Euler’s Method

A starting point and a gradient

A differential equation $\frac{dy}{dx}=f\left(x,y\right)$dy/dx = f(x, y) with a starting value $y=y_0$y = y₀ at $x=x_0$x = x₀ fixes one solution. Often no formula for it can be found: the variables will not separate, or they separate into a relation that cannot be solved for y.

One thing is always known, though. At the starting point the equation gives the gradient, $f\left(x_0,y_0\right)$f(x₀, y₀). So the direction in which the solution leaves that point is known, even when the solution is not.

One step along the tangent

Move a short width h along that tangent. The rise is the gradient times the width, so the new point is $x_1=x_0+h$x₁ = x₀ + h, $y_1=y_0+hf\left(x_0,y_0\right)$y₁ = y₀ + h f(x₀, y₀).

At the new point, ask the equation for the gradient again and take another step. In general $y_{n+1}=y_n+hf\left(x_n,y_n\right)$yₙ₊₁ = yₙ + h f(xₙ, yₙ) and $x_{n+1}=x_n+h$xₙ₊₁ = xₙ + h. Each step uses the gradient at its own start, so the walk keeps turning to follow the field.

Three steps for $\frac{dy}{dx}=y$dy/dx = y

Take $\frac{dy}{dx}=y$dy/dx = y with y = 1 at x = 0, and h = 0.5. Its exact solution is $y=e^x$y = eˣ, which gives something to check against.

At (0, 1) the gradient is 1, so the first step rises 0.5 × 1 = 0.5 and reaches (0.5, 1.5). There the gradient is 1.5, so the second step rises 0.5 × 1.5 = 0.75 and reaches (1, 2.25). There the gradient is 2.25, so the third step rises 1.125 and reaches (1.5, 3.375).

The exact value at x = 1.5 is $e^{1.5}\approx 4.482$e^1.5 ≈ 4.482, so the estimate 3.375 is low by 1.107. At x = 1 the estimate is 2.25 against $e\approx 2.718$e ≈ 2.718.

Three Euler steps of width 0.5 for $\frac{dy}{dx}=y$dy/dx = y from (0, 1), drawn as gold arrows on the slope field: to (0.5, 1.5), (1, 2.25) and (1.5, 3.375). The gold curve $y=e^x$y = eˣ runs above them and reaches the dot at (1.5, 4.482).

Why it lands low here

The curve $y=e^x$y = eˣ bends upward, so it lies above each of its tangents. Each step follows a tangent, and also uses the gradient at the start of the step, which is the smallest gradient anywhere on that step. Every step comes up short, and the shortfalls add.

Smaller steps miss by less. At x = 1.5, h = 0.25 (six steps) gives $1.25^6\approx 3.815$1.25⁶ ≈ 3.815, low by 0.667; h = 0.125 (twelve steps) gives $1.125^{12}\approx 4.110$1.125¹² ≈ 4.110, low by 0.372; h = 0.0625 gives 4.284, low by 0.197. Each halving of h cuts the error to 0.60, then 0.56, then 0.53 of what it was, closing on one half. The error is roughly proportional to h.

Euler’s method for $\frac{dy}{dx}=y$dy/dx = y from (0, 1) to x = 2 in N = 4 steps of h = 0.5. The walk ends at $1.5^4\approx 5.063$1.5⁴ ≈ 5.063 against $e^2\approx 7.389$e² ≈ 7.389, a miss of 2.33. Drag the handle at the end of the first step to shrink h: at N = 20 steps of 0.1 the walk ends at 6.727 and the miss is 0.66.

A downward bend lands high

Take $\frac{dy}{dx}=-x$dy/dx = −x with y = 1 at x = 0, whose exact solution is $y=1-\frac{x^2}{2}$y = 1 − x²/2, a curve bending downward. With h = 0.5: at (0, 1) the gradient is 0, so the first step stays level at (0.5, 1); there the gradient is −0.5, giving (1, 0.75); there it is −1, giving (1.5, 0.25).

The exact value at x = 1.5 is 1 − 1.125 = −0.125, so the estimate 0.25 is too high by 0.375. A curve bending downward lies below its tangents, and each step overshoots.

Euler steps of width 0.5 for $\frac{dy}{dx}=-x$dy/dx = −x from (0, 1): to (0.5, 1), (1, 0.75) and (1.5, 0.25). The gold curve $y=1-\frac{x^2}{2}$y = 1 − x²/2 bends down below the arrows and reaches the dot at (1.5, −0.125).

An equation that does not separate

$\frac{dy}{dx}=x+y$dy/dx = x + y cannot be split into a y side and an x side. Start at y = 1 when x = 0, with h = 0.5. At (0, 1) the gradient is 0 + 1 = 1, giving (0.5, 1.5). There it is 0.5 + 1.5 = 2, giving (1, 2.5).

The integrating factor, a later method, solves this equation exactly: $y=2e^x-x-1$y = 2eˣ − x − 1, which is $2e-2\approx 3.437$2e − 2 ≈ 3.437 at x = 1. The estimate 2.5 is low by 0.937. With h = 0.25 the estimate is 2.883, and with h = 0.1 it is 3.187, low by 0.249.

The usual mistakes

Leaving out the h. $y_{n+1}=y_n+f\left(x_n,y_n\right)$yₙ₊₁ = yₙ + f(xₙ, yₙ) takes a step one unit wide whatever h is. With h = 0.5 from (0, 1) for $\frac{dy}{dx}=y$dy/dx = y, the first step reaches 1.5, not 2.

Keeping the first gradient for every step. The gradient is worked out again at each new point; for $\frac{dy}{dx}=y$dy/dx = y the second step uses 1.5, not 1.

Dropping $y_n$yₙ. The rise is added to the height already reached, not to 0.

Calling the estimate exact. It is low wherever the solution bends upward and high where it bends downward, and halving h only halves the error.

Ammonia in a tank

In the application below, the equation separates into a relation that cannot be rearranged for the concentration, so it is stepped out instead. Halving the step shows which way Euler’s method errs and by how much.