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Finite-time Lyapunov exponents in the instantaneous limit and material transport

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Abstract

Lagrangian techniques, such as the finite-time Lyapunov exponent (FTLE) and hyperbolic Lagrangian coherent structures, have become popular tools for analyzing unsteady fluid flows. These techniques identify regions where particles transported by a flow will converge to and diverge from over a finite-time interval, even in a divergence-free flow. Lagrangian analyses, however, are time consuming and computationally expensive, hence unsuitable for quickly assessing short-term material transport. A recently developed method called Objective Eulerian Coherent Structures (OECSs) (Serra and Haller in Chaos Interdiscip J Nonlinear Sci 26(5):053110, 2016) rigorously connected Eulerian quantities to short-term Lagrangian transport. This Eulerian method is faster and less expensive to compute than its Lagrangian counterparts, and needs only a single snapshot of a velocity field. Along the same line, here we define the instantaneous Lyapunov Exponent, the instantaneous counterpart of the FTLE, and connect the Taylor series expansion of the right Cauchy-Green deformation tensor to the infinitesimal integration time limit of the FTLE. We illustrate our results on geophysical fluid flows from numerical models as well as analytical flows, and demonstrate the efficacy of attracting and repelling instantaneous Lyapunov exponent structures in predicting short-term material transport.

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Acknowledgements

We thank Hosein Foroutan for providing us with WRF model data, Nicole Abaid for assistance with the perturbation analysis, Gary Nave for fruitful conversations related to some of the ideas covered in this work during their nascent stage, and Siavash Ameli for developing and maintaining the TRACE server used for the geophysical particle integration calculations. M.S. acknowledges support from the Schmidt Science Fellowship and the Postdoc Mobility Fellowship from the Swiss National Science Foundation.

Funding

This research was supported in part by grants from the National Science Foundation (NSF) under grant numbers AGS 1520825 (Hazards SEES: Advanced Lagrangian Methods for Prediction, Mitigation and Response to Environmental Flow Hazards) and DMS 1821145 (Data-Driven Computation of Lagrangian Transport Structure in Realistic Flows). Any opinions, findings, and conclusions or recommendations expressed in this material are those of the authors and do not necessarily reflect the views of the sponsors.

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Appendices

Appendices

Expansion of the right Cauchy-Green tensor in the integration time

For tensor fields in what follows, the dependence on \({\mathbf {x}}_0\) and \(t_0\) will be notationally dropped for clarity, as it will be understood. For small integration time \(T=t-t_0\), the right Cauchy-Green tensor, \({\mathbf {C}}\), may be expanded, as in [65, 68, 77], in terms of the integration time T,

$$\begin{aligned} {\mathbf {C}}& = {\mathbf {C}}|_{T=0} +\left. \frac{\mathrm{d}{\mathbf {C}}}{\mathrm{d}T}\right| _{T=0}T + \frac{1}{2!} \left. \frac{\mathrm{d}^2{\mathbf {C}}}{\mathrm{d}T^2}\right| _{T=0}T^2\nonumber \\&+ \frac{1}{3!} \left. \frac{\mathrm{d}^3{\mathbf {C}}}{\mathrm{d}T^3}\right| _{T=0}T^3+ {\mathcal {O}}(T^4). \end{aligned}$$
(46)

where the dependence on the initial position and time is understood. Because all derivatives are evaluated at \(T=0\), \(\left. \tfrac{\mathrm{d}}{\mathrm{d}t}\right| _{t=t_0}=\left. \tfrac{\mathrm{d}}{\mathrm{d}T}\right| _{T=0}\). The first term on the right denotes the situation of no deformation, therefore, \({\mathbf {C}}|_{T=0}=\mathbb {1}\). The derivatives of the right Cauchy-Green tensor are given to any order by the Rivlin-Ericksen tensors [68, 77, 103],

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d}{\mathbf {C}}}{\mathrm{d}t}&= \nabla \frac{\mathrm{d}{\mathbf {x}}}{\mathrm{d}t}+\left( \nabla \frac{\mathrm{d}{\mathbf {x}}}{\mathrm{d}t}\right) ^{\top },&{~} \\ \frac{\mathrm{d}^k{\mathbf {C}}}{\mathrm{d}t^k}&= \nabla \frac{\mathrm{d}^k{\mathbf {x}}}{\mathrm{d}t^k} + \left( \nabla \frac{\mathrm{d}^k{\mathbf {x}}}{\mathrm{d}t^k}\right) ^{\top }\\&\quad + \sum _{i=1}^{k-1}\left( \!\begin{array}{c} k \\ i \end{array}\!\right) \left( \nabla \frac{\mathrm{d}^i{\mathbf {x}}}{\mathrm{d}t^i}\right) ^{\top } \nabla \frac{\mathrm{d}^{k-i}{\mathbf {x}}}{\mathrm{d}t^{k-i}},&k \ge 2. \end{aligned} \end{aligned}$$
(47)

For small \(|T|\ll 1\), the leading order behavior is given by the first Rivlin-Ericksen tensor \((\nabla {\mathbf {v}}+(\nabla {\mathbf {v}})^{\top })\), which is twice \({\mathbf {S}}\), from (7). The second-order term is,

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d}^2{\mathbf {C}}}{\mathrm{d}t^2}&= \nabla \frac{\mathrm{d}^2{\mathbf {x}}}{\mathrm{d}t^2} + \left( \nabla \frac{\mathrm{d}^2{\mathbf {x}}}{\mathrm{d}t^2}\right) ^{\top } + 2 \left( \nabla \frac{\mathrm{d}{\mathbf {x}}}{\mathrm{d}t}\right) ^{\top } \nabla \frac{\mathrm{d}{\mathbf {x}}}{\mathrm{d}t}, \\&= \nabla \frac{\mathrm{d}{\mathbf {v}}}{\mathrm{d}t} + \left( \nabla \frac{\mathrm{d}{\mathbf {v}}}{\mathrm{d}t}\right) ^{\top } + 2 \left( \nabla {\mathbf {v}}\right) ^{\top } \nabla {\mathbf {v}}, \\&= 2 {\mathbf {B}} \end{aligned} \end{aligned}$$
(48)

where \({\mathbf {B}}\) is the same as given in (11).

The third-order term is,

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d}^3{\mathbf {C}}}{\mathrm{d}t^3}&= \nabla \frac{\mathrm{d}^3{\mathbf {x}}}{\mathrm{d}t^3} + \left( \nabla \frac{\mathrm{d}^3{\mathbf {x}}}{\mathrm{d}t^3}\right) ^{\top } \\&\quad+ 3 \left[ \left( \nabla \frac{\mathrm{d}{\mathbf {x}}}{\mathrm{d}t}\right) ^{\top } \nabla \frac{\mathrm{d}^{2}{\mathbf {x}}}{\mathrm{d}t^{2}} + \left( \nabla \frac{\mathrm{d}^{2}{\mathbf {x}}}{\mathrm{d}t^{2}}\right) ^{\top } \nabla \frac{\mathrm{d}{\mathbf {x}}}{\mathrm{d}t} \right] , \\&= \nabla \frac{\mathrm{d}{\mathbf {a}}}{\mathrm{d}t} + \left( \nabla \frac{\mathrm{d}{\mathbf {a}}}{\mathrm{d}t}\right) ^{\top } + 3 \left[ \left( \nabla {\mathbf {v}}\right) ^{\top } \nabla {\mathbf {a}} + \left( \nabla {\mathbf {a}}\right) ^{\top } \nabla {\mathbf {v}} \right] , \\&= 3 {\mathbf {Q}}, \end{aligned} \end{aligned}$$
(49)

where \({\mathbf {Q}}\) is the same as given in (13).

