Dimerisation of a Unimolecular-decay Product (Consecutive First then Second-order Reaction)
Scheme:
`Astackrel(k_1) rarrB`
`2Bstackrel(k_2) rarrC`
Differential Equation:
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`A'(t)=-k_1A` `B'(t)=k_1A-k_2B^2` `C'(t)=k_2B^2` |
Mass balance equation:
`A_0=A+B+2C`
Initial conditions:
`A(0)=A_0` and `B(0)=C(0)=0`
Separable Equation:
`(dA)/A=-kt`
Solutions:
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`A(t)=A_0exp(-kt)` `B(t)=1/k_2 ([J_(1)(z)+alpha Y_(1)(z)]isqrt(A_0k_2k_1) exp((-k_1t)/2))/(J_0(z)+alpha Y_0(z))` `C(t)=1/2(A_0-A_0exp(-k_1t)-1/k_2 ([J_(1)(z)+alpha Y_(1)(z)]isqrt(A_0k_2k_1) exp((-k_1t)/2))/(J_0(z)+alpha Y_0(z)))` |
where `z=(2isqrt((A_0k_2)/k_1)) exp((-k_1t)/2)` and `alpha = -(J_(1)(2isqrt((A_0k_2)/k_1)) )/(Y_(1)(2isqrt((A_0k_2)/k_1)) )`
Derivation
Preamble
The solution to the kinetic system involving the dimerisation of a uni-molecular decay product, for example, dimerisation of the ground state produced by relaxation of an excited state, principally involves solving a Bessel-type differential equation. The time-dependance of starting material A(t), is easily found by solving the linear first-order autonomous equation for uni-molecular decay. Using this expential solution to A(t), we go about solving B'(t), which turns out to be a Bessel-type equation. Lastly, we produce an expression for C(t), given in terms of the mass-balance equation.
Once we have solutions for the concentration-time dependancies, we then need a means to evaluate the various Bessel-functions employed. Whilst for most arguments the Bessel-functions are are transcendental, they can be evaluated, in approximate form, algorithmically through the use of a series expansion. Such series can be evaluated through basic programming loops.
Solution to `A(t)`
The solution `A(t)` can be obtained directly by integration of the separable equation,
`int_(A_0)^(A_t)(dA)/A=-kt`
`[lnA]_(A_0)^(A_t)=-kt`
`lnA_t=lnA_0-kt`
Solution to `B(t)`
We now go about solving the `B'(t)`. Using the solution `A(t)`, the non-linear differential equation becomes:
`B'(t)=k_1A-k_2B^2=k_1A_0exp(-k_1t)-k_2B^2`
Following the solution to the special case of the Riccati equation, we use the substitution,
`B(t)=1/(k_2u(t)) (du(t))/(dt)`
which transforms the equation to,
`d/(dt)[1/(k_2u(t)) (du(t))/(dt)]=k_1A_0exp(-k_1t)-k_2(1/(k_2u(t)) (du(t))/(dt))^2`
Applying the product rule to the left hand side, expanding the brackets and cancelling the common terms we obtain the linear, second-order Bessel differential equation,
`(d^2u)/(dt^2)-A_0k_1k_2exp(-k_1t)u=0`
As shown here, a differential equation of the form,
`y''+aexp(kx)y=0`
has the solution,
`y=c_1J_0(z)+c_2Y_0(z)` with `z=(2sqrt(a))/k exp((kx)/2)`
Where `J_0` and `Y_0` are the Bessel functions of the First and Second Kind, respectively.
