Mathematical and Computer Modelling 53 (2011) 1154–1161
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Existence of solutions for nonlinear impulsive differential equations with Dirichlet boundary conditions✩ Dan Zhang, Binxiang Dai ∗ Department of Mathematics, Central South University, Changsha, 410075 Hunan, People’s Republic of China
article
abstract
info
Article history: Received 19 June 2010 Received in revised form 11 October 2010 Accepted 24 November 2010
In this paper, we consider the existence of solutions for a class of nonlinear impulsive problems with Dirichlet boundary conditions. We obtain some new existence theorems of solutions for the nonlinear impulsive problem by using critical point theory. We extend and improve some recent results. © 2010 Elsevier Ltd. All rights reserved.
Keywords: Impulsive differential equations Critical point theory Dirichlet boundary conditions
1. Introduction In this paper, we will consider the following problem:
′′ −u (t ) + g (t )u(t ) = f (t , u(t )), a.e. t ∈ [0, T ], 1u′ (tj ) = u′ (t + ) − u′ (tj− ) = Ij (u(tj )), j = 1, 2, . . . , p, u(0) = u(T ) =j 0,
(1.1)
where T > 0, f : [0, T ] × R → R is continuous, g ∈ L∞ [0, T ], p is a positive integer, 0 = t0 < t1 < t2 < · · · < tp < tp+1 = T , 1u′ (tj ) = u′ (tj+ ) − u′ (tj− ) = limt →t + u′ (t ) − limt →t − u′ (t ), Ij : R → R are continuous. j
j
Impulsive differential equations serve as basic models to study the dynamics of processes that are subject to sudden changes in their states. Recent development in this field has been motivated by many applied problems, such as control theory [1,2], population dynamics [3] and medicine [4–6]. For the general aspects of impulsive differential equations, we refer the reader to the classical monograph [7]. Some classical tools have been used to study such problems in the literature. These classical techniques include the coincidence degree theory of Mawhin [8], the method of upper and lower solutions [9,10], and some fixed point theorems [11,12]. For some general and recent works on the theory of impulsive differential equations, we refer the interested reader to [13–17]. A new approach via critical point and variational methods is proved to be very effective in studying the boundary problem for differential equations. For some general and recent works on the theory of critical point theory and variational methods we refer the reader to [18–30]. For a second order differential equation u′′ = f (t , u, u′ ), one usually considers impulses in the position u and the velocity ′ u . However, in the motion of spacecraft one has to consider instantaneous impulses depending on the position that result in jump discontinuities in velocity, but with no change in position [31–34].
✩ Supported by the National Natural Science Foundation of China (No. 10971229).
∗
Corresponding author. Tel.: +86 13117462355. E-mail addresses:
[email protected] (D. Zhang),
[email protected] (B. Dai).
0895-7177/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.mcm.2010.11.082
D. Zhang, B. Dai / Mathematical and Computer Modelling 53 (2011) 1154–1161
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More precisely, Zhou and Li [20] studied problem (1.1). They obtained the existence of infinitely many solutions for problem (1.1) under the assumption that the nonlinearity f is a superlinear case. The result is as follows. Theorem 1.1 ([20, Theorem 3.3]). Assume that the following conditions are satisfied. (h1) There exist aj , bj > 0 and γj ∈ [0, 1) such that |Ij (u)| ≤ aj + bj |u|γj for every u ∈ R, j = 1, 2, . . . , p. (h2) f (t , u) is odd in u. (h3) There exist r1 , r2 > 0 and α ∈ (1, +∞) such that f (t , u) ≤ r1 + r2 |u|α for every (t , u) ∈ [0, T ] × R. (h4) There exist R > 0 and θ > 2 such that 0 < θ F (t , u) ≤ uf (t , u) for every t ∈ [0, T ] and u ∈ R with |u| ≥ R, here u F (t , u) = 0 f (t , s)ds. Moreover, assume that f (t , u) = o(u) as u → 0 uniformly in t. (h5) Ij (j = 1, 2, . . . , p) are odd and nondecreasing. Then the problem (1.1) has an infinite number of nontrivial solutions. Soon after, Sun and Chen [21] extended the results in [20]. They obtained infinitely many solutions for problem (1.1) under the assumption that the nonlinearity f is a superlinear case or an asymptotically linear case. In order to obtain the existence of solutions for the nonlinear problem (1.1), in [20,21], the impulsive functions Ij are required to satisfy the sublinear condition (h1) or the corresponding sublinear condition. But so far, the case that the impulsive functions Ij are superlinear has been considered only by a few authors. Motivated by the above fact, in this paper, our aim is to revisit problem (1.1) and study the case that the impulsive functions Ij are superlinear when the nonlinearity f is a superlinear case or a sublinear case. We will use some critical theorems to obtain the existence and multiplicity of solutions for problem (1.1). Our main results extend the existing study. 2. Variational structure In the section, we first introduce some notations and some necessary definitions. Definition 2.1. Let E be a Banach space and ϕ : E → (−∞, +∞). ϕ is said to be sequentially weakly lower semi-continuous if limk→+∞ inf ϕ(xk ) ≥ ϕ(x) as xk ⇀ x in E. Definition 2.2 ([24, P 81]). Let E be a real reflexive Banach space. For any sequence {uk } ⊂ E, if {ϕ(uk )} is bounded and ϕ ′ (uk ) → 0 as k → 0 possesses a convergent subsequence, then we say ϕ satisfies the Palais–Smale condition (denoted by the PS condition for short). Theorem 2.1 ([25, Theorem 9.12]). Let E be an infinite dimensional real Banach space. Let ϕ ∈ C 1 (E , R) be an even functional which satisfies the PS condition, and ϕ(0) = 0. Suppose that E = V X , where V is infinite dimensional, and ϕ satisfies that (i) there exist α > 0 and ρ > 0 such that ϕ(u) ≥ α for all u ∈ X with ‖u‖ = ρ , (ii) for any finite dimensional subspace W ⊂ E there is R = R(W ) such that ϕ(u) ≤ 0 on W \ BR(W ) . Then ϕ possesses an unbounded sequence of critical values.
In the Sobolev space H01 (0, T ), consider the inner product
(u, v)H 1 (0,T ) =
∫
0
T
u′ (t )v ′ (t )dt ,
0
which induces the norm T
∫ ‖u‖H 1 (0,T ) = 0
(u (t )) dt ′
2
1/2
.
0
It is a consequence of Poincaré’s inequality that T
∫
(u(t ))2 dt 0
1/2
∫
1
≤ √ λ1
T
(u′ (t ))2 dt
1/2
.
0
2 Here, λ1 = πT 2 is the first eigenvalue of the Dirichlet problem
−u′′ (t ) = λu(t ), t ∈ [0, T ], u(0) = u(T ) = 0.
(2.1)
In this paper, we will assume that ess inft ∈[0,T ] g (t ) = m > −λ1 . We can also define the inner product
(u, v) =
T
∫
u′ (t )v ′ (t )dt +
T
∫
0
g (t )u(t )v(t )dt ,
0
which induces the equivalent norm T
∫
(u′ (t ))2 dt +
‖ u‖ = 0
T
∫
g (t )(u(t ))2 dt
1/2
.
0
Lemma 2.1 ([20, Lemma 2.1]). If ess inft ∈[0,T ] g (t ) = m > −λ1 , then the norm ‖ · ‖ and the norm ‖ · ‖H 1 (0,T ) are equivalent. 0
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D. Zhang, B. Dai / Mathematical and Computer Modelling 53 (2011) 1154–1161
Lemma 2.2 ([20, Lemma 2.2]). There exists C > 0 such that if u ∈ H01 (0, T ), then
‖u‖∞ ≤ C ‖u‖,
(2.2)
where ‖u‖∞ = maxt ∈[0,T ] |u(t )|. For u ∈ H 2 (0, T ), we have that u and u′ are both absolutely continuous, and u′′ ∈ L2 (0, T ), hence, 1u′ (tj ) = u′ (tj+ )−u′ (tj− )
for any t ∈ [0, T ]. If u ∈ H01 (0, T ), then u is absolutely continuous and u′ ∈ L2 (0, T ). In this case, the one-side derivatives u′ (tj+ ) and u′ (tj− ) may not exist. As a consequence, we need to introduce a different concept of solution. Suppose that u ∈ C [0, T ] satisfies the Dirichlet condition u(0) = u(T ) = 0. Assume that, for every j = 0, 1, . . . , p, uj = u|(tj ,tj+1 ) and uj ∈ H 2 (tj , tj+1 ). Let 0 = t0 < t1 < t2 < · · · < tp < tp+1 = T . Taking v ∈ H01 (0, T ) and multiplying the two sides of the equality
−u′′ (t ) + g (t )u(t ) − f (t , u(t )) = 0 by v and integrating between 0 and T , we have T
∫
[−u′′ (t ) + g (t )u(t ) − f (t , u(t ))]v(t )dt = 0.
