t)|; N t∈[0;r]
r ¿ 0:
(A.1)
Let J be a bounded subinterval of (0; ∞) such that +; + + h ⊂ J . Since f is locally Lipschitz on J , there exists r0 ; M ¿ 0 such that on [0; r0 ]; u; v ¿ 0 and | (r)| 6 M |u(r) − v(r)| r 6 M |h| + |u (s) − v (s)| ds : 0
(A.2)
We now estimate, ˜ (r) − v (r)); R(u (r)) − R(v (r)) = R (u(r))(u
(A.3)
where u(r) ˜ lies between the real numbers u (r) and v (r). Hence,
1 |w(r)|; r ∈ [0; r0 ]: |u (r) − v (r)| 6 sup ˜ r∈[0;r0 ] R (u(r)) That is, from (A.2),
| (r)| 6 M
|h| +
r 0
1 |w(s)| : sup ˜ s∈[0;r0 ] R (u(s))
(A.4)
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From (A.1) and (A.4) we get
r w(r) M 1 |w(s)| : sup r 6 N |h| + ˜ s∈[0;r0 ] R (u(s)) 0
877
(A.5)
Now we estimate R (u(r)) ˜ for some @xed r ∈ [0; r0 ]. If necessary, choose r0 ¿ 0 small so that f(u(r)) ¿ 12 f(u(0)) and f(v(r)) ¿ 12 f(v(0)) for 0 6 r 6 r0 . Without loss of generality, we assume u (r) ¿ v (r). Then, r t N −1 r f(u(t)) dt 6 − R(u(r)) ˜ 6 R(u (r)) = − f(+): (A.6) r 2N 0 By the de@nition of R, we have ∃%0 ¿ 0;
such that R (%) ¿ |R(%)| for all % ∈ (0; %0 ):
˜ ∈ (0; %0 )∀r ∈ (0; r0 ). Then, from (A.6), We can further shrink r0 if necessary so that u(r) r f(+); ∀r ∈ (0; r0 ): ˜ ¿ R (u(r)) 2N Similarly, if u (r) ¡ v (r) we obtain R (u(r)) ˜ ¿ (r=2N )f(+ + h). Hence, r max{f(+); f(+ + h)}; ∀r ∈ (0; r0 ): (A.7) ˜ ¿ R (u(r)) 2N Using the above inequality in (A.4) we get, r w(r) M 2N |w(s)| 6 |h| + ds ; r ∈ (0; r0 ): r N max{f(+); f(+ + h)} s 0 Using Grownwall’s inequality, w(r) M 2Mr 6 ; r ∈ (0; r0 ): |h| exp r N max{f(+); f(+ + h)}
(A.8)
From (A.3) and (A.8) we get, |u (r) − v (r)|
2M 2Mr ; r ∈ (0; r0 ) 6 |h| r exp max{f(+); f(+ + h)} max{f(+); f(+ + h)}
and hence |u(r) − v(r)| 6 C|h|;
r ∈ [0; r0 ]
for some constant C ¿ 0. Now uniqueness and continuous dependence on a small interval (0; r0 ) follows. References [1] Adimurthi, S.L. Yadava, An elementary proof of the uniqueness of positive radial solutions of a quasilinear Dirichlet problem, Arch. Rat. Mech. Anal. 127 (3) (1994) 219–229. [2] A. Ambrosetti, J.G. Azorero, I. Peral, Multiplicity results for some nonlinear elliptic equations, J. Funct. Anal. 137 (1996) 219–242.
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[3] A. Ambrosetti, H. Brezis, G. Cerami, Combined eMects of concave and convex nonlinearities in some elliptic problems, J. Funct. Anal. 122 (2) (1994) 519–543. [4] F.V. Atkinson, L.A. Peletier, Emden–Fowler equations involving critical exponents, Nonlinear Anal. TMA 10 (8) (1986) 755–776. [5] J.G. Azorero, I. Peral Alonso, Some results about the existence of a second positive solution in a quasilinear critical problem, Indiana Univ. Math. J. 43 (3) (1994) 941–957. [6] J.G. Azorero, I. Peral Alonso, J. Manfredi Juan, Sobolev versus Holder local minimizers and global multiplicity for some quasilinear elliptic equations, Comm. Contemp. Math. 2 (3) (2000) 385–404. [7] F. Brock, Radial Symmetry for nonnegative solutions of semilinear elliptic equations involving p-Laplacian, in: Pitman Research Notes, Mathematics Series, Vol. 383, Longman, Harlow, 1998. [8] L. Erbe, M. Tang, Uniqueness theorems for positive radial solutions of quasilinear elliptic equations in a ball, J. DiMerential Equations 138 (1997) 351–379. [9] S. Prashanth, K. Sreenadh, Multiplicity results in a ball for p-Laplace equation with positive nonlinearity, Adv. DiMerential Equations 7 (7) (2002) 877–896. [10] P.N. Srikanth, Uniqueness of solutions of nonlinear Dirichlet problems, DiMerential Integral Equations 6 (3) (1993) 663–670.
