A. Putze et al.: An MCMC technique to sample transport and source parameters of Galactic cosmic rays. II. A.1.1. Stable species A.1.2. Adding a β -decay source term: general solution As above, ϖ denotes the quantity evaluated without energy redistribution, whereas N $^{j}$(0) denotes the equilibrium flux at z = 0. To ensure N $^{j}$(0) also appears in the l.h.s. of the equation, we form the quantity which is solved numerically, N $^{m}$(0) being the solution when energy terms are discarded, i.e., Eq. (A.4). The terms a ( E ) and b ( E ) describing energy losses and gains are discussed in Sect. 2.1. When a radioactive contribution exists, the constant left to determine is ϖ from Eq. (A.7), which we denote now ϖ ∗ For the case of a mixed species m without radioactive contribution, the result is straightforward: the solution for the halo is still given by Eq. (A.3), but N $^{m}$(0) is now given by When energy redistributions are included, the solution N $^{h}$( z ) in the halo remains the same because our model assumes no energy redistributions in that region. Only the last step of the calculation changes (ensuring continuity during the disc crossing). The new solution is denoted N (0). The superscript on K indicates that the diffusion coefficient is to be evaluated at a rigidity calculated for the nucleus m . The latter can differ from one nucleus to another because, the calculation is performed at the same kinetic energy per nucleon for all the nuclei (hence at slightly different rigidi- ties for different nuclei). To compare with the data, the flux is calculated at z = 0 and where It is emphasised in Maurin et al. (2001) that the $^{10}$Be → $^{10}$B channel contributes up to 10% in the secondary boron flux at low energy and cannot be neglected. Although the spa- tial distribution of a radioactive nucleus decreases expo- nentially with z , we have to consider that the source term is emitted from the halo, complicating the solution. The equation to solve for the nucleus j , which is β -fed by its radioactive parent r is where N $^{r}$( z, E ) is given by Eq. (A.3). The solution is found following the same steps as above, although it has a more complicated form (due to a non-vanishing source term in the halo). If we take into account both the standard source term Q j $_{disc}$( E ) and the radioactive contribution of the nucleus N $^{r}$, we obtain: where For a mixed species, primary and secondary standard sources add up, so that, for a nucleus k with no radioactive contribution, the source term is rewritten as and the corresponding equation to solve is then We find the solution in the halo, apply the boundary con- dition N ( z = L ) = 0, and then ensure continuity between the disc and the halo, so that and The quantities S m and A m are defined as Q m $_{disc}$( E ) = q m 0 Q ( E ) + ∑ k>m Γ $^{k}$→m tot N $^{p}$( z = 0 , E ) , (A.2) L $^{m}$Nm = 2 hδ ( z ) · Q m $_{disc}$( E ) . N $^{m}$( z ) = N $^{m}$(0) · exp ( V$_{c}$z/ 2 K ) sinh( S $^{m}$( L − z ) / 2) sinh( S $^{m}$L/ 2) (A.3) N $^{m}$(0) = 2 hQ m $_{disc}$( E ) A m . (A.4) S m ≡ √ V 2 c K 2 + 4 Γ m rad K 2 ; (A.5) A m ≡ V$_{c}$ +2 h Γ m $_{tot}$+ KS $^{m}$coth ( S $^{m}$L 2 ) . (A.6) N $^{j}$( z ) = { ϖ · sinh( S $^{j}$( L − z ) / 2) sinh( S $^{j}$L/ 2) − ν ΘΛ · cosh( S $^{j}$z/ 2) cosh( S $^{j}$L/ 2) } × exp ( V$_{c}$z/ 2 K $^{j}$) + Θ { λ sinh ( S r 2 ( L − z ) + Λ cosh ( S r 2 ( L − z ) )} × exp ( V$_{c}$z/ 2 K $^{$^{r}$)}$, (A.7) Θ ≡ − Γ $^{k}$→j rad K $^{j}$( λ 2 − Λ $^{2}$) N $^{r}$(0) sinh( S $^{r}$L/ 2) (A.8) Ξ ≡ Θ [ λ sinh( S $^{r}$L 2 )+Λ cosh( S $^{r}$L 2 ) − ν Λ cosh S $^{j}$L 2 ] . (A.12) ϖ ∗ = ϖ − 2 h A j ( b ( E ) d N $^{j}$(0) dE + c ( E ) d $^{2}$N $^{j}$(0) dE 2 ) . (A.11) N $^{m}$(0)= N $^{m}$(0) − 2 h A m ( b ( E ) d N $^{m}$(0) dE + c ( E ) d $^{2}$N $^{m}$(0) dE 2 ) , N $^{j}$(0)= ϖ + Θ [ λ sinh( S $^{r}$L 2 )+Λ cosh( S $^{r}$L 2 ) − ν Λ cosh S $^{j}$L 2 ] . (A.10) κ ≡ 1 /K r − 1 /K j ν ≡ e κV$_{c}$L/ 2 Λ ≡ κ S $^{r}$V$_{c}$ 2 λ ≡ κV 2 c 2 K r + Γ r rad K r − Γ j rad K j . ϖ ≡ 2 hQ j disc A j + Θ A j × (A.9) { νa$_{i}$ cosh( S $^{j}$L/ 2) [ V$_{c}$ + 2 h Γ $^{j}$] − sinh( S $^{r}$L/ 2) a ( V$_{c}$ (2 − K j K $_{r}$) + 2 h Γ j ) + a$_{i}$K $^{j}$Sr ] − cosh( S $^{r}$L/ 2) [ a$_{i}$ V$_{c}$ (2 − K j K $_{r}$) + 2 h Γ j + aK $^{j}$Sr ]} , L $^{j}$Nj = Γ $^{r}$→j $_{decay}$N $^{r}$( z, E ) , A.1.3. Solution including energy redistributions 17