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A Taylor series-based continuation method for solutions of dynamical systems

Abstract

This paper describes a generic Taylor series-based continuation method, the so-called asymptotic numerical method, to compute the bifurcation diagrams of nonlinear systems. The key point of this approach is the quadratic recast of the equations as it allows to treat in the same way a wide range of dynamical systems and their solutions. Implicit differential-algebraic equations, forced or autonomous, possibly with time-delay or fractional-order derivatives are handled in the same framework. The static, periodic and quasi-periodic solutions can be continued and also transient solutions.

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Notes

  1. 1.

    Regular means that the Jacobian matrix \({\mathbf {J}}\) at \({\mathbf {U}}_0\) is of full rank.

  2. 2.

    This is possible thanks to the analytic version of the implicit function theorem [20].

  3. 3.

    In the case of a system with delay or fractional-order derivatives, it may be required to give the initial conditions over an interval of the form \(\left[ -T,0\right] \).

  4. 4.

    Following [42], we use here the name Hopf bifurcation. To our knowledge, this type of bifurcation has not been characterized theoretically but is analogous to a Hopf bifurcation point from a numerical point of view.

  5. 5.

    Manlab is available online on the dedicated website https://manlab.lma.cnrs-mrs.fr/.

  6. 6.

    If this is not the case, an auxiliary variable \(z(t) = {\dot{w}}(t)\) is added and the procedure explained here is applied recursively.

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Acknowledgements

This work has been carried out in the framework of the Labex MEC (ANR-10-LABX-0092) and of the A*MIDEX project (ANR-11-IDEX-0001-02), funded by the Investissements d’Avenir French Government program managed by the French National Research Agency (ANR).

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Correspondence to Louis Guillot.

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Appendices

Appendix A: Algebra of the HBM applied on transcendental functions

The case of an auxiliary variable v defined as \(v(t) = \mathrm {g}(u(t))\) is treated in detail. Let us suppose that the auxiliary variable \(w(t) = \dot{\mathrm {g}}(u(t))\) has been rewritten quadraticallyFootnote 6, the variable v can be defined as the unique solution of the differential system

$$\begin{aligned} \begin{aligned} v(0)&= \mathrm {g}(u(0)) \\ {\dot{v}}(t)&= w(t) {\dot{u}}(t) \end{aligned} \end{aligned}$$
(34)

which is the same as (16). The auxiliary variables \(u_0 = u(0)\), \(v_0 = v(0)\) and \(w_0 = w(0)\) are introduced to recast the first equation quadratically with the system :

$$\begin{aligned} \begin{aligned}&v_0 = \mathrm {g}(u_0) \\&\mathrm {d}v_0 = w_0 \mathrm {d}u_0 \end{aligned}. \end{aligned}$$
(35)

This system is treated with the formalism explained in [22] briefly recalled in Sect. 1.1.2. For the second equation, let us define the complex truncated Fourier series of u,v and w by:

$$\begin{aligned} u(t)= & {} \sum _{h=-H}^H {\mathbf {U}}_h \exp (ih\omega t)\; , \; \nonumber \\ v(t)= & {} \sum _{h=-H}^H {\mathbf {V}}_h \exp (ih\omega t) \; \mathrm {and} \nonumber \\ w(t)= & {} \sum _{h=-H}^H {\mathbf {W}}_h \exp (ih\omega t) \end{aligned}$$
(36)

The equation \({\dot{v}}(t) = w(t) {\dot{u}}(t)\) becomes :

$$\begin{aligned}&\sum _{h=-H}^H ih\omega {\mathbf {V}}_h \exp (ih\omega t) \nonumber \\&\quad = \left( \sum _{h=-H}^H {\mathbf {W}}_h \exp (ih\omega t) \right) \nonumber \\&\qquad \times \left( \sum _{h=-H}^H ih\omega {\mathbf {U}}_h \exp (ih\omega t) \right) \end{aligned}$$
(37)

which can be written

$$\begin{aligned}&\sum _{h=-H}^H ih\omega {\mathbf {V}}_h \exp (ih\omega t) \nonumber \\&\quad = \sum _{h=-2H}^{2H} \left( \sum _{\underset{-H \le k_1,k_2\le H}{k_1+k_2 = h}} {\mathbf {W}}_{k_1} ik_2\omega {\mathbf {U}}_{k_2} \right) \exp (ih\omega t).\nonumber \\ \end{aligned}$$
(38)

The right-hand-side series is truncated at order H, and the harmonics are balanced :

$$\begin{aligned} \forall h \in \llbracket -H , H \rrbracket , \quad ih \omega {\mathbf {V}}_h = \sum _{\underset{-H \le k_1,k_2\le H}{k_1+k_2 = h}} {\mathbf {W}}_{k_1} ik_2\omega {\mathbf {U}}_{k_2}.\nonumber \\ \end{aligned}$$
(39)

