Persistence in a Two-Dimensional Moving-Habitat Model

Abstract

Environmental changes are forcing many species to track suitable conditions or face extinction. In this study, we use a two-dimensional integrodifference equation to analyze whether a population can track a habitat that is moving due to climate change. We model habitat as a simple rectangle. Our model quickly leads to an eigenvalue problem that determines whether the population persists or declines. After surveying techniques to solve the eigenvalue problem, we highlight three findings that impact conservation efforts such as reserve design and species risk assessment. First, while other models focus on habitat length (parallel to the direction of habitat movement), we show that ignoring habitat width (perpendicular to habitat movement) can lead to overestimates of persistence. Dispersal barriers and hostile landscapes that constrain habitat width greatly decrease the population’s ability to track its habitat. Second, for some long-distance dispersal kernels, increasing habitat length improves persistence without limit; for other kernels, increasing length is of limited help and has diminishing returns. Third, it is not always best to orient the long side of the habitat in the direction of climate change. Evidence suggests that the kurtosis of the dispersal kernel determines whether it is best to have a long, wide, or square habitat. In particular, populations with platykurtic dispersal benefit more from a wide habitat, while those with leptokurtic dispersal benefit more from a long habitat. We apply our model to the Rocky Mountain Apollo butterfly (Parnassius smintheus).

Appendix

1.1 Nyström’s Method

1.1.1 Constructing Matrix $\mathbf {A}$

To discretize integral eigenvalue problem (7) using Nyström’s method, first define grid points

$$\begin{aligned} x_i&= -L/2 + i \cdot {\Delta } x,&i = 0, 1, \dots , n, \end{aligned}$$

(33)

$$\begin{aligned} y_j&= -W/2 + j \cdot {\Delta } y,&j = 0, 1, \dots , m, \end{aligned}$$

(34)

where ${\Delta } x = L/n$ and ${\Delta } y = W/m$. Create identical grid points $x'_k$ and $y'_{\ell }$ so that ${\Delta } x' = {\Delta } x$ and ${\Delta } y' = {\Delta } y$. Next, evaluate both sides of eigenvalue problem (7) only at grid points and approximate each integral on the right-hand side with a sum using, for example, the repeated trapezoidal rule. Because there are four sets of grid points, the resulting problem is, in a sense, four dimensional. We need to create additional indices to condense the system into a matrix eigenvalue problem with two dimensions. Define the indices

$$\begin{aligned} p(i,j) = i + 1 + (n+1)j, \quad q(k,\ell ) = k + 1 + (n+1)\ell . \end{aligned}$$

(35)

The mapping from (i, j) pairs to p-values effectively reshapes a two-dimensional array into a one-dimensional vector (and similarly for the mapping from $(k,\ell )$ pairs to q-values), making the problem two-dimensional. Integral equation (7) becomes

$$\begin{aligned} \lambda u_p = R_0 \sum _{q=1}^{(n+1)(m+1)} A_{pq}u_q, \end{aligned}$$

(36)

where $u_p$ is the p-th element of the reshaped vector u, and the element in row p, column q of matrix $\mathbf {A}$ is

$$\begin{aligned} A_{pq}= \alpha _1 \, \alpha _2 \, k(x_i + c - x'_k,\, y_j - y'_\ell ) \,{\Delta } y_1 {\Delta } y_2, \end{aligned}$$

(37)

where

$$\begin{aligned} \alpha _1 = {\left\{ \begin{array}{ll} \dfrac{1}{2} \qquad &{}\text {if}\ \quad k = 0 \ \text {or} \ n,\\ 1 &{}\text {otherwise;} \end{array}\right. } \qquad \alpha _2 = {\left\{ \begin{array}{ll} \dfrac{1}{2} \qquad &{}\text {if}\ \quad \ell = 0 \ \text {or} \ m, \\ 1 &{}\text {otherwise.} \end{array}\right. } \end{aligned}$$

(38)

System (36) is now a standard matrix eigenvalue problem.

