In this section, some existence results on anti-periodic solutions with symmetry of (1.3) and (1.4) will be given.

**Theorem 3.1**
*Assume that*

(H

_{1})

*the functions* $g(t,x)$ *and* $e(t)$ *are odd in* *t*,

*i*.

*e*.,

$g(-t,\cdot )=-g(t,\cdot ),\phantom{\rule{2em}{0ex}}e(-t)=-e(t),\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in \mathbb{R};$

(H

_{2})

*there exist non*-

*negative functions* ${\alpha}_{1},{\beta}_{1}\in C(\mathbb{R},{\mathbb{R}}^{+})$ *such that* $|g(t,x)|\le {\alpha}_{1}(t)|x|+{\beta}_{1}(t),\phantom{\rule{1em}{0ex}}\mathrm{\forall}t,x\in \mathbb{R};$

(H_{3}) ${\sum}_{i=1}^{m}|{a}_{i}|+{\parallel {\alpha}_{1}\parallel}_{0}-1<0$.

*Then* (1.3)

*has at least one even anti*-

*periodic solution* $x(t)$,

*i*.

*e*.,

$x(t)$ *satisfies* $x(t+\pi )=-x(t),\phantom{\rule{2em}{0ex}}x(-t)=x(t),\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in \mathbb{R}.$

*Proof* For making use of the Leray-Schauder degree theory to prove the existence of even anti-periodic solutions for (1.3), we consider the following homotopic equation of (1.3):

${x}^{(2m+1)}=-\lambda \sum _{i=1}^{m}{a}_{i}{x}^{(2i-1)}-\lambda g(t,x)+\lambda e(t),\phantom{\rule{1em}{0ex}}\lambda \in [0,1].$

(3.1)

Define the operator

${D}_{01}:{C}_{0}^{2m+1,\pi}\u27f6{C}_{1}^{0,\pi}$ by

$({D}_{01}x)(t)={x}^{(2m+1)}(t),\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in \mathbb{R}.$

Obviously, the operator

${D}_{01}$ is invertible. Let

${N}_{01}:{C}_{0}^{2m-1,\pi}\u27f6{C}^{0}$ be the Nemytskii operator

$({N}_{01}x)(t)=-\sum _{i=1}^{m}{a}_{i}{x}^{(2i-1)}(t)-g(t,x(t))+e(t),\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in \mathbb{R}.$

By hypothesis (H

_{1}), it is easy to see that

$({N}_{01}x)(t+\pi )\equiv -({N}_{01}x)(t),\phantom{\rule{2em}{0ex}}({N}_{01}x)(-t)\equiv -({N}_{01}x)(t),\phantom{\rule{1em}{0ex}}\mathrm{\forall}x\in {C}_{0}^{2m-1,\pi}.$

Thus, the operator

${N}_{01}$ sends

${C}_{0}^{2m-1,\pi}$ into

${C}_{1}^{0,\pi}$. Hence, the problem of even anti-periodic solutions for (3.1) is equivalent to the operator equation

${D}_{01}x=\lambda {N}_{01}x,\phantom{\rule{1em}{0ex}}x\in {C}_{0}^{2m+1,\pi}.$

From hypotheses (H

_{2}), (H

_{3}) and (5) in [

10], for the possible even anti-periodic solution

$x(t)$ of (3.1), there exists

*a prior bounds* in

${C}_{0}^{2m+1,\pi}$,

*i.e.*,

$x(t)$ satisfies

${\parallel x\parallel}_{{C}^{2m+1}}\le {T}_{1},$

(3.2)

where ${T}_{1}$ is a positive constant independent of *λ*. So, our problem is reduced to construct one completely continuous operator ${F}_{\lambda}$, which sends ${C}_{0}^{2m+1,\pi}$ into ${C}_{0}^{2m+1,\pi}$, such that the fixed points of operator ${F}_{1}$ in some open bounded set are the even anti-periodic solutions of (1.3).

