limSolve (version 1.4)

Solve.tridiag: Solution of a tridiagonal system of linear equations

Description

Solves the linear system of equations $$Ax=B$$ where A has to be square and tridiagonal, i.e with nonzero elements only on, one band above, and one band below the diagonal.

Usage

Solve.tridiag ( diam1, dia, diap1, B=rep(0,times=length(dia)))

Arguments

diam1

a vector with (nonzero) elements below the diagonal.

dia

a vector with (nonzero) elements on the diagonal.

diap1

a vector with (nonzero) elements above the diagonal.

B

Right-hand side of the equations, a vector with length = number of rows of A, or a matrix with number of rows = number of rows of A.

Value

• matrix with the solution, X, of the tridiagonal system of equations Ax=B, the number of columns of this matrix = number of columns of B.

Details

If the length of the vector dia is N, then the lengths of diam1 and diap1 should be equal to N-1

See Also

Solve.banded, the function to solve a banded system of linear equations.

Solve.block, the function to solve a block diagonal system of linear equations.

Solve the generalised inverse solution,

solve the R default

Examples

# create tridagonal system: bands on diagonal, above and below
nn   <- 20                          # nr rows and columns of A
aa   <- runif(nn)
bb   <- runif(nn)
cc   <- runif(nn)
# full matrix
A                       <- matrix(nrow=nn,ncol=nn,data=0)
diag(A)                 <- bb
A[cbind(1:(nn-1),2:nn)] <- cc[-nn]
A[cbind(2:nn,1:(nn-1))] <- aa[-1]
B<-runif(nn)

# solve as full matrix
solve(A,B)                           

# same,  now using tridiagonal algorithm
as.vector(Solve.tridiag(aa[-1],bb,cc[-nn],B))

# same, now with 3 different right hand sides
B3 <- cbind(B, B*2, B*3)
Solve.tridiag(aa[-1],bb,cc[-nn],B3)