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Multiplicity of positive solutions for p-Laplace equation with superlinear-type nonlinearity S. Prashanth∗ , K. Sreenadh TIFR Centre, Post Box No. 1234, IISc Campus, Bangalore 560012, India Received 5 December 2002; accepted 23 October 2003
Abstract In this work, we consider the problem ∗ −2p u = up + uq in BR (0); (PR ) u ¿ 0 u=0 on @BR (0); where p ¿ 2; p∗ = Np=(N − p) and BR (0) is the open ball of radius R centered about the origin in RN , N ¿ 3. Under the condition that p − 1 ¡ q ¡ p2 =(N − p) − 1, we show that there exists R∗ ¿ 0 such that (PR ) admits atleast two solutions for all R ∈ (R∗ ; ∞) and no solution if R ¡ R∗ . This extends the results of (Nonlinear Anal. TMA 10(8) (1986) 755) to the case of p-Laplace equation. ? 2003 Elsevier Ltd. All rights reserved. MSC: 35J65; 34B15; 35J60 Keywords: Multiplicity; Critical exponent; p-Laplace equation; Emden–Fowler equation
1. Introduction Let BR (0) ⊂ RN denote the open ball of radius R centered about the origin, N ¿ 3. Let p ∈ (1; ∞) and de@ne p∗ = Np=(N − p) if p ¡ N and p∗ = ∞ if p = N . In this work we are interested in studying the existence of multiple solutions to the following
∗
Corresponding author. E-mail addresses: [email protected] (S. Prashanth), [email protected] (K. Sreenadh).
0362-546X/$ - see front matter ? 2003 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2003.10.026
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quasilinear equation: −2p u = u + uq u ¿0 (PR ) u=0
in BR (0); on @BR (0);
where 2p u=div(|∇u|p−2 ∇u) is the usual p-Laplace operator, 0 ¡ q ¡ 6 p∗ −1. We brieGy recall the many multiplicity results known for problems of the type (PR ). A more easily understood situation is when we study (PR ) with 0 ¡ q ¡ p − 1 ¡ 6 p∗ − 1. This is the so called “sublinear” situation. In this case, due to the “sublinear” nature of the nonlinearity u +uq as u → 0, we can use the sub–super solution approach to obtain one solution with small L∞ norm to (PR ) and the other solution can be obtained using a variety of methods like degree theory [2], ODE shooting methods [9], or variational methods [3,5,6]. We summarize all these “sublinear” multiplicity results as below: Theorem 1. Let 0 ¡ q ¡ p − 1 ¡ 6 p∗ − 1, if p ¡ N and 0 ¡ q ¡ p − 1 ¡ ¡ ∞ if p = N . Then, there exists R∗ = R∗ (q; p; ; N ) ¿ 0 such that problem (PR ) admits atleast two solutions for all 0 ¡ R ¡ R∗ and no solution for R ¿ R∗ . Therefore it is surprising to @nd that even in the case q ¿ p − 1, when there is no “sublinear” eMect to take advantage of, [4] have shown that when p = 2; = 2∗ − 1 and N = 3 we may obtain multiple solutions to (PR ) for all large R. More precisely, they showed the following result. Theorem 2. Let N =3; p=2; =2∗ −1 and 1 ¡ q ¡ 3. Then there exists R∗ ¿ 0 such that problem (PR ) admits atleast two solutions for all R ∈ (R∗ ; ∞) and no solution for all R ∈ (0; R∗ ). We extend the above result to the case of the p-Laplace equation using the same methods to prove the following: Theorem 3. Let 2 ¡ p ¡ N; p − 1 ¡ q ¡ p2 =(N − p) − 1, = p∗ − 1. Then there exists R∗ = R∗ (p; q; N ) ¿ 0 such that problem (PR ) admits atleast two solutions for all R ∈ (R∗ ; ∞) and no solution for all R ∈ (0; R∗ ). Remark 1.1. The restriction p−1 ¡ p2 =(N −p)−1 implies the dimensional restriction N ¡ 2p. The above result also shows the limitations in the exponent q and the dimension N for the uniqueness results for (PR ) proved in [1,8,10]. In particular, the following uniqueness result can be inferred from [8]: The following two results (Corollaries 4 and 5) follow from Theorem 3. Corollary 4. Let = p∗ − 1 and p∗ − 2 6 q. Then problem (PR ) admits a unique solution for all R ¿ 0.
