# Matrix Formualtion of Interpolating Using Variational Methods, Lecture Notes - Advanced Calculus, Study notes for Calculus. University of California (CA) - UCLA

## Calculus

Description: Matrix Formulation, Mesh Points, Debugging, Example, Infinite Spline
Showing pages  1  -  4  of  5
Matrix Formulation of Variational Method
Interpolation
Adrian Down
March 03, 2006
1 Matrix formulation
1.1 Unequally-spaced mesh points
Recall the matrix formulation of the diﬀerential equation considered last
lecture,
2
h0
1
h0
1
h0
2
h0+2
h1
1
h1
.........
1
hn2
2
hn2+2
hn1
1
hn1
1
hn1
2
hn1
Y0(x0)
Y0(x1)
.
.
.
Y0(xn1)
Y0(xn)
= 3
1
h2
0
1
h2
0
1
h2
0
1
h2
01
h2
1
1
h2
1
.........
1
h2
n2
1
h2
n21
h2
n1
1
h2
n1
1
h2
n1
1
h2
n1
Y(x0)
Y(x1)
.
.
.
Y(xn1)
Y(xn)
where hj=xj+1 xj, for j { 0, . . . , n 1}. This is general formulation
for non-uniform node spacing.
1.2 Equally-spaced mesh points
Suppose that the mesh points are uniformly spaced, so that hi=hj=
h, i, j. The matrix simpliﬁes by multiplying by h. Note that all of the
1
diagonal elements in the interior of the second coeﬃcient matrix become 0.
Multiplying by the resulting matrix of coeﬃcients of Y(x),
2 1
1 4 1
.........
1 4 1
1 2
Y0(x0)
Y0(x1)
.
.
.
Y0(xn1)
Y0(xn)
=3
h
Y(x0) + Y(x1)
Y(x0) + Y(x2)
.
.
.
Y(xn2) + Y(xn)
Y(xn1) + Y(xn)
(1)
2 Debugging examples
2.1 Three points, symmetric about the center
We take the three nodes to be
(0,0) (1,1) (2,0)
In this case, we expect the interpolating polynomials to take an arch shape.
2.1.1 Construct the coeﬃcient matrices
First, identify the values of the function,
Y(x0) = 0 Y(x1) = 1 Y(x2) = 0
Substituting these values into (1),
2 1
141
1 2
Y0(x0)
Y0(x1)
Y0(x2)
=3
1
1
0
1
Solving the system yields,
Y0(x0)
Y0(x1)
Y0(x2)
=
3
2
0
3
2
This result is consistent with what we would expect by symmetry.
2
2.1.2 Restrictions on the second derivative
The second derivative is clearly continuous at the point (1,1) by symmetry.
We must still enforce that Y00(x0) = Y00 (xn) = 0. We could write the explicit
form of the interpolating polynomial and compute the derivative directly.
However, we can analyze the second derivative in a more clever method.
Recall that our interpolation method relies on two diﬀerent cubic poly-
nomials: one on (0,1) and the other on (1,2). The general form of a cubic
polynomial with an inﬂection point at the origin is
y=ax bx3
To be consistent with the other initial conditions, we must enforce that,
y(1) = ab= 1
y0(1) = a3b= 0
Solving for the coeﬃcients,
y(x) = 3
2x1
2x3
Hence the ﬁrst derivative of the proposed solution Y(x) is consistent with
the general form of a polynomial having second derivative equal to 0 at the
origin.
Note. A similar argument could be conducted on the interpolating polyno-
mial on (1,2).
2.2 Inﬁnite spline
Deﬁne some point x0as the origin of the coordinates. In this case, the
coeﬃcient matrix is still tri-diagonal. However, the matrix is now inﬁnite.
The calculation is essentially the same as that in the ﬁnite case neglecting
the endpoints of the interval,
.........
1 4 1
141
1 4 1
.........
.
.
.
Y0(x1)
Y0(x0)
Y0(x1)
.
.
.
=3
h
.
.
.
Y(x2) + Y(x0)
Y(x1) + Y(x1)
Y(x0) + Y(x2)
.
.
.
3
2.3 Y(x0) = 1, all other f(xk) = 0
2.3.1 Setup
We expect a curve that looks like a symmetric sinusoid that decays away
from the origin. Thus if we can ﬁnd a solution for the points xk,k0, we
can construct the solution for k < 0 by symmetry.
Using this symmetry,
Y0(x0) = 0
Y0(xk) = Y0(xk)k6= 0
For k= 1,
4Y0(x1) + Y0(x2) = 3 (2)
For k > 1,
Y0(xk1)+4Y0(xk) + Y0(xk+1) = 0 (3)
We also enforce the boundary condition that Y0(xk)0 as k→ ∞.
2.3.2 Solving diﬀerence equations
Elementary solution Solutions to iterative diﬀerence equations such as
(3) can be solved by assuming a solution of a special type. The elementary
solution to a diﬀerence equation of this type is
Y0(xk) = rk
Substituting this assumed form of the solution into (3),
rk1+ 4rk+rk+1 = 0
Dividing by rk1,
1+4r+r2= 0
Using the quadratic formula,
r=2±3
Hence we can construct two linearly independent solutions,
Y0
±(xk) = 2±3k
4
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