limSolve (version 1.5.5.3)

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 equals the number of columns of B.

Details

If the length of the vector dia is equal to 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

# NOT RUN {
# 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)

# }