Correction to: Periodic Solutions in Slowly Varying Discontinuous Differential Equations: A Non-Generic Case

Correction to: Journal of Dynamics and Differential Equations https://doi.org/10.1007/s10884-022-10155-0

1 Introduction

We correct a mistake on the Melnikov function given in (Battelli and Fečkan in J Dyn Differ Equ, 2022. https://doi.org/10.1007/s10884-022-10155-0) for the persistence of periodic solutions in perturbed slowly varying discontinuous differential equations.

In [1] the following piecewise $C^r$ system of differential equations

$$\begin{aligned} \begin{array}{l} {\dot{x}} = f(x,y) \\ \dot{y} = \varepsilon g(x,y,\varepsilon ), \quad \varepsilon \in {\mathbb {R}}\end{array} \end{aligned}$$

(1.1)

has been considered where

$$\begin{aligned} f(x,y) := \left\{ \begin{array}{ll} f_-(x,y) &{} \hbox {if }h(x,y)<0 \\ f_+(x,y) &{} \hbox {if }h(x,y)>0 \end{array} \right. \end{aligned}$$

(1.2)

and $f_\pm (x,y)$ and h(x, y) are $C^r$-functions, $r\ge 2$, $x\in {\mathbb {R}}^n$, $y\in {\mathbb {R}}^m$. Suppose that the unperturbed system

$$\begin{aligned} {\dot{x}} = f(x,\eta ), \; \eta \in {\mathbb {R}}^m \end{aligned}$$

has a family of ${\hat{T}}(\alpha ,\eta )$-periodic, continuous and piecewise $C^r$, solutions ${\bar{u}}(t,\alpha ,\eta )$, with $(\alpha ,\eta )\in {\mathbb {R}}^d\times {\mathbb {R}}^m$, where ${\hat{T}}(\alpha ,\eta )$ is a smooth function of $(\alpha ,\eta )\in {\mathbb {R}}^d\times {\mathbb {R}}^m$. Moreover there exist $t_*(\alpha ,\eta )<t^*(\alpha ,\eta )$ such that the following hold

• $h({\bar{u}}(t,\alpha ,\eta ),\eta )>0$ for $0\le t < t_*(\alpha ,\eta )$ and $t^*(\alpha ,\eta )< t \le {\hat{T}}(\alpha ,\eta )$

• $h({\bar{u}}(t,\alpha ,\eta ),\eta )<0$ for $t_*(\alpha ,\eta )< t < t^*(\alpha ,\eta )$

• $\bar{u}(t,\alpha ,\eta )$ intersects transversally the manifold $\{x | h(x,\eta )=0\}$ at $t=t_*(\alpha ,\eta )$, $t=t^*(\alpha ,\eta )$.

[Actually in [1] $t_*(\alpha ,\eta )$, $t^*(\alpha ,\eta )$ are denoted with ${\hat{t}}_*(\alpha ,\eta )$, ${\hat{t}}^*(\alpha ,\eta )$]. In [1] the following result has been stated:

Theorem 1.1

Suppose that conditions $(A_1){-}(A_5)$ in [1] hold and let $B_*, B^*, b_*, b^*$ be as in [1, Proposition 3.9]. Then the adjoint linear system

$$\begin{aligned} {\dot{x}} = -f_x({\bar{u}}(t,\alpha ,\eta ),\eta )^t x = \left\{ \begin{array}{ll} -f_{+,x}({\bar{u}}(t,\alpha ,\eta ),\eta )^t x &{} \hbox {if }h({\bar{u}}(t,\alpha ,\eta ),\eta )>0 \\ -f_{-,x}({\bar{u}}(t,\alpha ,\eta ),\eta )^t x &{} \hbox {if }h({\bar{u}}(t,\alpha ,\eta ),\eta )<0 \end{array}\right. \end{aligned}$$

has a d-dimensional space of ${\hat{T}}(\alpha ,\eta )$-periodic solutions $v(t,\alpha ,\eta )$ that are $C^r$ for $t\ne t_*(\alpha ,\eta ), t^*(\alpha ,\eta )$ with jumps at these points given by:

$$\begin{aligned} \begin{array}{l} v(t_*(\alpha ,\eta )^+,\alpha ,\eta ) = [B_*^{-1}]^tv(t_*(\alpha ,\eta )^-,\alpha ,\eta ), \\ v(t^*(\alpha ,\eta )^+,\alpha ,\eta ) = [B^{*-1}]^tv(t^*(\alpha ,\eta )^-,\alpha ,\eta ) \end{array} \end{aligned}$$

(1.3)

