Abstract
In this paper, we study the periodic solutions of high order differential delay equations with \(2k-1\) lags. The 4k-periodic solutions are obtained by using the variational method and the method of Kaplan–Yorke coupling system. These are new types of differential delay equations compared with all previous research. And it provides a precise counting method for the number of periodic solutions. Two examples are given to demonstrate our main results.
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1 Introduction
Given \(f\in C^{0}(R,R)\) with \(f(-x)=-f(x)\), \(xf(x)>0\), \(x\neq 0\). Kaplan and Yorke [13] studied the existence of 4-periodic and 6-periodic solutions to the differential delay equations
and
respectively. The method they applied is transforming the two equations into adequate ordinary differential equations by regarding the retarded functions \(x(t-1)\) and \(x(t-2)\) as independent variables. They guessed that the existence of \(2(n+1)\)-periodic solution to the equation
could be studied under the restriction
which was proved by Nussbaum [20] in 1978 by the use of a fixed point theorem on cones.
After that a lot of papers [3,4,5,6,7,8,9,10,11,12, 14,15,16,17,18] discussed the existence and multiplicity of \(2(n+1)\)-periodic solutions to Eq. (3) and its extension
where \(F\in C^{1}(R^{N},R)\), \(\nabla F(-x)=-\nabla F(x)\), \(F(0)=0\). But they all studied first order equations.
In this paper, we study the periodic orbits to two types of high order differential delay equations with \(2k-1\) lags in the form
and
which are different from (3) and can be regarded as a new extension of (3). The method applied in this paper is the variational approach in the critical point theory [1, 2, 19].
We suppose that
and there are \(\alpha ,\beta \in R\) such that
Let \(F(x)=\int_{0}^{x}f(s)\,ds\). Then \(F(-x)=F(x)\) and \(F(0)=0\). For convenience, we make the following assumptions:
- \((S_{1})\) :
- \((S_{2})\) :
-
there exist \(M>0\) and a function \(r\in C^{0}(R^{+},R^{+})\) satisfying \(r(s)\rightarrow \infty \), \(r(s)/s\rightarrow 0\) as \(s\rightarrow \infty \) such that
$$ \biggl\vert F(x)-\frac{1}{2}\beta x^{2} \biggr\vert >r \bigl( \vert x \vert \bigr)-M, $$ - \((S_{3}^{\pm })\) :
-
\(\pm [F(x)-\frac{1}{2}\beta x^{2}]>0\), \(|x|\rightarrow \infty \),
- \((S_{4}^{\pm })\) :
-
\(\pm [F(x)-\frac{1}{2}\alpha x^{2}]>0\), \(0<|x|\ll 1\).
In this paper, we need the following lemma as the base of our discussion.
Let X be a Hilbert space, \(L:X\rightarrow X\) be a linear operator, and \(\varPhi :X\rightarrow R\) be a differentiable functional.
Lemma 1.1
([3], Lemma 2.4)
Assume that there are two closed \(s^{1}\)-invariant linear subspaces, \(X^{+}\) and \(X^{-}\), and \(r>0\) such that
-
(a)
\(X^{+}\cup X^{-}\) is closed and of finite codimensions in X,
-
(b)
\(\widehat{L}(X^{-})\subset X^{-}\), \(\widehat{L}=L+P^{-1}A_{0}\) or \(\widehat{L}=L+P^{-1}A_{\infty } \),
-
(c)
there exists \(c_{0}\in R\) such that
$$ \inf_{x\in X^{+}}\varPhi (x)\geq c_{0}, $$ -
(d)
there is \(c_{\infty }\in R\) such that
$$\varPhi (x)\leq c_{\infty }< \varPhi (0)=0,\quad \forall x \in X^{-} \cap S_{r}=\bigl\{ x\in X^{-}: \Vert x \Vert =r\bigr\} , $$ -
(e)
Φ satisfies the \((P.S)_{c}\)-condition, \(c_{0}< c< c_{\infty }\), i.e., every sequence \(\{x_{n}\}\subseteq X\) with \(\varPhi (x_{n})\rightarrow c\) and \(\varPhi '(x_{n})\rightarrow 0\) possesses a convergent subsequence.
