Electronic Journal of Differential Equations, Vol. 2012 (2012), No. 236, pp. 1–16.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
MULTIPLICITY OF POSITIVE SOLUTIONS FOR A GRADIENT
SYSTEM WITH AN EXPONENTIAL NONLINEARITY
NASREDDINE MEGREZ, K. SREENADH, BRAHIM KHALDI
Abstract. In this article, we consider the problem
−∆u = λuq + f1 (u, v)
in Ω
−∆v = λv q + f2 (u, v)
in Ω
u, v > 0
u=v=0
in Ω
on ∂Ω,
where Ω is a bounded domain in R2 , 0 < q < 1, and λ > 0. We show that
there exists a real number Λ such that the above problem admits at least two
solutions for λ ∈ (0, Λ), and no solution for λ > Λ.
1. Introduction
In this article, we study the existence of multiple solutions of the system of
partial differential equations
−∆u = λuq + f1 (u, v)
in Ω
q
in Ω
−∆v = λv + f2 (u, v)
u, v > 0
u=v=0
(1.1)
in Ω
on ∂Ω,
2
where Ω is a bounded domain in R , 0 < q < 1, λ > 0, and fi , i = 1, 2 satisfy the
following conditions:
(H1) fi ∈ C 1 (R2 ), fi (t, s) > 0 for t > 0 and s > 0; fi (t, s) = 0 if t ≤ 0 or s ≤ 0.
(H2) The maps t 7→ fi (t, .) and s 7→ fi (., s) are nondecreasing for all t > 0, s > 0.
(H3) For all > 0,
lim
t2 +s2 →∞
2
fi (t, s)e−(1−)(t
+s2 )
= ∞,
lim
t2 +s2 →∞
2
fi (t, s)e−(1+)(t
+s2 )
= 0.
(H4) There exists λ such that
λtq + f1 (t, s) > λ1 t
and λsq + f2 (t, s) > λ1 s
for all λ > λ and s, t > 0, where λ1 is the first eigenvalue of −∆ on H01 (Ω).
2000 Mathematics Subject Classification. 35J50, 35J57, 35J60.
Key words and phrases. Gradient system; exponential nonlinearity; multiplicity.
c 2012 Texas State University - San Marcos.
Submitted June 16, 2011. Published December 26, 2012.
1
2
N. MEGREZ, K. SREENADH, B. KHALDI
EJDE-2012/236
(H5) Let F (t, s) be a C 2 function such that
∂F
∂F
= f1 (t, s),
= f2 (t, s),
∂t
∂s
F (t, s)
= 0 for some k > 1,
lim
(t,s)→(0,0) tk + sk
F (t, s)
lim
= 0, ∀ > 0.
2
2
t2 +s2 →∞ e(1+)(t +s )
(H6) There exists a constant κ ≥ 0 such that
F (t, s)
=κ
|t|,|s|→+∞ f1 (t, s) + f2 (t, s)
lim
(H7) For every > 0,
∂fi (t, s) −(1+)(t2 +s2 )
∂fi (t, s) −(1−)(t2 +s2 )
e
= ∞, 2 lim
e
= 0.
∂t
∂s
t +s2 →∞
As examples of a function satisfying the above assumptions, we have
(
2
2
(t2 + s2 )et +s if t > 0, s > 0
F (t, s) =
0
otherwise.
lim
t2 +s2 →∞
So
(
2
2
2t(t2 + s2 + 1)et +s
f1 (t, s) =
0
(
2
2
2s(t2 + s2 + 1)et +s
f2 (t, s) =
0
if t > 0, s > 0
otherwise,
if t > 0, s > 0
otherwise.
Starting from the work of Adimurthi [1], there are many results in the scalar case
for problems involving exponential growth, for example [8], [12]. Systems involving
exponential nonlinearities in two dimension have also been studied in [9]. Recently,
lot of interest has been shown for studying the multiplicity of positive solutions with
nonlinearities of sublinear growth at origin. In this direction we mention the works
of Ambrosetti-Brezis-Cerami [3] for higher dimensions, and Prashanth-Sreenadh
[21] in the case of R2 .
Our aim in this article is to generalize the result in [21] to the case of systems.
One of the motivations of this work is that parameter dependent systems with exponential nonlinearities have been recently shown to be very involved in relativistic
Abelian Chern-Simons model with two Higgs particles and two gauge fields, see
[7, 16, 17, 18]. Chern-Simons theories are applied in condensed matter physics,
anyon physics, superconductivity, quantum mechanics, and electro magnetic spin
density, to mention a few. This system may also be applied to heat transfer modeling in a nuclear fuel rod where the nonlinearities f1 and f2 represent the energy
production.
We shall find the weak solutions of the system (Pλ ) in the space
H := H01 (Ω) × H01 (Ω)
endowed with the norm
k(u, v)kH :=
hZ Ω
i1/2
|∇u|2 + |∇v|2 dx
.
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MULTIPLICITY OF POSITIVE SOLUTIONS
3
Throughout this paper, we denote by k · k1,2 the norm of Sobolev space H01 (Ω).
To handle exponential nonlinearity for dimension N = 2, the Moser-Trudinger
inequality [19, 22] plays the same role as the Sobolev imbedding Theorem for the
case of polynomial nonlinearity in dimension N ≥ 3. In this paper, we will use the
following adapted version of Moser-Trudinger inequality for the pair (u, v) [15]:
R
2
2
Lemma 1.1. Let (u, v) ∈ H, then Ω eγ(u +v ) dx < +∞ for any γ > 0. Moreover,
there exists a constant C = C(Ω) such that
Z
2
2
sup
eγ(u +v ) dx ≤ C, provided γ ≤ 4π.