Moreover if $\{v_1(t,\alpha ,\eta ),\ldots ,v_d(t,\alpha ,\eta )\}$ is a $C^{r-1}$-basis of the space of ${\hat{T}}(\alpha ,\eta )$-periodic solutions of the adjoint linear system satisfying (1.3) and the $(d+m)$-dimensional Melnikov vector

$$\begin{aligned} \mathcal{M}(\alpha ,\eta ) := \begin{pmatrix} \mathcal{M}_1(\alpha ,\eta ) \\ \mathcal{M}_2(\alpha ,\eta ) \end{pmatrix} \end{aligned}$$

where

$$\begin{aligned} \displaystyle \mathcal{M}_1(\alpha ,\eta )= & {} \Big [\int _0^{{\hat{T}}(\alpha ,\eta )} v_j(t,\alpha ,\eta )^t f_y({\bar{u}}(t,\alpha ,\eta ),\eta )g({\bar{u}}(t,\alpha ,\eta ),\eta ,0) dt \\{} & {} +\, v_j({\hat{t}}_*(\alpha ,\eta )^+,\alpha ,\eta )^t b_* + v_j({\hat{t}}^*(\alpha ,\eta )^+,\alpha ,\eta )^t b^*\Big ]_{j=1,\ldots d} \\ \displaystyle \mathcal{M}_2(\alpha ,\eta )= & {} \int _0^{{\hat{T}}(\alpha ,\eta )} g({\bar{u}}(t,\alpha ,\eta ),\eta ,0) dt, \end{aligned}$$

has a simple zero at $(\alpha ,\eta )=(\alpha _0,\eta _0)$, then there exists $\varepsilon _0>0$ such that for $|\varepsilon |<\varepsilon _0$ there exist $C^r$ functions $\alpha (\varepsilon ), \eta (\varepsilon )$ with $(\alpha (0),\eta (0))=(\alpha _0,\eta _0)$ such that system (1.1) has a unique ${\tilde{T}}(\varepsilon )$-periodic solution $(x(t,\varepsilon ),y(t,\varepsilon ))$ satisfying

$$\begin{aligned} \sup _{0\le t\le {\tilde{T}}(\varepsilon )} |(x(t,\varepsilon ) - {\bar{u}}(t,\alpha (\varepsilon ),\eta (\varepsilon ))| + |(y(t,\varepsilon ) - \eta (\varepsilon )| \rightarrow 0 \end{aligned}$$

as $\varepsilon \rightarrow 0$, and $\lim _{\varepsilon \rightarrow 0} {\tilde{T}}(\varepsilon )={\hat{T}}(\alpha _0,\eta _0)$.

We have now realised that the expression of $\mathcal{M}_1(\alpha ,\eta )$ given in the Theorem 1.1 above is not correct. Indeed Theorem 1.1 is obtained as an application of [1, Theorem 3.4] that correctly states

Theorem 1.2

Suppose that conditions $(A_1){-}(A_5)$ in [1] hold and let

$$\begin{aligned} J_{11}(\alpha ,\eta ) = \frac{\partial }{\partial \xi }[u(T(\xi ,\eta ),\xi ,\eta )-\xi ]_{\xi =\xi (\alpha ,\eta )} \end{aligned}$$

where $u(t,\xi ,\eta )$ and $T(\xi ,\eta )$, $\xi (\alpha ,\eta )$ have been defined in [1]. Let

$$\begin{aligned} \{\zeta _1(\alpha ,\eta ),\ldots \zeta _d(\alpha ,\eta )\} \end{aligned}$$

be a smooth basis of $\mathcal{N}J_{11}(\alpha ,\eta )^t$. Suppose, further, that $(\alpha ,\eta )=(0,0)$ is a simple solution of the system

$$\begin{aligned} \left\{ \begin{array}{l} \zeta _j(\alpha ,\eta )^t\mathcal{P}(\alpha ,\eta )x_\varepsilon ({\hat{T}}(\alpha ,\eta ),\xi (\alpha ,\eta ),\eta ,0) = 0, \quad j=1,\ldots , d \\ y_\varepsilon ({\hat{T}}(\alpha ,\eta ),\xi (\alpha ,\eta ),\eta ,0) = 0. \end{array}\right. \end{aligned}$$

(1.4)

where

$$\begin{aligned} \mathcal{P}(\alpha ,\eta )x = x - \frac{f(0,0)^tx}{f(0,0)^tf(\xi (\alpha ,\eta ),\eta )}f(\xi (\alpha ,\eta ),\eta ) \end{aligned}$$

is the projection with range $\{f(0,0)\}^\perp$ and kernel the span of $f(\xi (\alpha ,\eta ),\eta )$.

Then there exists $\varepsilon _0>0$ such that, for any $|\varepsilon |<\varepsilon _0$ equation (1.1) has a ${\tilde{T}}(\varepsilon )$-periodic solution $(x(t,\varepsilon ),y(t,\varepsilon ))$, with $\tilde{T}(\varepsilon )=T(\xi (\varepsilon ),\eta (\varepsilon ),\varepsilon )$ such that

$$\begin{aligned} \sup _{0\le t\le {\tilde{T}}(0)} |x(t,\varepsilon ) - u(t,0,0)| + |y(t,\varepsilon )-\eta (\varepsilon )| \rightarrow 0 \end{aligned}$$

as $\varepsilon \rightarrow 0$.

