On Conditions for Rate-induced Tipping in Multi-dimensional Dynamical Systems

Abstract

The possibility of rate-induced tipping (R-tipping) away from an attracting fixed point has been thoroughly explored in 1-dimensional systems. In these systems, it is impossible to have R-tipping away from a path of quasi-stable equilibria that is forward basin stable (FBS), but R-tipping is guaranteed for paths that are non-FBS of a certain type. We will investigate whether these results carry over to multi-dimensional systems. In particular, we will show that the same conditions guaranteeing R-tipping in 1-dimension also guarantee R-tipping in higher dimensions; however, it is possible to have R-tipping away from a path that is FBS even in 2-dimensional systems. We will propose a different condition, forward inflowing stability (FIS), which we show is sufficient to prevent R-tipping in all dimensions. The condition, while natural, is difficult to verify in concrete examples. Monotone systems are a class for which FIS is implied by an easily verifiable condition. As a result, we see how the additional structure of these systems makes predicting the possibility of R-tipping straightforward in a fashion similar to 1-dimension. In particular, we will prove that the FBS and FIS conditions in monotone systems reduce to comparing the relative positions of equilibria over time. An example of a monotone system is given that demonstrates how these ideas are applied to determine exactly when R-tipping is possible.

Appendix: Proofs of Lemmas from Sect. 3

The proof of Lemma 1 closely follows the proof of Theorem 2.2 in Ashwin et al. [3]:

Proof

Let us assume for the sake of simplicity that $X_- = 0$. Let

$$\begin{aligned} \omega (\epsilon )&:= \sup \{|d_x f(x,\varLambda (s)) - d_x f(0,\varLambda (s))| : s \in \mathbb {R}, |x|< \epsilon \} \\ \delta (S)&:= \max \left\{ \sup _{s< -S} |f(0,\varLambda (s))|, \sup _{s < -S} |d_x f(0,\varLambda (s)) - d_x f(0,\lambda _-)| \right\} . \end{aligned}$$

Note that $\omega (\epsilon ) \rightarrow 0$ as $\epsilon \rightarrow 0$ and $\delta (S) \rightarrow 0$ as $S \rightarrow \infty$.

By the linear stability of $X_-$, the eigenvalues of

$$\begin{aligned} A := d_x f(0,\lambda _-) \end{aligned}$$

have negative real parts, so there are $K > 0$ and $\alpha > 0$ such that $\left| e^{tA}\right| \le K e^{-\alpha t}$ for $t \ge 0$ (see Lemma 3.3.19 of [5]).

Now set $h(x,s) = f(x,\varLambda (s)) - Ax$ so that

$$\begin{aligned} \dot{x} = Ax + h(x,s) \end{aligned}$$

(7.1)

Then

$$\begin{aligned} d_x h(x,s)&= d_x f(x,\varLambda (s)) - A \\&= \left[ d_x f(x,\varLambda (s)) - d_x f(0,\varLambda (s))\right] + \left[ d_x f(0,\varLambda (s)) - d_x f(0,\lambda _-)\right] \end{aligned}$$

Therefore, if $s < -S$, we have

$$\begin{aligned} |h(0,s)| \le \delta (S), \ \ |d_x h(x,t)| \le \omega (|x|) + \delta (S). \end{aligned}$$

Consider the inequalities

$$\begin{aligned} \begin{aligned} 4K\alpha ^{-1} \omega (\epsilon )&\le 1 \\ 2K\alpha ^{-1} \delta (S)&\le \epsilon \\ 4K\alpha ^{-1} \delta (S)&\le 1, \end{aligned} \end{aligned}$$

(7.2)

for $\epsilon , S > 0$. If we choose $\epsilon$ sufficiently small, we can find some $S_0 > 0$ to satisfy (7.2). Now, we know that $\delta (S) \rightarrow 0$ as $S \rightarrow \infty$, so there is an $S_1 > 0$ such that $S \ge S_1$ implies that $\delta (S) \le \delta (S_0)$. Then if $S \ge S_1$, (7.2) is satisfied.

Now, fix any $r > 0$. We will show that the pullback attractor $x^r(t)$ to $X_- = 0$ satisfies $|x^r(t)| < \epsilon$ as long as $rt < -S_1$.