1.1.2 Avoiding Dispersal Kernel Singularities

If the dispersal kernel is singular at the origin, one way to avoid evaluating the singularity is to shift $y'$ grid points onto the midpoints of y grid points,

$$\begin{aligned} y_\ell ' = -W/2 + {\Delta } y/2 + \ell \cdot {\Delta } y', \qquad \ell = 0, 1, \dots , m-1. \end{aligned}$$

(39)

The action of the integral operator K in eigenvalue problem (7) is now the composition of two operators, ${K}_1$ and ${K}_2$. The first calculates values at (x, y) grid points, stored in some vector u. The second calculates values at $(x', y')$ grid points stored in another vector, v. Applying Nyström’s method to ${K}_1$ and ${K}_2$ as above yields a pair of matrices, $\mathbf {A}_1$ and $\mathbf {A}_2$, satisfying

$$\begin{aligned} u_{t+1} = \mathbf {A}_1 \,v_t, \qquad v_{t+1} = \mathbf {A}_2 \, u_t. \end{aligned}$$

(40)

Since $u_{t+2} = \mathbf {A}_1\mathbf {A}_2 \, u_t$, the critical speed occurs when $\rho (\mathbf {A}_1\mathbf {A}_2) = 1$.

1.2 Legendre-Expansion Method

1.2.1 Constructing Matrix $\mathbf {B}$

To construct matrix $\mathbf {B}$ in eigenvalue problem (15), we must first recover the Legendre expansion coefficients $b_{ijk\ell }$. Multiplying both sides of kernel expansion (14) by Legendre polynomials in each variable, integrating over the domain, and using orthogonality properties (11)–(12) gives

$$\begin{aligned} b_{ijk\ell }= & {} \left[ \dfrac{(2i+1)(2j+1)(2k+1)(2\ell +1)}{L^2W^2}\right] \nonumber \\&\times \int \limits ^{L/2}_{-L/2}\int \limits ^{L/2}_{-L/2}\int \limits ^{W/2}_{-W/2}\int \limits ^{W/2}_{-W/2} k(x+c - x', y - y')X_i(x)X_j(x')Y_k(y)Y_\ell (y')\nonumber \\&\times \, \mathrm{d}y' \,\mathrm{d}y \,\mathrm{d}x' \,\mathrm{d}x. \end{aligned}$$

(41)

Now, substitute kernel expansion (14) into integral eigenvalue problem (7).

$$\begin{aligned} \lambda u(x,y)&= R_0 \int ^{L/2}_{-L/2}\int ^{W/2}_{-W/2} \sum _{i,j,k,\ell = 0}^N b_{ijk\ell } \, X_i(x)X_j(x')Y_k(y)Y_\ell (y')\, u(x',y') \ \mathrm{d}y' \ \mathrm{d}x' \end{aligned}$$

(42)

$$\begin{aligned}&= R_0 \sum _{i,j,k,\ell = 0}^N b_{ijk\ell } \, X_i(x) Y_k(y) \left( \int ^{L/2}_{-L/2}\int ^{W/2}_{-W/2} X_j(x') Y_\ell (y') \ u(x',y') \, \mathrm{d}y' \, \mathrm{d}x'\right) \end{aligned}$$

(43)

$$\begin{aligned}&= R_0 \sum _{i,j,k,\ell = 0}^N b_{ijk\ell } \, X_i(x) Y_k(y) \, u_{j\ell }, \end{aligned}$$

(44)

where

$$\begin{aligned} u_{j\ell }&\equiv \int ^{L/2}_{-L/2}\int ^{W/2}_{-W/2} X_j(x') Y_\ell (y') \ u(x',y') \, \mathrm{d}y' \, \mathrm{d}x'. \end{aligned}$$

(45)

