In [1]:
%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:
In [2]:
# YOUR CODE HERE
#raise NotImplementedError()
def hat(x, a, b):
return -a*x**2+b*x**4
In [3]:
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 $a=5.0$:
In [4]:
a = 5.0
b = 1.0
In [10]:
# YOUR CODE HERE
#raise NotImplementedError()
x = np.linspace(-3,3,40)
plt.plot(x, hat(x,a,b))
plt.box(False)
plt.title('Hat Potential')
plt.xlabel('$x$')
plt.ylabel('$V(x)$');
In [11]:
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.
In [12]:
# YOUR CODE HERE
#raise NotImplementedError()
min1 = opt.minimize(hat, (-1.5), args=(a,b), bounds = [(-3,0)])
min2 = opt.minimize(hat, (1.5), args=(a,b), bounds = [(0,3)])
print ('The local minima are %f and %f' %(min1['x'][0], min2['x'][0]))
plt.plot(x, hat(x,a,b), label='Hat Potential')
plt.scatter(min1['x'][0], hat(min1['x'][0],a,b), color = 'r',label = 'local min 1')
plt.scatter(min2['x'][0], hat(min2['x'],a,b), color = 'r',label = 'local min 2')
plt.box(False)
plt.tick_params(axis = 'x', top = 'off')
plt.tick_params(axis = 'y', right = 'off')
plt.title('Hat Potential')
plt.legend();
In [8]:
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.
Our equation is: \begin{equation*} V(x) = -5x^2+x^4 \end{equation*}
To solve for maxima and minima values you must take the derivative of your function.
\begin{equation*} \frac{dV}{dx} = -10x+4x^3 \end{equation*}Set this derivative equal to zero and solve for x.
\begin{equation*}\\ 0 = x(-10+4x^2)\\ x = 0\\ x = \pm \sqrt{\frac{10}{4}}\\ \end{equation*}x = 0 by the graph is a local maxima.
$$\pm \sqrt{\frac{10}{4}} = \pm 1.5811388$$These are our minima!!