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In this paper we investigate a class of impulsive differential equations with Dirichlet boundary conditions. Firstly, we define new inner product of

This paper is mainly concerned with the following second order impulsive differential equations with Dirichlet boundary conditions

{ − ( p ( t ) x ′ ( t ) ) ′ + q ( t ) x ( t ) = f ( t , x ( t ) ) , t ∈ J , t ≠ t i , − Δ ( p ( t i ) x ′ ( t i ) ) = α i x ( t i ) , i = 1 , 2 , ⋯ , k , x ( 0 ) = x ( 1 ) = 0 , (1.1)

where J ⊂ [ 0 , 1 ] , p ∈ C 1 [ J , [ β , + ∞ ) ] , β > 0 , q ∈ C [ J , [ 0 , + ∞ ) ] , 0 = t 0 < t 1 < ⋯ < t k < t k + 1 = 1 , α 1 , α 2 , ⋯ , α k are nonnegative constants with ∑ i = 1 k α i < 4 β . − Δ ( p ( t i ) x ′ ( t i ) ) = p ( t i + ) x ′ ( t i + ) − p ( t i − ) x ′ ( t i − ) . Here x ′ ( t i + ) (respectively x ′ ( t i − ) ) denotes the right limit (respectively left limit) of x ′ ( t ) at t = t i , and f ( t , x ) ∈ C [ J × ℝ ] is locally Lipschitz continuous for x uniformly in t ∈ J \ { t 1 , t 2 , ⋯ , t k } .

The phenomena of sudden or discontinuous jumps are often seen in chemotherapy, population dynamics, optimal control, ecology, engineering, etc. The mathematical model that describes the phenomena is impulsive differential equations. Due to their significance, impulsive differential equations have been developing as an important area of investigation in recent years. For the theory and classical results, we refer to [

Motivated by the papers mentioned above, we study the existence of sign-changing solution for second order impulsive differential equations with Dirichlet boundary conditions. However, to the best of our knowledge, there are few papers concerned with the existence of sign-changing solution for impulsive differential equations. In this paper, we obtain the existence of a positive solution, a negative solution and a sign-changing solution of (1.1) by using critical point theory and variational methods. An example is presented to illustrate the application of our main result. In comparison with previous works such as [

Let H be a Hilbert space and E a Banach space such that E is imbedded in H. Let φ be a C 2 − 0 functional defined on H, that is, the differential φ ′ of φ is locally Lipschitz continuous from H to H. Assume that φ ′ ( x ) = x − B x and φ ′ is also locally Lipschitz continuous as an operator from E to E. Assume also that K = { x ∈ H | φ ′ ( x ) = 0 } ⊂ E . For x 0 ∈ E , consider the initial value problem both in H and in E:

{ d x d t = − x ( t ) + B x ( t ) , x ( 0 ) = x 0 . (1.2)

Let x ( t , x 0 ) and x ˜ ( t , x 0 ) be the unique solution of this initial value problem considered in H and E respectively, with [ 0 , η ( x 0 ) ) and [ 0 , η ˜ ( x 0 ) ) the right maximal interval of existence. Because of the imbedding E ↦ H , η ˜ ( x 0 ) ≤ η ( x 0 ) and x ˜ ( t , x 0 ) = x ( t , x 0 ) for 0 ≤ t < η ˜ ( x 0 ) . We assume that η ˜ ( x 0 ) = η ( x 0 ) and x ˜ ( t , x 0 ) = x ( t , x 0 ) for 0 ≤ t < η ( x 0 ) , and if lim t → η ( x 0 ) x ( t , x 0 ) = x * in H for some x * ∈ E then lim t → η ( x 0 ) x ( t , x 0 ) = x * in E.

