text:
### A Pluto.jl notebook ###
# v0.17.7

using Markdown
using InteractiveUtils

# This Pluto notebook uses @bind for interactivity. When running this notebook outside of Pluto, the following 'mock version' of @bind gives bound variables a default value (instead of an error).
macro bind(def, element)
    quote
        local iv = try Base.loaded_modules[Base.PkgId(Base.UUID("6e696c72-6542-2067-7265-42206c756150"), "AbstractPlutoDingetjes")].Bonds.initial_value catch; b -> missing; end
        local el = $(esc(element))
        global $(esc(def)) = Core.applicable(Base.get, el) ? Base.get(el) : iv(el)
        el
    end
end

# ╔═╡ 8c3e21c9-6514-4f14-8b1f-325d124681f5
using DifferentialEquations

# ╔═╡ ac95a2ae-5b1b-4399-8265-4b0bfb21162e
using ParameterizedFunctions

# ╔═╡ 2e6401b9-9c89-4b4b-8492-d0a83003579b
using PlutoUI,Plots, HypertextLiteral

# ╔═╡ 279eaa4f-41ca-4168-b492-f36543bb8204
md" ## Equilibrium And Stability

The model for Logistic Growth is simple enough that it can be solved analytically. Other more complex models cannot be solved, and therefore we have to use other tools to study them. The most common is the Linear Stability Analysis.

Linear stability analysis is a method that allows us to study how a system behaves near an equilibrium point. It will help us to know if equilibrium is stable or unstable, and the bifurcations that occur in these equilibrium points, based only on a simplified linearized version of the system of study:

Let's illustrate how linear stability analysis works this with the logistioc model,.

```math
\begin{align*}
\frac{\mathrm{d} N}{\mathrm{d} t}=\mu N(1 -\frac{N}{K}) \tag{17}
\end{align*}
```

The main equation can be written in the following generic form:

```math
\begin{eqnarray}
\frac{\partial N}{\partial t} = f(N)
\end{eqnarray}
```

where N is a variable and $f(N)$  is a functions which governs its temporal evolution.

The first step is to calculate the fixed points of Eq. \ref{base1}. In continuous systems, the steady states occur when there is no change in the amount of our quantity $N$

```math
\begin{eqnarray}
\frac{\partial \overline{N}}{\partial t} = 0 = f(\overline{N})
\end{eqnarray}
```

Where we denote $\overline{N}$ as the value of our variable in steady state."

# ╔═╡ 00037d0a-8be0-4c75-94bc-23c2c639e6e8
md"

Let's find the equilibrium points of our logistic model.

```math
\begin{align*}
\frac{\mathrm{d} N}{\mathrm{d} t}= \mu N(1 -\frac{N}{K}) \tag{17}
\end{align*}
```
By taking the derivate to zero:

```math
\begin{align*}
\frac{\mathrm{d} \overline{N}}{\mathrm{d} t}=0= \mu \overline{N}(1 -\frac{\overline{N}}{K}) \tag{17}
\end{align*}
```

 we can easily see that we have two equilibrium states:
```math
\begin{align*}
\overline{N}=0\\
1 -\frac{\overline{N}}{K} = 0 \Rightarrow \overline{N} = K
\end{align*}
```

"

# ╔═╡ 64dc539b-b268-43b6-a96e-2c4fa7b48474
md" The next step after finding the equilibrium points is to check if they are stable or unstable, or what type of equilibrium we have. We do that by introducing small perturbations.

Let’s take a fixed point $\overline{N}$ and perturb it by an infinitesimal amount n. We are interested in the dynamics of N = $\overline{N}$ + n and whether N will move away away or towards $\overline{N}$ as time progresses.

```math
\begin{align*}
\frac{\mathrm{d} N}{\mathrm{d} t}= \frac{\mathrm{d} \overline{N} + n}{\mathrm{d} t}= \frac{\mathrm{d} \overline{N} }{\mathrm{d} t} + \frac{\mathrm{d} n}{\mathrm{d} t}
\end{align*}
```

and since

```math
\begin{align*}
\frac{\mathrm{d} \overline{N} }{\mathrm{d} t} = 0
\end{align*}
```
by definition, we have that the change in time of the population is equivalent to the change in the time of the small perturbation.

```math
\begin{align*}
\frac{\mathrm{d} N}{\mathrm{d} t}= \frac{\mathrm{d} n}{\mathrm{d} t}
\end{align*}
```

Since $n$ is very small by definition, we can linearize the dynamics around the ficxed points $\overline{N}$, using a Taylor expansion
.

```math
\begin{align*}
\frac{\mathrm{d} N}{\mathrm{d} t}= \frac{\mathrm{d} n}{\mathrm{d} t} = f(\overline{N}+n)= f(\overline{N})+ f'(\overline{N})·n +···
\end{align*}
```

being $f'(\overline{N})$ the value of derivative of function $f(N)$ in the point of equilibrium $\overline{N}$

Since $f(\overline{N})$ = 0 by definition, we get

```math
\begin{align*}
 \frac{\mathrm{d} n}{\mathrm{d} t} = f'(\overline{N})·n +···
\end{align*}
```
"

