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Here we develop a method of eigenvalue analysis to find the stability of steady states of both the Amari model and the gap junction model with the decaying oscillatory coupling function in Equation (7.21) and the smooth firing rate function of

f(u) = 2e−r/(u−θ)2Θ(u−θ). (7.138)

Eigenvalue analysis was used to determine the stability of the steady states found numer- ically in Chapter 3.

Throughout the thesis, steady state solutions have been found using continuation methods. AUTO [23] has often been used to find solution curves of steady state solutions. AUTO uses pseudo-arclength continuation for following branches. A description of the pseudo-arclength continuation method is given in the Appendix. Various convergence criteria were set in AUTO and these criteria define the accuracy with which solutions are found. The relative convergence criterion of solution parameters and the relative convergence criterion for solution components in the Newton/Chord method were set at 1e−7. We also coded the pseudo-arclength continuation method in MATLAB [74]. The MATLAB code was successfully used to find solutions. Accuracy of the solutions found depended upon the finite difference approximations of the spatial and temporal derivatives. In general, solutions were of first order accuracy, as finite difference approximations were of first order accuracy.

Amari model

In Chapter 3, a fourth-order ODE was derived, solutions of which are steady state solutions of the time-dependent Amari model in Equation (7.1). The ODE is

u(iv)+ 2(1−b2)u00+ (b2+ 1)2u= 4b(b2+ 1)f(u). (7.139)

Using the ODE, the Amari model in Equation (7.1) can be written in the form

∂4 ∂x4 + 2(1−b 2) ∂2 ∂x2 + (b 2+ 1)2 1 + ∂ ∂t u(x, t) = 4b(b2+ 1)f(u(x, t)). (7.140)

Let u(x, t) = ¯u(x) +z(x, t) where ¯u is a steady state and z is a spatial temporal pertur- bation. Linearising about the steady state ¯u and Taylor expanding we have

f(u(x, t)) =f(¯u(x)) +z(x, t)f0(¯u(x)) +O(z2). (7.141)

So Equation (7.140) can be written as Lu(x) +¯ L 1 + ∂ ∂t z(x, t) = 4b(b2+ 1) f(¯u(x)) +z(x, t)f0(¯u(x)) (7.142) whereL= ∂ 4 ∂x4 + 2(1−b 2) ∂2 ∂x2 + (b

2+ 1)2. Bounded solutions of the formz(x, t) = ¯z(x)eλt are sought. Substituting for z(x, t) in Equation (7.142) results in

L(1 +λ)¯z(x) = 4b(b2+ 1)f0(¯u(x))¯z(x). (7.143) Ifλ≤0, then the perturbationz(x, t)→0 as t→ ∞and ¯u(x) is a stable steady state.

We wish to find the stability of steady states found using numerical methods such as numerical integration and continuation. To do this, we numerically evaluate the eigenvalues in Equation (7.143). Steady states are found over a finite domain Ω on R. For the numerical analysis here, we take Ω : [−10π,10π]. The domain is discretised by dividing Ω into nequal intervals of sizeh such that

h= 20π

n . (7.144)

Let xi = −10π +ih for i = 0,1, . . . , n. The steady state ¯u is discretised using the same discretisation as for the domain. Therefore ann×1 vector, v, is formed such that v= [v0;v1;. . .;vn−1] wherevi≈u(−10π¯ +ih) fori= 0,1, . . . , n−1. Using finite difference approximations, the n×n matrixL is created. From Equation (7.138), the derivative of f with respect to uis df du = 4r (u−θ)3e −r/(u−θ)2 Θ(u−θ). (7.145)

LetDf be then×nsparse diagonal matrix where the diagonal entries are df(v)

du . LetId be the n×n identity matrix. Then we have the equation

E=−Id+ 4b(b2+ 1)L−1Df (7.146) where E is an n×nmatrix. In computations, rather than calculating the inverse of the matrixL, Gaussian elimination is used as it is faster. We find the eigenvalues ofE. One eigenvalue will always be zero, as expected from the translational invariance of the model. The eigenvalue with maximum real part, therefore, dictates the stability. Consider the eigenvalue of E with maximum real part. If the real part is zero then the steady state ¯u is stable. If the real part is positive, then ¯u is unstable.

In Chapter 3, solution curves were found using continuation. Solution stability was determined using the method of eigenvalue analysis described above. Therefore the eigen- values were found for each solution point on the curve, that is, for a steady state associated with a particular value ofb. In practice, the top three eigenvalues are plotted for a solution curve. If all three eigenvalues are nonpositive for a particular value of b, then the steady state solution is stable. If one or more of the top three eigenvalues are positive, then the steady state solution is unstable. Slight variation was seen about zero for one eigenvalue. The number of points in the discretisation, n, was chosen so that the eigenvalue curves as a function of bappeared to be smooth curves. To find the most suitable discretisation, the eigenvalue analysis was performed for a particular value ofn. The number of points in the discretisation was then doubled to 2nand the eigenvalue analysis performed again. If the two eigenvalue analyses were qualitatively similar, then the discretisation of npoints was seen as sufficiently accurate for the stability analysis. In general, a discretisation of the domain [−10π,10π] into 1999 points was found to be sufficiently accurate.

Gap junction model

In Chapter 3, a sixth-order ODE was derived, solutions of which are steady state solutions of the time-dependent gap junction model in Equation (7.28). The ODE is

−κ2u(vi)+ 1 + 2κ2(b2−1)u(iv)− 2(b2−1) +κ2(b2+ 1)2u00

+(b2+ 1)2u= 4b(b2+ 1)f(u). (7.147)

For ease of reading, letc1= 1+2κ2(b2−1),c2 =− 2(b2−1)+κ2(b2+1)2

andc3= (b2+1)2. Then Equation (7.147) becomes

We begin by using the ODE to write the time-dependent gap junction model in Equa- tion (7.28) in the form

−κ2 ∂ 6 ∂x6 +c1 ∂4 ∂x4 +c2 ∂2 ∂x2 +c3 1 + ∂ ∂t u(x, t) = 4b(b2+ 1)f(u(x, t)). (7.149)

Following the analysis for the Amari model, we have the equation

E=−Id+ 4b(b2+ 1)L−1Df (7.150) where L is the discretised version of −κ2 ∂

6 ∂x6 +c1 ∂4 ∂x4 +c2 ∂2 ∂x2 +c3 and E is an n×n matrix. We find the eigenvalues ofE for a steady state for a particular value ofb. Again, one eigenvalue will be zero. If any eigenvalue is positive then the steady state is unstable. If all the eigenvalues are nonpositive, then the steady state is stable.

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