Lemma 1.1 [

h ( 0 ) ∈ D 1 \ D 2 , h ( 1 ) ∈ D 2 \ D 1 ,

and

inf x ∈ D ¯ 1 E ∩ D ¯ 2 E φ ( x ) > sup t ∈ [ 0 , 1 ] φ ( h ( t ) ) ,

then φ has at least four critical points, one in D 1 ∩ D 2 , one in D 1 \ D ¯ 2 E , one in D 2 \ D ¯ 1 E , and one in E \ ( D ¯ 1 E ∪ D ¯ 2 E ) . Here ∂ E D and D ¯ E mean respectively the boundary and the closure of D relative to E.

Now we state our main result

Theorem 1.1 Assume

(f_{1}) lim s → 0 | f ( t , s ) | ρ ( t ) | s | < λ 1 uniformly in t ∈ [ 0 , 1 ] , where ρ ∈ C [ J , ( 0 , + ∞ ) ] ,

(f_{2}) there exists a constant M ≥ 0 such that f ( t , s ) + M s is creasing in s,

(f_{3}) there exist μ > 2 and r > 0 such that

0 < μ F ( t , s ) ≤ s f ( t , s ) , ∀ t ∈ [ 0 , 1 ] , | s | ≥ r ,

where F ( t , s ) = ∫ 0 s f ( t , τ ) d τ .

Then problem (1.1) has at least three solutions: one positive, one negative, and one sign-changing.

Let H = H 0 1 [ 0 , 1 ] be the Sobolev space endowed the norm

‖ x ‖ 1 = ( ∫ 0 1 p ( t ) | x ′ ( t ) | 2 + q ( t ) | x ( t ) | 2 d t − ∑ i = 1 k α i x 2 ( t i ) ) 1 2 ,

which is equivalent to the usual norm ‖ x ‖ 0 = ( ∫ 0 1 | x ′ ( t ) | 2 d t ) 1 2 . Let

E = { x ∈ C [ 0 , 1 ] | x ( 0 ) = x ( 1 ) = 0 , ( p x ) ′ ( ⋅ ) ∈ C ( [ 0 , 1 ] \ { t 1 , t 2 , ⋯ , t k } ) , x ′ ( t i − ) = x ′ ( t i ) , x ′ ( t i + ) ∃ }

with the norm

‖ x ‖ E = max { max t ∈ [ 0 , 1 ] | x ( t ) | , sup t ∈ [ 0 , 1 ] | x ′ ( t ) | } .

Clearly E is a Banach space and densely embedded in H.

As is well known, for ρ ∈ C [ J , ( 0 , + ∞ ) ] , the following linear eigenvalue problem

{ − ( p ( t ) x ′ ( t ) ) ′ + q ( t ) x ( t ) = λ ρ ( t ) x ( t ) , t ∈ J , t ≠ t i , − Δ ( p ( t i ) x ′ ( t i ) ) = α i x ( t i ) , i = 1 , 2 , ⋯ , k , x ( 0 ) = x ( 1 ) = 0 ,

possesses a sequence of positive eigenvalue 0 < λ 1 < λ 2 < ⋯ < λ j < ⋯ → ∞ and the algebraic multiplicity of λ j is equal to 1. Moreover

λ 1 = inf x ∈ H \ { 0 } ∫ 0 1 p ( t ) | x ′ ( t ) | 2 + q ( t ) | x ( t ) | 2 d t − ∑ i = 1 k α i x 2 ( t i ) ∫ 0 1 ρ ( t ) | x ( t ) | 2 d t ,

the eigenfunction e 1 with respect to λ 1 satisfies ‖ e 1 ‖ 1 = 1 , e 1 > 0 in ( 0 , 1 ) and the eigenfunctions e j corresponding to λ j ( j ≥ 2 ) are sign-changing in ( 0 , 1 ) and ‖ e j ‖ 1 = 1 (see [

Let M be as in (f_{2}), we define new inner product of H as follows

( x , y ) = ∫ 0 1 p ( t ) x ′ ( t ) y ′ ( t ) + [ q ( t ) + M ] x ( t ) y ( t ) d t − ∑ i = 1 k α i x ( t i ) y ( t i ) . (2.1)

