Doubts about inner product operation of function.

Question

For a heat conduction equation： $$ \frac{\partial \varphi}{\partial t}-\Delta \varphi=0. $$ Take the inner product of both sides of this equation with ($ -\Delta \varphi $ ), how can I get： $$ \frac12\frac{d}{dt}\langle \nabla\varphi,\nabla\varphi\rangle + \Vert \Delta\varphi\Vert^2=0. \tag{1}$$ My main problem is the first term on the LHS of the equation (1).

Answer 1

The inner product is defined by :

\begin{align} \langle f,g\rangle = \int f^* g\;\text dx \end{align}

Therefore, integration by part gives : $\langle \nabla f,g\rangle = -\langle f,\nabla g \rangle$.

From this, you have : \begin{align} \frac 12 \frac{\text d}{\text dt}\langle \nabla \varphi,\nabla\varphi\rangle &= \left\langle \nabla \varphi,\frac{\partial}{\partial t}\nabla\varphi\right\rangle \\ &=\left\langle \nabla \varphi,\nabla\frac{\partial\varphi}{\partial t}\right\rangle \\ &=-\left\langle \Delta\varphi,\frac{\partial\varphi}{\partial t}\right\rangle\\ &=-\langle \Delta\varphi,\Delta\varphi\rangle\\ &=- \|\Delta\varphi\|^2 \end{align}