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.

  • $\begingroup$ When you try to apply these steps, where do you get stuck? $\endgroup$ – Tomás Aug 20 '14 at 23:51

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.

  • $\begingroup$ Gronwall's inequality asserts this, according to textbook's page 377: $$\max_{0 \le t \le T} \| \textbf{u}_m(t) \|_{L^2(U)}^2 \le C \left(\|g \|_{L^2(U)}^2 + \| \textbf{f} \|_{(0,T;L^2(U))}^2 \right).$$ Now, following your hint in the answer, can I say $$\sup_{0 \le t \le T} \int_{\mathbb{R}^n} |u_t(\cdot,s)|^2 \, dx=\sup_{0 \le t \le T} \| \textbf{u} \|_{L^2(\mathbb{R}^n)}^2 \le C \left(\|g \|_{L^2(\mathbb{R}^n)}^2 + \| \textbf{f} \|_{(0,T;L^2(\mathbb{R}^n))}^2 \right)$$ and place into your last equation (the one after $\text{(3)}$)? $\endgroup$ – Cookie Aug 23 '14 at 22:48
  • $\begingroup$ This is a just a application of Gronwall inequality. Refer to the appendix in Evans book, to find Gronwall inequality. Or type it in google. $\endgroup$ – Tomás Aug 23 '14 at 22:54
  • $\begingroup$ I thought about using the application of Gronwall's inequality because I was unsure on how to apply the appendix formula of the inequality itself, that is, applying (for all $0 \le t \le T$) $$\eta(t) \le e^{\int_0^t \phi(s) \, ds} \left[\eta(0) + \int_0^t \phi(s) \, dx \right]$$ to estimate $\displaystyle \sup_{0 \le t \le T} \int_{\mathbb{R}^n} |u_t(\cdot,t)|^2 dx$. $\endgroup$ – Cookie Aug 23 '14 at 23:00
  • $\begingroup$ I have added more information. Try to conclude now. If you get stuck again, please leave a comment. $\endgroup$ – Tomás Aug 24 '14 at 0:02
  • $\begingroup$ @Tomás I tried follow your hint, but still could not get (41) from your (T), basically, we want to show the rhs of (T) is the rhs of (41). Could you please provide some further detail to get to (41)? $\endgroup$ – math101 Oct 15 '15 at 9:32

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