The inner product induces the norm

‖ x ‖ = ( ∫ 0 1 p ( t ) | x ′ ( t ) | 2 + [ q ( t ) + M ] | x ( t ) | 2 d t − ∑ i = 1 k α i x 2 ( t i ) ) 1 2 (2.2)

Define the functional φ : H → ℝ by

φ ( x ) = 1 2 ∫ 0 1 p ( t ) | x ′ ( t ) | 2 + q ( t ) | x ( t ) | 2 d t − 1 2 ∑ i = 1 k α i x 2 ( t i ) − ∫ 0 1 F ( t , x ( t ) ) d t . (2.3)

It is clear that φ ∈ C 2 − 0 ( H , ℝ ) . By [

〈 φ ′ ( x ) , y 〉 = ∫ 0 1 p ( t ) x ′ ( t ) y ′ ( t ) + q ( t ) x ( t ) y ( t ) d t − ∑ i = 1 k α i x ( t i ) y ( t i ) − ∫ 0 1 f ( t , x ( t ) ) y ( t ) d t . (2.4)

Lemma 2.1 Two norms ‖ x ‖ and ‖ x ‖ 1 defined on H are equivalent, that is, there exist positive constants γ 1 , γ 2 such that

γ 1 ‖ x ‖ ≤ ‖ x ‖ 1 ≤ γ 2 ‖ x ‖ , ∀ x ∈ H . (2.5)

Proof. For any x ∈ H , by Lemma 2.3 in [

max t ∈ J x 2 ( t ) ≤ 1 4 ∫ 0 1 | x ′ ( t ) | 2 d t . (2.6)

From this it is easy to see that

‖ x ‖ 1 2 ≥ β ∫ 0 1 | x ′ ( t ) | 2 d t − 1 4 ∑ i = 1 k α i ∫ 0 1 | x ′ ( t ) | 2 d t ≥ ( β − 1 4 ∑ i = 1 k α i ) ‖ x ‖ 0 . (2.7)

Since p ∈ C 1 [ J , [ β , + ∞ ) ] , q ∈ C [ J , [ 0 , + ∞ ) ] , we may assume that β 1 = max t ∈ J p ( t ) , β 2 = max t ∈ J q ( t ) . From (2.6) we have

‖ x ‖ 2 ≤ β 1 ∫ 0 1 | x ′ ( t ) | 2 d t + ( β 2 + M + ∑ i = 1 k α i ) ∫ 0 1 | x ( t ) | 2 d t ≤ ( β 1 + 1 4 ( β 2 + M + ∑ i = 1 k α i ) ) ‖ x ‖ 0 . (2.8)

From (2.7) and (2.8), there exists a positive constant γ 1 such that γ 1 ‖ x ‖ ≤ ‖ x ‖ 1 . On the other hand, it is obvious that ‖ x ‖ 1 ≤ ‖ x ‖ . So, (2.5) holds.

Let G ( t , s ) be the Green’s function of

{ L x = 0 , t ∈ J , x ( 0 ) = x ( 1 ) = 0 , (2.9)

where L x ( t ) = − ( p ( t ) x ′ ( t ) ) ′ + [ q ( t ) + M ] x ( t ) . From [

Lemma 2.2 [

1) G ( t , s ) can be written by

G ( t , s ) = { m ( t ) n ( s ) ω , 0 ≤ t ≤ s ≤ 1 , m ( s ) n ( t ) ω , 0 ≤ s ≤ t ≤ 1.

2) m ( t ) ∈ C 2 ( [ 0 , 1 ] , ℝ ) is increasing and m ( t ) > 0 , t ∈ ( 0 , 1 ] .

3) L m ( t ) ≡ 0 , m ( 0 ) = 0 , m ′ ( 0 ) = 1 .

