Solving nonlinear reaction–diffusion problem in electrostatic interaction with reaction-generated pH change on the kinetics of immobilized enzyme systems using Taylor series method

Abstract

A mathematical model of electrostatic interaction with reaction-generated pH change on the kinetics of immobilized enzyme is discussed. The model involves the coupled system of non-linear reaction–diffusion equations of substrate and hydrogen ion. The non-linear term in this model is related to the Michaelis–Menten reaction of the substrate and non-Michaelis–Menten kinetics of hydrogen ion. The approximate analytical expression of concentration of substrate and hydrogen ion has been derived by solving the non-linear reactions using Taylor’s series method. Reaction rate and effectiveness factor are also reported. A comparison between the analytical approximation and numerical solution is also presented. The effects of external mass transfer coefficient and the electrostatic potential on the overall reaction rate were also discussed.

Appendices

Appendix 1: Relation between H(X) and S(X)

From Eqs. (2) and (3) it is observed that

$$\frac{{d^{2} H\left( X \right)}}{{dX^{2} }} = a\frac{{d^{2} S\left( X \right)}}{{dX^{2} }}$$

(14)

On integrating both sides we get,

$$\frac{dH\left( X \right)}{{dX}} = a\frac{dS\left( X \right)}{{dX}} + c_{1}$$

(15)

Using the boundary condition given in Eq. (4), the value of $c_{1} = 0$.

Hence Eq. (15) becomes,

$$\frac{dH\left( X \right)}{{dX}} = a\frac{dS\left( X \right)}{{dX}}$$

(16)

Again on integrating we get

$$H\left( X \right) = aS\left( X \right) + c_{2}$$

(17)

Since at $X = 0, S\left( 0 \right) = l, H\left( 0 \right) = m$, the value of $c_{2} = m - al$.

$$H\left( X \right) = m + a\left( {S\left( X \right) - l} \right)$$

(18)

where

$$a = - \frac{{D_{es} }}{{D_{eh} }}\frac{{s_{0} }}{{h_{0} }}$$

(19)

Appendix 2: Approximate analytical expression of concentration of hydrogen ion and substrate using Taylor series method

Taylor series method is used for solving the dimensionless Eqs. (2) and (3). We assume that the solution of Eqs. (2)–(6) can be expanded in a Taylor series about $X = 0$ given by

$$S\left( X \right) = \mathop \sum \limits_{r = 0}^{4} \left. {\frac{{d^{r} S}}{{dX^{r} }}} \right|_{X = 0} \frac{{X^{r} }}{r!} = S\left( 0 \right) + \left. {\frac{dS}{{dX}}} \right|_{X = 0} X + \left. {\frac{{d^{2} S}}{{dX^{2} }}} \right|_{X = 0} \frac{{X^{2} }}{2!} + \left. {\frac{{d^{3} S}}{{dX^{3} }}} \right|_{X = 0} \frac{{X^{3} }}{3!} + \left. {\frac{{d^{4} S}}{{dX^{4} }}} \right|_{X = 0} \frac{{X^{4} }}{4!}$$

(20)

$$H\left( X \right) = \mathop \sum \limits_{r = 0}^{4} \left. {\frac{{d^{r} H}}{{dX^{r} }}} \right|_{X = 0} \frac{{X^{r} }}{r!} = H\left( 0 \right) + \left. {\frac{dH}{{dX}}} \right|_{X = 0} X + \left. {\frac{{d^{2} H}}{{dX^{2} }}} \right|_{X = 0} \frac{{X^{2} }}{2!} + \left. {\frac{{d^{3} H}}{{dX^{3} }}} \right|_{X = 0} \frac{{X^{3} }}{3!} + \left. {\frac{{d^{4} H}}{{dX^{4} }}} \right|_{X = 0} \frac{{X^{4} }}{4!}$$

(21)

Let us assume that $S\left( 0 \right) = l \ and\ H\left( 0 \right) = m.$

We can now evaluate the derivatives at $X = 0$ by substituting ${\text{X}} = 0$ into Eqs. (2) and (3) and impose the boundary condition that $\frac{{\text{dS}}}{{\text{dX}}} = \frac{{{\text{dH}}}}{{{\text{dX}}}} = 0$ at ${\text{X}} = 0$. By imposing these conditions simultaneously, we get

$$\left. {\frac{{d^{2} S}}{{dX^{2} }}} \right|_{X = 0} { } = S^{\prime\prime}\left( 0 \right){ } = \frac{\phi S\left( 0 \right)}{{\left( {1 + \gamma S\left( 0 \right)} \right)\left( {1 + \alpha_{1} H\left( 0 \right) + \frac{1}{{H\left( 0 \right)\alpha_{2} }}} \right)}}$$

