# differential inequality involving p.s.d matrix

## Problem description

Let $$h(t) > 0$$ be a continuous real function and $$x(t) \in \mathbb{R}^{3}$$ be also a continuous function. Let $$T(t) = h + \dot{x}^T\dot{x}$$ It is known that $$\dot{T} = -\dot{x}^TP\dot{x}$$ where $$P$$ is a positive definite symmetric real matrix. Also we have: $$\ddot{x} = f(x,\dot{x})$$ We may also assume that $$\exists \lim_{t\to \infty} \dot{x}^T\dot{x} = 0$$. (We obtained this by the use of Barbalat's lemma and the fact that $$\ddot{T}$$ is bounded)

## Question

I am trying to obtain some estimates on the velocity of convergence to zero for $$\dot{x}^T\dot{x}$$.

## My work

This is what I've done so far:
$$-p_2\cdot \dot{x}^T\dot{x}\leq \dot{T} \leq -p_1\cdot \dot{x}^T\dot{x}$$ Furthermore, because $$\exists 0< T_1 \leq T \leq T_0$$ we have: $$\frac{-p_2}{T_1} \dot{x}^T\dot{x} \leq \frac{\dot{T}}{T} \leq -\frac{p_1}{T_0} \dot{x}^T\dot{x}$$ Integrating both sides, from $$0$$ to $$t$$ one obtains:

$$-\frac{p_2}{T_1} \int_{0}^t \dot{x}^T\dot{x} \leq \log(\frac{T}{T_0}) \leq -\frac{p_1}{T_0} \int_{0}^t \dot{x}^T\dot{x}$$ Let $$g(t) = \int_0^t \dot{x}^T\dot{x}; \hspace{0.5cm} \frac{p_2}{T_1} = k_2; \hspace{0.5cm} \frac{p_1}{T_0} = k_1$$, hence

$$T_0 e^{-k_2 g} \leq h + g' \leq T_0 e^{-k_1 g}$$

$$T_0 \leq h(t) e^{k_2 g} + e^{k_2g} g' ; \hspace{0.5cm} h(t) e^{k_1 g} + e^{k_1g} g' \leq T_0$$ Let $$u_1 = e^{k_1 g}$$ and $$u_2 = e^{k_2 g}$$ hence $$T_0 \leq h u_2 + \frac{1}{k_2}u_2'; \hspace{0.5cm} h u_1 + \frac{1}{k_1}u_1' \leq T_0$$ I think I should apply Gronwall's lemma here, but I am not very sure on how to proceed ...