The error of Euler’s method · Differential Equations

Intuition

Each Euler step follows a tangent, and the solution bends away from its tangent. Over one step the gap grows with the square of the step: halve the step and the gap is a quarter. But crossing an interval takes more steps when they are shorter — twice as many at half the size — so the gaps add up to something proportional to the step itself. Halve h and the final error halves. That is what it means for Euler’s method to be of order one, and it is why it is slow: ten times the accuracy costs ten times the work.

Rounding every line of a long bill. Each rounding is small, but a longer bill has more of them, and the error in the total depends on both the size of each rounding and how many there are.

One Euler step of length $h=1$ on $y'=y$ , started on the solution at $(0,1)$ . The step follows the tangent there and ends at $2$ , while the solution has bent away to $e\approx2.72$ . The gap is the local error, $\tfrac{h^{2}}{2}y''(\xi)$ for some $\xi$ in the step.

The error of one step, and of all of them

The local error of a step is what it would get wrong if it started on the solution: $\tau=y(x+h)-\bigl[y(x)+h\,f(x,y(x))\bigr]$ . Taylor’s theorem gives for some in the step, so . The global error is what the method gets wrong at after all its steps, each started from an estimate. A method is of order when its global error on a fixed interval is at most a constant times . Euler’s method is of order one. The order of a method has nothing to do with the order of the equation.

|e_{n}|\le C\,h\qquad\text{for}\ x_{0}\le x_{n}\le X

What the orders say

A local error of size $h^{2}$ , over about $\tfrac{X-x_{0}}{h}$ steps, makes a global error of size : one power of is lost to the number of steps.

Euler’s method on $y'=y$ to $x=1$ with two step sizes. Halving the step takes the error at $x=1$ from to , and further halvings take it to and then : each halving comes nearer to halving the error, the mark of a method of order one.

Euler’s method converges, with an error of order h

\begin{aligned}&\bigl|\partial f/\partial y\bigr|\le K,\ \ |y''|\le M\\&\Longrightarrow\ |e_{n}|\le\frac{Mh}{2K}\bigl(e^{K(x_{n}-x_{0})}-1\bigr)\end{aligned}

Compare one step of the method with the same step of the solution. The solution’s step is an Euler step from the true value plus a local error of at most M h²/2. The slopes at the true value and at the estimate differ by at most K times the difference of the values, by the mean value theorem, so in one step the error is multiplied by at most 1 + hK and gains at most M h²/2. Unwinding this from e₀ = 0 gives a geometric sum, and 1 + hK ≤ e^{hK} bounds it by the formula. On a fixed interval the bound is a constant times h: the method converges, with order one.

Proof steps

\begin{aligned}y(x_{n+1})&=y(x_{n})+h\,f\bigl(x_{n},y(x_{n})\bigr)+\tau_{n}\\|\tau_{n}|&\le\tfrac{M}{2}h^{2}\end{aligned}

Taylor’s theorem: the solution’s own step is an Euler step plus a local error.

\begin{aligned}e_{n+1}=e_{n}&+h\bigl[f\bigl(x_{n},y(x_{n})\bigr)-f(x_{n},y_{n})\bigr]\\&+\tau_{n}\end{aligned}

Subtract the method’s step $y_{n+1}=y_{n}+h\,f(x_{n},y_{n})$ .

|e_{n+1}|\le(1+hK)\,|e_{n}|+\tfrac{M}{2}h^{2}

By the mean value theorem in $y$ , the bracket is at most $K|e_{n}|$ .

\begin{aligned}|e_{n}|&\le\tfrac{M}{2}h^{2}\sum_{j=0}^{n-1}(1+hK)^{j}\\&=\frac{Mh}{2K}\bigl((1+hK)^{n}-1\bigr)\end{aligned}

Unwind from $e_{0}=0$ and sum the geometric series.

(1+hK)^{n}\le e^{nhK}=e^{K(x_{n}-x_{0})}

Since $1+u\le e^{u}$ and $nh=x_{n}-x_{0}$ . The bound is a constant times .

Applications

Practice

The Gap After One Step

Start a step on the solution itself. The step follows the tangent and the solution bends away from it; the gap at the end of the step is the local error.

\tau=y(x+h)-\bigl[y(x)+h\,y'(x)\bigr]=\tfrac{h^{2}}{2}y''(\xi)

Try it

One Euler step with $h=0.1$ on $y'=y$ from the exact value $y(0)=1$ gives $1.1$ . What is the local error $e^{0.1}-1.1$ ? Give four decimal places.