Lesson 1: Discretization and Euler's Method

1 Modelling context

A differential equation is a statement about rates of change. It tells you how fast a quantity is moving, but not where it will be at some later time. To answer that question, you integrate the rate over time — and in most applied problems, the integral cannot be evaluated in closed form.

Numerical methods replace integration with stepping. Starting from a known value at time t₀, they estimate the rate of change, take a small step forward, and repeat. The family of methods in this module all share the same structure; they differ in how they estimate the rate of change during each step.

Euler's method is the simplest. It assumes the rate of change is constant across each step, using the value at the start. This is often wrong, but the method is easy to implement and easy to analyse. More importantly, understanding exactly how Euler fails is the prerequisite for understanding why more sophisticated methods exist.

Discretization is a modelling decision

Choosing a step size is not a technical detail. It changes the numerical problem you are actually solving, and it determines the accuracy of the result. The step size is part of the model and must be documented alongside the parameters.

2 Learning objectives

By the end of this lesson you will be able to

Discretize an initial-value problem onto a uniform grid.
Derive Euler's method from the forward-difference approximation.
Implement Euler's method in a spreadsheet with documented inputs and step size.
Identify the qualitative failure modes of Euler's method on a convex or concave solution.
Explain the relationship between step size and global error for a first-order method.

3 Prerequisite review

Derivative as a rate of change: dy/dt is the instantaneous rate at which y changes with respect to t. In an ODE, this rate is expressed as a function of t and y.
Taylor expansion: A function can be expanded about a point as y(t + h) = y(t) + h y'(t) + (h²/2) y''(t) + …. Truncating after the first-order term gives Euler's method.
Forward-difference approximation: The derivative can be approximated as y'(t) ≈ ( y(t + h) − y(t) ) / h. Rearranging gives the Euler update rule.
Analytical solution: For a small class of ODEs, the solution can be written in closed form. When this is available, it is the gold standard against which the numerical solution is compared.

4 Mathematical formulation

Consider the initial-value problem

dy/dt = f(t, y),    y(t₀) = y₀

Choose a uniform step size h > 0 and define the grid

t_n = t₀ + n h,    n = 0, 1, 2, …

Derivation from the forward difference

The forward-difference approximation of the derivative at t_n is

dy/dt |_{t = t_n} ≈ ( y(t_{n+1}) − y(t_n) ) / h

Substituting the ODE dy/dt = f(t, y) and rearranging:

y(t_{n+1}) ≈ y(t_n) + h f(t_n, y(t_n))

Let y_n denote the numerical approximation to y(t_n). Euler's method is

y_{n+1} = y_n + h f(t_n, y_n),    y_0 given

Derivation from the Taylor expansion

Expand the exact solution about t_n:

y(t_n + h) = y(t_n) + h y'(t_n) + (h²/2) y''(t_n) + O(h³)

Since y'(t_n) = f(t_n, y(t_n)), this becomes

y(t_n + h) = y(t_n) + h f(t_n, y(t_n)) + (h²/2) y''(t_n) + O(h³)

Euler's method retains only the first two terms. The truncation error is therefore

τ_n = (h²/2) y''(t_n) + O(h³) = O(h²)

This is the local truncation error. It scales as O(h²), which means the method is of order 1 globally — a fact developed fully in Lesson 3.

Geometric interpretation

Euler's method advances along the tangent line to the solution at each grid point. When the solution is convex (y'' > 0), the tangent lies below the curve, and Euler underestimates. When the solution is concave (y'' < 0), the tangent lies above the curve, and Euler overestimates. The sign of the error is therefore predictable from the curvature of the exact solution.

Figure 2.3. Euler's method on a convex solution. The tangent-line approximation lies below the exact curve, and the error accumulates monotonically. Refining the step size reduces but does not eliminate the systematic bias.

5 Worked example

Exponential growth with a known analytical solution

Consider the simple growth model

dy/dt = y,    y(0) = 1

whose exact solution is y(t) = e^t. This example is deliberately simple so that the numerical error can be measured exactly.

Step 1 — Discretize.

