Make sense of norm notations and why is $\int_\Omega \nabla\theta\cdot\nabla\theta_t \ d\mathbf{x} = \frac{1}{2}\frac{d}{dt}|\theta|_1^2$?

Question

I have the heat equation

\begin{align} \dot{u}(x,t) -\Delta u(x,t) = f(x,t),& \quad x\in\Omega\subset\mathbb{R}^2 \\ u(x,t) = 0, & \quad x\in\Gamma = \partial\Omega, 0<t\leq T \\ u(x,0) = u_0(x),& \quad x\in\Omega \end{align}

In my book there is a proof about a stability estimate. In one of the lines they state that

$$a(u,u_t)=\int_\Omega \nabla u\cdot\nabla u_t \ d\mathbf{x} = \frac{1}{2}\frac{d}{dt}|u|_1^2.$$

How does the last equality follow? Need help getting through the arithmetic there.

What does the sub-index $1$ mean in the absolute value?

---

EDIT: Adding an attempt.

Attempt:

$$\int_\Omega \nabla\theta\cdot\nabla\theta_t \ dx = \int_\Omega\nabla\theta\cdot \nabla\left(\frac{d}{dt}\theta\right) \ dx$$

$$=\int_\Omega \nabla\theta\cdot\frac{d}{dt}\nabla\theta \ dx =\frac{1}{2}\frac{d}{dt}\int_\Omega(\nabla\theta)^2 \ dx =\frac{1}{2}\frac{d}{dt}||\nabla\theta||^2.$$

But why does the book have $\frac{1}{2}\frac{d}{dt}|\theta|_1^2$ instead?

Answer 1

Hint: Try proving the following using just the properties of the scalar product, Cauchy-Schwartz and the definition of the derivative

Lemma. Let $(\mathscr{H}, \langle \cdot, \cdot \rangle, \| \cdot \|)$ be a Hilbert space. Then $$\langle u'(x), u(x) \rangle = \frac{1}{2} \| u(x) \|^2$$ holds for all continuously differentiable $u$ with values in $\mathscr{H}$.

Proof

Let $\tau_h u := u(t + h)$. Then \begin{align} | \tau_h u |^2 - | u |^2 & = \langle \tau_h u - u, \tau_h u \rangle - \langle u, u - \tau_h u \rangle \\ & = \langle \tau_h u - u, \tau_h u \rangle + \langle \tau_h u - u, u \rangle. \end{align} holds. On one hand $$ \left| \langle u', u \rangle - \frac{\langle \tau_h u - u, u \rangle}{h} \right| = \left| \langle u' - \frac{\tau_h u - u}{h}, u \rangle \right| \le \left| u' - \frac{\tau_h u - u}{h} \right| \left| u \right| \xrightarrow{h \to 0} 0 $$ and therefore $$ \frac{\langle \tau_h u - u, u \rangle}{h} \xrightarrow{h \to 0} \langle u', u \rangle.$$ On the other hand\begin{align*} \left| \langle u', u \rangle - \frac{\langle \tau_h u - u, \tau_h u \rangle}{h} \right| & \le \left| \langle u', u - \tau_h u \rangle \right| + \left| \langle u' - \frac{\langle \tau_h u - u \rangle}{h}, \tau_h u \rangle \right| \\ & \le | u' | | u - \tau_h u | + \left| u' - \frac{\tau_h u - u}{h} \right| \left| \tau_h u \right| \xrightarrow{h \to 0} 0 \end{align*} and therefore $$ \frac{\langle \tau_h u - u, \tau_h u \rangle}{h} \xrightarrow{h \to 0} \langle u', u \rangle. $$

Edit This question is highly related and may answer your question.