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.


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

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

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

  • $\begingroup$ Which book is this from? $\endgroup$ Mar 12, 2020 at 13:50

1 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 $\mathscr{H}$.


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.

  • $\begingroup$ I had this in mind: $$\langle u',u \rangle = \int_I u'u \ dx = \int_I(\frac{d}{dx}u)u \ dx = \int_I\frac{1}{2}\frac{d}{dx}u^2 \ dx = \frac{1}{2}||u||^2.$$ $\endgroup$
    – Parseval
    Mar 12, 2020 at 11:30
  • $\begingroup$ @Parseval This is indeed the easiest solution when given the concrete Hilbert space, $L^2$. The general case is not that simple as I show in the proof. Feel free to post this as an answer. $\endgroup$ Mar 12, 2020 at 11:35
  • $\begingroup$ Why would it be an answer? My question is totally different. There I'm dealing with gradients since I'm in 2D. Also, I need to have $d/dt$ before $||u||^2$. In your example you give $\langle u',u\rangle$ but I need $\langle \nabla u, \nabla v \rangle$ where $v=\frac{d}{dt}u.$ Maybe I'm missing the point you tried to convey. $\endgroup$
    – Parseval
    Mar 12, 2020 at 11:41
  • $\begingroup$ I thought my lemma could easily be modified to be come your statement. I'll have another look at it later. $\endgroup$ Mar 12, 2020 at 11:47
  • $\begingroup$ Thanks a lot. I'll think throught this meanwhile and se how I can modify it. $\endgroup$
    – Parseval
    Mar 12, 2020 at 11:48

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