Explicit Euler Properties

Calculus Level 4

Consider a decaying exponential function of time.

$y(t) = e^{-t}$

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

$\begin{array} { r c l } y_k &= &y_{k-1} + \dot{y}_{k-1} \Delta t \\ \dot{y}_k &= &- y_k \end{array}$

In the above equation, $y_k$ is the present value of the function and $y_{k-1}$ is the previous value of the function. The simulated function is "monotonic" if $\frac{y_k}{y_{k-1}} > 0$ , and "oscillatory" if $\frac{y_k}{y_{k-1}} < 0$ . The function "converges" if $\Big| \frac{y_k}{y_{k-1}} \Big | < 1$ , and "diverges" if $\Big| \frac{y_k}{y_{k-1}} \Big | > 1$ . Different behaviors are exhibited for different values of the time step $\Delta t$ .

Subject to the constraints $\Big| \frac{y_k}{y_{k-1}} \Big | \neq 1$ , $\Big| \frac{y_k}{y_{k-1}} \Big | \neq 0$ , and $\Delta t > 0$ , which of the listed behaviors is impossible?

Inspiration

Monotonic convergence Oscillatory divergence Monotonic divergence Oscillatory convergence

1 solution

Mark Hennings
Nov 16, 2020

The recurrence relation is $y_k \; = \; y_{k-1} - y_{k-1}\Delta t \; = \; \big(1 - \Delta t)y_{k-1}$ and hence we deduce that $y_k \; = \; \big(1 - \Delta t\big)^k$ which converges to $0$ monotonically if $0 < \Delta t < 1$ , converges to $0$ in an oscillatory manner if $1 < \Delta t < 2$ , and which diverges in an oscillatory manner if $\Delta t > 2$ . We cannot have monotonic divergence, since that would require $\Delta t < 0$ , which is not possible.

Explicit Euler Properties

1 solution

0 pending reports