Multiplying both sides of Eq. (44) by Legendre polynomials $X_r(x) Y_s(y)$ and integrating,

$$\begin{aligned}&\lambda \int ^{L/2}_{-L/2}\int ^{W/2}_{-W/2} u(x,y) X_r(x) Y_s(y) \, \mathrm{d}x \, \mathrm{d}y \end{aligned}$$

(46)

$$\begin{aligned}&\quad = R_0 \sum _{i,j,k,\ell = 0}^N b_{ijk\ell } \ u_{j\ell } \int ^{L/2}_{-L/2}\int ^{W/2}_{-W/2} X_i(x) X_r(x) Y_k(y) Y_s(y) \, \mathrm{d}y \, \mathrm{d}x. \end{aligned}$$

(47)

Using definition (45) and orthogonality,

$$\begin{aligned} \lambda u_{ik} = R_0LW \sum _{j=0}^N \sum _{\ell = 0}^N \dfrac{b_{ijk\ell } \ u_{ik}}{(2i+1)(2k+1)}. \end{aligned}$$

(48)

We have reverted to indices (i, k) rather than (r, s).

System (48) has four sets of indices and is, in a sense, four dimensional. We need to create additional indices to condense the system into a matrix eigenvalue problem with two dimensions. Let

$$\begin{aligned} p(i,k) = i + 1 + (N+1)k, \qquad q(j,\ell ) = j + 1 + (N+1)\ell , \end{aligned}$$

(49)

where $i, j, k, \ell \in \{0, 1, \dots , N\}$. The mapping from (i, k) pairs to p-values effectively reshapes a two-dimensional array into a one-dimensional vector (and similarly for the mapping from $(j,\ell )$ pairs to q-values), making the problem two-dimensional. System (48) becomes

$$\begin{aligned} \lambda u_p = R_0LW \sum _{q=1}^{(N+1)^2} \text {B}_\text {pq} u_q, \end{aligned}$$

(50)

where $u_p$ is the p-th entry of the reshaped vector u. The element in row p, column q of matrix $\mathbf {B}$ is

$$\begin{aligned} \text {B}_\text {pq} = \dfrac{b_{ijk\ell }}{(2i+1)(2k+1)}. \end{aligned}$$

(51)

1.2.2 Taylor Expansion Coefficients

The coefficients in approximation (20) found by expanding the kernel in a Taylor series are

$$\begin{aligned} \alpha&\equiv \int \limits ^{L/2}_{-L/2}\int \limits ^{W/2}_{-W/2} k(x,y) \,\mathrm{d}y \,\mathrm{d}x, \end{aligned}$$

(52)

$$\begin{aligned} \beta&\equiv \int \limits ^{W/2}_{-W/2} [ k(L/2, y) - k(-L/2,y)] \,\mathrm{d}y, \end{aligned}$$

(53)

$$\begin{aligned} \gamma&\equiv \int \limits ^{W/2}_{-W/2} \left[ \dfrac{\partial k}{\partial x}(L/2, y) - \dfrac{\partial k}{\partial x} (-L/2,y) \right] \,\mathrm{d}y, \end{aligned}$$

(54)

$$\begin{aligned} \delta&\equiv \int \limits ^{L/2}_{-L/2} \left[ \dfrac{\partial k}{\partial y}(x,W/2) - \dfrac{\partial k}{\partial y}(x,-W/2)\right] \,\mathrm{d}x . \end{aligned}$$

(55)

1.3 Optimal Aspect Ratio for the Symmetrized Gaussian Kernel

We will demonstrate an instance of the mesokurtic portion of Conjecture 2, that $\theta _{\text {opt}} = 1$ for Gaussian dispersal kernels. We approximate the spectral radius $\rho (\theta )$ using geometric symmetrization and an eigenfunction-one Rayleigh quotient (22). Symmetrization does not give an exact value for $\rho (\theta )$, but it does give a tight lower bound and is analytically tractable. It may not be surprising that, should a maximal $\rho (\theta )$ exist for a symmetric kernel, it might occur at $\theta = 1$, when $L = W$. What is less obvious (and what we will prove) is that a maximum does exist, and that it is unique.