(1.2)
k(u,v)kH =1
Ω
kuk21,2
Proof. Let r1 =
Z
and r2 = kvk21,2 , be such that r1 + r2 = 1. Then, we have
Z
u2
v2
eγ r1 dx < C,
eγ r2 dx < C, for all γ ≤ 4π.
Ω
Ω
Hence, using Hölder inequality, (1.2) can be obtained as follows
Z
Z
Z
v2
u2
2
2
eγ(u +v ) dx ≤ ( eγ r1 dx)r1 ( eγ r2 dx)r2 ≤ C r1 +r2 = C.
Ω
Ω
Ω
It follows from the above inequality that the imbedding
(u, v) ∈ H 7→ e(|u|
α
+|v|α )
∈ L1 (Ω)
is compact for α < 2. Also, it can be shown using a class of functions called the
Moser functions, that the above imbedding is not compact for α = 2.
Weak solutions of (1.1) are the functions u, v ∈ H01 (Ω) such that
Z
Z
Z
∇u.∇φ dx = λ
uq φ dx +
f1 (u, v)φ dx,
Ω
Ω
Ω
Z
Z
Z
∇v.∇ψ dx = λ
v q ψ dx +
f2 (u, v)ψ dx,
Ω
1
H0 (Ω).
Ω
Ω
for all φ, ψ ∈
The main results of this article are given in the following theorem.
Theorem 1.2. There exists Λ > 0 such that (1.1) admits at least two solutions for
all λ ∈ (0, Λ), and no solution for λ > Λ.
Let us write H(u, v) as
H(u, v) =
λ q+1
|u|
+ |v|q+1 + F (u, v),
q+1
and let
∂H
= λuq + f1 (u, v),
∂u
Using (H6), for R sufficiently large
h1 (u, v) =
h2 (u, v) =
H(u, v) ≤ C(h1 (u, v) + h2 (u, v)),
∂H
= λv q + f2 (u, v).
∂v
for |u| > R, and |v| > R.
The functional associated with (1.1) is given by
Z Z
1
2
2
E(u, v) =
|∇u| + |∇v| dx −
H(u, v) dx.
2 Ω
Ω
(1.3)
4
N. MEGREZ, K. SREENADH, B. KHALDI
EJDE-2012/236
It is well defined on H and C 1 (H, R). Also, for all (φ, ψ) ∈ H, we have
Z
Z
Z
Z
E 0 (u, v)(φ, ψ) =
∇u.∇φ dx +
∇v.∇ψ dx −
h1 (u, v)φ dx −
h2 (u, v)ψ dx,
Ω
Ω
Ω
Ω
Our approach is to construct an H 1 local minimum (uλ , vλ ) for E for λ in the largest
interval of existence (0, Λ), and then use the generalized mountain-pass Theorem
of Ghoussoub-Priess [10] about (uλ , vλ ) to obtain a second solution.
2. Existence of a minimal solution and a local minimum for E
In this section, we prove the existence of a minimal solution (uλ , vλ ) of (1.1),
and then we show that this minimal solution is a local minimum for E. A solution
(uλ , vλ ) is said to be minimal if any other solution (u, v) of (1.1) satisfies u ≥ uλ
and v ≥ vλ in Ω.
Lemma 2.1. There exists λ0 > 0 such that (1.1) admits a solution for all λ ∈
(0, λ0 ).
Proof. From assumption (H5), for any > 0, we obtain C > 0, such that
2
2
|F (u, v)| ≤ C |u|k+1 + |v|k+1 e(1+)(u +v )
2π
, and by Hölder’s inequality and the MoserFor k(u, v)kH = ρ such that ρ2 ≤ 1+
Trudinger inequality (1.2), we obtain
Z
F (u, v)dx|
Ω
Z
2
2
≤C
|u|(k+1) + |v|(k+1) e(1+)(u +v ) dx
ZΩ
≤C
|u|(k+1) + |v|(k+1)
Ω
u2
v2
2
2
× exp (1 + )(
+
)(kuk1,2 + kvk1,2 ) dx
kuk21,2 + kvk21,2
kuk21,2 + kvk21,2
1/2 1/2 Z
Z
|v|2(k+1) dx
|u|2(k+1) dx
+
≤C
Ω
Ω
k+1
k+1
≤ C kuk1,2 + kvk1,2 ,
where C is a generic constant. Therefore, for k(u, v)kH = ρ, we have
λ 1
q+1
k+1
kukq+1
E(u, v) ≥ k(u, v)k2H − C kukk+1
−
q+1 + kvkLq+1
1,2 + kvk1,2
L
2
q+1
1
λ q+1
q+1
k+1
≥ k(u, v)k2H − C kukk+1
+
kvk
−
C
kuk
+
C
kvk
1
2
1,2
1,2
1,2
1,2
2
q+1
1
k+1
≥ k(u, v)k2H − C k(u, v)kk+1
H + k(u, v)kH
2
λ q+1
C1 k(u, v)kq+1
−
H + C2 k(u, v)kH
q+1
1 2
≥ ρ − 2Cρk+1 − C̃λρq+1 .
