Oscillations on one dimensional time dependent center manifolds: algebraic curves approach

Abstract

We consider a one dimensional time dependent ODE of degree $n\ge 2$ as the restriction of an arbitrary nonautonomous ODE to the associated one dimensional center manifold. Then, we present an algorithm for computing time dependent algebraic curves of m-th degree with $m\le n-1$. This computation leads us to a m dimensional time dependent bifurcation equation. We determine oscillatory behaviors of the system with the help of the bifurcation equation. Finally, we complete the method for a general parametric planar system and find periodic solutions. The method can be applied for a wide range of nonautonomous systems and does not have restrictions of classical methods such as the Poincaré map.

Appendix

Here, we give an illustration to show how the time dependent algebraic curves work. We assume $n=2$ and consider two cases $m =1<n$ and $n\le m=3$ (see Remark 1). Consider the equation

$$\begin{aligned} \dot{x}(t)=a_0(t)+a_1(t)x(t)+a_2(t)x^2(t), \end{aligned}$$

(42)

where $a_0(t)$ and $a_2(t)$ are not identically zero. We begin with the trivial case $m=1$ and try to find an algebraic curve as $F(x,t)=b_0(t)+x$. In this case, the cofactor curve K(x, t) is of the form $K(x,t)=c_0(t)+c_1(t)x(t)$, and we have

$$\begin{aligned} \frac{d}{dt}F\left( x(t),t\right) =K\left( x(t),t\right) F\left( x(t),t\right) \end{aligned}$$

or equivalently

$$\begin{aligned} \left( a_0(t)-b_0(t) c_0(t)+b_0'(t)\right) +\left( a_1(t)-b_0(t) c_1(t) -c_0(t)\right) x(t)+\left( a_2(t)-c_1(t)\right) x^2(t)=0. \end{aligned}$$

It implies that $c_0(t)=a_1(t)-c_1(t) b_0(t)$, $c_1(t)=a_2(t)$, and

$$\begin{aligned} b_0'(t)= -a_0(t)+a_1(t)b_0(t) -a_2(t) b^2_0(t). \end{aligned}$$

Note that, if $b_0(t)$ is a solution of the above equation, then $x(t)=-b_0(t)$ is a solution of (42); Especially, if $b_0(t)$ is periodic, then x(t) is periodic too.

Now, let $m=3$ and try for an algebraic curve of the form

$$\begin{aligned} F(x,t)=b_0(t)+b_1(t)x+b_2(t)x^2+x^3. \end{aligned}$$

From Remark 1, we consider (42) in the form

$$\begin{aligned} \dot{x}(t)=a_0(t)+a_1(t)x(t)+a_2(t)x^2(t)+a_3(t)x^3(t)+a_4(t)x^4(t), \end{aligned}$$

(43)

with $a_3(t),a_4(t)\equiv 0$ and the cofactor curve of the form

$$\begin{aligned} K(x,t)=c_0(t)+c_1(t)x+c_2(t)x^2+c_3(t)x^3. \end{aligned}$$

Thus,

$$\begin{aligned}&K\left( x(t),t\right) F\left( x(t),t\right) -\frac{d}{dt}F\left( x(t),t\right) \\&\quad =\left( b_0(t) c_0(t)-a_0(t) b_1(t)-b_0^{'}(t)\right) +c_3(t)x^6(t)\\&\qquad +\left( b_2(t) c_3(t)+c_2(t)\right) x^5(t)+\left( -3 a_2(t)+b_2(t) c_2(t) +b_1(t) c_3(t)+c_1(t)\right) x^4(t)\\&\qquad +\left( -2 a_2(t) b_2(t)-3 a_1(t)+b_2(t) c_1(t)+b_1(t) c_2(t) +b_0(t) c_3(t)+c_0(t)\right) x^3(t)\\&\qquad +\left( -a_2(t) b_1(t)-2 a_1(t) b_2(t)-3 a_0(t)+b_2(t) c_0(t) +b_1(t) c_1(t)+b_0(t) c_2(t)-b_2'(t)\right) x^2(t)\\&\qquad + \left( -a_1(t) b_1(t)-2 a_0(t) b_2(t) +b_1(t) c_0(t)+b_0(t) c_1(t)-b_1'(t)\right) x(t). \end{aligned}$$

It implies that $c_3(t)=c_2(t)\equiv 0$, $c_1(t)=3a_2(t)$, $c_0(t)=3 a_1(t)-a_2(t) b_2(t)$, and

$$\begin{aligned} \left\{ \begin{array}{l} b_0^{'}(t)=3 a_1(t) b_0(t)-a_2(t) b_2(t) b_0(t)-a_0(t) b_1(t),\\ b_1^{'}(t)=3 a_2(t) b_0(t)+2 a_1(t) b_1(t)+b_2(t) \left( -a_2(t) b_1(t)-2 a_0(t)\right) , \\ b_2^{'}(t)=-a_2(t) b_2(t){}^2+a_1(t) b_2(t)+2 a_2(t) b_1(t)-3 a_0(t). \end{array}\right. \end{aligned}$$

(44)

It is easy to check that the above equalities are in agreement with Remark 1.

In an especial case, let

$$\begin{aligned} a_0(t)= & {} \frac{128 \cos (t)}{-684 \cos (2 t)+81 \cos (4 t)+731},\\ a_1(t)= & {} \frac{378 \sin (2 t)-81 \sin (4 t)}{-684 \cos (2 t)+81 \cos (4 t)+731}, \\ a_2(t)= & {} \frac{288 \cos (t)}{-684 \cos (2 t)+81 \cos (4 t)+731}. \end{aligned}$$

It can be checked directly from (44) that

$$\begin{aligned} b_0(t)=\frac{1}{6} \sin (t) (3 \cos (2 t)-7), \ \ b_1(t)=\frac{2}{3}, \ \ b_2(t)=0 \end{aligned}$$

is a solution for (44), and thus a solution for (19). Finally, it is easy to see that $F(\sin (t),t)\equiv 0$, which means that $x(t)=\sin t$ is a solution of (42). This fact can also be verified by putting $x(t)=\sin t$ in (42).

