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Consider the Laplace operator over a 1D domain, with homogeneous Neumann boundary conditions. I have discretized this operator using the cell-centered Finite Volume method on a uniform grid of size $h$, employing a four-point central difference scheme in the bulk of the grid, while reverting to a two-point scheme near the boundaries. The result is the following matrix: \begin{equation} L_n = \frac{1}{24h} \begin{bmatrix} -24 & 24 & & & & &\\ 25 & -51 & 27 & -1 & & &\\ -1 & 28 & -54 & 28 & -1 &\\ &\ddots &\ddots&\ddots&\ddots&\ddots\\ & &-1 &28 &-54 &28 & -1\\ & & &-1 &27 &-51 &25\\ & & & & & 24 & -24 \end{bmatrix}\in\mathbb{R}^{n\times n}. \end{equation}

Since it is a discretization of a negative semi-definite operator, I would expect $L_n$ to have real, non-positive eigenvalues. This is indeed what I observe numerically (regardless of size), but I'm struggling to prove it due to the inconvenient properties. Matrix $L_n$ is singular, asymmetric and its Gershgorin disks don't confine its eigenvalues to the left complex half-plane (we get a disk of radius $\frac{58}{24h}$ centered around $-\frac{54}{24h}$). The things that I have been able to prove, are the following:

  • The vector $[1\dots 1]^T$ is an eigenvector of $L_n$, corresponding to eigenvalue 0.
  • The smallest version of $L_n$ that contains the repeating stencil, $L_5$, has demonstrably real, non-positive eigenvalues (an exact expression for the characteristic polynomial can be derived & plotted). \begin{equation} L_5 = \frac{1}{24h} \begin{bmatrix} -24 & 24 & & &\\ 25 & -51 & 27 & -1 &\\ -1 & 28 & -54 & 28 & -1\\ &-1 &27 &-51 &25\\ & & & 24 & -24 \end{bmatrix}. \end{equation}
  • The four-point scheme yields a negative semi-definite matrix when using periodic boundary conditions (an exact expression for eigenvalues can be derived, by leveraging the circularity of the resulting matrix $\tilde{L}_n$). \begin{equation} \tilde{L}_{n}= \frac{1}{24h} \begin{bmatrix} -54 & 28 & -1 & & & -1& 28\\ 28 & -54 & 28 & -1 & & & -1 \\ -1 & 28 & -54 & 28 & -1 &\\ &\ddots &\ddots&\ddots&\ddots&\ddots\\ & &-1 &28 &-54 &28 & -1\\ -1& & &-1 &28 &-54 &28\\ 28& -1& & & -1 & 28 & -54 \end{bmatrix}. \end{equation}

Does anyone have a suggestion for how I could go about proving the realness and non-positiveness of the eigenvalues of $L_n$? My hunch is that adding more rows with the repeating stencil somehow doesn't affect the bounds of the spectrum, but I don't know how I could prove this. I get the feeling that I might be missing a piece of theory here.

Thanks in advance!

Nicola
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  • maybe related. Elouafi, Mohamed. "An eigenvalue localization theorem for pentadiagonal symmetric Toeplitz matrices." Linear algebra and its applications 435.11 (2011): 2986-2998. – Yimin Jan 26 '24 at 23:09
  • You can split the matrix into a central petadiagonal matrix, which is negative definite, and top left part and right bottom part, then probably show adding the two parts will not affect much. However, there is cross-talk between them, though very weak. That probably makes things nasty. The core idea probably is to show $vL v^t \le 0$, and split the vector $v$ into 3 parts. – Yimin Jan 26 '24 at 23:20
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    Indeed, I've observed numerically that $\Re(v^* L v) \leq 0\ \forall v\in \mathbb{C}^n\ s.t. \lVert v\rVert = 1$. I'm currently constructing a proof for this and it seems doable. However, this still leaves the realness of the eigenvalues to be proven. – Nicola Jan 27 '24 at 10:51
  • I would be interested in seeing your proof for the positiveness. @Nicola – Yimin Jan 28 '24 at 04:40
  • In the process of constructing above said proof I found a very useful decomposition of $L_n$ into the product of a positive definite and negative semi-definite matrix. With this, the realness and non-positivity of the eigenvalues can be proven directly! I will add the proof as a solution to the question. – Nicola Jan 29 '24 at 00:31

1 Answers1

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I think I have found the answer to my own question! Let: \begin{align} M_{n} &= \begin{bmatrix} -1 & 1 & & \\ 1 & -2 & 1 \\ &\ddots & \ddots &\ddots & \\ & & 1 &-2 & 1\\ & & & 1& -1 \end{bmatrix}\in\mathbb{R}^{n\times n}. \end{align}

Then, $L_n$ can be decomposed as follows: \begin{align} L_{n} &= \frac{1}{h}M_n + \frac{1}{24h}\begin{bmatrix} 0 & & & & & &\\ 1 & -3 & 3 & -1 & & &\\ -1 & 4 & -6 & 4 & -1 &\\ &\ddots &\ddots&\ddots&\ddots&\ddots\\ & &-1 &4 &-6 &4 & -1\\ & & &-1 &-3 &3 &1\\ & & & & & & 0 \end{bmatrix}\\ &= \frac{1}{h}M_n + \frac{1}{24h} \begin{bmatrix} 0\\ & -M_{n-2}\\ & & 0 \end{bmatrix}M_n\\ &= \underbrace{\frac{1}{h}\left(I-\frac{1}{24}\begin{bmatrix} 0\\ & M_{n-2}\\ & & 0 \end{bmatrix}\right)}_{Q_n}M_n. \end{align}

Both $Q_n$ and $M_n$ are symmetric and the Gershgorin circle theorem tells us that $Q_n$ has eigenvalues of strictly positive and $M_n$ of strictly non-positive real part. Since 0 is an eigenvalue of $M_n$ (eigenvector $[1\dots1]^T$), we conclude that $Q_n$ is SPD and $M_n$ is SNSD. The eigenvalue problem for $L_n$ can thus be manipulated as follows:

\begin{align} L_n x = \lambda x\ &\leftrightarrow\ Q_n M_n x = \lambda x\\ &\leftrightarrow\ Q_n M_n Q_n z = \lambda Q_n z\\ &\leftrightarrow\ z^* Q_n M_n Q_n z = \lambda z^* Q_n z\\ &\leftrightarrow\ (Q_n z)^* M_n (Q_n z) = \lambda z^* Q_n z\\ &\leftrightarrow\ \frac{(Q_n z)^* M_n (Q_n z)}{z^* Q_n z} = \lambda. \end{align}

Since $M_n$ is SNSD and $Q_n$ is SPD, we conclude that $\lambda$ must be real and non-positive, QED.

Nicola
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  • pretty nice proof and seem can be generalized to other high order discretization of Laplace. – Yimin Jan 29 '24 at 03:47