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Remark 1.2. It is easy to check that the above restriction on q for the uniqueness always holds if 1 ¡ p ¡ 2; q ¿ p − 1 and N ¿ 6. Hence the hypothesis that p ¿ 2 in Theorem 3 is restrictive only when N = 3; 4; 5. We can also interpret the above multiplicity results in the language of bifurcation theory. For this, we set = Rp( −q)=( −p+1) and u(x) = 1=( −q) w(( −p+1)=p( −q) x). It can be easily checked that u solves problem (PR ) if and only if w solves the following problem: −2p w = w + wq in B1 (0); w ¿0 (P ) w=0 on @B1 (0): We may rewrite the multiplicity results for problem (PR ) given in Theorems 1–3 in terms of problem (P ) as below Corollary 5. (i) The sublinear case: Let 0 ¡ q ¡ p − 1 ¡ 6 p∗ − 1 if p ¡ N and 0 ¡ q ¡ p − 1 ¡ ¡ ∞ if p = N . Then, there exists ∗ = ∗ (q; p; ; N ) ¿ 0 such that problem (P ) admits atleast two solutions for all ∈ (0; ∗ ) and no solution for ¿ ∗ . (ii) Superlinear case: Let 2 6 p ¡ N; = p∗ − 1 and p − 1 ¡ q ¡ p2 =(N − p) − 1. Then there exists ∗ = ∗ (p; q; N ) ¿ 0 such that for all ¿ ∗ the problem (P ) has atleast two solutions and no solutions if ¡ ∗ . Therefore, the bifurcation diagrams in these two cases look like
|| u ||oo
|| u ||oo . . . .
. . . . . *
*
2. Preliminary lemmas By Gidas–Ni–Nirenberg type symmetry results (shown in [7] for example), we have that any solution of (PR ) is radial and radially decreasing about the origin. Since any solution of (PR ) is radial, it solves the following ODE (which we again denote by (PR )): −(r N −1 |u |p−2 u ) = r N −1 (u + uq ) in (0; R); (PR ) u ¿0 u (0) = 0; u(R) = 0:
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We prove Theorem 3 by using asymptotic analysis to analyse the ODE associated to problem (PR ). This was the approach used by Atkinson–Peletier in their elegant work [4] to prove Theorem 2. We will extend their results to the case of p-Laplace equation to prove Theorem 3. Because of the quasilinear nature of the problem, obtaining a lower bound for y in terms of z (see (2.3) and (2.4) below) is more delicate than in the semilinear case treated by Azorero and Peral Alonso [5]. We obtain such lower bound in Lemma 2.5. Similar delicate estimates are required also in Lemma 2.7. For simplicity of notations, we de@ne the following quantities: N −1 k −1 N −p ; k =p ; l= : (2.1) = p−1 N −p p−1 In what follows we @x = p∗ − 1; p − 1 ¡ q ¡ . Let u be a solution of (PR ). We now make the following transformation known as the Emden–Fowler transformation (below, r denotes the radial variable): t= ; y(t) = u(r): (2.2) r Then, it is easy to check that y solves the following Emden–Fowler equation: −(|y |p−2 y ) = t −k (y + yq ) in t ∈ (T; ∞); y ¿0 (PT ) y(T ) = 0; y (∞) = 0; T= : R To study the multiplicity of solutions to (PR ) it is natural to consider the following version of (PT ) which is parametrised by a parameter ¿ 0: −(|y |p−2 y ) = t −k (y + yq ); (P ) lim y(t) = ; y (∞) = 0: t→∞
The above “initial value” problem with the initial data prescribed at in@nity admits a unique solution (see the appendix). Let y(t; ) denote any solution of (P ). De@ne T () = inf {t ¿ 0; y(:; ) ¿ 0 in (t; ∞)}. Then it is easy to show that T () is the @rst zero of y(t; ) as t is decreased from in@nity. From continuous dependence on initial values (see Appendix) we note that T () is a continuous function on (0; ∞). It is also easy to see by integration that y(:; ) is a strictly increasing concave function in (T (); ∞). ∗ Let f(s) = sp −1 + sq , and de@ne for ¿ 0,
1=(p−1) 1=(1−l) f() l−1 −1 z(t; ) = t t + ; t ¿ 0: (2.3) k −1 It is easy to check that z solves the equation ∗
− (|z |p−2 z ) = t −k 1−p f()z p
∗
−1
;
lim z(t; ) = :
t→∞
We now prove several lemmas required for the proof of Theorem 3.