Given a Gaussian kernel,

$$\begin{aligned} k(x+c - x', y - y') = \dfrac{1}{2\pi \sigma ^2}\exp \left[ \dfrac{-(x+c-x')^2}{2\sigma ^2}\right] \exp \left[ \dfrac{-(y-y')^2}{2\sigma ^2}\right] , \end{aligned}$$

(56)

the symmetrized kernel, using formula (21), is

$$\begin{aligned} k_G(x+c - x', y - y') = \dfrac{1}{2\pi \sigma ^2} \exp \left[ \dfrac{-c^2}{2\sigma ^2}\right] \exp \left[ \dfrac{-(x-x')^2}{2\sigma ^2}\right] \exp \left[ \dfrac{-(y-y')^2}{2\sigma ^2}\right] . \end{aligned}$$

(57)

The Rayleigh quotient approximation for the spectral radius with constant eigenfunction, as a function of L and W, is

$$\begin{aligned} \rho (L,W) = \dfrac{\int ^{L/2}_{-L/2} \int ^{L/2}_{-L/2} \int ^{W/2}_{-W/2} \int ^{W/2}_{-W/2} k_G(x+c - x', y - y') \, \mathrm{d}y' \, \mathrm{d}y \, \mathrm{d}x' \, \mathrm{d}x}{LW}. \end{aligned}$$

(58)

The numerator of approximation (58) is separable, and we may write

$$\begin{aligned} \rho (L,W) = C I(L) I(W), \end{aligned}$$

(59)

where

$$\begin{aligned} C = \dfrac{1}{2\pi \sigma ^2} \exp \left( \dfrac{-c^2}{2 \sigma ^2} \right) , \end{aligned}$$

(60)

and

$$\begin{aligned} I(L) = \sqrt{2 \pi } \sigma \, \text {erf} \left( \dfrac{L}{\sqrt{2}\sigma }\right) - \dfrac{2\sigma ^2}{L} + \dfrac{2\sigma ^2}{L} \exp \left( \dfrac{-L^2}{2\sigma ^2}\right) . \end{aligned}$$

(61)

I(W) is identical to I(L) with W replacing L. (We have not absorbed the constant C into I(L) and I(W) to make future calculations easier.)

We now have a constrained optimization problem: to maximize $\rho (L,W)$ subject to the constraint $LW = A$. Before using the method of Lagrange multipliers, we will describe a few properties of the function I(L) necessary for the proof. Then, we will construct a Lagrangian function and find the unique critical point of the associated system. Finally, we will use the associated bordered Hessian matrix to show the critical point gives a maximum.

The first two derivatives of I(L) are

$$\begin{aligned} I'(L) = \dfrac{2\sigma ^2 - 2\sigma ^2 \exp \left( \dfrac{-L^2}{2\sigma ^2}\right) }{L^2} \, > \, 0 \quad \text {for} \quad L > 0, \end{aligned}$$

(62)

and

$$\begin{aligned} I''(L)&= \dfrac{2L^2 \exp \left( \dfrac{-L^2}{2\sigma ^2}\right) - 4\sigma ^2 + 4\sigma ^2 \exp \left( \dfrac{-L^2}{2\sigma ^2}\right) }{L^3} \end{aligned}$$

(63)

$$\begin{aligned}&= \dfrac{2\exp \left( \dfrac{-L^2}{2\sigma ^2}\right) - 2I'(L)}{L}. \end{aligned}$$

(64)

Using the above derivatives, we can show the following useful lemma.

Lemma 1

$$\begin{aligned} I'(L) < \dfrac{I(L)}{L} < 1, \quad \mathrm{for} \quad L > 0. \end{aligned}$$

(65)

Proof

It is sufficient to show that I(L) is concave down and that $\lim _{L\rightarrow 0^+} I'(L) = 1$. That is, if $I'(L)$ is decreasing and bounded from above by one, I(L) must satisfy the lemma.