4) n ( t ) ∈ C 2 ( [ 0 , 1 ] , ℝ ) is decreasing and n ( t ) > 0 , t ∈ [ 0 , 1 ) .

5) L n ( t ) ≡ 0 , n ( 1 ) = 0 , n ′ ( 1 ) = − 1 .

6) p ( t ) ( m ′ ( t ) n ( t ) − n ′ ( t ) m ( t ) ) ≡ ω is a positive constant.

7) G ( t , s ) is continuous and symmetrical over { ( t , s ) | 0 ≤ t ≤ s ≤ 1 } ∪ { ( t , s ) | 0 ≤ s ≤ t ≤ 1 } .

8) G ( t , s ) has continuously partial derivative over { ( t , s ) | 0 ≤ t ≤ s ≤ 1 } ∪ { ( t , s ) | 0 ≤ s ≤ t ≤ 1 } .

9) For fixed s ∈ [ 0 , 1 ] , G ( t , s ) satisfies L G ( t , s ) = 0 for t ≠ s , t ∈ [ 0 , 1 ] and G ( 0 , s ) = G ( 1 , s ) = 0 .

10) G ′ t has discontinuous point of the first kind at t = s , G ′ t ( s + 0 , s ) − G ′ t ( s − 0 , s ) = − 1 p ( s ) , s ∈ ( 0 , 1 ) .

Define an operator B : H → H by

B x ( t ) = ∫ 0 1 G ( t , s ) ( f ( s , x ( s ) ) + M x ( s ) ) d s + ∑ i = 1 k α i G ( t , t i ) x ( t i ) . (2.10)

Lemma 2.3 [

From Lemma 2.3, the critical point set K = { x ∈ H | φ ′ ( x ) = 0 } ⊂ E . Notice that f ( t , x ) ∈ C [ J × ℝ ] is locally Lipschitz continuous for x uniformly in t ∈ J \ { t 1 , t 2 , ⋯ , t k } . It is easy to obtain that B defined by (2.10) is locally Lipschitz continuous both as an operator from H to H and as one from E to E. Let x 0 ∈ E and consider the initial value problem (1.2) in both H and E.

Similar to Lemma 4.2 in [

Lemma 2.4 [

2) if lim t → η ( x 0 ) x ( t , x 0 ) = x * in H for some x * ∈ K then lim t → η ( x 0 ) x ( t , x 0 ) = x * in E.

Lemma 3.1 The gradient of φ at a point x ∈ H can be expressed as

g r a d φ ( x ) = x − B x .

This result is necessary, for the reader’s convenience we present the proof in the Appendix.

Lemma 3.2 Assume that (f_{3}) holds, then the functional φ satisfies (PS)-condition.

Proof. Suppose that { x n } is a (PS)-sequence, namely such that for some constant c > 0

| φ ( x n ) | ≤ c , and φ ′ ( x n ) → 0 as n → ∞ .

This implies that there is a constant C > 0 such that

| φ ( x n ) | ≤ C , and | 〈 φ ′ ( x n ) , x n 〉 | ≤ C ‖ x n ‖ 1 , ∀ n ∈ ℕ . (3.1)

Moreover, thanks to f ( t , x ) ∈ C [ J × ℝ ] we know that f ( t , x ) is bounded on J × [ − r , r ] . By (3.1) we have

C ( 1 + ‖ x n ‖ 1 ) ≥ φ ( x n ) − 1 μ 〈 φ ′ ( x n ) , x n 〉 = ( 1 2 − 1 μ ) ‖ x n ‖ 1 2 + ∫ 0 1 1 μ f ( t , x n ( t ) ) x n ( t ) − F ( t , x n ( t ) ) d t (3.2)

