Differential equations 3

Steady State Solutions and Mass Action

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YouTube lecture recording from October 2020

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Steady-state solutions

• It is often useful to find the steady state of a system described by ordinary differential equations

• This occurs when all rates (i.e. derivatives) are zero

• A steady state may be stable or unstable, depending on whether or not small deviations from the steady state tend to be corrected or amplified

• We can evaluate the stability of a steady state by considering the sign of the derivative nearby

Example 1: radioactive decay

• Here the steady state occurs when $N=0$, and is 'reached' after infinite time.

• For small positive $N$ the derivative $\;\displaystyle \frac{{\rm d}N}{{\rm d}t}\;$ is negative, moving the system towards the steady state.

• Physical considerations show that negative $N$ is impossible, hence the steady state is stable as we would expect.

Example 2: production and degradation of a protein

$\displaystyle \frac{{\rm d}s}{{\rm d}t} = v_1 - v_2$

Mass action

$\displaystyle \;\;\overset{v_1}{\rightarrow}S\overset{v_2}{\rightarrow}\;\;$

$\displaystyle \frac{{\rm d}s}{{\rm d}t} = k_1 - k_2 s.$

• If we move to the left of $s=1$, the gradient is positive, so we move back towards $s=1$

• If we move to the right of $s=1$, the gradient is negative, so we, again, move back towards $s=1$

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$\displaystyle \int {1\over k_1 - k_2 s}~{\rm d}s = \int~{\rm d}t$ $\displaystyle s(t) = Be^{-k_2 t} + {k_1\over k_2}$

$\displaystyle s(t) = s_0 e^{-k_2 t} + {k_1\over k_2}\left(1-e^{-k_2 t}\right)$

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• $v_1=k_1 s$

• $v_2=k_2 s^2$

$\displaystyle \frac{{\rm d}s}{{\rm d}t} = k_1s - k_2 s^2$

• $s=0$ or

• $s=k_1/k_2$

• The local behaviour near the fixed point as $s=1$ is the same as in the previous plot, so we can immediately see that it is a stable steady state

• The local behaviour near the fixed point at $s=0$ is the opposite: moving one way or the other, the gradient will take $s$ even further away

Non-graphical method

$\displaystyle \frac{{\rm d}s}{{\rm d}t} = k_1s - k_2 s^2$

• When $s$ is small (and positive, since it is a concentration) $s\gg s^2$ so the derivative $k_1 s - k_2 s^2$ will be positive, making $s=0$ an unstable steady state.

• For the other steady state, consider small deviations by $\varepsilon$ from the steady state, e.g. $s=k_1/k_2 + \varepsilon$.

$\displaystyle s^2=\frac{k_1^2}{k_2^2} + 2 \varepsilon\frac{k_1}{k_2} + \varepsilon^2\;$

$\displaystyle \frac{{\rm d}s}{{\rm d}t} = - k_1\varepsilon - k_2\varepsilon^2$.

Example: the logistic equation

$\displaystyle \frac{{\rm d}N}{{\rm d}t} = rN\left(1 - {N\over K}\right)$

$\displaystyle \frac{{\rm d}N}{{\rm d}t} = 0 \quad\Rightarrow\quad rN = {r\over K}N^2 \quad\Rightarrow\quad N = 0 {\rm~or~} N = K.$

• for small positive $N$, $rN>0$ and $N<K$ so $\displaystyle \frac{{\rm d}N}{{\rm d}t}>0$ and the population size will increase, meaning that $N=0$ is an unstable steady state.

• In fact the growth rate is positive for $0<N<K$ and negative for $N>K$, making $N=K$ a stable steady state.

$\int {1\over rKN-rN^2}~{\rm d}N = \int {1\over K}~{\rm d}t$

• With initial conditions $N(0)=N_0$ you get

  $\displaystyle N(t) = {K N_0 e^{rt}\over K + N_0(e^{rt}-1)}$

Summary of first order differential equations

1. Calculate the general solution

   1. Try to write it as a separable equation first

   2. Other methods (e.g. integrating factors) not covered in this course

2. This general solution will include an arbitrary constant this may be eliminated using initial conditions (if these are given)

3. Can check your solution numerically using Python

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Chemical reactions to equations: mass action

• that chemical reactions have uniform mixing

• that rates are proportional to the product of the masses of the reactants

• that predator/prey interactions, and epidemics, have similar rules

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1. A and B produce C. The reaction is governed by A meeting B:

   $$\displaystyle A+B\underset{^k}{\rightarrow} C$$

   Then we get the following differential equations:

   $\displaystyle \frac{{\rm d}[A]}{{\rm d}t} = -k[A] [B]$ > $\displaystyle \frac{{\rm d}[B]}{{\rm d}t} = -k[A] [B]$ > $\displaystyle \frac{{\rm d}[C]}{{\rm d}t} = k[A] [B]$

2. Predation of R by W

   $\displaystyle R+W\underset{^k}{\rightarrow} W$

   Gives the following differential equation:

   $\displaystyle \frac{{\rm d}R}{{\rm d}t} = -kRW$

3. Constant production (zeroth order)

   $\displaystyle 0 \underset{^k}{\rightarrow} A$

   Gives the following differential equation:

   $\displaystyle \frac{{\rm d}[A]}{{\rm d}t} = k$

4. Degradation/death

   $\displaystyle C \underset{^k}{\rightarrow} 0$

   Gives the following differential equation:

   $\displaystyle \frac{{\rm d}[C]}{{\rm d}t} = -k[C]$

Numerically solving differential equations

$\displaystyle \frac{{\rm d}y}{{\rm d}t} = \sin\left(\frac{y\ln(t\cos(y))}{\sqrt{1+y/e^t}}\right) + 3$

Euler's method

$\displaystyle \frac{{\rm d}y}{{\rm d}t} = f(y, t)$

$\displaystyle y_1 = y(t + \delta{t}) \approx y_0 + f(y, t) \cdot \delta{t}$

$\displaystyle y_2 = y(t + 2\delta{t}) \approx y_1 + f(y, t) \cdot \delta{t}$

1. We can calculate $f(y, t)$ (or approximate it with a computer)

2. We're patient enough to take really tiny steps $\delta{t}$

Introductory problems

Main problems

Extension problems