Implicit Euler Properties

Calculus Level pending

Consider a decaying exponential function of time.

y ( t ) = e − t y(t) = e^{-t}

Suppose we use the Implicit Euler integration method to discretely model this function.

y k = y k − 1 + y ˙ k Δ t y ˙ k = − y k y_k = y_{k-1} + \dot{y}_k \Delta t \\ \dot{y}_k = - y_k

In the above equation, y k y_k is the present value of the function and y k − 1 y_{k-1} is the previous value of the function. The simulated function is "monotonic" if y k y k − 1 > 0 \frac{y_k}{y_{k-1}} > 0 , and "oscillatory" if y k y k − 1 < 0 \frac{y_k}{y_{k-1}} < 0 . The function "converges" if ∣ y k y k − 1 ∣ < 1 \Big| \frac{y_k}{y_{k-1}} \Big | < 1 , and "diverges" if ∣ y k y k − 1 ∣ > 1 \Big| \frac{y_k}{y_{k-1}} \Big | > 1 .

Assuming Δ t > 0 \Delta t > 0 , which of the listed behaviors is possible ?

Oscillatory convergence Oscillatory divergence Monotonic convergence Monotonic divergence

Steven Chase by Steven Chase , 37, USA

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1 solution

Karan Chatrath
Nov 3, 2020

y ˙ = − e − t ⟹ y ˙ = − y \dot{y} = -\mathrm{e}^{-t} \implies \dot{y} = -y

y k = y k − 1 + Δ t y ˙ k y_k = y_{k-1} + \Delta t \dot{y}_k y k = y k − 1 − Δ t y k ∵ y ˙ k = − y k y_k = y_{k-1} - \Delta t y_k \ \ \because \dot{y}_k = -y_k y k y k − 1 = 1 1 + Δ t \frac{y_k}{y_{k-1}} = \frac{1}{1+\Delta t} ∵ Δ t > 0 \because \Delta t >0 y k y k − 1 = 1 1 + Δ t > 0 \frac{y_k}{y_{k-1}} = \frac{1}{1+\Delta t} >0 y k y k − 1 = 1 1 + Δ t < 1 \frac{y_k}{y_{k-1}} = \frac{1}{1+\Delta t} <1

Judging by the conditions satisfied, the only possibility is monotonic convergence. This ensures that the implicit Euler solver guarantees a stable solution despite a higher time-step, unlike the explicit Euler solver.

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