Based on Turing’s equation for reaction and diffusion, we can write the following system of partial differential equations:

∂y/∂t = f(y, x) + D∇^2y

∂x/∂t = g(y, x) + C∇^2x

where f(y, x) and g(y, x) are the reaction terms describing the population growth and the information density, respectively, and D and C are the diffusion coefficients for the population and the information density. The Laplacian operator (∇^2) represents the diffusion of the respective variable in space.

For the reaction terms, we can use a logistic growth model for the population:

f(y, x) = ry(1 - y/K)

where r is the intrinsic growth rate, K is the carrying capacity (maximum population density), and y/K is the effect of density-dependent regulation on the population growth.

For the information density term, we can use a linear model:

g(y, x) = ax

where a is a constant that determines the effect of information density on population growth.

Solving for the steady-states of the system (when ∂y/∂t = 0 and ∂x/∂t = 0) gives:

y = K

x = 0

This means that the population reaches its maximum density at the carrying capacity when there is no information in the environment. This is because there is no regulation of the population growth, so it grows until it reaches the available resources.

On the other hand, when there is a high density of information in the environment, the population density is lower and more stable. The behavior of small perturbations around this equilibrium depends on the diffusion coefficients, D and C. If D is much larger than C, the population is more diffusive and can respond more quickly to changes in the environment. If C is much larger than D, the information density is more diffusive, and changes in the environment may take longer to affect the population density.

Overall, this model suggests that the presence of information in the environment can regulate population growth, and the analysis of steady-states and perturbations can reveal important insights about the behavior of the system.

To perform a linear stability analysis, we need to assume that the population growth is near some stable equilibrium point, in which case small perturbations around the equilibrium point will grow or decay exponentially. We can write the equation above as:

∂x/∂t = D∇²x + f(x)

where ∇² is the Laplacian operator for the concentration and f(x) is a non-linear function describing the growth of the population in terms of the substance concentration.

To analyze the stability of the equilibrium point, we need to first set x = x0 + ε(x) where x0 is the equilibrium concentration and ε(x) is a small perturbation, and then linearize the equation above around the equilibrium point. The resulting equation is:

∂ε(x)/∂t = D∇²ε(x) + f’(x0)ε(x)

where f’(x0) is the derivative of the function f(x) evaluated at the equilibrium point x0. The term D∇²ε(x) is the diffusion term while f’(x0)ε(x) is the linearized reaction term.

If f’(x0) < 0, the system is stable and any perturbations will decay exponentially, whereas if f’(x0) > 0, the system is unstable and perturbations will grow exponentially.

In conclusion, the mathematical model for human population growth using Turing’s paper on reaction and diffusion is given by the partial differential equation ∂x/∂t = D(∂²x/∂x² + ∂²x/∂y²) + r, where x(x,y,t) is the concentration of an information substance and r is the rate at which this substance is produced by the population. To perform a linear stability analysis, we linearize the equation around an equilibrium point and analyze the sign of the linearized reaction term f’(x0) to determine the stability of the system.
To perform a linear stability analysis, we assume small perturbations around the steady state solution (y*, x*). We write:

y = y* + δy

x = x* + δx

We substitute these expressions into the system of equations, neglecting terms containing products of δy and δx. Linearizing the system around the steady state yields:

∂δy/∂t = (∂f/∂y)|* δy + (∂f/∂x)|* δx + D∇^2δy

∂δx/∂t = (∂g/∂y)|* δy + (∂g/∂x)|* δx + C∇^2δx

where the symbol |* denotes evaluation at the steady state. To simplify notation, we define:

A = (∂f/∂y)|

B = (∂f/∂x)|

C = (∂g/∂y)|

D = (∂g/∂x)|

We also make the ansatz:

δy(x,t) = Ψ(x)exp(λt)

δx(x,t) = Φ(x)exp(λt)

Substituting these ansatzes into the linearized system, we obtain the following equations for Ψ and Φ:

λΨ = -D∇^2Ψ - AΨ - BΦ

λΦ = -CΨ - D∇^2Φ

We can write these equations in matrix form:

(λ + D∇^2) | -A -B |

| -C (λ + D∇^2)| |Ψ| |0 |

|Φ|

We can then solve for the eigenvalues of this matrix by setting its determinant to zero:

(λ + D∇^2)(λ + D∇^2) + AC = 0

This equation can be rewritten as a second-order differential equation for λ^2:

D^2∇^4 + (2D^2AC - 2DΔ)∇^2 + ACD^2 = 0

where Δ = ∂^2/∂x^2 + ∂^2/∂y^2 is the Laplacian operator.

