2
$\begingroup$

Given the following coefficient matrix $A^h$, resulting from the finite difference approximation of the biharmonic equation on a mesh with mesh size $h$: \begin{equation*} A^h = \frac{1}{h^4} \begin{pmatrix} 5 & -4 & 1 & & \\ -4 & 6 & \ddots & \ddots & \\ 1 & \ddots & \ddots & \ddots & 1 \\ & \ddots & \ddots & 6 & -4 \\ & & 1 & -4 & 5 \end{pmatrix}. \end{equation*} Show that $A^h$ is symmetric, positive-definite.

Of course it is symmetric, since $(A^h)^T = A^h$. In order to show that it is also positive-definite, we need to let $\mathbf{u}\in \mathbb{R}^n\setminus\{\mathbf{0}\}$ and have that $\mathbf{u}^TA\mathbf{u} >0$. I wrote it out: \begin{align} \mathbf{u}^TA\mathbf{u} &= \frac{1}{h^4} \begin{pmatrix} u_1 & \cdots & u_n \end{pmatrix} \begin{pmatrix} 5 & -4 & 1 & & \\ -4 & 6 & \ddots & \ddots & \\ 1 & \ddots & \ddots & \ddots & 1 \\ & \ddots & \ddots & 6 & -4 \\ & & 1 & -4 & 5 \end{pmatrix} \begin{pmatrix} u_1\\ \\ \vdots\\ \\ u_n \end{pmatrix} \\[2ex] &= \frac{1}{h^4} \begin{pmatrix} u_1 & \cdots & u_n \end{pmatrix} \begin{pmatrix} 5u_1 - 4u_2 + u_3 \\ -4u_1 + 6u_2 -4u_3 + u_4 \\ u_1 - 4u_2 + 6 u_3 - 4u_4 + u5 \\ \vdots \\ u_{i-2} - 4u_{i-1} + 6u_i - 4u_{i+1} + u_{i+2} \\ \vdots \end{pmatrix} \\[2ex] &= \frac{1}{h^4}\left[5u_1^2 - 8u_1u_2 + 2u_1u_3 + 6u_2^2 - 8u_2u_3 + 2u_2u_4 + 6u_3^2 -8u_3u_4 + 2 u_3u_5 + \dots\right] \end{align} This where I get stuck. I know that I have to show that above can be expressed as the sum of squares, wich means it is positive. I've tried rewriting it using terms like $(u_1 - u_2)^2$ and $(u_1 + u_2 + u_3)^2$ etc, but I cannot see how it all fits together. Help would be greatly appreciated.

$\endgroup$
4
  • $\begingroup$ I suggest you use the Sylvester's Criterion to solve this problem. You may have a look here: en.wikipedia.org/wiki/Sylvester%27s_criterion $\endgroup$
    – Gino
    Oct 6, 2022 at 14:24
  • $\begingroup$ Otherwise, you may show that A has all positive eigenvalues. Since A is symmetric, it admits eigendecomposition. More precisely, there exist both an orthogonal matrix U, which means transpose(U) = inverse(U), and a diagonal matrix D whose entries are the eigenvalues of A, such that A=U D transpose(U). If you show that D has all positive entries, then A is pos def. $\endgroup$
    – Gino
    Oct 6, 2022 at 14:36
  • $\begingroup$ The determinants of the principal minors form an increasing sequence starting with 5, up to the order $n-1$. The last minor is $\det A = (n+1)^2$. (used $h=1$) $\endgroup$ Oct 6, 2022 at 14:41
  • 1
    $\begingroup$ @Violet you should amend your definition of positive definite... It should be $\bf{u} \in \mathbb{R}^n\setminus \{0\}$ $\endgroup$ Oct 6, 2022 at 14:47

1 Answer 1

3
$\begingroup$

Every symmetric, positive-definite matrix has a square root. In particular, the root may be asked to be symmetric and positive-definite as well, and then it is uniquely determined.

For the given discretisation matrix $A^h$, which is highly structured, (t-)his square root $$\sqrt{A^h} \;=\; \frac1{h^2} \begin{pmatrix} 2 & -1 & 0 & &\\ -1 & 2 & -1 & & & \\ 0 & -1 & \ddots & \ddots &\\ & & \ddots & \ddots & -1 & 0\\ & & & -1 & 2 & -1\\ & & & 0 & -1 & 2 \end{pmatrix}.$$ is not so far away, let's say by an educated guess which extends $$\begin{pmatrix} 5 & -4\\ -4 & 5 \end{pmatrix} \;=\; \begin{pmatrix} 2 & -1\\ -1 & 2 \end{pmatrix}^2\:.$$

And $\sqrt{A^h}$ is positive-definite:
Cf$\,$ https://en.wikipedia.org/wiki/Definite_matrix#Examples for the $3\times 3$ matrix. The method should generalise to higher dimensions.

And see here on this site: The full statement including proof of $(7.4.7)$ Theorem starts on page 537 in https://zhilin.math.ncsu.edu/TEACHING/MA580/Stoer_Bulirsch.pdf .


An explicit Sum of squares expression as looked for in the OP is given by $$(2u_1-u_2)^2 \,+\, \sum_{j=2}^{n-1}\,(-u_{j-1}+2u_j-u_{j+1})^2 \,+\, (2u_n-u_{n-1})^2\tag{SoS}$$

Having the knowledge of the square root $\sqrt{A^h}\,$ it results from expanding \begin{align} h^4\cdot\mathbf{u}^TA^h\,\mathbf{u} & \:=\: \left(h^2\cdot\mathbf{u}^T\sqrt{A^h}\right) \left(h^2\cdot\sqrt{A^h}\,\mathbf{u}\right) \\[2ex] & \:=\: \big(2u_1-u_2, \dots, -u_{n-1}+2u_n\big) \begin{pmatrix} 2u_1 -u_2\\ -u_1 +2u_2 -u_3\\ \vdots \\ -u_{n-2} +2u_{n-1} -u_n \\ -u_{n-1} +2u_n \end{pmatrix} \end{align} The expression $(\text{SoS})$ gets zero only if $\mathbf{u}= \mathbf{0}$.
This is equivalent to the linear system $\sqrt{A^h}\,\mathbf{u} = \mathbf{0}$ having only the trivial solution.

$\endgroup$
1
  • 1
    $\begingroup$ Ah, so I can say that the square root of $A^h$ is the discrete Laplacian matrix, for which I know that it is positive definite. Hence, $A^h$ is as well. Thanks! $\endgroup$
    – Ruby
    Oct 6, 2022 at 19:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .