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 $t=\epsilon.$ My goal is to estimate the energy $E(u\_t) = \int (u\_{tt})^2 + |\nabla u\_t|^2.$ I already know that $E(u) \leq C t^6$ and the constant depends on the function $f.$


As usual, we first obtain the equation satisfied by $\tilde{u} =u\_t$
$$\tilde{u}\_{tt} - \frac{2}{t}\tilde{u}\_{t}-\Delta \tilde{u} = g\_t -\frac{2}{t^2}\tilde{u}.$$
Thus the time derivative of the energy can be estimated as follows,
\begin{align\*}
\dot{E} &= 2\int \tilde{u}\_t (\tilde{u}\_{tt}-\Delta \tilde{u}) \\
&= 2\int \tilde{u}\_t (2t^{-1}\tilde{u}\_t+g\_t -2t^{-2}\tilde{u})\\
&\leq \frac{10}{t}E + \frac{t}{2}\|g\_t\|\_{L^2}^2 + \frac{1}{t^3} \|u\_t\|\_{L^2}^2 \\
&\leq \frac{10}{t}E + \frac{t}{2}\|g\_t\|\_{L^2}^2 + C t^{3}.
\end{align\*}
I am not sure how to proceed after this step. Ideally, I would apply Gronwall and get $E(t)\leq C t^{10}$ but the term with $t^{3-10}$ will create singularity as $\epsilon\to 0$.



How to estimate higher order regularity for wave type equation with time dependant coefficients?
$$ \tilde{u}\_{tt} - \frac{2}{t}\tilde{u}\_{t}-\Delta \tilde{u} = g\_t -\frac{2}{t^2}\tilde{u} $$


$$ \dot{E} = 2 \int \tilde{u}\_t (2 t^{-1} \tilde{u}\_t + g\_t - 2 t^{-2} \tilde{u} ) $$


$$ \dot{E} = 2 \int \tilde{u}\_t (2 t^{-1} \tilde{u}\_t + g\_t) - 2 t^{-2} \frac{d}{dt} \int \tilde{u}^2 $$


$$ \dot{E} + \frac{d}{dt} (2 t^{-2} \int \tilde{u}^2 ) = 2 \int \tilde{u}\_t (2 t^{-1} \tilde{u}\_t + g\_t) - 4 t^{-3} \int \tilde{u}^2 $$


If you are bounding with $\leq$, the final term on the RHS now has a good sign (is negative) and can be dropped. And the first two terms can be bounded by $\frac{6}{t} E + \frac{t}{2} \|g\_t\|\_{L^2}^2 $. Absorb the $\int \tilde{u}^2$ term into $E$ you can then Gronwall that instead.