This is a homogeneous equation with constant coefficients, which can be solved by assuming solutions of the form exp(kx), leading to a characteristic equation:

D^2k^4 + (2D^2AC - 2Dk^2)k^2 + ACD^2 = 0

This equation can be factored to yield the eigenvalues:

λ = ±ik^2, where k^2 is a root of the quadratic:

k^4 + (2AC/D)k^2 + C/D = 0

We can solve for the roots using the quadratic formula:

k^2 = (-2AC/D ± sqrt((2AC/D)^2 - 4(C/D))) / 2

k^2 = -AC/D ± sqrt((AC/D)^2 - C/D)

Since k^2 is real and positive, the square root can be written as a real number, and the eigenvalues are given by:

λ = ± i√k^2

λ = ± i√(-AC/D ± sqrt((AC/D)^2 - C/D))

These are the main results of the linear stability study. The sign of the real part of the eigenvalues determines the stability of the steady state solution. If the real part is negative, the perturbations decay and the steady state is stable. If the real part is positive, the perturbations grow and the steady state is unstable. If the real part is zero, further analysis is needed, because the perturbations neither grow nor decay.
To perform a linear analysis, we assume that the variables y and x are small perturbations around a steady state solution, such that y = y* + δy and x = x* + δx, where y* and x* are the steady state values. Substituting these expressions into the partial differential equations and neglecting terms of order higher than δy and δx yields:

∂δy/∂t = ry(1 - y/K)δy + D∇^2(δy)

∂δx/∂t = ay δx + C∇^2(δx)

We can now use the method of separation of variables to find solutions of the form δy(x, t) = Y(x)T(t) and δx(x, t) = X(x)T(t). Substituting these into the equations above and dividing by YT yields:

(1/T)dT/dt = ry(1 - y/K) + D(Y’‘/Y)

(1/T)dX/dt = ay* + C*(X’'/X)

Since the left-hand sides are functions of t only, and the right-hand sides are functions of x and y, we can set them equal to a constant λ to obtain two ordinary differential equations:

d^2Y/dx^2 + (λ/D)Y = 0

d^2X/dx^2 + (λ/C - ay)X = 0

These are Sturm-Liouville equations, whose solutions depend on the boundary conditions. For simplicity, we assume periodic boundary conditions on a one-dimensional interval of length L, such that Y(L) = Y(0) and X(L) = X(0). The general solutions are:

Y(x) = Asin(kx) + Bcos(kx)

X(x) = Cexp(-sx) + Dexp(sx)

where k = sqrt(λ/D) and s = sqrt(λ/C - ay). The constants A, B, C, and D can be determined from the boundary conditions.

We now impose the condition that the eigenvalue λ must be real and negative for stability. This means that the solutions δy and δx must decay exponentially in time, and any small perturbations must be damped out. From the equations above, we see that λ/D and λ/C - ay must both be positive for this condition to hold.

Therefore, the eigenvalues λ are given by:

λ = n^2 π^2 D/L^2, n = 1, 2, 3, …

for the population density, and

λ = (s^2 + ay) C

for the information density.

For stability, we require that λ < 0. The first condition is always satisfied, since D, L, and n^2 are positive. The second condition becomes:

s^2 + ay < 0

Since s^2 > 0, this requires ay < 0, which means that the sign of a and y must be opposite. In other words, if the information density has a positive effect on population growth (a > 0), then the steady state population density must be below the carrying capacity (y < K). Conversely, if the information density has a negative effect on population growth (a < 0), then the steady state population density must be above the carrying capacity (y* > K).

Therefore, the system is stable if the information density has a negative effect on population growth (a < 0), or if the effect is positive but the population density is below the carrying capacity (y* < K). If the effect is positive and the population density is above the carrying capacity (y* > K), the system is unstable and prone to oscillations or even population crashes.
explain the steps to writing code for matlab for this system