Let $\alpha, \beta > 0$ and $\Delta := \nabla^T \nabla$ be the discrete laplacian operator, $$\nabla: \mathbb{R}^{n_x \times n_y \times n_z} \rightarrow \mathbb{R}^{3 \times n_x \times n_y \times n_z}, \hspace{.5em}w \mapsto (\nabla_xw, \nabla_yw, \nabla_zw) $$ being the discrete spatial gradient operator in $3$D. You can see $\Delta$ as a linear operator of shape $p \times p$, where $p := n_xn_yn_z$.

My intuition is that the operator $(\alpha I + \beta \Delta)^{-1}$ corresponds to some kind of Gaussian smoothing. Is this true ? If yes, what's the width of the corresponding kernel in terms of $\alpha$ and $\beta$ ?

  • $\begingroup$ Wait; $\beta \Delta \in \mathbb K^{1\times N}$, but $I$ is inherently of $N\times N$ shape. How can you find an inverse of that operation? $\endgroup$ Feb 8, 2016 at 15:36
  • $\begingroup$ I think you should more explicitely define what $\nabla$ and $\Delta$ are, in your case. I don't think I know the definitions you're using. $\endgroup$ Feb 8, 2016 at 15:41
  • $\begingroup$ That inversion is well-defined. $\nabla$ is the discrete spatial gradient operator. $\endgroup$
    – dohmatob
    Feb 8, 2016 at 19:27
  • $\begingroup$ ha, I was assuming your $\nabla$ was the classical divergence operator in vector form, hence $\nabla_\text{Müller} = \left(\frac{\partial}{\partial x_1}, \dots, \frac{\partial}{\partial x_N}\right)$! $\endgroup$ Feb 8, 2016 at 19:29
  • $\begingroup$ Wouldn't the Taylor expansion of your operator show that? $\endgroup$ May 18 at 20:14


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