Then from (f_{3}) we get

∫ 0 1 1 μ f ( t , x n ( t ) ) x n ( t ) − F ( t , x n ( t ) ) d t = ∫ J 0 1 μ f ( t , x n ( t ) ) x n ( t ) − F ( t , x n ( t ) ) d t + ∫ J \ J 0 1 μ f ( t , x n ( t ) ) x n ( t ) − F ( t , x n ( t ) ) d t ≥ ∫ J 0 1 μ f ( t , x n ( t ) ) x n ( t ) − F ( t , x n ( t ) ) d t ≥ − κ , (3.3)

where κ > 0 and J 0 = { t ∈ J | | x n ( t ) ≤ r | } . By (3.2) and (3.3), one has

C ( 1 + ‖ x n ‖ 1 ) ≥ ( 1 2 − 1 μ ) ‖ x n ‖ 1 2 − κ ,

which of course implies that { x n } is bounded by means of Lemma 2.1.

Since H is a reflexive Banach space, we can assume that, up to a subsequence, there exists x ∈ H such that x n → w x . By (2.4) we have

〈 φ ′ ( x n ) − φ ′ ( x ) , x n − x 〉 = ∫ 0 1 p ( t ) | ( x n ( t ) − x ( t ) ) ′ | 2 + q ( t ) | x n ( t ) − x ( t ) | 2 d t − ∑ i = 1 k α i | x n ( t i ) − x ( t i ) | 2 − ∫ 0 1 [ f ( t , x n ( t ) ) − f ( t , x ( t ) ) ] ( x n ( t ) − x ( t ) ) d t = ‖ x n − x ‖ 1 2 − ∫ 0 1 [ f ( t , x n ( t ) ) − f ( t , x ( t ) ) ] ( x n ( t ) − x ( t ) ) d t . (3.4)

By φ ′ ( x n ) → 0 and x n → w x , we have

〈 φ ′ ( x n ) − φ ′ ( x ) , x n − x 〉 → 0 as n → ∞ . (3.5)

By x n → w x in H, we see that { x n } uniformly converges to x in C [ 0 , 1 ] . it is easy to obtain that

∫ 0 1 [ f ( t , x n ( t ) ) − f ( t , x ( t ) ) ] ( x n ( t ) − x ( t ) ) d t → 0 as n → ∞ . (3.6)

Then (3.4)-(3.6) and Lemma 2.1 yield that ‖ x n − x ‖ → 0 in H, that is, { x n } strongly converges to x in H.

Proof of Theorem 1.1. By (f_{1}), there exist 0 < σ < λ 1 and δ > 0 such that

− σ ρ ( t ) | s | ≤ f ( t , s ) ≤ σ ρ ( t ) | s | , ∀ t ∈ [ 0 , 1 ] , s ∈ [ − δ , δ ] \ { 0 } . (3.7)

Let v ( t ) = − δ e 1 ( t ) , w ( t ) = δ e 1 ( t ) , then

− ( p ( t ) v ′ ( t ) ) ′ + q ( t ) v ( t ) = − δ λ 1 ρ ( t ) e 1 ( t ) = λ 1 ρ ( t ) v ( t ) < σ ρ ( t ) v ( t ) ≤ f ( t , v ( t ) ) , t ∈ J , t ≠ t i ,

− Δ ( p ( t i ) v ′ ( t i ) ) = δ Δ ( p ( t i ) e ′ 1 ( t i ) ) = α i v ( t i ) , i = 1 , 2 , ⋯ , k .

So, we have

{ − ( p ( t ) v ′ ( t ) ) ′ + q ( t ) v ( t ) = λ 1 ρ ( t ) v ( t ) < f ( t , v ( t ) ) , t ∈ J , t ≠ t i , − Δ ( p ( t i ) v ′ ( t i ) ) = α i v ( t i ) , i = 1 , 2 , ⋯ , k , v ( 0 ) = v ( 1 ) = 0. (3.8)

Similarly, we also have

{ − ( p ( t ) w ′ ( t ) ) ′ + q ( t ) w ( t ) = λ 1 ρ ( t ) w ( t ) > f ( t , w ( t ) ) , t ∈ J , t ≠ t i , − Δ ( p ( t i ) w ′ ( t i ) ) = α i w ( t i ) , i = 1 , 2 , ⋯ , k , w ( 0 ) = w ( 1 ) = 0. (3.9)

Hence, v ( t ) and w ( t ) are lower and upper solutions to (1.1), respectively.

