# Bivariate Splines in Piecewise Constant Tension Kunimitsu Takahashi Masaru Kamada

```Bivariate Splines in Piecewise Constant Tension
Kunimitsu Takahashi
Science and Engineering
Ibaraki University
Hitachi, Ibaraki 316-8511, Japan
Email: 14nm714l@vc.ibaraki.ac.jp
Department of Computer
and Information Sciences
Ibaraki University
Hitachi, Ibaraki 316-8511, Japan
Abstract—An extension of the bivariate cubic spline on the
uniform grid is derived in this paper to have different tensions
in different square cells of the grid. The resulting function can
be interpreted also as a bivariate extension of the univariate
spline in piecewise constant tension which was applied to adaptive interpolation of digital images for their magnification and
rotation. The bivariate function will hopefully make it possible to
magnify and rotate images better and even to deform images into
any shapes. A locally supported basis, which is crucial for the
practical use of the bivariate functions, has not been constructed
at the moment and its construction is left for the next step of
study.
I. I NTRODUCTION
Interpolation of digital images for their magnification and
rotation is a classical problem in image processing [1]. Once
an image is represented by a bivariate function, we can resample it at any points in the domain to produce magnified and
rotated images and even those deformed into arbitrary shapes
[2], [3]. The bivariate cubic splines have been successfully
applied [4] to image interpolation. The only drawback of the
bivariate cubic splines is so-called “ringing” artifacts that may
appear around “edges” where brightness of the image changes
sharply.
In order to suppress ringing around the edges adaptively,
splines in piecewise constant tension have been derived [5] by
extending the splines in tension [6], which used to have the
tension fixed as a single constant over the entire domain [7], in
the univariate setting, and applied to image interpolation [8],
[9]. It was possible to magnify and rotate images by repeating
the univariate spline interpolation in piecewise constant tension
in the vertical and horizontal directions. But the quality of
magnified and rotated images may have been degraded because
the image was not represented by a bivariate function. Besides,
it was not possible to deform images into arbitrary shapes for
the same reason.
In this paper, we aim at deriving splines in piecewise
constant tension in the bivariate setting. The bivariate cubic
spline is reviewed in the context of functional minimization
in Section II. In Section III, the bivariate spline in piecewise
constant tension is formulated and derived as a minimizer
of a functional with piecewise constant tension incorporated.
Its representation in terms of state transitions reveals how to
c
978-1-4673-7353-1/15/\$31.00 ⃝2015
IEEE
evaluate the function for given tension and some Lagrange
multipliers. A locally supported basis, which is crucial for the
practical use, has not been obtained at the moment. A plan for
its construction is mentioned in Section IV.
II. B IVARIATE C UBIC S PLINES
The bivariate cubic spline on the uniform grid in the plane
with orthogonal x-y axes is usually represented in the literature
in terms of its basis given by the Cartesian product of the
cardinal cubic B-splines of x and those of y. In this section,
we review its formulation in terms of a minimum principle as
a preparation for its extension to allow for different tensions
in different square cells of the grid.
Let the bivariate cubic spline s(x, y) be a square-integrable
function having square-integrable first and second derivatives
sxy (x, y) and sxxyy (x, y). It should minimize
∫
∫
∞
−∞
∞
−∞
2
(sxxyy (x, y)) dxdy
(1)
subject to the interpolation constraints
s(k, l) = skl , k, l = 0, &plusmn;1, &plusmn;2, &middot; &middot; &middot;
(2)
for some fixed values skl at the grid points. Incorporating these
constraints by the Lagrange multipliers λkl and the identity
∫
∫
∞
∞
s(k, l) =
−∞
−∞
δ(x − k)δ(y − l)s(x, y)dxdy
(3)
valid for the Dirac delta function δ, we have the unconstrained
objective functional
∫
∞
∫
∞
2
Q[s] =
(sxxyy (x, y)) dxdy
∫
∞∫ ∞
∑∑
−
λkl
δ(x − k)δ(y − l)s(x, y)dxdy + skl .(4)
−∞
k
l
−∞
−∞ −∞
Denoted by ∆ the variational operator. Then the first variation
of Q[s] is reduced to
∫ ∞∫ ∞
∆Q[s] =
2 sxxyy (x, y)∆sxxyy (x, y)
−∞ −∞
∑∑
−
λkl δ(x − k)δ(y − l)∆s(x, y)dxdy
k
l
x→∞
[[2 sxxyy (x, y)∆sxy (x, y)]x→−∞ ]y→∞
y→−∞
=
x→∞
−[[2 sxxxyyy (x, y)∆s(x, y)]x→−∞ ]y→∞
y→−∞
∫ ∞∫ ∞
+
2 sxxxxyyyy (x, y)∆s(x, y)
−∞ −∞
∑∑
−
λkl δ(x − k)δ(y − l)∆s(x, y)dxdy
∫
=
∞
∫k
∞
l
(
2 sxxxxyyyy (x, y)
−∞
)
∑∑
−
λkl δ(x − k)δ(y − l) ∆s(x, y)dxdy (5)
l
by means of integration by parts and because ∆s(x, y) → 0
and ∆sxy (x, y) → 0 at x → &plusmn;∞ and y → &plusmn;∞ for the
square-integrable s(x, y) and sxy (x, y). Equating (5) to zero,
we have the Euler-Lagrange equation
∑∑
0 = 2 sxxxxyyyy (x, y) −
λkl δ(x − k)δ(y − l).
