How to estimate higher order regularity for wave type equation with time dependant coefficients?

Question

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 $\epsilon\to 0$.

Answer 1

$$ \tilde{u}_{tt} - \frac{2}{t}\tilde{u}_{t}-\Delta \tilde{u} = g_t -\frac{2}{t^2}\tilde{u} $$

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

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

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

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.