This page was generated from doc/euclidean/natural-non-uniform.ipynb. Interactive online version: Binder badge.

Non-Uniform Natural Splines#

The derivation is similar to the uniform case, but this time the parameter intervals can have arbitrary values.

[1]:
import sympy as sp
sp.init_printing(order='grevlex')
[2]:
from utility import NamedExpression, dotproduct
[3]:
t = sp.symbols('t')

Just like in the uniform case, we are considering two adjacent spline segments, but now we must allow arbitrary parameter values:

[4]:
t3, t4, t5 = sp.symbols('t3:6')
[5]:
b_monomial = t**3, t**2, t, 1
[6]:
coefficients3 = sp.symbols('a:dbm3')[::-1]
coefficients4 = sp.symbols('a:dbm4')[::-1]
[7]:
p3 = NamedExpression(
    'pbm3',
    dotproduct(b_monomial, coefficients3).subs(t, (t - t3)/(t4 - t3)))
p4 = NamedExpression(
    'pbm4',
    dotproduct(b_monomial, coefficients4).subs(t, (t - t4)/(t5 - t4)))
display(p3, p4)
$\displaystyle \boldsymbol{p}_{3} = \frac{\boldsymbol{d}_{3} \left(t - t_{3}\right)^{3}}{\left(- t_{3} + t_{4}\right)^{3}} + \frac{\boldsymbol{c}_{3} \left(t - t_{3}\right)^{2}}{\left(- t_{3} + t_{4}\right)^{2}} + \frac{\boldsymbol{b}_{3} \left(t - t_{3}\right)}{- t_{3} + t_{4}} + \boldsymbol{a}_{3}$
$\displaystyle \boldsymbol{p}_{4} = \frac{\boldsymbol{d}_{4} \left(t - t_{4}\right)^{3}}{\left(- t_{4} + t_{5}\right)^{3}} + \frac{\boldsymbol{c}_{4} \left(t - t_{4}\right)^{2}}{\left(- t_{4} + t_{5}\right)^{2}} + \frac{\boldsymbol{b}_{4} \left(t - t_{4}\right)}{- t_{4} + t_{5}} + \boldsymbol{a}_{4}$
[8]:
pd3 = p3.diff(t)
pd4 = p4.diff(t)
display(pd3, pd4)
$\displaystyle \frac{d}{d t} \boldsymbol{p}_{3} = \frac{3 \boldsymbol{d}_{3} \left(t - t_{3}\right)^{2}}{\left(- t_{3} + t_{4}\right)^{3}} + \frac{\boldsymbol{c}_{3} \cdot \left(2 t - 2 t_{3}\right)}{\left(- t_{3} + t_{4}\right)^{2}} + \frac{\boldsymbol{b}_{3}}{- t_{3} + t_{4}}$
$\displaystyle \frac{d}{d t} \boldsymbol{p}_{4} = \frac{3 \boldsymbol{d}_{4} \left(t - t_{4}\right)^{2}}{\left(- t_{4} + t_{5}\right)^{3}} + \frac{\boldsymbol{c}_{4} \cdot \left(2 t - 2 t_{4}\right)}{\left(- t_{4} + t_{5}\right)^{2}} + \frac{\boldsymbol{b}_{4}}{- t_{4} + t_{5}}$
[9]:
equations = [
    p3.evaluated_at(t, t3).with_name('xbm3'),
    p3.evaluated_at(t, t4).with_name('xbm4'),
    p4.evaluated_at(t, t4).with_name('xbm4'),
    p4.evaluated_at(t, t5).with_name('xbm5'),
    pd3.evaluated_at(t, t3).with_name('xbmdot3'),
    pd3.evaluated_at(t, t4).with_name('xbmdot4'),
    pd4.evaluated_at(t, t4).with_name('xbmdot4'),
    pd4.evaluated_at(t, t5).with_name('xbmdot5'),
]

We introduce a few new symbols to simplify the display, but we keep calculating with \(t_i\):