Then Φ has at least \(\frac{1}{2}[\dim (X^{+}\cap X^{-})- \operatorname{codim}_{X}(X^{+}\cup X^{-})]\) generally different critical orbits in \(\varPhi^{-1}([c_{0},c_{\infty }])\) if \([\dim (X^{+}\cap X^{-})- \operatorname{codim}_{X}(X^{+}\cup X^{-})]>0\).
Definition 1.2
We say that Φ satisfies the \((P.S)\)-condition if every sequence \(\{x_{n}\}\) with \(\varPhi (x_{n})\) is bounded and \(\varPhi '(x_{n})\rightarrow 0\) possesses a convergent subsequence.
Remark 1.3
We may use the \((P.S)\)-condition to replace condition (e) in Lemma 1.1 since the \((P.S)\)-condition implies that the \((P.S)_{c}\)-condition holds for each \(c\in R\).
2 Space X, functional Φ, and its differential \(\varPhi '\)
We are concerned with the 4k-periodic solutions to (5) and (6) and suppose
Let
and define \(P:X\rightarrow L^{2}\) by
Let
Then the inverse \(P^{-1}\) of P exists. For \(x\in X\), define
Therefore \((X,\|\cdot \|)\) is an \(H^{\frac{1}{2}}\) space.
For (5), define a functional \(\varPhi :X\rightarrow R\) by
where
Let \(m=k-1\) and
We have
If \(x_{i}(t)=a_{i}\cos \frac{(2i+1)\pi t}{2k}+b_{i}\sin \frac{(2i+1) \pi t}{2k}\in X(i)\), \(i\in N\), we have
Obviously, \(L|_{X(i)}:X(i)\rightarrow X(i)\) is invertible.
Based on the theorem given by Mawhin and Willem [19, Theorem 1.4] the differential of functional Φ is differentiable, and its differential is
where \(K(x)=P^{-1}f(x)\). It is easy to prove that \(K:(X,\|x\|^{2}) \rightarrow (X,\|x\|_{2}^{2})\) is compact.
Therefore, from (14) we have that if
then
On the other hand,
Therefore, we have
For (6), define the functional \(\varPsi :X\rightarrow R\) by
where
Let \(m=k-1\) and
We have
If \(x_{i}(t)=a_{i}\cos \frac{(2i+1)\pi t}{2k}+b_{i}\sin \frac{(2i+1) \pi t}{2k}\in X(i)\), \(i\in N\), we have
Obviously, \(L|_{X(i)}:X(i)\rightarrow X(i)\) is invertible.
Based on the theorem given by Mawhin and Willem [16, Theorem 1.4] the differential of functional Φ is differentiable, and its differential is
where \(K(x)=P^{-1}f(x)\). It is easy to prove that \(K:(X,\|x\|^{2}) \rightarrow (X,\|x\|_{2}^{2})\) is compact.
Therefore, from (14) we have that if
then
On the other hand,
Therefore, we have
Lemma 2.1
Each critical point of the functional Φ is a 4k-periodic classical solution of Eq. (5) satisfying (9).
Proof
Let x be a critical point of the functional Φ. Then \(x(t)\) satisfies
Consequently,
Calculating (21.1) + (21.2) + (21.3) + ⋯ + (21.(2k-1)), we can get
namely
which implies that x satisfies the above equation for all \(t\in [0,4k]\) since the function on the right-hand side is continuous, then x is a classical solution to (5). □
Lemma 2.2
Each critical point of the functional Ψ is a 4k-periodic classical solution of Eq. (6) satisfying (9).
Proof
Let x be a critical point of the functional Ψ. Then \(x(t)\) satisfies
Consequently,
Calculating (22.1) − (22.2) + (22.3) − ⋯ + (22.(2k-1)), we can get
namely
which implies that x satisfies the above equation for all \(t\in [0,4k]\) since the function on the right-hand side is continuous, then x is a classical solution to (6). □
3 Partition of space X and symbols
For (5), let
On the other hand,
Obviously, \(\dim X_{\infty }^{0}<\infty \) and \(\dim X_{0}^{0}< \infty \).
