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Non-Uniform Natural Splines

uniform

import sympy as sp
sp.init_printing(order='rev-lex')

utility.py

from utility import NamedExpression

t = sp.symbols('t')

t3, t4, t5 = sp.symbols('t3:6')

b_monomial = sp.Matrix([t**3, t**2, t, 1]).T
b_monomial

$\displaystyle \left[\begin{matrix}t^{3} & t^{2} & t & 1\end{matrix}\right]$

coefficients3 = sp.symbols('a:dbm3')[::-1]
coefficients4 = sp.symbols('a:dbm4')[::-1]

b_monomial.dot(coefficients3)

$\displaystyle \boldsymbol{d}_{3} t^{3} + \boldsymbol{c}_{3} t^{2} + \boldsymbol{b}_{3} t + \boldsymbol{a}_{3}$

p3 = NamedExpression(
    'pbm3',
    b_monomial.dot(coefficients3).subs(t, (t - t3)/(t4 - t3)))
p4 = NamedExpression(
    'pbm4',
    b_monomial.dot(coefficients4).subs(t, (t - t4)/(t5 - t4)))
display(p3, p4)

$\displaystyle \boldsymbol{p}_{3} = \frac{\boldsymbol{d}_{3} \left(- t_{3} + t\right)^{3}}{\left(t_{4} - t_{3}\right)^{3}} + \frac{\boldsymbol{c}_{3} \left(- t_{3} + t\right)^{2}}{\left(t_{4} - t_{3}\right)^{2}} + \frac{\boldsymbol{b}_{3} \left(- t_{3} + t\right)}{t_{4} - t_{3}} + \boldsymbol{a}_{3}$

$\displaystyle \boldsymbol{p}_{4} = \frac{\boldsymbol{d}_{4} \left(- t_{4} + t\right)^{3}}{\left(t_{5} - t_{4}\right)^{3}} + \frac{\boldsymbol{c}_{4} \left(- t_{4} + t\right)^{2}}{\left(t_{5} - t_{4}\right)^{2}} + \frac{\boldsymbol{b}_{4} \left(- t_{4} + t\right)}{t_{5} - t_{4}} + \boldsymbol{a}_{4}$

pd3 = p3.diff(t)
pd4 = p4.diff(t)
display(pd3, pd4)

$\displaystyle \frac{d}{d t} \boldsymbol{p}_{3} = \frac{3 \boldsymbol{d}_{3} \left(- t_{3} + t\right)^{2}}{\left(t_{4} - t_{3}\right)^{3}} + \frac{\boldsymbol{c}_{3} \left(- 2 t_{3} + 2 t\right)}{\left(t_{4} - t_{3}\right)^{2}} + \frac{\boldsymbol{b}_{3}}{t_{4} - t_{3}}$

$\displaystyle \frac{d}{d t} \boldsymbol{p}_{4} = \frac{3 \boldsymbol{d}_{4} \left(- t_{4} + t\right)^{2}}{\left(t_{5} - t_{4}\right)^{3}} + \frac{\boldsymbol{c}_{4} \left(- 2 t_{4} + 2 t\right)}{\left(t_{5} - t_{4}\right)^{2}} + \frac{\boldsymbol{b}_{4}}{t_{5} - t_{4}}$

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'),
]

to simplify the display, but we keep calculating with \(t_i\)

deltas = {
    t3: 0,
    t4: sp.Symbol('Delta3'),
    t5: sp.Symbol('Delta3') + sp.Symbol('Delta4'),
}

for e in equations:
    display(e.subs(deltas))

$\displaystyle \boldsymbol{x}_{3} = \boldsymbol{a}_{3}$

$\displaystyle \boldsymbol{x}_{4} = \boldsymbol{d}_{3} + \boldsymbol{c}_{3} + \boldsymbol{b}_{3} + \boldsymbol{a}_{3}$

$\displaystyle \boldsymbol{x}_{4} = \boldsymbol{a}_{4}$

$\displaystyle \boldsymbol{x}_{5} = \boldsymbol{d}_{4} + \boldsymbol{c}_{4} + \boldsymbol{b}_{4} + \boldsymbol{a}_{4}$

$\displaystyle \dot{\boldsymbol{x}}_{3} = \frac{\boldsymbol{b}_{3}}{\Delta_{3}}$

$\displaystyle \dot{\boldsymbol{x}}_{4} = \frac{3 \boldsymbol{d}_{3}}{\Delta_{3}} + \frac{2 \boldsymbol{c}_{3}}{\Delta_{3}} + \frac{\boldsymbol{b}_{3}}{\Delta_{3}}$