(6)
k
l
Its four-fold integration by each of x and y shows that s(x, y)
should be in the form
s(x, y)
=
∑∑ λkl
k
l
2
We apply the piecewise constant tension
p(x, y) = Pkl , (x, y) ∈ [k, k + 1) &times; [l, l + 1),
This functional falls back to the original (1) in the case
p(x, y) ≡ 0. On the other hand, as p(x, y) gets large, the
first term dominates in (10) and s(x, y) approaches the simple
polyhedral interpolation that is a minimizer of the integral of
squared first derivative under the interpolation constraints (2).
We seek such s(x, y) that minimizes (10) subject to (2)
among the square-integrable functions with square-integrable
first and second derivatives.
B. Euler-Lagrange equation
In the same way as Section II, we can incorporate the
constraints (2) into (10) to have the unconstrained objective
functional
∫ ∞∫ ∞
2
2
(p(x, y))2 (sxy (x, y)) +(sxxyy (x, y)) dxdy
−∞ −∞
∑∑ ∫ ∞ ∫ ∞
−
λkl
δ(x − k)δ(y − l)s(x, y)dxdy + skl . (11)
k
(◦ − •)+ =
◦−•
0
if ◦ &gt; •
otherwise
−∞ −∞
l
−∞
+g3 (x)y 3 + g2 (x)y 2 + g1 (x)y + g0 (x), (7)
{
−∞
Its first variation is reduced to
∫ ∞∫ ∞
2(p(x, y))2sxy (x, y)∆sxy (x, y)
(x − k)3+ (y − l)3+
+f3 (y)x3 + f2 (y)x2 + f1 (y)x + f0 (y)
where
(9)
which can be a different non-negative constant in each square
cell of the grid, to s(x, y) by extending the functional (1) to
∫ ∞∫ ∞
2
2
(p(x, y))2 (sxy (x, y)) + (sxxyy (x, y)) dxdy. (10)
−∞
−∞
k
A. Functional to be minimized
−∞
+2sxxyy (x, y)∆sxxyy (x, y)
∑∑
λkl δ(x − k)δ(y − l)∆s(x, y)dxdy
−
k
(8)
and fi (x) and gi (y), (i = 0, 1, 2, 3) are arbitrary four-times
differentiable functions of x and y, respectively. Choosing
cubic polynomials for fi (x) and gi (y), we have s(x, y) being
a cubic spline in x and also in y as one of the minimizers.
The bivariate cubic spline represented in (7) in terms of the
truncated power functions can also be represented in terms
of the basis constructed as the Cartesian product of the cubic
cardinal B-splines of x and those of y.
III. B IVARIATE S PLINES IN P IECEWISE T ENSION
We shall extend the functional (1) for the bivariate cubic
splines to include a term that incorporates different tensions
in different square cells of the grid. Then we will derive its
minimizer which is to be called bivariate splines in piecewise
constant tension.
