A Mathematical Biologist’s Guide to Absolute and Convective Instability

Abstract

Mathematical models have been highly successful at reproducing the complex spatiotemporal phenomena seen in many biological systems. However, the ability to numerically simulate such phenomena currently far outstrips detailed mathematical understanding. This paper reviews the theory of absolute and convective instability, which has the potential to redress this inbalance in some cases. In spatiotemporal systems, unstable steady states subdivide into two categories. Those that are absolutely unstable are not relevant in applications except as generators of spatial or spatiotemporal patterns, but convectively unstable steady states can occur as persistent features of solutions. The authors explain the concepts of absolute and convective instability, and also the related concepts of remnant and transient instability. They give examples of their use in explaining qualitative transitions in solution behaviour. They then describe how to distinguish different types of instability, focussing on the relatively new approach of the absolute spectrum. They also discuss the use of the theory for making quantitative predictions on how spatiotemporal solutions change with model parameters. The discussion is illustrated throughout by numerical simulations of a model for river-based predator–prey systems.

Appendix: Examples of Calculating Branch Points

In this Appendix, we show how to calculate the branch points for the Rosenzweig–MacArthur model (1a), (1b) and the Klausmeier model (2a), (2b). We present these calculations in some detail, with the aim of providing templates that readers can follow when performing corresponding calculations for their own models.

1.1 A.1 Branch Points for the Rosenzweig–MacArthur Model

Recall that the Rosenzweig–MacArthur model (1a), (1b) has a unique homogeneous co-existence steady state (h _s ,p _s ) where h _s =1/(aμ−μ) and p _s =(1−h _s )(1+μh _s )/μ. We begin by linearising (1a), (1b) about (h _s ,p _s ) giving

$$\begin{aligned} \tilde{p}_t &=\alpha\tilde{p}+\beta\tilde{h}+c\tilde{p}_x+d \tilde {p}_{xx}, \\ \tilde{h}_t &=\gamma\tilde{p}+\delta\tilde{h}+c \tilde{h}_x+\tilde{h}_{xx}, \end{aligned}$$

where $\tilde{p}=p-p_{s}$, $\tilde{h}=h-h_{s}$, and α, β, γ, δ are coefficients from linearisation and are given by

$$\begin{aligned} \alpha&=\frac{\mu h_s}{b(1+\mu h_s)}-\frac{1}{ab}, \\ \beta&=\frac{\mu p_s}{b(1+\mu h_s)^2}, \\ \gamma&=1-2h_s-\frac{\mu p_s}{(1+\mu h_s)^2}, \\ \delta&=-\frac{\mu h_s}{1+\mu h_s} . \end{aligned}$$

Substituting $(\tilde{p},\tilde{h})=(\bar{p},\bar{h})\exp (ikx+\lambda t)$ into (1a), (1b) and requiring $\bar{p}$ and $\bar{h}$ to be non-zero gives the dispersion relation

$$\begin{aligned} \mathcal{D}(\lambda,k) =& dk^4-cik^3(d+1) -k^2\bigl(\alpha-\lambda+d\gamma-d\lambda+c^2\bigr) \\ &{}+cik(\gamma+\alpha-2\lambda)+ (\alpha-\lambda) (\gamma-\lambda)-\delta\beta=0 . \end{aligned}$$

(9)

Branch points are double roots of the dispersion relation for k, and satisfy (9) and also

$$ \begin{aligned} 0&=\partial\mathcal{D}/\partial k \\ &= 4dk^3-3k^2(1+d)ci-2k \bigl(\gamma+d\alpha-(1+d)\lambda\bigr)+c^2+ ci(\alpha+\gamma-2 \lambda)\quad \Rightarrow \\ \lambda&= \frac{4dk^3-3cidk^2-2dk\gamma+\alpha ci-2c^2k-3cik^2-2\alpha k+ci\gamma}{-2(k+dk-ci)} . \end{aligned} $$

(10)

Substituting (10) into (9) gives the following hexic polynomial in k:

$$\begin{aligned} & \bigl[-4d(d-1)^2 \bigr]k^6+ \bigl[2ci(d+1) (d-1)^2 \bigr]k^5 + \bigl[(d-1) \bigl(c^2d+8d\alpha-8d\gamma-c^2\bigr) \bigr]k^4 \\ &\quad{} + \bigl[4ci(-\alpha+\gamma) (d-1) (d+1) \bigr]k^3 \\ &\quad{} + \bigl[2c^2(d-1) (\gamma-\alpha)-4\beta\delta(d+1)^2 -4d( \alpha-\gamma)^2 \bigr]k^2 \\ &\quad{} + \bigl[2i\bigl(4\beta\delta+\alpha^2-2\alpha\gamma + \gamma^2\bigr)c(d+1) \bigr]k+c^2\bigl(4\beta\delta + \alpha^2-2\alpha\gamma+\gamma^2\bigr)=0 . \end{aligned}$$

(11)

We must now proceed numerically and we fix a=1.3, b=4.0, c=−1, d=2, and μ=9. These parameter values satisfy μ>μ _crit, so that the coexistence steady state is unstable. Substituting these values into (11), we obtain six roots for k, two real and two pairs of complex conjugates. We then substitute each into (10) to find the corresponding value of λ. To determine whether these branch points belong to the absolute spectrum, we substitute each λ value into (9) and solve for k, giving the repeated roots found from (11) and two others.

