Skip to content

Backward Euler Method

Intuition

Forward Euler uses the slope where you are; backward Euler uses the slope where you will be:

\[\boxed{\,y_{n+1} = y_n + \Delta t\, f(t_{n+1}, y_{n+1})\,}\]

The unknown appears on both sides — an implicit method. That single change flips every property: unconditional stability, but energy dissipation on orbits.

Solving the implicit equation

For linear \(f\) you can solve algebraically. In practice the course uses a predictor–corrector approximation:

\[\tilde y_{n+1} = y_n + \Delta t\, f(t_n, y_n) \quad\text{(predict with forward Euler)}\]
\[y_{n+1} = y_n + \Delta t\, f(t_{n+1}, \tilde y_{n+1}) \quad\text{(correct)}\]
# predictor-corrector backward Euler
state += state_dot(state + state_dot(state)*dt) * dt

Properties

  • Order 1 — same accuracy class as forward Euler.
  • A-stable: for decaying problems, stable at any \(\Delta t\) — the tool of choice for stiff systems.
  • On orbits: lands slightly inside the circle each step → spirals inward, losing energy — overdamped, the mirror image of forward Euler's gain.

The averaging insight

Forward gains what backward loses, to leading order. Their average is Heun's method (trapezoidal rule):

def Heun_step(f, t, y, h):
    k1 = f(t, y)
    k2 = f(t + h, y + h * k1)
    return y + 0.5 * h * (k1 + k2)

— second-order accurate, with the leading energy errors cancelled. This "sample-and-average" idea, pushed further, becomes the whole Runge–Kutta family.

Common mistakes

  • Thinking stability = accuracy. Backward Euler never blows up — it serenely damps your plasma to zero. A stable wrong answer is still wrong.
  • Forgetting the implicit solve. The predictor–corrector shortcut is an approximation to true backward Euler; for stiff problems you may need a real (Newton) solve.

Knowledge graph position

Prerequisites: Forward Euler. Leads to: Leapfrog, Runge–Kutta.

Quiz

Q1 (computational). For \(\dot y = -\lambda y\), backward Euler gives \(y_{n+1} = y_n/(1 + \lambda\Delta t)\). Is this stable for large \(\Delta t\)?

Answer

Yes: \(|1/(1+\lambda\Delta t)| < 1\) for every \(\Delta t > 0\) — unconditional (A-) stability. Compare forward Euler's \(\Delta t \leq 2/\lambda\) limit.

Q2 (conceptual). Why does averaging the forward and backward Euler steps raise the order from 1 to 2?

Answer

Their leading error terms are equal and opposite (\(\pm\frac{\Delta t^2}{2}y''\)): the average is the trapezoidal rule, whose error starts at \(\mathcal{O}(\Delta t^3)\) per step — second order globally.

Q3 (MCQ). On the gyration benchmark, backward Euler produces a trajectory that:

  • (a) spirals outward (b) spirals inward (c) stays exactly circular (d) drifts linearly
Answer

(b). Numerical dissipation: the chord to the future point cuts inside the circle.