As another example, let $m=3$, $n=4$ with $a_4(t) \ne 0$ and consider the equation

$$\begin{aligned} \dot{x}(t)=a_0(t)+a_1(t)x(t)+ a_{2}(t)x^2(t)+ a_3(t)x^{3}(t) + a_4(t)x^{4}(t). \end{aligned}$$

(45)

In this case, the algebraic curve is of the form

$$\begin{aligned} F(t) = b_0(t) + b_1(t)x + b_2(t)x^{2} + x^{3} \end{aligned}$$

with the corresponding cofactor $K(t)=c_0(t)+c_1(t)x+c_{2}(t)x^{2} +c_3(t)x^{3}$. By using (14), (15), and (16), we find

$$\begin{aligned} c_0(t)= & {} -a_4(t) b_2(t){}^3+a_3(t) b_2(t){}^2-a_2(t) b_2(t) +3 a_4(t) b_1(t) b_2(t)\\&-3 a_4(t) b_0(t)-2 a_3(t) b_1(t)+3 a_1(t),\\ c_1(t)= & {} a_4(t) b_2(t){}^2-a_3(t) b_2(t)-2 a_4(t) b_1(t)+3 a_2(t), \\ c_2(t)= & {} 3 a_3(t)-a_4(t) b_2(t), \ \ \ \ \ c_3(t)=3 a_4(t). \end{aligned}$$

Substituting the above relations in the differential part of (13), we obtain

$$\begin{aligned} \left\{ \begin{array}{ll} b_0'(t) =&{} -a_4(t) b_0(t) b_2(t){}^3+a_3(t) b_0(t) b_2(t){}^2+b_0(t) b_2(t) \left( 3 a_4(t) b_1(t)-a_2(t)\right) \\ &{}-3 a_4(t) b_0(t){}^2-a_0(t) b_1(t)+b_0(t) \left( 3 a_1(t)-2 a_3(t) b_1(t)\right) , \\ b_1'(t) =&{} -a_4(t) b_1(t) b_2(t){}^3+b_2(t){}^2 \left( a_4(t) b_0(t)+a_3(t) b_1(t)\right) \\ &{} +b_2(t) \left( 3 a_4(t) b_1(t){}^2-a_2(t) b_1(t)-a_3(t) b_0(t)-2 a_0(t)\right) \\ &{}-2 a_3(t) b_1(t){}^2+2 a_1(t) b_1(t)+b_0(t) \left( 3 a_2(t)-5 a_4(t) b_1(t)\right) , \\ b_2'(t) =&{} -a_4(t) b_2(t){}^4+a_3(t) b_2(t){}^3+b_2(t){}^2 \left( 4 a_4(t) b_1(t) -a_2(t)\right) \\ &{}+b_2(t) \left( -4 a_4(t) b_0(t)-3 a_3(t) b_1(t)+a_1(t)\right) -2 a_4(t) b_1(t){}^2\\ &{}+3 a_3(t) b_0(t)+2 a_2(t) b_1(t)-3 a_0(t). \end{array}\right. \end{aligned}$$

(46)

In an especial case, let

$$\begin{aligned} a_0(t)= & {} \frac{16 \cos (t)}{\left( 9 \cos ^2(t)-17\right) \left( 9 \cos ^2(t)-11\right) },\\ a_1(t)= & {} \frac{\sin (t) \left( 3 \cos ^2(t)-5\right) \left( 81 \cos ^4(t)-252 \cos ^2(t) -90 \cos (t)+187\right) }{\left( 18 \cos ^2(t)-34\right) \left( 9 \cos ^2(t)-11\right) },\\ a_2(t)= & {} 1, \ \ \ \ \ \ \ \ a_3(t)=0,\\ a_4(t)= & {} \frac{243 \cos ^4(t)-756 \cos ^2(t)-108 \cos (t) +561}{\left( 18 \cos ^2(t)-34\right) \left( 9 \cos ^2(t)-11\right) }. \end{aligned}$$

Then $F(t) = b_0(t)+b_1(t)x+x^{3}$ is an algebraic curve for the equation (45) where

$$\begin{aligned} b_0(t) = \frac{1}{6} \sin (t) (3 \cos (2 t)-7) , \, b_1(t) =\frac{2}{3}, \ \ \ \ b_2(t)\equiv 0 \end{aligned}$$

is the solution of (46) and the corresponding cofactor curve $K(t) = c_0(t)+c_1(t)x+c_3(t)x^{3}$ is given by

$$\begin{aligned} c_0(t)= & {} -\frac{81 \sin (t) \cos (t) \left( 3 \cos ^2(t)-5\right) }{\left( 9 \cos ^2(t)-17\right) \left( 9 \cos ^2(t)-11\right) },\\ c_1(t)= & {} \frac{81 \cos ^4(t)-252 \cos ^2(t)+72 \cos (t)+187}{\left( 9 \cos ^2(t)-17\right) \left( 9 \cos ^2(t)-11\right) },\\ c_3(t)= & {} \frac{729 \cos ^4(t)-2268 \cos ^2(t)-324 \cos (t)+1683}{\left( 18 \cos ^2(t)-34\right) \left( 9 \cos ^2(t)-11\right) }. \end{aligned}$$

Finally, $x(t)=\sin (t)$ is a solution of $F(x(t),t) \equiv 0$, and thus (45) has a periodic orbit.