2
Now, we may fix ρ, λ0 > 0 small enough such that E(u, v) > 0 for all λ ∈ (0, λ0 ).
We note that E(tu, tv) < 0 for t > 0 small enough. So, inf k(u,v)kH ≤ρ E(u, v) < 0
EJDE-2012/236
MULTIPLICITY OF POSITIVE SOLUTIONS
5
and if this infimum is achieved at some (uλ , vλ ), then (uλ , vλ ) becomes a solution
of (1.1). Let {(un , vn )} ⊂ {k(un , vn )kH ≤ ρ} be a minimizing sequence and let
(un , vn ) * (uλ , vλ ) in H. Clearly,
k(uλ , vλ )kH ≤ lim inf k(un , vn )kH .
n→∞
Now, we can choose ρ < π so that {F (un , vn )} is bounded in Lr (Ω) for some r > 1.
Using this fact and Holders inequality, it is notR difficult to show that {F (un , vn )}
is equi-integrable family in L1 (Ω) and lim|A|→0 A |F (un , vn )|dx = 0. Therefore, by
Vitali’s convergence Theorem, we obtain
Z
Z
F (un , vn )dx →
F (uλ , vλ )dx.
Ω
Ω
Hence, (uλ , vλ ) is a minimizer of E(u, v). By the maximum principle, we obtain
uλ , vλ > 0 in Ω.
Lemma 2.2. Let Λ := sup{λ : (1.1) admits a solution}. Then 0 < Λ < ∞.
Proof. By Lemma 2.1, it is clear that Λ > 0. Suppose Λ = ∞. Then there exists a
sequence λn → ∞ such that (1.1) with λ = λn has a solution (uλn , vλn ). Hence
Z
Z
Z λ1
uλn φ1 dx =
∇uλn ∇φ1 dx =
λn uqλn + f1 (uλn , vλn ) φ1 dx.
(2.1)
Ω
Ω
Ω
where φ1 is the eigenfunction associated with the first eigenvalue λ1 of −∆ on
H01 (Ω).
Now, we choose λn > λ. By (H4) we have
λn tq + f1 (t, s) > λ1 t,
λn sq + f2 (t, s) > λ1 s.
From (2.1) and (2.2), we obtain
Z
Z
λ1
uλn φ1 dx > λ1
uλn φ1 dx
Ω
(2.2)
(2.3)
Ω
which is absurd. Hence, Λ is finite.
Lemma 2.3. For all λ ∈ (0, Λ), (1.1) admits a solution.
Proof. Suppose λ0 < λ < λ00 < Λ and (1.1) with λ = λ0 , and with λ = λ00 admit
solutions (uλ0 , vλ0 ), (uλ00 , vλ00 ) respectively. Then (uλ0 , vλ0 ) is a subsolution of (1.1),
and (uλ00 , vλ00 ) is a supersolution of (1.1), and hence, by the monotone iterative
procedure, there exists a solution of (1.1).
We recall the following well known comparison Theorem, whose proof can be
found in [20].
Lemma 2.4. Let f : [0, ∞) → [0, ∞) be such that f (t)/t is non-increasing for
t > 0. Let v, w ∈ W01,2 (Ω) be weak sub and super solutions (respectively) of
−∆u = f (u),
u=0
u>0
in Ω
on ∂Ω.
Then w ≥ v a.e. in Ω.
Lemma 2.5. For all λ ∈ (0, Λ), (1.1) admits a minimal solution (uλ , vλ ).
6
N. MEGREZ, K. SREENADH, B. KHALDI
EJDE-2012/236
Proof. Let v0 be the unique solution of the problem
−∆u = λuq ,
u>0
u=0
on ∂Ω
in Ω
(2.4)
Then, (v0 , v0 ) is a subsolution of (1.1). Now, let (u, v) be a solution of (1.1). Then,
u and v are supersolutions of (2.4), and by the above weak comparison Theorem,
u ≥ v0 , and v ≥ v0 . By the monotone iteration procedure with U = (v0 , v0 ), and
U = (u, v) as sub and super-solutions of (1.1), we obtain a solution (uλ , vλ ). It is
easy to see that (uλ , vλ ) is the minimal solution.
Lemma 2.6. (uλ , vλ ) is a local minimum of E in H.
Proof. We use Perron’s method as in [11] and [13]. Arguing by contradiction, let us
suppose that there exists λ ∈ (0, Λ) and a sequence (un , vn ) such that (un , vn ) →
(uλ , vλ ) strongly in H and E(un , vn ) < E(uλ , vλ ).
Let λ < λ0 < Λ and let (uλ0 , vλ0 ) be the minimal solution of (1.1) with λ = λ0 .
Let U = (u, v) = (uλ0 , vλ0 ), and let U = (u, v) = (v0 , v0 ), where v0 is the unique
solution of (2.4). Consider the following cut-off functions:


if un (x) ≤ u(x)
u(x)
y1,n (x) = un (x) if u(x) ≤ un (x) ≤ u(x)


u(x)
if un (x) ≥ u(x)


if vn (x) ≤ v(x)
v(x)
y2,n (x) = vn (x) if v(x) ≤ vn (x) ≤ v(x)


v(x)
if vn (x) ≥ v(x).
Also define
Wn = (w1,n , w2,n ) := ((un − u)+ , (vn − v)+ ),
Zn = (z1,n , z2,n ) := ((un − u)− , (vn − v)− ).
