Bedingungen fuer Kubische Spline-Interpolierende

Angenommen $S(x)$ ist eine kubische Spline-Interpolierende für $n$ Punkte, dann muss gelten:

$S_j(x_{j+1}) = S_{j+1}(x_{j+1})$
$S'_j(x_{j+1}) = S'_{j+1}(x_{j+1})$
$S''_j(x_{j+1}) = S''_{j+1}(x_{j+1})$
$S''_0(x_0) = S''_n(x_n)$ = 0

Aufgabenstellung: Testen ob eine Funktion eine kubische Spline-Interpolierende ist.

Gegeben: eine Funktion $S(x)$
Gesucht: ist $S$ eine kubische spline interpolierende?
Vorgehensweise: die Bedinungen testen



In [11]:

    
var('x')
# input: funktionen der einzelnens splines
S = [
    x/5 + 3/5 * x^3,
    14/5 - 41/5 * x + 42/5 * x^2 - 2*x^3,
    -122/5 + 151/5*x - 54/5* x^2 + 7/5*x^3
]
# input: definitionsbereich der spline funktionen
r = [
    (0, 1),
    (1, 2),
    (2, 3)
]

for i, s in enumerate(S):
    print "spline", i
    show(s)
    show(s.derivative(x))
    show(s.derivative(x, 2))
    print ""









    



spline 0






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}\frac{3}{5} \, x^{3} + \frac{1}{5} \, x$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}\frac{9}{5} \, x^{2} + \frac{1}{5}$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}\frac{18}{5} \, x$ 






    



spline 1






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}-2 \, x^{3} + \frac{42}{5} \, x^{2} - \frac{41}{5} \, x + \frac{14}{5}$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}-6 \, x^{2} + \frac{84}{5} \, x - \frac{41}{5}$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}-12 \, x + \frac{84}{5}$ 






    



spline 2






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}\frac{7}{5} \, x^{3} - \frac{54}{5} \, x^{2} + \frac{151}{5} \, x - \frac{122}{5}$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}\frac{21}{5} \, x^{2} - \frac{108}{5} \, x + \frac{151}{5}$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}\frac{42}{5} \, x - \frac{108}{5}$



In [12]:

    
eqns = []
for i in range(len(S) - 1):
    eqns.extend([
          (S[i] == S[i + 1]).subs(x=r[i][1]),
          (S[i].derivative(x) == S[i + 1].derivative(x)).subs(x=r[i][1]),
          (S[i].derivative(x, 2) == S[i + 1].derivative(x, 2)).subs(x=r[i][1])
        ])
eqns.extend([
        (S[0].derivative(x, 2) == 0).subs(x=r[0][0]),
        (S[-1].derivative(x, 2) == 0).subs(x=r[-1][1]),
        ])

for eqn in eqns:
    show(eqn)
    print "holds? ", eqn == True
    print ""









    




 $\newcommand{\Bold}[1]{\mathbf{#1}}\left(\frac{4}{5}\right) = 1$ 






    



holds?  False







    




 $\newcommand{\Bold}[1]{\mathbf{#1}}2 = \left(\frac{13}{5}\right)$ 






    



holds?  False







    




 $\newcommand{\Bold}[1]{\mathbf{#1}}\left(\frac{18}{5}\right) = \left(\frac{24}{5}\right)$ 






    



holds?  False







    




 $\newcommand{\Bold}[1]{\mathbf{#1}}4 = 4$ 






    



holds?  False







    




 $\newcommand{\Bold}[1]{\mathbf{#1}}\left(\frac{7}{5}\right) = \left(\frac{19}{5}\right)$ 






    



holds?  False







    




 $\newcommand{\Bold}[1]{\mathbf{#1}}\left(-\frac{36}{5}\right) = \left(-\frac{24}{5}\right)$ 






    



holds?  False







    




 $\newcommand{\Bold}[1]{\mathbf{#1}}0 = 0$ 






    



holds?  False







    




 $\newcommand{\Bold}[1]{\mathbf{#1}}\left(\frac{18}{5}\right) = 0$ 






    



holds?  False

Aufgabenstellung:

Gegeben: eine Funktion $S(x)$ mit reellen Konstanten $a,b,c,d \ldots$
Gesucht: Werte der Konstanten sodass $S(x)$ eine gültige kubische Spline-Interpolierende ist
Vorgehensweise: Basierend auf den Bedingungen ein Gleichunssystem aufstellen und lösen



In [6]:

    
var('x')
# input:
constants = var('a,b,c,d,e')
S = [
    a + b*(x - 1) + c*(x - 1)^2 + d * (x - 1)^3,
    (x - 1)^3 + e*x^2 - 1
]
r = [
    (0, 1),
    (1, 2)
]
for i, s in enumerate(S):
    print "S[{}], first, second derivative:".format(i)
    for j in range(3):
        show(s.derivative(x, j))
    print ""









    



S[0], first, second derivative:






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}d {\left(x - 1\right)}^{3} + c {\left(x - 1\right)}^{2} + b {\left(x - 1\right)} + a$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}3 \, d {\left(x - 1\right)}^{2} + 2 \, c {\left(x - 1\right)} + b$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}6 \, d {\left(x - 1\right)} + 2 \, c$ 






    



S[1], first, second derivative:






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}{\left(x - 1\right)}^{3} + e x^{2} - 1$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}3 \, {\left(x - 1\right)}^{2} + 2 \, e x$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}2 \, e + 6 \, x - 6$



In [7]:

    
eqns = []
for i in range(len(S) - 1):
    eqns.extend([
          (S[i] == S[i + 1]).subs(x=r[i][1]),
          (S[i].derivative(x) == S[i + 1].derivative(x)).subs(x=r[i][1]),
          (S[i].derivative(x, 2) == S[i + 1].derivative(x, 2)).subs(x=r[i][1])
        ])
eqns.extend([
        (S[0].derivative(x, 2) == 0).subs(x=r[0][0]),
        (S[-1].derivative(x, 2) == 0).subs(x=r[-1][1]),
        ])

for eqn in eqns:
    show(eqn.simplify())









    




 $\newcommand{\Bold}[1]{\mathbf{#1}}a = e - 1$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}b = 2 \, e$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}2 \, c = 2 \, e$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}2 \, c - 6 \, d = 0$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}2 \, e + 6 = 0$



In [8]:

    
for i in solve(eqns, *constants):
    show(i)









    




 $\newcommand{\Bold}[1]{\mathbf{#1}}\left[a = \left(-4\right), b = \left(-6\right), c = \left(-3\right), d = \left(-1\right), e = \left(-3\right)\right]$



In [ ]: