Find Stability for Euler Method: Absolute Stability of Second Order ODE

[evinda]

Hello! (Wave)

I am looking at the following exercise.We suppose that the explicit Euler method is applied at the differential equation of second order

$\left\{\begin{matrix}
x''(t)+(\lambda+1)x'(t)+ \lambda x(t)=0\\
x(0)=1\\
x'(0)= \lambda-2
\end{matrix}\right.$

$$|\lambda|>>1$$

What step $h$ do we have to pick so that we ensure absolute stability?We set $y_1=x$ , $y_2=x'$ and so we have:$$\binom{y_1}{y_2}'=\begin{pmatrix}
0 & 1\\
- \lambda & -(\lambda+1)
\end{pmatrix} \binom{y_1}{y_2} \ \ \text{ with } \binom{y_1(0)}{y_2(0)}= \binom{1}{\lambda-2}$$Applying Euler method we get

$$\binom{y_1^{n+1}}{y_2^{n+1}}= \begin{pmatrix}
1 & h\\
- \lambda h & 1-(\lambda+1)h
\end{pmatrix} \binom{y_1^n}{y_2^n}$$

which is equivalent to

$$\binom{y_1^n}{y_2^n}=
\begin{pmatrix}
2(1-h)^n-(1-h \lambda)^n \\
-2(1-h)^n+ \lambda (1-h \lambda)^n
\end{pmatrix}$$So does it have to hold that $|1-h \lambda|<1 \Rightarrow -1<1-h \lambda <1 \Rightarrow -2<-h \lambda \text{ and } -h \lambda<0 \Rightarrow h \lambda<2 \text{ and } -h \lambda<0$ ?In my notes, it is just said that in order Euler to be absolute stable we should have $h \lambda \in [-2,0) \Rightarrow h< - \frac{2}{\lambda}$.

But is this enough? (Thinking)Also, I found an other interval for $h \lambda$. So have I done something wrong?

 

[jvicens]

Hello! I would like to provide some insight into your question about ensuring absolute stability when applying the explicit Euler method to a second-order differential equation.

Firstly, your calculations and equations seem correct. In order for the explicit Euler method to be absolute stable, we need the eigenvalues of the coefficient matrix to lie within the unit circle in the complex plane. This ensures that the solution will not diverge as we take more and more steps.

In your case, the eigenvalues are $1$ and $1-\lambda h$, so we need to have $|1-\lambda h|<1$. This is equivalent to $-2<h\lambda<0$, which can also be written as $h<-\frac{2}{\lambda}$. So your notes are correct in stating that $h<-\frac{2}{\lambda}$ is the condition for absolute stability in this case.

However, as you mentioned, there is also another interval for $h\lambda$ that satisfies this condition. This is because the eigenvalue $1$ is always within the unit circle, so we only need to consider the eigenvalue $1-\lambda h$. This means that $h\lambda$ can also be in the interval $[-2,0)$, as you found in your calculations.

Overall, both intervals $h\lambda \in [-2,0)$ and $h<-\frac{2}{\lambda}$ are correct for ensuring absolute stability. It is up to you to choose which one to use based on your specific problem and preferences. I hope this helps clarify your doubts. Keep up the good work!