a) Using the definition (10.25) for the matter source densities π_1 (n_1, n_2) and π_2 (n_1, n_2) for the stoichiometric coefficients ν_{α1} = −1, ν_{α2} = 1, ν_{b1} = 1, ν_{b2} = −1 , we can write that,
\pi _A = \sum\limits_{\alpha =1}^{n}{\omega _\alpha } \nu _{\alpha A} (10.25)
π_1 (n_1, n_2) = Ω_{11} Δn_1 +Ω_{12} Δn_2 = ν_{α1} ω_α + ν_{b1} ω_b = −(ω_α − ω_b)
π_2 (n_1, n_2) = Ω_{21} Δn_1 +Ω_{22} Δn_2 = ν_{α2} ω_α + ν_{b2} ω_b = ω_α − ω_b
Using the definition (8.18) for the chemical affinities A_α and A_b ,
A_\alpha = -\sum\limits_{A=1}^{r}{\mu _A}\nu _{\alpha A} (8.18)
A_α = −μ_1 ν_{α1} − μ_2 ν_{α2} = μ_1 − μ_2
A_b = −μ_1 ν_{b1} − μ_2 ν_{b2} = −(μ_1 − μ_2) = −A_α
Since there is no expansion of the system, i.e. ∇ · ν = 0, the scalar linear empirical relations (11.6) reduce to,
\begin{cases}\omega _\alpha = \sum\limits_{b=1}^{n}{L_{\alpha b} A_b + L_{\alpha f} \nabla .\nu } & \forall \alpha =1,….,n \\ \tau ^{fr} = \sum\limits_{b=1}^{n}{L_{fb}A_b + L_{ff} \nabla .\nu } \end{cases} (11.6)
ω_α = L_{αα} A_α + L_{αb} A_b = (L_{αα} − L_{αb})A_α = (L_{αα} − L_{αb}) (μ_1 − μ_2)
ω_b = L_{bα} A_α + L_{bb} A_b = (L_{bα} − L_{bb})A_α = (L_{bα} − L_{bb}) (μ_1 − μ_2)
Since the temperature and the chemical potentials are homogeneous, i.e. ∇ T = 0 and ∇μ_1 = ∇μ_2 = 0 , the second law (10.88) reduces to,
\pi _s = \frac{1}{T} \left\{\sum\limits_{\alpha =1}^{n}{\omega _\alpha }A_\alpha +\tau ^{fr}(\nabla .\nu ) + j_s .(-\nabla T ) + \sum\limits_{A=1}^{r}{j_A} .(-\nabla \mu _A -q_A \nabla \varphi ) \right\} (10.88)
\pi _s = \frac{1}{T} (ω_α A_α + ω_b A_b) = \frac{A_α}{T}(ω_α − ω_b)
= \frac{A^2_α}{T} (L_{αα} − L_{αb} − L_{bα} + L_{bb}) = \frac{ W A^2_α}{T}≥ 0
which implies that W ≥ 0 since T > 0 and A^2_α≥ 0 . Thus,
Ω_{11} Δn_1 +Ω_{12} Δn_2 = −(ω_α − ω_b) = −W(μ_1 − μ_2)
Ω_{21} Δn_1 +Ω_{22} Δn_2 = ω_α − ω_b = W(μ_1 − μ_2)
Now, we have to express the chemical potentials μ_1 and μ_2 in terms of the concentration n_1/n and n_2/n of substances 1 and 2 using the relation (8.68) for a mixture of ideal gases,
μ_A (T, p, c_A) = μ_A (T, p) + RT \ln (c_A) (8.68)
\mu _1(T,n_1,n) = \mu _1(t,n_1 )+ R T\ln \bigl(\frac{n_1}{n}\bigr)
= μ_1 (T, n_1) + R T \ln \Bigl(\frac{n_{01}+\Delta n_1}{n}\Bigr)
\mu _2(T,n_2,n) = \mu _2(t,n_2 )+ R T\ln \bigl(\frac{n_2}{n}\bigr)
= μ_2 (T, n_2) + R T \ln \Bigl(\frac{n_{02}+\Delta n_2}{n}\Bigr)
which can be recast as,
μ_1 (T, n_1, n) = μ^0_1 (T, n_1, n) + R T \ln \Bigl(1+\frac{\Delta n_1}{n}\Bigr)
μ_2 (T, n_2, n) = μ^0_2 (T, n_1, n) + R T \ln \Bigl(1+\frac{\Delta n_2}{n}\Bigr)
where the chemical potentials at equilibrium are,
μ^0_1 (T, n_1, n) = μ_1 (T, n_1) + R T \ln (\frac{n_{01}}{n}\bigr)
μ^0_2 (T, n_2, n) = μ_2 (T, n_2) + R T \ln (\frac{n_{02}}{n}\bigr)
At equilibrium the chemical potentials are equal,
μ^0_1 = (T, n_1, n) = μ^0_2 =(T, n_2, n) ≡ μ^0 (T, n_1, n_2)
Thus, for small density perturbations, i.e. Δn_1 \ll 1 and Δn_2 \ll 1 , the chemical potentials become,
μ_1 (T, n_1, n) = μ^0 (T, n_1, n_2) + R T \frac{Δn_1}{n}
μ_2 (T, n_1, n) = μ^0 (T, n_1, n_2) + R T \frac{Δn_2}{n}
