Example for a continuous function $x \geq 0$ on $[0,\infty)$ so that $x(0)=0$ and $\left(x(t) \right)^2\leq 2+\int_{0}^{t}x(u)du,~~~\forall ~t\geq 0$ Q. Suppose $x:[0,\infty)\to [0,\infty)$ is continuous and $x(0)=0.$ If $$\left(x(t) \right)^2\leq 2+\int_{0}^{t}x(u)du,~~~\forall ~t\geq 0,$$ then which of the following is TRUE?

*

*$x(\sqrt{2})\in [0,2]$

*$x(\sqrt{2})\in [0,\frac{3}{\sqrt{2}}]$

*$x(\sqrt{2})\in [\frac{5}{\sqrt{2}},\frac{7}{\sqrt{2}}]$

*$x(\sqrt{2})\in [10,\infty)$
By letting $y(t)=2+\int_0^tx(u)du$, we have $\sqrt{x(t)^2}=x(t)=y'(t) \leq \sqrt{y(t)},$
and thereby considering the monotonicity of the function $g(t)=2\sqrt{y(t)}-t$, we get $x(\sqrt{2})\leq 3/\sqrt 2$.
Now, can you point out a function $x$ with $2<x(\sqrt 2) \leq 3/\sqrt 2$ to ignore option 1?
 A: Preliminary remarks: We know that $y(t) \le \left( \sqrt 2 + \frac 12 t \right)^2$ and therefore $x(t) \le  \sqrt 2 + \frac 12 t$. But (as you already observed) these upper bound does not satisfy the initial value $x(0) = 0$. So the idea is to construct a piecewise linear function. The first part “increases quickly” but satisfies $x(t)^2 \le 2$. The second part is the upper bound $\sqrt 2 + \frac 12 t$, but translated in the argument to make the function continuous.
This leads to the following construction. It shows that $x(\sqrt 2)$ can be arbitrarily close to $3/\sqrt 2$.
Consider the function $x:[0,\infty)\to [0,\infty)$ defined as
$$
 x(t) = \begin{cases}
\frac{\sqrt 2}{\epsilon} t & \text{ for } 0 \le t \le \epsilon \\
\sqrt 2 + \frac 12(t-\epsilon) & \text { for } t \ge \epsilon
\end{cases}
$$
where $\epsilon$ is some “small” positive real number to be chosen later.
$x$ is continuous with $x(0) = 0$ and
$$
 x(\sqrt 2) = \frac{3}{\sqrt 2} - \frac 12 \epsilon
$$
so that $x(\sqrt 2) > 2$ for sufficiently small $\epsilon$.
It remains to show that this function satisfies the integral inequality. For $0 \le t \le \epsilon$ this is surely the case:
$$
 x(t)^2 \le 2 \le 2 + \int_0^t x(u) \, du \, .
$$
And for $t > \epsilon $ we have
$$
2 + \int_0^t x(u) \, du = 
 2 + \int_0^\epsilon \frac{\sqrt 2}{\epsilon} u \, du 
+ \int_\epsilon^{t} \left( \sqrt 2 + \frac 12(u-\epsilon)\right) \, du \\
= 2 + \frac{\epsilon}{\sqrt 2} + \sqrt 2 (t-\epsilon) + \frac 14 (t-\epsilon)^2 \\
> 2 + \sqrt 2 (t-\epsilon) + \frac 14 (t-\epsilon)^2 \\
= \left( \sqrt 2 + \frac 12 (t-\epsilon)\right)^2 = x(t)^2 \, .
$$