# ╔═╡ f5faefd6-f496-40fb-8043-ef6b9d172ae0
md"

If the perturbation $n$ is very small, then linear and nonlinear evolution are in fact approximately the same. But as $n$ increases in size the nonlinear effects become increasingly more important and evolving y with the linearized dynamics or the full nonlinear dynamics is no longer equivalent. So, this means that if perturbations $n$ are small, we can forget about higher order terms and simply assume that this has a form of $\dot{n}= \lambda \cdot n$, so if we integrate we have a solution that is an exponential fucntion.

```math
\begin{align*}
n(t)=n(0) e^{\lambda t}
\end{align*}
```

So, depending on the sign of $\lambda$, this perturbation $n$ may increase ($\lambda >0$) or decrease ($\lambda <0$). As an example, let's test this for the logistic system.

```math
\begin{align*}
\frac{\mathrm{d} N}{\mathrm{d} t}=\mu N(1 -\frac{N}{K}) = \frac{\mathrm{d} (\mu \overline{N}(1 -\frac{\overline{N}}{K})}{\mathrm{d} t} n \tag{17}
\end{align*}
```
For simplicity, we will decompose the fucntion in to two terms:
```math
\begin{align*}
\mu N(1 -\frac{N}{K}) = g(N) \cdot h(N)
\end{align*}
```
being

```math
\begin{align*}
g(N) &= \mu N \\
h(N) &= 1 -\frac{N}{K}
\end{align*}
```

now we need the detivative
```math
\begin{align*}
\frac{\mathrm{d} [g(N) \cdot h(N)]}{\mathrm{d} t}= \frac{\mathrm{d} g(N) }{\mathrm{d} t}  h(N) + g(N) \frac{\mathrm{d} h(N) }{\mathrm{d} t}
\end{align*}
```

since
```math
\begin{align*}
\frac{\mathrm{d} g(N) }{\mathrm{d} t} &=\mu\\
\frac{\mathrm{d} h(N) }{\mathrm{d} t} &= -\frac{1}{K}
\end{align*}
```

so,

```math
\begin{align*}
\frac{\mathrm{d} (\mu N(1 -\frac{N}{K})}{\mathrm{d} t} =\mu (1 -\frac{N}{K}) - \frac{\mu \cdot N}{K}
\end{align*}
```

"

# ╔═╡ 32b3bc77-c4f4-40ab-a5a9-e035022189ac
md"
now we substitute our steady state values, for $\overline{N}=0$, we have

```math
\begin{align*}
\frac{\mathrm{d} (\mu \overline{N}(1 -\frac{\overline{N}}{K})}{\mathrm{d} t}\Biggr\rvert_{\overline{N}=0} =\mu (1 -\frac{0}{K}) - \frac{\mu \cdot 0}{K} = \mu
\end{align*}
```

So, as long as $\mu > 0$ this fixed point is unestable. If we have $N=0$, the systems remains there. As soon as we perturb the number (and we can only perturb by slightly increasing, these perturbations grow exponentially. For the second steady state.

```math
\begin{align*}
\frac{\mathrm{d} (\mu \overline{N}(1 -\frac{\overline{N}}{K})}{\mathrm{d} t}\Biggr\rvert_{\overline{N}=K} =\mu (1 -\frac{K}{K}) - \frac{\mu \cdot K}{K} = - \mu
\end{align*}
```

So, as long as $\mu > 0$ this fixed point is stable. Perturbations will allways decrease exponentially.

To observe this grafically, lets plot the function [$\mu N(1 -\frac{N}{K}$]

"

# ╔═╡ c70c7a06-d1aa-40a9-902b-4bd4b23a6392
begin
	T_slide = @bind T html"<input type=range min=5 max=15 step=1>"
	K_slide = @bind K html"<input type=range min=100 max=500 step=10>"
	md"""
	**Set the Cell Cycle Length and the carrying capacity**

	value of T: $(T_slide)

	value of K: $(K_slide)

	"""
end

# ╔═╡ 10ba443a-d7bc-40f2-b4f6-540328ce6e4b
N=collect(0:0.1:K);

# ╔═╡ 2cfd31c4-803a-4880-b2a4-5af6594c0975
plot(N, log(2)/T .* N .* (1 .- (N ./ K)),label="T= $T, K= $K",seriestype=:line,xlabel=("N"),ylabel=("f(N)"),ylims = (0,20),xlims = (0,500))

# ╔═╡ 2170f883-1b01-4d16-907e-49c72118e224
md"
Equilibrium points are the values where the function is zero. We see the two of them, the unsable $\overline{N}=0$ and the stable $\overline{N}=K$. In this system, every perturbation moves away from $\overline{N}=0$ towards $\overline{N}=K$, which is the carrying capacity of the system.

The slope represents how fast the change occurs, so increasing $T$, means that we reach the carrying capacity of the system faster.

So, this is a liner model, and we can solve it analitically, so performinng a perturbation analysis does not provide extra information (we now the full dynamics because we have an analitical solution). The advantage of this perturbation analysis (or linear stabilty analysis) is when we work with systems that cannot be solved anallyically, such as systems with multiple variables, and systems with nonlinearities.

"

# ╔═╡ 6c87979a-690a-400b-9ab8-7ef26b829195
md"### Stability analysis of nonlinear systems: Lotka-Volterra model

The Lotka-Volterra model, also known as Predator-Prey model describes the inteactions between two populations of species where one feeds into the other.  One can think of rabbits $x$ and foxes $y$, such that rabits multiply where there is no foxes (assuming an infinite amount of food for the rabbits). This is basically a first order production of rabbits, an autocalalitic system that results in exponetial growth of rabbits.

