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Weird observation about discretization of $\nabla^T\nabla$

In the previous question page, I saw "the adjoint of d/dx is -d/dx"

I think that it is related to integration by parts. But I don't know understand what it mean, clearly.

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jakeoung
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  • Do you know what the definition of the adjoint of an operator is? – Ben Grossmann Oct 04 '15 at 15:38
  • Yes, I know. Actually, I also heard that the adjoint of gradient operator is divergence operator. To preserve this property in discrete setting, some literatures implement gradient as forward difference and divergence as backward difference. To understand this, I am trying to understand. – jakeoung Oct 04 '15 at 17:16

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The adjoint of an operator, $L$, on an inner product space, $V$, written $L^{\ast}$ is the operator such that the following holds, for all $v,w \in V$.

$$\langle v, Lw\rangle=\langle L^{\ast}v,w\rangle$$

Let $f,g$ be $C^{\infty}(\Omega)$ with zero boundary conditions (a vector space) . We simply apply the definition and then integration by parts.

$$\left \langle f,\frac{d}{dx} g \right \rangle = \int_{\Omega} fg'dx=-\int_{\Omega}f'g dx=\left \langle \left (-\frac{d}{dx} \right )f,g \right \rangle $$

Ian
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