Let D 1 = { x ∈ E | x > v in ( 0 , 1 ) } and D 2 = { x ∈ E | x < w in ( 0 , 1 ) } , it is clear that D 1 and D 2 are open convex sets in E and D 1 ∩ D 2 ≠ ∅ . If x ∈ ∂ E D 1 , then by (f2)

B x ( t ) = ∫ 0 1 G ( t , s ) ( f ( s , x ( s ) ) + M x ( s ) ) d s + ∑ i = 1 k α i G ( t , t i ) x ( t i ) ≥ ∫ 0 1 G ( t , s ) ( f ( s , v ( s ) ) + M v ( s ) ) d s + ∑ i = 1 k α i G ( t , t i ) v ( t i ) = B v ( t ) . (3.10)

From Lemma 2.3 and (3.8), we easily know that

v ( t ) = ∫ 0 1 G ( t , s ) ( λ 1 ρ ( s ) v ( s ) + M v ( s ) ) d s + ∑ i = 1 k α i G ( t , t i ) v ( t i ) .

So,

B v ( t ) − v ( t ) = ∫ 0 1 G ( t , s ) ( f ( s , v ( s ) ) − λ 1 ρ ( s ) v ( s ) ) d s > 0. (3.11)

Since B defined by (2.10) is also an operator from E to E. Hence B x ∈ E , by (3.10) and (3.11), B x > v and B ( ∂ E D 1 ) ⊂ D 1 . Similarly, B ( ∂ E D 2 ) ⊂ D 2 . (f_{3}) implies that there exist two positive constants C 1 , C 2 such that

F ( t , s ) ≥ C 1 | s | μ − C 2 , ∀ t ∈ [ 0 , 1 ] , s ∈ ℝ .

Let E 1 = s p a n { e 1 , e 2 } , then E 1 is a finitely dimensional subspace of E, if x ∈ E 1 , then we have, for some C 3 > 0 ,

φ ( x ) = 1 2 ∫ 0 1 p ( t ) | x ′ ( t ) | 2 + q ( t ) | x ( t ) | 2 d t − 1 2 ∑ i = 1 k α i x 2 ( t i ) − ∫ 0 1 F ( t , x ( t ) ) d t ≤ 1 2 ‖ x ‖ 1 2 − C 1 | x | μ μ + C 2 ≤ 1 2 ‖ x ‖ 1 2 − C 3 ‖ x ‖ 1 μ + C 2 . (3.12)

We define h R : [ 0 , 1 ] → E by

h R ( s ) = R e 1 cos ( π s ) + R e 2 sin ( π s ) ,

where R will be determined later. Then we have ‖ h R ( s ) ‖ 1 2 = ‖ R e 1 cos ( π s ) ‖ 1 2 + ‖ R e 2 sin ( π s ) ‖ 1 2 = R 2 , by (3.12),

φ ( h R ( s ) ) ≤ 1 2 ‖ h R ( s ) ‖ 1 2 − C 3 ‖ h R ( s ) ‖ 1 μ + C 2 = 1 2 R 2 − C 3 R μ + C 2 .