Next, foxes $y$ feed on the rabbits and multiply due to the good food conditions. Then also foxes die at a constant rate (rabbits also die, but the mdoel assumes that their rate of birth is much higher than the rate of death). The scheme of interactions for this very simple system is the following:

```math
\begin{align}
 x &\overset{k_1}{\longrightarrow} 2 x   \\
 x + y &\overset{k_2}{\longrightarrow} 2 y \\
 y &\overset{k_3}{\longrightarrow} 0
 \end{align}
```

First, we find the differential equations that govern the dynamcis of the following system, and the steady state values. we start by writting the matrices of stoichiometric coefficients:

```math
A=\begin{bmatrix}
  1 & 0  \\
  1 & 1   \\
  0 & 1       \end{bmatrix} ;
B=\begin{bmatrix}
  2 & 0   \\
  0 & 2   \\
  0 & 0       \end{bmatrix}; \tag{8}
```

"

# ╔═╡ 04c14475-24cd-49c3-8a7b-83a9425664be
A = [1 0;1 1;0 1];B = [2 0;0 2;0 0];(B-A)'

# ╔═╡ 919bf3e4-6aa0-440b-b76f-34a4812c6752
md"in this particular case

```math
K=\begin{pmatrix}
 k_1 & 0 & 0  \tag{9}\\
 0 &  k_2 & 0  \\
 0 &  0 & k_3
\end{pmatrix}
```

and

```math
X^A=\begin{pmatrix}
X_1^1\cdot X_2^0  \\
X_1^1\cdot X_2^1  \\
X_1^0\cdot X_2^1
\end{pmatrix} = \begin{pmatrix}
 X_1 \\
 X_1 \cdot X_2\\
 X_2
\end{pmatrix} \tag{10}
```

so, the equations that define the system are

```math
\begin{align}
 \begin{bmatrix}
\frac{\mathrm{d} X_1}{\mathrm{d} t}\\ \frac{\mathrm{d} X_2}{\mathrm{d} t} \end{bmatrix}&
=  \begin{bmatrix} 1 & -1 & 0  \\ 0 & 1 &  -1  \end{bmatrix}
\begin{pmatrix}
 k_1 & 0 & 0  \tag{9}\\
 0 &  k_2 & 0  \\
 0 &  0 & k_3
\end{pmatrix}
 \begin{pmatrix}
 X_1 \\
 X_1 \cdot X_2\\
 X_2
\end{pmatrix}
\end{align}
```
and after multiplying the matrices
```math
\begin{align}
 \begin{bmatrix}
\frac{\mathrm{d} X_1}{\mathrm{d} t}\\ \frac{\mathrm{d} X_2}{\mathrm{d} t} \end{bmatrix}&
=  \begin{bmatrix} 1 & -1 & 0  \\ 0 & 1 &  -1  \end{bmatrix}
 \begin{pmatrix}
 k_1 \cdot X_1 \\
 k_2 \cdot X_1 \cdot X_2\\
 k_3 \cdot X_2
\end{pmatrix}
\end{align}
```

so finally,

```math
\begin{align}
 \begin{bmatrix}
\frac{\mathrm{d} X_1}{\mathrm{d} t}\\ \frac{\mathrm{d} X_2}{\mathrm{d} t}
\end{bmatrix}&=
 \begin{pmatrix}
   k_1 \cdot X_1 - k_2 \cdot X_1 \cdot X_2 \\
  k_2 \cdot X_1 \cdot X_2 - k_3 \cdot X_2
\end{pmatrix} \tag{11}
\end{align}
```

Therefore, the  equations for the evolution of `[x]` and `[y]` are as follows:

```math
\begin{align}
            \frac{ dx }{dt} &=  k_1 \cdot x - k_2 \cdot x \cdot y  \tag{5}\\
            \frac{ dy }{dt} &= k_2 \cdot x \cdot y - k_3 \cdot y  \tag{6}
            \end{align}
```

"

# ╔═╡ 855b8530-861e-427e-b6cc-c1b0283568ca
md"

which in general form, we can write as:

```math
\begin{eqnarray}
\frac{\partial x}{\partial t} = f(x, y) \\
\frac{\partial y}{\partial t} = g(x, y)
\end{eqnarray}
```

where $f(x, y)$ and $g(x, y)$ are nonlinear equations that govern the temporal evolution and couple the behavior of the two variables $x$ and $y$:

Next, we need to calculate the fixed points:

```math
\begin{eqnarray}
\frac{\partial \overline{x}}{\partial t} =0= f(\overline{x},\overline{y})  \\
\frac{\partial \overline{y}}{\partial t}= 0 = g(\overline{x},\overline{y})
\end{eqnarray}
```

For the particular case of the Lotcka-Volterra, we just set eqs. 5 and 6 to zero

```math
\begin{align}
            k_1 \cdot \overline{x} - k_2 \cdot \overline{x} \cdot \overline{y}  \tag{5} &= 0\\
            k_2 \cdot \overline{x} \cdot \overline{y} - k_3 \cdot \overline{y}  \tag{6} &= 0
            \end{align}
```

and solve for `x` and `y`. We obtain two solutions,  $\overline{x}=\overline{y}=0$ and

```math
\begin{align}
            \overline{x} &= \frac{k_3}{k_2}\tag{5} \\
            \overline{y} &= \frac{k_1}{k_2} \tag{6}
            \end{align}
```

