Supplementary Material of the Manuscript

Supplementary material of the manuscript

Part-2:Analytical Expressions for the Non-Steady State Concentrations of Glucose, Oxygen and Gluconic Acid in a Composite Membrane for Closed-Loop Insulin Delivery

N. Mehala1, L. Rajendran1,3*,V. Meena2

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Appendix A:Solution of equation (7) using Laplace Transform and complex inversion formula.

In this appendix we indicate how equation (13)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 (7) and in condition in (10) we obtained the solution of the equation (7) 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(17). (Here we used and )

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

(A9)

where is defined as in equation(18).

Appendix C :Matlab program to find the numerical solution of eqns. ( (7)-(9) )

function pdex10

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];

% ------

function u0 = 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];

Fig.1. The normalized three dimensional concentration profiles of glucose, oxygen and gluconic acid respectively a) for values of b) for values of c) for values of

Fig.2. Plot of concentration of profiles of glucose, oxygen and gluconic acid versus dimensionless length for the values of

Fig.3(a). Sensitivity analysis for evaluating the influence of the parameters in the concentration of glucoseusing equation (13).

Fig.3(b). Sensitivity analysis for evaluating the influence of the parameters in the concentration of oxygen using equation (14).

Fig.3(c).Sensitivity analysis for evaluating the influence of the parameters in the concentration of gluconic acid using equation(15).

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