Question:
Suppose that $w(t)$ is a non-negative continuous function, and there exists a constant $a \in \Bbb R$ such that
\begin{align} \frac{dw}{dt} + aw & \leq 0 \qquad \qquad \text{for }t \geq 0 \\ w(0) & = 0 \end{align}
Is it then true that $w(t)$ is a decreasing function (on the interval $[0,\infty)$)?
Attempt:
If in addition it is known that $a \geq 0$, then this would be easy, since we would have
$$\frac{dw}{dt} + aw \leq 0 \implies \frac{dw}{dt} \leq -aw \leq 0$$
However, I want to prove this for general constant $a$.
I have a feeling that I can somehow compare this to the exponential function $Ae^{-at}$ as this is the solution of the differential equation
$$\frac{dw}{dt} + aw =0$$
Gronwall's inequality also comes to mind, but I am not sure how to use it.
Any help would be much appreciated. Thanks!
Edit:
It seems to me that $w$ must in fact be the zero function. Thinking about it in an intuitive/heuristic way,
$$w'(0) + aw(0) \leq 0 \implies w'(0) \leq 0$$
since $w(0)=0$, i.e. $w$ is decreasing at $t=0$. However, $w$ is non-negative with $w(0)=0$, so in fact $w'(0)=0$. This implies that $w(0 + \delta) = 0$ and we can repeat the above argument.
Your intuition is correct and in fact the only non-increasing non-negative function $w$ with $w(0) = 0$ is the zero function.
To show this, set $v(t) = e^{at}w(t)$. Then $v(0) = 0, \, v(t) \ge 0$ for all $t$, and $ v'(t) = e^{at}(a w(t) + w'(t)) \le 0 \, . $ Therefore $0 \le v(t) \le v(0) = 0$ for all $t \ge 0$ and thus $w(t) = 0$ for all $t \ge 0$.