Zero-Hopf bifurcation in the Van der Pol oscillator with delayed position and velocity feedback

Abstract

In this paper, we consider the traditional Van der Pol oscillator with a forcing dependent on a delay in feedback. The delay is taken to be a nonlinear function of both position and velocity, which gives rise to many different types of bifurcations. In particular, we study the Zero-Hopf bifurcation that takes place at certain parameter values using methods of center manifold reduction of DDEs and normal form theory. We present numerical simulations that have been accurately predicted by the phase portraits in the Zero-Hopf bifurcation to confirm our numerical results and provide a physical understanding of the oscillator with the delay in feedback.

Proof of Theorem 1

First, let $V_j(\mathbb {C}^3\ \times$ Ker $\pi )$ be the space of homogeneous polynomials of degree $j$ in the variables $(x,\mu )$ with coefficients in $\mathbb {C}^3\ \times$ Ker $\pi$. Define $M_j: V_j(\mathbb {C}^3\ \times$ Ker $\pi ) \rightarrow V_j(\mathbb {C}^3\ \times$ Ker $\pi )$ such that

$$\begin{aligned} M_j(p,h) = (M^1_jp,M^2_jh) \end{aligned}$$

(64)

where

$$\begin{aligned} M_j^1p(x,\mu )&= D_xp(x,\mu )Jx - Jp(x,\mu ) \nonumber \\&= i\omega \left( \begin{array}{cc} x_1\dfrac{\partial p_1}{\partial x_1} - x_2\dfrac{\partial p_1}{\partial x_2} - p_1 \\ x_1\dfrac{\partial p_2}{\partial x_1} - x_2\dfrac{\partial p_2}{\partial x_2} + p_2 \\ x_1\dfrac{\partial p_3}{\partial x_1} - x_2 \dfrac{\partial p_3}{\partial x_2} \end{array} \right) , \\ M_j^2h(x,\mu )&= D_xh(x,\mu )Jx - \mathcal {A}_{Q^1}h(x,\mu ),\nonumber \end{aligned}$$

(65)

with $p(x,\mu ) \in V_j(\mathbb {C}^3)$ and $h(x,\mu )(\theta ) \in V_j(\hbox {Ker} \pi )$. We now have the decomposition $V_j(\mathbb {C}^3) = \mathrm{Im}(M^1_j) \otimes \mathrm{Ker}(M^1_j)$ for $j \ge 2$.

Introduce the nonlinear change of coordinates

$$\begin{aligned} x&= \hat{x} + U_2^1(\hat{x},\mu ) \nonumber \\ y&= \hat{y} + U_2^2(\hat{x},\mu ), \end{aligned}$$

(66)

where

$$\begin{aligned} U_2^1(x,\mu )&= (M_1^1)^{-1}Proj_{\mathrm{Im}(M_2^1)}f_2^1(x,0,\mu ), \nonumber \\ U_2^2(x,\mu )&= (M_2^2)^{-1}f_2^2(x,0,\mu ). \end{aligned}$$

(67)

Equation (40) now becomes

$$\begin{aligned}&\dot{x} =(I + D_xU_2^1(x,\mu ))^{-1}\big [Jx + JU_2^1(x,\mu ) \nonumber \\&\qquad +\sum _{j \ge 2}f_j^1(x+U_2^1(x,\mu ),y+U_2^2(x,\mu ))\big ] \nonumber \\&\frac{d}{dt}y = \mathcal {A}_{Q^1}y + \mathcal {A}_{Q^1}U_2^2(x,\mu ) - D_xU_2^2(x,\mu )\dot{x} \nonumber \\&\qquad \qquad +\sum _{j \ge 2} f_j^2(x+U_2^1(x,\mu ),y+U_2^2(x,\mu ))\nonumber \\ \end{aligned}$$

(68)

upon dropping the hats.

For $|x|$ small, we have that

$$\begin{aligned} (I + D_xU_2^1(x,\mu ))^{-1}&\approx I - D_xU_2^1(x,\mu )\nonumber \\&\quad + (D_xU_2^1(x,\mu ))^2, \end{aligned}$$

(69)

and using Taylor’s theorem, we obtain

$$\begin{aligned}&f_2^1(x+U_2^1(x,\mu ),y+U_2^2(x,\mu )) = f_2^1(x,y) \\&\quad +D_xf^1_2(x,y)U_2^1(x,\mu ) \\&\quad + D_yf^1_2(x,y)U_2^2(x,\mu ) + \mathrm {h.o.t.} \end{aligned}$$

Therefore, we now have the normal form on the center manifold given by

$$\begin{aligned} \dot{x} = Jx + g^1_2(x,0,\mu ) + \mathrm {h.o.t.} \end{aligned}$$

(70)

where $g^1_2$ are the following second-order terms in $(x,\mu )$:

$$\begin{aligned} g_2^1(x,0,\mu ) = Proj_{Ker(M_2^1)}f_2^1(x,0,\mu ) \end{aligned}$$

(71)

