solving series of algebric linear equations using for loop

Can anybody help me for writing a script file for solving these three alegberic equations (couple). I wrote a code, but I am interested to use a for loop for finding different ui. Right now the code just solve u1 u2 u3. I want something for finding ui (i=0:10). xi+1=xi+ih and h=0.5 (x0=0). In advance thank you so much!

Here is the code:

clc;

clear all;

close all;

syms u1 u2 u3

h=0.5;

x1=1.5;

x2=2;

x3=2.5;

x4=3;

u0=2;

u4=-1;

eqn1= u0-(2+h^2*(1-x1/5))*u1+u2-h^2*x1==0;

eqn2= u1-(2+h^2*(1-x2/5))*u2+u3-h^2*x2==0;

eqn3= u2-(2+h^2*(1-x3/5))*u3+u4-h^2*x3==0;

sol=solve([eqn1, eqn2, eqn3], [u1, u2, u3]);

u1sol=sol.u1

u2sol=sol.u2

u3sol=sol.u3

I have attached the problem to clarify the problem.

[Rik]

How have you tried to write the loop yourself?

[Arash Yahyazadeh]

No that is exactly what I need. Finding a loop code for finding ui (i=0:10)

Answer 1

[Torsten]

N = 4;
xleft = 1.0;
xright = 3.0;
x = linspace(xleft,xright,N+1);
h = (xright-xleft)/N;
A = zeros(N+1);
b = zeros(N+1,1);
A(1,1) = 1;
for i = 2:N
    A(i,i-1) = 1.0;
    A(i,i) = -(2+h^2*(1-x(i)/5));
    A(i,i+1) = 1.0;
end
A(N+1,N+1) = 1.0;
b(1) = 2.0;
b(2:N) = h^2*x(2:N);
b(N+1) = -1.0;
u = A\b

u = 5×1

2.0000 0.5520 -0.4244 -0.9644 -1.0000

plot(x,u)

[Arash Yahyazadeh]

Thank you Torsten. I had a problem for underestanding the code. ui are values and obtained by solving algebric equations. Can you help me to figure out? I have attached the screen shot of the problem.

[Torsten]

Yes, that's exactly what is implemented in the code.

You need to solve the linear system

A*u = b

for a tridiagonal matrix A with the elements

A(i,i-1) = 1.0;
A(i,i) = -(2+h^2*(1-x(i)/5));
A(i,i+1) = 1.0;

and the right-hand side vector

b(i) = h^2*x(i)

with the boundary values implemented as

A(1,1) = 1.0;
A(N+1,N+1) = 1.0;
b(1) = 2.0;
b(N+1) = -1.0;

which means u0 = 2 and u_(N+1)=-1.

I'm not sure whether the term x(i)/5 remains the same if N changes - maybe it becomes x(i)/N. But you better know the underlying problem of your algebraic system of equations.

[Arash Yahyazadeh]

Thank you so much for your time and effort.