Branch point k=0.676i.:

Substituting this value of k into (10) gives λ=1.380. Substituting this value of λ back into (9) gives a quartic polynomial for k whose roots are −1.727i, −1.125i, 0.676i, 0.676i. Recall that a branch point is in the absolute spectrum if the repeated roots are k ₂ and k ₃, when the roots k ₁, k ₂, k ₃ and k ₄ of (9) are labelled in increasing order of their imaginary parts. In this case, the repeated roots are k ₃ and k ₄ so that the branch point is not in the absolute spectrum.

Branch point k=−0.473−0.013i.:

Substituting this value of k into (10) gives λ=−0.255+0.483i. Substituting this value of λ back into (9) gives a quartic polynomial for k whose roots are 0.570−0.943i, 0.377−0.530i, −0.473−0.013i, −0.473−0.013i Therefore, the repeated roots are k ₃ and k ₄ so that the branch point is not in the absolute spectrum.

Branch point k=0.473−0.013i.:

Substituting this value of k into (10) gives λ=−0.255−0.483i. Substituting this value of λ back into (9) gives a quartic polynomial for k whose roots are −0.570−0.943i, −0.377−0.530i, 0.473−0.013i, 0.473−0.013i. Therefore, the repeated roots are k ₃ and k ₄ so that the branch point is not in the absolute spectrum.

Branch point k=0.001−0.334i.:

Substituting this value of k into (10) gives λ=−0.110−0.167i. Substituting this value of λ back into (9) gives a quartic polynomial for k whose roots are −0.314−0.816i, 0.001−0.334i, 0.001−0.334i, 0.312−0.0165i. Therefore, the repeated roots are k ₂ and k ₃ so that the branch point is in the absolute spectrum.

Branch point k=−0.001−0.334i.:

Substituting this value of k into (10) gives λ=−0.110+0.167i. Substituting this value of λ back into (9) gives a quartic polynomial for k whose roots are 0.314−0.816i, −0.001−0.334i, −0.001−0.334i, −0.312−0.0165i. Therefore, the repeated roots are k ₂ and k ₃ so that the branch point is in the absolute spectrum.

Branch point k=−0.732i.:

Substituting this value of k into (10) gives λ=0.000. Substituting this value of λ back into (9) gives a quartic polynomial for k whose roots are −0.732i, −0.732i, 0.160−0.017i, −0.160−0.017i. Therefore, the repeated roots are k ₁ and k ₂ so that the branch point is not in the absolute spectrum.

Therefore, of the six branch points, two are in the absolute spectrum, with the corresponding eigenvalues being −0.110±0.167i. Since these eigenvalues have negative real parts, the steady state (h _s ,p _s ) is absolutely stable. To determine whether the convective instability is of transient or remnant type, it is necessary to calculate the absolute spectrum. This can be done via numerical continuation of the generalised absolute spectrum, using the six branch points listed above as starting points, as discussed in Sect. 5. This shows that the branch points are the most unstable points in the absolute spectrum, so that the steady state has a transient convective instability.

1.2 A.2 Branch Points for the Klausmeier Model

For all parameters, the Klausmeier model (2a), (2b) has a “desert” steady state m=0, w=A. When A≥2B, there are two further steady states (m _±,w _±) where

$$m_\pm=\frac{2B}{A\pm\sqrt{A^2-4B^2}}, \qquad w_\pm=\frac{A\pm\sqrt{A^2-4B^2}}{2} . $$

Ecologically realistic values of B are relatively small, and in particular satisfy B<2 (Klausmeier 1999; Rietkerk et al. 2002). Under this constraint, (m ₋,w ₋) is stable as a solution of the kinetics odes, although it can be destabilised by the diffusion and advection terms, leading to spatial patterns (Klausmeier 1999; Sherratt 2005, 2010). However, (m ₊,w ₊) is unstable as a solution of the kinetic odes, and we will consider the nature of its instability as a solution of the pdes (2a), (2b).

We begin by linearising (2a), (2b) about (m ₊,w ₊), giving

$$\begin{aligned} \tilde{m}_t =&\tilde{\alpha}\tilde{m}+\tilde{\beta}\tilde{w}+\tilde{m}_{xx}, \end{aligned}$$

(12a)

$$\begin{aligned} \tilde{w}_t =&\tilde{\gamma}\tilde{m}+\tilde{\delta}\tilde{w}+\nu\tilde{w}_x, \end{aligned}$$

(12b)

where $\tilde{m}=m-m_{+}$, $\tilde{w}=w-w_{+}$, and the linear coefficients α, β, γ, δ are given by

$$\begin{aligned} \tilde{\alpha} =&B, \end{aligned}$$

(13a)

$$\begin{aligned} \tilde{\beta} =&\frac{A-\sqrt{A^2-4B^2}}{A+\sqrt{A^2-4B^2}}, \end{aligned}$$

