clear; clc; close all;
tau_0 = 0;%zero shear boundary condition
v_0 = 0;%Boundary layer condition
X0 = [v_0 tau_0];
global rho g mu tau0 mu0 n K
g = 9.81;
delta = 2e-3;%Thickness of layer
xspan = linspace(0,delta,100);

mu1 = 8.937E-4;
rho1 = 998;

mu2 = 10.211;
rho2 = 1300;

K3 = 0.01;
n3 = 0.5;
rho3 = 1123;

tau04 = 6;
mu04 = 0.2;
rho4 = 803;

%Case 1:Q(a),(b)
rho = rho1; mu = mu1;
fluidType = 'Newtonian';
[s1_x,s1_X] = ode45(@(t,X) fmshearsolve(t,X,fluidType), xspan, X0);
stra = 'Velocity Profile';
strb = 'Shear stress profile';
s1_str = 'Case 1(water like):';
str11 = [s1_str,' ',fluidType,' ',stra];
str12 = [s1_str,' ',fluidType,' ',strb];
figure('Name',str11);
plot(s1_x*1000,s1_X(:,1),'LineWidth',2,'Color',[1 0 0]);
xlabel('x(mm)'); ylabel('Velocity(m/s)'); title(str11);
figure('Name',str12);
plot(s1_x*1000,s1_X(:,2),'LineWidth',2,'Color',[1 0 0]);
xlabel('x(mm)'); ylabel('Shear stress(kg/m.s^2)'); title(str12);
s1_Vdot = trapz(s1_x,s1_X(:,1));
fprintf('\n\nThe volumetric flow rate for Case 1(water like) is computed as %10.8fm^3/s for unit depth.\n\n',s1_Vdot);

1. Using index notation, prove the following: 2. Consider the ve...