update

lu_tridiag.m

@@ -1,6 +1,28 @@
 
-function [ud,uo,lo]=lu_tridiag(e,f,g)
+function [ud,uo,lo]=lu_tridiag(e,f,g);
+  % lu_tridiag calculates the components for LU-decomposition of a tridiagonal matrix
+  % given its off-diagonal vectors, e (a) and g (b)
+  % and diagonal vector f (d)
+  % the output is 
+  % the diagonal of the Upper matrix, ud (dd)
+  % the off-diagonal of the Upper matrix, uo 
+  % and the off-diagonal of the Lower matrix, lo (bb)
+  % note: the diagonal of the Lower matrix is all ones
+[m n]=size(f);
+ud=zeros(m,n);
+uo=zeros(m,n);
+lo=zeros(m,n);
+%ud(1)=f(1);
+%lo(1)=0;
+%uo=e;
+
+
+%for i=2:n
+ %   lo(i)=g(i)/(ud(i-1));
+  %  ud(i)=f(i)-(lo(i)*e(i-1));
+%end 
+
+
 
-n=size(e);
 
 end

solve_tridiag.m

@@ -0,0 +1,31 @@
+
+function x=solve_tridiag(ud,uo,lo,b);
+  % solve_tridiag solves Ax=b  for x
+  % given 
+  % the diagonal of the Upper matrix, ud
+  % the off-diagonal of the Upper matrix, uo
+  % the off-diagonal of the Lower matrix, lo
+  % the vector b
+  % note: the diagonal of the Lower matrix is all ones
+
+[s n]=size(uo);
+z=zeros(s,n);
+z(1)=b(1);
+x=zeros(s,n);
+
+
+%solve Lz=b using forward subsitution
+for i= 2:n
+    z(i)=b(i)-(lo(i)*z(i-1));
+end 
+
+%solve Ux=z using backward substitution
+x(n)=z(n)/ud(n);
+
+for k= (n-1):-1:1 
+
+    x(k)=(z(k)-uo(k)*x(k+1))/ud(k);
+end
+end
+
+