How to find a initial condition such that the solution of a second ordered ODE is bounded?

Question

Let $A=\pmatrix{1&2\\1&1}$. The ODE is $X''= AX$. How to find an initial condition such that the solution of this second ordered differential equation is bounded?

Answer 1

The matrix has eigenvalues $1\pm\sqrt2$, one positive, one negative. The solutions on the positive eigenspace have terms $e^{\pm\sqrt{1+\sqrt2}t}$, the other $\sin(\sqrt{\sqrt2-1}t)$ and $\cos(\sqrt{\sqrt2-1}t)$. Depending on whether you want the solution bounded for $t>0$ or for all $t\in\Bbb R$, you have 3 or 2 bounded components and need to chose the initial values so that they avoid the eigen-spaces of the other 1 or 2 components.

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If only the trigonometric components are admissible, then the bounded solutions have $$ x_1=c_1\cos(\sqrt{\sqrt2-1}t)+c_2\sin(\sqrt{\sqrt2-1}t) $$ and from reading the first line of the system one also gets $$ x_2=\frac{x_1''-x_1}2=-c_1\frac{\sqrt2}2\cos(\sqrt{\sqrt2-1}t) - c_2\frac{\sqrt2}2\sin(\sqrt{\sqrt2-1}t) $$ from which one can read the space of admissible initial values $X(0), X'(0)$.