Hence,
Ω_{11} Δn_1 +Ω_{12}Δn_2 = −W(μ_1 − μ_2) = − \frac{R T W}{n} (Δ n_1 − Δn_2)
Ω_{21} Δn_1 +Ω_{22}Δn_2 = W(μ_1 − μ_2) = \frac{R T W}{n} (Δ n_1 − Δn_2)
which implies that the coefficients are given by,
Ω_{11} = Ω_{22} = −\frac{R T W}{n} ≤0 and Ω_{12} = Ω_{21} = \frac{R T W}{n}≥0
b) Using the matter current densities j_1 and j_2 , the matter continuity equations can be recast as,
\dot{n}_1 = D_1∇^2 n_1 + π_1 (n_1, n_2) and \dot{n}_2 = D_2∇^2 n_2 + π_2 (n_1, n_2)
where the Laplacian ∇^2 = ∇ · ∇ is a scalar operator. Introducing the scalar Ω ≡ Ω_{12} = Ω_{21} = −Ω_{11} = −Ω_{22} ≥ 0 and using the relations for the matter source densities π_1 (n_1, n_2) and π_2 (n_1, n_2) , the matter continuity equations become,
\dot{n}_1 = D_1∇^2 n_1 − Ω Δn_1 + Ω Δn_2 .
\dot{n}_2 = D_2∇^2 n_2 + Ω Δn_1 – Ω Δn_2 .
Since Δn_1 = n_1 − n_{01} and Δn_2 = n_2 − n_{02} where n_{01} and n_{02} are constants,
\dot{n}_1 = Δ\dot{n}_1 \dot{n}_2 = Δ\dot{n}_2 ∇^2 n_1 = ∇^2 (Δn_1) ∇^2 n_2 = ∇^2 (Δn_2)
Thus, the coupled time evolution equations for the density perturbations Δn_1 and Δn_2 yield,
\dot{n}_1 = D_1∇^2 (Δn_1) − Ω Δn_1 + Ω Δn_2
\dot{n}_2 = D_2∇^2 (Δn_2) +Ω Δn_1 – Ω Δn_2
c) The coupled time evolution equations can be written as a matrix system,
\Biggl(\begin{matrix} \Delta \dot{n}_1 \\ \Delta \dot{n}_2 \end{matrix} \Biggr) = \Biggl(\begin{matrix} -\Omega +D_1\nabla^2 & \Omega \\ \Omega & -\Omega +D_2\nabla^2 \end{matrix} \Biggr) \Biggl(\begin{matrix} \Delta n_1 \\ \Delta n_2 \end{matrix} \Biggr)
Replacing the solution in the coupled time evolution equations and then using the relations,
\dot{n}_1 = λΔn_1 and ∇^2 (Δn_1) = −k^2 Δn_1
\dot{n}_2 = λΔn_2 and ∇^2 (Δn_2) = −k^2 Δn_2
the matrix system can be recast as,
\Biggl(\begin{matrix} -\Omega -D_1k^2-\lambda & \Omega \\ \Omega & -\Omega -D_2k^2 -\lambda \end{matrix} \Biggr) \Biggl(\begin{matrix} Δn_1 \\ Δn_2 \end{matrix} \Biggr)=0
For non-trivial solutions, the determinant of this matrix vanishes,
(Ω+D_1 k^2 + λ)(Ω+D_2 k^2 + λ)− Ω^2 = 0
which is recast as,
λ^2 + 2 ω λ + α = 0
where,
ω ≡ \frac{1}{2}\Bigl(2\Omega +(D_1+D_2)k^2\Bigr)\geq 0
α ≡(Ω+D_1 k^2)(Ω+D_2 k^2)− Ω^2 ≥ 0
which implies that ω^2 − α ≥ 0 . The solutions of the quadratic equation in λ are,
\lambda _1= – \omega +\sqrt{\omega ^2-\alpha }
\lambda _2= – \omega -\sqrt{\omega ^2-\alpha }
These solutions are called the Lyapunov exponents of the system. Under the assumption of a closed system undergoing chemical reactions transforming substance 1 into substance 2 and vice versa, the Lyapunov exponents are negative, i.e. λ_1 < 0 and λ_2 < 0 , which corresponds to stable solutions. For the formation of instabilities, where the density perturbations grow exponentially, at least one of the Lyapunov exponents has to be positive, i.e. λ_1 > 0 or λ_2 > 0 . Thus, in a closed system, density perturbations cannot grow. In order to see the rise and growth of instabilities, which can lead to the formation of patterns called Turing patterns, there has to be a source for the substances 1 and 2, which is the environment.