Choose h = 0.1. The grid points are t_n = 0, 0.1, 0.2, …

Step 2 — Apply the update rule.

At n = 0: y_1 = y_0 + h · y_0 = 1 + 0.1 · 1 = 1.1.

At n = 1: y_2 = 1.1 + 0.1 · 1.1 = 1.21.

In general, the update rule for this problem is y_{n+1} = (1 + h) y_n, so the numerical solution after n steps is y_n = (1 + h)ⁿ.

Step 3 — Compare with the exact solution.

After 10 steps (t = 1):

Exact: y(1) = e¹ ≈ 2.71828.
Euler with h = 0.1: 1.1¹⁰ ≈ 2.59374.
Absolute error: 0.12454.

After 100 steps (t = 10):

Exact: y(10) = e¹⁰ ≈ 22026.47.
Euler with h = 0.1: 1.1¹⁰⁰ ≈ 13780.61.
Absolute error: 8245.86.

The absolute error grows dramatically over the interval because the solution itself grows. The relative error is a better measure: it is about 4.6% at t = 1 and about 37% at t = 10.

Step 4 — Reduce the step size.

Halve the step size to h = 0.05. The relative error at t = 10 becomes approximately 20%. Halve again to h = 0.025: the relative error becomes approximately 10%.

The pattern: halving h approximately halves the global error. This is the signature of a first-order method, and it motivates the higher-order methods of Lesson 2 and Unit 2.2.

The step size is not a free parameter. Reducing it improves accuracy but increases the number of steps required, and therefore the runtime. The tradeoff between accuracy and cost is the central theme of the module.

6 Verification check

Checks to perform

Reproduce a known case. For dy/dt = y, y(0) = 1, the exact solution is e^t. Confirm the spreadsheet's output at a few grid points against this.
Check the first step by hand. Compute y_1 = y_0 + h f(t_0, y_0) independently and compare against the spreadsheet.
Verify the step count. The number of steps from t_0 to t_end is (t_end − t_0) / h. A mismatch indicates a grid error.
Check units. If f has units of y per unit time, then h f has units of y, as required.
Test the limiting case f = 0. If the derivative is zero, the solution should remain constant at y_0 for all time.
Test the limiting case h → 0. As the step size decreases, the numerical solution should approach the exact solution.

7 Interpretation

Euler's method is a first-order method. This means that reducing the step size by a factor of k reduces the global error by a factor of approximately k. To gain one additional decimal digit of accuracy, you must reduce the step size by a factor of ten — and multiply the runtime by ten.

The method is rarely used in production work for this reason. Its value in this course is conceptual: it establishes the framework that all one-step methods share, and it demonstrates the systematic bias that arises from using a tangent-line approximation on a curved solution.

Reading the error pattern

The error of Euler's method is not random. On a convex solution, it is systematically negative — the method underestimates. On a concave solution, it is systematically positive. This bias is a property of the method, not of the data, and it persists no matter how small the step size becomes. It shrinks, but it does not disappear.

8 Uncertainty and limitations

Two sources of error limit the accuracy of Euler's method. Their relative importance depends on the step size.

Sources of error in Euler's method
Source	Scaling	Reduced by
Local truncation error	`O(h²)` per step	Smaller step size
Accumulated global error	`O(h)` over a fixed interval	Smaller step size; higher-order method
Roundoff	Grows with step count	Fewer steps; careful arithmetic

Truncation error decreases with the step size; roundoff error increases as the number of steps grows. There is a step size below which further reduction does not improve accuracy and may degrade it. The optimal step size depends on the problem and on the machine precision. For most practical problems with double-precision arithmetic, this limit is far below any step size you would choose for reasons of accuracy alone.

Smaller is not always better

Reducing the step size improves truncation error but increases the number of steps, and therefore roundoff accumulation and runtime. A carefully chosen moderate step size is usually preferable to an indiscriminately small one.