Now, the mistake in $\mathcal{M}_1(\alpha ,\eta )$ comes from the fact that $x_\varepsilon (t,\xi (\alpha ,\eta ),\eta ,0)$ is a piecewise $C^r$, bounded, solution of the linear inhomogeneous system

$$\begin{aligned} \begin{array}{l} {\dot{x}} = f_x(\bar{u}(t,\alpha ,\eta ),\eta )x + f_y(\bar{u}(t,\alpha ,\eta ),\eta )\int _0^t g({\bar{u}}(s,\alpha ,\eta ),\eta ,0)ds \\ x(0) = 0 \end{array} \end{aligned}$$

(1.5)

and not of

$$\begin{aligned} \begin{array}{l} {\dot{x}} = f_x(\bar{u}(t,\alpha ,\eta ),\eta )x + f_y(\bar{u}(t,\alpha ,\eta ),\eta )g(u(t,\xi ,\eta ),\eta ,0) \\ x(0) = 0 \end{array} \end{aligned}$$

(1.6)

as it was erroneously stated in [1, equation (3.32)].

As a matter of facts repeating the arguments given in [1, Section 3.2] we see that the correct statement of Theorem 1.1 is the following

Theorem 1.3

Suppose that conditions $(A_1){-}(A_5)$ in [1] hold and let $B_*, B^*, b_*, b^*$ be as in [1, Proposition 3.9]. Then the adjoint linear system

$$\begin{aligned} {\dot{x}} = -f_x({\bar{u}}(t,\alpha ,\eta ),\eta )^t x = \left\{ \begin{array}{ll} -f_{+,x}({\bar{u}}(t,\alpha ,\eta ),\eta )^t x &{} \hbox {if }h({\bar{u}}(t,\alpha ,\eta ),\eta )>0 \\ -f_{-,x}({\bar{u}}(t,\alpha ,\eta ),\eta )^t x &{} \hbox {if }h({\bar{u}}(t,\alpha ,\eta ),\eta )<0. \end{array}\right. \end{aligned}$$

has a d-dimensional space of ${\hat{T}}(\alpha ,\eta )$-periodic solutions $v(t,\alpha ,\eta )$ that are $C^r$ for $t\ne {\hat{t}}_*(\alpha ,\eta ,0), {\hat{t}}^*(\alpha ,\eta ,0)$ with jumps at these points given by:

$$\begin{aligned} \begin{array}{l} v(t_*(\alpha ,\eta )^+,\alpha ,\eta ) = [B_*^{-1}]^tv(t_*(\alpha ,\eta )^-,\alpha ,\eta ), \\ v(t^*(\alpha ,\eta )^+,\alpha ,\eta ) = [B^{*-1}]^tv(t^*(\alpha ,\eta )^-,\alpha ,\eta ) \end{array} \end{aligned}$$

(1.7)

Moreover if $\{v_1(t,\alpha ,\eta ),\ldots ,v_d(t,\alpha ,\eta )\}$ is a $C^{r-1}$-basis of the space of ${\hat{T}}(\alpha ,\eta )$-periodic solutions of the adjoint linear system satisfying (1.7) and the $(d+m)$-dimensional Melnikov vector

$$\begin{aligned} \mathcal{M}(\alpha ,\eta ) := \begin{pmatrix} \mathcal{M}_1(\alpha ,\eta ) \\ \mathcal{M}_2(\alpha ,\eta ) \end{pmatrix} \end{aligned}$$

where

$$\begin{aligned} \displaystyle \mathcal{M}_1(\alpha ,\eta )= & {} \Big [\int _0^{{\hat{T}}(\alpha ,\eta )} v_j(t,\alpha ,\eta )^t f_y({\bar{u}}(t,\alpha ,\eta ),\eta )\int _0^t g({\bar{u}}(s,\alpha ,\eta ),\eta ,0)ds\, dt \\{} & {} +\, v_j({\hat{t}}_*(\alpha ,\eta )^+,\alpha ,\eta )^t b_* + v_j({\hat{t}}^*(\alpha ,\eta )^+,\alpha ,\eta )^t b^*\Big ]_{j=1,\ldots d} \\ \displaystyle \mathcal{M}_2(\alpha ,\eta )= & {} \int _0^{{\hat{T}}(\alpha ,\eta )} g({\bar{u}}(t,\alpha ,\eta ),\eta ,0) dt , \end{aligned}$$

has a simple zero at $(\alpha ,\eta )=(\alpha _0,\eta _0)$, then there exists $\varepsilon _0>0$ such that for $|\varepsilon |<\varepsilon _0$ there exist $C^r$ functions $\alpha (\varepsilon ), \eta (\varepsilon )$ with $(\alpha (0),\eta (0))=(\alpha _0,\eta _0)$ such that system (1.1) has a unique ${\tilde{T}}(\varepsilon )$-periodic solution $(x(t,\varepsilon ),y(t,\varepsilon ))$ satisfying

$$\begin{aligned} \sup _{0\le t\le {\tilde{T}}(\varepsilon )} |(x(t,\varepsilon ) - {\bar{u}}(t,\alpha (\varepsilon ),\eta (\varepsilon ))| + |(y(t,\varepsilon ) - \eta (\varepsilon )| \rightarrow 0 \end{aligned}$$

as $\varepsilon \rightarrow 0$, and $\lim _{\varepsilon \rightarrow 0} {\tilde{T}}(\varepsilon )={\hat{T}}(\alpha _0,\eta _0)$.