We also define the following subsets:
S1,n = {x ∈ Ω : un (x) < u(x) and vn (x) < v(x)},
S2,n = {x ∈ Ω : un (x) < u(x) and v(x) ≤ vn (x) ≤ v(x)},
S3,n = {x ∈ Ω : un (x) < u(x) and vn (x) > v(x)},
S4,n = {x ∈ Ω : u(x) ≤ un (x) ≤ u(x) and vn (x) < v(x)},
S5,n = {x ∈ Ω : u(x) ≤ un (x) ≤ u(x) and vn (x) > v(x)},
S6,n = {x ∈ Ω : un (x) > u(x) and vn (x) < v(x)},
S7,n = {x ∈ Ω : un (x) > u(x) and v(x) ≤ vn (x) ≤ v(x)},
S8,n = {x ∈ Ω : un (x) > u(x) and vn (x) > v(x)}.
Then,
(un , vn ) = (y1,n , y2,n ) − (z1,n , z2,n ) + (w1,n , w2,n ),
(y1,n , y2,n ) ∈ M := (u, v) ∈ H : u ≤ u ≤ u and v ≤ v ≤ v ,
E(un , vn ) = E(y1,n , y2,n ) +
8
X
i=1
(2.5)
Ai,n ,
EJDE-2012/236
MULTIPLICITY OF POSITIVE SOLUTIONS
7
where
A1,n =
Z
1
|∇un |2 − |∇u|2 dx +
|∇vn |2 − |∇v|2 dx
2 S1,n
S
Z 1,n −
H(un , vn ) − H(u, v) dx,
1
2
Z
S1,n
A2,n =
A3,n
1
2
Z
Z
|∇un |2 − |∇u|2 dx −
H(un , vn ) − H(u, vn ) dx,
S2,n
S2,n
1
=
2
Z
2
2
|∇un | − |∇u|
dx +
S3,n
Z
−
1
2
Z
|∇vn |2 − |∇v|2 dx
S3,n
H(un , vn ) − H(u, v) dx,
S3,n
A4,n
1
=
2
Z
Z
A5,n
1
=
2
A6,n
2
2
|∇vn | − |∇v|
Z
dx −
S4,n
H(un , vn ) − H(un , v) dx,
H(un , vn ) − H(un , v) dx,
S4,n
|∇vn |2 − |∇v|2 dx −
Z
S5,n
S5,n
1
=
2
Z
|∇un |2 − |∇u|2 dx +
S6,n
Z
−
1
2
Z
|∇vn |2 − |∇v|2 dx
S6,n
H(un , vn ) − H(u, v) dx,
S6,n
A7,n =
A8,n
1
2
Z
Z
|∇un |2 − |∇u|2 dx −
S7,n
H(un , vn ) − H(u, vn ) dx,
S7,n
Z
1
|∇un | − |∇u| dx +
|∇vn |2 − |∇v|2 dx
2 S8,n
S
Z 8,n H(un , vn ) − H(u, v) dx.
−
1
=
2
Z
2
2
S8,n
Following Perron’s method as in the proof of [2, Proposition 2.2], one can states
that E(uλ , vλ ) = inf E(u, v), and then concludes that
M
E(un , vn ) ≥ E(uλ , vλ ) +
8
X
Ai,n .
i=1
Now, since (un , vn ) → (uλ , vλ ) strongly in H, u < uλ < u and v < vλ < v in Ω, we
have meas(Si,n )i=1−8 → 0 as n → ∞. Therefore,
kWn kH , kZn kH → 0
as n → ∞.
Using (2.5), mean value Theorem, and (H2), we obtain for some 0 < θ < 1:
8
Z
X
1
Ai,n ≥
kWn k2H + kZn k2H −
∇u∇z1,n dx
2
Ω
i=1
Z
Z
+
h1 (u − θz1,n , v − θz2,n )z1,n dx −
∇v∇z2,n dx
S1,n ∪S2,n ∪S3,n
Ω
Z
Z
h2 (u − θz1,n , v − θz2,n )z2,n dx +
+
S1,n ∪S4,n ∪S6,n
∇u∇w1,n dx
Ω
8
N. MEGREZ, K. SREENADH, B. KHALDI
EJDE-2012/236
Z
Z
−
∇v∇w2,n dx
h1 (u + θw1,n , v + θw2,n )w1,n dx +
S6,n ∪S7,n ∪S8,n
Ω
Z
−
h2 (u + θw1,n , v + θw2,n )w2,n dx.
S3,n ∪S5,n ∪S8,n
Since (u, v)(resp. (u, v)) is a supersolution (resp. subsolution ) of (1.1),
8
X
Ai,n ≥
i=1
1
kWn k2H + kZn k2H
2
Z
h1 (u − θz1,n , v − θz2,n ) − h1 (u, v) z1,n dx
+
S1,n ∪S2,n ∪S3,n
Z
+
h2 (u − θz1,n , v − θz2,n ) − h2 (u, v) z2,n dx
h1 (u + θw1,n , v + θw2,n ) − h1 (u, v) w1,n dx
h2 (u + θw1,n , v + θw2,n ) − h2 (u, v) w2,n dx
S1,n ∪S4,n ∪S6,n
Z
−
S6,n ∪S7,n ∪S8,n
Z
−
S3,n ∪S5,n ∪S8,n
≥
1
2
Z
kWn k2H + kZn k2H −
∂h
S1,n ∪S2,n ∪S3,n
1
∂t
(u − θ0 z1,n , v − θ0 z2,n )
∂h1
2
(u − θ0 z1,n , v − θ0 z2,n ) z1,n
dx
+
Z∂s
∂h
2
−
(u − θ0 z1,n , v − θ0 z2,n )
∂t
S1,n ∪S4,n ∪S6,n
∂h2
2
(u − θ0 z1,n , v − θ0 z2,n ) z2,n
dx
−
∂s
Z
∂h
1
−
(u + θ0 w1,n , v + θ0 w2,n )
∂t
S6,n ∪S7,n ∪S8,n
∂h1
2
+
dx
(u + θ0 w1,n , v + θ0 w2,n ) w1,n
∂t
Z
∂h
2
−
(u + θ0 w1,n , v + θ0 w2,n )
∂t
S3,n ∪S5,n ∪S8,n
∂h2
2
+
(u + θ0 w1,n , v + θ0 w2,n ) w2,n
dx.
