Understanding Transcendental Equations in Multicompartment Systems

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SUMMARY

This discussion focuses on the analysis of a transcendental equation related to diffusion in multicompartment systems, specifically the equation 2*u*(cos(k)-cos(q))-k*sin(k) = 0. The roots of this equation, denoted as kj, are shown to depend periodically on the parameter q, with the analysis constrained to the range 0 < q < π. For small values of u (u << 1), the roots approximate to jπ, with k0 approximately equal to 2*(u^0.5)*sin(q/2) and kj approximately equal to jπ + ((2*u)/(jπ))*(1-(-1)^j cos(q)). The discussion highlights the confusion regarding the periodic dependence of roots on q while q is treated as a fixed constant.

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stoucha
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Hi,

I am reading a paper where part of the solution to an equation of diffusion in a multicompartment system includes the "sum of all nonnegative roots kj of the following transcendental equation,

2*u*(cos(k)-cos(q))-k*sin(k) = 0.

Then the authors of the paper say: "Note that the roots periodically depend on the parameter q, and, for their analysis, it is sufficient ot consider 0<q<pi. In the case of small u (i.e. uMM1), all roots are close to j*pi:

k0 approximately equals 2*(u^0.5)*sin(q/2), j=0

kj approximately equals j*pi + ((2*u)/(j*pi))*(1-(-1)^j cos(q))

end quote.

Can somebody help me with this derivation? The whole statement about roots periodically depending on q is confusing because q is fixed constant of the system. Further it seems to me that if we consider the case were u<<1, then the equation simplifies to -ksin(k) = 0 whith the roots not having q or u in them at all.

What am I missing?

Thanks for any help.
 
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We have ##2u\cos k -k\sin k = 2u \cos q##. The right hand side is periodic in ##q## and the left hand side doesn't depend on ##q## at all. Thus it is sufficient to consider all possible values of ##\cos q##, that is ##0 < q < \pi##, where I assume ##q\neq 0## for other reasons.

Next we have ##\cos k -\dfrac{k}{2u}\sin k = \cos q##. If ##u << 1## then the quotient is large, i.e. ##\sin k## has to be close to zero, for otherwise we would leave the range ##[-1,1]## given by the right hand side. Now the zeroes of ##\sin k## are the points ##j\pi##.

Since you haven't said what k0 and kj are, I can't explain those.
 

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