If $x:[0,1]\rightarrow \mathbb{R}$ verifies $0\leq x(t)\leq \int_0^te^sx(s)^2ds \ \forall t\in [0,1]$ then $x(t)=0 \ \forall t\in[0,1]$

Question

As part of my differential equations homework we have this problem:

If $x:[0,1]\rightarrow \mathbb{R}$ verifies $0\leq x(t)\leq \int_0^te^sx(s)^2ds \ \forall t\in [0,1]$ then $x(t)=0 \ \forall t\in[0,1]$

I tried bounding the integral using that $x$ is continuous and $[0,1]$ compact but I gained no useful info. Reasoning by contradiction doesn't seems useful neither, and I couldn't use any result from this unit as it was all about solving linear ODEs of order $\geq 2$.

Any help or hint is appreciated

Answer 1

A common technique is to reformulate the integral inequality as a differential inequality. Here we can define $$ f(t) = \int_0^te^sx(s)^2 \, ds \, . $$ Then $$ f'(t) = e^t x(t)^2 \le e^t f(t)^2 \, . $$ Note that $f$ is nonnegative and $f(0) = 0$.

Now we want to divide by $f(t)^2$ and integrate. That is only possible on intervals where $f$ is not zero. So let us assume that $f$ is not identically zero. Then we can find an interval $[a, b]$ with $f(a) = 0$ and $f(t) > 0$ for $a < t \le b$. For $a < c < b$ is $$ \frac{1}{f(c)} - \frac{1}{f(b)} = \int_c^b \frac{f'(t)}{f(t)^2} \, dt \le \int_c^b e^t \, dt = e^b - e^c < e^b \, . $$ But for $c \to a^+$ the left-hand side converges to $+\infty$ whereas the right-hand side stays bounded. This is a contradiction.

So $f$ is identically zero, and since $0 \le x(t) \le f(t)$, $x$ is identically zero as well.

Some remarks:

• As @FShrike observed, the exponential function can be replaced by any continuous positive function in the differential inequality and the same conclusion still holds.

• Also the domain $[0, 1]$ can be replaced by an arbitrary closed interval $[x_0, x_1]$ or $[x_0, \infty)$.

• As @ OliverDíaz said, this is a special case of Grönwall's inequality in its integral form: $$ u(t)\leq \alpha (t)+\int _{a}^{t}\beta (s)u(s)\,ds $$ with $u(t) = x(t)$, $\alpha(t) = 0$, $\beta(s) = e^{s} u(s)$.