The expansion of the right Cauchy-Green tensor (46) can be written as,

$$\begin{aligned} \begin{aligned} {\mathbf {C}}&= \mathbb {1} + 2T{\mathbf {S}} + T^2 {\mathbf {B}} + \tfrac{1}{2}T^3 {\mathbf {Q}} + {\mathcal {O}}(T^4), \\&=\mathbb {1} + 2T \left( {\mathbf {S}} + \tfrac{1}{2}T {\mathbf {B}} + (\tfrac{1}{2}T)^2 {\mathbf {Q}} + {\mathcal {O}}(T^3) \right) , \end{aligned} \end{aligned}$$
(50)

which is a form convenient for matrix perturbation analysis, as in Appendix “Eigenvalues of the Taylor-expanded right Cauchy-Green tensor”.

Details of approximating the FTLE to second-order in integration time

Note the following general result for the eigenvalues,

$$\begin{aligned} \lambda _{-}({\mathbf {A}})=\lambda _{1}({\mathbf {A}})\le \cdots \le \lambda _{n}({\mathbf {A}})=\lambda _{+}({\mathbf {A}}), \end{aligned}$$
(51)

of \(n \times n\) real symmetric matrices \({\mathbf {A}}\). Here, we use \(\lambda _{-}({\mathbf {A}})\) and \(\lambda _{+}({\mathbf {A}})\) as shorthand for \(\lambda _{\mathrm{min}}({\mathbf {A}})\) and \(\lambda _{\mathrm{max}}({\mathbf {A}})\), the minimum and maximum eigenvalues of \({\mathbf {A}}\), respectively. For scalar \(c \ne 0\),

$$\begin{aligned} \lambda _{\pm }(\mathbb {1} + c {\mathbf {A}}) = 1 + \lambda _{\pm }(c {\mathbf {A}}), \end{aligned}$$
(52)

where,

$$\begin{aligned} \lambda _{\pm }(c {\mathbf {A}}) = {\left\{ \begin{array}{ll} c \lambda _{\pm }({\mathbf {A}}), &{} \text {for}\ c>0, \\ c \lambda _{\mp }({\mathbf {A}}), &{} \text {for}\ c<0. \end{array}\right. } \end{aligned}$$
(53)

See Appendix “Proof of equation (52)” for the proof.

In (6), \(\lambda _{n} = \lambda _{+}({\mathbf {C}}_{t_{0}}^{t}({\mathbf {x}}))\). For small \(T>0\), where the \({\mathcal {O}}(T^2)\) and higher terms can be neglected,

$$\begin{aligned} \lambda _{+}({\mathbf {C}}_{t_{0}}^{t}({\mathbf {x}})) = 1 + 2T\lambda _{+}({\mathbf {S}}(\mathbf{x} ,t_0)) + {\mathcal {O}}(T^2). \end{aligned}$$
(54)

Thus,

$$\begin{aligned} \log (\lambda _n) = \log (1 + 2T\lambda _{+}({\mathbf {S}}(\mathbf{x} ,t_0)))= 2T\lambda _{+}({\mathbf {S}}(\mathbf{x} ,t_0)) = 2Ts_n(\mathbf{x} ,t_0), \end{aligned}$$
(55)

in the limit of small T using the Taylor expansion, \(\log (1+\delta ) = \delta + {\mathcal {O}}(\delta ^2)\) for small \(|\delta |\).

From (6), and noting that \(|T|=T\) for \(T>0\),

$$\begin{aligned} \sigma _{t_{0}}^{t}({\mathbf {x}}) = \frac{1}{2|T|}\log (\lambda _{n}) =\frac{1}{2T}2Ts_n (\mathbf{x} ,t_0) =s_n(\mathbf{x} ,t_0) \end{aligned}$$
(56)

Therefore, the maximum eigenvalue of \({\mathbf {S}}(\mathbf{x} ,t_0)\) is the limit of the FTLE value for forward time as \(T \rightarrow 0^{+}\).

For \(T<0\) with small T,

$$\begin{aligned} \lambda _{+}({\mathbf {C}}_{t_{0}}^{t}({\mathbf {x}})) = 1 + 2T\lambda _{-}({\mathbf {S}}(\mathbf{x} ,t_0)) + {\mathcal {O}}(T^2). \end{aligned}$$
(57)

Thus,

$$\begin{aligned} \log (\lambda _n) = 2T\lambda _{-}({\mathbf {S}}(\mathbf{x} ,t_0)) = 2Ts_1(\mathbf{x} ,t_0), \end{aligned}$$
(58)

in the limit of small T.

From (6), and noting that \(|T|=-T\) for \(T<0\),

$$\begin{aligned} \sigma _{t_{0}}^{t}({\mathbf {x}}) = \frac{1}{2|T|}\log (\lambda _{n}) =-\frac{1}{2T}2Ts_1(\mathbf{x} ,t_0) =-s_1(\mathbf{x} ,t_0). \end{aligned}$$
(59)

Therefore, the negative of the minimum eigenvalue of \({\mathbf {S}}(\mathbf{x} ,t_0)\) is the limit of the FTLE value for backward time as \(T \rightarrow 0^{-}\).

Consider now the third term, the order \(T^2\) term, in the expansion (10) of the right Cauchy-Green tensor. Then, (54) becomes,

$$\begin{aligned} \lambda _{+}({\mathbf {C}}_{t_{0}}^{t}({\mathbf {x}})) = 1 + 2T\lambda ^{+}\Big ({\mathbf {S}}(\mathbf{x} ,t_0) +\tfrac{1}{2}T {\mathbf {B}}(\mathbf{x} ,t_0) \Big ) + {\mathcal {O}}(T^3). \end{aligned}$$
(60)

Note that \({\mathbf {B}}(\mathbf{x} ,t_0)\), like \({\mathbf {S}}(\mathbf{x} ,t_0)\), is symmetric.

Below, we adopt the notation of \(s_-\) and \(s_+\) for \(s_1\) and \(s_n\), respectively, as in the main text.

It can be shown via matrix perturbation techniques (see Appendix “Eigenvalues of the Taylor-expanded right Cauchy-Green tensor”) that,

$$\begin{aligned} \lambda _{+}\Big ({\mathbf {S}}(\mathbf{x} ,t_0) +\tfrac{1}{2}T {\mathbf {B}}(\mathbf{x} ,t_0) \Big ) = s_+ + \tfrac{1}{2}T \varvec{e}_{+}^{\top } {\mathbf {B}} \varvec{e}_{+} + {\mathcal {O}}(T^2). \end{aligned}$$
(61)

Using the Taylor expansion \(\log (1+\delta ) = \delta -\tfrac{1}{2}\delta ^2 + \tfrac{1}{3}\delta ^3 +{\mathcal {O}}(\delta ^4)\) for small \(|\delta |\), by a similar argument as before, for small T,

$$\begin{aligned} \begin{aligned}&\log (\lambda _{+}({\mathbf {C}}_{t_{0}}^{t}({\mathbf {x}}))) = \log \Big (1 + 2T \Big [ s_+ + \tfrac{1}{2}T \varvec{e}_{+}^{\top } {\mathbf {B}} \varvec{e}_{+} + {\mathcal {O}}(T^2) \Big ] \Big ), \\&\quad = 2T \Big [ s_+ + \tfrac{1}{2}T \varvec{e}_{+}^{\top } {\mathbf {B}} \varvec{e}_{+} + {\mathcal {O}}(T^2) \Big ] - \tfrac{1}{2} 4T^2 s_+^2 + {\mathcal {O}}(T^3), \\&\quad = 2T \Big [ s_+ + T \Big ( - s_+^2 + \tfrac{1}{2} \varvec{e}_{+}^{\top } {\mathbf {B}} \varvec{e}_{+} \Big ) + {\mathcal {O}}(T^2) \Big ]. \end{aligned} \end{aligned}$$
(62)

Therefore, for \(T>0\) with small |T|,

$$\begin{aligned}&\sigma _{t_{0}}^{t}({\mathbf {x}}) = s_+(\mathbf{x} ,t_0)\nonumber \\&+ T \Big ( - s_+(\mathbf{x} ,t_0)^2 + \tfrac{1}{2} \varvec{e}_{+}(\mathbf{x} ,t_0)^{\top } {\mathbf {B}}(\mathbf{x} ,t_0) \varvec{e}_{+}(\mathbf{x} ,t_0) \Big ) + {\mathcal {O}}(T^2). \end{aligned}$$
(63)

And similarly, for \(T<0\) with small |T|,

$$\begin{aligned}&\sigma _{t_{0}}^{t}({\mathbf {x}}) = -s_-(\mathbf{x} ,t_0)\nonumber \\&- T \Big (- s_-(\mathbf{x} ,t_0)^2 + \tfrac{1}{2} \varvec{e}_{-}(\mathbf{x} ,t_0)^{\top } {\mathbf {B}}(\mathbf{x} ,t_0) \varvec{e}_{-}(\mathbf{x} ,t_0) \Big ) + {\mathcal {O}}(T^2). \end{aligned}$$
(64)

If we continue this procedure to obtain the approximate forward and backward FTLE through second order in T, we get (14).

Proof of equation (52)

Let \({\mathbf {A}}\) be an \(n \times n\) matrix, \(\lambda \) an eigenvalue of \({\mathbf {A}}\), \(\varvec{\xi }\) the corresponding eigenvector of \({\mathbf {A}}\), \(\mathbb {1}\) the \(n \times n\) identity matrix and \(\zeta \in {\mathbb {C}}\). By the definition of an eigenvalue \({\mathbf {A}}\varvec{\xi }\) = \(\lambda \varvec{\xi }\), we have,

$$\begin{aligned} \left( \zeta \,\mathbb {1}+{\mathbf {A}}\right) \varvec{\xi } = \zeta \,\varvec{\xi }+{\mathbf {A}}\varvec{\xi }=\zeta \,\varvec{\xi }+\lambda \varvec{\xi }=\left( \zeta +\lambda \right) \varvec{\xi }. \end{aligned}$$
(65)

Therefore, if \(\lambda \) is an eigenvalue of \({\mathbf {A}}\) with eigenvector \(\varvec{\xi }\), then \((\zeta +\lambda )\) is an eigenvalue of \(\zeta \,\mathbb {1}+{\mathbf {A}}\) with the same eigenvector \(\varvec{\xi }\). In particular, this holds when \(\zeta =1\), as in (52).

Equality of the eigenvectors of S and C as integration time goes to zero

Let \(T>0\) be small enough that the relationships in (10) and (54) hold and \({\mathcal {O}}(T^2)\) terms are negligible. As before, let \(\varvec{e}_{i}\) be the eigenvector of \({\mathbf {S}}\) associated with \(s_{i}\), then,

$$\begin{aligned}&{\mathbf {S}}\,\varvec{e}_{i}=s_{i}\varvec{e}_{i}, \end{aligned}$$
(66)
$$\begin{aligned}&2 T {\mathbf {S}}\,\varvec{e}_{i}+\varvec{e}_{i}= 2 T s_{i}\varvec{e}_{i}+\varvec{e}_{i},\end{aligned}$$
(67)
$$\begin{aligned}&\left( 2 T {\mathbf {S}}+\mathbb {1}\right) \varvec{e}_{i}= \left( 2 T s_{i}+1\right) \varvec{e}_{i},\end{aligned}$$
(68)
$$\begin{aligned}&{\mathbf {C}}\,\varvec{e}_{i}=\lambda _{i}\varvec{e}_{i}, \end{aligned}$$
(69)

where the dependence on \({\mathbf {x}}\) and \(t_0\) is understood and we used the order-T approximation for \({\mathbf {C}}\). But from (4) and (5),

$$\begin{aligned} {\mathbf {C}}\,\varvec{\xi }_{\lambda _i}=\lambda _{i}\varvec{\xi }_{\lambda _i} \end{aligned}$$
(70)

thus,

$$\begin{aligned} \varvec{e}_i = \varvec{\xi }_{\lambda _i} \end{aligned}$$
(71)

that is, if \(\varvec{e}_{i}\) is an eigenvector of \({\mathbf {S}}\), then it is also an eigenvector of \({\mathbf {C}}\) in the limit as \(T\rightarrow 0\). Now, assuming that \(\varvec{\xi }_{\lambda _i}\) is the eigenvector of \({\mathbf {C}}\) associated with \(\lambda _{i}\), and working through (66-69) in reverse proves that if \(\varvec{\xi }_{\lambda _i}\) is an eigenvector of \({\mathbf {C}}\), then it is also an eigenvector of \({\mathbf {S}}\) in the limit as T goes to 0. For \(T<0\) an analogous argument holds using (57) in place of (54) and with the ordering of the eigenvalues opposed, i.e., \(\lambda _i \sim s_{n-i+1},\,i\in \{1,\ldots ,n\}\).

Therefore, in the limit as |T| goes to 0, the eigenvectors of \({\mathbf {S}}\) and \({\mathbf {C}}\) are equal. For small |T|, we can also use the perturbation expansion of \({\mathbf {C}}\), to get the estimated eigenvectors of \({\mathbf {C}}\) from (73) in Appendix “Eigenvalues of the Taylor-expanded right Cauchy-Green tensor”, which provides the eigenvectors through order \(T^2\) using only the velocity field \({\mathbf {v}}\) from (1) evaluated at \({\mathbf {x}}\) and time \(t_0\) as well as appropriate derivatives.

Eigenvalues of the Taylor-expanded right Cauchy-Green tensor

Let \({\mathbf {S}}\) be a real, symmetric \(n \times n\) matrix with n distinct eigenvalues, and let \({\mathbf {B}}\) and \({\mathbf {Q}}\) also be real, symmetric \(n \times n\) matrices. We seek the eigenvalues of,

$$\begin{aligned} {\mathbf {S}}_{\varepsilon } = {\mathbf {S}} + \varepsilon {\mathbf {B}} + \varepsilon ^2{\mathbf {Q}}, \end{aligned}$$
(72)

a perturbation of \({\mathbf {S}}\), where \(|\varepsilon |\) is a small scalar. In our case of interest, from (50), the small parameter is \(\varepsilon = \tfrac{1}{2}T\).

Consider the eigenvalue \(\mu _0\) of \({\mathbf {S}}\) with corresponding normalized eigenvector \(\varvec{\xi }_0\). Let’s refer to the perturbed eigenvalue and corresponding perturbed eigenvector of \({\mathbf {S}}_{\varepsilon }\) as \(\mu _{\varepsilon }\) and \(\varvec{\xi }_{\varepsilon }\). One can expand \(\varvec{\xi }_{\varepsilon }\) and \(\mu _{\varepsilon }\) in powers of \(\varepsilon \) as

$$\begin{aligned}&\varvec{\xi }_{\varepsilon } = \varvec{\xi }_{0} + \varepsilon \varvec{\xi }_{1} + \varepsilon ^2 \varvec{\xi }_{2} + {\mathcal {O}}(\varepsilon ^3), \end{aligned}$$
(73)
$$\begin{aligned}&\mu _{\varepsilon } = \mu _{0} + \varepsilon \mu _{1} + \varepsilon ^2 \mu _{2} + {\mathcal {O}}(\varepsilon ^3). \end{aligned}$$
(74)

The eigenvector equation, \({\mathbf {S}}_{\varepsilon }\varvec{\xi }_{\varepsilon } = \mu _{\varepsilon } \varvec{\xi }_{\varepsilon }\), can be approximated as

$$\begin{aligned}&({\mathbf {S}} + \varepsilon {\mathbf {B}} + \varepsilon ^2{\mathbf {Q}}) (\varvec{\xi }_{0} + \varepsilon \varvec{\xi }_{1} + \varepsilon ^2 \varvec{\xi }_{2}) \nonumber \\&\quad = (\mu _{0} + \varepsilon \mu _{1} + \varepsilon ^2 \mu _{2})(\varvec{\xi }_{0} + \varepsilon \varvec{\xi }_{1} + \varepsilon ^2 \varvec{\xi }_{2}), \end{aligned}$$
(75)

which leads to the following three expressions, corresponding to the order one terms, order \(\varepsilon \), and order \(\varepsilon ^2\) terms, respectively,

$$\begin{aligned}&{\mathbf {S}} \varvec{\xi }_{0} = \mu _{0} \varvec{\xi }_{0}, \end{aligned}$$
(76)
$$\begin{aligned}&{\mathbf {S}} \varvec{\xi }_{1} + {\mathbf {B}} \varvec{\xi }_{0} = \mu _{0} \varvec{\xi }_{1} + \mu _1 \varvec{\xi }_{0}, \end{aligned}$$
(77)
$$\begin{aligned}&{\mathbf {S}} \varvec{\xi }_{2}+{\mathbf {B}} \varvec{\xi }_{1}+{\mathbf {Q}}\varvec{\xi }_{0} = \mu _{0} \varvec{\xi }_{2} + \mu _1 \varvec{\xi }_{1} + \mu _2 \varvec{\xi }_{0}. \end{aligned}$$
(78)

Multiply (77) by \(\varvec{\xi }_{0}^{\top }\) to get,

$$\begin{aligned} \varvec{\xi }_{0}^{\top }{\mathbf {S}} \varvec{\xi }_{1} + \varvec{\xi }_{0}^{\top }{\mathbf {B}} \varvec{\xi }_{0} = \mu _{0} \varvec{\xi }_{0}^{\top }\varvec{\xi }_{1} + \mu _1 \varvec{\xi }_{0}^{\top }\varvec{\xi }_{0}, \end{aligned}$$
(79)

Since \(\varvec{\xi }_{0}\) is normalized, \(\varvec{\xi }_{0}^{\top }\varvec{\xi }_{0}=1\). Also, since \({\mathbf {S}}\) is symmetric,

$$\begin{aligned} \begin{aligned} \varvec{\xi }_{0}^{\top }{\mathbf {S}} \varvec{\xi }_{1}&= (\varvec{\xi }_{1}^{\top }{\mathbf {S}} \varvec{\xi }_{0})^{\top } , \\&= (\varvec{\xi }_{1}^{\top }\mu _0 \varvec{\xi }_{0})^{\top } , \\&= \mu _0 \varvec{\xi }_{0}^{\top }\varvec{\xi }_{1}, \end{aligned} \end{aligned}$$
(80)

where (76) was used. Now (79) is,

$$\begin{aligned} \mu _0 \varvec{\xi }_{0}^{\top }\varvec{\xi }_{1} + \varvec{\xi }_{0}^{\top }{\mathbf {B}} \varvec{\xi }_{0} = \mu _{0} \varvec{\xi }_{0}^{\top }\varvec{\xi }_{1} + \mu _1. \end{aligned}$$
(81)

Thus,

$$\begin{aligned} \mu _1 = \varvec{\xi }_{0}^{\top }{\mathbf {B}} \varvec{\xi }_{0}, \end{aligned}$$
(82)

which, since \({\mathbf {B}}\) is symmetric, represents a quadratic form.

A bound can be put on the term \(\varvec{\xi }_{0}^{\top }{\mathbf {B}} \varvec{\xi }_{0}\), noting that \(\varvec{\xi }_{0}\) is a unit vector. If \(b_n\) is the maximum eigenvalue of \({\mathbf {B}}\), then,

$$\begin{aligned} \max _{\varvec{\xi }_{0}} \varvec{\xi }_{0}^{\top }{\mathbf {B}} \varvec{\xi }_{0} = b_n. \end{aligned}$$
(83)

Similarly, if \(b_1\) is the minimum eigenvalue of \({\mathbf {B}}\), then,

$$\begin{aligned} \min _{\varvec{\xi }_{0}} \varvec{\xi }_{0}^{\top }{\mathbf {B}} \varvec{\xi }_{0} = b_1. \end{aligned}$$
(84)

So,

$$\begin{aligned} \mu _1 = \varvec{\xi }_{0}^{\top }{\mathbf {B}} \varvec{\xi }_{0} \in [b_1,b_n]. \end{aligned}$$
(85)

So (74) becomes,

$$\begin{aligned} \mu _{\varepsilon } = \mu _0 + \varepsilon \mu _1 + {\mathcal {O}}(\varepsilon ^2), \end{aligned}$$
(86)

where \(\mu _1\) is from (82).

With \(\mu _1\) in hand, \(\varvec{\xi }_{1}\) can also be determined as the solution of the following re-arranged version of (77),

$$\begin{aligned} ({\mathbf {S}} - \mu _0 \mathbb {1})\varvec{\xi }_{1} = - ({\mathbf {B}} - \mu _1 \mathbb {1}) \varvec{\xi }_{0}. \end{aligned}$$
(87)

Note that \(({\mathbf {S}} - \mu _0 \mathbb {1})\) is not invertible as it has zero determinant, since \(\mu _0\) is an eigenvalue of \({\mathbf {S}}\). The null space of \(({\mathbf {S}} - \mu _0 \mathbb {1})\) is span\(\{\varvec{\xi }_{0}\}\). Note that (87) is of the form \({\mathbf {A}}{\mathbf {x}}={\mathbf {b}}\) with a square matrix \({\mathbf {A}}\) of nullity 1 and a vector \({\mathbf {b}}\) which is in the image of \({\mathbf {A}}\), as shown below.

Note that, as a consequence of (82), the vector \({\mathbf {B}}\varvec{\xi }_{0}\) can be written as,

$$\begin{aligned} {\mathbf {B}}\varvec{\xi }_{0}= \mu _1 \varvec{\xi }_{0} + d \varvec{\xi }_{0}^{\prime \perp }, \end{aligned}$$
(88)

where \(d \in \mathbb {R}\) and \(\varvec{\xi }_{0}^{\prime \perp }\) is, in general, a vector in \(\mathrm{im}({\mathbf {S}} - \mu _0 \mathbb {1})\). This equation can be re-arranged to yield,

$$\begin{aligned} - {\mathbf {B}}\varvec{\xi }_{0} + \mu _1 \varvec{\xi }_{0} =-d \varvec{\xi }_{0}^{\prime \perp }. \end{aligned}$$
(89)

The left-hand side of (89) is the same as the right-hand side of (87), which means the right-hand side of (87) is a vector \({\mathbf {b}}\) which is in \(\mathrm{im}({\mathbf {S}} - \mu _0 \mathbb {1})\), which we will use below.

One can determine \(\mu _2\) by multiplying (78) by \(\varvec{\xi }_{0}^{\top }\) to get, by a similar procedure as before,

$$\begin{aligned} \mu _0 \varvec{\xi }_{0}^{\top } \varvec{\xi }_{2}+\varvec{\xi }_{0}^{\top }{\mathbf {B}} \varvec{\xi }_{1}+\varvec{\xi }_{0}^{\top }{\mathbf {Q}}\varvec{\xi }_{0} = \mu _{0} \varvec{\xi }_{0}^{\top }\varvec{\xi }_{2} + \mu _1 \varvec{\xi }_{0}^{\top }\varvec{\xi }_{1} + \mu _2. \end{aligned}$$
(90)

Canceling the identical terms on both sides, we get,

$$\begin{aligned} \mu _2= \varvec{\xi }_{0}^{\top }{\mathbf {Q}}\varvec{\xi }_{0} +\varvec{\xi }_{0}^{\top }{\mathbf {B}} \varvec{\xi }_{1} - \mu _1 \varvec{\xi }_{0}^{\top }\varvec{\xi }_{1}. \end{aligned}$$
(91)

But take the transpose and,

$$\begin{aligned} \mu _2 = \varvec{\xi }_{0}^{\top }{\mathbf {Q}}\varvec{\xi }_{0} +\varvec{\xi }_{1}^{\top }( {\mathbf {B}} - \mu _1 \mathbb {1}) \varvec{\xi }_{0}, \end{aligned}$$
(92)
$$\begin{aligned} = \varvec{\xi }_{0}^{\top }{\mathbf {Q}}\varvec{\xi }_{0} -\varvec{\xi }_{1}^{\top }({\mathbf {S}} - \mu _0 \mathbb {1})\varvec{\xi }_{1}, \end{aligned}$$
(93)

where (87) was used. One can write \(\varvec{\xi }_{1}\) as,

$$\begin{aligned} \varvec{\xi }_{1} = a\varvec{\xi }_{0} + b \varvec{\xi }_{0}^{\perp }, \end{aligned}$$
(94)

where \(a,b \in \mathbb {R}\) and \(\varvec{\xi }_{0}^{\perp } \in \mathrm{im}({\mathbf {S}} - \mu _0 \mathbb {1})\), which is, in general, not equal to \(\varvec{\xi }_{0}^{\prime \perp }\) from (89). Hence,

$$\begin{aligned} \mu _2 = \varvec{\xi }_{0}^{\top }{\mathbf {Q}}\varvec{\xi }_{0} -b^2 \varvec{\xi }_{0}^{\perp {\top }} ({\mathbf {S}} - \mu _0 \mathbb {1}) \varvec{\xi }_{0}^{\perp }. \end{aligned}$$
(95)

Therefore, the only part of \(\varvec{\xi }_{1}\) which contributes to \(\mu _2\) is the part which is in \(\mathrm{im}({\mathbf {S}} - \mu _0 \mathbb {1})\).

When dealing with a two-dimensional flow field, \(\mathrm{im}({\mathbf {S}} - \mu _0 \mathbb {1})\) is just a 1-dimensional subspace of \(\mathbb {R}^2\), and thus \(\varvec{\xi }_{0}^{\prime \perp }\) in (89) is parallel to \(\varvec{\xi }_{0}^{\perp }\) in (94). Without loss of generality, they can be taken to be equal unit vectors, \(\varvec{\xi }_{0}^{\perp }=\varvec{\xi }_{0}^{\prime \perp }\). Thus, (87) becomes,

$$\begin{aligned} b({\mathbf {S}} - \mu _0 \mathbb {1})\varvec{\xi }_{0}^{\perp } = - d \varvec{\xi }_{0}^{\perp }, \end{aligned}$$
(96)

or, assuming \(b\ne 0\),

$$\begin{aligned} ({\mathbf {S}} - \mu _0 \mathbb {1})\varvec{\xi }_{0}^{\perp } = - \tfrac{d}{b} \varvec{\xi }_{0}^{\perp }, \end{aligned}$$
(97)

which is an eigenvector equation for the matrix \(({\mathbf {S}} - \mu _0 \mathbb {1})\) with the eigenvector \(\varvec{\xi }_{0}^{\perp }\) and corresponding eigenvalue \({\bar{\mu }}=- \tfrac{d}{b}\). Note that if \(b=0\), then \(d=0\) also, from (96).

For two-dimensional flows, from \(\varvec{\xi }_{0}\), one can easily obtain \(\varvec{\xi }_{0}^{\perp }\) from a \(90^{\circ }\) counterclockwise rotation,

$$\begin{aligned} \varvec{\xi }_{0}^{\perp } = {\mathbf {R}}\varvec{\xi }_{0}, \end{aligned}$$
(98)

where,

$$\begin{aligned} {\mathbf {R}} = \begin{bmatrix} 0 &{} -1 \\ 1 &{} ~~ 0 \end{bmatrix}. \end{aligned}$$
(99)

Now, \(\varvec{\xi }_{0}^{\perp }\) can be used to obtain \({\bar{\mu }}\) from (97) for the case \(d\ne 0\). With (98) in (97), (97) becomes the following eigenvector equation for \({\mathbf {R}}^{\top }({\mathbf {S}} - \mu _0 \mathbb {1}){\mathbf {R}}\) with eigenvector \(\varvec{\xi }_{0}\),

$$\begin{aligned} {\mathbf {R}}^{\top }({\mathbf {S}} - \mu _0 \mathbb {1}){\mathbf {R}}\varvec{\xi }_{0}= {\bar{\mu }} \varvec{\xi }_{0}, \end{aligned}$$
(100)

Therefore \({\bar{\mu }}\) is obtained by taking the dot product with \(\varvec{\xi }_{0}\),

$$\begin{aligned} {\bar{\mu }} = \varvec{\xi }_{0}^{\top }{\mathbf {R}}^{\top }({\mathbf {S}} - \mu _0 \mathbb {1}){\mathbf {R}}\varvec{\xi }_{0}, \end{aligned}$$
(101)

and d is obtained from (89), noting that \(\varvec{\xi }_{0}^{\perp {\top }}\varvec{\xi }_{0}=0\),

$$\begin{aligned} \begin{aligned} d&= \varvec{\xi }_{0}^{\perp {\top }} {\mathbf {B}}\varvec{\xi }_{0},\\&= \varvec{\xi }_{0}^{\top }{\mathbf {R}}^{\top } {\mathbf {B}}\varvec{\xi }_{0}. \end{aligned} \end{aligned}$$
(102)

Thus, (95), for two-dimensional systems, simplifies to,

$$\begin{aligned} \mu _2 = \left\{ \begin{array}{ll} \varvec{\xi }_{0}^{\top }{\mathbf {Q}}\varvec{\xi }_{0}, &{\text {if}}\; d=0 \\ \varvec{\xi }_{0}^{\top }{\mathbf {Q}}\varvec{\xi }_{0} -\tfrac{d^2}{{\bar{\mu }}}, &{\text {if}}\; d\ne 0 \end{array}\right. \end{aligned}$$
(103)

where \({\bar{\mu }}\) and d are from (101) and (102), respectively.

Details for the examples

1.1 Details for nonlinear saddle example

Writing the \(\log \) term of (35) as follows, using Taylor series approximations for small |T|, we have,

$$\begin{aligned} \begin{aligned} \log&(e^{4T}) - \log [ (1 - y_0^2)e^{2T} + y_0^2)^3], \\&= 4T - 3 \log [ (1 - y_0^2)(1 + 2T + \tfrac{1}{2!}(2 T)^2 + \tfrac{1}{3!}(2 T)^3 + {\mathcal {O}}(T^4)) + y_0^2], \\&= 4T - 3 \log [ 1 + (1 - y_0^2)2T + (1 - y_0^2)2 T^2 + (1 - y_0^2)\tfrac{4}{3}T^3 + {\mathcal {O}}(T^4) ], \\&= 4T - 3 [ (1 - y_0^2)2T +y_0^2(1-y_0^2)2T^2 - y_0^2(1 - y_0^2)(1 - 2y_0^2)\tfrac{4}{3} T^3 + {\mathcal {O}}(T^4) ], \\&= 4T - (1 - y_0^2)6T - y_0^2(1 - y_0^2)6T^2 +4T^3y_0^2(1 - y_0^2)(1 - 2y_0^2) + {\mathcal {O}}(T^4), \\&= -2T [(1 - 3y_0^2) + 3 y_0^2 (1 - y_0^2)T - 2y_0^2(1 - y_0^2)(1 - 2y_0^2)T^2 + {\mathcal {O}}(T^3)]. \end{aligned} \end{aligned}$$
(104)

So the backward FTLE is expanded in T as follows, obtained by dividing by \(-2T\),

$$\begin{aligned} \sigma _0^T({\mathbf {x}}_0) = (1 - 3y_0^2) + 3 y_0^2 (1 - y_0^2) T - 2y_0^2(1 - y_0^2)(1 - 2y_0^2)T^2 + {\mathcal {O}}(T^3). \end{aligned}$$
(105)

which is the same as (36).

The FTLE can be approximated by the first, second, and third terms (the zeroth-order, first-order, and second-order in T, respectively) using the procedure outlined in Sect. 3. The gradient of the velocity is,

$$\begin{aligned} \nabla {\mathbf {v}}({\mathbf {x}}_0) = \begin{bmatrix} 1 &{} 0 \\ 0 &{} (- 1 + 3y_0^2) \end{bmatrix}, \end{aligned}$$
(106)

which is also \({\mathbf {S}}({\mathbf {x}}_0)\), since the gradient is diagonal. This has a minimum eigenvalue \(s_{-} = - 1 + 3y_0^2\), the negative of which matches the first term of (105), as prescribed by (14). To calculate the second term of (105), the term first-order in T, the acceleration field needs to be calculated and then \({\mathbf {B}}({\mathbf {x}}_0)\). The acceleration field is, following (12),

$$\begin{aligned} \begin{aligned} \ddot{x} = \tfrac{d}{dt}\dot{x}&= x, \\ \ddot{y} = \tfrac{d}{dt}\dot{y}&= y - 4 y^3 + 3y^5. \end{aligned} \end{aligned}$$
(107)

Therefore (11) gives,

$$\begin{aligned} \begin{aligned} {\mathbf {B}}({\mathbf {x}}_0)&= \begin{bmatrix} 1 &{} 0 \\ 0 &{} (1 - 12y_0^2 + 15y_0^4) \end{bmatrix} + \begin{bmatrix} 1 &{} 0 \\ 0 &{} (1 - 6y_0^2 + 9y_0^4) \end{bmatrix}, \\&= \begin{bmatrix} 2 &{} 0 \\ 0 &{} (2 - 18y_0^2 + 24y_0^4) \end{bmatrix}. \end{aligned} \end{aligned}$$
(108)

The normalized eigenvector of \({\mathbf {S}}({\mathbf {x}}_0)\) corresponding to \(s_-\) is simply \(\varvec{e}_{-}=[ 0, 1]^{\top }\), which, via (16), yields,

$$\begin{aligned} \begin{aligned} \mu _{1-} = \varvec{e}_{-}^{\top } {\mathbf {B}}(\mathbf{x} _0) \varvec{e}_{-}&= \begin{bmatrix} 0 &{} 1 \\ \end{bmatrix} \begin{bmatrix} 2 &{} 0 \\ 0 &{} (2 - 18y_0^2 + 24y_0^4) \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \\&= 2 - 18y_0^2 + 24y_0^4, \end{aligned} \end{aligned}$$
(109)

hence,

$$\begin{aligned} \begin{aligned} -s_-^2 + \tfrac{1}{2} \mu _{1-}&=-(1 - 6y_0^2 + 9y_0^4) + 1 -9y_0^2 + 12y_0^4, \\&=-3y_0^2 (1 - y_0^2), \end{aligned} \end{aligned}$$
(110)

the negative of which matches the T coefficient of the second term of (105), as prescribed by (14).