∂s
It follows from (H7), (1.2), Hölder’s and Sobolev’s inequalities that for n sufficiently
large,
8
X
i=1
Ai,n ≥
1
kWn k2H + kZn k2H
2
Z
2
2
2
dx
− C1
e(1+ε)(u +v ) z1,n
S1,n ∪S2,n ∪S3,n
Z
− C2
e(1+ε)(u
2
+v 2 ) 2
z2,n dx
S1,n ∪S4,n ∪S6 ,n
Z
− C3
S6,n ∪S7,n ∪S8,n
e(1+ε)((u+w1,n )
2
+(v+w2,n )2 )
2
w1,n
dx
EJDE-2012/236
MULTIPLICITY OF POSITIVE SOLUTIONS
Z
e(1+ε)((u+w1,n )
− C4
2
+(v+w2,n )2 )
9
2
w2,n
dx
S3,n ∪S5,n ∪S8,n
1
kWn k2H + kZn k2H − o(1) kWn k2H + kZn k2H .
2
Hence E(un , vn ) ≥ E(uλ , vλ ) which is a contradiction.
≥
3. Existence of a second solution
Throughout this section, we fix λ ∈ (0, Λ) and we denote by (uλ , vλ ) the local
minimum of E obtained in the previous section as the minimal solution of (1.1).
Using min-max methods and Mountain pass lemma around a closed set, we prove
the existence of a second solution (uλ , v λ ) of (1.1) such that uλ ≥ uλ and v λ ≥ vλ
in Ω. Let T = {(u, v) : u ≥ uλ , v ≥ vλ a.e. in Ω}.
We note that limt→+∞ E(uλ +tu, vλ +tv) = −∞ for any (u, v) ∈ H \{0}. Hence,
we may fix (ũ, ṽ) ∈ H \{0} such that E(uλ + ũ, vλ + ṽ) < 0. We define the mountain
pass level
ρ0 = inf sup E(γ(t)),
(3.1)
γ∈Γ t∈[0,1]
where Γ = {γ : [0, 1] → H : γ ∈ C, γ(0) = (uλ , vλ ), and γ(1) = (uλ + ũ, vλ + ṽ)}.
It follows that ρ0 ≥ E(uλ , vλ ). If ρ0 = E(uλ , vλ ), we obtain that inf{E(u, v) :
k(u, v) − (uλ , vλ )kH = R} = E(uλ , vλ ) for all R ∈ (0, R0 ) for some R0 small.
We now let F = T if ρ0 > E(uλ , vλ ), and F = T ∩ {k(u − uλ , v − vλ )kH = R20 }
if ρ0 = E(uλ , vλ ). We have the following upper bound on ρ0 .
Lemma 3.1. With ρ0 defined as in (3.1), we have ρ0 < E(uλ , vλ ) + 2π.
Proof. Without loss of generality, we assume that 0 ∈ Ω. Define the sequence
 1
1/2
if 0 ≤ |x| ≤ n1

 2√π (log n)
ψ̃n (x) = 2√1 π log(1/|x|)
if n1 ≤ |x| ≤ 1
(log n)1/2


0
if|x| ≥ 1
Now, consider (ψ̃n , ψ̃n ) ∈ H. Then k(ψ̃n , ψ̃n )kH = 1. We now choose δ > 0 such
that Bδ (0) ⊂ Ω and let ψn (x) = ψ̃n ( xδ ). Then, ψn has support in Bδ (0) and
(ψn , ψn ) is such that k(ψn , ψn )kH = 1 for all n. Now, suppose ρ0 ≥ E(uλ , vλ ) + 2π
and we derive a contradiction. This means that for some tn , sn > 0:
E(uλ + tn ψn , vλ + sn ψn ) = sup E(uλ +tψn , vλ +sψn ) ≥ E(uλ , vλ ) + 2π,
∀n.
t,s>0
Since E(uλ + tu, vλ + sv) → −∞ as t, s → +∞, we obtain that (tn , sn ) is bounded
in R2 . Then, using k(ψn , ψn )kH = 1, we obtain
Z t2n + s2n
+
tn ∇uλ ∇ψn + sn ∇vλ ∇ψn dx
4
Ω
Z ≥
H(uλ + tn ψn , vλ + sn ψn ) − H(uλ , vλ ) dx + 2π
Ω
Now, using the fact that (uλ , vλ ) is a solution, we obtain
Z h
t2n + s2n
≥
H(uλ + tn ψn , vλ + sn ψn )
4
Ω
i
− H(uλ , vλ ) − ψn tn h1 (uλ , vλ ) + sn h2 (uλ , vλ ) dx + 2π.
