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.
Best Answer
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.