(22)

$$\left. {\frac{{d^{2} H}}{{dX^{2} }}} \right|_{X = 0} = H^{\prime\prime}\left( 0 \right) = \frac{a\phi S\left( 0 \right)}{{\left( {1 + \gamma S\left( 0 \right)} \right)\left( {1 + \alpha_{1} H\left( 0 \right) + \frac{1}{{H\left( 0 \right)\alpha_{2} }}} \right)}} = a S^{\prime\prime}\left( 0 \right)$$

(23)

$$\left. {\frac{{d^{3} S}}{{dX^{3} }}} \right|_{X = 0} = \left. {\frac{{d^{3} H}}{{dX^{3} }}} \right|_{X = 0} = 0$$

(24)

$$\left. {\frac{{d^{4} S}}{{dX^{4} }}} \right|_{X = 0} = S^{4} \left( 0 \right) = - \frac{{\phi \alpha_{2} \left[ {S\left( 0 \right)\left( {\alpha_{1} \alpha_{2} H\left( 0 \right)^{2} + \alpha_{2} H\left( 0 \right) + 1} \right)\left( {\alpha_{1} \alpha_{2} H\left( 0 \right)^{2} - 1} \right)\left( {\gamma S\left( 0 \right) + 1} \right)^{2} H^{\prime\prime}\left( 0 \right) - H\left( 0 \right)\left( {\alpha_{1} \alpha_{2} H\left( 0 \right)^{2} + \alpha_{2} H\left( 0 \right) + 1} \right)^{2} \left( {\gamma S\left( 0 \right) + 1} \right)S^{\prime\prime}\left( 0 \right)} \right]}}{{\left( {\gamma S\left( 0 \right) + 1} \right)^{3} \left( {\alpha_{1} \alpha_{2} H\left( 0 \right)^{2} + \alpha_{2} H\left( 0 \right) + 1} \right)^{3} }}$$

(25)

$$\left. {\frac{{d^{4} H}}{{dX^{4} }}} \right|_{X = 0} = a S^{4} \left( 0 \right)$$

(26)

These derivative terms evaluated at $X = 0$ allow us to express the Taylor series solution Eqs. (20) and (21) as

$$S\left( X \right) = S\left( 0 \right) + S^{\prime\prime}\left( 0 \right)\frac{{X^{2} }}{2!} + S^{\left( 4 \right)} \left( 0 \right)\frac{{X^{4} }}{4!}$$

(27)

$$H\left( X \right) = H\left( 0 \right) + H^{\prime\prime}\left( 0 \right)\frac{{X^{2} }}{2!} + H^{\left( 4 \right)} \left( 0 \right)\frac{{X^{4} }}{4!}$$

(28)

Equations (27) and (28) can be written as

$$S\left( X \right) = l + S_{1} X^{2} + S_{2 } X^{4}$$

(29)

$$H\left( X \right) = m + aS_{1} X^{2} + aS_{2 } X^{4}$$

(30)

where

$$S_{1} = \phi l\left[ {2\left( {1 + \gamma l} \right)\left[ {1 + \alpha_{1} m + \left( {m\alpha_{2} } \right)^{ - 1} } \right]} \right]^{ - 1}$$

(31)

$$S_{2} = - \frac{{\phi \alpha_{2} \left[ {l\left( {\alpha_{1} \alpha_{2} m^{2} + \alpha_{2} m + 1} \right)\left( {\alpha_{1} \alpha_{2} m^{2} - 1} \right)\left( {\gamma l + 1} \right)^{2} H^{\prime\prime}\left( 0 \right) - m\left( {\alpha_{1} \alpha_{2} m^{2} + \alpha_{2} m + 1} \right)^{2} \left( {\gamma l + 1} \right)S^{\prime\prime}\left( 0 \right)} \right]}}{{24\left( {\gamma l + 1} \right)^{3} \left( {\alpha_{1} \alpha_{2} m^{2} + \alpha_{2} m + 1} \right)^{3} }}$$

