One inequality connected with the linear second order ODE

Question

Is the following statement true?

Let $ a>0, b>0, h>0 $, $x(t)$ be the solution of the differential equation

$ \ddot{x}+a \dot{x}+bx=h$

with initial conditions $x(0)=u<0 , \dot{x}(0)=v$ ($\dot{x}(0)$ may be any real number).

Let $ H : [0,\infty) \rightarrow R $ be a continuous function, $ \forall t \geq 0 H(t) \geq h $ and $y(t)$ be the solution of the differential equation

$\ddot{y}+a \dot{y}+by=H(t)$

with the same initial conditions: $y(0)=u<0, \dot{y}(0)=v$.

Denote $ \tau = \inf ⁡ \left\lbrace t \geq 0 : x(t)=0 \right\rbrace $ , $T=\inf \left\lbrace t \geq 0: y(t)=0 \right\rbrace $

Then the inequality $ \dot{y}(T) \geq \dot{x}(\tau) $ occurs.

In what publication this problem was researched? Thanks.

Answer 1

For the subcase $v \geq 0$ this can be approached via the hodograph transform.

First, observe that in this case $\dot{x} > 0$ on $(0,\tau]$, and $\dot{y} > 0$ on $(0,T]$. This follows by noting that $$ \frac{d}{dt}(e^{at} \dot{x}) = e^{at} (h - b x) > 0 $$ and similarly for $y$. Hence on the relevant intervals the function $t\mapsto x(t)$ and $t \mapsto y(t)$ are invertible.

Denote, for $\sigma\in (u,0]$, $$ \eta(\sigma) = \dot{y}\circ y^{-1}(\sigma), \qquad \xi(\sigma) = \dot{x} \circ x^{-1}(\sigma) $$ A direct computation shows that $$ \frac{d}{d\sigma}(\eta^2) = \ddot{y}\circ y^{-1} $$ and similarly $\xi$. So we have $$ \frac{d}{d\sigma} (\eta^2 - \xi^2) + a (\eta - \xi) = H\circ y^{-1} - h \geq 0$$ So we get, using integrating factors $$ \frac{d}{d\sigma} \left[ (\eta^2 - \xi^2) \exp \left( \int_u^\sigma \frac{a}{\eta(s) + \xi(s)} ds\right) \right] \geq 0 $$ Since we already know that both $\eta, \xi$ are non-negative on the region, and that $\eta(u) = \xi(u) = v \geq 0$ by assumption, we conclude $\eta^2(0) - \xi^2(0) \geq 0$ as desired.

Answer 2

When $v < 0$, I don't think the result is true in general.

Let $$ y(t) = \begin{cases} -1 + (t - \frac{1}2)^2, & t \in [0,\frac12] \\ 1 - 2 \cos(t - \frac12), & t \geq \frac12 \end{cases} $$ $y$ is twice continuously differentiable, and $$ \ddot{y} + y = \begin{cases} 1 + (t - 1/2)^2, & t \in [0,1/2] \\ 1, & t \geq 1/2\end{cases} $$ and so $H(t)$ is continuous and at least 1.

Note that $y(0) = -3/4$, and $\dot{y}(0) = -1$. The value $T = 1/2 + \pi/3$, where $\dot{y}(T) = \sqrt{3} \approx 1.732$.

Next, let's solve $$ \ddot{x} + x = 1 $$ with initial data $x(0) = -3/4$ and $\dot{x}(0) = -1$. This is given by $$ x(t) = 1 - \frac{7}4 \cos(t) - \sin(t) $$ which can be rewritten for some $\theta$ as $$ x(t) = 1 - \sqrt{1 + (7/4)^2} \sin(\theta + t)$$ which means that $$ \dot{x}(\tau) = 7/4 \approx 1.75$$ and hence we have $$ \dot{y}(T) \not\geq \dot{x}(\tau) $$ in this instance.

In the OP you asked for equations where $a > 0$. If we take some positive $a$ that is $\ll 1$, by continuous dependence of solutions of ODE on coefficients, we can conclude that for all sufficiently small $a$ the inequality $$ \dot{y}(T) < \dot{x}(\tau) $$ would still hold.