(3.2)
10
N. MEGREZ, K. SREENADH, B. KHALDI
EJDE-2012/236
Using (H2) we have that h1 , h2 are non-decreasing, then there exist θn ∈ (0, 1)
such that
Z h
i
H(uλ + tn ψn , vλ + sn ψn ) − H(uλ , vλ ) − ψn tn h1 (uλ , vλ ) + sn h2 (uλ , vλ ) dx
Ω
Z
h ∂h
∂h2
1
=
(uλ + θn tn ψn , vλ + θn sn ψn ) + s2n
(uλ + θn tn ψn , vλ + θn sn ψn )
ψn2 t2n
∂u
∂v
Ω
i
∂h1
∂h2
+ tn sn
(uλ + θn tn ψn , vλ + θn sn ψn ) + tn sn
(uλ + θn tn ψn , vλ + θn sn ψn ) dx
∂v
∂u
≥ 0.
Now, by (3.2), we see that
t2n + s2n ≥ 8π,
for all n
(3.3)
Since (tn , sn ) is a critical point of E(uλ + tψ, vλ + sψ), we obtain
E 0 (uλ + tψn , vλ + sψn )|(t,s)=(tn ,sn ) = 0.
Then
t2n
+
s2n
=
Z h
h1 (uλ + tn ψn , vλ + sn ψn ) − h1 (uλ , vλ ) tn
Ω
i
+ h2 (uλ + tn ψn , vλ + sn ψn ) − h2 (uλ , vλ ) sn ψn dx.
Since tn ψn → ∞, sn ψn → ∞ on {|x| ≤ δ/n}, we obtain
Z
2
2
2
2
2
tn + sn ≥
e(tn +sn )ψn (tn + sn )ψn dx
Ω∩{|x|≤δ/n}
πδ 2 (t2n +s2n ) log n
4π (t
e
n
2n2
√
=
√
=
+ sn )(log n)1/2
(3.4)
πδ 2 ( t2n +s2n −2) log n
e 4π
(tn + sn )(log n)1/2
2
This and (3.3) imply that
t2n + s2n → 8π,
(3.5)
and by (3.4) we obtain
t2n + s2n ≥ (tn + sn )(log n)1/2 .
This in turn implies that t2n + s2n → ∞ as n → ∞, which contradicts (3.5).
Definition 3.2. Let F ⊂ H be a closed set. We say that a sequence (un , vn ) ⊂ H
is a Palais-Smale sequence for E at level ρ around F, and we denote (P S)F ,ρ , if
lim dist (un , vn ), F = 0,
lim E(un , vn ) = ρ,
n→+∞
n→+∞
0
lim kE (un , vn )kH−1 = 0.
n→+∞
Lemma 3.3. Let F ⊂ H be a closed set and ρ ∈ R. Let {(un , vn )} ⊂ H be
a (P S)F ,ρ sequence. Then there exists (u0 , v0 ) such that, up to a subsequence,
un * u0 and vn * v0 in H01 (Ω), and
Z
Z
lim
h1 (un , vn )dx =
h1 (u0 , v0 )dx,
n→∞ Ω
Z
ZΩ
lim
h2 (un , vn )dx =
h2 (u0 , v0 )dx
n→∞
Ω
Ω
EJDE-2012/236
MULTIPLICITY OF POSITIVE SOLUTIONS
Proof. We have the following relations as n → +∞
Z
Z
Z
1
1
|∇un |2 dx +
|∇vn |2 dx −
H(un , vn )dx = ρ + on (1)
2 Ω
2 Ω
Ω
Z
Z
∇un .∇ϕ dx −
h1 (un , vn )ϕ dx ≤ on (1)kϕk, ∀ϕ ∈ H01 (Ω)
Ω
Ω
Z
Z
∇vn .∇ϕ dx −
h2 (un , vn )ϕ dx ≤ on (1)kϕk, ∀ϕ ∈ H01 (Ω)
Ω
11
(3.6)
(3.7)
(3.8)
Ω
Step 1: We claim that
sup kun kH01 (Ω) + kvn kH01 (Ω) < +∞,
n
Z
h1 (un , vn )un dx < +∞,
sup
n
Ω
Z
sup
h2 (un , vn )vn dx < +∞.
n
Ω
We note that for all ε > 0, there exists sε such that
h(t, s) ≤ ε h1 (t, s)t + h2 (t, s)s , for |s|, |t| ≥ sε .
From (3.6), we obtain
Z Z 1
|∇un |2 + |∇vn |2 dx ≤ Cε + ε
h1 (un , vn )un + h2 (un , vn )vn dx
2 Ω
Ω
(3.9)
From (3.7) with ϕ = un , and (3.8) with ϕ = vn , we obtain
Z
Z
h1 (un , vn )un dx ≤
|∇un |2 dx + o(1)kun kH01 (Ω) ,
Ω
Ω
Z
Z
h2 (un , vn )vn dx ≤
|∇vn |2 dx + o(1)kvn kH01 (Ω) .
Ω
Ω
From (3.9) we obtain
Z h1 (un , vn )un + h2 (un , vn )vn dx
Ω
Z ≤ 2Cε + 2ε
h1 (un , vn )un + h2 (un , vn )vn dx + o(1)
Ω
2Cε
≤
+ o(1) kun k + kvn k .
