Equilibria of reversible processes
First, simple example
!mathematical-methods-lecturenotes, p.50 x K − x =
d x d t = − α x + β ( K − x ) = f ( x ) f ( x ) = β K − ( α + β ) x Equilibria
FInd the equilibria (f ( x ) = 0
β K = ( α + β ) x $ $ \hat{x} = {\beta \over \alpha + \beta}K
You can't use 'macro parameter character #' in math mode There is only one equilibrium here, since there is only one solution to $f(x)$ (linear function). ### Stability $f'(x)=-(\alpha + \beta)$ This is always negative for all possible values of x, so the only equilibrium is globally stable. ## Second example !mathematical-methods-lecturenotes, p.51 ### Equilibria $\hat{c}$ is the solution to the quadratic equation - no easy way to do this, needs the quadratic formula. We will just not solve it but know that's what it is. ### Stability $f'(c) = -k_1(a_0-c) - k_1(b_0-c) - k_2 \lt0$ This is always negative, since $a_0 - c$ and $b_0 - c$ are both concentrations and must be positive, and the rates $k_1,k_2$ are both positive. This means that any equilibrium that exists must be stable, since the derivative is always negative Since equilibria always must alternate between stable and unstable, there must only be one true equilibrium. This is a problem, since the quadratic formula will give us two solutions. Let's plot the function to examine: $a_0=0.5, b_0=0.6, k_1 = 0.8, k_2 = 0.4$ ```desmos-graph left=-0.3;right=1.5; bottom=-1;top=1 --- y = 0.8(a-x)(b-x) - 0.4x a=0.5 b=0.6 (0, 0.8*a*b)|label: k_1*a_0*b_0 (a, -(0.4 * a))|label: The actual maximum value of c ``` Looks like theres two solutions, but due to conservation of mass, A + B = C must be true. So c must be less than or equal to the mnimum of So actually, the second (higher) root around $c = 1.4$ isn't real, since c must be less than or equal to the minimum of A and B. $$c \le min(a_0,b_0) There is only one equilibrium here, since there is only one solution to $f(x)$ (linear function). ### Stability $f'(x)=-(\alpha + \beta)$ This is always negative for all possible values of x, so the only equilibrium is globally stable. ## Second example !mathematical-methods-lecturenotes, p.51 ### Equilibria $\hat{c}$ is the solution to the quadratic equation - no easy way to do this, needs the quadratic formula. We will just not solve it but know that's what it is. ### Stability $f'(c) = -k_1(a_0-c) - k_1(b_0-c) - k_2 \lt0$ This is always negative, since $a_0 - c$ and $b_0 - c$ are both concentrations and must be positive, and the rates $k_1,k_2$ are both positive. This means that any equilibrium that exists must be stable, since the derivative is always negative Since equilibria always must alternate between stable and unstable, there must only be one true equilibrium. This is a problem, since the quadratic formula will give us two solutions. Let's plot the function to examine: $a_0=0.5, b_0=0.6, k_1 = 0.8, k_2 = 0.4$ ```desmos-graph left=-0.3;right=1.5; bottom=-1;top=1 --- y = 0.8(a-x)(b-x) - 0.4x a=0.5 b=0.6 (0, 0.8*a*b)|label: k_1*a_0*b_0 (a, -(0.4 * a))|label: The actual maximum value of c ``` Looks like theres two solutions, but due to conservation of mass, A + B = C must be true. So c must be less than or equal to the mnimum of So actually, the second (higher) root around $c = 1.4$ isn't real, since c must be less than or equal to the minimum of A and B. $$c \le min(a_0,b_0) In this case, the minimum of these two is a 0 = 0.5
Predator-prey dynamics with functional resonse
!mathematical-methods-lecturenotes, p.55
Equilibria:
Trivial equilibrium N ^ = 0 N ^ = − 1 / β T N ^ = K f ( x ) = β P
Stable equilibrium at the intersection of this line and the parabola when
Special equilibrium
If there was a very precise number of predators such that β P P F
Bifurcation Diagram
This can be used to show the stability of the equilibria:.
We plot N as a function of P:mathematical-methods-lecturenotes, p.57 P T = 0