I have the series


where $m$ is an integer. Is it possible to compute the coefficients $a_{n}(\nu)$? An exact solution is found for $\nu=\pi$:

$a_0=1, a_{n}=2\times(-1)^n$.

It is related to a well known problem of Pierce's electrodes in physics. My problem also deals with the Pierce electrodes but in a different geometry. I found a numerical solution for the most wanted value $\nu=2\pi/3$ using SVD pseudoinverse but is not satisfactory in a certain sense. Since the above equation for $a_{n}$ looks simple it is hoped that a simple solution to the problem might exist.

Attached sequence of numerical solutions of the truncated system


seems to indicate that the numerical solution slowly converges to an oscillating function.

Numerical solution for $N=60$ equations: Numerical solution for <span class=$N=60$ equations">

Numerical solution for $N=120$ equations: Numerical solution for <span class=$N=120$ equations">

Numerical solution for $n=240$ equations: Numerical solution for <span class=$N=240$ equations">

  • $\begingroup$ Are there some additional conditions on $a_n(\nu)$ or am I misreading something? Why couldn't you just take all $a_n(\nu)$ to be zero except $a_0$. $\endgroup$ – j.c. Oct 19 '13 at 7:57
  • $\begingroup$ @j.c.:Indeed, that is the obvious solution: $a_n=\delta_{n,0}$ with $\nu\rightarrow 0$. $\endgroup$ – Jon Oct 19 '13 at 12:04
  • 1
    $\begingroup$ @j.c: $a_n$ may not depend on $m$. It is assumend that $m$ runs from 1 to $\infty$, so we have infinite number of equations. $\endgroup$ – Igor Kotelnikov Oct 20 '13 at 11:44
  • $\begingroup$ Looks like weakly-* converging to something strange... For further numerical investigation I would suggest not to use the pure pseudo-inverse but some regularization in the solution of the truncated linear system. Probably ordinary Tikhonov-regularization, i.e. solving $(A^*A + \alpha I)a = A^*b$ (where $A$ is the coefficient matrix and $b$ is the right hand side $b_m = 1/m$) with some small $\alpha$, would be helpful to see what the limit may be. Moreover, one may want to couple $\alpha$ to $N$ in some clever way. Also: Is it clear that a solution exists? $\endgroup$ – Dirk Oct 21 '13 at 7:25
  • $\begingroup$ @Dirk, I thought that SVD pseudoinverse is a sort of regularization method so that Tikhonov-regularization is not needed. $\endgroup$ – Igor Kotelnikov Oct 23 '13 at 8:48

Here is an approach that leads to an integral equation for $$A(\nu,t) := \sum_{n=0}^{\infty}(-1)^{n}\frac{a_{n}(\nu)}{\nu} e^{Int}.$$

First we notice that $$(-1)^{n}\frac{a_{n}(\nu)}{\nu} = \frac{1}{2\pi} \int_0^{2\pi} A(\nu,t) e^{-Int}\,dt.$$

The given identity multiplied by $e^{2mz}$ can be restated as \begin{split} \frac{e^{2mz}}{m} &= \sum_{n=0}^{\infty}(-1)^{n}a_{n} (\nu)\frac{\sin[\nu\,(m-n)]}{\nu\,(m^2-n^2)}e^{2mz} \\ &= (-1)^{m}\frac{a_m(\nu)e^{2mz}}{2m} + \sum_{n=0}^{m-1} (-1)^{n}\frac{a_{n} (\nu)}{\nu}\sin(\nu\,(m-n)) \frac{e^{(m-n)z}}{m-n}\frac{e^{(m+n)z}}{m+n} + \sum_{n=m+1}^{\infty} (-1)^{n}\frac{a_{n} (\nu)}{\nu}\sin(\nu\,(n-m)) \frac{e^{-(n-m)z}}{n-m}\frac{e^{(m+n)z}}{m+n}. \end{split}