1 − 2ε
Substituting this in (3.9), we obtain
sup kun kH01 (Ω) + kvn kH01 (Ω) < +∞,
n
which implies
Z
n
Z
h1 (un , vn )un dx < ∞,
sup
h2 (un , vn )vn dx < ∞.
sup
n
Ω
Ω
Step 2: We claim that
Z
lim
n→+∞
ZΩ
lim
n→+∞
Z
h1 (un , vn )dx =
h2 (un , vn )dx =
Ω
h1 (u0 , v0 )dx,
(3.10)
h2 (u0 , v0 )dx,
(3.11)
ZΩ
Ω
12
N. MEGREZ, K. SREENADH, B. KHALDI
Z
Z
lim
H(un , vn )dx =
n→+∞
EJDE-2012/236
H(u0 , v0 )dx
Ω
(3.12)
Ω
Let |A| denote the Lebesgue measure of A ⊂ R2 . We show that {h1 (un , vn )} and
{h2 (un , vn )} are equi-integrable in L1 , and then, (3.10) and (3.11) follow from Vitali’s convergence Theorem. (3.12) follows from (1.3) and the Lebesgue dominated
convergence Theorem. We claim that for all ε > 0, there exists δ > 0 such that for
any A ⊂ Ω with |A| < δ, we have
Z
sup
|h2 (un , vn )|dx ≤ ε.
n
A
R
Let C1 = supn A |h2 (un , vn )vn |dx. By step 1, C1 < +∞. Since {un } and {vn } are
bounded in H01 , by (3.7) and (3.8) we have
Z
Z
|h2 (un , vn )un |dx < ∞,
|h1 (un , vn )vn |dx < ∞.
Ω
Let C2 = supn
R
Ω
n
o
|h
(u
,
v
)u
|dx,
and
let
C
=
max
C
,
C
. Define
2
n
n
n
1
2
A
n
o
|h2 (t, s)| .
µε =
max
3C
|t|≤ 3C
ε , |s|≤ ε
Then, for any A ⊂ Ω with |A| ≤ 3µε ε , we obtain
Z
|h2 (un , vn )| dx
A
Z
Z
≤
|h2 (un , vn )| dx +
A∩{|un |, |vn |≤ 3C
ε }
Z
+
A∩{|un |≥ 3C
ε
}
A∩{|vn |≥ 3C
ε }
|h2 (un , vn )vn |
dx
vn
|h2 (un , vn )un |
dx
un
ε ε
≤ |A|µε + + ≤ ε.
3 3
which shows the equi-integrability of {h2 (un , vn )}. In a similar way, we can show
the equi-integrability of {h1 (un , vn )}. This completes step 2 and the proof of
Lemma 3.3.
We will also use the following version of Lion’s Lemma [14].
Lemma 3.4. Let {(un , vn )} be a sequence in H such that k(un , vn )kH = 1, for
all n and un * u, vn * v in H01 for some (u, v) 6= (0, 0). Then, for 4π < p <
4π(1 − kuk21,2 − kvk21,2 )−1 ,
Z
2
2
sup
ep(un +vn ) dx < ∞
n≥1
Ω
Proof. It is easy to see that
lim k(un − u, vn − v)k2H = lim kun − uk21,2 + kvn − vk21,2 = 1 − kuk21,2 − kvk21,2 ,
n→∞
n→∞
u2n ≤ (un − u)2 + 2u2n + C u2
Then
Z
2
)
p(u2n +vn
e
Ω
Z
dx ≤
Ω
ep((un −u)
2
for small.
2
+(vn −v)2 ) p(u2n +vn
) (u2 +v 2 )C
e
e
dx.
EJDE-2012/236
MULTIPLICITY OF POSITIVE SOLUTIONS
13
Now, using Hölder’s inequality with r1 , r2 , r3 such that r11 + r12 + r13 = 1, we obtain
Z
Z
1/r1 Z
1/r2
2
2
2
2
2
2
ep(un +vn ) dx ≤
epr1 ((un −u) +(vn −v) ) dx
epr2 (un +vn ) dx
Ω
Ω
Z
1/r3
2
2
×
epr3 (u +v )C dx
.
The second and third integrals are finite for small and using inequality (1.2). For
the first integral we have
Z
2
2
epr1 ((un −u) +(vn −v) ) dx
Ω
Z
vn −v
un −u
)2 +( k(u −u,v
)2 k(un −u,vn −v)k2H
pr (
n
n −v)kH
dx.
=
e 1 k(un −u,vn −v)kH
Ω
We can choose r1 > 1 and close to 1 such that pr1 (1 − kuk21,2 − kvk21,2 ) < 4π by
the hypothesis and the first equation of this proof. Hence, this is bounded again
thanks to inequality (1.2).
Now, we prove our main result.
Theorem 3.5. For λ ∈ (0, Λ), problem (1.1) has a second nontrivial solution
(uλ , v λ ) such that uλ ≥ uλ > 0 and v λ ≥ vλ > 0 in Ω.
Proof. Let {(un , vn )} be a Palais-Smale sequence for E at the level ρ0 around F.
Existence of a such sequence can be obtained using Ekeland Variational principle
on F ([2, 10]). Then, by Lemma 3.3, there exist (uλ , v λ ) and a subsequence denoted
again by (un , vn ), such that un * uλ and vn * v λ in H01 (Ω). It is easy to verify
that (uλ , v λ ) is a solution of (1.1).