2 An example

It turns out, (see Appendix) that the correct Melnikov function for the example given in [1] does not have a zero. Because of this here we give another example, similar to the example in [1], using the correct result. To simplify matter, here we consider the 3-dimensional equation

$$\begin{aligned} \begin{array}{l} {\dot{x}}_1= -x_2 \\ {\dot{x}}_2= x_1-y\,{\textrm{sgn}}({x_1}) \\ \dot{y} = \varepsilon Y(x_1,x_2,y,\varepsilon ), \end{array} \end{aligned}$$

(2.1)

whose associated unperturbed system is the 2-dimensional system depending on the parameter $y\in {\mathbb {R}}$:

$$\begin{aligned} \begin{array}{l} {\dot{x}}_1 = - x_2,\\ {\dot{x}}_2 = x_1-y\,{\textrm{sgn}}({x_1}). \end{array} \end{aligned}$$

(2.2)

As in [1, Proposition 4.1] we prove that, for $x_1>2y$, (2.2) has the family of ${\hat{T}}(\alpha )$-periodic solutions

$$\begin{aligned} u(t,\alpha ,y) = \left\{ \begin{array}{ll} y\left( {\begin{matrix} {(\alpha - 1)\cos t + 1} \\ {(\alpha - 1)\sin t} \end{matrix}} \right) &{} \hbox {if }-t_1\le t\le t_1 \\ \\ y \left( {\begin{matrix} {(1-\alpha )\cos (t-2t_1)-1} \\ {(1-\alpha )\sin (t-2t_1)} \end{matrix}} \right) &{} \hbox {if }t_1\le t\le 3t_1 \\ \\ \end{array}\right. \end{aligned}$$

(2.3)

where

$$\begin{aligned} \frac{1}{4}{\hat{T}}(\alpha ) = t_1 := \arccos \left( \frac{1}{1-\alpha }\right) , \quad \alpha = \frac{x_1}{y}, \quad x_1>0. \end{aligned}$$

(2.4)

Note that $x_1>2y \Leftrightarrow \alpha >2$, so in the remaining part of this note we take $\alpha >2$. As in [1] we see that

$$\begin{aligned} u(t+2t_1,\alpha ,y) = -u(t,\alpha ,y). \end{aligned}$$

(2.5)

Arguing as is [1] we prove that all assumptions $(A_1){-}(A_5)$ are satisfied and using the correct form of the Melnikov vector for the existence of periodic solutions of system (2.2) given here, we obtain

$$\begin{aligned} \begin{array}{l} \displaystyle \mathcal{M}_1(\alpha ,y) = \int _{-t_1}^{3t_1}v(t,\alpha ,y)^t \begin{pmatrix} 0 \\ \mp 1 \end{pmatrix} \int _0^t Y(u(s,\alpha ,y),y,0)ds dt \\ \displaystyle \mathcal{M}_2(\alpha ,y) = \int _{-t_1}^{3t_1} Y(u(t,\alpha ,y),y,0) dt \end{array} \end{aligned}$$

(2.6)

where we take $-1$ in $[-t_1,t_1]$ and $+1$ in $[t_1,3t_1]$. As in [1] it can be proved that

$$\begin{aligned} v(t,\alpha ,y) = \left\{ \begin{array}{l} \begin{pmatrix} \cos t \\ \sin t \end{pmatrix} \quad -t_1\le t\le t_1 \\ -\begin{pmatrix} \cos (t-2t_1) \\ \sin (t-2t_1) \end{pmatrix} \quad t_1\le t\le 3t_1 \\ \end{array} \right. \end{aligned}$$

Hence we obtain, instead of [1, eq. (4.18)],

$$\begin{aligned}{} & {} \mathcal{M}_1(\alpha ,y) = \displaystyle \int _{-t_1}^{t_1} \begin{pmatrix} \cos t \\ \sin t \end{pmatrix}^t \begin{pmatrix} 0 \\ -1 \end{pmatrix} \int _0^t Y(u(s,\alpha ,y),y,0)ds dt \\{} & {} \quad -\, \displaystyle \int _{t_1}^{3t_1} \begin{pmatrix} \cos (t-2t_1) \\ \sin (t-2t_1) \end{pmatrix}^t \begin{pmatrix} 0 \\ 1 \end{pmatrix} \int _0^t Y(u(s,\alpha ,y),y,0)ds \\{} & {} \qquad = \displaystyle - \int _{-t_1}^{t_1} \sin t \int _0^t Y(u(s,\alpha ,y),y,0)ds dt - \int _{t_1}^{3t_1} \sin (t-2t_1) \int _0^t Y(u(s,\alpha ,y),y,0) ds dt \\{} & {} \qquad = \displaystyle -\int _{-t_1}^{t_1} \sin t \int _0^t Y(u(s,\alpha ,y),y,0)ds dt - \int _{-t_1}^{t_1} \sin t \int _0^{t+2t_1} Y(u(s,\alpha ,y),y,0)ds dt\\{} & {} \qquad = \displaystyle -\int _{-t_1}^{t_1} \sin t \int _0^t Y(u(s,\alpha ,y),y,0)ds dt- \int _{-t_1}^{t_1} \sin t \int _{2t_1}^{t+2t_1} Y(u(s,\alpha ,y),y,0) ds dt \\{} & {} \qquad = \displaystyle - \int _{-t_1}^{t_1} \sin t \int _0^t [Y(u(s,\alpha ,y),y,0) + Y(u(s+2t_1,\alpha ,y),y,0)]ds dt \\{} & {} \qquad =\displaystyle - 2 \int _{-t_1}^{t_1} \sin t \int _0^t Y_{ev}(u(s,\alpha ,y),y,0)ds dt \end{aligned}$$

where we used (2.5) and set

$$\begin{aligned} Y_{ev}(x,y,0) = \frac{1}{2}[Y(x,y,0)+Y(-x,y,0)]. \end{aligned}$$