For the first sum, we have

\begin{split} &\sum_{n=0}^{m-1} \frac{1}{2\pi}\int_0^{2\pi}dt\, A(\nu,t)e^{-Int} \Im e^{I\nu(m-n)} \int_{-\infty}^z dx\,e^{(m-n)x} \int_{-\infty}^z dy\,e^{(m+n)y} \\ =&\frac{1}{2\pi}\Im \int_0^{2\pi}dt\, A(\nu,t) \int_{-\infty}^z dx\, \int_{-\infty}^z dy\, e^{(I\nu+x+y)m} \sum_{n=0}^{m-1} e^{(-It-I\nu-x+y)n}\\ =&\frac{1}{2\pi}\Im \int_0^{2\pi}dt\, A(\nu,t) \int_{-\infty}^z dx\, \int_{-\infty}^z dy\, e^{(I\nu+x+y)m} \frac{1-e^{(-It-I\nu-x+y)m}}{1-e^{-It-I\nu-x+y}}\\ =&\frac{1}{2\pi}\Im \int_0^{2\pi}dt\, A(\nu,t) \int_{-\infty}^z dx\, \int_{-\infty}^z dy\, \frac{e^{(I\nu+x+y)m}-e^{(-It+2y)m}}{1-e^{-It-I\nu-x+y}}. \end{split}

Similarly, for the second sum \begin{split} &\sum_{n=m+1}^{\infty} \frac{1}{2\pi}\int_0^{2\pi}dt\, A(\nu,t)e^{-Int} \Im e^{I\nu(n-m)} \int_{-\infty}^{-z} dx\,e^{(n-m)x} \int_{-\infty}^z dy\,e^{(m+n)y} \\ =&\frac{1}{2\pi}\Im \int_0^{2\pi}dt\, A(\nu,t) \int_{-\infty}^{-z} dx\, \int_{-\infty}^z dy\, e^{(-I\nu-x+y)m} \sum_{n=m+1}^{\infty} e^{(-It+I\nu+x+y)n}\\ =&\frac{1}{2\pi}\Im \int_0^{2\pi}dt\, A(\nu,t) \int_{-\infty}^{-z} dx\, \int_{-\infty}^z dy\, e^{(-I\nu-x+y)m} \frac{e^{(-It+I\nu+x+y)(m+1)}}{1-e^{-It+I\nu+x+y}}\\ =&\frac{1}{2\pi}\Im \int_0^{2\pi}dt\, A(\nu,t) \int_{-\infty}^{-z} dx\, \int_{-\infty}^z dy\, e^{-It+I\nu+x+y} \frac{e^{(-It+2y)m}}{1-e^{-It+I\nu+x+y}}. \end{split}

Finally, $$(-1)^{m}\frac{a_m(\nu)e^{2mz}}{2m} = \frac{\nu}{2\pi}\int_0^{2\pi}dt\, A(\nu,t) \int_{-\infty}^z dy\, e^{(-It+2y)m}.$$

Summing over $m=1,2,\dots$, we get \begin{split} -\log(1-e^{2z}) &= \frac{\nu}{2\pi}\int_0^{2\pi}dt\, A(\nu,t) \int_{-\infty}^z dy\, \frac{e^{-It+2y}}{1-e^{-It+2y}} \\ &+\frac{1}{2\pi}\Im \int_0^{2\pi}dt\, A(\nu,t) \int_{-\infty}^z dx\, \int_{-\infty}^z dy\, \frac{ \frac{e^{I\nu+x+y}}{1-e^{I\nu+x+y}}-\frac{e^{-It+2y}}{1-e^{-It+2y}}}{1-e^{-It-I\nu-x+y}}\\ &+\frac{1}{2\pi}\Im \int_0^{2\pi}dt\, A(\nu,t) \int_{-\infty}^{-z} dx\, \int_{-\infty}^z dy\, \frac{e^{-2It+I\nu+x+3y}}{(1-e^{-It+I\nu+x+y})(1-e^{-It+2y})}. \end{split}

Evaluating integrals over $x$ and $y$ (which is a techincal task) will give an integral equation for $A(\nu,t)$.


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