Lemma 3.1
Under assumptions \((S_{1})\) and \((S_{2})\), there is \(\sigma >0\) such that
Proof
First, we have that, for \(\beta \geq 0\) and \(s=\mathrm{even}\), \((-1)^{s+1}=-1\), \(i \in \{0,1,\ldots ,m\}\),
where \(l^{+}(i)=\max \{l\in N:- (\frac{(4lk+2i+1)\pi }{2k} )^{2s+1}\tan \frac{(2i+1)\pi }{4k} +\beta >0 \}\) and
where \(l^{-}(i)=\min \{l\in N:- (\frac{(4lk+2i+1)\pi }{2k} )^{2s+1}\tan \frac{(2i+1)\pi }{4k} +\beta <0 \}\).
In this case, we may choose
and let \(\sigma =\min \{\sigma_{0},\sigma_{1},\ldots ,\sigma_{m}\}>0\).
Second, we have that, for \(\beta \geq 0\) and \(s=\mathrm{odd}\), \((-1)^{s+1}=1\), \(i \in \{0,1,\ldots ,m\}\),
where \(l^{+}(i)=\max \{l\in N:- (\frac{(4lk+4k-2i-1)\pi }{2k} )^{2s+1}\tan \frac{(2i+1)\pi }{4k} +\beta >0 \}\) and
where \(l^{-}(i)=\min \{l\in N:- (\frac{(4lk+4k-2i-1)\pi }{2k} )^{2s+1}\tan \frac{(2i+1)\pi }{4k} +\beta <0 \}\).
In this case, we may choose
and let \(\sigma =\min \{\sigma_{0},\sigma_{1},\ldots ,\sigma_{m}\}>0\). The proof for the case \(\beta <0\) is similar. We omit it. The inequalities in (23) are proved. □
Lemma 3.2
Under conditions \((S_{1})\) and \((S_{2})\), the functional Φ defined by (11) satisfies the \((P.S)\)-condition.
Proof
Let Π, \(\mathbb{N}\), \(\mathbb{Z}\) be the orthogonal projections from X onto \(X_{\infty }^{+}\), \(X_{\infty }^{-}\), \(X_{\infty }^{0}\), respectively. From the second condition in (8) it follows that
for some \(M>0\).
Assume that \(\{x_{n}\}\subset X\) is a subsequence such that \(\varPhi '(x_{n})\rightarrow 0\) and \(\varPhi (x_{n})\) is bounded. Let \(w_{n}=\varPi x_{n}\), \(y_{n}=\mathbb{N}x_{n}\), \(z_{n}=\mathbb{Z}x_{n}\). Then we have
From
and (25), we have
and then, by (23), we have
which, together with \(\varPi \varPhi '(x_{n})\rightarrow 0\), implies the boundedness of \(w_{n}\). Similarly we have the boundedness of \(y_{n}\). At the same time, \((S_{2})\) yields
Then the boundedness of \(\varPhi (x)\) implies that \(\|z_{n}\|_{2}\) is bounded. Consequently \(\|z_{n}\|\) is bounded since \(X_{\infty }^{0}\) is finite-dimensional. Therefore, \(\|x_{n}\|\) is bounded.
It follows from (15) that
From the compactness of operator K and the boundedness of \(x_{n}\), we have that \(K(x_{n})\rightarrow u\). Then
The finite-dimensionality of \(X_{\infty }^{0}\) and the boundedness of \(z_{n}=\mathbb{Z}x_{n}\) imply \(z_{n}\rightarrow \varphi \in X_{\infty }^{0}\). Therefore,
which implies the \((P.S)\)-condition.
For (6), let
On the other hand,
Obviously, \(\dim X_{\infty }^{0}<\infty \) and \(\dim X_{0}^{0}< \infty \). □
Lemma 3.3
Under assumptions \((S_{1})\) and \((S_{2})\), there is \(\sigma >0\) such that
Proof
The proof is similar to Lemma 3.1, we omit it. □
Lemma 3.4
Under conditions \((S_{1})\) and \((S_{2})\), the functional Ψ defined by (16) satisfies the \((P.S)\)-condition.
