Backward Euler Method - Introduction to Scientific Computing

Big Idea

Backward Euler evaluates $f$ at the next time step instead of the current one, making it implicit: $u_{n+1}$ appears on both sides, requiring a solve at each step. This extra cost buys unconditional stability: the method is stable for any step size when $\operatorname{Re}(\lambda) < 0$ . For stiff problems, where explicit methods waste enormous effort satisfying stability constraints, this is essential.

Derivation¶

Integrate $u' = f(t,u)$ over one time step:

u(t_{n+1}) = u(t_n) + \int_{t_n}^{t_{n+1}} f(s, u(s))\,ds

(1)

Forward Euler (Definition 1) approximates this integral by freezing the integrand at the left endpoint $t_n$ (a left rectangle rule). Backward Euler freezes it at the right endpoint $t_{n+1}$ instead:

\int_{t_n}^{t_{n+1}} f(s, u(s))\,ds \approx h\,f(t_{n+1}, u(t_{n+1}))

(2)

This gives the backward Euler (or implicit Euler) method:

Definition 1 (Backward Euler Method)

u_{n+1} = u_n + h f(t_{n+1}, u_{n+1})

(3)

The value $u_{n+1}$ , the next approximation of the solution, appears on both sides: it is defined implicitly as the solution to

u_{n+1} - u_n - h f(t_{n+1}, u_{n+1}) = 0

(4)

The Cost of Implicitness¶

Each step of backward Euler requires solving an equation for $u_{n+1}$ .

Linear problems. For $f(t, u) = \lambda u + g(t)$ , we can solve algebraically:

u_{n+1} = u_n + h(\lambda u_{n+1} + g(t_{n+1})) \quad\implies\quad u_{n+1} = \frac{u_n + hg(t_{n+1})}{1 - h\lambda}

(5)

No iteration needed.

Nonlinear problems. For general $f$ , we must solve the nonlinear equation

G(u_{n+1}) = u_{n+1} - u_n - hf(t_{n+1}, u_{n+1}) = 0

(6)

at every time step, using any of the root-finding methods from the nonlinear equations chapter (e.g., Newton’s method or a fixed-point iteration). The previous time step $u_n$ provides a natural initial guess, and for small $h$ convergence is typically rapid (1--2 iterations).

See the Euler’s method notebook for an interactive implementation.

Error Analysis and Convergence¶

Following the same analysis as for forward Euler, we obtain the same accuracy order.

Theorem 1 (Local Truncation Error of Backward Euler)

For backward Euler applied to $u' = f(t,u)$ with $u \in C^2$ , the local truncation error is

\tau_n = -\frac{h}{2}u''(t_{n+1}) + O(h^2) = O(h)

(7)

Backward Euler has consistency order $p = 1$ .

Proof 1

Theorem 2 (Convergence of Backward Euler)

Under the same Lipschitz conditions as Theorem 1, the global error of backward Euler satisfies

|E_n| \leq \frac{C}{L}\left(e^{L(t_n - t_0)} - 1\right) h

(10)

Backward Euler converges with order 1.

Proof 2

Stability Analysis¶

Apply backward Euler to the test equation $u' = \lambda u$ :

u_{n+1} = u_n + h\lambda\,u_{n+1} \quad\implies\quad u_{n+1} = \frac{1}{1 - h\lambda}\,u_n

(11)

The stability function is $R(z) = \frac{1}{1-z}$ where $z = h\lambda$ . Compare forward Euler’s $R(z) = 1 + z$ (Definition 6).

Definition 2 (Stability Region of Backward Euler)

The stability region of backward Euler is

\mathcal{S} = \left\{z \in \mathbb{C} : \left|\frac{1}{1-z}\right| \leq 1\right\} = \{z \in \mathbb{C} : |1-z| \geq 1\}

(12)

This is everything outside the disk of radius 1 centered at $(1, 0)$ . In particular, it contains the entire left half-plane $\{\operatorname{Re}(z) \leq 0\}$ .

Stability region of backward Euler: everything outside the disk |1 - z| \geq 1. The entire left half-plane is included, so backward Euler is stable for all h > 0 whenever \operatorname{Re}(\lambda) < 0. — Stability region of backward Euler: everything **outside** the disk $|1 - z| \geq 1$ . The entire left half-plane is included, so backward Euler is stable for all $h > 0$ whenever $\operatorname{Re}(\lambda) < 0$ .

For $\operatorname{Re}(\lambda) < 0$ , backward Euler is stable for all $h > 0$ : unconditionally stable.

Stiffness¶

Stiff problems (Definition 7) arise whenever the ODE has widely separated time scales. The fast components demand small steps for explicit stability, even after those components have decayed to negligible levels. This is where backward Euler’s unconditional stability pays off.