l
2
y→∞
= [[2 (p(x, y)) sxy (x, y)∆s(x, y)]x→∞
x→−∞ ]y→−∞
∫ ∞∫ ∞
(
)
∂2
−
(p(x, y))2 sxy (x, y) ∆s(x, y)dxdy
2 ∂x∂y
−∞ −∞
y→∞
+[[2 sxxyy (x, y)∆sxy (x, y)]x→∞
x→−∞ ]y→−∞
y→∞
−[[2 sxxxyyy (x, y)∆s(x, y)]x→∞
x→−∞ ]y→−∞
∫ ∞∫ ∞
+
2 sxxxxyyyy (x, y)∆s(x, y)
−∞ −∞
∑∑
−
λkl δ(x − k)δ(y − l)∆s(x, y)dxdy
∫
=
k
∞ ∫ ∞(
−∞ −∞
l
(
)
∂2
−2 ∂x∂y
(p(x, y))2 sxy (x, y) ∆s(x, y)
+2 sxxxxyyyy (x, y)∆s(x, y)
)
∑∑
−
λkl δ(x − k)δ(y − l) ∆s(x, y)dxdy(12)
k
l
by means of integration by parts and because ∆s(x, y) → 0
and ∆sxy (x, y) → 0 at x → &plusmn;∞ and y → &plusmn;∞ for the
square-integrable s(x, y) and sxy (x, y). That yields the EulerLagrange equation
(
)
∂2
2sxxxxyyyy (x, y)−2 ∂x∂y
(p(x, y))2sxy (x, y)
∑∑
=
λkl δ(x − k)δ(y − l).
(13)
k
l
This differential equation includes p(x, y) being constant
inside each cell of the grid and discontinuous at the cell
borders. We shall handle this equation in the two kinds of
domains: the open domain within each cell and the infinitesimal domains across the borders of adjacent cells, in the
following subsections. We take a state-space representation
of the solution inside the cells because of its compatibility
with boundary conditions written in terms of derivatives at
the borders.
C. State transition inside each cell
In each open square (k + x, l + y) for 0 &lt; x, y &lt; 1, the
differential equation (13) falls back to the homogeneous one
2
sxxxxyyyy (k+x, l+y)−Pkl
sxxyy (k+x, l+y) = 0
(14)
since p(k + x, l + y) = Pkl according to (9) and because the
delta functions in (13) are identically zero anywhere other than
the grid lines. Denote
w(k+x, l+y) = sxxyy (k+x, l+y).
(15)
Then (14) becomes
2
wxxyy (k+x, l+y)−Pkl
w(k+x, l+y) = 0
(16)
of which the general solution is represented in the form
w(k + x, l + y)
(√
)
(√
)
= akl cosh Pkl (x + y) +bkl cosh Pkl (x − y)
)
(√
)
(√
+ ckl sinh Pkl (x + y) +dkl sinh Pkl (x − y) . (17)
Integrating (17) twice by each of x and y with the simplest
choice of integral constants, we have a solution for (14) as
s(k + x, l + y)
(√
)
(√
)
1 {
= 2 akl cosh Pkl (x + y) + bkl cosh Pkl (x − y)
Pkl
(√
)
(√
)}
+ ckl sinh Pkl (x + y) + dkl sinh Pkl (x − y)
+αkl xy + βkl x + γkl y + θkl .
(18)
The eight coefficients akl , bkl , ckl , dkl , αkl , βkl , γkl , and
θkl determine s(k + x, l + y) within the open cell and also on
its boundary since s(x, y) is assumed to be twice differentiable
over the entire domain. Across the cell borders, the function
should be connected to the next ones via the boundary values
of s(k + x, l + y) and seven of its derivatives constituting a
state vector.
Let such a state vector and the coefficient vector be composed by




sxxxyyy (x+k, y+l)
akl

 bkl 
sxyy (x+k, y+l) 






 ckl 
sxxy (x+k, y+l) 



 sxxyy (x+k, y+l) 


 and ckl = dkl  , (19)
s(k+x, l+y)=




sxy (x+k, y+l) 

 αkl 




sx (x+k, y+l) 

 βkl 



sy (x+k, y+l)
γkl 
s(x+k, y+l)
θkl
respectively. Then they are linearly related to each other by
s(k+x, l+y) = Ψkl (x, y)ckl ,
(20)
where the matrix Ψkl (x, y) and its inverse Ψ−1
kl (x, y) are
calculated with labor as follows:

√
Pkl cosh(√Pkl (x + y))

√ 1 sinh( Pkl (x + y))

Pkl
√

√ 1 sinh( Pkl (x + y))

Pkl
√


cosh(√Pkl (x + y))

1
Ψkl (x, y) = 
Pkl cosh(√Pkl (x + y))

1
 √ 3 sinh( Pkl (x + y))
 ( Pkl )
√
 √1
 ( Pkl )3 sinh(√Pkl (x + y))
1
Pkl (x + y))
2 cosh(
Pkl
√
√
−Pkl cosh(√Pkl (x − y))
Pkl sinh(√Pkl (x + y))
√ 1 sinh( Pkl (x − y))
√ 1 cosh( Pkl (x + y))
Pkl
Pkl
√
√
1
√ 1 cosh( Pkl (x + y))
− √P sinh( Pkl (x − y))
P
kl
kl
√
√
sinh(√Pkl (x + y))
cosh(√Pkl (x − y))
1
− P1kl cosh( Pkl (x − y))
Pkl sinh(√Pkl (x + y))
√
1
1
√
sinh( Pkl (x − y)) (√P )3 cosh( Pkl (x + y))
( Pkl )3
kl
√
√
1
√
− ( P )3 sinh( Pkl (x − y)) (√P1 )3 cosh( Pkl (x + y))
kl
kl
√
√
1
1
Pkl (x − y))
Pkl (x + y))
2 cosh(
2 sinh(
Pkl
Pkl

√
−Pkl sinh(√Pkl (x − y)) 0 0 0 0
√ 1 cosh( Pkl (x − y))
0 0 0 0 

Pkl
√
1
− √P cosh( Pkl (x − y)) 0 0 0 0 

kl
√

sinh(√Pkl (x − y)) 0 0 0 0 

(21)
− P1kl sinh( Pkl (x − y)) 1 0 0 0 
√

1
√
cosh( Pkl (x − y)) y 1 0 0 
( Pkl )3

√
− (√P1 )3 cosh( Pkl (x − y)) x 0 1 0 

kl
√
1
sinh(
P
(x
−
y))
xy
x
y
1
2
kl
P
kl







−1
Ψkl (x, y) = 






√
1
Pkl (x + y))
2Pkl cosh( √
1
− 2Pkl cosh( Pkl (x − y))
√
− 2P1kl sinh( Pkl (x + y))
√
1
2Pkl sinh( Pkl (x − y))
− P12
kl
y
2
Pkl
x
2
Pkl
− Pxy2
kl
√
√
− √P2 kl sinh( Pkl (x + y))
√
−√ P2 kl sinh( Pkl (x − y))
√
Pkl
cosh( Pkl (x + y))
√2
√
Pkl
2 cosh( Pkl (x − y))
0
− P1kl
0
√
Pkl
Pkl (x + y))
2 sinh(
√
Pkl
sinh(
P
(x − y))
√2
√ kl
Pkl
cosh( Pkl (x + y))
√2
√
− P2 kl cosh( Pkl (x − y))
x
Pkl
y
Pkl
√
√
−√
0
0
0
0
0
− P12
kl
0
0
0
0
1
−y
−x
xy
0
0
0
0
0
1
0
−x
0
0
0
0
0
0
1
−y
0
0
0
0
0
0
0
1












s(k+x, l+y)=Ψkl (x, y)Ψ−1
kl (x0 , y0 )s(k+x0 , l+y0 ),
1 ∑ ∑
λij+f k+g(l+y)+h(26)
2
sxxxyyy (k + 0, l + y)
1∑ ∑
2
= Pkl
sxy (k, l + y) +
λij+f k+g(l+y)+h. (27)
2
i≤k j≤l
(22)
s(k+x, l+y)=Ψkl (x, y)Ψ−1
kl (0, 0)s(k+0, l+0)
Subtracting (26) from (27), we have
sxxxyyy (k + 0, l + y) − sxxxyyy (k − 0, l + y)
(
)
1∑
2
2
λkj
=
Pkl
− P(k−1)l
sxy (k, l + y) +
2
(28)
j≤l
(23)
where 0 &lt; x, y, x0 , y0 &lt; 1. Given the state vector s(k + 0, l +
0) = limx0 →+0,y0 →+0 s(k + x0 , l + y0 ) at the left bottom
corner (k + 0, l + 0) of the cell, for example, we can evaluate
the state at anywhere inside the same cell by
(24)
for 0 &lt; x, y &lt; 1. Here we used the fact that every entry of the
matrix Ψ−1
kl (x, y) is a continuous function of x and y so that
Ψ−1
(+0,
+0) = Ψ−1
kl
kl (0, 0). The same is true of Ψkl (x, y) so
that we can evaluate the state even at x = 1 − 0 and y = 1 − 0
by using Ψkl (1, y) and Ψkl (x, 1) instead of Ψkl (1 − 0, y) and
Ψkl (x, 1 − 0), respectively.
