Related to the heat equation: \begin{equation}\tag{1} \begin{cases} u_t-\Delta u &= f \text{ in $\mathbb{R}^n\times(0,T]$}\\ u &= g \text{ on $\mathbb{R}^n\times\{t=0\}$} \end{cases} \end{equation} Taking the derivative w.r.t. $t$ and setting $\hat{u}:=u_t$ \begin{equation}\tag{2} \begin{cases} \hat{u}_t-\Delta \hat{u} &= \hat{f} \text{ in $\mathbb{R}^n\times(0,T]$}\\ \hat{u} &= \hat{g} \text{ on $\mathbb{R}^n\times\{t=0\}$} \end{cases} \end{equation} for $\hat{f}:=f_t, \hat{g}:=u_t(\cdot,0)=f(\cdot,0)+\Delta g$. Then multiplying this by $\hat{u}$, integration by parts and applying Gronwall's Lemma will leads to the following result: \begin{equation}\tag{3} \sup_{0\leq t\leq T} |u_t|^2 dx+ \int_0^T\int_{\mathbb{R}^n} |Du_t|^2 dxdt \leq C\Big(\int_0^T\int_{\mathbb{R}^n}f_t^2 dxdt+\int_{\mathbb{R}^n}[|D^2g|^2+f(\cdot,0)^2] dx \Big) \end{equation} Where The gronwoll briefly gives as: for $\eta(\cdot)$, non-negative continuous function on $[0,T]$, which satisfies $\eta^{'}(t)\leq \phi(t)\eta(t)+\xi(t)$, for $\phi,\xi$ non-negative. Then $\eta(t)\leq e^{\int_0^T\phi(s) ds}\Big[\eta(0)+\int_0^T\xi(s) ds\Big]$.
I did not quite get exactly how the Gronwall's Lemma applied here, Can someone explain to me how? since I tried to get the inequality for $\hat{u}$ by directly applying Lemma to $(\Delta\hat{u})\hat{u}+f\hat{u}$ in r.h.s but I could not reach the desired result. Can someone explain to me how it applies here?
What I did to get (3) was:
- multiply (2) by $\hat{u}$ and integrate \begin{equation}\tag{4} \int_{\mathbb{R}^n}\hat{u}_t\hat{u}-\int_{\mathbb{R}^n}(\Delta \hat{u})\hat{u} = \int_{\mathbb{R}^n}\hat{f}\hat{u} \end{equation}
- reformulate $\int_{\mathbb{R}^n}\hat{u}_t\hat{u}$ as \begin{equation} \int_{\mathbb{R}^n}\hat{u}_t\hat{u} = \frac{1}{2}\int_{\mathbb{R}^n}\frac{d}{dt}\hat{u}^2 \end{equation}
- Use integration by parts to $\int_{\mathbb{R}^n}(\Delta \hat{u})\hat{u}$ to obtain: \begin{equation} \int_{\mathbb{R}^n}(\Delta \hat{u})\hat{u} = -\int_{\mathbb{R}^n}\nabla\hat{u}\cdot\nabla\hat{u}= -\int_{\mathbb{R}^n}|D\hat{u}|^2 \end{equation}
- Then I applied the Gronwall to (4) as follows: \begin{equation} \frac{1}{2}|\hat{u}|^2 \leq e^{\int_0^T\hat{f}(s) ds}\Big[|\hat{u}(\cdot,0)|^2+\int_0^T\int_{\mathbb{R}^n}|D\hat{u}|^2 dx ds \Big] \end{equation}
And I can apply triangular inequality to $|\hat{u}(\cdot,0)|^2$. But after that I cannot figure out a way to reach $(3)$.