Suppose we have a system of ODE's: $a' = -a – 2b$ and $b' = 2a-b$ with initial conditions $a(0)=1$ and $b(0)=-1$.
How can we find the maximum value of the step size such that the norm a solution of the system goes to zero (if we apply the forward Euler formula)?
Edit: the main part is to calculate the eigenvalues of the following matrix, based on the Euler method, this becomes
\begin{pmatrix}
-1-h & -2-2h \\
2+2h & -1-h
\end{pmatrix}
The eigenvalues are $(-1+2i)(1+h)$ and $(-1-2i)(1+h)$
Best Answer
The Euler discretizations of a differential system $$\left\{\begin{eqnarray}a'(t)&=&f(a(t),b(t))\\b'(t)&=&g(a(t),b(t))\end{eqnarray}\right.$$ are based on the difference systems $$\left\{\begin{eqnarray}a_{n+1}&=&a_{n}+h\cdot f(a_n,b_n)\\ b_{n+1}&=&b_n+h\cdot g(a_n,b_n)\end{eqnarray}\right.$$ for some positive step size $h$. In the present case, this reduces to $$\begin{pmatrix}a_{n+1}\\b_{n+1}\end{pmatrix}=M_h\cdot\begin{pmatrix}a_{n}\\b_{n}\end{pmatrix},$$ where $$M_h=\begin{pmatrix}1-h&-2h\\2h&1-h\end{pmatrix}.$$ The eigenvalues of $M_h$ are $$1-h\pm2\mathrm ih,$$ hence the square of their common modulus is $$(1-h)^2+(2h)^2=1-h(2-5h).$$ When both eigenvalues of $M_h$ have modulus less than $1$, then $(a_n,b_n)\to(0,0)$ for every starting point $(a_0,b_0)$. When this modulus is at least $1$, then $(a_n,b_n)\to(0,0)$ never happens except when $(a_0,b_0)=(0,0)$ (this is because in the present situation both eigenvalues have the same modulus).
Thus, $(a_n,b_n)\to(0,0)$ for every starting point $(a_0,b_0)$ when $$0\lt h\lt\frac25.$$ Note that the eigenvalues of the linear differential system are $\lambda=-1\pm2\mathrm i$ such that $\Re\lambda=-1$ and $|\lambda|^2=5$. More generally, for a linear differential system with eigenvalues $\lambda$ such that $\Re\lambda\lt0$ for every $\lambda$, Euler discretizations yield sequences with limit $0$ for every starting point and every positive step size $h$ such that $$ h\lt\min_\lambda\left(-2\frac{\Re\lambda}{|\lambda|^2}\right). $$