9 Common implementation errors

Watch for these

Using the updated value of t when evaluating f. Euler is an explicit method: the increment uses f(t_n, y_n), not f(t_{n+1}, y_n). Reading the wrong row from a spreadsheet is a common source of this error.
Filling the formula with a relative reference that shifts incorrectly. The reference to h should be absolute so that it does not change as the formula is copied down.
Off-by-one in the step count. The number of steps is (t_end − t_0) / h. Verify the last row reaches t_end exactly, or document the endpoint.
Mixing units between h and f. If f has units of per-hour, then h must be in hours.
Assuming the error is small because the output looks right. A first-order method can produce a solution that appears visually correct while carrying substantial error. The comparison against an analytical solution or a refined step size is required.
Applying Euler to a stiff problem. See Unit 2.2, Lesson 3. The step size required for stability can be far smaller than the step size required for accuracy.

10 Independent modelling task

Your turn

Choose an initial-value problem with a known analytical solution. Good candidates: dy/dt = −y, dy/dt = y − t, or dy/dt = cos(t).
Implement Euler's method in a spreadsheet, with inputs (initial condition, step size, endpoint) in a clearly labelled input region.
Compute the numerical solution at h = 0.2, h = 0.1, and h = 0.05.
For each step size, compute the absolute error against the analytical solution at the endpoint.
Tabulate the errors against the step sizes. Compute the ratio of error between successive step sizes. Confirm the ratio is close to 2, the signature of a first-order method.
Plot the numerical and analytical solutions on the same axes for the largest and smallest step sizes. Describe the qualitative difference.

This exercise is the basis for Laboratory 3 and feeds the ODE component of Assignment 2.

11 Reflection prompt

Reflect before continuing

In the worked example, the relative error grew substantially over the integration interval, even though the local truncation error per step remained the same. What does this suggest about the relationship between local and global error?

Two or three sentences. Lesson 3 develops this relationship formally.

12 Lesson summary

Key takeaways

Euler's method advances along the tangent line at each step. It assumes the derivative is constant across the step, which is exact only for linear solutions.
The update rule is y_{n+1} = y_n + h f(t_n, y_n). It is explicit: the increment uses values at the start of the step.
The local truncation error is O(h²). The global error is O(h), so the method is first-order.
The error is systematic, not random. On a convex solution, Euler underestimates; on a concave solution, it overestimates.
Halving the step size halves the global error. Gaining one decimal digit of accuracy requires a tenfold increase in runtime. This motivates the higher-order methods that follow.

Coming next

Lesson 2 introduces second-order corrections — modified Euler and midpoint methods — that use a trial step to improve accuracy per step.

§ Self-quiz, practice, and homework

Practice exercises

Apply Euler's method by hand to dy/dt = −2y, y(0) = 1 with h = 0.1 for three steps. Compute the exact value at each step and the error.
Repeat exercise 1 with h = 0.05. Confirm the error ratio is approximately 2.
Write the Euler update rule for dy/dt = t − y. Compute three steps from (t₀, y₀) = (0, 1) with h = 0.1.
For dy/dt = y sin(t), write the Euler update and explain why the method cannot be simplified to a closed-form recurrence.

Homework problems

Euler implementation and analysis. Implement Euler's method in a spreadsheet for the initial-value problem dy/dt = −y + sin(t), y(0) = 1. Run at h = 0.2, 0.1, 0.05. For each, compute the numerical solution at t = 1, 2, 3 and compare against the analytical solution. Tabulate the errors, compute the observed order, and produce a log-log plot. Write a 500-word report.
Error sign analysis. For the problem dy/dt = t², y(0) = 0, compute the exact solution and the Euler approximation at t = 1 for h = 0.1. Determine the sign of the error and explain it geometrically.
Boundary case. Consider dy/dt = 0 with y(0) = 5. Verify that Euler's method is exact for this problem. Explain why.

Mini project

Choose an ODE model of a physical, biological, or economic process with a known analytical solution. Implement Euler's method in a spreadsheet, run a convergence study, and produce a technical report that documents the grid, the initial condition, the parameter values, the observed order of convergence, and a claim-support statement. This is the basis for Laboratory 3 and part of Assignment 2.