I have a rectangle of 1.3 mm * 0.2 mm, take the origin to the lower left corner of the rectangle. Say, I have a circular arc defined by the (0.3,0),(0.2732,0.1) & (0.3,0.2). It has a radius of 0.2 mm & center of the circular arc is (0.4732,0.1)
Now, imagine that the circular arc has a thickness of $\epsilon = 0.0002$
I want to define a signed distance function such that $\phi$ is 0 to the left of the left interface (or the arc), smoothly varies between 0 and 1 within the interface, and 1 beyond the right interface.
This is what I tried:
$d(x,y) = \sqrt{(x-x_{center})^2 + (y-y_{center})^2} - radius$
$\phi = \frac{1}{1+\exp(\frac{d}{\epsilon})}$
I am getting the following variation:
$\phi$ as per above Equation" />
However, I want this to be the result
$\phi$" />How do I approach this differently to get the target function ?
Code for the plot(s):
import numpy as np
def level_set_function(x, y, x_center, y_center,radius,epsilon):
d = np.sqrt((x-x_center)**2 + (y-y_center)**2) - radius
phi = 1/(1+np.exp(d/epsilon))
return phi
# Define constants
epsilon = 0.0002
x_center = 0.4732
y_center = 0.1
radius = 0.2
x = np.linspace(0, 1.3, 100)
y = np.ones_like(x) * y_center
phi = level_set_function(x, y, x_center, y_center,radius, epsilon)
# phi[x>0.4] = 1
import matplotlib.pyplot as plt
fig, ax = plt.subplots(figsize=(8, 4))
ax.plot(x, phi)
ax.set_xlabel('x ')
ax.set_ylabel('ϕ')
ax.set_title('ϕ along y=0.1')
plt.show()