# Ordinary Differential Equations Exercise 1

## Imports



In [11]:

%matplotlib inline
import matplotlib.pyplot as plt
import numpy as np
import seaborn as sns
from scipy.integrate import odeint
from IPython.html.widgets import interact, fixed



## Euler's method

Euler's method is the simplest numerical approach for solving a first order ODE numerically. Given the differential equation

$$\frac{dy}{dx} = f(y(x), x)$$

with the initial condition:

$$y(x_0)=y_0$$

Euler's method performs updates using the equations:

$$y_{n+1} = y_n + h f(y_n,x_n)$$$$h = x_{n+1} - x_n$$

Write a function solve_euler that implements the Euler method for a 1d ODE and follows the specification described in the docstring:



In [12]:

def solve_euler(derivs, y0, x):
"""Solve a 1d ODE using Euler's method.

Parameters
----------
derivs : function
The derivative of the diff-eq with the signature deriv(y,x) where
y and x are floats.
y0 : float
The initial condition y[0] = y(x[0]).
x : np.ndarray, list, tuple
The array of times at which of solve the diff-eq.

Returns
-------
y : np.ndarray
Array of solutions y[i] = y(x[i])
"""
y=np.zeros_like(x)
h=x[1]-x[0]
y[0]=y0
for i in range(len(x)-1):
y[i+1]=y[i]+h*derivs(y[i],x[i])
return y




In [13]:

assert np.allclose(solve_euler(lambda y, x: 1, 0, [0,1,2]), [0,1,2])



The midpoint method is another numerical method for solving the above differential equation. In general it is more accurate than the Euler method. It uses the update equation:

$$y_{n+1} = y_n + h f\left(y_n+\frac{h}{2}f(y_n,x_n),x_n+\frac{h}{2}\right)$$

Write a function solve_midpoint that implements the midpoint method for a 1d ODE and follows the specification described in the docstring:



In [14]:

def solve_midpoint(derivs, y0, x):
"""Solve a 1d ODE using the Midpoint method.

Parameters
----------
derivs : function
The derivative of the diff-eq with the signature deriv(y,x) where y
and x are floats.
y0 : float
The initial condition y[0] = y(x[0]).
x : np.ndarray, list, tuple
The array of times at which of solve the diff-eq.

Returns
-------
y : np.ndarray
Array of solutions y[i] = y(x[i])
"""
y=np.zeros_like(x)
h=x[1]-x[0]
y[0]=y0
for i in range(len(x)-1):
y[i+1]=y[i]+h*derivs(y[i]+(h/2)*derivs(y[i],x[i]),x[i]+(h/2))
return y




In [15]:

assert np.allclose(solve_midpoint(lambda y, x: 1, 0, [0,1,2]), [0,1,2])



You are now going to solve the following differential equation:

$$\frac{dy}{dx} = x + 2y$$

which has the analytical solution:

$$y(x) = 0.25 e^{2x} - 0.5 x - 0.25$$

First, write a solve_exact function that compute the exact solution and follows the specification described in the docstring:



In [16]:

def solve_exact(x):
"""compute the exact solution to dy/dx = x + 2y.

Parameters
----------
x : np.ndarray
Array of x values to compute the solution at.

Returns
-------
y : np.ndarray
Array of solutions at y[i] = y(x[i]).
"""
y=np.zeros_like(x)
y=0.25*np.exp(2*x)-0.5*x-0.25
return y




In [17]:

assert np.allclose(solve_exact(np.array([0,1,2])),np.array([0., 1.09726402, 12.39953751]))



In the following cell you are going to solve the above ODE using four different algorithms:

1. Euler's method
2. Midpoint method
3. odeint
4. Exact

Here are the details:

• Generate an array of x values with $N=11$ points over the interval $[0,1]$ ($h=0.1$).
• Define the derivs function for the above differential equation.
• Using the solve_euler, solve_midpoint, odeint and solve_exact functions to compute the solutions using the 4 approaches.

Visualize the solutions on a single figure with two subplots:

1. Plot the $y(x)$ versus $x$ for each of the 4 approaches.
2. Plot $\left|y(x)-y_{exact}(x)\right|$ versus $x$ for each of the 3 numerical approaches.

Your visualization should have legends, labeled axes, titles and be customized for beauty and effectiveness.

While your final plot will use $N=10$ points, first try making $N$ larger and smaller to see how that affects the errors of the different approaches.



In [18]:

x=np.linspace(0,1,11)

def derivs(y,x):
return x+2*y

plt.figure(figsize=(10,9))

euler_diff=abs(solve_euler(derivs,0,x)-solve_exact(x))

midpt_diff=abs(solve_midpoint(derivs,0,x)-solve_exact(x))

ode_diff=np.empty(len(solve_euler(derivs,0,x)))
for i in range(len(solve_euler(derivs,0,x))):
ode_diff[i]=abs(odeint(derivs,0,x)[i]-solve_exact(x)[i])

plt.subplot(211);
plt.plot(x,solve_euler(derivs,0,x), label='Euler');
plt.plot(x,solve_midpoint(derivs,0,x), label='Midpoint');
plt.plot(x,solve_exact(x), 'o', label='Exact');
plt.plot(x,odeint(derivs,0,x), label='ODE');
plt.legend(bbox_to_anchor=(1.05, 1), loc=2);
plt.title('Differential Methods'),plt.xlabel('x'),plt.ylabel('y(x)');

plt.subplot(212);
plt.plot(x,euler_diff, label='Euler');
plt.plot(x,midpt_diff, label='Midpoint');
plt.plot(x,ode_diff,label='ODE');
plt.legend(bbox_to_anchor=(1.05, 1), loc=2);
plt.title('Differential Methods Difference'),plt.xlabel('x'),plt.ylabel('Difference');

plt.tight_layout();







In [19]:

assert True # leave this for grading the plots