Stability and Stability Functions¶

Copyright (C) 2026 Andreas Kloeckner

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In [1]:
import numpy as np
import numpy.linalg as la
import matplotlib.pyplot as pt
In [2]:
A = np.array([[0, 1], [-1, 0]])
def harmonic_rhs(yvec):
    return A@yvec

Write out the step functions for forward and backward Euler as functions step_XXX(y, h, f):

In [3]:
def step_fw(y, h, f):
    return y + h*f(y)

    
def step_bw(y, h, f):
    return la.solve(np.eye(2) - h*A, y)

Now write Heun and RK4: $$ \begin{array}{c|cc} 0 & & \\ 1 & 1 & \\ \hline & \tfrac{1}{2} & \tfrac{1}{2} \end{array} \qquad \begin{array}{c|cccc} 0 & & & & \\ \tfrac{1}{2} & \tfrac{1}{2} & & & \\ \tfrac{1}{2} & 0 & \tfrac{1}{2} & & \\ 1 & 0 & 0 & 1 & \\ \hline & \tfrac{1}{6} & \tfrac{1}{3} & \tfrac{1}{3} & \tfrac{1}{6} \end{array} $$

In [4]:
def step_heun(y, h, f):
    k1 = f(y)
    k2 = f(y + h*k1)
    return y + h/2*(k1 + k2)

def step_rk4(y, h, f):
    k1 = f(y)
    k2 = f(y + h/2*k1)
    k3 = f(y + h/2*k2)
    k4 = f(y + h*k3)
    return y + h/6*(k1 + 2*k2 + 2*k3 + k4)

Now write out the stability functions: (see below for Heun, RK4 and trapezoidal):

In [5]:
def stab_fw(z):
    return 1 + z
    
def stab_bw(z):
    return 1/(1 - z)

def stab_heun(z):
    return (1/2)*z**2 + z + 1

def stab_rk4(z):
    return (1/24)*z**4 + (1/6)*z**3 + (1/2)*z**2 + z + 1

def stab_trapezoidal(z):
    return -z**2/(2*z - 4) + (1/2)*z - z/(z - 2) + 1
In [26]:
if 0:
    step = step_fw
    stab = stab_fw
elif 0:
    step = step_bw
    stab = stab_bw
elif 0:
    step = step_heun
    stab = stab_heun
elif 1:
    step = step_rk4
    stab = stab_rk4
elif 0:
    step = None
    stab = stab_trapezoidal
In [27]:
times = [0]
ys = [np.array([0.1, 0.9])]

dt = 0.01

while times[-1] < 6*np.pi:
    y = ys[-1]
    ynext = step(y, dt, harmonic_rhs)
    ys.append(ynext)
    times.append(times[-1] + dt)

ys = np.array(ys)
times = np.array(times)
In [28]:
pt.plot(times, ys[:, 0])
Out[28]:
[<matplotlib.lines.Line2D at 0x7fedf08d6e40>]
No description has been provided for this image
In [29]:
ext = 3
re, im = np.mgrid[-ext:ext:400j, -ext:ext:400j]
z = re + 1j*im

pt.imshow(np.log10(np.abs(stab(z))[..., ::-1].T), extent=(-ext, ext, -ext, ext))
pt.colorbar()
pt.contour(np.log10(np.abs(stab(z))[..., ::-1].T), extent=(-ext, ext, -ext, ext), levels=[0], colors=["r"])
pt.gca().set_aspect("equal")
No description has been provided for this image

Why are these symmetric about the $x$ axis?

Order stars¶

Plot the order star for each method:

In [30]:
order_star = stab(z)/np.exp(z)

pt.imshow(np.log10(np.abs(order_star)[..., ::-1].T), extent=(-ext, ext, -ext, ext))
pt.colorbar()
pt.contour(np.log10(np.abs(order_star)[..., ::-1].T), extent=(-ext, ext, -ext, ext), levels=[0], colors=["r"])
pt.gca().set_aspect("equal")
No description has been provided for this image

Deriving stability functions¶

In [118]:
import sympy as sp
from sympy.printing.numpy import NumPyPrinter

h = sp.Symbol("h")
y = sp.Symbol("y")
ynext = sp.Symbol("y_n")
lam = sp.Symbol("lamda")
k1, k2 = sp.symbols("k1, k2")
z = sp.Symbol("z")

gen_code = NumPyPrinter().doprint

def rhs(y):
    return lam*y

Explicit¶

Use .expand() and .subs(h*lam, z) to find the stability function. Make sure to divide by y.

Do this for both Heun and RK4.

In [119]:
heun(y, h, rhs).expand().subs(h*lam, z)
Out[119]:
$\displaystyle \frac{y z^{2}}{2} + y z + y$
In [120]:
print(gen_code((step_heun(y, h, rhs)/y).expand().subs(h*lam, z)))
(1/2)*z**2 + z + 1
In [121]:
print(gen_code((step_rk4(y, h, rhs)/y).expand().subs(h*lam, z)))
(1/24)*z**4 + (1/6)*z**3 + (1/2)*z**2 + z + 1

Do you notice something about these stability functions?

They're truncations of $e^x$.

Implicit¶

Derive the stability function for the trapezoidal rule:

$$ \begin{array}{c|cc} 0 & 0 & 0 \\ 1 & \tfrac{1}{2} & \tfrac{1}{2} \\ \hline & \tfrac{1}{2} & \tfrac{1}{2} \end{array} $$

Set up the equations for the stage values:

In [124]:
equations = [
    k1 - rhs(y),
    k2 - rhs(y + h/2*(k1+k2))
]
equations
Out[124]:
[k1 - lamda*y, k2 - lamda*(h*(k1 + k2)/2 + y)]

Solve for ynext:

In [126]:
soln = sp.solve(equations, [k1, k2])
soln
Out[126]:
{k1: lamda*y, k2: (-h*lamda**2*y - 2*lamda*y)/(h*lamda - 2)}

Assemble the overall scheme:

In [128]:
trapezoidal = y + h/2*(soln[k1] + soln[k2])
trapezoidal
Out[128]:
$\displaystyle \frac{h \left(\lambda y + \frac{- h \lambda^{2} y - 2 \lambda y}{h \lambda - 2}\right)}{2} + y$

And find the stability function as above:

In [129]:
print(gen_code((trapezoidal/y).expand().subs(h*lam, z)))
-z**2/(2*z - 4) + (1/2)*z - z/(z - 2) + 1
In [ ]: