Question:
Consider the differential equation $y'(t) = \lambda y(t)$ with y(0) given, where $\lambda$ is complex. Write down the analytical solution. How does this solution behave if the real part of $\lambda$ is positive, negative, or zero? Next, consider Euler's method and the backward Euler method for this problem. Under what conditions are each of these methods zero-stable? Under what conditions are each of these methods absolutely stable? Which of these methods is A-stable? Which of these methods is A(0)-stable? Which of these methods is A_0-stable? If the real part of $\lambda$ is negative, do we want the numerical method to be absolutely stable? If the real part of $\lambda$ is positive, do we want the numerical method to be absolutely stable?
Discussion:
(1)Firstly the analytical solution to the differential equation $y'(t) = \lambda y(t)$ with y(0) given is $ y(t)=y(0)e^{\lambda t}.$
This solution grows exponentially with oscillation if $Re(\lambda)>0$, and it decays exponentially with oscillation if $Re(\lambda)<0$. When $Re(\lambda)=0$ the solution keeps constant as y(0) if $Im(\lambda) = 0$, or oscillates about y(0) if $Im(\lambda) \neq 0$
(2)The Euler's method for this problem is $ y_{n+1} = y_n + \lambda h y_n,$ while the backward Euler method is $ y_{n+1} = y_n + \lambda h y_{n+1}. $
(3)For the Euler's method the first characteristic polynomial is $ p(\zeta) = \zeta -1$ and the second characteristic polynomial is $ q(\zeta) = 1.$
The first characteristic polynomial for the backward Euler method is $ p(\zeta) = \zeta -1 $ and the second characteristic polynomial $q(\zeta) = \zeta. $
Since the above two methods have first characteristic polynomial with the only zero $ \zeta = 1$, both of them are zero-stable.
(4) For the Euler's method $ p(\zeta) - h\lambda q(\zeta) = \zeta - 1 - h\lambda $ has its root at $ \zeta = 1 + h\lambda. $ Absolute stability requires the root to satisfy $ | \zeta | \le 1.$ Therefore $ |1 + h\lambda| \le 1 $, the absolute stability region for Euler's method is the a unit circle with center -1.
The backward Euler method takes $ p(\zeta) - h\lambda q(\zeta) = \zeta - 1 - h\lambda \zeta $ with root at $ \zeta = \frac{1}{1-\lambda h}. $ $ | \zeta | \le 1 $ leads to $ |\lambda h -1| \ge 1.$ Hence the absolute stability region for the backward Euler method is the exterior of the unit circle with center 1.
(5) Since the negative half-plane is contained in the region of absolute stability of the backward Euler method, the backward Euler method is A-stable and therefore A(0) stable and $A_0$ stable. On the other hand, the Euler's method only has its region of absolute stability in the unit circle with center -1, thus it is neither $A_0$ stable, A(0) stable nor A-stable.
(6) Finally if $Re(\lambda)<0$ we want the numerical method to be absolutely stable since the analytical solution decays along with the time. But if $Re(\lambda)>0$ we don't need the numerical method to be absolutely stable as the analytical solution grows exponentially and any perturbation in its initial value will result in a great change in $y(t)$ when t is large enough.
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