Simplifying further, and using (2.4) we get

$$\begin{aligned} \displaystyle \mathcal{M}_1(\alpha ,y)= & {} 2 \int _{-t_1}^{t_1} \left[ \frac{d}{dt}\cos t\right] \int _0^t Y_{ev}(u(s,\alpha ,y),y,0) ds dt \\ \displaystyle= & {} 2 \left[ \cos t_1 \int _0^{t_1} Y_{ev}(u(s,\alpha ,y),y,0) ds - \cos (- t_1) \int _0^{-t_1} Y_{ev}(u(s,\alpha ,y),y,0) ds \right] \\{} & {} \displaystyle -\, 2 \int _{-t_1}^{t_1} Y_{ev}(u(t,\alpha ,y),y,0)\cos t dt \\ \displaystyle= & {} \frac{2}{1-\alpha } \int _{-t_1}^{t_1} Y_{ev}(u(t,\alpha ,y),y,0) dt - 2\int _{-t_1}^{t_1} Y_{ev}(u(t,\alpha ,y),y,0)\cos t dt \\ \displaystyle= & {} - 2\int _{t_1}^{t_1} Y_{ev}(u(t,\alpha ,y),y,0) \left( \frac{1}{\alpha -1} + \cos t \right) dt . \end{aligned}$$

Next, it is easy to check that

$$\begin{aligned} \mathcal{M}_2(\alpha ,y) = 2\int _{-t_1}^{t_1} Y_{ev}(u(t,\alpha ,y),y,0) dt. \end{aligned}$$

Note also that, if $Y(x,y,\varepsilon )$ is an odd function with respect to x, then $\mathcal{M}_1(\alpha ,y) = \mathcal{M}_2(\alpha ,y) = 0$. To give a concrete example of application of Theorem 1.1 we take

$$\begin{aligned} Y(x_1,x_2,y,0)=x_2^2 + yx_1^2 - 1 \end{aligned}$$

Note that $Y_{ev}(x,y)=Y(x,y)$ and

$$\begin{aligned} Y(x_1,x_2,y,0) = Y(x_1,-x_2,y,0). \end{aligned}$$

(2.7)

As

$$\begin{aligned}{} & {} u_1(-t,\alpha ,y) = y(\alpha -1)\cos (-t) + y = y(\alpha -1)\cos t + y = u_1(t,\alpha ,y) \\{} & {} u_2(-t,\alpha ,y) = y(\alpha -1)\sin (- t) = - y(\alpha -1)\sin t = - u_2(t,\alpha ,y) \end{aligned}$$

we see that

$$\begin{aligned}\begin{array}{l} Y(u(-t,\alpha ,y),y,0) = Y(u_1(-t,\alpha ,y),u_2(-t,\alpha ,y),y,0) \\ \quad = Y(u_1(t,\alpha ,y),-u_2(t,\alpha ,y),y,0) = Y(u(t,\alpha ,y),y,0) . \end{array}\end{aligned}$$

Then, replacing $\alpha$ with $\alpha +1$ (and then $t_1=\arccos \left( -\frac{1}{\alpha }\right)$) we get:

