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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 $(-1-2i)(1+h)$

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  • $\begingroup$ The Euler matrix is incorrect. $\endgroup$ – Did Jun 1 '14 at 8:19
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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 $$ h\lt\min_\lambda\left(-2\frac{\Re\lambda}{|\lambda|^2}\right). $$

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You need to compute the eigenvalues of the system. As a shortcut consider that $$ (a+ib)'=(2i-1)(a+ib) $$

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  • $\begingroup$ So the initial conditions are not important here? $\endgroup$ – ABC May 30 '14 at 20:36
  • $\begingroup$ Not here. In general it could be important to check that the initial vector is in the sum of the correct eigenspaces. But if that were decisive, then the convergence to zero is numerically instable, every random perturbation could generate an exponentially increasing component. $\endgroup$ – LutzL May 30 '14 at 20:41

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