So in particular, we obtain the solution,
`u(t)=c_1J_0(z)+c_2Y_0(z)`
where `z=(2isqrt((A_0k_2)/k_1)) exp((-k_1t)/2)`
Referring back to our definition of `B(t)`, cf:
`B(t)=1/(k_2u(t)) (du(t))/(dt)=1/k_2 (u'(t))/(u(t))`
We have evidently calculated the denominator, `u(t)` in the form the Bessel functions, `J_0` and `Y_0`. Next we need to calculate the numerator, which involves the derivative, `u'(t)`. To do this we exploit the differentiation, recurrence relationships `J_0(x)'` and `Y_0(x)'`, in particular,
`J'_v(x)=1/2[J_(v-1)(x)-J_(v+1)(x)]`
and
`Y'_v(x)=1/2[Y_(v-1)(x)-Y_(v+1)(x)]`
Thus we need to evaluate,
`u'(t)=(c_1J_0'(z)+c_2Y_0'(z))`
We will first evaluate Y0'(x), and later evaluate J0'(x).
We note that z is a function of t, thus we have to use the chain-rule to obtain the derivative, ie., following the method outlined here,
`d/(dt) Y_0(z) = d/(dz) Y_0(z) (dz)/(dt)`
where as before, `z=(2isqrt((A_0k_2)/k_1)) exp((-k_1t)/2)`
First we evaluate `z'(t)`,
`(dz)/(dt)=(2isqrt((A_0k_2)/k_1)) exp((-k_1t)/2)`
Which we perform again using u-substitution, with `u=(-k_1t)/2`, and so, `u'(t)=(-k_1)/2`, thus,
`(dz)/(dt)=(2isqrt((A_0k_2)/k_1)) d/(du) exp(u) (du)/(dt)=(2isqrt((A_0k_2)/k_1)) exp(u) (-k_1)/2`
On reversing the substitution for u, and simplifying, we obtain,
`(dz)/(dt)=-isqrt(A_0k_2k_1) exp((-k_1t)/2)`
Now we evaluate `Y_0'(z)`, which we do using the recurrance relation,
`Y'_v(x)=1/2[Y_(v-1)(x)-Y_(v+1)(x)]`
Where v=0, thus,
`Y'_0(x)=1/2[Y_(-1)(x)-Y_(1)(x)]`
Next we simplify the Y-1(x) term using the symmetry relations,
`Y_v(-x)=-Y_(-v)(x)=(-1)^vY_v(x)`
and so, `Y_(-1)(z)=(-1)^(1)Y_(1)(z)` and `Y_(-1)(z)=-(-1)^1Y_1(z)`, which leads us to,
`d/(dz) Y_0(z)=1/2[-Y_(1)(z)-Y_(1)(z)]=-Y_(1)(z)`
Plugging in all these results in the chain-rule we obtain,
`d/(dt) Y_0(z) = Y_(1)(z) isqrt(A_0k_2k_1) exp((-k_1t)/2)`
`d/(dt) Y_0(z)=Y_(1)((2isqrt((A_0k_2)/k_1)) exp((-k_1t)/2)) isqrt(A_0k_2k_1) exp((-k_1t)/2)`
ie.,
`d/(dt) Y_0(z)=Y_(1)(z) isqrt(A_0k_2k_1) exp((-k_1t)/2)`
Now we will turn our attention to the derivative, `J_0'(z)`. Up to this point, `J_0'(z)` and `Y_0'(z)` behave similarly: they have the same arguments, obey the same recurrence relations and give the same derivative (other than swapping Yv(...) with Jv(...). Hence,
`d/(dt) J_0(z)=J_(1)((2isqrt((A_0k_2)/k_1)) exp((-k_1t)/2)) isqrt(A_0k_2k_1) exp((-k_1t)/2)`
However, with the J function, we can go a little further. We can simplify the expression through the relationship between the funtions of the First Kind and the Modified First Kind, namely, `I_v(z)=i^(-v)J_v(iz)`, where v=1, thus,
`iI_1(z)=J_1(iz)`
This enables us to remove the imaginary number from the argument, by replacing the J1 with the I1 function. The newly introduced coefficient of i, multiplied by the existing instance of i, changes the sign of the function, hence our expression becomes,