(2.4)
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Lemma 2.1. Let y solve (P ); ¿ 0. Then y(t; ) ¡ z(t; ) for all t ∈ (T (); ∞). Proof. See [4] for the case p = 2 and [9] for general p. Lemma 2.2. Let y solve (P ); ¿ 0. Then, the 7rst zero T () of y is positive. Proof. See again [5,9]. Lemma 2.3. Let y solve (P ). Then, T () = O((q−1+p)=(k−p) )
as → 0:
Proof. Integrating the ODE in (P ) twice, we obtain, y(t) = −
∞
t
s
∞
1=(p−1)
−k f(y( )) d
ds:
(2.5)
We estimate the integral on the right of Eq. (2.5) as below: t
∞
s
∞
−k
1=(p−1) f(y( )) d
6 [f()] =
1=(p−1)
∞
t
∞
s
(p − 1)(f())1=(p−1) (k − 1)1=(p−1) (p − k)
−k
1=(p−1) d ds
t (p−k)=(p−1)
= K(k; p)(f())1=(p−1) t (p−k)=(p−1) :
(2.6)
Hence, using (2.6) in (2.5) we obtain y(t) ¿ − K(k; p)(f())1=(p−1) t (p−k)=(p−1) ;
t ¿ T ():
(2.7)
Hence, letting t = T () in (2.7) we obtain the estimate 0 ¿ − K(k; p)(f())1=(p−1) (T ())(p−k)=(p−1) :
(2.8)
Noting that f()1=(p−1) ∼ q=(p−1) as → 0, we obtain from (2.8) that T () = O((q−p+1)=(k−p) ) as → 0. This proves the lemma.
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Lemma 2.4. Let ! = p=(p − 1). We have the following estimates for the solution y of (P ) and the function z given in (2.3): (i) There exists k1 ¿ 0 such that y(t; ) ¡ k1 1−! t for t ¿ T (). (ii) For every a ¿ 0, we have z(a! ; ) ¿ for all ¿ 0. (iii) z(t; ) ¿ 1−! t for T () 6 t 6 ! . Proof. (i) By an easy estimate we obtain k1 ¿ 0 such that z(t; ) 6 k1 1−! t for all t ¿ 0. Now (i) follows by using Lemmas 2.1 and 2.2. (ii) Let a ¿ 0. Then, we compute
1=(p−1) 1=(1−l) f() −1 ! !+1 l−1 !(l−1) a + z(a ; ) = a k −1 ¿ a!+1 [al−1 !(l−1) ]1=(1−l) = : This proves (ii). (iii) Since t 6 ! , we obtain by a simple estimate,
1=(p−1) 1=(1−l) f() !(l−1) −1 + z(t; ) ¿ t k −1 T () 6 t 6 ! :
¿ 1−! t; This proves (iii).
