Deriving an estimate in regularity theory of the heat equation I have another question from PDE Evans 2nd edition, this time from pages 380-381. It's about a step in the formal derivation of estimates.
Given the initial-value problem for the heat equation

\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\} \tag{37} \end{cases}

this estimate was derived:

$$\sup_{0 \le t \le T} \int_{\mathbb{R}^n} |Du|^2 \, dx + \int_0^T \int_{\mathbb{R}^n} u_t^2 + |D^2 u|^2 \, dx dt \le C \left(\int_0^T \int_{\mathbb{R}^n} f^2 \, dxdt+\int_{\mathbb{R}^n} |Dg|^2 \, dx \right) \tag{38}$$

The book now proceeds to say, in its part (ii):

$\quad$(ii) Next differentiate the PDE with respect to $t$ and set $\tilde{u} := u_t$. Then $$\begin{cases} \tilde{u_t} - \Delta \tilde{u} = \tilde{f} & \text{in }\mathbb{R}^n \times (0,T] \\ \tilde{u} = \tilde{g} & \text{on } \mathbb{R}^n \times \{t=0\} \end{cases} \tag{40}$$ for $\tilde{f} :=f_t, \tilde{g}:=u_t(\cdot,0)=f(\cdot,0)+\Delta g$. Multiplying by $\tilde{u}$, integrating by parts and invoking Gronwall's inequality, we deduce $$\sup_{0 \le t \le T} \int_{\mathbb{R}^n} |u_t|^2 \, dx + \int_0^T \int_{\mathbb{R}^n} |Du_t|^2 \, dx dt \le C\left(\int_0^T \int_{\mathbb{R}^n} f_t^2 \, dx dt + \int_{\mathbb{R}^n} |D^2g|^2+f(\cdot,0)^2 \, dx \right) \tag{41}$$

I was having trouble working it out myself when following this procedure of "multiplying by $\tilde{u}$, integrating by parts and invoking Gronwall's inequality". Is it possible if someone can show this step? None of it was actually written in the book, as you can see above.
 A: Multiply $(40)$ by $u_t$ and integrate: $$\int_{\mathbb{R}^N}u_t(u_t)_t-\int_{\mathbb{R}^N}u_t\Delta u_t=\int_{\mathbb{R}^N}f_tu_t.\tag{1}$$
Use integration by parts in $(1)$ combined with $(u_t^2)_t/2=u_t(u_t)_t$ to conclude that $$\frac{1}{2}\int_{\mathbb{R}^N} (u_t^2)_t+\int_{\mathbb{R}^N} |Du_t|^2=\int_{\mathbb{R}^N} f_tu_t. \tag{2}$$
Now integrate $(2)$ from $0$ to $T$ to get 
\begin{eqnarray}
\int_0^T\int_{\mathbb{R}^N} |Du_t|^2 &=& \int_0^T\int_{\mathbb{R}^N}f_tu_t+\frac{1}{2}\left(\int_{\mathbb{R}^N}u_t^2(\cdot, 0)-\int_{\mathbb{R}^N}u_t^2(\cdot,T)\right)      \nonumber \\
   &=& \int_0^T\int_{\mathbb{R}^N}f_tu_t+\frac{1}{2}\left(\int_{\mathbb{R}^N}(f(\cdot, 0)+\Delta g)^2-\int_{\mathbb{R}^N}u_t^2(\cdot,T)\right). \tag{3}\end{eqnarray}
From $(3)$  $$\sup_{s\in[0,T]}\int_{\mathbb{R}^N}|u_t(\cdot,s)|^2+\int_0^T\int_{\mathbb{R}^N} |Du_t|^2=\sup_{s\in[0,T]}\int_{\mathbb{R}^N}|u_t(\cdot,s)|^2+\int_0^T\int_{\mathbb{R}^N}f_tu_t+\frac{1}{2}\left(\int_{\mathbb{R}^N}(f(\cdot, 0)+\Delta g)^2-\int_{\mathbb{R}^N}u_t^2(\cdot,T)\right).\tag{T}$$
From here you can proceed as follows. We combine $(2)$ with inequality $2ab\le a^2+b^2$ to obtain $$\int_{\mathbb{R}^N} (u_t^2)_t\le \int_{\mathbb{R}^N}(f_t^2+u_t^2). \tag{4}$$
Let $$\eta(s)=\int_{\mathbb{R}^N} u_t^2(x,s)dx,\ \ \phi(s)=\int_{\mathbb{R}^N}f_t^2(x,s),\ s\in [0,T],$$
and note that from $(4)$ $$\eta'(s)\le \eta(s)+\phi(s).$$
Now apply Gronwall inequality and use $(T)$ to conclude.
