Nonlinear Analysis 46 (2001) 591 – 600
www.elsevier.com/locate/na
Multiplicity of nontrivial solutions for an
asymptotically linear nonlocal Tricomi problem
Daniela Lupo 1 , Kevin R. Payne2;∗;†
Dipartimento di Matematica, Politecnico di Milano, Piazza Leonardo da Vinci, 32 20133 Milano, Italy
Received 7 December 1998; accepted 16 December 1999
Keywords: Mixed type equations; Dual variational method; Linking theorem
1. Introduction and statement of the result
In this work, we are concerned with the multiplicity of nontrivial solutions for the
following nonlocal semilinear Tricomi problem, introduced in [7]:
Tu ≡ −yuxx − uyy = Rf(u) in 7
u = 0 on AC ∪ ;
−y@2x
@2y
(NST)
2
where T ≡
−
is the Tricomi operator on R ; R is the re9ection operator
induced by composition with the map : : R2 → R2 de;ned by :(x; y)=(−x; y); f(u)
is an asymptotically linear term such that f(0) = 0, and 7 is a symmetric admissible
Tricomi domain (cf. De;nition 2:1 of Lupo and Payne [7]). That is, 7 is a bounded
region in R2 that is symmetric with respect to the y-axis and has a piecewise C 2
boundary @7 = AC ∪ BC ∪ , where is a symmetric with respect to the y-axis C 2 arc
in the elliptic region y ¿ 0, with endpoints on the x-axis at A = (−x0 ; 0) and B = (x0 ; 0)
and AC and BC are two characteristic arcs for the Tricomi operator in the hyperbolic
region of negative and positive slope issuing from A and B, respectively, which meet
at the point C on the y-axis. It should be noted that u ≡ 0 is always a solution to
(NST). We refer the reader to [7] for the rationale concerning this problem, the interest
∗ Corresponding author. Tel.: +39-02-26602280; fax: +39-02-70630346.
E-mail addresses: [email protected] (D. Lupo), [email protected] (K.R. Payne).
1
Supported by MURST, Project “Metodi Variazionali ed Equazioni DiHerenziali Non Lineari”.
2
Supported by a CNR grant; program for foreign mathematicians.
address: Dipartimento di Matematica, UniversitJa di Milano, Via C. Saldini 50, 20133 Milano,
Italy.
† Current
0362-546X/01/$ - see front matter ? 2001 Elsevier Science Ltd. All rights reserved.
PII: S 0 3 6 2 - 5 4 6 X ( 0 0 ) 0 0 1 3 5 - 8
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D. Lupo, K.R. Payne / Nonlinear Analysis 46 (2001) 591 – 600
in variational methods for mixed elliptic–hyperbolic type equations, and for references
to relevant literature on the Tricomi operator and its importance.
First of all, let us specify in which sense we are looking for generalized solutions.
For N ⊂ 7, we denote by CN∞ (7) the set of all smooth functions on 7 such that u ≡ 0
on N, by WN1 the closure of CN∞ (7) with respect to the W 1; 2 (7) norm, and by WN−1
the dual of WN1 .
1
Denition 1.1. One says that u ∈ WAC∪
is a generalized solution of (NST) if
Tu = Rf(u)
in L2 (7);
∞
and there exists a sequence {uj } ⊂ CAC∪
(7) such that
1
lim uj − uWAC∪
=0
j→∞
and
1
lim Tuj − Rf(u)WBC∪
= 0:
j→∞
∞
1
and
(7) into WAC∪
Let TAC be the unique continuous extension of T from CAC∪
1
1
note that R induces an isometric isomorphism between WAC∪
and WBC∪
. In [7],
following ideas in [2], it is shown that, given a symmetric admissible domain 7, the
re9ected Tricomi operator
1
RTAC : D(RTAC ) = WA ⊂ WAC∪
→ L2 (7);
admits a continuous left inverse
1
K = (RTAC )−1 : L2 (7) → WA ⊂ WAC∪
for which K is compact as an operator on L2 (7). Denote by k+ the sequence of
positive eigenvalues of K decreasing to zero (cf. Sections 3 and 2 of [6] for more
details). Suppose that
(f 1) f ∈ C 0 (R; R); f is invertible and f(0) = 0.