(2.3)
0
Moreover, combining u′ (0) − u′ (T ) = 0, one has T
∫
u′′ (t )v(t )dt = −
− 0
p ∫ −
=−
u′′ (t )v(t )dt
tj
j=0 p −
tj+1
t
−
j+1 u (t )v(t ) | + + t
′
j
j=0
=− −
p −
T
∫
u′ (t )v ′ (t )dt 0
1u (tj )v(tj ) − u (0)v(0) + u (T )v(T ) + ′
′
′
p −
1u′ (tj )v(tj ) +
=
T
∫
u′ (t )v ′ (t )dt 0
j=1 p −
u′ (t )v ′ (t )dt 0
j=1
=
T
∫
Ij (u(tj ))v(tj ) +
T
∫
u′ (t )v ′ (t )dt . 0
j=1
Combining (2.3), we get T
∫
u (t )v (t )dt + ′
′
T
∫
0
g (t )u(t )v(t )dt −
T
∫
0
f (t , u(t ))v(t )dt + 0
p −
Ij (u(tj ))v(tj ) = 0.
j =1
Definition 2.3. A function u ∈ H01 (0, T ) is said to be a weak solution of (1.1), if u satisfies T
∫
u′ (t )v ′ (t )dt +
T
∫
0
g (t )u(t )v(t )dt − 0
T
∫
f (t , u(t ))v(t )dt + 0
p −
Ij (u(tj ))v(tj ) = 0,
j =1
for any v ∈ H01 (0, T ). Consider ϕ : H01 (0, T ) → R defined by
ϕ(u) =
1 2
‖ u‖ 2 −
T
∫
F (t , u(t ))dt + 0
p ∫ − j =1
u(tj )
Ij (s)ds,
(2.4)
0
u
where F (t , u) = 0 f (t , s)ds. Using the continuity of f and Ij , j = 1, 2, . . . , p, we obtain the continuity and differentiability of ϕ and ϕ ∈ C 1 (H01 (0, T ), R). For any v ∈ H01 (0, T ), one has
ϕ ′ (u)v =
T
∫
u′ (t )v ′ (t )dt + 0
∫ 0
T
g (t )u(t )v(t )dt −
T
∫
f (t , u(t ))v(t )dt + 0
Thus, the solutions of problem (1.1) are the corresponding critical points of ϕ .
p − j =1
Ij (u(tj ))v(tj ).
(2.5)
D. Zhang, B. Dai / Mathematical and Computer Modelling 53 (2011) 1154–1161
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Lemma 2.3. If u ∈ H01 (0, T ) is a weak solution of (1.1), then u is a classical solution of (1.1). Proof. Evidently, u(0) = u(T ) = 0 since u ∈ H01 (0, T ). By the definition of weak solution, one has T
∫
T
∫
u′ (t )v ′ (t )dt +
g (t )u(t )v(t )dt −
f (t , u(t ))v(t )dt + 0
0
0
T
∫
p −
Ij (u(tj ))v(tj ) = 0.
(2.6)
j =1
For j ∈ {0, 1, 2, . . . , p}, choose v ∈ H01 (0, T ) with v(t ) = 0 for every t ∈ [0, tj ] ∪ [tj+1 , T ]. Then tj+1
∫
u (t )v (t )dt + ′
′
tj+1
∫
g (t )u(t )v(t )dt =
f (t , u(t ))v(t )dt .
tj
tj
tj
tj+1
∫
By the definition of weak derivative, the above equality implies that
− u′′ (t ) + g (t )u(t ) = f (t , u(t )),
a.e. t ∈ (tj , tj+1 ).
