Let $z \in \mathbb R\backslash \left\{2 \right\}$ then I would like to understand the following:

Consider the dynamical system with $x_i \in \mathbb C^2:$

$$ x_{i} = \left(\begin{matrix} z &&-1 \\ 1 && 0 \end{matrix} \right)^ix_0.$$

I would like to understand whether one can obtain a sharp bound on $\Vert x_i \Vert$ for $N \ge i \ge 0$ just in terms of

$\Vert x_0 \Vert$, $\Vert x_N \Vert$ and $z \neq 2.$

Observations: It seems that this dynamical system is rather simple in the sense that the matrix $A=\left(\begin{matrix} z &&-1 \\ 1 && 0 \end{matrix} \right)$ is diagonalizable for $z \neq 2.$

More precisely, the eigenvalues are $$\tfrac{1}{2} \left(z \pm \sqrt{z^2-4 }\right)$$ with eigenvectors $$\left(\tfrac{1}{2} \left(z \pm \sqrt{z^2-4 }\right), 1\right).$$

It is also worth noticing that this system is invertible since $\operatorname{det}(A)=1$

so it is believable that once the boudary norms for $x_0$ and $x_N$ are known. Everything else should be fixed as well.

Of course there are trivial bounds like $\Vert x_i \Vert \le \Vert A \Vert^i \Vert x_0 \Vert$ but I am looking for something more refined.

Motivation: This is the discrete analogue of $-y''(x)=zy(x),$ see for details and in the continuous world it is almost trivial to bound $\vert y(x_{\text{middle}}) \vert$ in terms of $\vert y(x_0) \vert$ and $\vert y(x_1)\vert$ where $x_0 \le x_{\text{middle}} \le x_1.$

  • $\begingroup$ Is $\|x_i\|^N \leqslant C(z) \|x_0\|^{N-i} \|x_N\|^i$ something that you are looking for? This follows quite simply: write $u$, $v$ for the eigenvectors and $\alpha_i$, $\beta_i$ for the coefficients of the eigenvector expansion of $x_i$. Then $\alpha_i^N = \alpha_0^{N-i} \alpha_N^i$, $\beta_i^N = \beta_0^{N-i} \beta_N^i$, and $C_1(z)(|\alpha_i|+|\beta_i|) \|x_i\| \leqslant C_2(z)(|\alpha_i|+|\beta_i|)$. $\endgroup$ – Mateusz Kwaśnicki Mar 5 at 10:34
  • $\begingroup$ sorry, so what is this $C(z)$ precisely in your notation? $\endgroup$ – J.Doe Mar 5 at 12:20
  • $\begingroup$ A "constant" that depends on $z$. $\endgroup$ – Mateusz Kwaśnicki Mar 5 at 12:22
  • $\begingroup$ sorry, I mean is it explicit? $\endgroup$ – J.Doe Mar 5 at 12:23
  • $\begingroup$ That was too long for a comment, I posted some details as an answer. $\endgroup$ – Mateusz Kwaśnicki Mar 5 at 12:49

This is an extension and correction of my comment, made more explicit.

Denote the normalised eigenvectors of $A$ by $u$ and $v$, and the corresponding eigenvalues by $\lambda$ and $\mu$. Since all norms on $\mathbb{R}^2$ are equivalent, we have $$ C_1(z) \max\{|\alpha|, |\beta|\} \leqslant \|x\| \leqslant C_2(z) \max\{|\alpha|, |\beta|\} $$ for some constants $C_1(z)$, $C_2(z)$ that can be estimated explicitly (see below).

Recall that $x_i = A^i x_0$, and write $x_i = \alpha_i u + \beta_i v$, so that $\alpha_i = \lambda^i \alpha_0$ and $\beta_i = \mu^i \beta_0$. It follows that $$ |\alpha_i| = |\alpha_0|^{1 - i/N} |\alpha_N|^{i/n} , \qquad |\beta_i| = |\beta_0|^{1 - i/N} |\beta_N|^{i/n} . $$ Thus, $$ \begin{aligned} \|x_i\| & \leqslant C_2(z) \max\{|\alpha_i|, |\beta_i|\} \\ & = C_2(z) \max\{|\alpha_0|^{1 - i/N} |\alpha_N|^{i/n}, |\beta_0|^{1 - i/N} |\beta_N|^{i/n}\} \\ & \leqslant C_2(z) \times (\max\{|\alpha_0|, |\beta_0|\})^{1 - i/N} \times (\max\{|\alpha_N|, |\beta_N|\})^{i/n} \\ & \leqslant \frac{C_2(z)}{C_1(z)} \, \|x_0\|^{1 - i/N} \|x_N\|^{i/n} . \end{aligned} $$ That is, $$ \|x_i\| \leqslant C(z) \|x_0\|^{1 - i/N} \|x_N\|^{i/N} . $$

Evaluation of optimal values of $C_1(z)$ and $C_2(z)$, especially in higher dimensions, is not completely obvious. Rougher estimates in dimension 2, however, are quite simple: clearly, $$ \|\alpha u + \beta v\| \leqslant |\alpha| + |\beta| \leqslant 2 \max\{|\alpha|, |\beta|\} ,$$ so that $C_2(z) = 2$ does the job. On the other hand, if we denote the dot product of $u$ and $v$ by $p = u \cdot v$, then $$ \begin{aligned} \|\alpha u + \beta v\|^2 & = |\alpha|^2 + |\beta|^2 - 2 p \operatorname{Re}(\alpha \bar\beta) \geqslant |\alpha|^2 + |\beta|^2 - 2 |p| |\alpha| |\beta| \\ & \geqslant (1 - |p|) (|\alpha|^2 + |\beta|^2) \ge (1 - |p|) (\max\{|\alpha|, |\beta|\})^2 , \end{aligned} $$ and so $C_1(z) = \sqrt{1 - |p|}$ works well. It follows that we may take $C(z) = 2 / \sqrt{1 - |p|}$.


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