For the term second-order in T, note that, as prescribed by (13),

$$\begin{aligned} \begin{aligned} {\mathbf {Q}}({\mathbf {x}}_0)&= \frac{2}{3} \begin{bmatrix} 1 &{} 0 \\ 0 &{} (-1 + 39y_0^2 - 135 y_0^4 + 105y_0^6) \end{bmatrix}\\&\quad + 2\begin{bmatrix} 1 &{} 0 \\ 0 &{} (1 - 12y_0^2 + 15y_0^4) \end{bmatrix} \begin{bmatrix} 1 &{} 0 \\ 0 &{} (- 1 + 3y_0^2) \end{bmatrix}, \\&= \begin{bmatrix} \tfrac{2}{3} &{} 0 \\ 0 &{} (-\tfrac{2}{3} + 26y_0^2 - 90 y_0^4 + 70y_0^6) \end{bmatrix}\\&\quad + \begin{bmatrix} 2 &{} 0 \\ 0 &{} (-2 + 30y_0^2 - 102y_0^4 + 90y_0^6) \end{bmatrix}, \\&= \begin{bmatrix} \tfrac{8}{3} &{} 0 \\ 0 &{} (-\tfrac{8}{3} + 56y_0^2 - 192 y_0^4 + 160 y_0^6) \end{bmatrix}, \end{aligned} \end{aligned}$$
(111)

and since (87) implies that \(\varvec{\xi }_{1-}\) is parallel to \(\varvec{e}_-\), (16) yields,

$$\begin{aligned} \mu _{2-} = \varvec{e}_{-}^{\top } {\mathbf {Q}}(\mathbf{x} _0) \varvec{e}_{-} = -\tfrac{8}{3} + 56y_0^2 - 192 y_0^4 + 160 y_0^6. \end{aligned}$$
(112)

According to (14), the second-order term is,

$$\begin{aligned} \begin{aligned} -T^2&[ -\tfrac{4}{3}(1-3y_0^2)(1 - 6y_0^2 + 9y_0^4) \\&~~~+ (1-3y_0^2) (2 - 18y_0^2 + 24y_0^4) \\&~~~+ \tfrac{1}{4} (-\tfrac{8}{3} + 56y_0^2 - 192 y_0^4 + 160 y_0^6)] \\ = -T^2&[ ( -\tfrac{4}{3} + 8y_0^2 - 12y_0^4 + 4y_0^2 - 24y_0^4 +36 y_0^6)\\&~~~+ (2 - 18y_0^2 + 24y_0^4 -6y_0^2 + 54y_0^4 -72y_0^6) \\&~~~+ (-\tfrac{2}{3} + 14y_0^2 - 48y_0^4 + 40 y_0^6) ], \\ = -T^2&[ ( -\tfrac{4}{3} + 12y_0^2 - 36y_0^4 + 36 y_0^6)\\&~~~+ (2 - 24y_0^2 + 78y_0^4 - 72 y_0^6) \\&~~~+ (-\tfrac{2}{3} + 14y_0^2 - 48y_0^4 + 40 y_0^6) ], \\ = -T^2&[ 2 y_0^2 - 6 y_0^4 + 4 y_0^6],\\ = -T^2&y_0^2(1-y_0^2)(1-2y_0^2) \end{aligned} \end{aligned}$$
(113)

which matches the \(T^2\) term of the true FTLE field (105).

1.2 Details for the time-varying double-gyre example

The gradient tensor for the double-gyre velocity field (40) is,

$$\begin{aligned} \begin{aligned}&\nabla {\mathbf {v}} = \begin{bmatrix} \frac{\partial u}{\partial x} &{} \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} &{} \frac{\partial v}{\partial y} \end{bmatrix}, \\&= \begin{bmatrix} -\pi ^2 A\cos (\pi f)\cos (\pi y)\frac{\partial f}{\partial x} &{} \pi ^2 A\sin (\pi f)\sin (\pi y) \\ -\pi ^2 A\sin (\pi f)\sin (\pi y) \frac{\partial f}{\partial x} + \pi A \cos (\pi f)\sin (\pi y)\frac{\partial ^2 f}{\partial x^2} &{} \pi ^2 A\cos (\pi f)\cos (\pi y)\frac{\partial f}{\partial x} \end{bmatrix}. \end{aligned} \end{aligned}$$
(114)

The acceleration field, \({\mathbf {a}}=\tfrac{d}{dt}{\mathbf {v}}=(a_x,a_y)\), for the double-gyre, (40), is given by,

$$\begin{aligned} \begin{aligned} a_x =&-\pi ^2 A \cos (\pi f)\cos (\pi y)\tfrac{\partial f}{\partial t} +\tfrac{1}{2}\pi ^3 A^2 \sin (2\pi f)\tfrac{\partial f}{\partial x}, \\ a_y =&~~\pi ^2 A \Big [ - \sin (\pi f)\sin (\pi y)\tfrac{\partial f}{\partial x}\tfrac{\partial f}{\partial t} +\tfrac{1}{\pi } \cos (\pi f)\sin (\pi y)\tfrac{\partial ^2 f}{\partial x \partial t} \Big ] \\ +\tfrac{1}{2}&\pi ^3 A^2 \sin (2\pi y) \Big [ \sin ^2(\pi f) \tfrac{\partial f}{\partial x} + \cos ^2(\pi f)( \tfrac{\partial f}{\partial x} )^2 -\tfrac{1}{2\pi }\sin (2\pi f)\tfrac{\partial ^2 f}{\partial x^2} \Big ], \end{aligned} \end{aligned}$$
(115)

where the dependence of the function f, from (39), is understood.