In addition, suppose that a and b are real numbers such that a ∈ ((RTAC )−1 )
+
; j+ ) for some j ≥ 1. After writing f(s)=s=a+f∞ (s)=s=b+f0 (s),
and b ∈ (j+1
suppose that the perturbations f∞ ; f0 ∈ C 0 (R; R) satisfy
(f 2) f∞ is bounded and lims→0 f0 (s)=s = 0, and
(f 3) −s2 =b ≤ sf0 (s) ≤ 0; s ∈ R.
In [6] we have proved the following result.
Theorem A. Let 7 be a symmetric admissible Tricomi domain and assume that f
satis0es (f 1) – (f 3). Then; if a ¿ j+ ; the problem (NST) admits at least one nontrivial
generalized solution.
In this paper we want to show how, if the nonlinearity “interacts” more with the
spectrum of the inverse operator K = (RTAC )−1 , then one obtains more nontrivial solutions. More precisely, the following result holds.
Theorem B. Let 7 be a symmetric admissible Tricomi domain and assume that f
satis0es (f 1) – (f 3). Suppose that a ∈ ((RTAC )−1 ) and that for some j ≥ 2;
D. Lupo, K.R. Payne / Nonlinear Analysis 46 (2001) 591 – 600
593
+
+
+
j+1
¡b¡j+ ¡j−1
¡a. Then there exists a value b∗ ∈ (j+1
; j+ ) such that for every
b ∈ (b∗ ; j+ ) the problem (NST) admits at least two nontrivial solutions.
The result will be obtained utilizing the dual variational method, as in [6,7], combined with the following critical point theorem of Marino, Micheletti, and Pistoia on
the existence of distinct linking critical points. More precisely, consider a Hilbert space
H decomposed into H = W ⊕ Z, with dim Z ¡ + ∞. Within Z consider a proper subspace Z1 and two elements e1 ∈ Z1 and e2 ∈ (Z Z1 ) with e1 = e2 = 1. De;ne,
for any Y subspace of H ,
B (Y ):={u ∈ Y | u ≤ }
and by @B (Y ) the boundary of B (Y ) relative to Y . Furthermore, for any e ∈ H ,
consider
QR (Y; e):={u + ae ∈ Y ⊕ Re | u ∈ Y a ≥ 0; u + ae ≤ R}
and denote by @QR (Y; e) its boundary relative to Y ⊕ Re.
Theorem 1.2 (cf. Theorem 8:4 of Marino et al. [8]). Suppose that I ∈ C 1 (H; R) satis0es the (PS) condition. In addition; assume that there exist i ; Ri ; for i = 1; 2; such
that 0 ¡ i ¡ Ri and
inf
I ¿ sup I;
inf
I ¿ − ∞;
@QR1 (W; e1 )
QR1 (W; e1 )
@B1 (Z)
inf
@QR2 (W ⊕Z1 ; e2 )
inf
QR2 (W ⊕Z1 ; e2 )
I¿
sup
@B2 (ZZ1 )
I;
I ¿ − ∞:
If R1 ¡ R2 , then there exist at least three critical levels of I. Moreover; the critical
levels satisfy the following inequalities:
sup I ≥ c1 ≥
B1 (Z)
≥
inf
@QR1 (W; e1 )
inf
@QR2 (W ⊕Z1 ; e2 )
I ¿ sup I ≥ c2 ≥
I¿
@B1 (Z)
sup
@B2 (ZZ1 )
inf
QR1 (W; e1 )
I ≥ c3 ≥
inf
I
QR2 (W ⊕Z1 ; e2 )
I:
Remark 1.3. (1) Since dim Z ¡ + ∞, one has supB (Z) I ¡ + ∞.
1
(2) Theorem 2:1 of [5] gives (utilizing a result of [4]) the analog to Theorem 1.2
above in the case dim Z = + ∞; (cf. also [3, Theorem 2:1] for a simpler statement for
Z1 = Re1 ).