We first show that $I''(L)$ is negative. From expression (63),

$$\begin{aligned} I''(L) = \dfrac{1}{L^3}\exp \left( \dfrac{-L^2}{2\sigma ^2}\right) \left[ 2L^2 - 4\sigma ^2\exp \left( \dfrac{L^2}{2\sigma ^2}\right) + 4\sigma ^2\right] . \end{aligned}$$

(66)

Since $\exp (L) > L + 1$ when $L \ne 0$,

$$\begin{aligned} I''(L) < \dfrac{1}{L^3}\exp \left( \dfrac{-L^2}{2\sigma ^2}\right) \left[ 2L^2 - 4\sigma ^2\left( \dfrac{L^2}{2\sigma ^2}+ 1\right) + 4\sigma ^2\right] = 0. \end{aligned}$$

(67)

Then, using L’Hôpital’s rule twice on expression (62), the desired limit quickly follows. $\square$

With the above lemma available, we now construct the Lagrangian function with objective function $\rho (L,W)$ and constraint $g(L,W) = LW - A = 0$.

$$\begin{aligned} {\varLambda }(L,W,\lambda ) = \rho (L,W) - \lambda g(L,W). \end{aligned}$$

(68)

The necessary first-order conditions for critical points are

$$\begin{aligned} {\varLambda }_L&= C I'(L) I(W) - \lambda W = 0, \end{aligned}$$

(69)

$$\begin{aligned} {\varLambda }_W&= C I(L) I'(W) - \lambda L = 0, \end{aligned}$$

(70)

$$\begin{aligned} {\varLambda }_\lambda&= LW - A = 0 \end{aligned}$$

(71)

(Marsden and Tromba 2012). Here, subscripts indicate partial differentiation. From (69) and (70),

$$\begin{aligned} \lambda = \dfrac{C I'(L) I(L)}{W} = \dfrac{C I(L) I'(W)}{L}. \end{aligned}$$

(72)

Rearranging the last equation,

$$\begin{aligned} \dfrac{ I'(L) L}{I(L)} = \dfrac{I'(W) W}{I(W)}. \end{aligned}$$

(73)

If we can show that the function

$$\begin{aligned} J(L) \equiv \dfrac{ I'(L) L}{I(L)} \end{aligned}$$

(74)

is one-to-one for $L > 0$, then Eq. (73) implies $L = W$, and with the constraint $LW = A$, the only critical point is $(L,W) = (\sqrt{A},\sqrt{A})$. We will show J(L) is one-to-one by showing it is monotone decreasing.

From expressions (62) and (63) for $I'(L)$ and $I''(L)$,

$$\begin{aligned} J'(L) = \dfrac{1}{L}\left[ I I''+ \dfrac{I I'}{L} - (I')^2 \right] . \end{aligned}$$

(75)

(We assume differentiation is with respect to L on the right-hand side.) Substituting expression (64) for $I''$,

$$\begin{aligned} J'(L)&= \dfrac{1}{L} \left[ I \left( \dfrac{2}{L}\exp \left( \dfrac{-L^2}{2\sigma ^2}\right) - \dfrac{2}{L}I'\right) + \dfrac{I I'}{L} - (I')^2\right] \end{aligned}$$

(76)

$$\begin{aligned}&= \dfrac{1}{L} \left[ \dfrac{2I}{L}\exp \left( \dfrac{-L^2}{2\sigma ^2}\right) - \dfrac{I I'}{L} - (I')^2\right] . \end{aligned}$$

(77)

The first and second inequalities in Lemma 1 now give, one after the other,

$$\begin{aligned} J'(L)&< \dfrac{1}{L}\left[ 2\exp \left( \dfrac{-L^2}{2\sigma ^2}\right) - \dfrac{I I'}{L} - (I')^2\right] \end{aligned}$$

(78)

$$\begin{aligned}&< \dfrac{1}{L}\left[ 2\exp \left( \dfrac{-L^2}{2\sigma ^2}\right) - 2(I')^2\right] \end{aligned}$$

