Thomas algorithm - tridiagonal matrix

Is there any other way to code and solve the tridiagonal matrix? the idea would be to try to get the plot shown. Matlab beginner, so, no sure how to do it. Any help will be greatly appreciated. Thanks

clear
cm=1/100;
delta = 1*cm;
nu=1e-6;
Uinf=1;
H=2*delta;
y=linspace(0,H,40);
u=Uinf*erf(3*y/delta);
dy=mean(diff(y));
dx=100*dy;

Q=nu*dx/Uinf/dy^2;
a=-Q;
c=-Q;
b=1+2*Q;
N=length(y)-2;
M = diag(b*ones(1,N)) + diag(c*ones(1,N-1),1) + diag(a*ones(1,N-1),-1);
%Constant Ue
Ue = @(x) Uinf;
u = u(2:end-1);
x=0;
uall=[0,u,Ue(x)];

Answer 1

[Torsten]

clear
cm=1/100;
delta = 1*cm;
nu=1e-6;
Uinf=1;
H=2*delta;
y=linspace(0,H,40);
u=Uinf*erf(3*y/delta);
dy=mean(diff(y));
dx=100*dy;

Q=nu*dx/Uinf/dy^2;
a=-Q;
c=-Q;
b=1+2*Q;
N = length(y);
M = diag(b*ones(1,N)) + diag(c*ones(1,N-1),1) + diag(a*ones(1,N-1),-1);
M(1,:) = [1,zeros(1,N-1)];
M(end,:) = [zeros(1,N-1),1];
%Constant Ue
Ue = @(x) Uinf;
u = u(2:end-1);
x=0;
uall=[0,u,Ue(x)];
sol = M\uall.';
plot(y,sol)
grid on

[Cesar Cardenas]

thanks a lot, could you describe a bit the changes you made? thank you

[Torsten]

These are the changes I made:

N = length(y);
M = diag(b*ones(1,N)) + diag(c*ones(1,N-1),1) + diag(a*ones(1,N-1),-1);
M(1,:) = [1,zeros(1,N-1)];
M(end,:) = [zeros(1,N-1),1];

instead of

N=length(y)-2;
M = diag(b*ones(1,N)) + diag(c*ones(1,N-1),1) + diag(a*ones(1,N-1),-1);

Think about what they mean.

[Cesar Cardenas]

Right thanks much, just a final question, is there any other way to write the this line, the tridiagonal matrix? Thanks much,

M = diag(b*ones(1,N)) + diag(c*ones(1,N-1),1) + diag(a*ones(1,N-1),-1);

[Torsten]

You can assign values to certain elements in a matrix by using a loop. But if the above line to define M is correct, it's elegant, isn't it ? Why do you want to define it differently ?

[Cesar Cardenas]

Right thank you, just wanted to have a different version. Thanks