Numerical Methods for Engineers · Problem type

Step size for a stable Euler method

Use this for “For dy/dt = −50y, is Euler’s method with h = 0.05 stable? What step size is?”

Chart

y(t) = e−50t012345−10−5051015tyy₆ = 11.39
Euler, 6 steps of h = 0.05: y = 11.39 at t = 0.3; solid, the exact y = 3.059 × 10⁻⁷, far off: the steps run away from it
This is an example. Type your own values to start a new problem.
g
=
1
+
(
k
)
×
h
hₘₐₓ
=
2
÷
|
k
|
y₀
Starting value
example
n
Steps
example
yₙ
Euler’s value after n steps
worked out (not typed)

Formulas

g
=
1
+
k
×
h
−1.5 = 1 + (−50) × 0.05
hₘₐₓ
=
2
÷
|
k
|
0.04 = 2 ÷ |−50|
yₙ
=
y₀
×
g
n
11.3906
=
1
×
(−1.5)
6

Assumptions

•
dy/dt = ky with k < 0 decays to 0, but each Euler step multiplies y by g = 1 + kh.
•
Stable when |g| ≤ 1, so h ≤ 2 ÷ |k|; past it the steps flip sign and grow.
•
A large |k| (a stiff problem) forces a tiny h; implicit methods avoid that limit.

Step-by-step

WE KNOW
k = −50, h = 0.05, y₀ = 1, n = 6
FIND
growth factor per step (g), largest stable step (hₘₐₓ), Euler’s value after n steps (yₙ)
Step 1 · Find growth factor per step (g)
Use g = 1 + k × h
An
Euler
step
adds
h
×
ky
to
y,
which
multiplies
y
by
1
+
kh.
g
=
1
+
(−50)
×
0.05
g
=
1
+
(−2.5)
g
=
−1.5
Step 2 · Find largest stable step (hₘₐₓ)
Use hₘₐₓ = 2 ÷ |k|
g
=
−1
at
h
=
2
÷
|
k
|;
any
larger
h
makes
g
<
−1.
hₘₐₓ = 2 ÷ |−50|
hₘₐₓ
=
2
50
hₘₐₓ
=
0.04
→
h
=
0.05
is
past
hₘₐₓ:
unstable;
any
h
≤
0.04
is
stable.
Step 3 · Find Euler’s value after n steps (yₙ)
Use yₙ = y₀ × gⁿ
n
steps
multiply
y₀
by
g
,
n
times.
yₙ
=
1
×
(−1.5)
6
yₙ = 11.3906
Check
Put the numbers back in. Both sides should match.
−1.5 = 1 + (−50) × 0.05 ✓
0.04 = 2 ÷ |−50| ✓
11.3906
=
1
×
(−1.5)
6
✓
Topic check
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