(79)

$$\begin{aligned}&= \dfrac{2}{L}\left[ \exp \left( \dfrac{-L^2}{4\sigma ^2}\right) + I'\right] \left[ \exp \left( \dfrac{-L^2}{4\sigma ^2}\right) - I'\right] \end{aligned}$$

(80)

$$\begin{aligned}&= \dfrac{2}{L^3} \left[ \exp \left( \dfrac{-L^2}{4\sigma ^2}\right) + I'\right] \left[ L^2 \exp \left( \dfrac{-L^2}{4\sigma ^2}\right) + 2\sigma ^2 \exp \left( \dfrac{-L^2}{4\sigma ^2}\right) - 2\sigma ^2\right] . \end{aligned}$$

(81)

Since $I'(L) > 0$, we only need to show that

$$\begin{aligned} M(L) \equiv L^2 \exp \left( \dfrac{-L^2}{4\sigma ^2}\right) + 2\sigma ^2 \exp \left( \dfrac{-L^2}{4\sigma ^2}\right) - 2\sigma ^2 < 0 \end{aligned}$$

(82)

when $L > 0$ to prove J(L) is monotone decreasing. We have that $M(0) = 0$, and

$$\begin{aligned} M'(L)&= -\dfrac{L^3}{2\sigma ^2} \exp \left( \dfrac{-L^2}{4\sigma ^2}\right) + 2L\exp \left( \dfrac{-L^2}{4\sigma ^2}\right) - 2L \exp \left( \dfrac{-L^2}{2\sigma ^2}\right) \end{aligned}$$

(83)

$$\begin{aligned}&= L\exp \left( \dfrac{-L^2}{4\sigma ^2}\right) \left[ -\dfrac{L^2}{2\sigma ^2} + 2 - 2 \exp \left( \dfrac{-L^2}{4\sigma ^2}\right) \right] \end{aligned}$$

(84)

$$\begin{aligned}&< L\exp \left( \dfrac{-L^2}{4\sigma ^2}\right) \left[ -\dfrac{L^2}{2\sigma ^2} + 2 - 2\left( -\dfrac{L^2}{4\sigma ^2} + 1\right) \right] \, = \, 0. \end{aligned}$$

(85)

Thus, $M(L) < 0$, which makes J(L) monotone decreasing and one-to-one, which makes $L = W$ the only solution of Eq. (73). Together with the area constraint, the unique critical point is $(L, W) = (\sqrt{A}, \sqrt{A})$.

To show our critical point maximizes the spectral radius, we will use the bordered Hessian matrix

$$\begin{aligned} H&= \begin{pmatrix} 0 &{}\quad -g_L &{}\quad -g_W \\ -g_L &{}\quad {\varLambda }_{LL} &{}\quad {\varLambda }_{LW}\\ -g_W &{}\quad {\varLambda }_{WL} &{}\quad {\varLambda }_{WW} \end{pmatrix} \end{aligned}$$

(86)

$$\begin{aligned}&= \begin{pmatrix} 0 &{}\quad -W &{}\quad -L\\ -W &{}\quad C I''(L)I(W) &{}\quad C I'(L)I'(W) - \lambda \\ -L &{}\quad C I'(L)I'(W) - \lambda &{}\quad C I(L) I''(W) \end{pmatrix}. \end{aligned}$$

(87)

(Recall that C is a constant.) For our $3 \times 3$ matrix, a sufficient condition for the critical point to be a maximum is $|H| > 0$, where the matrix is evaluated at the critical point (Marsden and Tromba 2012). $| \cdot |$ denotes the determinant.

$$\begin{aligned} |H| = -CW^2I(L)I''(W) - CL^2I''(L)I(W) + 2CLWI'(L)I'(W) - 2C\lambda LW. \end{aligned}$$

(88)

At the critical point, $L = W$, so we will use L to represent both quantities.

