Neural Fields: Localised States with Piece-Wise Constant Interactions

Abstract

Neural field models are typically cast as continuum integro-differential equations for describing the idealised coarse-grained activity of populations of interacting neurons. For smooth Mexican hat kernels, with short-range excitation and long-range inhibition, these non-local models can support various localised states in the form of spots in two-dimensional media. In recent years, there has been a growing interest in the mathematical neuroscience community in studying such models with a Heaviside firing rate non-linearity, as this often allows substantial insight into the stability of stationary solutions in terms of integrals over the kernels. Here we consider the use of piece-wise constant kernels that allow the explicit evaluation of such integrals. We use this to show that azimuthal instabilities are not possible for simple piece-wise constant Top Hat interactions, whilst they are easily realised for piece-wise constant Mexican hat interactions.

Appendix: Circular Geometry for a Top Hat Kernel

Consider a portion of a disk whose upper boundary is an (circular) arc and whose lower boundary is a chord making a central angle ϕ ₀ < π, illustrated as the shaded region in Fig. 5a. The area A = A(r ₀, ϕ ₀) of the (shaded) segment is then simply given by the area of the circular sector (the entire wedge-shaped portion) minus the area of an isosceles triangle, namely

$$\displaystyle \begin{aligned} A(r_0,\phi_0) = \frac{1}{2} r_0^2 \left ( \phi_0 - \sin \phi_0\right ) . \end{aligned} $$

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Fig. 5

The area of the total shaded segment is $r_0^2 (\phi _0 -\sin \phi _0)/2$ (a). Overlap of two circles shows the area of active region (b)

The area of the overlap of two circles, as illustrated in Fig. 5b, can be constructed as the total area of A(r ₀, ϕ ₀) + A(r ₁, ϕ ₁). To determine the angles ϕ _0,1 in terms of the centers, (x ₀, y ₀) and (x ₁, y ₁), and radii, r ₀ and r ₁, of the two circles we use the cosine formula that relates the lengths of the three sides of a triangle formed by joining the centers of the circles to a point of intersection. We denote distance between the two centers by d where d ² = (x ₀ − x ₁)² + (y ₀ − y ₁)² so that

$$\displaystyle \begin{aligned} r_1^2 = r_0^2 + d^2 - 2 r_0 d \cos (\phi_0/2) , \qquad r_0^2 = r_1^2 + d^2 - 2 r_1 d \cos (\phi_1/2) . \end{aligned} $$

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Hence the angles are given by

$$\displaystyle \begin{aligned} \phi_0= 2 \cos^{-1} \left ( \frac{r_0^2 + d^2 - r_1^2}{2 r_0 d} \right), \qquad \phi_1= 2 \cos^{-1} \left ( \frac{r_1^2 + d^2 - r_0^2}{2 r_1 d} \right) . \end{aligned} $$

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