Since

inf x ∈ D ¯ 1 E ∩ D ¯ 2 E φ ( x ) > − ∞ ,

we see that

h R ( 0 ) ∈ D 1 \ D 2 , h R ( 1 ) ∈ D 2 \ D 1 ,

and

inf x ∈ D ¯ 1 E ∩ D ¯ 2 E φ ( x ) > sup s ∈ [ 0 , 1 ] φ ( h R ( s ) ) ,

if R is sufficiently large. Applying Lemma 2.4 and Lemma 1.1, problem (1.1) has at least four solutions, x 1 ∈ D 1 ∩ D 2 , x 2 ∈ D 1 \ D ¯ 2 E , x 3 ∈ D 2 \ D ¯ 1 E , and x 4 ∈ E \ ( D ¯ 1 E ∪ D ¯ 2 E ) . It is clear that x 2 is positive, x 3 is negative, and x 4 is sign-changing.

To illustrate the application of our main result we present the following example.

Example 4.1 Consider the following second order impulsive differential equations with Dirichlet boundary condition:

{ x ″ ( t ) = 10 x 3 − 2 x , t ∈ [ 0 , 1 ] , t ≠ 1 2 , − Δ x ′ ( 1 2 ) = x ( 1 2 ) , x ( 0 ) = x ( 1 ) = 0 , (4.1)

Then (4.1) has at least three solutions: one positive, one negative, and one sign-changing.

Proof. It is clear that (4.1) has the form of (1.1). Let p ( t ) = ρ ( t ) = 1 , q ( t ) = 0 , f ( t , x ) = 10 x 3 − 2 x , k = 1 , α 1 = 1 in (1.1). By (1.4) and (2.6), we can obtain that λ 1 = inf x ∈ H \ { 0 } ∫ 0 1 | x ′ ( t ) | 2 d t − x 2 ( 1 2 ) ∫ 0 1 | x ( t ) | 2 d t ≥ 3 .

Taking M = 2 , μ = 3 and r = 1 , by simple calculations, the conditions in Theorem 1.1 are satisfied.

Hence, (4.1) has at least three solutions: one positive, one negative, and one sign-changing.

Supported by the National Natural Science Foundation of China (61803236), Natural Science Foundation of Shandong Province (ZR2018MA022).

The authors declare no conflicts of interest regarding the publication of this paper.

Wang, H.H., Lu, D. and Lu, H.Q. (2021) Multiplicity Results for Second Order Impulsive Differential Equations via Variational Methods. Engineering, 13, 82-93. https://doi.org/10.4236/eng.2021.132007

In this Appendix , for the reader’s convenience we give the proof of Lemma 3.1.

Proof of Lemma 3.1. For any h ∈ H ,

φ ( x + τ h ) − φ ( x ) τ = 1 2 ∫ 0 1 p ( t ) | x ′ ( t ) + τ h ′ ( t ) | 2 − | x ′ ( t ) | 2 τ + q ( t ) | x ( t ) + τ h ( t ) | 2 − | x ( t ) | 2 τ d t − 1 2 ∑ i = 1 k α i | x ( t i ) + τ h ( t i ) | 2 − | x ( t i ) | 2 τ − ∫ 0 1 F ( t , x ( t ) + τ h ( t ) ) − F ( t , x ( t ) ) τ d t .

By the Lagrange Theorem there exists θ with 0 < θ < 1 such that

φ ( x + τ h ) − φ ( x ) τ = ∫ 0 1 p ( t ) x ′ ( t ) h ′ ( t ) + q ( t ) x ( t ) h ( t ) d t − ∑ i = 1 k α i x ( t i ) h ( t i ) − ∫ 0 1 f ( t , x ( t ) + θ τ h ( t ) ) h ( t ) d t + τ 2 ∫ 0 1 p ( t ) | h ′ ( t ) | 2 + q ( t ) | h ( t ) | 2 d t − τ 2 ∑ i = 1 k α i h 2 ( t i ) . (A.1)