$$\begin{aligned} |H|&= -2CL^2I''(L) I(L) + 2CL^2 [I'(L)]^2 - 2CL^2\lambda \end{aligned}$$

(89)

$$\begin{aligned}&= 2CL^2\left( -I''(L) I(L) + [I'(L)]^2 - \dfrac{I'(L) I(L)}{L}\right) . \end{aligned}$$

(90)

However, we have already shown from steps (75)–(85) that

$$\begin{aligned} I''(L) I(L) - [I'(L)]^2 + \dfrac{I'(L) I(L)}{L} < 0 \end{aligned}$$

(91)

when $L > 0$. The opposite quantity must be positive. Therefore, from expression (90), $|H| > 0$ and the unique critical point $(L,W) = (\sqrt{A},\sqrt{A})$ maximizes the spectral radius. $\theta _{\text {opt}} = 1$ for the symmetrized Gaussian kernel.

1.4 Effect of Switching Length and Width for the Symmetrized Generalized Gaussian Kernel

One way to illustrate Conjecture 1 is to plot the level curves of shifted kernels with different kurtosis values. If the level curves are stretched along the x-axis (the dimension of habitat length), a habitat with larger length than width is better because it captures more of the kernel’s probability mass (and, thus, more propagules). If level curves are stretched along the y-axis (the dimension of width), a habitat with larger width is preferable. We will use the geometric symmetrization approximation (21) on the generalized Gaussian kernel (26) and show kurtosis determines the direction the level curves are stretched. Our result provides evidence that, for the broad class of generalized Gaussian kernels, Conjecture 1 is true.

To simplify our approach, we will consider the square of the geometrically symmetrized kernel,

$$\begin{aligned} k_G^2(x+c-x',y-y') = = k(x + c - x', y - y') k(x' + c - x, y' - y). \end{aligned}$$

(92)

Evaluating (92) for the generalized Gaussian kernel (26) gives a distribution of the form

$$\begin{aligned}&k_G^2(x+c-x',y-y') \nonumber \\&\quad = \gamma \exp \left\{ -\dfrac{1}{2} \left[ (x+c-x')^2 + (y-y')^2\right] ^\beta \right\} \nonumber \\&\qquad \times \exp \left\{ -\dfrac{1}{2} \left[ (x'+c-x)^2 + (y'-y)^2\right] ^\beta \right\} , \end{aligned}$$

(93)

where $\gamma$ is a normalizing constant. Since $x'$ and $y'$ simply shift the kernel, we will evaluate approximation (93) at $(x', y') = 0$.

$$\begin{aligned} k_G^2(x+c,y) = \gamma \exp \left\{ -\dfrac{1}{2} \left[ (x+c)^2 + y^2\right] ^\beta \right\} \exp \left\{ -\dfrac{1}{2} \left[ (x-c)^2 + y^2\right] ^\beta \right\} . \end{aligned}$$

(94)

As the speed of climate change c increases from zero, the level curves of kernel (94) stretch along one axis or the other when $\beta \ne 1$ (Fig. 7). To measure the effect of the stretch, we will slice kernel (94) along each axis and compare the corresponding functions.

$$\begin{aligned} f_1(x)&\equiv k_G^2(x+c,0) = \gamma \exp \left\{ -\dfrac{1}{2} \left[ (x+c)^2\right] ^\beta -\dfrac{1}{2} \left[ (x-c)^2 \right] ^\beta \right\} , \end{aligned}$$

(95)

$$\begin{aligned} f_2(y)&\equiv k_G^2(0 + c,y) = \gamma \exp \left[ -(c^2+y^2)^\beta \right] . \end{aligned}$$

(96)

Fig. 7

Level curves of symmetrized generalized Gaussian kernels show the effect of increasing the speed of climate change, c. Plots (a), (c), and (e) used $c = 0$. In plots (b), (d), and (f), we increased c to 1 km/year. We used $\beta = 2$ to obtain a platykurtic kernel for (a, b), $\beta = 1$ to obtain a mesokurtic kernel for (c, d), and $\beta = 1/2$ to obtain a leptokurtic kernel for (e, f). The orientation of the level curves determines whether it is better to have a larger habitat length (for leptokurtic kernels) or larger width (for platykurtic kernels). For the Gaussian kernel, length and width have the same effect on persistence