Let S() be de@ned by 1=(p−1) f() l−1 −1 : S () = k −1 Lemma 2.5. For each 7xed a ¿ 0 we may 7nd a positive function da such that da () → 0 as → ∞ and the following inequality is true: y(t; ) ¿ (1 − da ()) z(t; )
for t ¿ aS():
Proof. Integrating the ODE in (P ) we obtain, ∞ ∞ −k p∗ −1 p−1 s y (s) ds + s−k yq (s) ds: (y ) (t) = t
t
Since by Lemma 2.1, y(t) ¡ z(t) for t ¿ T (), we get, ∞ ∞ ∗ s−k z p −1 (s) ds + s−k z q (s) ds (y )p−1 (t) 6 t
= I1 (t) + I2 (t):
t
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Now a simple computation gives that, z(aS(); )=%(a) where %(a)=a(1+al−1 )−1=(l−1) . Since t ¿ aS() and z is increasing, we get ∞ p∗ −q−1 I1 (t) ¿ (%(a)) s−k z q (s) ds t
= (%(a))p
∗
−q−1
I2 (t):
Therefore, (y (t))p−1 6 (1 + (%(a))q−p
∗
+1
)
∞
0
s−k z p
∗
−1
(s) ds:
(2.9)
We note that z satis@es the diMerential equation −((z )p−1 (t)) = t −k −p
∗
+1
f()z p
∗
−1
(t):
Integrating the above diMerential equation from t to ∞, we obtain ∞ ∗ ∗ (z )p−1 (t) = (1 + q−p +1 ) s−k z p −1 (s) ds: t
(2.10)
Therefore, letting &(c; ) =
1 ; (1 + (c)q−p∗ +1 )
we obtain from (2.9) and (2.10) that &(%(a); )(y )p−1 − &(1; )(z )p−1 6 0. Therefore, 1=(p−1) &(1; ) y (t) 6 z (t); ∀t ¿ max{aS(); T ()}: &(%(a); ) Integrating the above inequality once more between t and ∞, 1=(p−1) &(1; ) − y(t) 6 ( − z(t)): &(%(a); )
(2.11)
Now we note that [&(1; )=&(%(a); )]1=(p−1) ¡ 1 for a small enough. Let da () = 1 − [&(1; )=&(%(a); )]1=(p−1) . Then (2.11) implies y(t) ¿ (1 − da ()) z(t) + da () ¿ (1 − da ()) z(t). It is now easy to see that da () → 0 as → ∞ for @xed a ¿ 0 which proves the lemma. Lemma 2.6. T () = o(! ) as → ∞. Proof. Clearly, it is suPcient to show that for any a ¿ 0; y(t) ¿ 0 for all t ¿ a! . But this follows from Lemma 2.5 by noting that S() ¿ C! for some C ¿ 0. Lemma 2.7. Let p ¿ 2. Then as → ∞, the following hold: (i) If 0 ¡ q ¡ k − p; T () = O((p−q−1)=(k−q−1) ). 2 (ii) If q = k − p; T () = O((p−1−q)=(p−1) log ). (iii) If k − p ¡ q ¡ p∗ − 1; T () = O(1=(p−1)(!p−!k+q+1) ).
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Proof. (i) q ¡ k − p: Integrating the ODE satis@ed by y from t to ∞ and using Lemma 2.1, we get ∞ ∗ (y )p−1 (t) 6 s−k (z p + z q )(s) ds: t
Now using the fact that z solves the ODE given in (2.4), from the above inequality we obtain p∗ −1 ∞ p−1 p−1 (z ) (t) + s−k z q (s) ds: (y ) (t) 6 f() t We note that 1=(p − 1) ¡ 1 since we assumed p ¿ 2. We now apply the elementary inequality (a + b)1=(p−1) 6 a1=(p−1) + b1=(p−1) to the last inequality to obtain p∗ −1 1=(p−1) ∞ 1=(p−1) y (t) 6 z (t) + s−k z q (s) ds : f() t Integrating the above inequality once more between t and ∞, we obtain as → ∞, 1=(p−1) ∞ ∞ − y(t) 6 − z(t) + −k z q ( ) d ds: (2.12) t
s
We now note the following simple estimates for the function z: ∃k1 ¿ 0
such that z(t) ¿ k1 1−! t for t ∈ [T (); ! ];
(2.13)
∃k2 ¿ 0
such that z(t) 6 k2 1−! t for all t ¿ T ():
(2.14)
Using (2.13) and (2.14) in (2.12) and using the fact that q ¡ k − p, we obtain for all t ∈ [T (); ! ], as → ∞, − y(t) 6 − k1 1−! t + C(1−!)q=(p−1) t (p+q−k)=(p−1) ;
C ¿ 0:
Rearranging the terms, we get as → ∞, k1 1−! t − C(1−!)q=(p−1) t (p+q−k)=(p−1) 6 y(t):
(2.15)
It can be checked easily that if we choose t = (p−1−q)=(p−1)(k−q−1) with large enough but @xed, then LHS in (2.15) becomes positive. Hence, T () = O((p−1−q)=(p−1)(k−q−1) ). This proves (i). (ii) q = k − p: We proceed as in case (i) to derive inequality (2.12). But we evaluate the integral term in this inequality diMerently as below: 1=(p−1) 1=(p−1) ∞ ∞ ! ∞ −k q −k q z ( ) d ds 6 z ( ) d ds t
s
s
t
+
∞
!
= I1 + I2 :
s
∞
−k q
z ( ) d
1=(p−1) ds (2.16)
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It is easy to see (using (2.14)) that I1 6 C1
−q=(p−1)2
!
t
(q−k+1)=(p−1)
d = C1
(p−k)=(p−1)2
! log t
:
Now we use the fact that z(t) 6 ; t ¿ T () to obtain, I2 6 C2
q=(p−1)
∞
!
∞
s
−k
1=(p−1) d
2
ds = C2 (p−k)=(p−1) :
Therefore, substituting these estimates for I1 and I2 in (2.12) and rearranging the terms, we obtain for all t ∈ [T (); ! ] as → ∞, ! 2 1−! (p−k)=(p−1)2 k1 t − C1 − C2 (p−k)=(p−1) 6 y(t): log t 2
It is easy to check that if we choose t = (p−q−1)=(p−1) log for large enough, then 2 the LHS in the above inequality is positive. Hence T () = O((p−q−1)=(p−1) log ). (iii) q ¿ k − p: Again, in this case we proceed as in case (i) to derive inequality (2.12). We split the integral term in this inequality as in (2.16) and evaluate I1 and I2 : I1 6 C1
−q=(p−1)2
I2 6 C2
q=(p−1)
t
∞
!
!
(q−k+1)=(p−1) d = 6 C1 1=(p−1)(!p−!k+q) :
s
∞
−k
1=(p−1) d
ds 6 C2 1=(p−1)(!p−!k+q) :
Therefore, we obtain that for all t ∈ [T (); ! ] as → ∞, k1 1−! t − (C1 + C2 )[1=(p−1)](!p−!k+q) 6 y(t): In the above inequality, it is easy to check that if we choose t= [1=(p−1)](!p−!k+q)+!−1 , with large enough, then the LHS in the inequality is positive. Therefore, we get T () = O([1=(p−1)(!p−!k+q)+!−1 ) in this case. This proves the lemma.
3. Proof of Theorem 3 If we assume q ¡ p2 =(N − p) − 1, then thanks to Lemma 2.7 we obtain that lim→0 T () = lim→∞ T () = 0. Let T ∗ = sup¿0 T (). Then, for any T ¡ T ∗ we may @nd atleast two values 1 ; 2 ¿ 0 such that T (1 ) = T (2 ) = T . Also, if T ¿ T ∗ there does not exist any ¿ 0 such that T () = T . This proves the theorem.
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Appendix Cauchy problem Let p ¿ 2 and N ¿ 1 be an integer. Consider the Cauchy problem (|u |p−2 u ) + N − 1 |u |p−2 u + f(u) = 0; r ¿ 0; r (P) u(0) ¿ 0; u (0) = 0: Theorem A.1. Let f : [0; ∞) → [0; ∞) be a locally Lipschitz function on (0; ∞) with f(0) = 0 and f(s) ¿ 0 for all s ¿ 0. Then there exists r0 ¿ 0 and a unique solution denoted by u(:; +) of (P) on (0; r0 ) which depends continuously on the initial data +. Proof. Let h = 0 be small so that + + h ¿ 0. Let u and v be solutions of (P) with initial conditions u(0) = +, v(0) = + + h. De@ne R(%) = |%|p−2 % and w = R(u ) − R(v ). Then w satis@es w + where
N −1 w = (r); r
(r) = f(v(r)) − f(u(r)). Then we have r t N −1 w(r) = (t) dt; r ¿ 0; r 0
and hence, |w(r)| 6
r sup | (t)|; N t∈[0;r]
r ¿ 0:
(A.1)
Let J be a bounded subinterval of (0; ∞) such that +; + + h ⊂ J . Since f is locally Lipschitz on J , there exists r0 ; M ¿ 0 such that on [0; r0 ]; u; v ¿ 0 and | (r)| 6 M |u(r) − v(r)| r 6 M |h| + |u (s) − v (s)| ds : 0
(A.2)
We now estimate, ˜ (r) − v (r)); R(u (r)) − R(v (r)) = R (u(r))(u
(A.3)
where u(r) ˜ lies between the real numbers u (r) and v (r). Hence,
1 |w(r)|; r ∈ [0; r0 ]: |u (r) − v (r)| 6 sup ˜ r∈[0;r0 ] R (u(r)) That is, from (A.2),
| (r)| 6 M
|h| +
r 0
1 |w(s)| : sup ˜ s∈[0;r0 ] R (u(s))
