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I have:

$$y' = -22*y+3*sin(3*x)$$ $$0 \le x \le 3 $$ $$y(0) = 4$$

as my initial value problem.

The question is:

What is the largest step size h for which the Euler method is stable, when applied to the initial value problem above?

I believe the region of stability for Euler is something like:

$$|1+h*\lambda| < 1$$

Don't really get what $\lambda$ is meant to be?

I also have a follow-up question:

When applying the same theoretical analysis to the midpoint method, what is the largest step size h for which the method is stable?

PS. $\lambda$ is a constant in the complex numbers and I believe $Re \lambda < 0$.

  • In the case of this simple situation $\lambda$ is just $-22$; in practice $\lambda$ can't be nicely determined. – Ian Mar 06 '19 at 17:34
  • That gives me the domain 0 < h < 1/11. But then what is the largest step-size h? Surely it's not 1/11. It's 0.00000000000000001 less than that. Is there no exact answer? –  Mar 06 '19 at 17:42
  • Technically the question is probably asking for the supremum over all $h$ that give stability. Generally the stability at the exact crossover point is generally sensitive to other details, so we don't worry about it. – Ian Mar 06 '19 at 18:09

1 Answers1

2

We have

$$ \frac{y_{k+1}-y_k}{h} + 22 y_k = 3\sin(3 k h) $$

The method stability is related to the stability for the recurrence

$$ \frac{y_{k+1}-y_k}{h} + 22 y_k = 0 $$

Solving the homogeneous linear recurrence equation we have

$$ y_k = C_0(1-22h)^{k-1} $$

which is stable for $|1-22h| < 1$ or $-1 < 1-22h < 1$ or

$$ 0 < h <\frac{1}{11} $$

NOTE

Solving the recurrence

$$ \frac{y_{k+1}-y_k}{h} + 22 y_k = 3\sin(3 k h) $$

with $C_0 = 0$ we have

$$ y_k = \frac{3 h \left(\sin (3 h) \left((1-22 h)^k-\cos (3 h k)\right)+(22 h+\cos (3 h)-1) \sin (3 h k)\right)}{44 h (11 h-1)+(44 h-2) \cos (3 h)+2} $$

Follows a plot showing the result for $h = \frac{1}{11}-0.0002$ (red) and $h = \frac{1}{11}-0.02$ (blue). Both solutions are stable but the blue shows better behavior being more resistant to limited precision and round-off errors.

enter image description here

Cesareo
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  • I am asked for an exact answer, though, not a boundary. I managed to work out this domain, but not sure what the exact answer I'm meant to give is. –  Mar 06 '19 at 21:20
  • The largest is $\frac{1}{11}-\epsilon$ where $\epsilon$ is the machine precision. – Cesareo Mar 06 '19 at 21:47