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Forward Euler Method

Intuition

The most obvious idea in numerical analysis: stand at \((t_n, y_n)\), measure the slope, walk straight along the tangent for time \(\Delta t\). Repeat. It's the numerical method you'd invent by accident — and its failure mode on plasma orbits (slow, systematic energy gain) motivates everything that follows.

The scheme

\[\boxed{\,y_{n+1} = y_n + \Delta t\, f(t_n, y_n)\,}\]
def Euler_step(f, t, y, h):
    return y + h * f(t, y)

Properties

  • Order 1: local truncation error \(\mathcal{O}(\Delta t^2)\), global error \(\mathcal{O}(\Delta t)\) — halving the step halves the error.
  • Cost: one evaluation of \(f\) per step (cheapest possible).
  • Stability: conditional; for oscillatory systems, effectively never stable — the amplitude grows every step.

Why it spirals outward on orbits

For circular gyration, the exact motion curves; Euler walks along the tangent, which always lands slightly outside the circle. Each step the radius (and kinetic energy) grows by a factor \(\sqrt{1 + (\omega\Delta t)^2}\). On the cross-field benchmark, the trajectory visibly spirals outward — see the live comparison widget.

Backward Euler makes the mirror-image error (lands inside, spirals inward); their average (Heun's method) cancels the leading error and jumps to order 2 — the first hint that clever slope sampling pays.

When to use it

Prototyping, non-oscillatory dissipative problems, and as the "kick" building block inside better schemes (kick–drift–kick leapfrog). Never for long-time orbital or wave problems.

Common mistakes

  • "It's converging, so it's fine." It converges as \(\Delta t \to 0\) at fixed \(T\), but at fixed \(\Delta t\) the energy error grows exponentially in time for orbits.
  • Blaming the physics. An outward-spiraling particle in a uniform magnetic field is a numerical artifact — magnetic forces do no work.

Knowledge graph position

Prerequisites: ODE integration. Leads to: Backward Euler, Runge–Kutta.

Quiz

Q1 (computational). For \(\dot y = -y\), \(y_0 = 1\), \(\Delta t = 0.1\): what is \(y\) after two Euler steps, and the exact answer?

Answer

\(y_1 = 1 - 0.1 = 0.9\), \(y_2 = 0.9(1 - 0.1) = 0.81\). Exact: \(e^{-0.2} = 0.8187\). Error ≈ 0.009 — first-order behavior.

Q2 (conceptual). For \(\dot y = -\lambda y\) (\(\lambda > 0\)), forward Euler gives \(y_{n+1} = (1 - \lambda\Delta t)y_n\). For which \(\Delta t\) is it stable?

Answer

Need \(|1 - \lambda\Delta t| \leq 1 \Rightarrow \Delta t \leq 2/\lambda\). Beyond that the iterates oscillate with growing amplitude — numerical explosion of a decaying solution.

Q3 (MCQ). On a charged-particle gyration problem, forward Euler:

  • (a) conserves energy (b) loses energy (c) gains energy (d) preserves the orbit radius
Answer

(c). Tangent steps always exit the circle outward: systematic energy gain.