$$\begin{aligned} \begin{array}{l} \displaystyle \mathcal{M}_1(\alpha +1,y) = -4\int _{0}^{t_1} Y(u(t,\alpha +1,y),y,0)\left( \frac{1}{\alpha } + \cos t \right) dt \\ \displaystyle \quad = -\frac{4}{\alpha }\int _0^{t_1} \left( y^2\alpha ^2\sin ^2 t + y^3(\alpha \cos t + 1)^2 - 1\right) \left( \alpha \cos t +1 \right) dt \\ \displaystyle \quad = -4y^2\alpha \int _0^{t_1}\sin ^2 t dt - 4y^2\alpha ^2 \int _0^{t_1} \sin ^2 t \cos t \, dt \\ \displaystyle \qquad + \frac{4}{\alpha }\int _0^{t_1} (\alpha \cos t + 1)\, dt - \frac{4}{\alpha }\int _0^{t_1} y^3(\alpha \cos t + 1)^3\, dt \\ \displaystyle \quad =-2y^2\alpha (t_1-\sin t_1\cos t_1) - \frac{4}{3}y^2\alpha ^2\sin ^3 t_1 + 4\sin t_1 + \frac{4t_1}{\alpha } \\ \displaystyle \qquad -4y^3\sin t_1 \left( \frac{3}{2}\alpha \cos t_1-\frac{1}{3}\alpha ^2\sin ^2 t_1 + \alpha ^2+3 \right) -\frac{2y^3}{\alpha }(3\alpha ^2+2)t_1 \\ \displaystyle \quad = - 2y^2\alpha \left( \arccos \left( -\frac{1}{\alpha }\right) + \frac{\sqrt{\alpha ^2-1}}{\alpha ^2}\right) - \frac{4}{3}y^2\frac{(\alpha ^2-1)^{3/2}}{\alpha } + 4\frac{\sqrt{\alpha ^2-1}}{\alpha } + \frac{4}{\alpha }\arccos \left( -\frac{1}{\alpha }\right) \\ \displaystyle \qquad -\,4y^3\frac{\sqrt{\alpha ^2-1}}{\alpha } \left( -\frac{3}{2} - \frac{1}{3}(\alpha ^2-1) + \alpha ^2 + 3 \right) -\frac{2y^3}{\alpha }(3\alpha ^2+2)\arccos \left( -\frac{1}{\alpha }\right) \\ \displaystyle \quad = \left[ \frac{4}{\alpha }(1-y^3) -2\alpha y^2(1+3y)\right] \arccos \left( -\frac{1}{\alpha }\right) - 2y^2 \frac{\sqrt{\alpha ^2-1}}{\alpha } - \frac{4}{3}y^2\frac{(\alpha ^2-1)^{3/2}}{\alpha } \\ \displaystyle \qquad +\, 4\frac{\sqrt{\alpha ^2-1}}{\alpha } - 4y^3\frac{\sqrt{\alpha ^2-1}}{3\alpha } \left( \frac{11}{2} + 2\alpha ^2 \right) = \left[ \frac{4}{\alpha }(1-y^3) -2\alpha y^2(1+3y)\right] \arccos \left( -\frac{1}{\alpha }\right) \\ \displaystyle \qquad -\, \frac{2\sqrt{\alpha ^2-1}}{3\alpha } \left[ y^2(2\alpha ^2+1) + y^3(4\alpha ^2+11) - 6\right] . \end{array} \end{aligned}$$

Similarly

$$\begin{aligned} \begin{array}{l} \displaystyle \mathcal{M}_2(\alpha +1,y) = 2\int _{-t_1}^{t_1}Y(u(t,\alpha +1,y),y,0) dt \\ \displaystyle \quad = 4\int _{0}^{t_1}Y(u(t,\alpha +1,y),y,0)\, dt \quad = 4\int _{0}^{t_1}\, [(\alpha \cos t +1)^2y^3+\alpha ^2y^2\sin ^2 t - 1]\, dt \\ \quad = 4y^3\left( \frac{\alpha ^2}{2} \sin t_1\cos t_1 + 2\alpha \sin t_1 +\left( 1+\frac{\alpha ^2}{2}\right) t_1\right) + 2\alpha ^2y^2( t_1 -\sin t_1\cos t_1) - 4t_1 \\ \quad = 4y^3 \left( -\frac{1}{2}\sqrt{\alpha ^2-1} + 2\sqrt{\alpha ^2-1} + \left( 1+\frac{\alpha ^2}{2}\right) \arccos \left( -\frac{1}{\alpha }\right) \right) \\ \qquad + \,(2\alpha ^2y^2-4)\arccos \left( -\frac{1}{\alpha }\right) + 2y^2\sqrt{\alpha ^2-1} \\ \quad = 2y^2(3y+1)\sqrt{\alpha ^2-1} + 2[(\alpha ^2+2)y^3+\alpha ^2y^2-2]\arccos \left( -\frac{1}{\alpha }\right) . \end{array} \end{aligned}$$

So

$$\begin{aligned} \begin{array}{l} \mathcal{M}(\alpha +1,y) = \\ \left( \begin{array}{c} \left[ \frac{4}{\alpha }(1-y^3) -2\alpha y^2(1+3y)\right] \arccos \left( -\frac{1}{\alpha }\right) - \frac{2\sqrt{\alpha ^2-1}}{3\alpha } \left[ y^2(2\alpha ^2+1) + y^3(4\alpha ^2+11) - 6\right] \\ 2y^2(3y+1)\sqrt{\alpha ^2-1} + 2[(\alpha ^2+2)y^3+\alpha ^2y^2-2]\arccos \left( -\frac{1}{\alpha }\right) \end{array}\right) \end{array}\nonumber \\ \end{aligned}$$

(2.8)

with $\alpha >1$. Using Newton method we find the following approximate solution of equation $\mathcal{M}(\alpha +1,y)=0$

$$\begin{aligned} \alpha \sim 2.237893508862806, \; y \sim 0.4392165858210144 \end{aligned}$$

and the Jacobian matrix at this point is invertible, its determinant being approximately equal to 37.25014840155305. Numerical computation shows that

$$\begin{aligned} \mathcal{M}(3.237893508862806,0.4392165858210144) = \begin{pmatrix} 0.999200722162641 \\ -1.332267629550188 \end{pmatrix}10^{-15}. \end{aligned}$$

In Figures -- we plot the periodic solution of the perturbed equation (2.1) with $Y(x_1,x_2,y,\varepsilon )=x_2^2+yx_1^2-1$.

