Vector Field and Euler Method Plotter
This Python script provides functionality to visualize vector fields and numerically solve first-order ordinary differential equations (ODEs) using the Euler method. The script utilizes the matplotlib and numpy libraries for plotting and mathematical operations.
Features
The vectorfield function in the script allows you to plot a vector field representing the normalized gradient vectors of a given function. You can customize the function, axis ranges, vector scaling, root plotting, and more.
Plotting the vectorfield for
import numpy as np
import matplotlib.pyplot as plt
from vectorfields import vectorfield
# Define a vector field function
a = 0.1
my_ode = lambda y, x: -x + a*y + x**2 * y
# Plot the vector field
vectorfield(my_ode, y_min=0, y_max=2.5, y_step=0.1, x_min=0, x_max=2.5, x_step=0.1,
title="$y(x)'= -x + a \cdot y + x^2 \cdot y$" , vector_scale=.1,
plot_root=True, xlabel='X-axis', ylabel='Y-axis', show=True, color='b')
The euler_method function allows you to numerically solve a first-order ODE using the Euler method. You can specify the ODE function, step size, initial values, and the x-value at which the solution is approximated. Additionally, the solution can be plotted. The method returns a pandas dataframe.
Plotting the euler approximation for
Analytical exact solution is
import numpy as np
import matplotlib.pyplot as plt
from vectorfields import euler_method
function1 = lambda t,x: t*x
a1 = euler_method(function1, step_h=0.4, initial_x=0, initial_y=0.1, approx_x=2, plot=True, label='h=0.4')
a2 = euler_method(function1, step_h=0.2, initial_x=0, initial_y=0.1, approx_x=2, plot=True, label='h=0.2')
a3 = euler_method(function1, step_h=0.1, initial_x=0, initial_y=0.1, approx_x=2.1, plot=True, label='h=0.1')
# manually build the exact solution
exact_sol = lambda t: 0.1 * np.e**(1/2*t**2)
x_values = np.linspace(0,2,100)
y_values = [(exact_sol)(t) for t in x_values]
# add the exact solution
fig, ax = plt.gcf(), plt.gca()
ax.plot(x_values,y_values, label='exact')
ax.legend()
# annotate the solutions
ax.annotate(str(round(a1.iloc[-1,2],2)), (2, a1.iloc[-1,2]))
ax.annotate(str(round(a2.iloc[-1,2],2)), (2, a2.iloc[-1,2]))
ax.annotate(str(round(a3.iloc[-1,2],2)), (2, a3.iloc[-1,2]))
ax.annotate(str(round(y_values[-1],2)), (2, y_values[-1]))
plt.show()
Plotting the euler approximation for
import numpy as np
import matplotlib.pyplot as plt
from vectorfields import vectorfield, euler_method
function1 = lambda t,x: t*x
vectorfield(function1, vector_scale=0.2,
x_min=-1, x_max=3, x_step=0.25,
y_min=-1, y_max=3, y_step=0.25,
xlabel='t', ylabel='x', plot_root=True,
title="$x'(t)=t \\cdot x(t)$", color='b')
a1 = euler_method(function1, step_h=0.4, initial_x=0, initial_y=0.1, approx_x=2, plot=True, label='h=0.4')
a2 = euler_method(function1, step_h=0.2, initial_x=0, initial_y=0.1, approx_x=2, plot=True, label='h=0.2')
a3 = euler_method(function1, step_h=0.1, initial_x=0, initial_y=0.1, approx_x=2.1, plot=True, label='h=0.1')
# manually build the exact solution
exact_sol = lambda t: 0.1 * np.e**(1/2*t**2)
x_values = np.linspace(0,2,100)
y_values = [(exact_sol)(t) for t in x_values]
# add the exact solution
fig, ax = plt.gcf(), plt.gca()
ax.plot(x_values,y_values, label='exact')
ax.legend()
# annotate the solutions
ax.annotate(str(round(a1.iloc[-1,2],2)), (2, a1.iloc[-1,2]))
ax.annotate(str(round(a2.iloc[-1,2],2)), (2, a2.iloc[-1,2]))
ax.annotate(str(round(a3.iloc[-1,2],2)), (2, a3.iloc[-1,2]))
ax.annotate(str(round(y_values[-1],2)), (2, y_values[-1]))
plt.show()
Heun's method is a numerical technique used to approximate the solution of first-order ordinary differential equations (ODEs). It is an iterative method that improves upon the Euler method by using a predictor-corrector approach. This method is also known as the improved Euler method.
Consider a first-order ODE in the form: dy/dx = f(x, y), where y is the dependent variable and f(x, y) is a given function.
Heun's method involves the following iterative steps:
-
Predictor Step (k_1):
- Compute the slope at the current point (x, y) using the function f(x, y). This is denoted as k_1.
- Predict the value of the dependent variable at the next time step using the Euler method: y_pred = y + h * k_1.
-
Corrector Step (k_2):
- Use the predicted value (x + h, y_pred) to compute the slope k_2.
- Average the slopes k_1 and k_2 to obtain a more accurate estimate of the slope over the interval.
- Update the dependent variable using the averaged slope.
-
Iteration:
- Repeat the predictor-corrector steps until the desired approximation point is reached.
The heuns_method function takes the following parameters:
function_of_xy: A callable representing the ODE, accepting x and y as arguments.step_h: Step size for numerical integration.initial_x,initial_y: Initial values of the independent and dependent variables.approx_x: Value of the independent variable where the solution is approximated.plot(optional): If True, plot the numerical solution. Default is False.show(optional): If True andplotis True, display the plot. Default is False.**kwargs: Additional keyword arguments for customizing the plot.
import matplotlib.pyplot as plt
# Define the ODE dx/dt = t - x
ode_function = lambda x, t: t - x
# Set initial conditions and parameters
initial_x_value = 0
initial_t_value = 1
step_size = 0.001
target_x_value = 2
# Apply Heun's method
func = lambda t, x: t - x
vectorfield(func,
x_min=-0.25, x_max=2.25, x_step=0.1,
y_min=0.5, y_max=1.5, y_step=0.1,
plot_root=True, root_size=3, show=False,
xlabel='t', ylabel='x', title="$x'(t) = t - x$",
vector_scale=0.07, color='b')
a3 = heuns_method(function_of_xy=func, step_h=step_size,
initial_x=initial_x_value, initial_y=initial_t_value,
approx_x=target_x_value, plot=True, show=False, color='g')
plt.show()
print(a3.tail())
# Returns:
# k x_k y_k k_1 k_2 h/2*(k_1 + k_2)
# 1997 1997.0 1.997 1.268484 0.728516 0.728788 0.000729
# 1998 1998.0 1.998 1.269213 0.728787 0.729059 0.000729
# 1999 1999.0 1.999 1.269941 0.729059 0.729329 0.000729
# 2000 2000.0 2.000 1.270671 0.729329 0.729600 0.000729
# 2001 2001.0 2.001 1.271400 0.729329 0.729600 0.000729