 # 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)? musicbachv7 2022-08-10 Answered
Suppose we have a system of ODE's: a${a}^{\prime }=-a-2b$ and ${b}^{\prime }=2a-b$ with initial conditions $a\left(0\right)=1$ and $b\left(0\right)=-1$.

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
$\left(\begin{array}{cc}-1-h& -2-2h\\ 2+2h& -1-h\end{array}\right)$
The eigenvalues are $\left(-1+2i\right)\left(1+h\right)$ and $\left(-1-2i\right)\left(1+h\right)$
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The Euler discretizations of a differential system
$\left\{\begin{array}{rcl}{a}^{\prime }\left(t\right)& =& f\left(a\left(t\right),b\left(t\right)\right)\\ {b}^{\prime }\left(t\right)& =& g\left(a\left(t\right),b\left(t\right)\right)\end{array}$
are based on the difference systems
$\left\{\begin{array}{rcl}{a}_{n+1}& =& {a}_{n}+h\cdot f\left({a}_{n},{b}_{n}\right)\\ {b}_{n+1}& =& {b}_{n}+h\cdot g\left({a}_{n},{b}_{n}\right)\end{array}$
for some positive step size $h$. In the present case, this blackuces to
$\left(\begin{array}{c}{a}_{n+1}\\ {b}_{n+1}\end{array}\right)={M}_{h}\cdot \left(\begin{array}{c}{a}_{n}\\ {b}_{n}\end{array}\right),$
where
${M}_{h}=\left(\begin{array}{cc}1-h& -2h\\ 2h& 1-h\end{array}\right).$
The eigenvalues of ${M}_{h}$ are
$1-h±2\mathrm{i}h,$
hence the square of their common modulus is
$\left(1-h{\right)}^{2}+\left(2h{\right)}^{2}=1-h\left(2-5h\right).$
When both eigenvalues of ${M}_{h}$ have modulus less than 1, then $\left({a}_{n},{b}_{n}\right)\to \left(0,0\right)$ for every starting point $\left({a}_{0},{b}_{0}\right)$. When this modulus is at least 1, then $\left({a}_{n},{b}_{n}\right)\to \left(0,0\right)$ never happens except when $\left({a}_{0},{b}_{0}\right)=\left(0,0\right)$ (this is because in the present situation both eigenvalues have the same modulus).

Thus, $\left({a}_{n},{b}_{n}\right)\to \left(0,0\right)$ for every starting point $\left({a}_{0},{b}_{0}\right)$ when
$0
Note that the eigenvalues of the linear differential system are $\lambda =-1±2\mathrm{i}$ such that $\mathrm{\Re }\lambda =-1$ and $|\lambda {|}^{2}=5$. More generally, for a linear differential system with eigenvalues $\lambda$ such that $\mathrm{\Re }\lambda <0$ for every $\lambda$, Euler discretizations yield sequences with limit 0 for every starting point and every positive step size $h$ such that
$h<\underset{\lambda }{min}\left(-2\frac{\mathrm{\Re }\lambda }{|\lambda {|}^{2}}\right).$

We have step-by-step solutions for your answer! Nica2t
You need to compute the eigenvalues of the system. As a shortcut consider that
$\left(a+ib{\right)}^{\prime }=\left(2i-1\right)\left(a+ib\right)$

We have step-by-step solutions for your answer!