Reference request: Gronwall's inequality with negative sign(s)

Question

The following claim is a consequence of Gronwall's theorem

Let $x \colon [0,\infty) \to \mathbb R$ with $x(0) = 0$ be a continuously differentiable function, whose derivative satisfies $$ \dot x(t) \le a(t)x(t) + C$$ with a constant $C \ge 0$ and an integrable function $a \colon [0,\infty) \to (-\infty,0]$. Then we have $$ x(t) \le tC $$

See e.g. this blog post.

Gronwall's theorem can be found in many text books; however, all those that I looked at only considered the case of nonnegative $x$ and $a$, whereas I need $x$ to be arbitrary and $a$ to be nonpositive.

Q: Is there a (standard) text book (which can be cited) that treats this general case?

Answer 1

After searching several books on ODEs I think you could refer to a stronger result, the so called comparison lemma which appears in appendix E.4 of

[1] W.J. Terrell, Stability and stabilization, Princeton University Press, 2009.

Lemma E.4 [1] (Comparison Lemma)

Suppose $w(t, u)$ is continuous on a connected open set $\Omega$ in $R^2$ and the scalar equation $$\dot{u} = w(t, u)\qquad\qquad(E.3)$$ has a unique solution for each initial condition $(t_0, u_0)$. If $u(t)$ is a solution of (E.3) on $[a, b]$ and $v(t)$ is a solution of the differential inequality $$\dot{v}\leq w(t,v)$$ on $[a, b)$ with $v(a) \leq u(a)$, then $$ v(t) \leq u(t),\qquad a \leq t < b.$$

In your case $$\dot{x}\leq a(t)x(t)+C\leq C$$ as $a(t)x(t)\leq 0$ for all $t$ by definition. The solution of $\dot{u}=C$ is $u(t)=u(0)+Ct$ and for $u(0)=0\geq x(0)$ the conditions of Lemma E.4 of [1] hold true and therefore $$x(t)\leq u(t)=Ct.$$

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The above analysis holds true as long as $x(t)\geq 0$ for all$t\in[0,\infty)$. If $x$ is negative over some intervals then the above Lemma cannot be directly applied but needs a modification.

We consider two cases:

a) If $x(t)<0$ then $x(t)<0\leq Ct$ since $C\geq 0$.

b) For $x(t)>0$ due to continuity (and starting from $0$) there exists a time instant $$t^*=\inf\{s\in[0,t):x(t^*)=0\}$$ which is the nearest time up to $t$ for which $x(t^*)=0$. Obviously $x(s)\geq 0$ for all $s\in[t^*,t]$ and we can apply Lemma E.4 as described before in $[t^*,t]$ to prove that $$x(t)\leq C(t-t^*)\leq Ct$$

Answer 2

Here is the usual proof of (the differential form of) Gronwall's lemma. One assumes that some differentiable function $x$ is such that $$x'(t)\leqslant a(t)x(t).$$ Then one considers $A(t)=\displaystyle\int_0^ta(s)\mathrm ds$ and one notes that $$(\mathrm e^{-A(t)}x(t))'=\mathrm e^{-A(t)}\cdot(x'(t)-a(t)x(t))\leqslant0, $$ hence, for every $t\geqslant0$, $$\mathrm e^{-A(t)}x(t)\leqslant\mathrm e^{-A(0)}x(0), $$ that is, $$ x(t)\leqslant x(0)\exp\left(\int_0^ta(s)\mathrm ds\right). $$ This was Gronwall's lemma. Now, the task is to adapt the brilliant idea used in its proof to the setting you are interested in! Your hypotheses are that $$ x'(t)\leqslant a(t)x(t)+C, $$ and $x(0)=0$, hence...