	"

# ╔═╡ c112e62c-9c38-48dc-8294-42911b02ccc5
md" The next step is to find the characteristics of the steady states. We do that by following the same rationalle of the logistic model, i.e., to expand our equations as Taylor series around the steady state $(\overline{x},\overline{y})$., but now for multi-variable equations.

```math
\begin{eqnarray}
\frac{\partial x}{\partial t} = M_{11} \cdot x + M_{12} \cdot y + ... \\
\frac{\partial y}{\partial t} = M_{21} \cdot x + M_{22} \cdot y + ...
\end{eqnarray}
```

Where $M_{ij}$ are the components of the Jacobian matrix, evaluated at the steady state $(\overline{x},\overline{y})$.

```math
\begin{align}
 J=\begin{bmatrix}
 M_{11} & M_{12} \\
 M_{21} & M_{22}
 \end{bmatrix}_{\overline{x},\overline{y}}= \begin{bmatrix}
\frac{\partial  f(x,y)}{\partial x}\Biggr\rvert_{\overline{x},\overline{y}} & \frac{\partial  f(x,y)}{\partial y}\Biggr\rvert_{\overline{x},\overline{y}} \\
\frac{\partial  g(x,y)}{\partial x}\Biggr\rvert_{\overline{x},\overline{y}} & \frac{\partial  g(x,y)}{\partial y}\Biggr\rvert_{\overline{x},\overline{y}}
 \end{bmatrix}
 \end{align}
```

which for the particular case of the Lotka-Volterra is

```math
\begin{align}
 J=\begin{bmatrix}
 M_{11} & M_{12} \\
 M_{21} & M_{22}
 \end{bmatrix}_{\overline{x},\overline{y}} = \begin{bmatrix}
 k_1 - k_2 \cdot \overline{y}  &  - k_2 \cdot \overline{x}   \\
k_2  \cdot \overline{y}  & k_2 \cdot \overline{x}  - k_3
 \end{bmatrix}
 \end{align}
```

Next, to investigate the stability, we check solutions in the form of small perturbations as follows:

```math
\begin{eqnarray}
(x,y) = (X_0,Y_0) e^{\lambda t}
\end{eqnarray}
```
Here, $\lambda$ is the growth rate of the perturbations, also refered as eigenvalue. Each steady state will behave differently in terms of the dynamcis of the perturbations. Therefore, each steady state will have an associated eigenvalue. To find the eigen values, we solve the characteristic polynomial $det[J-\lambda I]=0$.

```math
\begin{eqnarray}
Det \left(\begin{array}{cc}M_{11}-\lambda & M_{12} \\M_{21}& M_{22}-\lambda \end{array}\right) =0
\end{eqnarray}
```
which gives us the corresponding equation:
```math
\begin{eqnarray}
\lambda^{2} -\lambda Tr(M) + Det(M)=0
\end{eqnarray}
```
where:
```math
\begin{eqnarray}
Tr(M)= M_{11}+M_{22} \\
Det(M)= M_{11}M_{22}-M_{12}M_{21}
\end{eqnarray}
```

"

# ╔═╡ 6c96da07-b5ea-4096-a14c-5cd8f0f47c04
md"for the lotka-volterra case:

```math
\begin{eqnarray}
Tr(M)= k_{1} + k_2 (\overline{x}-\overline{y}) - k_3  \\
Det(M)= (k_1- k_2 \cdot \overline{y})(k_2 \cdot \overline{x} - k_3) - (k_2 \cdot\overline{y})(-k_2 \cdot \overline{x})  )
\end{eqnarray}
```

calculating

```math
\begin{eqnarray}
Tr(M)= k_{1} + k_2 (\overline{x}-\overline{y}) - k_3  \\
Det(M)= - k_1 \cdot k_3 - k_2^2 \cdot \overline{y} \cdot \overline{x} + k_2  \cdot k_3 \cdot \overline{y} + k_1 \cdot k_2 \cdot \overline{x}+ k_2^2 \cdot\overline{y} \cdot \overline{x}
\end{eqnarray}
```
and

```math
\begin{eqnarray}
Tr(M)= k_{1} + k_2 (\overline{x}-\overline{y}) - k_3  \\
Det(M)= k_2  \cdot k_3 \cdot \overline{y} + k_1 \cdot k_2 \cdot \overline{x} - k_1 \cdot k_3
\end{eqnarray}
```

so the chracteristic equation becomes:
```math
\begin{eqnarray}
\lambda^{2} -\lambda (k_{1} + k_2 (\overline{x}-\overline{y}) - k_3) + k_2  \cdot k_3 \cdot \overline{y} + k_1 \cdot k_2 \cdot \overline{x} - k_1 \cdot k_3=0
\end{eqnarray}
```

So, the polynomium for the fixed point $\overline{x},\overline{y}=[0,0]$ is

```math
\begin{eqnarray}
\lambda^{2} +\lambda (k_3 - k_{1})  - k_1 \cdot k_3=0
\end{eqnarray}
```

"

# ╔═╡ 906a0931-e86f-48f3-be11-eb692639e8cc
md" Again, since $\lambda$ is the exponent that sets the dynamcis of the perturbations, depending on its value, the steady state is stable or unstable.