Let $\mathcal {P}$ be space of continuous functions x(t) defined for $t < -\frac{S_1}{r}$ such that $|x(t)| \le \epsilon$. We define ${{\hat{x}}}$ for $x \in \mathcal {P}$ by

$$\begin{aligned} {{\hat{x}}}(t) = \int _{-\infty }^t e^{(t-u)A} h(x(u),ru) \ du. \end{aligned}$$

Then if $x \in \mathcal {P}$, ${{\hat{x}}} = x$ if and only if x(t) is a solution of (7.1). Also,

$$\begin{aligned} |{{\hat{x}}}(t)| \le \epsilon , \end{aligned}$$

so $x \mapsto {{\hat{x}}}$ is a map from $\mathcal {P}$ to itself. Furthermore, if $x_1,x_2 \in \mathcal {P}$, then

$$\begin{aligned} ||{{\hat{x}}}_1 - {{\hat{x}}}_2|| \le \frac{1}{2} ||x_1 - x_2||, \end{aligned}$$

where $||x|| := \sup _{t < -S_1/r} |x(t)|$. Thus, $x \mapsto {{\hat{x}}}$ is a contraction mapping on $\mathcal {P}$, so there is a unique $x(t) \in \mathcal {P}$ such that $x(t) = {{\hat{x}}}(t)$. This x(t) is a solution to (7.1) and satisfies $|x(t)| \le \epsilon$ for all $t < -S_1/r$. This x(t) must be the pullback attractor to $X_- = 0$, since the pullback attractor is the only trajectory of (1.3) that stays within a small neighborhood of $X_-$ for all backward time. (See Theorem 2.2 of [3].) $\square$

Proof of Lemma 2

Fix $r > 0$. For each $n \in {\mathbb {N}}$, there exists some $N_n \in {\mathbb {N}}$ such that $m \ge N_n$ implies $|y \cdot _{\lambda _+} s_m - z| < \frac{1}{n}$. (If $n \ge 2$, choose $N_n > N_{n - 1}$.) Since $\varLambda (rt) \rightarrow \lambda _+$ as $t \rightarrow \infty$, there exists some $T_n > 0$ and $\delta _n > 0$ such that if $|x(t) - y| < \delta _n$ and $t > T_n$, then $|x(t + s_{N_n}) - y \cdot _{\lambda _+} s_{N_n}| < \frac{1}{n}$. There exists some $M_n \in {\mathbb {N}}$ such that $t_{M_n} > T_n$ and if $m \ge M_n$, then $|x(t_m) - y| < \delta _n$. (Again, if $n \ge 2$, choose $M_n > M_{n - 1}$.)

Now, set $u_n = s_{N_n} + t_{M_n}$. Then,

$$\begin{aligned} |x(u_n) - z|&= |x(t_{M_n}+ s_{N_n}) - z| \\&\le |x(t_{M_n}+s_{N_n}) - y \cdot _{\lambda _+} s_{N_n}| + |y \cdot _{\lambda _+} s_{N_n} - z| \\&< \frac{1}{n} + \frac{1}{n} \\&= \frac{2}{n} \end{aligned}$$

Therefore, $x(u_n) \rightarrow z$ as $n \rightarrow \infty$. $\square$

Proof of Lemma 3

Let us assume for the sake of simplicity that $X_+ = 0$ and $\lambda _+ = 0$. Since $(X_+,\lambda _+) = (0,0)$ is attracting, all eigenvalues of $A = d_x f(0,0)$ have negative real part, so there is some $k > 0$ such that $\text {Re}(\mu ) < -k$ for every eigenvalue $\mu$ of A. We can choose an inner product $\langle , \rangle$ on U such that $\langle Ax, x \rangle \le -k \langle x,x \rangle$ for all $x \in U$ (see the lemma in Chapter 7, Section 1 of [6]). This defines a norm $||x|| = \langle x,x \rangle ^{1/2}$. By Taylor’s formula in several variables we can write

$$\begin{aligned} f(x,\lambda )&= Ax + \alpha (x,\lambda ) + \beta (x), \end{aligned}$$

where $||\alpha (x,\lambda )|| \le \gamma (x,\lambda )|\lambda |$ for a positive continuous $\gamma$, and $||\beta (x)|| \le \delta (x)||x||$, where $\delta$ is positive, continuous and $\delta (x) \rightarrow 0$ as $x \rightarrow 0$. Then we can write (1.3) as

$$\begin{aligned} \frac{dx}{dt}&= Ax + \alpha (x,\varLambda (s)) + \beta (x) \\ \frac{ds}{dt}&= r \end{aligned}$$

