Numerical Stability of Euler's Method

Calculus Level 5

Suppose we want to use Euler's Method to numerically (discretely) compute the time-equation below through a sequence of iterative steps:

y = e − A t y k = y k − 1 + y k − 1 ′ Δ t \large{y = e^{-A t} \\ y_k = y_{k-1} + y'_{k-1} \Delta t}

In the above equation, Δ t = 1 1000 \Delta t = \frac{1}{1000} . y k y_k denotes the present value of the function, y k − 1 y_{k-1} denotes the previous value of the function, and y k − 1 ′ y'_{k-1} denotes the previous value of the function's time-derivative. The parameter A A is an arbitrary positive constant.

Suppose y 0 = 1 y_0 = 1 and y 0 ′ = − A y'_0 = -A .

For 0 < A < A 1 0 < A < A_1 , the system undergoes monotonic decay toward zero.
For A 1 < A < A 2 A_1 < A < A_2 , the system undergoes damped oscillation.
For A > A 2 A > A_2 , the system undergoes divergent oscillation.

Determine A 1 + A 2 A_1 + A_2 . Assume no rounding or truncation errors (we have a perfect computer).

Note: This version of Euler's Method is more specifically known as the "forward" or "explicit" version.


The answer is 3000.

Steven Chase by Steven Chase , 37, USA

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1 solution

Steven Chase
Mar 31, 2017

Consider the time-function and its derivative.

y = e − A t y ′ = − A e − A t = − A y \large{y = e^{-At} \\ y' = -Ae^{-At} = -Ay }

It is apparent therefore that y 0 y_0 and y 0 ′ y'_0 are associated with t = 0 t = 0 . Plug these expressions into the Euler equation:

y k = y k − 1 + y k − 1 ′ Δ t y k = y k − 1 − A y k − 1 Δ t = y k − 1 ( 1 − A Δ t ) = y k − 1 ( 1 − A 1000 ) \large{y_k = y_{k-1} + y'_{k-1} \Delta t \\ y_k = y_{k-1} -A y_{k-1} \Delta t = y_{k-1}(1 - A \Delta t) = y_{k-1}(1 - \frac{A}{1000})}

So we see that there is a definite relationship between y k y_k and y k − 1 y_{k-1} . For 0 < A < 1000 0 < A < 1000 , y k y_k is equal to y k − 1 y_{k-1} multiplied by a positive constant with a magnitude less than one. This results in monotonic decay, as shown below for A = 300 A = 300 . Therefore, A 1 = 1000 A_1 = 1000 .

For 1000 < A < 2000 1000 < A < 2000 , y k y_k is equal to y k − 1 y_{k-1} multiplied by a negative number with a magnitude less than one. This results in damped oscillation, as shown below for A = 1800 A = 1800 . Therefore, A 2 = 2000 A_2 = 2000 .

For A > 2000 A > 2000 , y k y_k is equal to y k − 1 y_{k-1} multiplied by a negative number with a magnitude greater than one. This results in divergent oscillation, as shown below for A = 2200 A = 2200 . As a final result, A 1 + A 2 = 3000 A_1 + A_2 = 3000 .

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