$\max_{t\in[0,T]}|x(t)|^2 \le |x(0)|^2e^{\int_0^T|y|^2\,\mathrm ds}.$ using gronwall's lemma?

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 $[0,T]$

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

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

We have $$x(t)^2 - x(0)^2 = 2\int_0^t x'(s)x(s) \mathrm ds = 2\int_0^t y(s)x(s)^3\mathrm d s - 2\int_0^tx(s)^4\mathrm ds \underbrace{\le}_{\text{using } 2ab \le a^2 +b^2} \int_0^t \left( y(s)^2x(s)^2 + x(s)^4 \right) \mathrm d s -2\int_0^tx(s)^4\mathrm ds = \int_0^t y(s)^2x(s)^2 \mathrm d s$$ Now you use the Gronwall Lemma and you have your result.