With this in mind, let us define the set as follows:

${\mathrm{\Omega}}_{01}=\{x\in {C}_{0}^{2m+1,\pi}:{\parallel x\parallel}_{{C}^{2m+1}}<{T}_{1}+1\}.$

Obviously, the set

${\mathrm{\Omega}}_{01}$ is a open bounded set in

${C}_{0}^{2m+1,\pi}$ and zero element

$\theta \in {\mathrm{\Omega}}_{01}$. Define the completely continuous operator

${F}_{\lambda}:\overline{{\mathrm{\Omega}}_{01}}\u27f6{C}_{0}^{2m+1,\pi}$ by

${F}_{\lambda}x=\underset{2m+1}{\underset{\u23df}{{J}_{1}{J}_{0}\cdots {J}_{0}{J}_{1}}}\lambda {N}_{01}x=\lambda {D}_{01}^{-1}{N}_{01}x,\phantom{\rule{1em}{0ex}}\lambda \in [0,1].$

Let us define the completely continuous field

${h}_{\lambda}(x):\overline{{\mathrm{\Omega}}_{01}}\times [0,1]\u27f6{C}_{0}^{2m+1,\pi}$ by

${h}_{\lambda}(x)=x-{F}_{\lambda}x.$

By (3.2), we get that zero element

$\theta \notin {h}_{\lambda}(\partial \mathrm{\Omega})$ for all

$\lambda \in [0,1]$. So, the following Leray-Schauder degrees are well defined and

$\begin{array}{rcl}deg(id-{F}_{1},\mathrm{\Omega},\theta )& =& deg({h}_{1},\mathrm{\Omega},\theta )\\ =& deg({h}_{0},\mathrm{\Omega},\theta )=deg(id,\mathrm{\Omega},\theta )=1\ne 0.\end{array}$

Consequently, the operator ${F}_{1}$ has at least one fixed point in ${\mathrm{\Omega}}_{01}$ by using Lemma 2.1. Namely, (1.3) has at least one even anti-periodic solution. The proof is complete. □

**Theorem 3.2**
*Assume that*

(H

_{4})

*the function* $g(t,x)$ *is even in* *t*,

*x* *and* $e(t)$ *is even in* *t*,

*i*.

*e*.,

$g(-t,-x)=g(t,x),\phantom{\rule{2em}{0ex}}e(-t)=e(t),\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in \mathbb{R}$

*and the assumptions* (H_{2}), (H_{3}) *are true*.

*Then* (1.3)

*has at least one odd anti*-

*periodic solution* $x(t)$,

*i*.

*e*.,

$x(t)$ *satisfies* $x(t+\pi )=-x(t),\phantom{\rule{2em}{0ex}}x(-t)=-x(t),\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in \mathbb{R}.$

*Proof* We consider the homotopic equation (

3.1) of (1.3). Define the operator

${D}_{11}:{C}_{1}^{2m+1,\pi}\u27f6{C}_{0}^{0,\pi}$ by

$({D}_{11}x)(t)={x}^{(2m+1)}(t),\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in \mathbb{R}.$

Let

${N}_{11}:{C}_{1}^{2m-1,\pi}\u27f6{C}^{0,\pi}$ be the Nemytskii operator

$({N}_{11}x)(t)=-\sum _{i=1}^{m}{a}_{i}{x}^{(2i-1)}(t)-g(t,x(t))+e(t),\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in \mathbb{R}.$

By hypothesis (H

_{4}), it is easy to see that

$({N}_{11}x)(-t)\equiv ({N}_{11}x)(t),\phantom{\rule{1em}{0ex}}\mathrm{\forall}x\in {C}_{1}^{2m-1,\pi}.$

Thus, the operator

${N}_{11}$ sends

${C}_{1}^{2m-1,\pi}$ into

${C}_{0}^{0,\pi}$. Hence, the problem of odd anti-periodic solutions for (3.1) is equivalent to the operator equation

${D}_{11}x=\lambda {N}_{11}x,\phantom{\rule{1em}{0ex}}x\in {C}_{1}^{2m+1,\pi}.$

Our problem is reduced to construct one completely continuous operator

${G}_{\lambda}$, which sends

${C}_{1}^{2m+1,\pi}$ into

${C}_{1}^{2m+1,\pi}$, such that the fixed points of operator

${G}_{1}$ in some open bounded set are the odd anti-periodic solutions of (1.3). With this in mind, let us define the following set:

