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The following recurrent relation is posed

$\begin{cases} b_{i+1} = (1-\frac2N)b_i+2(1-\frac1N)a_i,\\ a_{i+1} = -\frac2Nb_i + (1-\frac2N)a_i \end{cases}$

Now I would like to understand how the author of the paper I referenced below (Section 4.3, page 8 of pdf, page 472 of journal, http://www.complex-systems.com/abstracts/v11_i06_a03.html) solved this problem. I already contacted him, but I received no reply. I understand that he determines the solution by first transforming the recurrent relations to a difference between two steps

$\begin{cases} b_{i+1} - b_i = \frac2Nb_i+2(1-\frac1N)a_i,\\ a_{i+1} - a_i = -\frac2Nb_i + -\frac2Na_i \end{cases},$

then extends it to difference equations

$\begin{cases} \dot{b} \delta = \frac2Nb_i+2(1-\frac1N)a_i,\\ \dot{a} \delta = -\frac2Nb_i + -\frac2Na_i \end{cases}$

for $\delta$ small as required. Now I do not understand what he does in the next step, when he derives

$$\ddot{b} + \frac4{\delta^2 N}b+O(\frac{b}{\delta^2N^2})+\ddot{b}O(\frac1N)+\dot{b}O(\frac1{N\delta})=0$$

Also it seems for me strange to keep the two terms $\frac4{\delta^2 N}b+O(\frac{b}{\delta^2N^2})$, since $\frac4{\delta^2 N}b+O(\frac{b}{\delta^2N^2})\in O(\frac{b}{\delta^2N^2})$

The way I know to solve a linear system of differential equations is to formulate the equations into a linear system, determine eigenvalues and -vectors, and insert them into a linear combination of eigenevectors times exponential functions like it is described in tutorial.math.lamar.edu/Classes/DE/RealEigenvalues.aspx (cannot post link because of reputation system), but I do not have any clue how the author comes to this equation.

Ozhigov, Yuri, Speedup of iterated quantum search by parallel performance, Complex Syst. 11, No.6, 465-486 (1997). ZBL0955.68036.

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  • $\begingroup$ yes thanks, updated $\endgroup$ – Alex Go Jun 22 '17 at 16:26
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I am not sure if you need to convert the recurrence system into a differential equation to solve. Define $c_i = \begin{pmatrix}b_i\\a_i\end{pmatrix}$.

Let $\delta = 1-{2\over N}$, and $A = \begin{pmatrix}\delta \ \ \ \ \ \delta+1 \\\delta-1 \ \ \ \ \ \delta \end{pmatrix}$.

Then,

$$c_{i+1}=Ac_i, $$ or $$c_k=A^kc_0.$$

To compute $A^k$, you can try decomposing the matrix using eigenvectors, $v$, such that $(A - \lambda I) v = 0$

It is a bit of algebra, but the eigenvectors and eigenvalues are: $$v_1 = \begin{pmatrix}i\sqrt{\frac{1+\delta}{1-\delta}}\\1\end{pmatrix}, \lambda_1=\delta-i\sqrt{1-\delta^2}$$ $$v_2 = \begin{pmatrix}-i\sqrt{\frac{1+\delta}{1-\delta}}\\1\end{pmatrix}, \lambda_2=\delta+i\sqrt{1-\delta^2}$$

So, $A^k = V\Lambda^k V^{-1}$, where $\Lambda=\begin{pmatrix}\lambda_1\ 0\\0 \ \ \lambda_2\end{pmatrix}$, and $V=[v1, v2]$

Addendum: This simplifies further (again, with more algebra). Let $\phi = \tan ^{-1}\frac{\sqrt{1-\delta^2}}{\delta}$ and $\beta = \sqrt{\frac{1+\delta}{1-\delta}}=\sqrt{N-1}.$

Then, $$c_k=\begin{pmatrix}\cos\phi k \ \ \ \ \ \ \beta\sin \phi k \\ -\beta^{-1}\sin \phi k \ \ \ \ \cos\phi k\end{pmatrix}c_0$$

For $N>>1$, $\beta\sin \phi k \approx 2k\frac{N-1}{N-2}$. As a sanity check, in the limit as $N$ approaches $\infty$, notice the recurrence becomes $a_n = a_0$, and $b_{n+1}=b_n+2a_0$. Thus, $b_n=2na_0+b_0$

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  • $\begingroup$ Yes you definitely do not need to do it with differential equations. It is just the author uses the same method to calculate a bigger linear system with 4 dimensions, and there it is a bit harder to do it (but probably also possible). I also feel that there is more material for solving Cauchy problems online than for recurrence relations, but probably I am just missing a lot of literature since I do not know about it. $\endgroup$ – Alex Go Jun 23 '17 at 6:43
  • $\begingroup$ Could you explain the step where you substitute $\phi = \tan ^{-1}\frac{\sqrt{1-\delta^2}}{\delta}$. I do not see how you can substitute it. When I calculate the $A^k = V\Lambda^k V^{-1}$, I obtain $A^k = \frac12 \begin{pmatrix}\lambda_1^k + \lambda_2^k \ \ \ \ \ \ i\beta (\lambda_1^k - \lambda_2^k ) \\ \frac1{i\beta} (\lambda_1^k - \lambda_2^k ) \ \ \ \ \lambda_1^k + \lambda_2^k \end{pmatrix}$. I do not see how to substitute it there and before. $\endgroup$ – Alex Go Jun 23 '17 at 6:47
  • $\begingroup$ $\lambda=e^{\pm i\phi}$ by Euler's relation. I don't have access to the paper, but I suspect you can still perform the same analysis. $\endgroup$ – player100 Jun 23 '17 at 10:20
  • $\begingroup$ ah I see, cant you access the paper with this link? complex-systems.com/pdf/11-6-3.pdf $\endgroup$ – Alex Go Jun 23 '17 at 10:22
  • $\begingroup$ Perused the paper. The authors essentially arrived at the same solution. They assert, without reference, that their method is "more universal." If you look at the recurrence relation on pg 476 (not the matrix Z), you can derive a similar constant matrix relationship as above. For ease of computation, the authors simply neglect multiplying factors $(1-1/N)$ or $(1-2/N)$ but leave the terms $1/N$ or $2/N$. That is how the matrix Z is derived. Any number of symbolic solvers can then help you with the analytic solution. $\endgroup$ – player100 Jun 23 '17 at 10:45

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