$\displaystyle \dot{\boldsymbol{x}}_{4} = \frac{\boldsymbol{b}_{4}}{\Delta_{4}}$

$\displaystyle \dot{\boldsymbol{x}}_{5} = \frac{3 \boldsymbol{d}_{4}}{\Delta_{4}} + \frac{2 \boldsymbol{c}_{4}}{\Delta_{4}} + \frac{\boldsymbol{b}_{4}}{\Delta_{4}}$

coefficients = sp.solve(equations, coefficients3 + coefficients4)

for c, e in coefficients.items():
    display(NamedExpression(c, e.subs(deltas)))

$\displaystyle \boldsymbol{a}_{3} = \boldsymbol{x}_{3}$

$\displaystyle \boldsymbol{b}_{3} = \Delta_{3} \dot{\boldsymbol{x}}_{3}$

$\displaystyle \boldsymbol{c}_{3} = 3 \boldsymbol{x}_{4} - 3 \boldsymbol{x}_{3} - \Delta_{3} \dot{\boldsymbol{x}}_{4} - 2 \Delta_{3} \dot{\boldsymbol{x}}_{3}$

$\displaystyle \boldsymbol{d}_{3} = - 2 \boldsymbol{x}_{4} + 2 \boldsymbol{x}_{3} + \Delta_{3} \dot{\boldsymbol{x}}_{4} + \Delta_{3} \dot{\boldsymbol{x}}_{3}$

$\displaystyle \boldsymbol{a}_{4} = \boldsymbol{x}_{4}$

$\displaystyle \boldsymbol{b}_{4} = \dot{\boldsymbol{x}}_{4} \left(\Delta_{4} + \Delta_{3}\right) - \Delta_{3} \dot{\boldsymbol{x}}_{4}$

$\displaystyle \boldsymbol{c}_{4} = - \dot{\boldsymbol{x}}_{5} \left(\Delta_{4} + \Delta_{3}\right) - 2 \dot{\boldsymbol{x}}_{4} \left(\Delta_{4} + \Delta_{3}\right) + 3 \boldsymbol{x}_{5} - 3 \boldsymbol{x}_{4} + \Delta_{3} \dot{\boldsymbol{x}}_{5} + 2 \Delta_{3} \dot{\boldsymbol{x}}_{4}$

$\displaystyle \boldsymbol{d}_{4} = \dot{\boldsymbol{x}}_{5} \left(\Delta_{4} + \Delta_{3}\right) + \dot{\boldsymbol{x}}_{4} \left(\Delta_{4} + \Delta_{3}\right) - 2 \boldsymbol{x}_{5} + 2 \boldsymbol{x}_{4} - \Delta_{3} \dot{\boldsymbol{x}}_{5} - \Delta_{3} \dot{\boldsymbol{x}}_{4}$

pdd3 = pd3.diff(t)
pdd4 = pd4.diff(t)
display(pdd3, pdd4)

$\displaystyle \frac{d^{2}}{d t^{2}} \boldsymbol{p}_{3} = \frac{3 \boldsymbol{d}_{3} \left(- 2 t_{3} + 2 t\right)}{\left(t_{4} - t_{3}\right)^{3}} + \frac{2 \boldsymbol{c}_{3}}{\left(t_{4} - t_{3}\right)^{2}}$

$\displaystyle \frac{d^{2}}{d t^{2}} \boldsymbol{p}_{4} = \frac{3 \boldsymbol{d}_{4} \left(- 2 t_{4} + 2 t\right)}{\left(t_{5} - t_{4}\right)^{3}} + \frac{2 \boldsymbol{c}_{4}}{\left(t_{5} - t_{4}\right)^{2}}$

sp.Eq(pdd3.expr.subs(t, t4), pdd4.expr.subs(t, t4))

$\displaystyle \frac{3 \boldsymbol{d}_{3} \left(2 t_{4} - 2 t_{3}\right)}{\left(t_{4} - t_{3}\right)^{3}} + \frac{2 \boldsymbol{c}_{3}}{\left(t_{4} - t_{3}\right)^{2}} = \frac{2 \boldsymbol{c}_{4}}{\left(t_{5} - t_{4}\right)^{2}}$

_.subs(coefficients).subs(deltas).simplify()

$\displaystyle \frac{2 \left(- 3 \boldsymbol{x}_{4} + 3 \boldsymbol{x}_{3} + 2 \Delta_{3} \dot{\boldsymbol{x}}_{4} + \Delta_{3} \dot{\boldsymbol{x}}_{3}\right)}{\Delta_{3}^{2}} = \frac{2 \left(3 \boldsymbol{x}_{5} - 3 \boldsymbol{x}_{4} - \Delta_{4} \dot{\boldsymbol{x}}_{5} - 2 \Delta_{4} \dot{\boldsymbol{x}}_{4}\right)}{\Delta_{4}^{2}}$

generalize by substituting 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*}

End Conditions

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

Natural

notebook about “natural” end conditions

\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

See notebook about uniform “natural” splines