${\mathrm{\Omega}}_{11}=\{x\in {C}_{1}^{2m+1,\pi}:{\parallel x\parallel}_{{C}^{2m+1}}<{T}_{1}+1\}.$

Define the completely continuous operator

${G}_{\lambda}:\overline{{\mathrm{\Omega}}_{11}}\u27f6{C}_{1}^{2m+1,\pi}$ by

${G}_{\lambda}x=\underset{2m+1}{\underset{\u23df}{{J}_{0}{J}_{1}\cdots {J}_{1}{J}_{0}}}\lambda {N}_{11}x=\lambda {D}_{11}^{-1}{N}_{11}x,\phantom{\rule{1em}{0ex}}\lambda \in [0,1].$

The remainder of the proof work is quite similar to the proof of Theorem 3.1, so we omit the details. The proof is complete. □

**Theorem 3.3**
*Assume that*

(H

_{5})

*the functions* $g(t,x)$ *and* $e(t)$ *are even in* *t*,

*i*.

*e*.,

$g(-t,\cdot )=g(t,\cdot ),\phantom{\rule{2em}{0ex}}e(-t)=e(t),\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in \mathbb{R}$

*and the assumptions* (H_{2}), (H_{3}) *are true*.

*Then* (1.4) *has at least one even anti*-*periodic solution*.

*Proof* We consider the homotopic equation of (1.4) as follows:

${x}^{(2m+2)}=-\lambda \sum _{i=1}^{m}{a}_{i}{x}^{(2i)}-\lambda g(t,x)+\lambda e(t),\phantom{\rule{1em}{0ex}}\lambda \in [0,1].$

(3.3)

Define the operator

${D}_{02}:{C}_{0}^{2m+2,\pi}\u27f6{C}_{0}^{0,\pi}$ by

$({D}_{02}x)(t)={x}^{(2m+2)}(t),\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in \mathbb{R}.$

Let

${N}_{02}:{C}_{0}^{2m,\pi}\u27f6{C}^{0,\pi}$ be the Nemytskii operator

$({N}_{02}x)(t)=-\sum _{i=1}^{m}{a}_{i}{x}^{(2i)}(t)-g(t,x(t))+e(t),\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in \mathbb{R}.$

By hypothesis (H

_{5}), it is easy to see that

$({N}_{02}x)(-t)\equiv ({N}_{02}x)(t),\phantom{\rule{1em}{0ex}}\mathrm{\forall}x\in {C}_{0}^{2m,\pi}.$

Thus, the operator

${N}_{02}$ sends

${C}_{0}^{2m,\pi}$ into

${C}_{0}^{0,\pi}$. Hence, the problem of even anti-periodic solutions for (3.3) is equivalent to the operator equation

${D}_{02}x=\lambda {N}_{02}x,\phantom{\rule{1em}{0ex}}x\in {C}_{0}^{2m+2,\pi}.$

Our problem is reduced to construct one completely continuous operator

${L}_{\lambda}$, which sends

${C}_{0}^{2m+2,\pi}$ into

${C}_{0}^{2m+2,\pi}$, such that the fixed points of operator

${L}_{1}$ in some open bounded set are the even anti-periodic solutions of (1.4). With this in mind, let us define the following set:

${\mathrm{\Omega}}_{02}=\{x\in {C}_{0}^{2m+2,\pi}:{\parallel x\parallel}_{{C}^{2m+2}}<{T}_{2}+1\},$

where

${T}_{2}$ is a positive constant independent of

*λ*. Define the completely continuous operator

${L}_{\lambda}:\overline{{\mathrm{\Omega}}_{02}}\u27f6{C}_{0}^{2m+2,\pi}$ by

${L}_{\lambda}x=\underset{2m+2}{\underset{\u23df}{{J}_{1}{J}_{0}\cdots {J}_{1}{J}_{0}}}\lambda {N}_{02}x=\lambda {D}_{02}^{-1}{N}_{02}x,\phantom{\rule{1em}{0ex}}\lambda \in [0,1].$

The remainder of the proof work is quite similar to the proof of Theorem 3.1, so we omit the details. The proof is complete. □

**Theorem 3.4**
*Assume that*

(H

_{6})

*the function* $g(t,x)$ *is odd in* *t*,

*x* *and* $e(t)$ *is odd in* *t*,

*i*.