[10]:
deltas = {
    t3: 0,
    t4: sp.Symbol('Delta3'),
    t5: sp.Symbol('Delta3') + sp.Symbol('Delta4'),
}
[11]:
for e in equations:
    display(e.subs(deltas))
$\displaystyle \boldsymbol{x}_{3} = \boldsymbol{a}_{3}$
$\displaystyle \boldsymbol{x}_{4} = \boldsymbol{a}_{3} + \boldsymbol{b}_{3} + \boldsymbol{c}_{3} + \boldsymbol{d}_{3}$
$\displaystyle \boldsymbol{x}_{4} = \boldsymbol{a}_{4}$
$\displaystyle \boldsymbol{x}_{5} = \boldsymbol{a}_{4} + \boldsymbol{b}_{4} + \boldsymbol{c}_{4} + \boldsymbol{d}_{4}$
$\displaystyle \dot{\boldsymbol{x}}_{3} = \frac{\boldsymbol{b}_{3}}{\Delta_{3}}$
$\displaystyle \dot{\boldsymbol{x}}_{4} = \frac{\boldsymbol{b}_{3}}{\Delta_{3}} + \frac{2 \boldsymbol{c}_{3}}{\Delta_{3}} + \frac{3 \boldsymbol{d}_{3}}{\Delta_{3}}$
$\displaystyle \dot{\boldsymbol{x}}_{4} = \frac{\boldsymbol{b}_{4}}{\Delta_{4}}$
$\displaystyle \dot{\boldsymbol{x}}_{5} = \frac{\boldsymbol{b}_{4}}{\Delta_{4}} + \frac{2 \boldsymbol{c}_{4}}{\Delta_{4}} + \frac{3 \boldsymbol{d}_{4}}{\Delta_{4}}$
[12]:
coefficients = sp.solve(equations, coefficients3 + coefficients4)
[13]:
for c, e in coefficients.items():
    display(NamedExpression(c, e.factor().subs(deltas).simplify()))
$\displaystyle \boldsymbol{a}_{3} = \boldsymbol{x}_{3}$
$\displaystyle \boldsymbol{a}_{4} = \boldsymbol{x}_{4}$
$\displaystyle \boldsymbol{b}_{3} = \Delta_{3} \dot{\boldsymbol{x}}_{3}$
$\displaystyle \boldsymbol{b}_{4} = \Delta_{4} \dot{\boldsymbol{x}}_{4}$
$\displaystyle \boldsymbol{c}_{3} = - 2 \Delta_{3} \dot{\boldsymbol{x}}_{3} - \Delta_{3} \dot{\boldsymbol{x}}_{4} - 3 \boldsymbol{x}_{3} + 3 \boldsymbol{x}_{4}$
$\displaystyle \boldsymbol{c}_{4} = - 2 \Delta_{4} \dot{\boldsymbol{x}}_{4} - \Delta_{4} \dot{\boldsymbol{x}}_{5} - 3 \boldsymbol{x}_{4} + 3 \boldsymbol{x}_{5}$
$\displaystyle \boldsymbol{d}_{3} = \Delta_{3} \dot{\boldsymbol{x}}_{3} + \Delta_{3} \dot{\boldsymbol{x}}_{4} + 2 \boldsymbol{x}_{3} - 2 \boldsymbol{x}_{4}$
$\displaystyle \boldsymbol{d}_{4} = \Delta_{4} \dot{\boldsymbol{x}}_{4} + \Delta_{4} \dot{\boldsymbol{x}}_{5} + 2 \boldsymbol{x}_{4} - 2 \boldsymbol{x}_{5}$
[14]:
pdd3 = pd3.diff(t)
pdd4 = pd4.diff(t)
display(pdd3, pdd4)
$\displaystyle \frac{d^{2}}{d t^{2}} \boldsymbol{p}_{3} = \frac{3 \boldsymbol{d}_{3} \cdot \left(2 t - 2 t_{3}\right)}{\left(- t_{3} + t_{4}\right)^{3}} + \frac{2 \boldsymbol{c}_{3}}{\left(- t_{3} + t_{4}\right)^{2}}$
$\displaystyle \frac{d^{2}}{d t^{2}} \boldsymbol{p}_{4} = \frac{3 \boldsymbol{d}_{4} \cdot \left(2 t - 2 t_{4}\right)}{\left(- t_{4} + t_{5}\right)^{3}} + \frac{2 \boldsymbol{c}_{4}}{\left(- t_{4} + t_{5}\right)^{2}}$
[15]:
sp.Eq(pdd3.expr.subs(t, t4), pdd4.expr.subs(t, t4))
[15]:
$\displaystyle \frac{3 \boldsymbol{d}_{3} \left(- 2 t_{3} + 2 t_{4}\right)}{\left(- t_{3} + t_{4}\right)^{3}} + \frac{2 \boldsymbol{c}_{3}}{\left(- t_{3} + t_{4}\right)^{2}} = \frac{2 \boldsymbol{c}_{4}}{\left(- t_{4} + t_{5}\right)^{2}}$
[16]:
_.subs(coefficients).subs(deltas).simplify()
[16]:
$\displaystyle \frac{2 \left(\Delta_{3} \dot{\boldsymbol{x}}_{3} + 2 \Delta_{3} \dot{\boldsymbol{x}}_{4} + 3 \boldsymbol{x}_{3} - 3 \boldsymbol{x}_{4}\right)}{\Delta_{3}^{2}} = \frac{2 \left(- 2 \Delta_{4} \dot{\boldsymbol{x}}_{4} - \Delta_{4} \dot{\boldsymbol{x}}_{5} - 3 \boldsymbol{x}_{4} + 3 \boldsymbol{x}_{5}\right)}{\Delta_{4}^{2}}$

Like in the uniform case, we can generalize by renaming index \(4\) to \(i\):

\begin{equation*} \frac{1}{\Delta_{i-1}} \dot{\boldsymbol{x}}_{i-1} + \left(\frac{2}{\Delta_{i-1}} + \frac{2}{\Delta_i}\right) \dot{\boldsymbol{x}}_i + \frac{1}{\Delta_i} \dot{\boldsymbol{x}}_{i+1} = \frac{3 (\boldsymbol{x}_i - \boldsymbol{x}_{i-1})}{{\Delta_{i-1}}^2} + \frac{3 (\boldsymbol{x}_{i+1} - \boldsymbol{x}_i)}{{\Delta_i}^2} \end{equation*}

We are not showing the full matrix here, because it would be quite a bit more complicated and less instructive than in the uniform case.

End Conditions#

Like in the uniform case, we can come up with a few end conditions in order to define the missing matrix rows.

The Python class splines.Natural uses “natural” end conditions by default.

“Natural” end conditions are derived in a separate notebook, yielding these expressions:

\begin{align*} 2 \Delta_0 \dot{\boldsymbol{x}}_0 + \Delta_0 \dot{\boldsymbol{x}}_1 &= 3 (\boldsymbol{x}_1 - \boldsymbol{x}_0) \\ \Delta_{N-2} \dot{\boldsymbol{x}}_{N-2} + 2 \Delta_{N-2} \dot{\boldsymbol{x}}_{N-1} &= 3 (\boldsymbol{x}_{N-1} - \boldsymbol{x}_{N-2}) \end{align*}

Other end conditions can be derived as shown in the notebook about uniform “natural” splines.