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.