Consider the basis $\{\mu ^px^qe_k: k=1,2,3, p \in \mathbb {N}^2_0, q \in \mathbb {N}^3_0, |p| + |q| = j\}$ of $V_j(\mathbb {C}^3)$, where $e_1=(1,0,0)^T$, $e_2 = (0,1,0)^T$, and $e_3 = (0,0,1)^T$. Then, for $j=2$, upon finding the images of each basis element under $M_2^1$, we find that Ker$(M_2^1)$ is spanned by

$$\begin{aligned}&\mu _1x_1e_1, \mu _2x_1e_1, x_1x_3e_1,\\&\mu _1x_2e_2, \mu _2x_2e_2, x_2x_3e_2,\\&x_1x_2e_3, \mu _1x_3e_3, \mu _2x_3e_3, \mu _1^2e_3, \mu _2^2e_3, \mu _1\mu _2e_3, x_3^2e_3. \end{aligned}$$

To see the images of each of these individual basis elements under the operator $M_2^1$ see [33],

we now compute $g_2^1(x,0,\mu )$ in (70). We have that

$$\begin{aligned}&g_2^1(x,0,\mu ) = Proj_{\mathrm{Ker}(M_2^1)}f_2^1(x,0,\mu ) + \mathcal {O}(|\mu |^2) \nonumber \\&\quad = \begin{pmatrix} (a_{11}\mu _1 + a_{12}\mu _2)x_1 + a_{13}x_1x_3 \\ (\bar{a}_{11}\mu _1 + \bar{a}_{12}\mu _2)x_2 + \bar{a}_{13}x_2x_3 \\ (a_{21}\mu _1 + a_{22}\mu _2)x_3 + a_{23}x_1x_2 + a_{24}x_3^2 \end{pmatrix}\nonumber \\&\quad + \mathcal {O}(|\mu |^2). \end{aligned}$$

(72)

Now, expanding out the terms in (41), we get

$$\begin{aligned}&F^1_2 = \mu _2(i\omega _0x_1 - i\omega _0x_2 + y_2(0)), \ \ \ \ \ F^1_3 = 0, \nonumber \\&F^2_2 = -\mu _2(x_1 + x_2 + x_3 + y_1(0))\nonumber \\&\quad +\varepsilon \mu _2(i\omega _0x_1 - i\omega _0x_2\nonumber \\&\quad + y_2(0)) +a\mu _2(i\omega _0e^{-i\omega _0\tau _0}x_1\nonumber \\&\qquad -i\omega _0e^{i\omega _0\tau _0}x_2 + y_2(-1))\nonumber \\&\quad +\tau _0\mu _1(e^{-i\omega _0\tau _0}x_1 + e^{i\omega _0\tau _0}x_2+ x_3 +y_1(-1))\nonumber \\&\quad +\mu _2(e^{-i\omega _0\tau _0}x_1+ e^{i\omega _0\tau _0}x_2 +x_3+ y_1(-1)) \nonumber \\&\quad +\frac{1}{2}\tau _0g_{u_1u_1}(0,0)(e^{-i\omega _0\tau _0}x_1+e^{i\omega _0\tau _0}x_2 \nonumber \\&\qquad + x_3 + y_1(-1))^2\nonumber \\&\quad +\tau _0g_{u_1u_2}(0,0)(e^{-i\omega _0\tau _0}x_1+e^{i\omega _0\tau _0}x_2 + x_3 \nonumber \\&\quad +y_1(-1))(i\omega _0e^{-i\omega _0\tau _0}x_1 - i\omega _0e^{i\omega _0\tau _0}x_2+ y_2(-1)) \nonumber \\&\quad +\frac{1}{2}\tau _0g_{u_2u_2}(0,0)(i\omega _0e^{-i\omega _0\tau _0}x_1\nonumber \\&\qquad - i\omega _0e^{i\omega _0\tau _0} x_2 + y_2(-1))^2. \end{aligned}$$

(73)

Therefore, recalling that the characteristic equation gives us $(1+ia\omega _0)e^{-i\omega _0\tau _0} = -\omega _0^2 - i\varepsilon \omega _0 + 1$, we compute the coefficients in (72) to be

$$\begin{aligned} a_{11}&= \tau _0\bar{D}\bar{\sigma }e^{-i\omega _0\tau _0}, \ \ \ a_{12} = \bar{D}(i\omega _0 - \bar{\sigma }\omega _0^2), \nonumber \\ a_{13}&= \tau _0\bar{D}\bar{\sigma }(g_{u_1u_1}(0,0)e^{-i\omega _0\tau _0}\nonumber \\&\quad +i\omega _0g_{u_1u_2}(0,0)e^{-i\omega _0\tau _0}), \nonumber \\ a_{21}&= \frac{\tau _0}{\tau _0 - \varepsilon - a}, \ \ a_{22} = 0, \ \ a_{24} = \frac{1}{2}a_{21}g_{u_1u_1}(0,0) \nonumber \\ a_{23}&= a_{21}(g_{u_1u_1}(0,0) + \omega _0^2g_{u_2u_2}(0,0)). \end{aligned}$$

(74)

We thus get the desired conclusion. As mentioned earlier, the computations required to get the formulae for the cubic coefficients of the normal form are lengthy, but straightforward and standard. We omit the details here because we only require the quadratic order normal form (50) for our analysis, but the details can be found in [4].