D. State transition across the cell borders
Integrating the differential equation (13) once by each of x
and y, we have
2
= P⌊x⌋⌊y⌋
sxy (x, y) +
2
= P(k−1)l
sxy (k, l + y) +
and
Using the above state transition matrices, we can evaluate
the state vector at any point (k + x, l + y) inside the cell from
the state vector at any other point (k + x0 , l + y0 ) inside the
same cell by
sxxxyyy (x, y)
sxxxyyy (k − 0, l + y)
i≤k−1 j≤l
− P1kl
1
2 cosh(√Pkl (x + y))
1
2 cosh(√Pkl (x − y))
− 12 sinh(√Pkl (x + y))
− 12 sinh( Pkl (x − y))
The two ways of crossing the borders are investigated in
the following. For 0 &lt; y &lt; 1, (25) gives
1 ∑ ∑
λkl + f x + gy + h,(25)
2
k≤⌊x⌋ l≤⌊y⌋
where f , g and h are integral constants and ⌊•⌋ denotes
the largest integer not exceeding •. We recall sxy (x, y) is
continuous over the entire domain because s(x, y) is assumed
to be twice differentiable with respect to x and y. Then
(25) means that sxxxyyy (x, y) is a continuous function over
the entire domain except at the cell borders where it is
2
discontinuous with a finite height of jump caused by P⌊x⌋⌊y⌋
∑
∑
and k≤⌊x⌋ l≤⌊y⌋ λkl . That also means that all the other
state variables in s(x, y) being integrals of sxxxyyy (x, y) are
continuous over the entire domain. So we have only to take
care of the change of sxxxyyy (x, y) in the state transitions
across the cell borders.
which tells how to update the state variable sxxxyyy (k−0, l+y)
to sxxxyyy (k + 0, l + y) when crossing the border x = k from
left to right. In the same way, for 0 &lt; x &lt; 1, subtracting
sxxxyyy (k + x, l − 0)
2
= Pk(l−1)
sxy (k + x, l) +
1∑ ∑
λij+f (k+x)+gl+h(29)
2
i≤k j≤l−1
from
sxxxyyy (k + x, l + 0)
1∑ ∑
2
λij+f (k+x)+gl+h, (30)
= Pkl
sxy (k + x, l) +
2
i≤k j≤l
we have
sxxxyyy (k + x, l + 0) − sxxxyyy (k + x, l − 0)
(
)
1∑
2
2
=
Pkl
− Pk(l−1)
sxy (k + x, l) +
λil
2
(31)
i≤k
which tells how to update the state variable sxxxyyy (k + x, l −
0) to sxxxyyy (k + x, l + 0) when crossing the border y = l
upward.
By the notation


1
1 0 0 0 P 0 0 0
2
0 1 0 0 0 0 0 0
0


 
0 0 1 0 0 0 0 0
0


 
0 0 0 1 0 0 0 0
0


 
Φ(P ) = 
 and u =  0  , (32)
0
0
0
0
1
0
0
0


 
0 0 0 0 0 1 0 0
0


 
0 0 0 0 0 0 1 0
0
0 0 0 0 0 0 0 1
0
we can write the state transitions instructed by (28) and (31)
as follows:
∑
2
2
s(k+0, l+y) = Φ(Pkl
−P(k−1)l
)s(k−0, l+y)+u
λkj ,(33)
j≤l
∑
2
2
s(k+x, l+0) = Φ(Pkl
−Pk(l−1)
)s(k+x, l−0)+u
i≤k
λil . (34)
E. Evaluation of bivariate splines in piecewise tension
Making use of the state transitions inside the cell (24) and
those across the cell borders (33) and (34), we can evaluate
the state s(k+x, l+y) in the given piecewise constant tension
Pij (i, j = 0, &plusmn;1, &plusmn;2, &middot; &middot; &middot; ) for a given series of parameters
λij (i, j = 0, &plusmn;1, &plusmn;2, &middot; &middot; &middot; ). The function value s(k +x, l +y)
is included in the state s(k+x, l+y) at its bottom row.