(2.7)
Hence uj ∈ H (tj , tj+1 ) and u satisfies the equation in (1.1) a.e. on [0, T ]. By integrating (2.6), one has 2
−
p −
1u′ (tj )v(tj ) + u′ (T )v(T ) − u′ (0)v(0) +
j=1
p −
Ij (u(tj ))v(tj ) +
T
∫
[−u′′ (t ) + g (t )u(t ) − f (t , u(t ))]v(t )dt = 0. 0
j =1
Combining this with (2.7), we get p −
1u′ (tj )v(tj ) =
p −
j =1
Ij (u(tj ))v(tj ).
j =1
Hence 1u′ (tj ) = Ij (u(tj )), for every j = 1, 2, . . . , p and the impulsive condition in (1.1) is satisfied. This completes the proof. Lemma 2.4. If ess inft ∈[0,T ] g (t ) = m > −λ1 , then the functional ϕ is sequentially weakly lower semi-continuous. Proof. Let {uk } be a weakly convergent sequence to u in H01 (0, T ), then ‖u‖ ≤ limk→∞ inf ‖uk ‖. We have that {uk } converges uniformly to u on C [0, T ]. Then
lim inf ϕ(uk ) = lim
k→∞
k→∞
≥
1 2
1 2
2
T
∫
F (t , uk (t ))dt +
‖ uk ‖ − 0
‖u‖2 −
∫
T
F (t , u(t ))dt +
0
j =1 u(tj )
p ∫ − j =1
uk (tj )
p ∫ −
Ij (s)ds
0
Ij (s)ds
0
= ϕ(u). This completes the proof.
3. Main results In this paper, we make the following assumptions: (H1) Ij (j = 1, 2, . . . , p) satisfy
u 0
Ij (s)ds ≥ 0, for all u ∈ R.
(H2) There exist µ > 2, δj > 0, j = 1, 2, . . . , p such that
u
u 0
Ij (s)ds ≤ δj |u|µ , for u ∈ R \ {0}.
(H3) Ij (u)u ≤ µ 0 Ij (s)ds, for u ∈ R \ {0}. (H4) There exists a constant β > µ such that 0 < β F (t , u) ≤ uf (t , u) for every t ∈ [0, T ] and u ∈ R \ {0}, where F (t , u) = u f (t , s)ds. 0 The main results are the following theorems. Theorem 3.1. Suppose that (H1) holds. Assume that the following condition is satisfied: (H5) f is sublinear, i.e. there exist constants a > 0, b > 0 and γ ∈ [0, 1) such that |f (t , u)| ≤ a + b|u|γ for every (t , u) ∈ [0, T ] × R. Then the impulsive problem (1.1) has at least one solution.
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D. Zhang, B. Dai / Mathematical and Computer Modelling 53 (2011) 1154–1161
Proof. According to the conditions (H1) and (H5), we can imply that
ϕ(u) = ≥ ≥
1 2 1 2 1 2
‖ u‖ 2 −
T
∫
F (t , u(t ))dt +
u(tj )
p ∫ −
0
j=1
Ij (s)ds
0
‖u‖2 − T [a‖u‖∞ + b‖u‖γ∞+1 ] ‖u‖2 − T [aC ‖u‖ + bC γ +1 ‖u‖γ +1 ]
for all u ∈ H01 (0, T ). This implies that lim‖u‖→∞ ϕ(u) = ∞, and ϕ is coercive. By Theorem 1.1 of [24], ϕ has a minimum which is a critical point of ϕ . Hence, problem (1.1) has at least one solution. We readily have the following corollary. Corollary 3.1. Suppose that (H1) holds. If f is bounded, then the impulsive problem (1.1) has at least one solution. Example 3.1. Let T = 3, a1 > 0, b1 > 0, k > 0, t1 = 1. Consider the following problem:
2 −u′′ (t ) + t 3 u(t ) = a1 sin(ln |u(t )|) + b1 u 5 (t ) cos(u(t )); a.e. t ∈ [0, 3], ′ 2 u(t1 ) (3.1) 1u (t1 ) = (u (t1 ) + 2u(t1 ))e , u(0) = u(T ) = 0. u It is easy to know that I1 (u) = (u2 + 2u)eu , 0 I1 (s)ds = u2 eu ≥ 0, the condition (H1) holds. And |f (t , u(t ))| = 2
2
|a1 sin(ln |u(t )|) + b1 u 5 (t ) cos(u(t ))| ≤ a1 + b1 |u(t )| 5 , the condition (H5) holds. So problem (3.1) has at least one solution by Theorem 3.1. Remark 3.1. I1 (u) = (u2 + 2u)eu is superlinear, so Example 3.1 cannot obtain the existence of solutions in [20,21]. Theorem 3.2. Suppose that (H1)–(H4) hold. If f (t , u) and Ij are odd about u, then the impulsive problem (1.1) has infinitely many solutions. Remark 3.2. The impulsive functions Ij as some superlinear functions can satisfy the conditions (H2) and (H3), though the conditions (H2) and (H3) cannot guarantee that the impulsive functions Ij are superlinear. For example, take Ij (u) = u2j+1 , j = 1, 2, . . . , p. Obviously, Ij (u) are superlinear and satisfy the conditions (H2) and (H3). Lemma 3.1. If (H1)–(H4) hold, then ϕ(u) satisfies the PS condition. Proof. Let {uk } be a sequence in H01 (0, T ) such that {ϕ(uk )} is bounded and ϕ ′ (uk ) → 0 as k → ∞. First, we prove that {uk } is bounded. By (2.4) and (2.5), one has
βϕ(uk ) − ϕ ′ (uk )uk =
β
T
∫ − 1 ‖uk ‖2 − β
2
F (t , uk (t ))dt 0
uk (tj )
p ∫ −
+β
Ij (s)ds +
0
j =1
T
∫
f (t , uk (t ))uk (t )dt − 0
p −
Ij (uk (tj ))uk (tj ).
j =1
By (H3) and (H4), we can deduce that
β
uk (tj )
p ∫ − j =1
Ij (s)ds −
0
p −
Ij (uk (tj ))uk (tj ) ≥ 0,
T
∫
(3.2)
j =1
f (t , uk (t ))uk (t )dt − β 0
T
∫
F (t , uk (t ))dt ≥ 0.
(3.3)
0
By (3.2) and (3.3), we conclude that
βϕ(uk ) − ϕ (uk )uk ≥ ′
β 2
− 1 ‖ uk ‖ 2 .
Since β > 2 it follows that {uk } is bounded in H01 (0, T ). Going if necessary to a subsequence, we can assume that there exists u ∈ H01 (0, T ) such that uk ⇀ u,
weakly in H01 (0, T ),
uk → u,
strongly in C ([0, T ], R),
D. Zhang, B. Dai / Mathematical and Computer Modelling 53 (2011) 1154–1161
1159
as k → +∞. Hence
(ϕ ′ (uk ) − ϕ ′ (u))(uk − u) → 0, ∫ T [f (t , uk (t )) − f (t , u(t ))] (uk (t ) − u(t ))dt → 0, 0 p − [Ij (uk (tj )) − Ij (u(tj ))](uk (tj ) − u(tj )) → 0, j =1
as k → +∞. Moreover, an easy computation shows that
(ϕ (uk ) − ϕ (u))(uk − u) = ‖uk − u‖ − ′
′
2
T
∫
[f (t , uk (t )) − f (t , u(t ))](uk (t ) − u(t ))dt 0
+
p − [Ij (uk (tj )) − Ij (u(tj ))](uk (tj ) − u(tj )). j =1
So ‖uk − u‖ → 0 as k → +∞. That is, {uk } converges strongly to u in H01 (0, T ). Therefore, ϕ satisfies the Palais–Smale condition. Lemma 3.2 ([19, Lemma 2.2]). Assume that (H4) holds. Then, for every t ∈ [0, T ], the following inequalities hold: F (t , x) ≤ F F (t , x) ≥ F
t,
x
|x|
t,
x
|x|
|x|β ,
if 0 < |x| ≤ 1,
(3.4)
|x|β ,
if |x| ≥ 1.