2. Setting of the problem
Following ideas contained in [6,7], we translate the problem of ;nding nontrivial
solutions of (NST) into the problem of ;nding nontrivial critical points of the dual
594
D. Lupo, K.R. Payne / Nonlinear Analysis 46 (2001) 591 – 600
action functional associated to (NST); the rough idea behind the use of a dual variational method is the following. Let f be an invertible function and denote by g = f−1
its inverse. The statement
1
there exists u0 ∈ WAC∪
such that u0 = 0 and TAC u0 = Rf(u0 ) in L2 (7)
is equivalent to
1
there exists u0 ∈ WAC∪
such that u0 = 0 and RTAC u0 = f(u0 ) in L2 (7):
Hence, if we are able to ;nd a v0 ∈ L2 (7) such that
g(v0 ) = Kv0 ;
where K =(RTAC )−1 is compact on L2 (7) (cf. [Section 2 of 7] for a precise description
of the linear theory), then u0 = Kv0 is a solution of our problem.
Therefore, our goal will be to set-up a variational formulation of the dual problem.
To this end, de;ne J : L2 (7) → R as
1
J (v) = G(v) d x dy −
vKv d x dy;
(2.1)
2 7
7
v
where G(v) = 0 g(t) dt denotes the primitive of g such that G(0) = 0. If g satis;es
the growth condition (for well-known properties of Nemistkii operators cf. [1])
|g(t)| ≤ K1 + K2 |t|
(2.2)
it is easy to check that J ∈ C 1 (L2 (7); R) and that
J (v)[w] = g(v)w d x dy − wKv d x dy:
7
7
(2.3)
Therefore, critical points of J will be weak solutions of the dual problem; let us remark
explicitly that the key point in proving (2.3) is the symmetry of K.
3. The critical point result
We consider the case of nonlinearity f which is asymptotically linear, both in zero
and at in;nity, where the slope in zero is diHerent from the slope at in;nity. More
precisely, ;rst we recall that since K =(RTAC )−1 is a compact linear operator on L2 (7)
which is injective, nonsurjective, and self-adjoint the spectrum of K is {0}∪{k }, where
k is a sequence of eigenvalues of ;nite multiplicity whose only possible accumulation
point is zero. Furthermore, it is shown in Proposition 2:1 of [6] that there exist in;nitely
many positive and in;nitely many negative eigenvalues, which we will denote by k± .
As pointed out in Remark 2:3 of [6], it is possible to associate to k± a complete
orthonormal basis in L2 (7).
The solutions will be found as the preimage of two distinct linking critical points
which are obtained by applying Theorem 1.2 to the dual action functional J . Hence,
we have to show that J satis;es an appropriate geometrical condition (cf. Lemmas 3.2
and 3.4) and a suitable compactness condition (cf. Lemma 3.1).
To begin with, one can check that the hypotheses (fi); i = 1; 2; 3, imply that the
inverse g = f−1 satis;es
D. Lupo, K.R. Payne / Nonlinear Analysis 46 (2001) 591 – 600
595
(g1) g ∈ C 0 (R; R); g is invertible, g(0) = 0; g(t) = at + g∞ (t) = bt + g0 (t),
where the perturbations g0 and g∞ belong to C 0 (R; R) and satisfy
(g2) g∞ is bounded and limt→0 g0 (t)=t = 0; and
(g3) tg0 (t) ≥ 0; t ∈ R.
t
One then notes that (g1) – (g3) imply that the primitives G0 (t)= 0 g0 (,) d, and G∞ (t)=
t
g (,) d, satisfy, for every t ∈ R,
0 ∞
|G∞ (t)| ≤ M̃ |t| and
0 ≤ G0 (t) ≤ 12 (a − b)t 2 + M̃ |t|;
(3.1)
where M̃ = aM∞ (denoting by M∞ the positive constant such that |f∞ (s)| ≤ M∞ , for
every s ∈ R), and
G0 (t) = 12 (a − b)t 2 + G∞ (t):
(3.2)
Furthermore, one has
G0 (t)
(b − a) 2
= 0 and G∞ (t) ≥
t ; t ∈ R;
lim
t→0 t 2
2
and
2G(t)
2G(t)
= a and lim 2 = b:
lim
t→±∞
t→0
t2
t
Let us now consider the dual action functional J de;ned in (2.1)
1
vKv d x dy
J (v) = G(v) d x dy −
2 7
7
b
1
= G0 (v) d x dy +
v2 d x dy −
vKv d x dy
2 7
2 7
7
a
1
2
= G∞ (v) d x dy +
v d x dy −
vKv d x dy;
2 7
2 7
7
(3.3)
(3.4)
where the use of one of the last two equivalent decompositions will be made as suits
the situation. It is clear that the growth condition on g∞ implies (2.2) and hence
J ∈ C 1 (L2 (7); R).
