Cubic splines

Splines were formulated to relieve these oscillations by piecing together a series of lower-order polynomials and requiring smoothness. Consider a polynomial over the interval between \(x_i\) and \(x_{i+1}\), and assert:

• \(y(x_i) = y_i\)

• \(y(x_{i+1}) = y_{i+1}\)

• \(y'(x_i)\) be continuous

• \(y''(x_i)\) be continuous

with these 4 constraints, it is clear we are looking for cubic functions, and therefore these splines are piecewise cubic curves.

We will be describing the splines in terms of the knots, \(k_i\) which parameterize the curves. For these splines, these knots are the second derivatives at a point \(x_i\).

To find the coefficients of the cubic splines, consider that the second derivative is linear and represent it with a 2-point Lagrange interpolation: \begin{align} y’’{i, i+1} &= k_i P_i(x) + k{i+1} P_{i+1}(x) \ &= \frac{k_i [x-x_{i+1}] + k_{i+1} [x-x_i]}{x_i-x_{i+1}} \end{align}

Using the constraints above we end up with:

\[k_{i-1}[x_{i-1}-x_i] + 2 k_i [x_{i-1} - x_{i+1}] + k_{i+1}[x_i-x_{i+1}] = 6\left[ \frac{y_{i-1}-y_i}{x_{i-1}-x_i} - \frac{y_{i}-y_{i+1}}{x_{i}-x_{i+1}} \right]\]

which is a tridiagonal matrix!

\[\begin{split} \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 4 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 4 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 4 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 4 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 4 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 4 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 4 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 4 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 4 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} k_1 \\ k_2\\ k_3\\ k_4\\ k_5\\ k_6\\ k_7\\ k_8\\ k_9\\ k_{10}\\ k_{11} \end{bmatrix} = \begin{bmatrix} 0\\ -0.424 \\ -1.052 \\ -0.891 \\ 1.138 \\ 2.654 \\ 1.138 \\ -0.891 \\ -1.052 \\ -0.424 \\ 0 \end{bmatrix} \end{split}\]

# prompt: Do a cubic spline of x_d and y_d and plot against the original function from -5.5 to 5.5

import matplotlib.pyplot as plt
import numpy as np
from scipy.interpolate import CubicSpline

# Create a cubic spline interpolation
cs = CubicSpline(x_d, y_d)

# Create x values for plotting the interpolated spline
x_interp = np.linspace(-6, 6, 200)
y_interp = cs(x_interp)

# Plot the original curve, sampled points, and interpolated spline
plt.plot(x_toy, y_toy, label='exp(-(x/2)^2)')
plt.scatter(x_d, y_d, color='red', label='Sampled points')
plt.plot(x_interp, y_interp, label='Cubic Spline Interpolation')
plt.xlabel('x')
plt.ylabel('y')
plt.legend()
plt.title('Function, Sampled Points, and Cubic Spline Interpolation')
plt.grid(True)
plt.show()

---------------------------------------------------------------------------
ModuleNotFoundError                       Traceback (most recent call last)
Cell In[1], line 5
      3 import matplotlib.pyplot as plt
      4 import numpy as np
----> 5 from scipy.interpolate import CubicSpline
      7 # Create a cubic spline interpolation
      8 cs = CubicSpline(x_d, y_d)

ModuleNotFoundError: No module named 'scipy'

Analysis of cubic splines

We note:

• Cubic splines are stiffer in that they don’t have high-frequency oscillations (thus avoiding Runge’s phenomenon).

• The concept of smoothness is easy in 1D, but what does it mean for 2D+? How would you ensure continuity along an edge?

• Specifying smoothness as part of the goals going in suggest this is more of a global scheme. This requires simultaneous linears systems to be solved.