Using gronwall's lemma to prove inequalities involving max$x(t)$.

Question

Given $\frac{\partial x}{\partial t}+x^3=yx^2$, $x(0)=x_{0}$, where $x$ : $\mathbb{R}^+ \mapsto \mathbb{R}$ and $y$ : $\mathbb{R}^+ \mapsto \mathbb{R} $ with $\int_{0}^{t} y^2(s) ds < \infty$ for all $t$

if $x(t)$ is continuous function satisfying the above equation in $[0,T]$

$$ \max_{t\in[0,T]}|x(t)|^2 \le |x(0)|^2e^{\int_0^T|y|^2\,\mathrm ds}.$$

$$ \int_0^T|x|^4\,\mathrm ds <|x(0)|^2+\max_{s\in[0,T]}|x(s)|^2\int_0^Ty^2\,\mathrm ds.$$ It seems like it has something to do with Gronwalls lemma.

My first approach was to set $x(t)'<y(t)x(t)^2$ and find bound for $x(t)$, but it did not work nicely.

Can anyone help me?

Answer 1

Looking at the answer to this question : $\max_{t\in[0,T]}|x(t)|^2 \le |x(0)|^2e^{\int_0^T|y|^2\,\mathrm ds}.$ using gronwall's lemma? you have $$x(t)^2 - x(0)^2 \le \int_{0}^t y(s)^2x(s)^2 \mathrm d s - \int_0^t x(s)^4 \mathrm d s$$ Then $$\int_0^t x(s) ^4\mathrm d s \le x(0)^2 - x(t)^2 + \int_0^t \left(\max_{u\in[0,T]} x(u)^2\right) y(s)^2 \mathrm d s$$