j
As a ;nal preparation, let L2 (7) = V ⊕ V ⊥ , where V = k=1 Vk is a direct sum of
eigenspaces Vk = {v ∈ L2 (7) | Kv = k+ v}, while V ⊥ is the orthogonal complement of
V in L2 (7).
From now on, we will denote by v = vL2 (7) . Keeping in mind the ordering
0 ¡ · · · ¡ j+ ¡ · · · ¡ 1+ , it is easy to see that one has
− vKv d x dy ≤ −j+ v2 for every v ∈ V;
(3.5)
7
−
and
7
−
7
vKv d x dy ≥ −1+ v2
+
vKv d x dy ≥ −j+1
v2
for every v ∈ V
for every v ∈ V ⊥ :
(3.6)
(3.7)
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D. Lupo, K.R. Payne / Nonlinear Analysis 46 (2001) 591 – 600
Lemma 3.1. Let 7 be a symmetric admissible Tricomi domain and assume that g
satis0es (g1); (g2) and (g3). Then J satis0es the Palais–Smale condition on L2 (7).
Proof. Consider a Palais–Smale sequence; that is {vn } such that
g∞ (vn ) + avn − Kvn → 0
in L2 (7)
(3.8)
and
|J (vn )| ≤ M:
One has that {vn } must be bounded in L2 (7). If not, one has that vn → +∞. Then,
one has g∞ (vn )=vn → 0 in L2 (7) since (g2) yields
2
2
M̃ d x dy
g (v ) d x dy
M̃ |7|
7 ∞ n
7
:
≤
≤
vn 2
vn 2
vn 2
Now, since vn =vn is a norm-bounded sequence, by the compactness of K, there is
a w ∈ L2 (7) \ {0} such that K(vn =vn ) → w in L2 (7), where one passes to another
subsequence if needed. Then by (3.8), one must have avn =vn → w in L2 (7), and
thus by the continuity of K we have aw − Kw = 0 in L2 (7). This gives a contradiction
since a ∈ ((RTAC )−1 ). Hence, vn must be bounded. Then, keeping in mind that (3.8)
can be written as
g(vn ) − Kvn → 0
in L2 (7)
and that K is compact, one has that, at least for a subsequence, g(vnk ) → z in L2 (7) for
some z ∈ L2 (7). Therefore, f(g(vnk )) = vnk → f(z) in L2 (7) since f is a continuous
Nemitskii operator and hence the result.
We now consider the geometrical conditions. Denoting by
@B (V ) = {v ∈ V | v = }
and, letting ej+ ∈ Vj such that ej+ = 1, by
@QR (V ⊥ ; ej+ ) = {w + 1ej+ | 1 ¿ 0 and w + 1ej+ = R}
∪{w ∈ V ⊥ | w ≤ R};
the following lemmas provide a geometrical structure suitable for applying Theorem
1.2 to the dual functional J .
Lemma 3.2. Let 7 be a symmetric admissible Tricomi domain and assume that g
satis0es (g1) – (g3). If a ¿ j+ ; there exist 1 ¿ 0 and R1 ¿ 1 such that the following
inequality holds:
sup J (v) ¡
@B1 (V )
inf
@QR1 (V ⊥ ; ej+ )
J (v):
(3.9)
Proof. We give a brief sketch of the proof which is contained in [7]. We want to
show that it is possible to choose a 1 small enough to make the sup strictly negative
and to choose R1 large enough to make the inf nonnegative. Then, choosing R1 even
D. Lupo, K.R. Payne / Nonlinear Analysis 46 (2001) 591 – 600
597
larger, if need be so that R1 ¿ 1 , will complete the lemma. To estimate the sup from
above, we begin by noting that inequality (3.5) applied to arbitrary v ∈ V yields
1
b
2
vKv d x dy
v d x dy + G0 (v) d x dy −
J (v) =
2 7
2 7
7
1
≤ (b − j+ )v2 + G0 (v) d x dy:
2
7
If we show that
G0 (v) d x dy
= 0;
(3.10)
lim 7
v2
v→0
v∈V
then we will obtain the claim: there exists a 1 ¿ 0 such that
1
J (v) ≤ (b − j+ )v2 ¡ 0 for any v ∈ V; v = 1 :
4
In fact, it is shown in Lemma 4:5 of [7], that
G (v) d x dy
7 0
lim
= 0;
v2W 1; 2
v 1;2 →0
(3.11)
W
v∈Vj
and, since all norms are equivalent on V with dim V ¡ + ∞, the desired limit property
on G0 follows.
