Consider the following heat equation on $\Omega\subset\mathbb{R}^d$ with smooth boundary and $v(x)\in L^2(\Omega)$ $$ \begin{cases} u_t - \Delta u = 0 & \text{in } \Omega\times(0,+\infty) \\ u(x,t) = 0 & \text{on }\partial\Omega\times(0,+\infty)\\ u(x,0) = v(x) & \text{in }\Omega\times\{t=0\} \end{cases} $$ Prove that there exists a constant $C=C(\Omega, d)>0$ such that $$ \Vert \nabla u\Vert^2 \leq \dfrac{C}{t}\Vert v\Vert^2 $$ where $\nabla = \nabla_x$ and $$ \Vert f\Vert^2 := \int_{\Omega}|f|^2dx \quad (L^2(\Omega) \text{ norm}) $$
Here is my attempt on this problem: Using energy function, it is not difficult to show that for all $t>0$, $$ \Vert u\Vert \leq \Vert v\Vert $$ We are going to prove that $$ t\Vert \nabla u\Vert^2 \leq C\Vert u\Vert^2 $$ for some constant $C(\Omega,d)>0$. To show this is true, take the derivative with respect to $t$ of both sides, we have $$ \int_{\Omega}|\nabla u|^2 + 2t\int_{\Omega}\nabla u\cdot\nabla u_t dx \leq 2C\int_{\Omega} uu_tdx \tag{1} $$ Using some derivative manipulation along with Green's identity, we have \begin{align*} \int_{\Omega}uu_tdx &= \Vert u\Vert\dfrac{d}{dt}(\Vert u\Vert)\\ % \int_{\Omega}|\nabla u|^2dx &= \int_{\Omega}u\dfrac{\partial u}{\partial\mathbf{n}}dS - \int_{\Omega}u\Delta udx = -\int_{\Omega}uu_tdx \\ &= -\Vert u\Vert\dfrac{d}{dt}(\Vert u\Vert)\\ % \int_{\Omega}\nabla u\cdot\nabla u_t &= \Vert\nabla u\Vert\dfrac{d}{dt}(\Vert\nabla u\Vert) \end{align*} Plug the above three manipulations to (1), we have $$ 2t\Vert\nabla u\Vert\dfrac{d}{dt}(\Vert\nabla u\Vert) \leq (2C + 1)\Vert u\Vert\dfrac{d}{dt}(\Vert u\Vert) \tag{2} $$ Now I'm going to prove (2) is true by changing $t$ to $s$, then integrate both side with respect to $s$ on range $[0, t]$, $$ 2\int_0^t s\Vert\nabla u(\cdot,s)\Vert\dfrac{d}{ds}(\Vert\nabla u(\cdot,s)\Vert)ds \leq (2C + 1)\int_0^t \Vert u(\cdot,s)\Vert\dfrac{d}{ds}(\Vert u(\cdot,s)\Vert)ds $$ It can be shown using integration by part that $$ LHS = \dfrac{1}{2}t\Vert\nabla u(\cdot,t)\Vert^2 - \int_0^t\Vert\nabla u(\cdot,s)\Vert^2 ds $$ and $$ RHS = \left(C + \dfrac{1}{2}\right) (\Vert u(\cdot,t)\Vert^2 - \Vert v(\cdot)\Vert^2) $$ Now I'm struggling on proving $LHS\leq RHS$, maybe my method doesn't hold true since it envolve a lot of integration and derivation of inequality. What should I do next?