B-spline collocation matrix

`colmat = spcol(knots,k,tau) `

colmat = spcol(knots,k,tau,arg1,arg2,...)

`colmat = spcol(knots,k,tau) `

returns the matrix, with `length(tau)`

rows and `length(knots)-k`

columns,
whose (*i*,*j*)th entry is

$${D}^{m}{}^{(i)}{B}_{j}(\text{tau}(i))$$

This is the value at tau(*i*) of the *m*(*i*)th
derivative of the *j*th B-spline of order `k`

for
the knot sequence `knots`

.
Here, `tau`

is a sequence of sites, assumed to be *nondecreasing*,
and *m* = knt2mlt(tau), i.e., *m*(*i*)
is #{*j* < *i*:tau(*j*) = tau(*i*)}, all *i*.

`colmat = spcol(knots,k,tau,arg1,arg2,...) `

also returns that matrix, but gives you the opportunity to specify
some aspects.

If one of the `argi`

is a character vector
with the same first two letters as in `'slvblk'`

,
the matrix is returned in the almost block-diagonal format (specialized
for splines) required by `slvblk`

(and
understood by `bkbrk`

).

If one of the `argi`

is a character vector
with the same first two letters as in `'sparse'`

,
then the matrix is returned in the `sparse`

format of MATLAB^{®}.

If one of the `argi`

is a character vector
with the same first two letters as in `'noderiv'`

,
multiplicities are ignored, i.e., *m*(*i*)
is taken to be 1 for all *i*.

To solve approximately the non-standard second-order ODE

$${D}^{2}y(t)=5\cdot (y(t)-\mathrm{sin}(2t))$$

on the interval [0..π], using cubic splines with 10 polynomial
pieces, you can use `spcol`

in the following way:

tau = linspace(0,pi,101); k = 4; knots = augknt(linspace(0,pi,11),k); colmat = spcol(knots,k,brk2knt(tau,3)); coefs = (colmat(3:3:end,:)/5-colmat(1:3:end,:))\(-sin(2*tau).'); sp = spmak(knots,coefs.');

You can check how well this spline satisfies the ODE by computing
and plotting the residual, *D*^{2}*y*(*t*)
– 5· (*y*(*t*) –
sin(2*t*)), on a fine mesh:

t = linspace(0,pi,501); yt = fnval(sp,t); D2yt = fnval(fnder(sp,2),t); plot(t,D2yt - 5*(yt-sin(2*t))) title(['residual error; to be compared to max(abs(D^2y)) = ',... num2str(max(abs(D2yt)))])

The statement `spcol([1:6],3,.1+[2:4])`

provides
the matrix

ans = 0.5900 0.0050 0 0.4050 0.5900 0.0050 0 0.4050 0.5900

in which the typical row records the values at 2.1, or 3.1,
or 4.1, of all B-splines of order 3 for the knot sequence `1:6`

.
There are three such B-splines. The first one has knots 1,2,3,4, and
its values are recorded in the first column. In particular, the last
entry in the first column is zero since it gives the value of that
B-spline at 4.1, a site to the right of its last knot.

If you add the character vector `'sl'`

as an
additional input to `spcol`

, then you can ask `bkbrk`

to extract detailed information
about the block structure of the matrix encoded in the resulting output
from `spcol`

. Thus, the statement `bkbrk(spcol(1:6,3,.1+2:4,'sl'))`

gives:

block 1 has 2 row(s) 0.5900 0.0050 0 0.4050 0.5900 0.0050 next block is shifted over 1 column(s) block 2 has 1 row(s) 0.4050 0.5900 0.0050 next block is shifted over 2 column(s)

The sequence `tau`

is assumed to be nondecreasing.

Was this topic helpful?