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Fixed-time error slopes for Euler, midpoint, and classical RK4

Example

Write $e^{x} : = \exp (x)$ for the power-series exponential of The real exponential function and the number $e$ by a power series. Apply the three methods of The explicit Euler, midpoint, and classical fourth-order Runge–Kutta tableaus to $y^{'} = - y$ , $y (0) = 1$ , through the fixed final time $T = 1$ . For $N$ equal steps put $h = 1 / N$ and define the global grid error $E_{N} : = ∣ y_{N} - e^{- 1} ∣$ . The claimed slopes mean $E_{N} \sim c h^{p}$ with a nonzero constant $c$ ; the table below records both constants and orders.

Method	$p$	$c$ in $E_{N} \sim c h^{p}$
Forward Euler	1	$e^{- 1} / 2$
Explicit midpoint	2	$e^{- 1} / 6$
Classical RK4	4	$e^{- 1} / 120$

Facts & Assumptions

Given: The scalar IVP $y^{'} = - y$ , $y (0) = 1$ , the three tableaus, and the uniform grid $h = 1 / N$ .

[F1]

The explicit Euler, midpoint, and classical fourth-order Runge–Kutta tableaus: Euler, explicit midpoint, and classical RK4 have the displayed explicit tableaus and stage updates.

[F2]

Orders and scalar stability functions of Euler, midpoint, and classical RK4: for $y^{'} = λ y$ , their stability polynomials are $1 + z$ , $1 + z + z^{2} / 2$ , and $1 + z + z^{2} / 2 + z^{3} / 6 + z^{4} / 24$ , respectively.

[F3]

Multivariable Taylor formula with a Lagrange remainder along a line segment and Higher derivatives and the classes $C^{k}$ and $C^{\infty}$ : a smooth scalar function equals its Taylor polynomial through degree $k$ plus a remainder bounded by a constant times the next power, when the next derivative is bounded on the segment.

[F4]

The exponential function is smooth and $(\exp)^{'} = \exp$ , The real exponential function and the number $e$ by a power series, and The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ : $\exp$ is smooth with $\exp^{'} = \exp$ , its power series gives $\exp (0) = 1$ , and the chain rule gives $(\exp (- t))^{'} = - \exp (- t)$ .

[F5]

$1 + x \leq \exp (x)$ for every real $x$ , hence $(1 - p)^{m} \leq \exp (- m p)$ : $1 + x \leq \exp (x)$ for every real $x$ .

[F6]

The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ : $\exp (x + y) = \exp (x) \exp (y)$ for all real $x, y$ .

Verification

1.1F1F2F3F4F5algebra

By [F4]–[F5], $y (t) = \exp (- t)$ solves $y^{'} = - y$ with $y (0) = 1$ . For one step, the tableaus [F1] and factors [F2] give $R$ ; put $d_{h} : = R - e^{- h}$ . Taylor's theorem [F3] applied to $e^{- h}$ gives, for $p = 1, 2, 4$ in the respective methods, $d_{h} = c_{0} h^{p + 1} + O (h^{p + 2}), c_{0} = - \frac{1}{2}, \frac{1}{6}, \frac{1}{120} .$ The first omitted exponential terms after the Euler, midpoint, and RK4 polynomials in [F2] are respectively $+ h^{2} / 2$ , $- h^{3} / 6$ , and $- h^{5} / 120$ ; subtracting them from those polynomials gives the listed defect coefficients.

2.1F5F6step 1.1algebra

For each method and sufficiently small $h$ , set $q_{h} : = R e^{h} = 1 + e^{h} d_{h}$ , using $e^{- h} e^{h} = e^{0} = 1$ from [F6]. Thus $∣ q_{h} - 1 ∣ \leq C h^{p + 1}$ for a constant independent of $N$ . For $m \leq N = 1 / h$ , factoring $q_{h}^{m} - 1$ gives $∣ q_{h}^{m} - 1 ∣ \leq m ∣ q_{h} - 1 ∣ \max (1, ∣ q_{h} ∣)^{m - 1} \leq C h^{p} e^{C} .$ The last bound follows from $∣ q_{h} ∣ \leq 1 + C h^{p + 1} \leq e^{C h^{p + 1}}$ by [F5], so the factor of powers is at most $e^{C N h^{p + 1}} \leq e^{C}$ .

3.1F3F4F6step 2.1algebra

Repeated use of [F6] gives $(e^{- h})^{N} = e^{- N h} = e^{- 1}$ and $e^{- h (N - 1)} = e^{- 1 + h}$ . Since $R = q_{h} e^{- h}$ , the finite-factor identity yields $R^{N} - e^{- 1} = (R - e^{- h}) \sum_{k = 0}^{N - 1} R^{N - 1 - k} e^{- k h} = d_{h} e^{- h (N - 1)} \sum_{m = 0}^{N - 1} q_{h}^{m} .$ By step 2.1 each summand is $1 + O (h^{p})$ uniformly in $m$ , and $e^{- 1 + h} = e^{- 1} (1 + O (h))$ by [F3]–[F4]. Thus the sum factor is $N e^{- 1} (1 + O (h))$ .

4.1step 1.1step 3.1algebra∎

Substituting the expansion from step 1.1 and $N = 1 / h$ into step 3.1 gives $R^{N} - e^{- 1} = c_{0} e^{- 1} h^{p} (1 + O (h)) .$ Because $c_{0} \neq 0$ , taking absolute values gives the three positive constants and slopes in the Example.

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Sources

Ernst Hairer, course notes, Chapter III: Équations Différentielles Ordinaires, §III.2 (standard reference, not scraped)
MIT OpenCourseWare, 18.330 Introduction to Numerical Analysis, Chapter 5 (standard reference, not scraped)