Fig. 1

In red the $(x_1(t),x_2(t))$-component of the periodic solution of (2.1) with $\varepsilon =0.02$, in blue the periodic solution of the unperturbed equation ($\varepsilon =0$)

Fig. 2

y(t)-component of the periodic solution of (2.1) with $\varepsilon =0.02$

Fig. 3

$(x_1(t),x_2(t),y(t))$ versus time of the periodic solution of (2.1) with $\varepsilon =0.02$

3 Conclusion

We have corrected a mistake in the Melnikov function given in [1] for the existence of periodic solutions of equation (1.1). The error in [1] comes from the fact that in [1, equation (3.32)] we missed an integral sign and we continued to use the wrong formula in the proof of the theorem. However we emphasize that the arguments in [1] give the true Melnikov condition when using the correct formula for the differential equation that $x_\varepsilon (t,\xi (\alpha ,\eta ),\eta ,0)$ must satisfy, i.e. (1.5) instead of (1.6).

Appendix

In [1] the following equation has been studied as an application of the bifurcation Theorem:

$$\begin{aligned} \begin{array}{l} {\dot{x}}_1= -\omega x_2 \\ {\dot{x}}_2= \omega (x_1-y\,{\textrm{sgn}}({x_1})) \\ {\dot{\omega }} = \varepsilon \Omega (x_1,x_2,\omega ,y,\varepsilon ) \\ \dot{y} = \varepsilon Y(x_1,x_2,\omega ,y,\varepsilon ). \end{array} \end{aligned}$$

(4.1)

The unperturbed system ($\varepsilon =0$) of (4.1) has the family of periodic solutions

$$\begin{aligned} u(t,\alpha ,y) = \left\{ \begin{array}{ll} y\left( {\begin{matrix} {(\alpha - 1)\cos \omega t + 1} \\ {(\alpha - 1)\sin \omega t} \end{matrix}} \right) &{} \hbox {if }-t_1\le t\le t_1 \\ \\ y \left( {\begin{matrix} {(1-\alpha )\cos \omega (t-2t_1)-1} \\ {(1-\alpha )\sin \omega (t-2t_1)} \end{matrix}} \right)&\hbox {if }t_1\le t\le 3t_1 \end{array}\right. \end{aligned}$$

(4.2)

where

$$\begin{aligned} \omega t_1 = \arccos \left( \frac{1}{1-\alpha }\right) . \end{aligned}$$

Recall that $u(t+2t_1,\alpha ,\omega ,y) = -u(t,\alpha ,\omega ,y)$ for all $t\in {\mathbb {R}}$. Next, with reference to Theorem 1.3, we have

$$\begin{aligned} v(t,\alpha ,\omega ,y) = \left\{ \begin{array}{ll} \left( {\begin{matrix} {\cos \omega t} \\ {\sin \omega t} \end{matrix}} \right) &{} \hbox {if }-t_1\le t\le t_1 \\ \\ -\left( {\begin{matrix} {\cos \omega (t-2t_1)} \\ {\sin \omega (t-2t_1)} \end{matrix}} \right)&\hbox {if }t_1\le t\le 3t_1. \end{array}\right. \end{aligned}$$

As in [1] we have

$$\begin{aligned} \begin{array}{l} \frac{1}{2}\mathcal{M}_2(\alpha ,\omega ,y) = \begin{pmatrix} \int _{-t_1}^{t_1} \Omega _{ev}(u(t,\alpha ,\omega ,y), \omega , y, 0) dt \\ \int _{-t_1}^{t_1} Y_{ev}(u(t,\alpha ,\omega ,y), \omega , y, 0) dt \end{pmatrix} \\ \end{array} \end{aligned}$$

then, since for eq. (4.1) $b_*=b^*=0$ (see [1]), we get, skipping some steps as they are quite similar to those in the previous section