For any h ∈ H , by (2.1), we have

( x − B x , h ) = ( x − ∫ 0 1 G ( t , s ) ( f ( s , x ( s ) ) + M x ( s ) ) d s + ∑ i = 1 k α i G ( t , t i ) x ( t i ) , h ) = ∫ 0 1 p ( t ) x ′ ( t ) h ′ ( t ) + [ q ( t ) + M ] x ( t ) h ( t ) d t − ∑ i = 1 k α i x ( t i ) h ( t i ) − ∫ 0 1 p ( t ) ( ∫ 0 1 G ′ t ( t , s ) ( f ( s , x ( s ) ) + M x ( s ) ) d s ) h ′ ( t ) d t − ∫ 0 1 [ q ( t ) + M ] ( ∫ 0 1 G ( t , s ) ( f ( s , x ( s ) ) + M x ( s ) ) d s ) h ( t ) d t

− ∫ 0 1 p ( t ) ∑ i = 1 k α i G ′ t ( t , t i ) x ( t i ) h ′ ( t ) d t − ∫ 0 1 [ q ( t ) + M ] ∑ i = 1 k α i G ( t , t i ) x ( t i ) h ( t ) d t + ∑ i = 1 k α i ( ∫ 0 1 G ( t i , s ) ( f ( s , x ( s ) ) + M x ( s ) ) d s ) h ( t i ) + ∑ i = 1 k α i ∑ j = 1 k α j G ( t i , t j ) x ( t j ) h ( t i ) . (A.2)

By integrating by parts and Lemma 2.2, we can obtain immediately

∫ 0 1 p ( t ) ( ∫ 0 1 G ′ t ( t , s ) ( f ( s , x ( s ) ) + M x ( s ) ) d s ) h ′ ( t ) d t = ∑ i = 1 k α i ( ∫ 0 1 G ( t i , s ) ( f ( s , x ( s ) ) + M x ( s ) ) d s ) h ( t i ) − ∫ 0 1 ( p ( t ) ( ∫ 0 1 G ′ t ( t , s ) ( f ( s , x ( s ) ) + M x ( s ) ) d s ) ) ′ h ( t ) d t (4.3)

and

∫ 0 1 p ( t ) ∑ i = 1 k α i G ′ t ( t , t i ) x ( t i ) h ′ ( t ) d t = ∑ i = 1 k α i ∑ j = 1 k α j G ( t i , t j ) x ( t j ) h ( t i ) − ∫ 0 1 ( p ( t ) ∑ i = 1 k α i G ′ t ( t , t i ) x ( t i ) ) ′ h ( t ) d t . (A.4)

By Lemma 2.2

− ∫ 0 1 ( p ( t ) ( ∫ 0 1 G ′ t ( t , s ) ( f ( s , x ( s ) ) + M x ( s ) ) d s ) ) ′ h ( t ) d t − ∫ 0 1 ( p ( t ) ∑ i = 1 k α i G ′ t ( t , t i ) x ( t i ) ) ′ h ( t ) d t + ∫ 0 1 [ q ( t ) + M ] ( ∫ 0 1 G ( t , s ) ( f ( s , x ( s ) ) + M x ( s ) ) d s + ∑ i = 1 k α i G ( t , t i ) x ( t i ) ) h ( t ) d t = ∫ 0 1 h ( t ) L ( ∫ 0 1 G ( t , s ) ( f ( s , x ( s ) ) + M x ( s ) ) d s + ∑ i = 1 k α i G ( t , t i ) x ( t i ) ) d t = ∫ 0 1 ( f ( t , x ( t ) ) + M x ( t ) ) h ( t ) d t . (A.5)

Substituting (A.3), (A.4) into (A.2), and using (A.5), one has

( x − B x , h ) = ∫ 0 1 p ( t ) x ′ ( t ) h ′ ( t ) + q ( t ) x ( t ) h ( t ) d t − ∑ i = 1 k α i x ( t i ) h ( t i ) − ∫ 0 1 f ( t , x ( t ) ) h ( t ) d t (A.6)

From (A.1) and (A.6)

| φ ( x + τ h ) − φ ( x ) τ − ( x − B x , h ) → 0 as τ → 0 |

From Definition 1.1 and Remarks 1.2 in [