Issue 
A&A
Volume 567, July 2014



Article Number  A33  
Number of page(s)  10  
Section  Astrophysical processes  
DOI  https://doi.org/10.1051/00046361/201322996  
Published online  07 July 2014 
Online material
Appendix A: Derivation of Green’s function for Eq. (1)
The Green’s function G(r,r′,p,p′) for Eq. (1) satisfies (A.1)where G(u) ≡ G(r,r′,u,p′). In rectangular coordinates, and assuming the sources are on the Galactic plane, i.e., z′ = 0, we can write the above equation as
(A.2)After taking Fourier transform of Eq. (A.2) with respect to x and y, we have
For z ≠ 0, Eq. (A.3) reduces to the following simple differential equation: (A.4)Solving the above equation by using the boundary condition that the particle density, and hence , vanishes at z = ± H, the solution of Eq. (A.3) for regions above (z> 0) and below (z< 0) the Galactic plane is obtained as, (A.5)where . The function can be determined using the continuity relation at z = 0 as follows. Integrating Eq. (A.3) over z around z = 0, we get
(A.6)where . Substituting for at z = 0 ±, and rearranging the terms, Eq. (A.5) reduces into a first order linear differential equation in p as (A.7)where, (A.8)The integrating factor of Eq. (A.6) is: (A.9)With this, the general solution of Eq. (A.6) is obtained as, (A.10)where we have written,
(A.11)The first term on the right hand side of Eq. (A.8) can be integrated by parts taking E(p_{1}) as the first function and the derivative part as the second function as follows: (A.12)where H[m] is the Heaviside step function, which has the property H[m] = 1(0) for m> 0( < 0). Substituting for E(p), which is defined by Eq. (A.9), into Eq. (A.10), we get, (A.13)Imposing the boundary condition that there are no particles at p = 0, Eq. (A.11) becomes,
(A.14)In the present study, as we assume that particles are injected with a finite momentum p′ ≥ p_{0} where p_{0} is the lowmomentum cutoff we have introduced so as to approximate the ionization losses (see Sect. 2), the delta function and the Heaviside step function in the first two terms on the right hand side of the above equation becomes zero. This gives I_{0} = 0, and the general solution of Eq. (A.6) becomes, (A.15)Proceeding further, we have,
(A.16)Then, E(p′) defined by Eq. (A.9) becomes,
(A.17)Substituting for C(p′) as defined by Eq. (A.7) into Eq. (A.14), we get (A.18)Differentiating Eq. (A.15) with respect to p′, we have
(A.19)where we have substituted in the last expression. Analogous to Eq. (A.13), we also obtain, (A.20)Substituting Eqs. (A.16) and (A.17) into Eq. (A.12), we get (A.21)where we have written, Substituting for from Eq. (A.18) into Eq. (A.4), the Green’s function for Eq. (1) can be obtained using the relation,
(A.22)In cylindrical coordinates (r,r′,z,z′), where x − x′ = (r − r′)cosθ, y − y′ = (r − r′)sinθ, and k_{x} = kcosφ, k_{y} = ksinφ, we obtain
(A.23)where J_{0} is a Bessel function of order 0.
© ESO, 2014
Current usage metrics show cumulative count of Article Views (fulltext article views including HTML views, PDF and ePub downloads, according to the available data) and Abstracts Views on Vision4Press platform.
Data correspond to usage on the plateform after 2015. The current usage metrics is available 4896 hours after online publication and is updated daily on week days.
Initial download of the metrics may take a while.