*e*.,

$g(-t,-x)=-g(t,x),\phantom{\rule{2em}{0ex}}e(-t)=-e(t),\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in \mathbb{R}$

*and the assumptions* (H_{2}), (H_{3}) *are true*.

*Then* (1.4) *has at least one odd anti*-*periodic solution*.

*Proof* We consider the homotopic equation (

3.3) of (1.4). Define the operator

${D}_{12}:{C}_{1}^{2m+2,\pi}\u27f6{C}_{1}^{0,\pi}$ by

$({D}_{12}x)(t)={x}^{(2m+2)}(t),\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in \mathbb{R}.$

Let

${N}_{12}:{C}_{1}^{2m,\pi}\u27f6{C}^{0,\pi}$ be the Nemytskii operator

$({N}_{12}x)(t)=-\sum _{i=1}^{m}{a}_{i}{x}^{(2i)}(t)-g(t,x(t))+e(t),\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in \mathbb{R}.$

By hypothesis (H

_{6}), it is easy to see that

$({N}_{12}x)(-t)\equiv -({N}_{12}x)(t),\phantom{\rule{1em}{0ex}}\mathrm{\forall}x\in {C}_{1}^{2m,\pi}.$

Thus, the operator

${N}_{12}$ sends

${C}_{1}^{2m,\pi}$ into

${C}_{1}^{0,\pi}$. Hence, the problem of odd anti-periodic solutions for (3.3) is equivalent to the operator equation

${D}_{12}x=\lambda {N}_{12}x,\phantom{\rule{1em}{0ex}}x\in {C}_{1}^{2m+2,\pi}.$

Our problem is reduced to construct one completely continuous operator

${P}_{\lambda}$ which sends

${C}_{1}^{2m+2,\pi}$ into

${C}_{1}^{2m+2,\pi}$, such that the fixed points of operator

${P}_{1}$ in some open bounded set are the odd anti-periodic solutions of (1.4). With this in mind, let us define the set as follows:

${\mathrm{\Omega}}_{12}=\{x\in {C}_{1}^{2m+2,\pi}:{\parallel x\parallel}_{{C}^{2m+2}}<{T}_{2}+1\}.$

Define the completely continuous operator

${P}_{\lambda}:\overline{{\mathrm{\Omega}}_{12}}\u27f6{C}_{1}^{2m+2,\pi}$ by

${P}_{\lambda}x=\underset{2m+2}{\underset{\u23df}{{J}_{0}{J}_{1}\cdots {J}_{0}{J}_{1}}}\lambda {N}_{12}x=\lambda {D}_{12}^{-1}{N}_{12}x,\phantom{\rule{1em}{0ex}}\lambda \in [0,1].$

The remainder of the proof work is quite similar to the proof of Theorem 3.1, so we omit the details. The proof is complete. □

When $g(t,x)=g(x)$, we can remove the assumption (H_{2}) in Theorem 3.1, Theorem 3.2 and obtain the following results.

**Theorem 3.5**
*Assume that*

(H_{7}) ${\sum}_{i=1}^{m}|{a}_{i}|-1<0$ *and the assumption* (H_{1}) *is true*.

*Then* (1.3) ($g(t,x)=g(x)$) *has at least one even anti*-*periodic solution*.

**Theorem 3.6** *Suppose that the assumptions* (H_{4}), (H_{7}) *are true*. *Then* (1.3) ($g(t,x)=g(x)$) *has at least one odd anti*-*periodic solution*.

Basing on the proof of Theorem 2 in [

10], for the possible anti-periodic solution

$x(t)$ of (3.1) (

$g(t,x)=g(x)$), the hypothesis (H

_{7}) yields that there exists

*a prior bounds* in

${C}^{2m+1,\pi}$,

*i.e.*,

$x(t)$ satisfies

${\parallel x\parallel}_{{C}^{2m+1}}\le {T}_{3},$

where ${T}_{3}$ is a positive constant independent of *λ*. The remainder of the proof work of Theorem 3.5 and Theorem 3.6 is quite similar to the proof of Theorem 3.1 and Theorem 3.2, so we omit the details.