`d/(dt) J_0((2isqrt((A_0k_2)/k_1)) exp((-k_1t)/2))=-I_(1)((2sqrt((A_0k_2)/k_1)) exp((-k_1t)/2)) sqrt(A_0k_2k_1) exp((-k_1t)/2)`
Ie.,
`d/(dt) J_0(z)=-I_(1)(-iz) sqrt(A_0k_2k_1) exp((-k_1t)/2)`
Combining our results for the derivatives `Y_0'(z)` and `J_0'(z)`, we obtain and expression for `u'(t)=(c_1J_0'(z)+c_2Y_0'(z))`. Thus,
`u'(t)=c_1(-I_(1)(-iz) sqrt(A_0k_2k_1) exp((-k_1t)/2))+c_2(Y_(1)(z) isqrt(A_0k_2k_1) exp((-k_1t)/2))`
We then go on to simplify. First factorising,
`u'(t)=[-c_1I_(1)(-iz)+c_2iY_(1)(z)]sqrt(A_0k_2k_1) exp((-k_1t)/2)`
We can now reconstruct our definition of `B(t)`,
`B(t)=1/k_2 (u'(t))/(u(t))=1/k_2 ([-c_1I_(1)(-iz)+c_2iY_(1)(z)]sqrt(A_0k_2k_1) exp((-k_1t)/2))/(c_1J_0(z)+c_2Y_0(z))`
where as before, `z=(2isqrt((A_0k_2)/k_1)) exp((-k_1t)/2)`
...or alternatively, using only J0 and Y0, (both are valid solutions),
`B(t)=1/k_2 ([c_1J_(1)(z)+c_2Y_(1)(z)]isqrt(A_0k_2k_1) exp((-k_1t)/2))/(c_1J_0(z)+c_2Y_0(z))`
We will proceed using the later of these two solutions. We factorise the numerator and denominator by c1, introducing a new constant, `alpha=c_2/c_1`, and obtain the final expression for `B(t)`,
`B(t)=1/k_2 ([J_(1)(z)+alpha Y_(1)(z)]isqrt(A_0k_2k_1) exp((-k_1t)/2))/(J_0(z)+alpha Y_0(z))`
We now have to solve the initial value problem and obtain a expression for `alpha`. To do this we use the initial condition, B(0)=0 at t=0. Thus,
`0=1/k_2 ([J_(1)(z)+alpha Y_(1)(z)]isqrt(A_0k_2k_1) exp((-k_1t)/2))/(J_0(z)+alpha Y_0(z))`
Multiplying both sides by the denominator, we obtain,
`0=[J_(1)(z)+alpha Y_(1)(z)]isqrt(A_0k_2k_1) exp((-k_1t)/2)`
Divide both sides by the exponential term, by i, and by `sqrt(A_0k_2k_1)`,
`0=J_(1)(z)+alpha Y_(1)(z)`
We make `alpha` the subject, and reverse the substitution for z,
`alpha =-(J_(1)(z))/(Y_(1)(z))=-(J_(1)((2isqrt((A_0k_2)/k_1)) exp((-k_1t)/2)))/(Y_(1)((2isqrt((A_0k_2)/k_1)) exp((-k_1t)/2)))`
Settting t=0, we obtain a value for `alpha`,
`alpha = -(J_(1)(2isqrt((A_0k_2)/k_1)) )/(Y_(1)(2isqrt((A_0k_2)/k_1)) )`
Now that we have a complete expression for `B(t)`, we aim to find the an expression for `C(t)`. To do this, we exploit the mass balance equation,
`A_0=A+B+2C`
Solving it for `C(t)`,
`C=1/2(A_0-A-B)`
Using this solutions for `A(t)` and `B(t)`,
`C(t)=1/2(A_0-A_0exp(-k_1t)-1/k_2 ([J_(1)(z)+alpha Y_(1)(z)]isqrt(A_0k_2k_1) exp((-k_1t)/2))/(J_0(z)+alpha Y_0(z)))`
Evaluating the Bessel Functions
The Bessel-functions are transcendental, like the Lerch-function (found in the solution to the parallel-consecutive bimolecular mechanism kinetics), cannot be expressed through a fininite number of elementary functions. However, as detailed here, the Bessel-functions series expansions can be used to approximate their evaluation.