It remains to show that (uλ , v λ ) 6≡ (uλ , vλ ). We suppose that uλ ≡ uλ and
v λ ≡ vλ and we derive a contradiction:
Case 1: ρ0 = E(uλ , vλ ). In this case, we recall that
R0
},
F = {(u, v) ∈ T : k(u − uλ , v − vλ )kH =
2
Z
Z
Z
1
1
E(uλ , vλ ) + o(1) = E(un , vn ) =
|∇un |2 dx +
|∇vn |2 dx −
H(un , vn )dx.
2 Ω
2 Ω
Ω
R
R
From Lemma 3.3, (equation (3.8)) we have Ω H(un , vn )dx → Ω H(uλ , vλ ). Thus,
k(un − uλ , vn − vλ )kH = o(1), which contradicts the fact that (un , vn ) ∈ F.
Case 2: ρR0 6= E(uλ , vλ ). In this case ρ0 − E(uλ , vλ ) ∈ (0, 2π) and E(un , vn ) → ρ0 .
Let β0 = Ω H(uλ , vλ )dx. Then from Lemma 3.3,
Z
Z
1
1
2
|∇un | dx +
|∇vn |2 dx → (ρ0 + β0 ) as n → ∞
(3.13)
2 Ω
2 Ω
Also, by Fatou’s lemma we have that E(uλ , vλ ) ≤ lim inf n→+∞ E(un , vn ). If
{(un , vn )} does not converge strongly in H, then E(uλ , vλ ) < ρ0 . By Lemma
3.1, for small, we have
(1 + )(ρ0 − E(uλ , vλ )) < 2π.
Hence, from (3.13) we have
(1 + )k(un , vn )k2H < 4π
ρ0 + β0
ρ0 − E(uλ , vλ )
14
N. MEGREZ, K. SREENADH, B. KHALDI
EJDE-2012/236
ρ0 + β0
ρ0 + β0 − 21 k(uλ , vλ )k2H
1 k(uλ , vλ )k2H −1
< 4π 1 − (
)
2
ρ0 + β0
−1
vλ
uλ
k1,2 − k p
k1,2
.
< 4π 1 − k p
2(ρ0 + β0 )
2(ρ0 + β0 )
< 4π
Now, choose p > 4π such that
uλ
vλ
(1 + )k(un , vn )k2H ≤ p < 4π(1 − k p
k1,2 − k p
k1,2 )−1 .
2(ρ0 + β0 )
2(ρ0 + β0 )
Since
un
k(un ,vn )kH
* √
uλ
2(ρ0 +β0 )
Lemma 3.4, we have
Z
sup
exp p
n
Ω
and
vn
k(un ,vn )kH
* √
vλ
2(ρ0 +β0 )
weakly in H01 (Ω), by
2
2 vn
un
+
dx < ∞
k(un , vn )kH
k(un , vn )kH
(3.14)
From the definition of h1 , for any δ > 0, there exists a constant C > 0 such that
2
2
sup h1 (un , vn ) ≤ Ce(1+δ)(un +vn ) .
n
Now, it is not difficult to show that h1 (un , vn ) ∈ Lq (Ω) for some q > 1. Indeed,
taking δ close to zero and q close to 1 such that q(1 + δ) < 1 + ,
Z
Z
q
2
2
h1 (un , vn ) dx ≤ C
eq(1+δ)(un +vn ) dx
Ω
Ω
h
2 2 i
Z
un
q(1+δ)
+ k(u v,vn )k
k(un ,vn )k2H
k(un ,vn )kH
n n H
≤C
e
dx
Ω
h
2 2 i
Z
un
p
+ k(u v,vn )k
k(un ,vn )kH
n n H
≤C
e
dx
Ω
Now, using (3.14), we obtain that h1 ∈ Lq (Ω). So, by Hölder inequality we have
and the assumption that uλ = uλ , v λ = vλ , we have
Z
Z
h1 (un , vn )un dx −→
h1 (uλ , vλ ) as n → ∞.
Ω
Ω
Hence,
Z
Z
1
2
o(1) = E (un , vn )(un , 0) =
|∇un | dx −
h1 (un , vn )un dx
2 Ω
ZΩ
Z
1
=
|∇un |2 dx −
|∇uλ |2 + o(1).
2 Ω
Ω
R
R
2
2
Similarly, we obtain Ω |∇vn | dx = Ω |∇vλ | + o(1). This is a contradiction to the
assumption that ρ0 6= E(uλ , vλ ).
0
Acknowledgements. The authors would like to thank the anonymous referees for
theirs comments an suggestions. Nasreddine Megrez would like to thank Professor
Fraser Forbes for interesting discussions about the applicability of this problem to
heat transfer.
EJDE-2012/236
MULTIPLICITY OF POSITIVE SOLUTIONS
15
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Nasreddine Megrez
Chemical and Material Engineering Department, University of Alberta, 9107 - 116
Street, Edmonton (Alberta) T6G 2V4, Canada
E-mail address: nmegrez@gmail.com
16
N. MEGREZ, K. SREENADH, B. KHALDI
EJDE-2012/236
K. Sreenadh
Department of Mathematics, Indian Institute of Technology Delhi Hauz Khaz, New
Delhi-16, India
E-mail address: sreenadh@gmail.com
Brahim Khaldi
Departement of Sciences, University of Bechar, PB 117, Bechar 08000, Algeria
E-mail address: khaldibra@yahoo.fr