Ordinary Differential Equations Exercise 1

Imports



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









    



:0: FutureWarning: IPython widgets are experimental and may change in the future.

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:



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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])
    """
    h=x[-1]-x[-2]
    data = [y0]
    for t in x[1:]:
        data.append(data[-1]+h*derivs(data[-1],t))
    return data



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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 [4]:

    
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])
    """
    h=x[-1]-x[-2]
    data = [y0]
    for t in x[1:]:
        data.append(data[-1]+h*derivs(data[-1]+h/2*derivs(data[-1],t),t+h/2))
    return data



In [5]:

    
assert np.allclose(solve_euler(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 [6]:

    
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]).
    """
    data = np.array(.25*np.exp(2*x)-.5*x-.25)
    return data



In [7]:

    
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:

Euler's method
Midpoint method
odeint
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 sigle figure with two subplots:

Plot the $y(x)$ versus $x$ for each of the 4 approaches.
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 [46]:

    
x = np.linspace(0,1,11)
def derivs(y, x):
    dy = x+2*y
    return dy
euler_error = np.array(solve_euler(derivs, 0, x))-solve_exact(x)
midpoint_error = np.array(solve_midpoint(derivs, 0, x))-solve_exact(x)
odeint_error = np.array(odeint(derivs, 0, x)).flatten()-solve_exact(x)

f = plt.figure(figsize = (9,6))
ax = plt.subplot(211)
ax.spines['top'].set_visible(False)
ax.spines['right'].set_visible(False)
ax.yaxis.set_ticks_position('left')
ax.xaxis.set_ticks_position('bottom')
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), label="Exact")

plt.plot(x,odeint(derivs, 0, x), label="ODEInt")
plt.ylabel("y(x)")
plt.xlabel("x")
plt.title(r"Numerical Solutions to $\frac{dy}{dx}=x+2y$")
plt.legend(loc = "best")

ax = plt.subplot(212)
ax.spines['top'].set_visible(False)
ax.spines['right'].set_visible(False)
ax.yaxis.set_ticks_position('left')
ax.xaxis.set_ticks_position('bottom')
plt.plot(x,abs(euler_error), label = "Euler Error")

plt.plot(x,abs(midpoint_error), label = "Midpoint Error")

plt.plot(x,abs(odeint_error), label = "ODEInt Error")
plt.ylabel("Errors")
plt.xlabel("x")
plt.title(r"Errors of numerical solutions to $\frac{dy}{dx}=x+2y$")
plt.legend(loc = "best")
plt.tight_layout()



In [9]:

    
assert True # leave this for grading the plots



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