Erratum to: Existence and uniqueness of anti-periodic solutions for prescribed mean curvature Rayleigh equations

Abstract

In this paper, we give a complementary proof on the paper ‘Existence and uniqueness of anti-periodic solutions for prescribed mean curvature Rayleigh equations’.

1 Introduction

In [1], the authors were concerned with the existence and uniqueness of anti-periodic solutions of the following prescribed mean curvature Rayleigh equation:

$${\left(\frac{x^{\prime }}{\sqrt{1+x^{\prime 2}}}\right)}^{\prime }+f(t,x^{\prime }(t))+g(t,x(t))=e(t),$$

(1.1)

where $e\in C(R,R)$ is T-periodic, and $f,g\in C(R\times R,R)$ are T-periodic in the first argument, T is a constant.

The paper mentioned above obtained the main result by using Mawhin’s continuation theorem in the coincidence degree theory. Unfortunately, the proof of main result Theorem 3.1 (see [1]) has a serious problem: $F_{\mu }(x)=\mu L(Q_1(t,x_1,x_2))$ where $Q_1$ depends on $\psi (x_2)$ and $\psi (x)=\frac{x}{\sqrt{1-x^2}}$ which is only defined for $|x|<1$ and cannot be continuously extended; therefore, $F_{\mu }$ should not be defined on $\overline{\mathrm{\Omega }}=\{x\in X:\parallel x\parallel <M\}$ since $|x_2(t)|>1$ can occur, where $\parallel x\parallel =max\{{\parallel x_1\parallel }_{\mathrm{\infty }},{\parallel x_2\parallel }_{\mathrm{\infty }}\}$ and $M=1+max\{D_1,D_2\}$.

In this paper, we shall give a complementary proof to correct the errors.

2 Complementary proof

Rewrite (1.1) in the equivalent form as follows:

$$\{\begin{array}{c}{x}_1^{\prime }(t)=\psi (x_2(t))=\frac{x_2(t)}{\sqrt{1-x_2^2(t)}},\hfill \\ x_2^{\prime }(t)=-f(t,\psi (x_2(t)))-g(t,x_1(t))+e(t),\hfill \end{array}$$

(2.1)

where $\psi (x)=\frac{x}{\sqrt{1-x^2}}$. In [1], the authors embed (2.1) into a family of equations with one parameter $\lambda \in (0,1]$,

$$\{\begin{array}{c}{x}_1^{\prime }(t)=\lambda \frac{x_2(t)}{\sqrt{1-x_2^2(t)}}=\lambda \psi (x_2(t)),\hfill \\ x_2^{\prime }(t)=-\lambda f(t,\psi (x_2(t)))-\lambda g(t,x_1(t))+\lambda e(t).\hfill \end{array}$$

(2.2)

They have proved that there exists a constant $D_1>0$ such that

$$|x_1^{\prime }{|}_2\le D_1,\text{and}|x_1{|}_{\mathrm{\infty }}\le D_1,$$

(2.3)

and there exists $\eta \in [0,T]$ such that $x_2(\eta )=0$.

In fact, to use the continuation theorem, it suffices to prove that there exists a positive constant $0<{\epsilon }_0\ll 1$ such that, for any possible solution ($x_1(t),x_2(t)$) of (2.2), the following condition holds:

$$|x_2(t)|<1-{\epsilon }_0.$$

(2.4)

In what follows, we shall give a complementary proof for the main result in [1] by giving a proof of (2.4).

In [1], the authors assume that

(H₁): $(g(t,x_1)-g(t,x_2))(x_1-x_2)<0$, for all $t,x_1,x_2\in R$ and $x_1\ne x_2$;

(H₂): there exists $l>0$ such that

$$|g(t,x_1)-g(t,x_2)|\le l|x_1-x_2|\text{for all }t,x_1,x_2\in R;$$

(H₃): there exists β, γ such that

$$\gamma \le \underset{|x|\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{f(t,x)}{x}\le \underset{|x|\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{f(t,x)}{x}\le \beta ,\text{uniformly in }t\in R;$$

(H₄): for all $t,x\in R$,

$$f(t+\frac{T}{2},-x)=-f(t,x),g(t+\frac{T}{2},-x)=-g(t,x),e(t+\frac{T}{2})=-e(t).$$

Under the conditions mentioned above, we prove that (2.4) holds.

Since ${|x_1|}_{\mathrm{\infty }}<D_1$ and g, e are continuous, we find that there exists $M_3>0$ such that

$$-M_3<-g(t,x_1(t))+e(t)<M_3,\mathrm{\forall }t\in R.$$

(2.5)

By (H₃), there exists a positive constant $M_4>0$ such that

$$f(t,x)\ge \gamma x-M_4,\mathrm{\forall }x>0\text{ and }\mathrm{\forall }t\in R.$$

(2.6)

Next, we shall prove that

$$x(t)\le \frac{M_3+M_4}{\sqrt{{(M_3+M_4)}^2+{\gamma }^2}},\mathrm{\forall }t\in R.$$

Assume by contradiction that there exist $t_2^{\ast }>t_1^{\ast }>\eta$ such that

$$x_2\left(t_1^{\ast }\right)=\frac{M_3+M_4}{\sqrt{{(M_3+M_4)}^2+{\gamma }^2}},x_2\left(t_2^{\ast }\right)>\frac{M_3+M_4}{\sqrt{{(M_3+M_4)}^2+{\gamma }^2}},$$

and

$$x_2(t)>\frac{M_3+M_4}{\sqrt{{(M_3+M_4)}^2+{\gamma }^2}},\text{for }t\in (t_1^{\ast },t_2^{\ast }).$$

Noticing that $\lambda \in (0,1]$, we have, $\mathrm{\forall }t\in (t_1^{\ast },t_2^{\ast })$,

$$x_2^{\prime }(t)=\lambda (-f(t,\psi (x_2(t)))-g(t,x_1(t))+e(t))<0,$$

which is a contradiction.

By (H₃), there exists a positive constant $M_5>0$ such that

$$f(t,x)\le \beta x+M_5,\mathrm{\forall }x<0\text{ and }\mathrm{\forall }t\in R.$$

By using a similar argument, we can prove that

$$x_2(t)\ge -\frac{M_3+M_5}{\sqrt{{(M_3+M_5)}^2+{\beta }^2}},\text{for }t\in R.$$

Therefore, we get from the continuity of $x_2(t)$, for any solution ($x_1(t),x_2(t)$) of (2.2),

$$-\frac{M_3+M_5}{\sqrt{{(M_3+M_5)}^2+{\beta }^2}}\le x_2(t)\le \frac{M_3+M_4}{\sqrt{{(M_3+M_4)}^2+{\gamma }^2}},\mathrm{\forall }t\in R.$$

Consequently, (2.4) holds.

Putting

$$\mathrm{\Omega }=\{x=(x,x)\in C_T^{0,\frac{1}{2}}(R,R^2)=X:\parallel x\parallel <M,|x_2(t)|<1-{\epsilon }_0\},$$

we can use Mawhin’s continuation theorem on Ω.