Let $u:\mathbb{R}^N\times (t_1,t_2)\to(0,\infty)$ be a measurable function such that $w(x,t)=(h-u)_{+}(x,t)$ and let $p>2$. Let $B_r$ be a ball of radius $r$ and onsider the following integral for $(a,b)\subset(t_1,t_2)$ defined by $$ I=\int_{a}^{b}\int_{B_r}\frac{\partial}{\partial t}(u^{p-1})\theta^p(x)\eta^p(t)w(x,t)\,dx\,dt, $$ where $\eta\in C_c^{\infty}((t_1,t_2))$ such that $0\leq\eta\leq 1$ and $\eta(b)\neq 0$ and $\eta(a)=0$ and $\theta\in C_c^{\infty}(B_r)$ such that $0\leq\theta\leq 1$. Then is the following estimate true? $$ I\leq -c\int_{B_r}w^p(x,b)\theta^p(x)\,dx+p(p-1)\int_{a}^{b}\int_{B_r}\sum_{d=0}^{\infty}h^{p-2-d}\binom {p-2}{d}\frac{w^{d+2}}{d+2}\eta^{p-1}\eta_t\,dx\,dt. $$ I am finding difficulty to prove the above estimate. In fact the second term in the right hand side comes by splitting $u$ as $u=h-(h-u)$ and then writing $$\frac{\partial}{\partial u}(u^{p-1})=(p-1)u^{p-2}u_t\\=(p-1)h^{p-2}(1-\frac{h-u}{h})^{p-2}u_t=(p-1)\sum_{d=0}^{\infty}h^{p-2-d}(-1)^d (h-u)^d u_t.$$
Plugging this term into $I$, we obtain after using integration by parts $$ I=(p-1)\int_{a}^{b}\int_{B_r}\sum_{d=0}^{\infty}(-1)^{d+1}h^{p-2-d}\binom{p-2}{d}(h-u)^d \theta^p(x)\eta^p(t)w(x,t)w_t(x,t)\,dx\,dt\\ =(p-1)\int_{a}^{b}\int_{\{0<u<h\}}\sum_{d=0}^{\infty}(-1)^{d+1}\binom{p-2}{d}[\frac{w^{d+2}}{d+2}]_t\theta^p(x)\eta^p(t)\,dx\,dt\\ =(p-1)\int_{\{0<u<h\}}\sum_{d=0}^{\infty}(-1)^{d+1}\binom{p-2}{d}h^{p-2-d}\frac{w^{d+2}(x,b)}{d+2}\theta^p(x)\,dx-p(p-1)\int_{a}^{b}\int_{\{0<u<h\}}\sum_{d=0}^{\infty}(-1)^dh^{p-2-d}\binom{p-2}{d}\frac{w^{d+2}}{d+2}\theta^p(x)\eta^{p-1}\eta_t\,dx\,dt\\ \leq (p-1)\int_{\{0<u<h\}}\sum_{d=0}^{\infty}(-1)^{d+1}\binom{p-2}{d}h^{p-2-d}\frac{w^{d+2}(x,b)}{d+2}\theta^p(x)\,dx+p(p-1)\int_{a}^{b}\int_{B_r}\sum_{d=0}^{\infty}h^{p-2-d}\binom {p-2}{d}\frac{w^{d+2}}{d+2}\eta^{p-1}\eta_t\,dx\,dt. $$ Now I got stuck to derive the estimate for the first term in the above inequality. In fact my question is it somehow possible to bound $I$ by the first term in the rhs as mentioned above? Thank you very much.