Stiffness in Initial Value Problems¶

In [1]:
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: $y'(t)=1-100ce^{-100t}$.

In [2]:
def f(t, y):
return -100*y+100*t + 101

In [3]:
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")

In [4]:
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)

/usr/local/lib/python3.5/dist-packages/IPython/core/formatters.py:92: DeprecationWarning: DisplayFormatter._ipython_display_formatter_default is deprecated: use @default decorator instead.
def _ipython_display_formatter_default(self):
/usr/local/lib/python3.5/dist-packages/IPython/core/formatters.py:669: DeprecationWarning: PlainTextFormatter._singleton_printers_default is deprecated: use @default decorator instead.
def _singleton_printers_default(self):


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:

In [38]:
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")


Using an Explicit Method¶

First, implement forward_euler_step(tk, yk, h):

In [44]:
def forward_euler_step(tk, yk, h):
return yk + h*f(tk, yk)

In [45]:
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)

• What's the main challenge here?

Using an Implicit Method¶

Next, implement backward_euler_step(tk, yk, h):

In [46]:
def backward_euler_step(tk, yk, h):
tkp1 = tk+h
return (yk + h*(100*tkp1 + 101))/(1+100*h)

In [48]:
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()

Out[48]:
<matplotlib.legend.Legend at 0x7f35508265c0>