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A spectral representation of the linear multivelocity transport problem

  • Mariam Avalishvili EMAIL logo and Dazmir Shulaia
Published/Copyright: August 18, 2016
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Abstract

The transformation of the original characteristic equation of the multivelocity linear transport theory was carried out by expanding the scattering function for the problem to be solved as a spectral integral over a complete set of eigenfunctions for the previously solved transport problem. The obtained equation represents a singular integral equation containing a spectral integral over the spectrum of the solved problem, whose kernel depends on the difference between the scattering of the problem to be solved and that of the already solved problem. We consider also the examples illustrating the validity of such a transformation. M. Kanal and J. Davies made a similar transformation of the characteristic equation of the one-velocity transport theory.

A Appendix

Suppose that the kernel f(μ,E;μ,E), with eigenfunctions ψω,(ζ)(μ,E), is of the form

(A.1)f(μ,E;μ,E)=s=0N(2s+1)Ps(μ)qs(E,E)Ps(μ).

The kernel f0(μ,E,μ,E), with eigenfunctions {φν,(ξ)(μ,E)}, is of the form

(A.2)f0(μ,E;μ,E)=s=0N0(2s+1)Ps(μ)rs(E,E)Ps(μ).

Let us denote

gs(ω,(ζ),E)=-1+1𝑑μPs(μ)ψω,(ζ)(μ,E)

and

hs(ν,(ξ),E)=-1+1𝑑μPs(μ)φν,(ξ)(μ,E).

Using the orthogonality and recursion properties of the Legendre polynomials from equation (2.2) yields

(s+1)gs+1(ω,(ζ),E)+sgs-1(ω,(ζ),E)=(2s+1)ω-1(E)(gs(ω,(ζ),E)-cE0E1𝑑Eqs(E,E)gs(ω,(ζ),E))

for s0. Analogous recursion properties are also valid for hs(ν,(ξ),E).

From equation (3.8) we get

(A.3)S0𝑑ρ(ν,(ξ))hs(ν,(ξ),E)K(ν,(ξ);ω,(ζ))=gs(ω,(ζ),E).

An important special case of equation (A.3) occurs when qs=rs in equations (A.1) and (A.2) for all sn and h0=g0. Then for all sn+1, we have hs=gs and

S0dρ(ν,(ξ))hs(ν,(ξ),E))K(ν,(ξ);ω,(ζ))=hs(ω,(ζ),E).

A corollary of the latter is that

S0𝑑ρ(ν,(ξ))νsh0(ν,(ξ),E)K(ν,(ξ);ω,(ζ))=ωsh0(ω,(ζ),E)

for all sn+1. Taking into account that q0=r0 for all kernels considered by us, we have

g1(ω,(ζ),E)=ω-1(E)(g0(ω,(ζ),E)-cE0E1𝑑Eq0(E,E)g0(ω,(ζ),E))

and

h1(ω,(ζ),E)=ω-1(E)(h0(ω,(ζ),E)-cE0E1𝑑Eq0(E,E)h0(ω,(ζ),E)),

respectively. From equation (A.3) we obtain

S0𝑑ρ(ν,(ξ))νh0(ν,(ξ),E)K(ν,(ξ);ω,(ζ))=ωg0(ω,(ζ),E).

References

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Received: 2014-12-29
Accepted: 2015-4-14
Published Online: 2016-8-18
Published in Print: 2016-9-1

© 2016 by De Gruyter

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