Appendix A: Solution of Equation (12) Using Laplace Transform and Complex Inversion Formula

Appendix A: Solution of equation (12) using Laplace Transform and complex inversion formula.

In this appendix we indicate how equation (18) is derived. By solving a differential equation of second order with constant coefficients by using new homotopy approach and also by applying Laplace transform in equation (12) and in condition in (15-17) we obtained the solution of the equation (12) as

(A1)

In this appendix we indicate how equation (A1) may be inverted using the complex inversion formula. If represents the Laplace transform of a function, then according to the complex inversion formula we can state that

(A2)

where the integration in equation (A2) is to be performed along a line in the complex plane where The real number is chosen such that lies to the right of all the singularities, but is otherwise assumed to be arbitrary. In practice, the integral is evaluated by considering the contour integral presented on the right-hand side of equation (A2), which is then evaluated using the so-called Bromwich contour. The contour integral is then evaluated using the residue theorem which states for any analytic function

(A3)

where, the residues are computed at the poles of the function

Hence from eq. (A3), we note that

(A4)

From the theory of complex variables we can show that the residue of a function at a simple pole at is given by

(A5)

Hence in order to invert equation (A1), we need to evaluate

The poles are obtained from = 0. Hence there is a simple pole at = 0 and there are infinitely many poles given by the solution of the equation = 0 and

so where n = 0,1,2,…….

Hence we note that

(A6)

The first residue in equation (A6) is given by

=

= (A7)

The second residue in equation (A6) is given by

=

=

=-(A8)

where is defined as in equation(16). (Here we used and )

From (A6), (A7) and (A8) we conclude that

(A9)

where is defined as in equation(21).

Appendix B: Approximate Analytical Solution of equations (8) and (9) from the relation between the concentrationsand

From the equations (12) and (13), we can obtain the following equation.

(B1)

Let (B2)

Then (B1) becomes

(B3)

By using the boundary conditions in eqns. (15-17), the boundary conditions for will be

; when (B4)

; when (B5)

; when (B6)

Now by applying Laplace transform in (B3) and by using complex inversion formula and proceeding as in Appendix A the solution of (B3) will be

(B7)

From (B2), we have (B8)

The analytical expression for the concentration can be obtained by substituting (A9) and (B7) in (B8). Similarly the relation between and from the equations (12) and (14) is got by

(B9)

Let (B10)

Then (B9) becomes

(B11)

By using the boundary conditions in (15) to (17), the boundary conditions for will be

; when (B12)

; when (B13) when (B14)

Now by applying Laplace transform in (B11), (B12), (B13) and (B14) and by using complex inversion formula (Appendix A ), the solution of (B11) becomes as

(B15)

From (B10), we have (B16)

Using the above equation, we can obtain the eqn. (20) (the analytical expression for the concentration of in the text.

Appendix C: Approximate Analytical Solution of equations (22-24) the release kinetics of ethyl ferulate from immobilized catalyst

from (11)

from (11)

from (14 )

from ( )

From the equation (24) in the text we can understand the release kinetics of ethyl ferulate from immobilized enzyme.

Appendix D : Mat lab program to find the numerical solution of eqns. (12-14)

functionpdex10

m = 0;

x = linspace(0,1);

t=linspace(0,100);

sol = pdepe(m,@pdex4pde,@pdex4ic,@pdex4bc,x,t);

u1 = sol(:,:,1);

u2 = sol(:,:,2);

u3 = sol(:,:,3);

%------

figure

plot(x,u1(end,:))

title('u1(x,t)')

xlabel('Distance x')

ylabel('u1(x,1)')

%------

figure

plot(x,u2(end,:))

title('u2(x,t)')

xlabel('Distance x')

ylabel('u2(x,2)')

% ------

figure

plot(x,u3(end,:))

title('u3(x,t)')

xlabel('Distance x')

ylabel('u3(x,3)')

%------

function [c,f,s] = pdex4pde(x,t,u,DuDx)

c = [0; 0; 0];

f = [1; 1; 1] .* DuDx;

g1=1;g2=1;g3=15;a=30; b=0.39;

F1= - g1*u(1)*u(2)/((u(1)/b)+(u(2)/a)+(u(1)*u(2)));

F2=- g2*u(1)*u(2)/((u(1)/b)+(u(2)/a)+(u(1)*u(2)));

F3= g3*u(1)*u(2)/((u(1)/b)+(u(2)/a)+(u(1)*u(2)));

s=[F1; F2; F3];

% ------

functionu0 = pdex4ic(x)

u0 = [1; 0; 0];

% ------

function [pl,ql,pr,qr]=pdex4bc(xl,ul,xr,ur,t)

pl = [0; 0; 0];

ql = [1; 1; 1];

pr = [ur(1)-1; ur(2)-1; ur(3)-0];

qr = [0; 0; 0];