\(\omega \) is the angular frequency of the periodic solution and thus remains positive. It can be simplified to obtain

$$\begin{aligned} \forall h \in \llbracket -H , H \rrbracket , \quad h {\mathbf {V}}_h = \sum _{\underset{-H \le k_1,k_2\le H}{k_1+k_2 = h}} {\mathbf {W}}_{k_1} k_2 {\mathbf {U}}_{k_2}. \end{aligned}$$
(40)

This expression is quadratic with respect to the Fourier coefficients. Hence, the transcendental variable v can be written with a quadratic formalism in the frequency domain in an automated way. The operators \({\mathbf {V}}\mapsto \mathrm {d}{\mathbf {L}}({\mathbf {V}})\) and \(({\mathbf {U}},{\mathbf {W}}) \mapsto \mathrm {d}{\mathbf {Q}}({\mathbf {W}},{\mathbf {U}})\) are, respectively, the left hand side and the right hand side of the equations (40). \({\mathbf {U}}\), \({\mathbf {V}}\) and \({\mathbf {W}}\) are the vectors of Fourier coefficients of u, v and w, respectively. It is easy to see that \(\mathrm {d}{\mathbf {L}}\) and \(\mathrm {d}{\mathbf {Q}}\) are, respectively, linear and bilinear in their argument(s).

Appendix B: Time integration of the pendulum with a class of Euler schemes

The ODE equations of the pendulum in polar coordinates (20)

$$\begin{aligned} \left\{ \begin{array}{rl} {\dot{\theta }} &{} = \phi \\ {\dot{\phi }} &{} = - \sin (\theta ) \end{array}\right. \end{aligned}$$
(41)

are solved using a generalized Euler scheme on the discretized grid \(\left\{ 0 , h , 2h , \dots , (N-1)h \right\} \). The standard notation \(\theta _n = \theta (nh)\) and \(\phi _n = \phi (nh)\) is used. The initial conditions are \(\theta _0 = \frac{\pi }{2}\) and \(\phi _0 = 0\). The discretized system is then

$$\begin{aligned} \left\{ \begin{array}{rl} \theta _{n+1} &{} = \theta _{n} + h( (1-\lambda ) \phi _n + \lambda \phi _{n+1} ) \\ \phi _{n+1} &{} = \phi _n - h \sin ( (1-\lambda ) \theta _n + \lambda \theta _{n+1}) \end{array}\right. \end{aligned}$$
(42)

where the parameter \(\lambda \) is the continuation parameter and interpolates an explicit Euler scheme (for \(\lambda = 0\)) and an implicit Euler scheme (for \(\lambda = 1\)). The centered Euler scheme (\(\lambda = \frac{1}{2}\)) is known to be a simple symplectic scheme [32]. It is then well suited to solve these conservative equations. The auxiliary variables \(\vartheta _n = (1-\lambda ) \theta _n + \lambda \theta _{n+1}\) and \(F_n = \sin ( \vartheta _n)\) are added to the system (42) with the companion variables \(G_n = \cos (\vartheta _n)\). The final quadratic recast with the differentiated forms of the auxiliary variables is

$$\begin{aligned} \forall n \in \llbracket 0 , N-1 \rrbracket , \left\{ \begin{array}{lllll} \theta _{n+1} &{} = \theta _{n} + h( (1-\lambda ) \phi _n + \lambda \phi _{n+1} ) &{} &{} &{}\\ \phi _{n+1} &{} = \phi _n - h F_n &{} &{} &{}\\ \vartheta _n &{} = (1-\lambda ) \theta _n + \lambda \theta _{n+1} &{} &{} &{}\\ F_n &{} = \sin (\vartheta _n) &{} \text {and} &{}\quad \mathrm {d}F_n &{} = G_n \mathrm {d}\vartheta _n \\ G_n &{} = \cos (\vartheta _n) &{} \text {and} &{} \quad \mathrm {d}G_n &{} = -F_n \mathrm {d}\vartheta _n \\ \end{array}\right. \end{aligned}$$
(43)
Fig. 9
figure9

The top figure shows a waterfall of the trajectories of (42). The bottom figure shows the evolution of the discrete energy (in color) with respect to time (y axis) for all the values of \(\lambda \) (x axis). From \(\lambda =0\) to \(\lambda =1\), the scheme evolves from an explicit Euler to an implicit Euler. The value \(\lambda =0.5\) is plotted in black to enhance the trajectory obtained with this scheme, which is known to be symplectic [32]. It is clear in the bottom figure that the scheme either creates or dissipates energy when \(\lambda \) is, respectively, smaller or greater than 0.5. (Color figure online)

The top Fig. 9 shows the trajectories obtained for \(N=500\) and \(h=\frac{2\pi }{100}\), that is over around 5 periods of the analytic solution. The bottom Fig. 9 shows the discrete energy \(H_n = \frac{\phi _n^2}{2} - 1 - \cos (\theta _n)\) in color for \(0\le \lambda \le 1\). It is clear that energy is created when \(\lambda <0.5\) while energy is dissipated when \(\lambda >0.5\). The trajectory obtained for \(\lambda =0.5\) is represented in black showing the stability of the energy over time, as expected for this symplectic scheme.

