Consider the following wave-type equation,
$$u_{tt}-\frac{2}{t}u_t-\Delta u=g(t,x)$$
where $(t,x)\in [\epsilon, 1]\times \mathbb{R}^3$ for some $\epsilon>0.$ Furthermore assume that $(u,u_{t})=(0,0)$ at $t=\epsilon.$ Then define the energy of the solution $u$ as follows,
$$E(t) = E(u(t))=t^2 \int (u_t)^2+|\nabla u|^2 dx.$$
My **goal** is to derive a time decay estimates on the energy $E$.

As usual, we compute the time derivative of $E$. Thus, $$\frac{d}{dt}E(t) = \frac{2E}{t} + 2t^2 \int u_t u_{tt} + \nabla u\cdot \nabla u_t.$$ Using integration by parts we get, $$\frac{d}{dt}E(t) = \frac{2E}{t} + 2t^2 \int u_t \left(g+\Delta u + \frac{2}{t}u_t\right) - u_t\Delta u \\ =\frac{2E}{t} + 2t^2 \int u_t g + 4t \int (u_t)^2.$$ For the third term, we can simply control it as follows, $$4t \int (u_t)^2 \leq 4t\int (u_t)^2 + |\nabla u|^2\leq \frac{4E(t)}{t}.$$ For the second term, using the above estimate, we get $$ 2t^2 \int u_t g\leq 2t^2 \|u_t\|_{L^2}\|g\|_{L^2}\leq 2 t \sqrt{E(t)}\|g\|_{L^2}$$ which implies, $$\frac{dE}{dt}\leq \frac{6E}{t} + 2t \sqrt{E}\|g\|_{L^2}.$$

I am not sure how to proceed further. I would like to apply Gronwall's inequality but the right-hand side does not seem very nice. I started with $g\in L^2$, but perhaps I need more regularity for the function $g.$ Any ideas here will be much appreciated.

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