Suppose a population \(P\) grows at an instantaneous rate of \(0.1P\) individuals per year. At the moment when \(P=100\), its rate is 10 individuals per year. Euler’s method assumes that current rate stays constant for one short time step:
\[ P_{\text{new}}=P_{\text{old}}+\underbrace{(0.1P_{\text{old}})}_{\text{rate at the old value}}\Delta t. \]
The rate is the slope of the population curve at the starting point. Euler follows a straight tangent line for one step, then finds a new slope and repeats. It does not know the future slope in advance.
Start at \(P(0)=100\) and jump all the way to year 8 in one step. What would Euler predict? The starting slope is \(0.1(100)=10\) individuals/year, so
\[ P_{\text{Euler}}(8)=100+10(8)=180. \]
For this particular model, we also know the exact solution \(P(t)=100e^{0.1t}\). It gives about 222.55 at year 8. The large Euler step falls short because the true curve bends upward while Euler keeps the starting slope of 10 for all eight years.
initial_population <- 100
growth_fraction <- 0.1 # per year
end_time <- 8 # years
delta_t <- 8 # one very large step
old_population <- initial_population
growth_rate <- growth_fraction * old_population
euler_estimate <- old_population + growth_rate * delta_t
exact_value <- initial_population * exp(growth_fraction * end_time)
euler_estimate[1] 180exact_value[1] 222.5541abs(euler_estimate - exact_value)[1] 42.55409The blue curve is the exact solution. The red segment shows the one-step Euler estimate. Where do the two paths begin to separate?
smooth_time <- seq(0, end_time, by = 0.1)
exact_curve <- initial_population * exp(growth_fraction * smooth_time)
plot(smooth_time, exact_curve, type = "l", lwd = 2, col = "steelblue",
xlab = "Years", ylab = "Population", ylim = c(90, 230))
segments(0, initial_population, end_time, euler_estimate,
col = "firebrick", lwd = 2)
points(c(0, end_time), c(initial_population, euler_estimate),
col = "firebrick", pch = 19)
legend("topleft", c("Exact curve", "One Euler step"),
col = c("steelblue", "firebrick"), lty = 1, lwd = 2, bty = "n")
Now use four steps of two years each. Before running the loop, predict the first two estimates. The first is \(100+0.1(100)(2)=120\). For the second, use the new starting value of 120 to find a new slope: \(120+0.1(120)(2)=144\).
delta_t <- 2
time <- seq(0, end_time, by = delta_t)
population <- numeric(length(time))
population[1] <- initial_population
for (i in 2:length(time)) {
old_population <- population[i - 1]
growth_rate <- growth_fraction * old_population
population[i] <- old_population + growth_rate * delta_t
}
data.frame(time, population) time population
1 0 100.00
2 2 120.00
3 4 144.00
4 6 172.80
5 8 207.36At year 8, this estimate is 207.36: closer to 222.55 than the one-step estimate, but still low. Each two-year segment starts with a fresh slope.
exact_at_times <- initial_population * exp(growth_fraction * time)
plot(time, exact_at_times, type = "l", lwd = 2, col = "steelblue",
xlab = "Years", ylab = "Population", ylim = c(90, 230))
lines(time, population, type = "b", col = "firebrick", pch = 19)
legend("topleft", c("Exact values", "Two-year Euler steps"),
col = c("steelblue", "firebrick"), lty = 1,
pch = c(NA, 19), lwd = 2, bty = "n")
Change only delta_t in the loop above to 8, 1, and 0.5. Keep the same initial population, rate, and final time. Record the final estimate and its absolute error against the exact value. The book’s Module 6.2 asks us to compare solutions this way.
| \(\Delta t\) (years) | Steps to year 8 | Euler estimate at year 8 | Absolute error, approximately |
|---|---|---|---|
| 8 | 1 | 180.000 | 42.554 |
| 2 | 4 | 207.360 | 15.194 |
| 1 | 8 | 214.359 | 8.195 |
| 0.5 | 16 | 218.287 | 4.267 |
Smaller steps improve this particular approximation, but require more calculations. Halving the step from 1 to 0.5 roughly halves the error here—an example of Euler’s first-order behavior. That pattern is not a promise for every model or every step size. It also says nothing by itself about whether \(0.1P\) is a good model of a real population.
Euler’s update is not limited to proportional growth. In the book’s Quick Review Question 1, the rate is \(10+P/5\), the initial population is 500, and \(\Delta t=0.1\). Find the rate at the old value, multiply by the step, and add it. Predict the first estimate before running R.
delta_t <- 0.1
time <- seq(0, 0.2, by = delta_t)
population <- numeric(length(time))
population[1] <- 500
for (i in 2:length(time)) {
old_population <- population[i - 1]
rate_now <- 10 + old_population / 5
population[i] <- old_population + rate_now * delta_t
}
data.frame(time, population) time population
1 0.0 500.00
2 0.1 511.00
3 0.2 522.22The first estimate is 511; the second is 522.22. The only part of the loop that changed was the formula for the rate. That is the core of Euler’s method: start with the current value, estimate a change using the current slope, then step forward.