For a given $\epsilon > 0$ and $S > 0$, define $N_{\epsilon , S} = B_\epsilon (0) \times [S,\infty )$, where $B_\epsilon (0) = \{x \in U: ||x || < \epsilon \}$. Note that

$$\begin{aligned} \frac{d}{dt} \left( ||x||^2\right)&= 2 \langle \dot{x},x \rangle \\&= 2\langle Ax,x \rangle + 2\langle \alpha (x,\varLambda (s)),x\rangle + 2 \langle \beta (x), x \rangle \\&\le -2 k \langle x,x \rangle + 2 ||\alpha (x,\varLambda (s))|| \cdot ||x|| + 2 ||\beta (x)|| \cdot ||x|| \\&\le -2 k ||x||^2 + 2\gamma (x,\varLambda (s))|\varLambda (s)| \cdot ||x|| + 2 \delta (x) \cdot ||x||^2 \\&= 2 ||x||^2 \left( -k + 2\gamma (x,\varLambda (s)) \frac{|\varLambda (s)|}{||x||} + \delta (x) \right) \end{aligned}$$

Choose $\epsilon > 0$ such that if $||x|| \le \epsilon$, $\delta (x) < \frac{k}{2}$. Then choose $S > 0$ such that if $s \ge S$, $|\varLambda (s)| < \frac{k \epsilon }{4M}$ where $M > \sup _{||x|| \le \epsilon , \lambda \in [\lambda _-,\lambda _+]} \gamma (x,\lambda )$. Thus, if $||x|| = \epsilon$ and $s \ge S$,

$$\begin{aligned} \frac{d}{dt} \left( ||x||\right) ^2&< 2 \epsilon ^2 \left( -k + 2\gamma (x,\varLambda (s)) \left( \frac{k \epsilon }{4M\epsilon }\right) + \frac{k}{2} \right) \\&< 2 \epsilon ^2 \left( -k + \frac{k}{2} + \frac{k}{2} \right) \\&= 0 \end{aligned}$$

Therefore, for sufficiently small $\epsilon > 0$ there exists an $S>0$ such that the vector field of (1.3) points into $N_{\epsilon , S}$ on its boundary, so $N_{\epsilon , S}$ is forward invariant. $\square$

Proof of Lemma 4

Pick an $\epsilon > 0$ sufficiently small for Lemma 3. Make $\epsilon$ smaller if necessary so that $\overline{B_\epsilon (X_+)} \subset \mathbb {B}(X_+,\lambda _+)$. Then by Lemma 3, there exists an $S > 0$ such that if $x(t) \in B_\epsilon (X_+)$ for $rT > S$, then $x(t) \in B_\epsilon (X_+)$ for all $t \ge T$. Now fix $r > 0$. Since $\overline{B_\epsilon (X_+)}$ is compact, there is some $y \in \overline{B_\epsilon (X_+)}$ such that $x(t_n) \rightarrow y$ as $t_n \rightarrow \infty$. But $y \in \mathbb {B}(X_+,\lambda _+)$ by assumption, so $y \cdot t \rightarrow X_+$ in the autonomous system (1.1). Therefore, by Lemma 2, there exists a $\{u_n\} \rightarrow \infty$ such that $x(u_n) \rightarrow X_+$.

Now pick any $\delta \in (0,\epsilon )$. Then by Lemma 3, there exists some $S_\delta > 0$ such that if $x(T) \in B_\delta (X_+)$ for $rT > S_\delta$, then $x(T) \in B_\delta (X_+)$ for all $t \ge T$. By the previous paragraph, there is a $u_{n_\delta } > S_\delta /r$ such that $|x(u_{n_\delta }) - X_+| < \delta$. Therefore, $|x(t) - X_+| < \delta$ for all $t \ge u_{n_\delta }$. Hence $x(t) \rightarrow X_+$ as $t \rightarrow \infty$. $\square$

Proof of Lemma 5

By Lemma 4, there is an $\epsilon > 0$ and an $S_1 > 0$ such that if $|x(t) - X_+| < \epsilon$ for $rt > S_1$, then $x(t) \rightarrow X_+$ as $t \rightarrow \infty$. Since $K \subset \mathbb {B}(X_+,\lambda _+)$ is compact, there is some $T_0 > 0$ such that $y \cdot _{\lambda _+} t \in B_{\epsilon /2}(X_+)$ for any $y \in K$ and $t \ge T_0$. Also, there is some $S_2 > 0$ such that if $x(T) = y_0 \in K$ for $rT > S_2$, then $|x(T + T_0) - y_0 \cdot _{\lambda _+} T_0| < \epsilon /2$ for any $y_0 \in K$.

Take $S = \max \{S_1,S_2\}$. Then, suppose $x(T) \in K$ for $rT > S$. If $x(T) = y_0$, then

$$\begin{aligned} |x(T + T_0) - X_+|&\le |x(T + T_0) - y_0 \cdot _{\lambda _+} T_0| + |y_0 \cdot _{\lambda _+} T_0 - X_+| \\&< \epsilon /2 + \epsilon /2 \\&= \epsilon \end{aligned}$$

Therefore, $x(t) \rightarrow X_+$ as $t \rightarrow \infty$. $\square$