Proof
The proof is similar to Lemma 3.2, we omit it. □
4 Notations and main results of this paper
We first give some notations.
For (5), if \((-1)^{s+1}=-1\), denote
And
Alternatively, if \((-1)^{s+1}=1\), denote
And
For (6), if \((-1)^{s+1}=-1\), denote
And
Alternatively, if \((-1)^{s+1}=1\), denote
And
Now we give the main results of this paper.
Theorem 4.1
Suppose that \((S_{1})\) and \((S_{2})\) hold. Then Eq. (5) possesses at least
4k-periodic solutions satisfying \(x(t-2k)=-x(t)\) provided that \(n > 0\).
Theorem 4.2
Suppose that \((S_{1})\), \((S_{2})\), \((S_{3}^{+})\), and \((S_{4}^{-})\) hold. Then Eq. (5) possesses at least
4k-periodic solutions satisfying \(x(t-2k)=-x(t)\) provided that \(n > 0\).
Theorem 4.3
Suppose that \((S_{1})\), \((S_{2})\), \((S_{3}^{-})\), and \((S_{4}^{+})\) hold. Then Eq. (5) possesses at least
4k-periodic solutions satisfying \(x(t-2k)=-x(t)\) provided that \(n > 0\).
Theorem 4.4
Suppose that \((S_{1})\) and \((S_{2})\) hold. Then Eq. (6) possesses at least
4k-periodic solutions satisfying \(x(t-2k)=-x(t)\) provided that \(n > 0\).
Theorem 4.5
Suppose that \((S_{1})\), \((S_{2})\), \((S_{3}^{+})\), and \((S_{4}^{-})\) hold. Then Eq. (6) possesses at least
4k-periodic solutions satisfying \(x(t-2k)=-x(t)\) provided that \(n > 0\).
Theorem 4.6
Suppose that \((S_{1})\), \((S_{2})\), \((S_{3}^{-})\), and \((S_{4}^{+})\) hold. Then Eq. (6) possesses at least
4k-periodic solutions satisfying \(x(t-2k)=-x(t)\) provided that \(n > 0\).
5 Proof of main results of this paper
Proof of Theorem 4.1
Suppose without loss of generality that
Let \(X^{+}=X_{\infty }^{+}\) and \(X^{-}=X_{0}^{-}\). Then
Obviously,
which implies that condition (a) in Lemma 1.1 holds. Let \(A_{\infty }= \beta \). Then condition (b) in Lemma 1.1 holds since, for each \(j\in N\), we have that \(x\in X(j)\) yields \((L+P^{-1}\beta )x\in X(j)\).
At the same time, Lemma 3.2 gives the \((P.S)\)-condition.
Now it suffices to show that conditions (c) and (d) in Lemma 1.1 hold under assumptions \((S_{1})\) and \((S_{2})\).
In fact, we have shown in Lemma 3.1 that there is \(\sigma >0\) such that \(\langle (L+P^{-1}\beta )x,x \rangle >\sigma \|x\|^{2}\), \(x \in X_{\infty }^{+}\). And the second condition in (8) implies that \(|F(x)-\frac{1}{2}\beta x^{2}|<\frac{1}{4}\sigma |x|^{2}+M_{1}\), \(x \in R\) for some \(M_{1}>0\).
Then
if \(x\in X^{+}\). Clearly, there is \(c_{0}\in R\) such that
On the other hand, we have shown in Lemma 3.1 that there is \(\sigma >0\) such that \(\langle (L+P^{-1}\beta )x,x \rangle <- \sigma \|x\|^{2}\), \(x\in X_{\infty }^{-}\). And we can show that there are \(r, \sigma >0\) such that \(|F(x)-\frac{1}{2}\beta x^{2}|< \frac{1}{4} \sigma |x|^{2}\), \(\|x\|=r\). So
That is, there are \(r>0\) and \(c_{\infty }<0\) such that
Our last task is to compute the value of
By computation we get that, for each \(i\in \{0,1,\ldots ,m\}\),
and
Therefore,
-
\(X_{\infty }^{+}(2lk+i)=X_{\infty }^{+}\cap X(2lk+i)=\emptyset \),
-
\(X_{\infty }^{+}(2lk+2k-i-1)=X_{\infty }^{+}\cap X(2lk+2k-i-1)=X(2lk+2k-i-1)\),
-
\(X_{0}^{-}(2lk+i)=X_{0}^{-}\cap X(2lk+i)=X(2lk+i)\),
-
\(X_{0}^{-}(2lk+2k-i-1)=X_{0}^{-}\cap X(2lk+2k-i-1)=\emptyset \)
if \(i\in \{0,1,\ldots ,m\}\) and \((-1)^{s+1}=1\) and \(l\geq 0\) is large enough, which means that there is \(M>0\) such that \(\dim (X_{\infty } ^{+}(j)\cap X_{0}^{-}(j))-\operatorname{codim}_{X}(X_{\infty }^{+}(j)+X_{0} ^{-}(j))=0\), \(j>M\), from which it follows that
Then we have
and
Therefore
The proof for the case \((-1)^{s+1}=-1\) is similar.
Theorem 4.1 is proved. □
Proof of Theorem 4.2 and Theorem 4.3
Since the proof for the two theorems is similar, we prove only Theorem 4.2.
Let \(X^{+}=X_{\infty }^{+}+X_{\infty }^{0}\), \(X^{-}=X_{-}^{0}+X_{0} ^{0}\). Then, as in the proof of Theorem 4.1, we check conditions (a), (b), (c), (d), and (e). In the present case, we may suppose that (28) still holds for some \(M>0\). Let \(X_{\infty }^{0}(i)=X_{\infty }^{0} \cap X(i)\), \(X _{0}^{0}(i)=X_{0}^{0}\cap X(i)\). Then
□
Proof of Theorem 4.4, Theorem 4.5, and Theorem 4.6
Since the proof of Theorem 4.4 is similar to that of Theorem 4.1, and the proofs of Theorems 4.5 and 4.6 are similar to that of Theorem 4.2, we omit it. Our proof is completed. □
6 Examples
Example 6.1
Suppose that \(f\in C^{0}(R,R)\) satisfies
We are to discuss the multiplicity of 8-periodic solutions of the equation
In this case, \(s=1\), \((-1)^{s+1}=1\), \(k=2\), \(m=1\), \(\alpha =-\pi^{3}\), \(\beta =5\pi^{3}\). This yields that
and \(N_{1}^{0}(\alpha_{+})=N_{1}^{0}(\beta_{-})=N_{1}^{0}(\alpha_{-})=N _{1}^{0}(\beta_{+})=0\).
Applying Theorem 4.2, we conclude that Eq. (29) possesses at least three different 8-periodic orbits satisfying \(x(t-4)=-x(t)\).
Example 6.2
Suppose that \(f\in C^{0}(R,R)\) satisfies
We are to discuss the multiplicity of 12-periodic solutions of the equation
In this case, \(s=0\), \((-1)^{s+1}=-1\), \(k=3\), \(m=2\), \(\alpha =\pi \), \(\beta =3\pi \). This yields that
and \(N_{2}^{0}(\alpha_{+})=N_{2}^{0}(\beta_{-})=N_{2}^{0}(\alpha_{-})=N _{2}^{0}(\beta_{+})=0\).
Applying Theorem 4.5, we conclude that Eq. (30) possesses at least five different 12-periodic orbits satisfying \(x(t-6)=-x(t)\).
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Acknowledgements
The present research is supported by the National Natural Science Foundations of China (No. 61179031). The authors thank the referees for carefully reading the manuscript and for their valuable suggestions which have significantly improved the paper.
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Li, L., Sun, H. & Ge, W. Multiple periodic solutions of high order differential delay equations with \(2k-1\) lags. Adv Differ Equ 2019, 3 (2019). https://doi.org/10.1186/s13662-018-1928-9
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DOI: https://doi.org/10.1186/s13662-018-1928-9