Consider the viscous Burgers equation on the torus $\mathbb{T} = \mathbb{R}/(2\pi\mathbb{Z})$, with $\nu > 0$ and $f$, $u_0$ in $L^{2}(\mathbb{T})$ which are assumed to be time-independent and mean-zero. Specifically we consider:
$$u_{t} + uu_{x} = \nu u_{xx} + f$$ $$u(0, x) = u_{0}(x)$$
with $t$ in $(0, \infty)$ and $x$ in $\mathbb{T}$.
We want to show that the energy $\|u\|_{L^{2}(T)}^{2}$ is uniformly bounded in time for any $t$.
Work so far:
My work so far only gives a weaker bound, for $t$ in $[0, T]$ and not general $t$:
by multiplying the PDE by u and integrating over $\mathbb{T}$, we get:
$$\int_{\mathbb{T}}u_{t}u\,dx + \int_{\mathbb{T}}u^{2}u_{x}\,dx = \nu \int_{\mathbb{T}} u u_{xx}\,dx + \int_{\mathbb{T}}uf\,dx$$
We have the first term is $\frac{1}{2}\|u\|_{L^2}^{2}$. For the second term we get zero by using integration by parts. The third term is $-\nu\int_{\mathbb{T}}u_{x}u_{x}\,x = -\nu \int_{\mathbb{T}}(u_{x})^{2}\,dx$.
Then we consider $\int_{\mathbb{T}}uf\,dx$, and since $f$ is time-independent, we have:
$\|uf\|_{L^2} \leq \|u\|_{L^2} \|f\|_{L^2} = C\|u\|_{L^2}$
for some constant $C$
Then in total we have:
$$\frac{1}{2}\frac{d}{dt}\|u\|_{L^2}^{2} \leq -\nu \int_{\mathbb{T}}(u_{x})^{2}\,dx + C\|u\|_{L^2} \leq C\|u\|_{L^2}$$
Then we consider the differential form of Gronwall's inequality, which is that $u_t \leq \alpha(t)u(t) + \beta(t)$, and then for all $t \geq 0$, $u(t) \leq u(0)e^{\int_{0}^{t}\alpha(s)\,ds} + \int_{0}^{f}\beta(s)e^{\int_{s}^{t}\alpha(u)\,dx}\,ds$.
Then if we set $\alpha = 0$, $\beta = C\|u\|_{L^2}$, we should get a bound for any $t$ in $[0, T]$. But this still doesn't give a uniform bound for all $t$ in $(0, \infty)$.