Suppose that we know the state s(k +0, l+0) at the left
bottom corner of the cell (k, k + 1) &times; (l, l + 1). Then the state
at (k + x, l + y) for 0 &lt; x, y &lt; 1 inside the cell can be
evaluated by (24), i.e.,
s(k+x, l+y)=Ψkl (x, y)Ψ−1
kl (0, 0)s(k+0, l+0).
As a special case, the state at the right bottom corner is
given by
s(k+1−0, l+0)=Ψkl (1, 0)Ψ−1
kl (0, 0)s(k+0, l+0).
In order to cross the border x = k + 1 to the right, we use the
state transition (33) to evaluate
∑
2
2
s(k+1+0, l+0)= Φ(P(k+1)l
−Pkl
)s(k+1−0, l+0)+u
λ(k+1)j
j≤l
to arrive at the left bottom corner of the next cell (k + 1, k +
2) &times; (l, l + 1) on the right.
In the same way, the state at the left top corner is given by
s(k+0, l+1−0)=Ψkl (0, 1)Ψ−1
kl (0, 0)s(k+0, l+0).
In order to cross the border y = l + 1 upward, we use the state
transition (34) to evaluate
∑
2
2
λi(l+1)
s(k+0, l+1+0)= Φ(Pk(l+1)
−Pkl
)s(k+0, l+1−0)+u
i≤k
to arrive at the left bottom corner of the next upper cell (k, k+
1) &times; (l + 1, l + 2).
The state at the right top corner given by
−1
s(k+1−0, l+1−0)=Ψkl (1, 1)Ψkl
(0, 0)s(k+0, l+0)
is moved to
s(k+1+0, l+1−0)
2
2
= Φ(P(k+1)l
−Pkl
)s(k+1−0, l+1−0)+u
∑
λ(k+1)j
j≤l+1
by (33) to cross the border x = k + 1, and then to
s(k+1+0, l+1+0)
2
2
= Φ(P(k+1)(l+1)
−P(k+1)l
)s(k+1+0, l+1−0)+u
∑
λi(l+1)
i≤k+1
by (34) to cross the border y = l + 1 to reach the state at the
left bottom corner of the next upper right cell (k + 1, k + 2) &times;
(l + 1, l + 2).
Applying the above transitions successively for larger k and
l, we can evaluate s(k +x, l+y) for any k, l, x and y if we are
given the series of parameters λkl . However, λkl will never be
given in practice but must be determined so that s(k, l) = skl
for the given data skl .
IV. T OWARD LOCALLY SUPPORTED BASIS
It is crucial for practical use of the splines to have a locally
supported basis like the Cartesian product of the cubic cardinal
B-splines of x and those of y for the bivariate cubic splines.
It was possible in the univariate case to construct a locally
supported basis for the splines in piecewise tension by making
use of the state transitions [9].
Construction of a locally supported basis in the bivariate
case has not yet been made but only planned as follows:
Consider the problem of constructing a basis function with
the support [0, 4] &times; [0, 4] by choosing a set of 25 possibly
non-zero λkl for (k, l) ∈ [0, 4] &times; [0, 4]. We start from the
zero state s(−0, −0) = 0 just outside the support. Then the
2
state transitions (33) and (34) give s(+0, +0) = Φ(P00
−
2
2
2
P0 −1 )Φ(P0 −1 − P−1 −1 )s(−0, −0) + uλ00 = uλ00 . Successive application of the state transitions described in the
last subsection for four times will give the terminal states
s(4 + 0, +0), s(+0, 4 + 0) and s(4 + 0, 4 + 0) represented
as linear combinations of some eight-dimensional vectors
with the 25 coefficients λkl . Fixing λ00 = 1 and requesting
s(4 + 0, +0) = 0, s(+0, 4 + 0) = 0 and s(4 + 0, 4 + 0) = 0,
we may determine the other 24 coefficients λkl so that the
states be identically zero outside the support [0, 4] &times; [0, 4].
V. C ONCLUSION
We derived the expression of bivariate splines in piecewise
constant tension in terms of state transitions of the characterizing Euler-Lagrange equation. The derived function is an
extension of the bivariate cubic splines to piecewise constant
tensions and is also an extension of the univariate spline in
piecewise constant tension to the bivariate case. A plan for
constructing a locally supported basis was briefly described.
ACKNOWLEDGMENT
This work was partially supported by the JSPS grant-in-aid
#26420409.
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