(32)

$$H^{\prime\prime}\left( 0 \right) = a S^{\prime\prime}\left( 0 \right) = a\phi \left[ {\left( {1 + \gamma l} \right)\left( {1 + \alpha_{1} m + \left( {m\alpha_{2} } \right)^{ - 1} } \right)} \right]^{ - 1}$$

(33)

By applying the remaining boundary conditions (Eqs. (5) and (6)), we get

$$2S_{1} + 4S_{2} = \beta_{s} \left( {1 - l - S_{1} - S_{2 } } \right)$$

(34)

$$2aS_{1} + 4aS_{2} = b\left( {1 - \left( {m + aS_{1} + aS_{2 } } \right)P^{ - 1} } \right)$$

(35)

In order to find the unknown constants l and m, substitute the values of the parameter (refer Table 1) in (34) and (35). Using computational knowledge engine like Wolfram alpha we can get the values easily. The values of the unknown constants are $l = 0.1094, m = 1.2410$.

Hence,

$$S\left( X \right) = 0.1094 + 0.5108X^{2} + 0.3554X^{4}$$

(36)

$$H\left( X \right) = 1.2410 - 0.1277X^{2} - 0.0888X^{4}$$

(37)

The dimensionless reaction rate is

$$V\left( {pH_{0} } \right) = 2S_{1} + 4S_{2}$$

(38)

The effectiveness factor is

$$\eta \left( \phi \right) = \frac{1}{\phi }\left( {1 + \gamma } \right)\left( {1 + \alpha_{1} + \alpha_{2}^{ - 1} } \right)\left( {2S_{1} + 4S_{2} } \right)$$

(39)

Appendix 3: Maple coding to find dimensionless reaction rate

$$\begin{aligned} & > p : = 20;\,k : = 0.25;\,beta : = 100;\,M : = 1;\,g: = 1;\,P : = \exp \left( 0 \right);\,so : = 10^{ - 4} ;\,c : = 1; \\ & a1 : = 0.1; a2: = 1;\,ph0 : = 4; \\ & > ho : = 10^{ - ph0} ;a : = - k \cdot \frac{so}{{ho}}\,;b : = M \cdot beta \cdot k \cdot c; \\ & > S1 : = p \cdot \frac{l}{{2 \cdot \left( {g \cdot l + 1} \right) \cdot \left( {a1 \cdot m + 1 + \frac{1}{m \cdot a2}} \right)}}: \\ & > S2 : = a2^{2} \cdot p^{2} \cdot l \cdot m \cdot \frac{{\left( {a1 \cdot a2 \cdot m^{3} + a2 \cdot \left( {a \cdot a1 \cdot g \cdot l^{2} + a \cdot a1 \cdot l + 1} \right) \cdot m^{2} + m - a \cdot l \cdot \left( {g \cdot l + 1} \right)} \right)}}{{24 \cdot \left( {g \cdot l + 1} \right)^{3} \cdot \left( {a1 \cdot a2 \cdot m^{2} + a2 \cdot m + 1} \right)^{3} }}: \\ & > S : = S2.X^{4} + S1.X^{2} + l: \\ & > H : = S2 \cdot X^{4} \cdot a + S1 \cdot X^{2} \cdot a + m: \\ & > SD : = diff\left( {S, X} \right): \\ & > HD : = diff\left( {H, X} \right): \\ & > f1 : = simplify\left( {subs\left( {X = 1, SD} \right)} \right) = beta \cdot \left( {1 - simplify\left( {subs\left( {X = 1, S} \right)} \right)} \right): \\ & > f2 : = simplify\left( {subs\left( {X = 1, HD} \right)} \right) = b \cdot \left( {1 - \frac{{simplify\left( {subs\left( {X = 1, H} \right)} \right)}}{P}} \right): \\ & > solve\left( {\left\{ {f1, f2} \right\}, \left\{ {l, m} \right\}} \right); \\ & \left\{ {l = 0.1093690396, m = 1.240981202} \right\} \\ & > V : = simplify\left( {subs\left( {X = 1, \left\{ {l = 0.1093690396, m = 1.240981202} \right\}, SD} \right)} \right); \\ & V : = 2.443128278 \\ \end{aligned}$$