For our Locka Volterra case, we can evaluate first the eigenvalues for the first steady state  ,

"

# ╔═╡ c374e1bd-ec3d-4cf7-aff1-072d6ba60b27
function quadratic(a, b, c)
          discr = b^2 - 4*a*c
          discr >= 0 ?   ( (-b + sqrt(discr))/(2a), (-b - sqrt(discr))/(2a) ) : error("Only complex roots")
        end

# ╔═╡ 2b91a06c-3803-4877-ac9a-1669fbe58d30
begin
	k1_slide = @bind k1 html"<input type=range min=1 max=5 step=.1>"
	k2_slide = @bind k2 html"<input type=range min=1 max=5 step=.1>"
	k3_slide = @bind k3 html"<input type=range min=1 max=5 step=.1>"
	md"""
	**Set the values of the kinetic constants**

	value of k1: $(k1_slide)

	value of k2: $(k2_slide)

	value of k3: $(k3_slide)

	"""
end

# ╔═╡ e8ca5682-cce0-4b2c-ba18-47f2e38192c7
begin
	a= 1
	b= k3 - k1
	c= - k1 * k3
	quadratic(a,b,c)
end

# ╔═╡ 17e34ce6-a1fa-4193-9201-3e61a188b48f
md" For this steady state and for any combination of parameters, one eigen value is positive and the other is negative. It means that perturbations in one variable grow while perturbations in the other variable decay.

The stability of this fixed point [0,0] is of importance. If it both ewigenvalues are negative, the point will be stable, and non-zero populations might be attracted towards it, and as such the dynamics of the system might lead towards the extinction of both species for many cases of initial population levels.

However, as the steady state at the origin is unstable in one of the variables, we find that the extinction of both species is difficult in the model. In fact, in teh absence of foxes and rabits, a small increase in the foxes will lead to extintion (no food), while a small increase in the amount of rabbits will read to exponential increase (the unstable branch).

These type of points are called a saddle node (a minimum in one variabel and a maximum in the other).

For the other solution $[\overline{x},\overline{y}]=[\frac{k_3}{k_2},\frac{k_1}{k_2}]$

so the chracteristic equation becomes:
```math
\begin{eqnarray}
\lambda^{2} -\lambda (k_{1} + k_2 (\frac{k_3}{k_2}-\frac{k_1}{k_2}) - k_3) + k_2  \cdot k_3 \cdot \frac{k_1}{k_2} + k_1 \cdot k_2 \cdot \frac{k_3}{k_2} - k_1 \cdot k_3=0
\end{eqnarray}
```

```math
\begin{eqnarray}
\lambda^{2}  + \cdot k_3 \cdot k_1 =0
\end{eqnarray}
```

"

# ╔═╡ 87df7c88-7e16-4c50-bd80-df5bfea26ee8
begin

	aa=1
	bb= 0
	cc=- k1  + k2 *( k1 * k3)
	quadratic(aa,bb,cc)
end

# ╔═╡ 9433f4d1-3abf-4dea-8107-1070ee5feb2e
md"The two values are purely imaginary so we cannot say much about the stability. A small perturbation will not experience repulsion or atraction towards this steady state. There is no stable state (no atractor), and trajectories circulate about the fixed point in a stable orbit. This is called a _center_.

The solutions travel periodically around the level sets in the counterclockwise direction

To test this, we solve numerically the system

We assume as initial conditions:

```math
\begin{align}
            x (0) &= 1 \tag{7} \\
            y (0) &= 1   \tag{8}
\end{align}
```

"

# ╔═╡ e6f51a60-9701-4ba6-a6e6-9703f7c9ebf9
https://cs.carleton.edu/faculty/awb/cs111/f19/worksheets/lotka-volterra/Lotka-Volterra%20equation.pdf

# ╔═╡ 2b9e43fd-d00d-40f1-ac98-07d48c53d861
md" Similarly to what we did in the previous case, we would try to see the fixed points graphically. To do that in two dimensional systems, we find the functions where $\dot{x} = 0$ and $\dot{y} = 0$. These lines will represent the boundaries  between increase and decrease in $x$ and $y$.

These curves are called the  nullclines. The method of nullclines is a technique for determining the global behavior of solutions of competing species models. This method provides an effective means of finding trapping regions for some differential equations. In a competition model, if a species population x is above a certain level, the fact of limited resources will cause x to decrease.

Let's illustrate this again with the Lotka-Volterra. The functions that satisfy that the defivatives of $x$ and $y$ are zero are:

```math
\begin{align}
            k_1 \cdot \overline{x} - k_2 \cdot \overline{x} \cdot \overline{y}  \tag{5} &= 0\\
            k_2 \cdot \overline{x} \cdot \overline{y} - k_3 \cdot \overline{y}  \tag{6} &= 0
            \end{align}
```
In this particuular case, the lines are very simple, just constant values.

```math
\begin{align}
            \overline{x} &= \frac{k_3}{k_2}\tag{5} \\
            \overline{y} &= \frac{k_1}{k_2} \tag{6}
            \end{align}
```

"

# ╔═╡ a5b38e90-cdbe-442f-a078-dd8598a0a4c2

md"

More concretely, if $Re(\lambda) < 0$, the perturbation decays in time and the steady state ($\overline{x},\overline{y}$) is stable.

On the contrary, when $Re(\lambda) > 0$, the perturbation grows exponentially and the steady sate is unstable.