Functions (95) and (96) are the cross sections of kernel (94) along the x- and y-axis, respectively. To compare the functions, let us, for example, set $f_2$ as a function of x.

$$\begin{aligned} f_2(x) \equiv \gamma \exp \left[ -(c^2+x^2)^\beta \right] . \end{aligned}$$

(97)

We wish to find conditions under which, for example, $f_1(x) > f_2(x)$, leading to a stretch along the x-axis.

$$\begin{aligned} \gamma \exp \left\{ -\dfrac{1}{2} \left[ (x+c)^2\right] ^\beta -\dfrac{1}{2} \left[ (x-c)^2 \right] ^\beta \right\} \, > \, \gamma \exp \left[ -(c^2+x^2)^\beta \right] , \end{aligned}$$

(98)

or

$$\begin{aligned} -\dfrac{1}{2}[(x+c)^2]^\beta - \dfrac{1}{2}[(x-c)^2]^\beta \, > \, -(c^2 + x^2)^\beta . \end{aligned}$$

(99)

Switching sides,

$$\begin{aligned} (c^2 + x^2)^\beta \, > \, \dfrac{1}{2}[(x+c)^2]^\beta + \dfrac{1}{2}[(x-c)^2]^\beta . \end{aligned}$$

(100)

The observation that

$$\begin{aligned} c^2 + x^2 = \dfrac{1}{2}(x+c)^2 + \dfrac{1}{2}(x-c)^2 \end{aligned}$$

(101)

suggests we can use a convexity argument with the function $g(z) \equiv z^\beta$. In particular, Jensen’s inequality states that if (and only if) g(z) is concave down, then for any two values $z_1$ and $z_2$,

$$\begin{aligned} g\left( \dfrac{1}{2} z_1 + \dfrac{1}{2} z_2\right) > \dfrac{1}{2}g(z_1) + \dfrac{1}{2}g(z_2). \end{aligned}$$

(102)

g(z) is concave down only when $\beta < 1$. Setting

$$\begin{aligned} z_1 = (x+c)^2, \quad z_2 = (x-c)^2, \end{aligned}$$

(103)

we see that inequality (100) is only true when $\beta < 1$. The same argument applies for $\beta > 1$, resulting in $f_1(x) < f_2(x)$. In general,

$$\begin{aligned} f_1(x) {\left\{ \begin{array}{ll} > f_2(x) \quad \text { if} \quad \beta < 1,\\ = f_2(x) \quad \text { if} \quad \beta = 1,\\ < f_2(x) \quad \text { if} \quad \beta >1. \end{array}\right. } \end{aligned}$$

(104)

The parameter $\beta$ controls the kurtosis of the generalized Gaussian distribution. $\beta = 1$ gives a standard Gaussian. Increasing c has no effect on the Gaussian (Fig. 7c, d), so there is no preference for habitat length or width. $\beta >1$ gives platykurtic kernels, for which increasing c stretches the kernel along the y-axis (Fig. 7a, b), favoring larger width. Finally, $\beta < 1$ gives leptokurtic kernels, for which increasing c stretches the kernel along the x-axis (Fig. 7e, f), favoring larger length.

The use of convexity (102) above suggests a possible way to prove Conjecture 1. Recent definitions of kurtosis have centered around partial orderings of distributions that preserve certain ordering properties. One such ordering involves convexity (Wang 2009). If it can be shown that the partial ordering of two shifted kernels corresponds to an ordering of the importance of length and width (e.g., kernels with higher kurtosis favor habitat length over width), Conjecture 1 is proven. We will not attempt such a proof here, but convexity ordering provides a promising approach to our conjecture.