(13b)

$$\begin{aligned} \tilde{\gamma} =&-2B, \end{aligned}$$

(13c)

$$\begin{aligned} \tilde{\delta} =&\frac{-2A}{A+\sqrt{A^2-4B^2}} . \end{aligned}$$

(13d)

Substituting $(\tilde{m},\tilde{w} )= (\overline{m},\overline{w} ) \exp(ikx+\lambda t)$ into (12a), (12b) and requiring $\overline{m}$ and $\overline{w}$ to be non-zero gives the dispersion relation

$$ {\tilde{\mathcal{D}}}(\lambda,k)= \lambda^2+\lambda \bigl(k^2-\tilde{\alpha}-ik\nu -\tilde{ \delta} \bigr) +\bigl(\tilde{\alpha}-k^2\bigr) (ik\nu+\tilde{\delta})- \tilde{\beta}\tilde {\gamma}=0 . $$

(14)

Branch points are double roots (for k) of the dispersion relation, and satisfy (14) and also

$$\begin{aligned} \begin{aligned} 0&=\partial{\tilde{\mathcal{D}}}/\partial k= \lambda (2k-i\nu ) - \bigl(3ik^2\nu+2\tilde{\delta}k-i\nu\tilde{\alpha} \bigr)\quad \Rightarrow \\ \lambda&= \bigl(3ik^2\nu+2\tilde{\delta}k-i\nu \tilde{\alpha} \bigr) / (2k-i\nu ) . \end{aligned} \end{aligned}$$

(15)

Substituting (15) into (14) gives a quintic polynomial in k:

$$\begin{aligned} &\bigl(3ik^2\nu+2\tilde{\delta}k-i\nu\tilde{\alpha} \bigr)^2 + (2k-i\nu ) \bigl(3ik^2\nu+2\tilde{\delta}k-i\nu \tilde{\alpha} \bigr) \bigl(k^2-\tilde{\alpha}-ik\nu-\tilde{\delta} \bigr) \\ &\quad{} + (2k-i\nu )^2 \bigl[\bigl(\tilde{\alpha}-k^2\bigr) (ik \nu+\tilde {\delta}) -\tilde{\beta}\tilde{\gamma} \bigr]=0 . \end{aligned}$$

(16)

We must now proceed numerically, and we will fix the parameter values to be A=2, B=0.5 and ν=20. These parameters satisfy the condition A>2B, but otherwise they are chosen arbitrarily. The value ν=20 is too small for ecological realism: The formula for the dimensionless parameter ν involves the ratio of the advection rate of water and the (square root of the) plant diffusion coefficient (Klausmeier 1999; Sherratt 2005), so that ν is relatively large, with Klausmeier’s (1999) estimate being 182.5. However, we use the smaller value to improve the clarity of the numerical calculations. Substituting the parameter values into (16) gives five distinct roots for k, two complex and three pure imaginary. For each, we substitute into (15) to find the corresponding value of λ. We then substitute this value of λ into (14) to determine whether the branch point is in the absolute spectrum. We performed all of the various calculations using the software package maple with 20 decimal places, but for clarity we give results to 3 decimal places.

Branch point k=−0.099+i0.012.:

Substituting this value of k into (15) gives λ=0.473+i0.022. Substituting this value of λ back into (14) gives a cubic polynomial for k whose roots are −0.099+i0.012, −0.099+i0.012, 0.199−i0.101. Recall from Sect. 5 that the branch point is in the absolute spectrum if the repeated roots are k ₂ and k ₃, when the roots k ₁, k ₂, k ₃ of (14) are labelled in increasing order of their imaginary parts. Therefore, in this case the branch point is in the absolute spectrum.

Branch point k=−i19.950.:

Substituting this value of k into (15) gives λ=398.112. Substituting this value of λ back into (14) gives a cubic polynomial for k whose roots are −i19.950, −i19.950 and i19.94. Therefore, the repeated roots are k ₁ and k ₂, so that this branch point is not in the absolute spectrum.

Branch point k=−i19.892.:

Substituting this value of k into (15) gives λ=396.587. Substituting this value of λ back into (14) gives a cubic polynomial for k whose roots are −i19.892, −i19.892 and i19.902. Therefore, the repeated roots are k ₁ and k ₂, so that this branch point is not in the absolute spectrum.

Branch point k=−i0.181.:

Substituting this value of k into (15) gives λ=0.569. Substituting this value of λ back into (14) gives a cubic polynomial for k whose roots are −i0.181, −i0.181, and i0.281. Therefore, the repeated roots are k ₁ and k ₂, so that this branch point is not in the absolute spectrum.

Branch point k=0.099+i0.012.:

Substituting this value of k into (15) gives λ=0.473−i0.022. Substituting this value of λ back into (14) gives a cubic polynomial for k whose roots are −0.199−i0.101, 0.099+i0.012, and 0.099+i0.012. Therefore, the repeated roots are k ₂ and k ₃, so that this branch point is in the absolute spectrum.

Therefore, of the five branch points, two are in the absolute spectrum, with the corresponding eigenvalues being 0.473±i0.022. Since these eigenvalues have positive real part, the steady state (m ₊,w ₊) is absolutely unstable.