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Consider this function $$\frac{k^{2}-\xi^{2}}{k^{2}+1}$$ which has singularities at $k=\pm i$, the strips where it is analytic are $$ -1<k^{\prime \prime}<0 \quad \text { or } \quad 0<k^{\prime \prime}<1 $$ Let us concentrate on the strip $-1<k^{\prime \prime}<0 .$ The decomposition in this strip yields $$ 1-\hat{K}(k)=\frac{k^{2}-\xi^{2}}{k^{2}+1}=\frac{A(k)}{B(k)}=\frac{k^{2}-\xi^{2}}{k-i} \frac{1}{k+i} $$ where $$ \begin{array}{c} A(k)=\frac{k^{2}-\xi^{2}}{k-i} \\ B(k)=k+i \end{array} $$ such that now $A(k)$ is analytic for $k^{\prime \prime}<0$ and $B(k)$ is analytic for $k^{\prime \prime}>-1$. I worked out it just based on trial and error(Or I would say it is easy to find out such factors in this case).

Now a paper that I am reading claims that for $1+\widetilde{K}(p)=2 e^{-(c / 2)|p|} \cosh \left(\frac{c}{2} p\right)$ the decomposition $$ 1+\tilde{K}(p)=\frac{K_{+}\left(\frac{c}{2 \pi} p\right)}{K_{-}\left(\frac{c}{2 \pi} p\right)} $$ in such a way that $K_{+}\left(K_{-}\right)$ has singularities only in the upper (lower) half plane of the complex $p$ plane is given by

\begin{aligned} K_{+}(q) &=K_{-}^{-1}(-q)=\frac{(2 \pi)^{1 / 2}}{\Gamma\left(\frac{1}{2}+i q\right)} \exp \left[-i q\left(1+\frac{i \pi}{2}-\ln (-q+i \varepsilon)\right]\right] \end{aligned}

Is there a general process by which they got it?

Also, I have another problem where $c$ is now a complex variable instead of real as in this decomposition. How does changing $c$ to $c_r+i c_i$ change the above factorization?

The necessity for such a factorization appears in study of integral equations that are solvable by Wiener-Hopf method.

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  • $\begingroup$ Looks like what you are after is the Wiener–Hopf factorisation (which even has its Wikipedia page). But unfortunately I find your question very unclear. To start: what is $k''$? What are $\hat K$, $A$ and $B$? $\endgroup$ Jan 30 '21 at 18:50
  • $\begingroup$ Hi Yeah I am looking for Weiner-Hopf factorization. A and B are the factors such that A is analytic in upper plane and B is analytic in the lower place. $k''$ are the ranges of k over with those functions are analytic. $\endgroup$
    – user824530
    Jan 30 '21 at 19:07
  • $\begingroup$ Then, indeed, there is a general method to find the Wiener–Hopf factors, which is essentially: take the exp of the Cauchy integral of the log of your kernel. In some cases one can easily guess the Wiener–Hopf factors, and this seems to be the case here: the cosh term has zeroes along the imaginary axis, which correspond to the poles of the gamma function in the expression for $K_\pm$, and the exp terms seem to somehow correspond to the $\exp(-|p|)$ part. But this is just a guess, as your question is really very unclear. For example, what is $\epsilon$ in the expression for $K+\pm$? $\endgroup$ Jan 30 '21 at 20:38
  • $\begingroup$ That $\epsilon$ is $0^+$ and I am not sure why that is required there. $\endgroup$
    – user824530
    Jan 30 '21 at 21:50

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