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%matplotlib inline
import matplotlib.pyplot as plt
import numpy as np
import scipy.optimize as opt
The following potential is often used in Physics and other fields to describe symmetry breaking and is often known as the "hat potential":
$$ V(x) = -a x^2 + b x^4 $$Write a function hat(x,a,b)
that returns the value of this function:
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# YOUR CODE HERE
# raise NotImplementedError()
def hat(x, a, b):
c = x**2
d = x**4
p = -a * c + b * d
q = np.array(p)
return q
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assert hat(0.0, 1.0, 1.0)==0.0
assert hat(0.0, 1.0, 1.0)==0.0
assert hat(1.0, 10.0, 1.0)==-9.0
Plot this function over the range $x\in\left[-3,3\right]$ with $b=1.0$ and $a=5.0$:
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a = 5.0
b = 1.0
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# YOUR CODE HERE
# raise NotImplementedError()
x = np.arange(-3.0,3.0,0.1)
plt.plot(x,hat(x,a,b))
plt.xlim(-3, 3)
plt.ylim(-7, 35)
plt.box(False)
plt.xlabel('Position')
plt.ylabel('Hat Potential')
plt.title('Hat Potential versus Position')
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assert True # leave this to grade the plot
Write code that finds the two local minima of this function for $b=1.0$ and $a=5.0$.
scipy.optimize.minimize
to find the minima. You will have to think carefully about how to get this function to find both minima.
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# YOUR CODE HERE
# raise NotImplementedError()
# x0 = np.arange(-3.0, 0.0, 0.2)
minima_1 = opt.minimize(hat, -1.5, args = (5.0, 1.0))
minima_2 = opt.minimize(hat, 1.5, args = (5.0, 1.0))
x = np.arange(-3.0,3.0,0.1)
plt.plot(x,hat(x,a,b))
plt.xlim(-3, 3)
plt.ylim(-7, 35)
plt.box(False)
plt.scatter(minima_1.x, hat(minima_1.x, 5.0, 1.0), color = 'r')
plt.scatter(minima_2.x, hat(minima_2.x, 5.0, 1.0), color = 'r')
plt.xlabel('Position')
plt.ylabel('Hat Potential')
plt.title('Hat Potential versus Position')
plt.legend('Vm')
Out[7]:
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assert True # leave this for grading the plot
To check your numerical results, find the locations of the minima analytically. Show and describe the steps in your derivation using LaTeX equations. Evaluate the location of the minima using the above parameters.
With a = 5.0 and b = 1.0 $$ V'(x) = -2 a x + 4 b x^3 $$ $$ V'(x) = -10 x + 4 x^3 $$ Root are x = -1.581, x = 0, 1.581
$$ V''(x) = -2 a + 12 b x^2 $$$$ V''(x) = -10 + 12 x^2 $$V''(-1.581) = 19.994, V''(0) = -10, V''(1.581) = 19.994 By second derivative test x=-1.585 and x=1.581 are minima
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