More concretely, the steady state is stable if the following conditions are fulfilled:
```math
\begin{eqnarray}
Tr(M) < 0 \\
Det(M) >0
\end{eqnarray}
```

We can write the eigenvalue expression separating real and imaginary part:
```math
\begin{eqnarray}
\lambda=\mu \pm i \omega
\end{eqnarray}
```
where

```math
\begin{eqnarray}
\mu=\frac{1}{2} Tr(M)\\
\omega=\sqrt{-\frac{1}{4} Tr(M)^2 + Det(M)}
\end{eqnarray}
```

The Hopf bifurcation takes place when $Det(M) > (1 / 4) Tr(M)^2 $ and $Tr(M) > 0$. In this case the eigenvalue has nonzero imaginary part and the solution of the system is oscillatory. In the Hopf threshold ($M_{11}=-M_{22}$) the complex part of the eigenvalue becomes:

```math
\begin{eqnarray}
\omega^2= \omega_{c}^{2}=-M_{11}^{2}-M_{12}M_{21}>0 \\
M_{12}M_{21}>M_{11}^{2}  (>0)
\end{eqnarray}
```
One of the values must be positive, and the other negative. We choose $M_{11}>0$ $\longrightarrow$ $M_{22}<0$ and $M_{12}>0$ $\longrightarrow$ $M_{21}<0$. This way, we can write Eq. \ref{lineal1} and  Eq. \ref{lineal2} as follows:

```math
\begin{eqnarray}
\frac{\partial x}{\partial t} = M_{11} x - |M_{12}| y + ... \\
\frac{\partial y}{\partial t} = M_{21} x - |M_{22}| y + ...
\end{eqnarray}
```
"

# ╔═╡ 3349c942-15a8-4b1e-8b4a-7f5154f48b12
lv! = @ode_def LotkaVolterra begin
  dx = k1*x - k2*x*y
  dy =  k2*x*y - k3*y
    end k1 k2 k3

# ╔═╡ c3828c90-0559-4686-9f56-41b5b6d48176
begin
	x₀=1
	y₀=1
	t₀=0.0
	final_time=10.0;
	prob = ODEProblem(lv!,[x₀,y₀],(t₀,final_time),(k1,k2,k3))
	sol = solve(prob)
	plot(sol,ylims = (0, 5))

	title!("Lotka-Volterra ")
	xlabel!("time [a.u.]")
    ylabel!("Amplitude [a.u.]")

end

# ╔═╡ ff0c4a06-de7b-4cf2-b920-cd84ac9927ea
[u[1] for (u,t) in tuples(sol)]

# ╔═╡ f8238a6e-1e5e-4629-ac20-2c9cb964c53e
tuples(sol)

# ╔═╡ df81eb5c-16ef-4bf6-9ad0-2cc899c44cb6
begin
	vline([k3/k2],ylims = (0, 5),xlims = (0, 10));
	hline!([k1/k2],ylims = (0, 5),xlims = (0, 10));
	title!("Null-Clines of the Lotka Volterra ")
	xlabel!("x [a.u.]")
    ylabel!("y [a.u.]")
	plot!([u[1] for (u,t) in tuples(sol)],[u[2] for (u,t) in tuples(sol)],ylims = (0, 15))

prob2 = ODEProblem(lv!,[x₀*2,y₀*2],(0.0,10.0),(k1,k2,k3))
	sol2 = solve(prob2)
	plot!([u[1] for (u,t) in tuples(sol2)],[u[2] for (u,t) in tuples(sol2)],ylims = (0, 15))

end

# ╔═╡ cf72446a-a512-11ec-2b47-ef706c91c6a0
md" ## Linear stability analysis

To study the behavior of nonlinear systems, a set of mathematical tools is  commonly used. Here, we will outline its main aspects from a simplified point of view, trying to introduce the reader to the mathematics inside the nonlinear pattern formation field. In addition we will try to illustrate the problem using a very simple autocatalitic model: the _Cubic Autocatalor Model_.

### Stability of zero-Dimensional Systems

The main equation which governs the aspects of pattern formation systems is the following nonlinear equations:

```math
\begin{eqnarray}
\frac{\partial u}{\partial t} = f(\mu, u, v) \\
\frac{\partial v}{\partial t} = g(\mu, u, v)
\end{eqnarray}
```

where $u$ and $v$ correspond to the concentration of activator and inhibitor.
Here, $f(\mu, u, v)$ and $g(\mu, u, v)$ are nonlinear functions which govern the temporal evolution of the variables.

The first step is to calculate the fixed points of Eq. \ref{base1} and \ref{base2}, i.e., the values of the variables where the null-clines are in coincidence and equal to zero. This defines the steady state for the variables in a zero dimensional system.

```math
\begin{eqnarray}
f(\mu, u, v)=0  \\
g(\mu, u, v)=0
\end{eqnarray}
```

An example of the null-clines for the \textit{Cubic Autocatalor} Model can be seen in Fig. \ref{nullclines_cubic}. The equations for this specific model are:

```math
\begin{eqnarray}
\frac{\partial u}{\partial t} = u^2 v -u \\
\frac{\partial v}{\partial t} = \mu - u^2 v
\end{eqnarray}
```

The steady state for this model is ($u_0,v_0$)= ($\mu, 1/\mu$)."