The components of the symmetric \({\mathbf {B}}\) matrix are,

$$\begin{aligned} B_{xx}& = -A\pi ^2 \cos (\pi f) \cos (\pi y) \tfrac{\partial ^2 f}{\partial x \partial t} +\tfrac{1}{2}A\pi ^3 \sin (2 \pi f) \tfrac{\partial f}{\partial x}\tfrac{\partial f}{\partial t}\nonumber \\&\quad +A^2\pi ^3 \sin (2 \pi f)\tfrac{\partial ^2 f}{\partial x^2} \Big ( \tfrac{1}{2} - \sin ^2(\pi y)(\tfrac{\partial f}{\partial x} )^2 \Big ) + A^2\pi ^4 \cos (2 \pi f)(\tfrac{\partial f}{\partial x} )^2 \nonumber \\&\quad + A^2\pi ^4 \sin ^2(\pi f)\sin ^2(\pi y)(\tfrac{\partial f}{\partial x} )^4 + A^2\pi ^2\cos ^2(\pi f)\sin ^2(\pi y) \tfrac{\partial ^2 f}{\partial x^2} \nonumber \\&\quad + A^2\pi ^4\cos ^2(\pi f) \cos ^2(\pi y) (\tfrac{\partial f}{\partial x} )^2, \end{aligned}$$
(116)
$$\begin{aligned} B_{xy} &= \tfrac{1}{2} A \pi \cos (\pi f) \sin (\pi y) \Big [ \tfrac{\partial ^3 f}{\partial x^2 \partial t} + \pi ^2 \Big ( 1 - (\tfrac{\partial f}{\partial x} )^2 \Big ) \Big ] \nonumber \\&\quad -A \pi ^2 \sin (\pi f) \sin (\pi y) \Big ( \tfrac{\partial f}{\partial x}\tfrac{\partial ^2 f}{\partial x \partial t} - \tfrac{1}{2} \tfrac{\partial ^2 f}{\partial x ^2} \tfrac{\partial f}{\partial t} \Big ) \nonumber \\&\quad - \tfrac{1}{4}A^2 \pi ^4 \sin (2 \pi f) \sin (2 \pi y) \Big [ \tfrac{\partial f}{\partial x} \Big ( 1 + (\tfrac{\partial f}{\partial x} )^2 \Big ) +\tfrac{1}{2}\tfrac{\partial ^3 f}{\partial x ^3} \Big ], \end{aligned}$$
(117)
$$\begin{aligned} B_{yy}& = A \pi ^2 \cos (\pi f) \cos (\pi y) \tfrac{\partial ^2 f}{\partial x \partial t} - A \pi ^3 \sin (\pi f) \cos (\pi y) \tfrac{\partial f}{\partial x} \tfrac{\partial f}{\partial t} \nonumber \\&\quad - \tfrac{1}{2}A^2 \pi ^3 \sin (2 \pi f)\sin (2 \pi y) \tfrac{\partial ^2 f}{\partial x^2} \nonumber \\&\quad + A^2 \pi ^4 \cos ^2(\pi f) \cos ^2(\pi y) (\tfrac{\partial f}{\partial x} )^2 + A^2 \pi ^4 \sin ^2(\pi f) \sin ^2(\pi y) \nonumber \\&\quad + A^2 \pi ^4 (\tfrac{\partial f}{\partial x} )^2 \Big ( \cos ^2(\pi f) - \sin ^2(\pi y) \Big ). \end{aligned}$$
(118)

The eigenvalue \(s_-({\mathbf {x}}_0,t_0)\) of \({\mathbf {S}}({\mathbf {x}}_0,t_0)\) is,

$$\begin{aligned} \begin{aligned} s_- = -\frac{1}{2}\pi ^2 A \Bigg [&\Big ( \sin (\pi f) \sin (\pi y_0)\Big (1-\frac{\partial f}{\partial x} \Big ) + \tfrac{1}{\pi }\cos (\pi f) \sin (\pi y_0)\frac{\partial ^2 f}{\partial x^2} \Big )^2 \\&~~ + 4 \Big ( \cos (\pi f) \cos (\pi y_0) \frac{\partial f}{\partial x} \Big )^2 ~\Bigg ]^{1/2}. \end{aligned} \end{aligned}$$
(119)

The normalized eigenvector of \({\mathbf {S}}({\mathbf {x}}_0,t_0)\) corresponding to the eigenvalue \(s_-({\mathbf {x}}_0,t_0)\) is given by,

$$\begin{aligned} \varvec{e}_{-}= \begin{bmatrix} e_x \\ e_y \end{bmatrix} = \frac{1}{N} \begin{bmatrix} {\bar{s}}_- - \beta \\ \tfrac{1}{2}\alpha \end{bmatrix}, \end{aligned}$$
(120)

where,

$$\begin{aligned}&{\bar{s}}_- = \frac{s_-}{\pi ^2 A} = -\tfrac{1}{2} \sqrt{\alpha ^2 + 4\beta ^2},\nonumber \\&N = \sqrt{ \tfrac{1}{4}\alpha ^2 + ( {\bar{s}}_1 - \beta )^2 }, \nonumber \\&\alpha = \sin (\pi f) \sin (\pi y) \Big ( 1 - \tfrac{\partial f}{\partial x} \Big ) + \tfrac{1}{\pi } \cos (\pi f)\sin (\pi y)\tfrac{\partial ^2 f}{\partial x^2}, \nonumber \\&\beta = \cos (\pi f) \cos (\pi y) \tfrac{\partial f}{\partial x}. \end{aligned}$$
(121)

The coefficient of T in the approximation of the backward-time FTLE for the double-gyre is thus given by \(s_-^2 - \tfrac{1}{2} \varvec{e}_{-}^{\top } {\mathbf {B}} \varvec{e}_{-}\) which can be expressed in terms of,

$$\begin{aligned} a_-(\mathbf {x_0},t_0) = -s_-^2 +\tfrac{1}{2} (B_{xx} e_x^2 + 2 B_{xy} e_x e_y + B_{yy} e_y^2), \end{aligned}$$
(122)

using the above formulas. This yields a backward-time FTLE approximation for small backward times \(T<0\) of,

$$\begin{aligned} \sigma _{t_{0}}^{t_0 + T}(\mathbf {x_0}) = s_-(\mathbf {x_0},t_0) - a_-(\mathbf {x_0},t_0)T + {\mathcal {O}}(T^2). \end{aligned}$$
(123)

Note that the first and second terms have explicit dependence on both initial position and initial time.

1.3 Details for the ABC flow example

For the ABC velocity field (45), the characteristic polynomial for the rate-of-strain tensor \({\mathbf {S}}\) for this system is,

$$\begin{aligned} s^3 +a_1 s + a_0 = 0, \end{aligned}$$
(124)

where

$$\begin{aligned} \begin{aligned} a_0&= -\tfrac{1}{4}( B\cos (x)-C\sin (y))( C\cos (y)-A\sin (z))(-B\sin (x)+A\cos (z)),\\ a_1&=-\tfrac{1}{4}\Big [ ( B\cos (x)-C\sin (y))^2 \\&\quad +( C\cos (y)-A\sin (z))^2 +(-B\sin (x)+A\cos (z))^2 \Big ] \end{aligned} \end{aligned}$$
(125)

The repulsion and attraction rate fields, \(s_+\) and \(s_-\), are given by

$$\begin{aligned} \begin{aligned} s_+&= 2 \rho ^{1/3} \cos \big (\tfrac{\theta }{3}\big ) >0, \\ s_-&= -\tfrac{1}{2}s_+ - \sqrt{3}\rho ^{1/3} \sin \big (\tfrac{\theta }{3}\big ) <0, \end{aligned} \end{aligned}$$
(126)

where the dependence on initial position \({\mathbf {x}}\) is understood and \(\rho \) and \(\theta \) are given by,

$$\begin{aligned} \begin{aligned} \rho&= \sqrt{ q^2 + |p|}, \\ \theta&= \tan ^{-1}\Big (\tfrac{\mathrm{Im}(\sqrt{p})}{q}\Big ), \end{aligned} \end{aligned}$$
(127)

where,

$$\begin{aligned} \begin{aligned} q&=-\tfrac{1}{2}a_0,\\ p&=\frac{1}{27}a_1^3 + \frac{1}{4}a_0^2. \end{aligned} \end{aligned}$$
(128)

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Nolan, P.J., Serra, M. & Ross, S.D. Finite-time Lyapunov exponents in the instantaneous limit and material transport. Nonlinear Dyn 100, 3825–3852 (2020). https://doi.org/10.1007/s11071-020-05713-4

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