On the other hand, to estimate the inf from below, we begin by observing that
inequality (3.7) and the lower bound on G0 in (3.1) yields, for arbitrary w ∈ V ⊥
b
1 +
J (w) ≥ G0 (w) d x dy +
w2 d x dy − j+1
w2
2 7
2
7
1
+
≥ (b − j+1
)w2 ≥ 0:
(3.12)
2
Furthermore, for arbitrary w + 1ej+ ∈ V ⊥ ⊕ Rej+ , the inequality (3.7) together with
the lower bound on G∞ in (3.1) yields
a
a
J (w + 1ej+ ) = w2 + 12 + G∞ (w + 1ej+ ) d x dy
2
2
7
1
12
−
wKw d x dy − j+
2 7
2
1
12
+
≥ (a − j+1
)w2 + (a − j+ )
2
2
−M̃ |7|1=2 (w + 1ej+ ):
ej+ ,
(3.13)
R2 = w + 1ej+ 2 = w2 + 12
we can set
Now, by recalling that w is orthogonal to
+
and from (3.13) taking into account that a ¿ j+ ¿ j+1
, we get
J (w + 1ej+ ) ≥
1
(a − j+ )R2 − M̃ |7|1=2 R:
2
(3.14)
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D. Lupo, K.R. Payne / Nonlinear Analysis 46 (2001) 591 – 600
It is clear that it is possible to choose R suRciently large to make the right-hand
side of (3.14) strictly positive, and hence to choose, if need be, a bigger R1 so that
R1 ¿ 1 and hence the inequality (3.9) follows.
Remark 3.3. Inequality (3.9) of Lemma 3.2 gives the ;rst inequality in the hypotheses
of Theorem 1.2 with the choices H = L2 (7); Z = V; W = V ⊥ and e1 = ej+ . Moreover, inequalities (3.12) and (3.13) yield the second inequality in the hypotheses of
Theorem 1.2 and hence Lemma 3.1 completes the proof of Theorem A.
Now we will show that if two eigenvalues fall between a and b, then one can
construct a second linking geometry that is suitable for the application of Theorem 1.2.
Lemma 3.4. Let 7 be a symmetric admissible Tricomi domain and assume that g
+
+
¡ b ¡ j+ ¡ j−1
¡ a then there exist
satis0es (g1) – (g3). If for some j ≥ 2; j+1
+
+
∗
2 ¿ 0; R2 ¿ 2 and a value b ∈ (j+1 ; j ) such that for every b ∈ (b∗ ; j+ ) the
following inequality holds:
sup
@B2 (V Vj )
J (v) ¡
inf
+
)
@QR1 (V ⊥ ⊕Vj ;ej−1
J (v):
(3.15)
j−1
Proof. Since V Vj = k=1 Vk , one can choose a 2 , independently of b, such that
for every v ∈ V Vj with v = 2 one has
1 +
+
)22 ¡ 0:
( − j−1
4 j
Indeed, for every v ∈ V Vj one has
b
1
2
J (v) =
v d x dy + G0 (v) d x dy −
vKv d x dy
2 7
2 7
7
1
+
2
≤ (b − j−1 )v + G0 (v) d x dy
2
7
1
+
≤ (j+ − j−1
)v2 + G0 (v) d x dy;
2
7
J (v) ≤
(3.16)
and hence, by using the same reasoning that leads to the inequality (3:11), one can ;x
a 2 independently of b so that (3:16) holds.
