Stability theory: Every solution of the scalar equation: $\ddot{x}+\left [a+b(t) \right ]x=0$ is bounded in $\left [t_0, +\infty \right )$.

Question

EDITED

Prove that: If $a>0$ and $$\int_{t_0}^{\infty} |b(t_1)|\mathrm{d}t_1<+\infty$$ then every solution of the scalar equation: $$\ddot{x}+\left [a+b(t) \right ]x=0$$ is bounded in $\left [t_0, +\infty \right )$.

I have thought about my problem, but I still no solution :( . Can anyone have an idea or a solution? Any help will be appreciated! Thanks.

Answer 1

Noting that solutions of $\ddot{y} + a y = 0$ have $\dot{y}^2 + a y^2$ constant, I suspect you might start by looking at $E(t) = \dot{x}^2 + a x^2$ for solutions of your equation.

EDIT: With $E(t)$ defined as above, you get $$\dot{E}(t) = 2 \dot{x} (\ddot{x} + a x) = - 2 b(t) x \dot{x} \le \frac{|b(t)|}{\sqrt{a}} (\dot{x}^2 + a x^2) = \frac{|b(t)|}{\sqrt{a}} E(t)$$ Thus $$\dfrac{d}{dt} \log E(t) = \dfrac{\dot{E}(t)}{E(t)} \le \dfrac{|b(t)|}{\sqrt{a}}$$ Can you take it from there?

Answer 2

Write as a 2-dimensional differential equation in matrix form by defining $y = \dot x$. Then Use the Gronwall Lemma (or the version known as the Bellman-Gronwall Lemma) to treat the time-dependent "small" part that depends on $b(t)$ as a perturbation of the time-invariant stable system. (A textbook in adaptive control such as Sastry & Bodson, Ioannou & Sun, or Narendra & Annaswamy well set you in the right direction on the use of the Bellman-Gronwall for stability analysis of your problem).

Then let me know whether your system is indeed stable for any $b(t)$. I suspected not, because the time-invariant unperturbed system is merely stable, not asymptotically so, and the time-varying perturbation could be chosen to resonate with its modes, which are $i\sqrt{a}$ and $-i\sqrt{a}$, but Artem's argument shows that it is indeed stable. Thanks!

Answer 3

EDITED

• By leting: $u_1=\dot{x}$, $u_2=\dot{u_1}$ then we have: $$\begin{cases} & \text{ } \dot{u_1}=u_2 \\ & \text{ } \dot{u_2}=-[a+b(t)]u_1 \end{cases}$$ Hence, $$\begin{bmatrix} \dot{x}\\ \ddot{x}\\ \end{bmatrix}=\begin{bmatrix} &0&1\\ &-(a+b(t)& 0 \end{bmatrix}\begin{bmatrix} x\\ \dot{x}\\ \end{bmatrix} \implies \frac{dv}{dt}=A(t)v, (2)$$.

Now, We need to show that all the solutions of the scalar equation:$$\ddot{x}+\left [a+b(t) \right ]x=0, (1)$$ are bounded.

• We have $\dfrac{dv}{dt}=A(t)v$ can be rewrite: $$\begin{bmatrix} &0&1\\ &-(a+b(t)& 0 \end{bmatrix} =\begin{bmatrix} &0&1\\ &-a& 0 \end{bmatrix} +\begin{bmatrix} &0&0\\ &-b(t)& 0 \end{bmatrix}=A+B(t)$$. The solution $v(t)$ of (2) such that: $$v(t)=e^{At}\cdot v(t_0)+\int_{0}^{t} e^{(t-s)A}\cdot B(s) \cdot v(s)\mathrm{d}s$$.

• Now, since Artem's hints, implies $v(t)$ is bounded!

I did exactly as what you say! Is it correct?

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Here's my solution:

Aplying the Theorem:

Assume that:

• The system: $$\dfrac{dx}{dt}=Ax$$ is stable.

• And $$\int_{0}^{\infty}\|B(s)\|\mathrm{d}s<\infty$$ Whence, system: $$\dfrac{dy}{dt}=[A+B(t)]y$$ is stable.

Now, $A=\begin{bmatrix} &0&1\\ &-a& 0 \end{bmatrix}$; and $B=\begin{bmatrix} &0&0\\ &-b(t)& 0 \end{bmatrix}$. QED.

Answer 4

I will give you a general result.

Your equation can be rewritten as a system of two first order equations, in the general form $$ \dot u=(A+B(t))u,\quad u=(u_1,u_2)^{\top}. $$ Assume that the system $$ \dot u=Au $$ has the fundamental matrix solution $$ \Phi(t)=\exp(At). $$ Since the eigenvalues of $A$ have zero real parts, it can be shown that $$ \|\Phi(t)\|\leq C $$ for some constant $C$.

Now look for a solution of the original system in the form $u=\Phi(t)z$. After simplifications you should be able to find $$ \dot z=\Phi^{-1}(t)B(t)\Phi(t)z. $$ Hence the solution to the system can be written as $$ u(t)=\Phi(t)u_0+\int_0^t\Phi(t-\tau)B(\tau)u(\tau)d\tau. $$ From the last, we have $$ \|u\|\leq C\|u_0\|+\int_{0}^tC\|B(\tau)\|\|u(\tau)\|d\tau. $$ By Grownwall's inequality this is the same as $$ \|u\|\leq C\|u_0\|\exp( C\int\|B(\tau)\|d\tau), $$ which proves that the solutions are bounded.