If all the solutions of $\frac{dy}{dt}=A(t)y$ are bounded, then all the solutions of $\frac{dz}{dt}=[A(t)+B(t)]z$ are bounded

Question

In the book by Richard Bellman, Stability theory of differential equations

Theorem 6 (p.43)

If all the solutions of $\frac{dy}{dt}=A(t)y$ are bounded, then all the solutions of $\frac{dz}{dt}=[A(t)+B(t)]z$ are bounded, provided that:

• (a),$\int_{t_0}^{\infty}\left \|B(t) \right \|\mathrm{d}t<\infty$
• (b),$\lim_{t \to \infty}\int_{t_0}^{t}TrA(t_1)\mathrm{d}t_1>-\infty$

or, in particular, that

• (b') $Tr(A)=0$.

Condition (b') is relevant to the important equation: $$\boxed{\ddot{u}+a(t)u=0}$$ which is equivalent to a two-dimensional system satisfying (b')

I don't understand "Condition (b') is relevant to the important equation: $$\boxed{\ddot{u}+a(t)u=0}$$ which is equivalent to a two-dimensional system satisfying (b')".

I don't see the equation $\ddot{u}+a(t)u=0$ makes sense with the theorem above, if you understand this theorem, you can explain to me? anyone? Thanks!

Answer 1

$A(t)$ and $B(t)$ are matrices, not just ordinary functions. To turn a second-order differential equation into a first-order differential equation with matrices, let $v$ be the column vector $\left[\begin{array}{cc} u\\ \dot{u}\end{array}\right]$. The derivative of that is $\left[\begin{array}{c}\dot{u}\\ \ddot{u}\end{array}\right]=\left[\begin{array}{cc} 0&1\\-a(t)&0\end{array}\right]\left[\begin{array}{c}u\\ \dot{u}\end{array}\right] =A(t)v$