+
+
= 1. Then, for every w ∈
On the other hand, let ej−1
∈ Vj−1 be such that ej−1
+
+
⊥
¡ j+ ¡ j−1
and that w; v
V ; v ∈ Vj and for every 2 ∈ R, keeping in mind that j+1
+
+
2
2
and ej−1 are mutually orthogonal, by choosing R =w+v+2ej−1 =w2 +v2 +22 ,
one gets
a
a
a
+
+
) = w2 + v2 + 22 + G∞ (w + v + 2ej−1
) d x dy
J (w + v + 2ej−1
2
2
2
7
−
1
2
7
wKw d x dy −
j+
1
+
v2 − 22 j−1
2
2
D. Lupo, K.R. Payne / Nonlinear Analysis 46 (2001) 591 – 600
599
1
1
22
+
+
≥ (a − j+1
)w2 + (a − j+ )v2 + (a − j−1
)
2
2
2
+
−M̃ |7|1=2 w + 1ej+ + 2ej−1
and hence
1
+
(a − j−1
)R2 − M̃ |7|1=2 R:
(3.17)
2
Hence, one can choose an R large enough to ensure that the right-hand side of (3.17)
strictly positive. Therefore, with R2 ¿ max{R; R1 ; 2 }, one has
+
)≥
J (w + v + 2ej−1
+
)¿0
J (w + v + 2ej−1
(3.18)
+
= R2 .
for every (w; v) ∈ V ⊥ ⊕ Vj and for every 2 ∈ R such that w + v + 2ej−1
⊥
On the other hand, for every w ∈ V and every v ∈ Vj , taking into account (3.3)
+
and j+1
¡ j+ , one has
a
(w + v)2 d x dy
J (w + v) = G∞ (w + v) d x dy +
2
7
7
− (w + v)K(w + v) d x dy
7
+
j+1
j+
(b − a)
a
2
2
2
2
(w + v ) + (w + v ) −
v2
≥
w2 −
2
2
2
2
1
≥ (b − j+ )[w2 + v2 ]:
2
Therefore, for every w + v ∈ V ⊥ ⊕ Vj with 0 ≤ w2 + v2 ≤ R22 , one gets
1
1
(3.19)
(b − j+ )[w2 + v2 ] ≥ (b − j+ )R22 :
2
2
From (3.19), since b − j+ 0 when b j+ , it is possible to ;x a value b∗ ∈
+
; j+ ) such that for every b ∈ (b∗ ; j+ ) and for every w + v ∈ V ⊥ ⊕ Vj , with
(j+1
w + v ≤ R2 , one has
J (w + v) ≥
1 +
+
)22 :
( − j−1
6 j
From (3.16), (3.18) and (3.20) one gets
J (w + v) ≥
sup
@B2 (V Vj )
J (v) ≤
(3.20)
1 +
1
+
+
)22 ¡ (j+ − j−1
)22 ≤
inf + J (v);
(j − j−1
4
6
@QR1 (V ⊥ ⊕Vj ;ej−1
)
and hence the result.
Remark 3.5. Inequality (3.15) gives the third inequality in the hypotheses of Theorem
+
.
1.2 with the choices H = L2 (7); Z = V; W = V ⊥ ; Z1 = Vj ; e1 ∈ Vj and e2 = ej−1
Moreover, the fourth inequality in the hypotheses of Theorem 1.2 follows from the
estimates of J from below that lead to inequalities (3.17) – (3.19).
600
D. Lupo, K.R. Payne / Nonlinear Analysis 46 (2001) 591 – 600
+
+
Proof of Theorem B. Suppose that, for some j ≥ 2; j+1
¡ b ¡ j+ ¡ j−1
¡ a and
−1
that f satis;es (f1) – (f3). Then g = f satis;es (g1) – (g3); hence, the functional J
satis;es, by Lemmas 3.1, 3.2 and 3.4 and Remarks 3.3 and 3.5, the hypotheses of
Theorem 1.2. Therefore, there will exist two distinct nontrivial critical points v1 and v2
1
for i = 1; 2 are distinct nontrivial generalized solutions
of J . Hence ui = Kvi ∈ WAC∪
to (NST).
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