Systems of differential equations 2

1. $x + 3y = 0$ gives $x = -3y$. Substituting into $-6x + 5y = 0$ gives $18y + 5y = 0$, so $y = 0$ and $x = 0$. The fixed point is $(0, 0)$.
2. Multiply the first equation by 6 and add it to the second:

   $$(6x + 18y + 24) + (-6x + 5y - 1) = 23y + 23 = 0$$

   So, $y = -1$, and then $x = -4 - 3y = -1$. The fixed point is $(-1, -1)$.
3. In the same way, $(6x + 18y + 6) + (-6x + 5y) = 23y + 6 = 0$, so $y = -6/23$, and then $x = -1 - 3y = -5/23$. The fixed point is $(-5/23, -6/23)$.

1. $\dot{y} = 0$ gives $x^2 = 4y^2$, so $x = 2y$ or $x = -2y$. Substitute each into $\dot{x} = 0$.

   • $x = 2y$: $-4y + 4y^2 - 8 = 0$, so $y^2 - y - 2 = (y - 2)(y + 1) = 0$. This gives $y = 2$, $x = 4$ and $y = -1$, $x = -2$.
   • $x = -2y$: $-4y - 4y^2 - 8 = 0$, so $y^2 + y + 2 = 0$. This has no real solutions, because $1^2 - 4\times 2 < 0$.

   The fixed points are $(4, 2)$ and $(-2, -1)$.
2. $\dot{x} = 0$ gives $y = x^2 - 2$, and $\dot{y} = 0$ gives $y = x$ or $y = -x$.

   • $y = x$: $x^2 - x - 2 = (x - 2)(x + 1) = 0$, which gives $(2, 2)$ and $(-1, -1)$.
   • $y = -x$: $x^2 + x - 2 = (x + 2)(x - 1) = 0$, which gives $(-2, 2)$ and $(1, -1)$.

   There are four fixed points: $(2, 2)$, $(-1, -1)$, $(-2, 2)$ and $(1, -1)$.

1. A is produced at the constant rate $k_0$, turns into B at rate $k_1 a$, and is made from B at rate $k_{-1}b$. B is removed at rate $k_2 b$. So,

   $$\begin{aligned} \frac{da}{dt} &= k_0 - k_1 a + k_{-1} b \\ \frac{db}{dt} &= k_1 a - k_{-1} b - k_2 b \end{aligned}$$
2. In steady state $db/dt = 0$, so $k_1 a = (k_{-1} + k_2) b$, and

   $$\frac{a}{b} = \frac{k_{-1} + k_2}{k_1}$$

   We can also find the concentrations themselves. Adding the two equations gives $k_0 - k_2 b = 0$, so $b^* = k_0/k_2$, and then $a^* = k_0(k_{-1} + k_2)/(k_1 k_2)$.

1. On $y = x$, $\dot{x} = 0$ (this is an $x$-nullcline), so the flow crosses it vertically. $\dot{y} = (1-x)x$, so the flow is upwards for $x < 1$ and downwards for $x > 1$.
2. On $x = 0$, $\dot{x} = 0$ and $\dot{y} = y > 0$, so the flow is up the $y$ axis. With no parasites, the hosts grow exponentially.
3. On $y = 0$, $\dot{y} = 0$ (a $y$-nullcline). The equations are not defined on the line itself, because of $x/y$, but just above it $x/y$ is very large, so $\dot{x}$ is large and negative and the flow moves sharply to the left. With almost no hosts, the parasites die out.
4. On $x = 1$, $\dot{y} = 0$ (a $y$-nullcline), so the flow crosses it horizontally. $\dot{x} = \alpha(1 - 1/y)$, so the flow is to the right for $y > 1$ and to the left for $y < 1$.
5. On $y = \beta x$, $\dot{x} = \alpha x(1 - 1/\beta)$ and $\dot{y} = \beta x(1 - x)$. For $\beta > 1$ (above the line $y = x$) the flow is to the right, and for $\beta < 1$ (below it) the flow is to the left. In both cases it is upwards for $x < 1$ and downwards for $x > 1$.

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import numpy as np
from matplotlib import pyplot as plt

alpha = 1
x, y = np.meshgrid(np.linspace(0.01, 2.5, 30), np.linspace(0.01, 2.5, 30))
xdot = alpha * x * (1 - x / y)
ydot = (1 - x) * y

plt.streamplot(x, y, xdot, ydot, color='grey')
plt.plot([0, 2.5], [0, 2.5], color='#0072bd', label='x-nullcline y = x')
plt.axvline(0, color='#0072bd', label='x-nullcline x = 0')
plt.axvline(1, color='#F39200', label='y-nullcline x = 1')
plt.axhline(0, color='#F39200', label='y-nullcline y = 0')
plt.xlabel('x (parasites)')
plt.ylabel('y (hosts)')
plt.legend()

1. Let $F(T)$ be the number of fish and $A(T)$ the number of anglers at time $T$. Then

   $$\begin{aligned} \frac{dF}{dT} &= aF\left(1 - \frac{F}{K}\right) - bFA \\ \frac{dA}{dT} &= cF - eA \end{aligned}$$

   where all the constants are positive:

   • $a$ is the growth rate of the fish, and $K$ is the largest number of fish the lake can support (its carrying capacity);
   • $b$ measures how quickly anglers catch fish;
   • $c$ measures how strongly the fish attract anglers;
   • $e$ is the rate at which anglers are discouraged and leave.
2. As in the lecture, let $F = \theta x$, $A = \phi y$ and $T = \tau t$, so that $d/dT = (1/\tau)\,d/dt$. Then

   $$\begin{aligned} \dot{x} &= \tau a x\left(1 - \frac{\theta x}{K}\right) - \tau b\phi xy \\ \dot{y} &= \frac{\tau c\theta}{\phi}x - \tau e y \end{aligned}$$

   Choosing $\theta = K$, $\tau = 1/e$ and $\phi = e/b$ gives

   $$\dot{x} = \frac{a}{e}x(1 - x) - xy \qquad\text{and}\qquad \dot{y} = \frac{bcK}{e^2}x - y$$

   which is the model given, with $r = a/e$ and $\beta = bcK/e^2$.

1. Each step uses up one species and makes the next, so

   $$\begin{aligned} \frac{ds}{dt} &= -k_1 s, \\ \frac{di}{dt} &= k_1 s - k_2 i, \\ \frac{dp}{dt} &= k_2 i. \end{aligned}$$
2. Adding the three equations,

   $$\frac{d}{dt}(s + i + p) = -k_1 s + k_1 s - k_2 i + k_2 i = 0$$

   so $s + i + p = T$, where $T$ is a constant: the total amount of material. Then $p = T - s - i$, and the system reduces to the two ODEs for $s$ and $i$.
3. If $k_2 \gg k_1$, the intermediate is used up almost as soon as it is made, so $i$ quickly reaches a quasi-steady state. Solving $di/dt = 0$:

   $$k_1 s - k_2 i^* = 0 \qquad\Rightarrow\qquad i^* = \frac{k_1 s}{k_2}$$

   This is small, because $k_1/k_2$ is small, and it moves slowly with $s$. Substituting it into the equation for $p$,

   $$\frac{dp}{dt} = k_2 i^* \approx k_1 s$$
4. In this approximation, every molecule of $I$ is turned into $P$ almost straight away, so $P$ is made as fast as $I$ is made, which is $k_1 s$. The first step is the slow one, so it limits the rate of the whole chain. This is often called the rate-limiting step.