(3.5)
In view of Lemma 3.2, (H4) implies that, for every t ∈ [0, T ], F (t , x) ≤ M |x|β ,
if |x| ≤ 1,
(3.6)
F (t , x) ≥ m|x|β ,
if |x| ≥ 1
(3.7)
where M = maxt ∈[0,T ],|x|=1 F (t , x), m = mint ∈[0,T ],|x|=1 F (t , x). Thanks to (H4), one has M > 0 and m > 0. Since F (t , x) − m|x|β is continuous on [0, T ] × [−1, 1], there exists a constant C2 such that F (t , x) ≥ m|x|β − C2 ,
(t , x) ∈ [0, T ] × [−1, 1].
(3.8)
So it follows from (3.7) and (3.8) that F (t , x) ≥ m|x|β − C2 ,
(t , x) ∈ [0, T ] × R.
(3.9)
From Lemma 5 of [22], we have that for any finite dimensional subspace W ⊂ H01 (0, T ) and any u ∈ W , there exists a constant d > 0 such that T
∫
|u(t )| dt p
‖ u‖ p =
1p ≥ d‖u‖,
p ≥ 1.
0
Then, for each u ∈ W , there exists a constant C3 such that T
∫
β
F (t , u(t ))dt ≥ m‖u‖β − C2 T
0
β
≥ mC3 ‖u‖β − C2 T .
(3.10)
Proof of Theorem 3.2. Using the continuity of f and Ij , j = 1, 2, . . . , p, we obtain that ϕ(u) is continuous and differentiable. In view of (2.4), it is obvious that ϕ(u) is even and ϕ(0) = 0. By Lemma 3.1, ϕ(u) satisfies the PS condition. It is clear that ‖u‖ ≤ √2 implies ‖u‖∞ ≤ 1. Due to (3.6), one has T
T
∫
F (t , u(t ))dt ≤ M
T
∫
0
‖u‖β∞ dt
0
√ β ≤ MT
T
2
‖ u‖ β ,
2
‖ u‖ ≤ √ . T
(3.11)
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D. Zhang, B. Dai / Mathematical and Computer Modelling 53 (2011) 1154–1161
Combining (3.11) and (H1), one has
ϕ(u) ≥
1 2
‖u‖2 − MTC β ‖u‖β ,
2
‖ u‖ ≤ √ , T
√
here C = 2T , which implies that we can choose ρ > 0 small enough such that ϕ(u) ≥ α > 0 with ‖u‖ = ρ . By (H2) and (3.9), we can imply that for any finite dimensional subspace W ⊂ H01 (0, T ),
ϕ(u) ≤
1 2
β
‖u‖2 − mC3 ‖u‖β + C2 T +
p −
δ j C µ ‖ u‖ µ ,
∀u ∈ W .
j=1
Noting that β > µ > 2, the above inequality implies that ϕ(u) → −∞ as u ∈ W and ‖u‖ → ∞. Therefore, there exists R = R(W ) such that ϕ(u) ≤ 0 on W \ BR(W ) . According to Theorem 2.1, the functional ϕ(u) possesses infinitely many critical points, i.e. the impulsive problem (1.1) has infinitely many solutions. Example 3.2. Let T = 3, a(t ), b(t ) ∈ C ([0, T ], (0, +∞)), k > 0, t1 =
−u′′ (t ) + t (1 + t )u(t ) = a(t )u7 (t ) + b(t )u9 (t ); 1u′ (t1 ) = ku5 (t1 ), u(0) = u(T ) = 0.
1 . 4
Consider the following problem:
a.e. t ∈ [0, T ], (3.12)
Let β = 8, µ = 6, δ1 = 61 k, the conditions (H1)–(H4) are satisfied. f (t , u) and I1 (u) are odd about u, so problem (3.12) has infinitely many solutions. Remark 3.3. It is easy to know that I1 (u) = ku5 is superlinear, so Example 3.2 cannot obtain the existence of solutions in [20,21]. But in this paper, we get the existence of infinitely many solutions. Acknowledgements The authors are grateful to the referees for their useful suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26]
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