(A.4)
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From (A.1) and (A.4) we get
r w(r) M 1 |w(s)| : sup r 6 N |h| + ˜ s∈[0;r0 ] R (u(s)) 0
877
(A.5)
Now we estimate R (u(r)) ˜ for some @xed r ∈ [0; r0 ]. If necessary, choose r0 ¿ 0 small so that f(u(r)) ¿ 12 f(u(0)) and f(v(r)) ¿ 12 f(v(0)) for 0 6 r 6 r0 . Without loss of generality, we assume u (r) ¿ v (r). Then, r t N −1 r f(u(t)) dt 6 − R(u(r)) ˜ 6 R(u (r)) = − f(+): (A.6) r 2N 0 By the de@nition of R, we have ∃%0 ¿ 0;
such that R (%) ¿ |R(%)| for all % ∈ (0; %0 ):
˜ ∈ (0; %0 )∀r ∈ (0; r0 ). Then, from (A.6), We can further shrink r0 if necessary so that u(r) r f(+); ∀r ∈ (0; r0 ): ˜ ¿ R (u(r)) 2N Similarly, if u (r) ¡ v (r) we obtain R (u(r)) ˜ ¿ (r=2N )f(+ + h). Hence, r max{f(+); f(+ + h)}; ∀r ∈ (0; r0 ): (A.7) ˜ ¿ R (u(r)) 2N Using the above inequality in (A.4) we get, r w(r) M 2N |w(s)| 6 |h| + ds ; r ∈ (0; r0 ): r N max{f(+); f(+ + h)} s 0 Using Grownwall’s inequality, w(r) M 2Mr 6 ; r ∈ (0; r0 ): |h| exp r N max{f(+); f(+ + h)}
(A.8)
From (A.3) and (A.8) we get, |u (r) − v (r)|
2M 2Mr ; r ∈ (0; r0 ) 6 |h| r exp max{f(+); f(+ + h)} max{f(+); f(+ + h)}
and hence |u(r) − v(r)| 6 C|h|;
r ∈ [0; r0 ]
for some constant C ¿ 0. Now uniqueness and continuous dependence on a small interval (0; r0 ) follows. References [1] Adimurthi, S.L. Yadava, An elementary proof of the uniqueness of positive radial solutions of a quasilinear Dirichlet problem, Arch. Rat. Mech. Anal. 127 (3) (1994) 219–229. [2] A. Ambrosetti, J.G. Azorero, I. Peral, Multiplicity results for some nonlinear elliptic equations, J. Funct. Anal. 137 (1996) 219–242.
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[3] A. Ambrosetti, H. Brezis, G. Cerami, Combined eMects of concave and convex nonlinearities in some elliptic problems, J. Funct. Anal. 122 (2) (1994) 519–543. [4] F.V. Atkinson, L.A. Peletier, Emden–Fowler equations involving critical exponents, Nonlinear Anal. TMA 10 (8) (1986) 755–776. [5] J.G. Azorero, I. Peral Alonso, Some results about the existence of a second positive solution in a quasilinear critical problem, Indiana Univ. Math. J. 43 (3) (1994) 941–957. [6] J.G. Azorero, I. Peral Alonso, J. Manfredi Juan, Sobolev versus Holder local minimizers and global multiplicity for some quasilinear elliptic equations, Comm. Contemp. Math. 2 (3) (2000) 385–404. [7] F. Brock, Radial Symmetry for nonnegative solutions of semilinear elliptic equations involving p-Laplacian, in: Pitman Research Notes, Mathematics Series, Vol. 383, Longman, Harlow, 1998. [8] L. Erbe, M. Tang, Uniqueness theorems for positive radial solutions of quasilinear elliptic equations in a ball, J. DiMerential Equations 138 (1997) 351–379. [9] S. Prashanth, K. Sreenadh, Multiplicity results in a ball for p-Laplace equation with positive nonlinearity, Adv. DiMerential Equations 7 (7) (2002) 877–896. [10] P.N. Srikanth, Uniqueness of solutions of nonlinear Dirichlet problems, DiMerential Integral Equations 6 (3) (1993) 663–670.