# ╔═╡ 67091261-3b6c-4a94-9a70-86519a1eed76
md"
The following step to study the evolution of the system is to linearize Eq. \ref{base1} and \ref{base2} around the steady state $(u_{0},v_{0})$.

```math
\begin{eqnarray}
\frac{\partial u}{\partial t} = M_{11} u + M_{12} v + f_{2}(\mu, u, v) +   f_{3}(\mu, u, v) + ... \\
\frac{\partial v}{\partial t} = M_{21} u + M_{22} v + g_{2}(\mu, u, v) + g_{3}(\mu, u, v) + ...
\end{eqnarray}
```
where $M_{ij}$ is calculated in the steady state ($u_{0},v_{0}$) as follows:
```math
\begin{eqnarray}
M_{11} = \frac{\partial  f(\mu, u, v)}{\partial u}\\
M_{12} = \frac{\partial  f(\mu, u, v)}{\partial v} \\
M_{21} = \frac{\partial  g(\mu, u, v)}{\partial u}\\
M_{22} = \frac{\partial  g(\mu, u, v)}{\partial v}
\end{eqnarray}
```
To investigate the stability, we check solutions in the form of small perturbations as follows:
```math
\begin{eqnarray}
(u,v) = (U,V) e^{\lambda t}
\end{eqnarray}
```
Here, $\lambda$ is the growth rate of the perturbations. The next step is to solve the eigenvalue problem, resulting of the introduction of Eq. \ref{solucion1} in the linearized system:
```math
\begin{eqnarray}
Det \left(\begin{array}{cc}M_{11}-\lambda & M_{12} \\M_{21}& M_{22}-\lambda \end{array}\right) =0
\end{eqnarray}
```
and the corresponding equation:
```math
\begin{eqnarray}
\lambda^{2} -\lambda Tr(M) + Det(M)=0
\end{eqnarray}
```
where:
```math
\begin{eqnarray}
Tr(M)= M_{11}+M_{22} \\
Det(M)= M_{11}M_{22}-M{12}M_{21}
\end{eqnarray}
```

Depending on the value of $\lambda$, the solution \ref{solucion1} is stable or unstable. If $Re(\lambda) < 0$,the perturbation decays in time and the steady state ($u_{0},v_{o}$) calculated from \ref{base3} and \ref{base4} is stable. More concretely, the steady state is stable if the following conditions are fulfilled:
```math
\begin{eqnarray}
Tr(M) < 0 \\
Det(M) >0
\end{eqnarray}
```
On the contrary, when $Re(\lambda) > 0$, the perturbation grows exponentially and the steady sate is unstable. We can write the eigenvalue expression separating real and imaginary part:
```math
\begin{eqnarray}
\lambda=\mu \pm i \omega
\end{eqnarray}
```
where

```math
\begin{eqnarray}
\mu=\frac{1}{2} Tr(M)\\
\omega=\sqrt{-\frac{1}{4} Tr(M)^2 + Det(M)}
\end{eqnarray}
```

The Hopf bifurcation takes place when $Det(M) > (1 / 4) Tr(M)^2 $ and $Tr(M) > 0$. In this case the eigenvalue has nonzero imaginary part and the solution of the system is oscillatory. In the Hopf threshold ($M_{11}=-M_{22}$) the complex part of the eigenvalue becomes:

```math
\begin{eqnarray}
\omega^2= \omega_{c}^{2}=-M_{11}^{2}-M_{12}M_{21}>0 \\
M_{12}M_{21}>M_{11}^{2}  (>0)
\end{eqnarray}
```
One of the values must be positive, and the other negative. We choose $M_{11}>0$ $\longrightarrow$ $M_{22}<0$ and $M_{12}>0$ $\longrightarrow$ $M_{21}<0$. This way, we can write Eq. \ref{lineal1} and  Eq. \ref{lineal2} as follows:

```math
\begin{eqnarray}
\frac{\partial u}{\partial t} = M_{11} u - |M_{12}| v + ... \\
\frac{\partial v}{\partial t} = M_{21} u - |M_{22}| v + ...
\end{eqnarray}
```
$u$ plays the role of the activator due to the autocatalitic behavior (represented in Eq. \ref{lineal3})   and $v$ is the inhibitor, which decays by itself.

### Stability of Spatial Systems

The next step is to to consider the spatial dimensions of the system in Eq. \ref{base1} and Eq. \ref{base2}.

```math
\begin{eqnarray}
\frac{\partial u}{\partial t} = f(\mu, u, v)  + D_u \frac{\partial^{2} u}{\partial \vec{r}^{2}}\\
\frac{\partial v}{\partial t} = g(\mu, u, v) + D_v \frac{\partial^{2} v}{\partial \vec{r}^{2}}
\end{eqnarray}
```