Appendix C: Computation of Taylor series coefficients

Some examples (taken from Sect. 1.1) of practical computation of the Taylor series coefficients are shown. However, this procedure is completely automatized from the quadratic formulation of the system of equations. It is detailed here for pedagogical purposes. The following computations illustrate how easy and straightforward it is to derive Taylor series coefficients when a quadratic recast is given.

C.1 Rational case

The system of equations is :

$$\begin{aligned} \left\{ \begin{array}{ll} y - u - xv &{} = 0\\ ux - 1 &{} = 0\\ v - x^2 &{} = 0 \end{array}\right. \end{aligned}$$
(44)

Assume that \((x_0,y_0,u_0,v_0)\) is a solution of the equations. The n-th Taylor series coefficient of the unknown x is written \(x_n\), such that:

$$\begin{aligned} x(a)&= \sum \limits _{n=0}^{+\infty } x_n a^n \end{aligned}$$
(45)

Replacing the variables by their Taylor series in the equations gives:

$$\begin{aligned} \left\{ \begin{array}{ll} \sum \limits _{n=0}^{+\infty } y_n a^n - \sum \limits _{n=0}^{+\infty } u_n a^n - \left( \sum \limits _{n=0}^{+\infty }x_n a^n \right) \left( \sum \limits _{n=0}^{+\infty }v_n a^n \right) &{} = 0 \\ \left( \sum \limits _{n=0}^{+\infty }u_na^n \right) \left( \sum \limits _{n=0}^{+\infty }x_na^n \right) - 1 &{} = 0\\ \sum \limits _{n=0}^{+\infty }v_n a^n - \left( \sum \limits _{n=0}^{+\infty }x_n a^n \right) \left( \sum \limits _{n=0}^{+\infty }x_n a^n \right) &{} = 0 \end{array}\right. \nonumber \\ \end{aligned}$$
(46)

Developing the products and equating the terms of same order in a finally gives, for \(n\ge 0\):

$$\begin{aligned} \left\{ \begin{array}{ll} y_n - u_n - \sum \nolimits _{i=0}^{n}x_i v_{n-i} &{} = 0\\ \sum \nolimits _{i=0}^{n} u_i x_{n-i} - 1\delta _{n=0} &{} = 0\\ v_n - \sum \nolimits _{i=0}^{n} x_i x_{n-i} &{} = 0 \end{array}\right. \end{aligned}$$
(47)

where \(\delta _{n=0}\) is the Kronecker symbol, its value is 1 if and only if \(n=0\).

This system for \(n=0\) gives

$$\begin{aligned} \left\{ \begin{array}{ll} y_0 - u_0 - x_0v_0 &{} = 0\\ u_0x_0 - 1 &{} = 0\\ v_0 - x_0^2 &{} = 0 \end{array}\right. \end{aligned}$$
(48)

which is verified from the assumption that \((x_0,y_0, u_0,v_0)\) is a solution of the equations (44).

For \(n\ge 1\), the terms of higher order are put apart to make the recurrence relations appear :

$$\begin{aligned} \left\{ \begin{array}{ll} y_n - u_n - x_0 v_n - x_n v_0 &{} = \sum \nolimits _{i=1}^{n-1}x_i v_{n-i}\\ u_0 x_n + u_n x_0 &{} = - \sum \nolimits _{i=1}^{n-1} u_i x_{n-i} \\ v_n - 2 x_0 x_n &{} = \sum \nolimits _{i=1}^{n-1} x_i x_{n-i} \end{array}\right. \end{aligned}$$
(49)

which is equivalent to the linear system

$$\begin{aligned} {\mathbf {J}}_0 \times \begin{bmatrix} x_n \\ y_n \\ u_n \\ v_n \end{bmatrix} = \begin{bmatrix} \sum \nolimits _{i=1}^{n-1}x_i v_{n-i}\\ - \sum \nolimits _{i=1}^{n-1} u_i x_{n-i} \\ \sum \nolimits _{i=1}^{n-1} x_i x_{n-i} \end{bmatrix} \end{aligned}$$
(50)

where \({\mathbf {J}}_0 = \begin{bmatrix} -v_0&1&- 1&x_0 \\ u_0&0&x_0&0 \\ - 2 x_0&0&0&1 \end{bmatrix}\) is the Jacobian matrix of the system at the solution point \((x_0,y_0,u_0,v_0)\). From the knowledge of the initial solution point \((x_0,y_0,u_0,v_0)\), it is then possible to deduce the whole Taylor series of the variables xyu and v.

