import numpy as np
import matplotlib.pyplot as pt
Consider $y'=-100y+100t + 101$.
Exact solution: $y(t)=1+t+ce^{-100t}$.
Exact solution derivative: $y'(t)=1-100ce^{-100t}$.
def f(t, y):
return -100*y+100*t + 101
t_end = 0.2
def plot_solution(t0, y0):
c = (y0-1-t0)/np.exp(-100*t0)
t_mesh = np.linspace(t0, t_end, 1000)
solution = 1+t_mesh+c*np.exp(-100*t_mesh)
pt.plot(t_mesh, solution, label="exact")
pt.plot(t0, y0, "ko")
plot_solution(t0=0, y0=1)
plot_solution(t0=0, y0=1.2)
plot_solution(t0=0, y0=-0.5)
plot_solution(t0=0.05, y0=-0.5)
Here's a helper function that uses a time stepper in the form of a step_function
to numerically solve an ODE and plot the numerical solution:
def integrate_ode(step_function, t0, y0, h):
times = [t0]
ys = [y0]
while times[-1] <= t_end + 1e-14:
t = times[-1]
ys.append(step_function(t, ys[-1], h))
times.append(t + h)
pt.plot(times, ys, label=step_function.__name__)
pt.xlim([t0, t_end])
pt.ylim([-1, 2])
pt.legend(loc="best")
First, implement forward_euler_step(tk, yk, h)
:
def forward_euler_step(tk, yk, h):
return yk + h*f(tk, yk)
t0 = 0.05
y0 = -0.5
h = 0.008 # start this at 0.001, then grow
plot_solution(t0=t0, y0=y0)
integrate_ode(forward_euler_step, t0=t0, y0=y0, h=h)
Next, implement backward_euler_step(tk, yk, h)
:
def backward_euler_step(tk, yk, h):
tkp1 = tk+h
return (yk + h*(100*tkp1 + 101))/(1+100*h)
t0 = 0.05
y0 = -0.5
h = 0.05 # start this at 0.001, then grow
plot_solution(t0=t0, y0=y0)
integrate_ode(backward_euler_step, t0=t0, y0=y0, h=h)
pt.xlim([t0, t_end])
pt.ylim([-1, 2])
pt.legend()