Here, $D_u$ and $D_v$ are the diffusion coefficients of activator and inhibitor and $ \vec{r}$ is the spatial coordinate. We will scale the diffusion coefficients in a way that we can reduce to a variable which only takes account of the ratio between the diffusion coefficients: $d=D_{v}/D_{u}$. Now we have to check solutions with the spatial part:
```math
\begin{eqnarray}
(u,v) = (U,V) e^{\lambda t + i \vec{k}\vec{r}}
\end{eqnarray}
```
The Jacobian matrix $M$ of the system is:
```math
\begin{eqnarray}
M=\left(\begin{array}{cc}M_{11} - k^{2} & M_{12} \\M_{21}& M_{22} - d k^{2} \end{array}\right)
\end{eqnarray}
```
If we solve the eigenvalue problem ($Det (M-\lambda I)=0$), as in the previous case without spatial dimensions (unstable steady state) some other conditions are required to get positive eigenvalues. The equation is:
```math
\begin{eqnarray}
\lambda^2+\lambda(k^2(1+d)-Tr(M))+Det(M)=0
\end{eqnarray}
```
The solution is in the form:
```math
\begin{eqnarray}
\lambda=\frac{1}{2}(-k^{2}(1+d)+Tr (M) \pm
 \sqrt{(k^{2}(1+d)-Tr (M))^{2}-4 B}
 \end{eqnarray}
```
 where

```math
 \begin{eqnarray}
 d &=& \frac{D_{v}}{D_{u}} \\
 Tr (M) &=& M_{11} + M_{22} \\
 Det(M) &=& M_{11}  M_{22} - M_{21}  M_{12} \\
 B &=& d k^{4} - d k^{2} M_{11} - k^{2} M_{22} + Det (M)
\end{eqnarray}
```
So, the system will be unstable if one of the following conditions is fulfilled:
```math
\begin{eqnarray}
k^2(1+d)-Tr(M) < 0 \\
 Det(M) < 0
\end{eqnarray}
```
In addition, if the eigenvalues are positive and real, which means that $k^2(1+d)-Tr(M))^{2}> 4 Det(M)$, the system will grow exponentially (Turing bifurcation) The system, now with spatial dimensions, develops steady periodic patterns. There is a window of unstable wavelengths which the system may exhibit ($k$ with Re[$\lambda_{1,2}] > 0$). But there is one with maximum growth rate, which can be easily calculated by solving Eq. \ref{eigen1}:
```math
\begin{eqnarray}
\frac{\partial \lambda}{\partial k}=0
\end{eqnarray}
```

Fig.~\ref{Re_dispersion} is a plot of the real part of one of the the eigenvalues $\lambda_{1}$ which has a region of positive growth for some wavenumbers in the  Lengyel-Epstein model (see Sec.~\ref{sec:LE_model}). This means that a perturbation with a wavenumber with positive eigenvalue will grow exponentially in time. The other eigenvalue is negative, so it does not influence the behavior of the system. In addition Fig.~\ref{Im_dispersion} shows the imaginary part of both eigenvalues. Positive imaginary values of the growth rate are outside of the regime of positive real values in Fig.~\ref{Re_dispersion}, so the periodic pattern (with wavenumber $k$) is steady in time."

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"""

# ╔═╡ Cell order:
# ╟─279eaa4f-41ca-4168-b492-f36543bb8204
# ╟─00037d0a-8be0-4c75-94bc-23c2c639e6e8
# ╟─64dc539b-b268-43b6-a96e-2c4fa7b48474
# ╟─f5faefd6-f496-40fb-8043-ef6b9d172ae0
# ╟─32b3bc77-c4f4-40ab-a5a9-e035022189ac
# ╟─c70c7a06-d1aa-40a9-902b-4bd4b23a6392
# ╠═10ba443a-d7bc-40f2-b4f6-540328ce6e4b
# ╠═2cfd31c4-803a-4880-b2a4-5af6594c0975
# ╟─2170f883-1b01-4d16-907e-49c72118e224
# ╟─6c87979a-690a-400b-9ab8-7ef26b829195
# ╠═04c14475-24cd-49c3-8a7b-83a9425664be
# ╟─919bf3e4-6aa0-440b-b76f-34a4812c6752
# ╟─855b8530-861e-427e-b6cc-c1b0283568ca
# ╟─c112e62c-9c38-48dc-8294-42911b02ccc5
# ╟─6c96da07-b5ea-4096-a14c-5cd8f0f47c04
# ╟─906a0931-e86f-48f3-be11-eb692639e8cc
# ╠═c374e1bd-ec3d-4cf7-aff1-072d6ba60b27
# ╠═2b91a06c-3803-4877-ac9a-1669fbe58d30
# ╠═e8ca5682-cce0-4b2c-ba18-47f2e38192c7
# ╟─17e34ce6-a1fa-4193-9201-3e61a188b48f
# ╠═87df7c88-7e16-4c50-bd80-df5bfea26ee8
# ╟─9433f4d1-3abf-4dea-8107-1070ee5feb2e
# ╠═c3828c90-0559-4686-9f56-41b5b6d48176
# ╠═e6f51a60-9701-4ba6-a6e6-9703f7c9ebf9
# ╟─2b9e43fd-d00d-40f1-ac98-07d48c53d861
# ╠═ff0c4a06-de7b-4cf2-b920-cd84ac9927ea
# ╠═f8238a6e-1e5e-4629-ac20-2c9cb964c53e
# ╠═df81eb5c-16ef-4bf6-9ad0-2cc899c44cb6
# ╠═a5b38e90-cdbe-442f-a078-dd8598a0a4c2
# ╠═8c3e21c9-6514-4f14-8b1f-325d124681f5
# ╠═ac95a2ae-5b1b-4399-8265-4b0bfb21162e
# ╠═3349c942-15a8-4b1e-8b4a-7f5154f48b12
# ╠═cf72446a-a512-11ec-2b47-ef706c91c6a0
# ╠═67091261-3b6c-4a94-9a70-86519a1eed76
# ╠═2e6401b9-9c89-4b4b-8492-d0a83003579b
# ╟─00000000-0000-0000-0000-000000000001
# ╟─00000000-0000-0000-0000-000000000002

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