$$\begin{aligned} \begin{array}{l} \mathcal{M}_1(\alpha ,\omega ,y) \\ = \displaystyle \int _{-t_1}^{t_1} \begin{pmatrix} \cos \omega t \\ \sin \omega t\end{pmatrix}^t \begin{pmatrix} -u_2(t,\alpha ,\omega ,y) &{} 0 \\ u_1(t,\alpha ,\omega ,y) - y &{} -1 \end{pmatrix} \begin{pmatrix} \int _0^t \Omega ((u(s,\alpha ,\omega ,y), \omega , y, 0) ds \\ \int _0^t Y((u(s,\alpha ,\omega ,y), \omega , y, 0) ds \end{pmatrix} \, dt \\ \displaystyle - \int _{t_1}^{3t_1} \begin{pmatrix} \cos \omega (t-2t_1) \\ \sin \omega (t-2t_1) \end{pmatrix}^t \begin{pmatrix} -u_2(t,\alpha ,\omega ,y) &{} 0 \\ u_1(t,\alpha ,\omega ,y) + y &{} 1 \end{pmatrix} \begin{pmatrix} \int _0^t \Omega ((u(s,\alpha ,\omega ,y), \omega , y, 0) ds \\ \int _0^t Y((u(s,\alpha ,\omega ,y), \omega , y, 0) ds \end{pmatrix} \, dt \\ \displaystyle = \int _{-t_1}^{t_1} -\sin \omega t \int _0^t Y(u(s,\alpha ,\omega ,y), y, 0) ds \, dt \\ \displaystyle \qquad - \int _{t_1}^{3t_1} \sin \omega (t-2t_1) \int _0^t Y(u(s,\alpha ,\omega ,y), \omega , y, 0) ds \, dt \\ \displaystyle = - 2\int _{-t_1}^{t_1} \sin \omega t \int _0^t [Y_{ev}(u(s,\alpha ,\omega ,y), \omega , y, 0) ds] \, dt \\ \displaystyle = \frac{2}{\omega } \left[ \cos \omega t \int _0^t [Y_{ev}(u(s,\alpha ,\omega ,y), \omega , y, 0) ds \right] _{-t_1}^{t_1}\\ - \frac{2}{\omega } \int _{-t_1}^{t_1} Y_{ev}(u(t,\alpha ,\omega ,y), \omega , y, 0) \cos \omega t dt \\ \displaystyle = \frac{2}{\omega } \int _0^{t_1} [Y_{ev}(u(t,\alpha ,\omega ,y), \omega , y, 0) ds \,{+} Y_{ev}(u(-t,\alpha ,\omega ,y), \omega , y, 0)]\left( \cos \omega t {-} \cos \omega t_1\right) dt . \end{array} \end{aligned}$$

Now both examples given in [1] are of the following kind:

$$\begin{aligned} \begin{array}{l} \Omega (x_1,x_2,\omega ,y,\varepsilon ) =\Omega _0(\omega ,y) , \\ Y(x_1,x_2,\omega ,y,\varepsilon ) = \omega (x_2^2-1) + x_1x_2(y-1) \end{array} \end{aligned}$$

(4.3)

then, since $u_1(-t,\alpha ,\omega ,y)=u_1(t,\alpha ,\omega ,y)$ and $u_2(-t,\alpha ,\omega ,y)=-u_2(t,\alpha ,\omega ,y)$, we have

$$\begin{aligned} \begin{array}{l} \displaystyle \mathcal{M}_1(\alpha +1,\omega ,y) = 4 \int _0^{t_1}(u_2(t,\alpha +1,\omega ,y)^2-1) \left( \cos \omega t - \cos \omega t_1\right) dt \\ \quad \displaystyle = \frac{4}{\omega } \int _0^{\omega t_1}(\alpha ^2y^2\sin ^2 t-1) \left( \cos t - \cos \omega t_1\right) dt \\ \quad \displaystyle = \frac{2\sqrt{\alpha ^2-1}}{3\alpha \omega }\left[ y^2(2\alpha ^2+1) - 6\right] + \frac{2}{\alpha \omega } (\alpha ^2y^2-2)\arccos \left( -\frac{1}{\alpha }\right) \end{array} \end{aligned}$$

and

$$\begin{aligned} \mathcal{M}_2(\alpha +1,\omega ,y) = \begin{pmatrix} \frac{4}{\omega }\Omega _0(\omega ,y)\arccos \left( -\frac{1}{\alpha }\right) \\ (\alpha ^2y^2-2)\arccos \left( -\frac{1}{\alpha }\right) + y^2\sqrt{\alpha ^2-1} \end{pmatrix}. \end{aligned}$$

Solving $\mathcal{M}(\alpha +1,\omega ,y)=0$ is equivalent to $\Omega _0(\omega ,y)= 0$ with $\omega \ne 0$, and

$$\begin{aligned} \begin{array}{l} \sqrt{\alpha ^2-1}\left[ y^2(2\alpha ^2+1) - 6\right] + 3 (\alpha ^2y^2-2)\arccos \left( -\frac{1}{\alpha }\right) = 0 \\ (\alpha ^2y^2-2)\arccos \left( -\frac{1}{\alpha }\right) + y^2\sqrt{\alpha ^2-1} = 0 . \end{array} \end{aligned}$$

From the second equation we get

$$\begin{aligned} \arccos \left( -\frac{1}{\alpha }\right) = - \frac{y^2\sqrt{\alpha ^2-1}}{(\alpha ^2y^2-2)} \end{aligned}$$

that we plug in the first equation to see that it is equivalent to:

$$\begin{aligned} y^2(\alpha ^2-1) - 3 = 0 . \end{aligned}$$

So $\mathcal{M}(\alpha ,\omega ,y)=0$ is equivalent to

$$\begin{aligned} \begin{array}{l} \Omega _0(\omega ,y) = 0 , \; \omega \ne 0 \\ y^2 = \frac{3}{\alpha ^2-1} \\ \arccos \left( -\frac{1}{\alpha }\right) = - \frac{ \frac{3}{\alpha ^2-1}\sqrt{\alpha ^2-1}}{(\alpha ^2 \frac{3}{\alpha ^2-1}-2)} = - \frac{ 3\sqrt{\alpha ^2-1}}{\alpha ^2+2} . \end{array} \end{aligned}$$

which has no solutions since $\frac{\pi }{2}<\arccos (\theta )\le \pi$, for $-1<\theta <0$.