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

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 $$\epsilon\to 0$$.

$$\tilde{u}_{tt} - \frac{2}{t}\tilde{u}_{t}-\Delta \tilde{u} = g_t -\frac{2}{t^2}\tilde{u}$$
$$\dot{E} = 2 \int \tilde{u}_t (2 t^{-1} \tilde{u}_t + g_t - 2 t^{-2} \tilde{u} )$$
$$\dot{E} = 2 \int \tilde{u}_t (2 t^{-1} \tilde{u}_t + g_t) - 2 t^{-2} \frac{d}{dt} \int \tilde{u}^2$$
$$\dot{E} + \frac{d}{dt} (2 t^{-2} \int \tilde{u}^2 ) = 2 \int \tilde{u}_t (2 t^{-1} \tilde{u}_t + g_t) - 4 t^{-3} \int \tilde{u}^2$$
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.
• Thank you very much for your detailed answer! I just had one question what do you mean absorb the $\int \tilde{u}^2$ term in $E$? Nov 10, 2021 at 19:32
• Redefine $\tilde{E} = E + 2t^{-2} \|\tilde{u}\|_2^2$ and you will have $\frac{d}{dt} \tilde{E} \leq \frac{6}t \tilde{E} + \frac{t}{2} \|g_t\|_2^2$. Nov 11, 2021 at 13:57