C.2 Transcendental case

The system of equations and its differential form are given by :

$$\begin{aligned} \left\{ \begin{array}{ll} y - \sin (x) = 0 &{} \hbox {and} \quad \mathrm {d}y - z \mathrm {d}x = 0\\ z - \cos (x) = 0 &{} \hbox {and} \quad \mathrm {d}z + y \mathrm {d}x = 0 \end{array}\right. \end{aligned}$$
(51)

Assume that \((x_0,y_0,z_0)\) is a solution of the equations. Thus, \(y_0 = \sin (x_0)\) and \(z_0 = \cos (x_0)\). Contrarily to the previous section, the differential form of the equations is used to compute the higher-order coefficients of the Taylor series by writing

$$\begin{aligned} \left\{ \begin{array}{cc} \frac{\partial y}{\partial a} - z(a) \frac{\partial x}{\partial a} = 0 \\ \frac{\partial z}{\partial a} + y(a) \frac{\partial x}{\partial a} = 0 \end{array}\right. \end{aligned}$$
(52)

The derivative \(\frac{\partial x}{\partial a}\) of the Taylor series of the unknown x with respect to a is :

$$\begin{aligned} \frac{\partial x}{\partial a}(a)&= \sum _{n=0}^{+\infty } (n+1) x_{n+1} a^n \end{aligned}$$
(53)

Replacing the variables by the corresponding series in the equations (52) gives:

$$\begin{aligned} \left\{ \begin{array}{cc} \sum \nolimits _{n=0}^{+\infty } (n+1) y_{n+1} a^n - \left( \sum \nolimits _{n=0}^{+\infty } z_n a^n\right) \left( \sum \nolimits _{n=0}^{+\infty } (n+1) x_{n+1} a^n \right) = 0 \\ \sum \nolimits _{n=0}^{+\infty } (n+1) z_{n+1} a^n + \left( \sum \nolimits _{n=0}^{+\infty } y_n a^n \right) \left( \sum \nolimits _{n=0}^{+\infty } (n+1) x_{n+1} a^n \right) = 0 \end{array}\right. \end{aligned}$$
(54)

Developing the products and equating the terms of same order in a finally gives, for \(n\ge 0\):

$$\begin{aligned} \left\{ \begin{array}{cl} (n+1) y_{n+1} - \sum \nolimits _{i=0}^{n} z_i (n+1-i) x_{n+1-i} &{} = 0\\ (n+1) z_{n+1} + \sum \nolimits _{i=0}^{n} y_i (n+1-i) x_{n+1-i} &{} = 0 \end{array}\right. \end{aligned}$$
(55)

The terms of higher order are put apart and everything is divided by \((n+1)\) to make the recurrence relations appear :

$$\begin{aligned} \left\{ \begin{array}{cl} y_{n+1} - z_0 x_{n+1} &{} = \sum \nolimits _{i=1}^{n} z_i \frac{n+1-i}{n+1} x_{n+1-i} \\ z_{n+1} + y_0 x_{n+1} &{} = - \sum \nolimits _{i=1}^{n} y_i \frac{n+1-i}{n+1} x_{n+1-i} \\ \end{array}\right. \end{aligned}$$
(56)

which is equivalent to the linear systems

$$\begin{aligned} {\mathbf {J}}_0 \times \begin{bmatrix} x_{n+1} \\ y_{n+1} \\ z_{n+1} \end{bmatrix} = \begin{bmatrix} \sum \nolimits _{i=1}^{n} z_i (n+1-i) x_{n+1-i} \\ - \sum \nolimits _{i=1}^{n} y_i (n+1-i) x_{n+1-i} \end{bmatrix} \end{aligned}$$
(57)

where \({\mathbf {J}}_0 = \begin{bmatrix} - z_0&1&0 \\ y_0&0&1 \end{bmatrix}\) is the Jacobian matrix of the system at the solution point \((x_0,y_0,z_0)\). From the knowledge of the initial solution point \((x_0,y_0,z_0)\), it is then possible to deduce the whole Taylor series of the variables xy and z.

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Guillot, L., Cochelin, B. & Vergez, C. A Taylor series-based continuation method for solutions of dynamical systems. Nonlinear Dyn 98, 2827–2845 (2019). https://doi.org/10.1007/s11071-019-04989-5

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Keywords

  • Nonlinear dynamics
  • Numerical continuation
  • Quadratic recast
  • Asymptotic numerical method
  • Taylor series
  • Dynamical systems