SME
Differential Equations · Problem type
One mode of the heat equation on a rod
Use this for “A 1 m rod with its ends at 0 °C starts at u = 100 sin(πx). With k = 0.01 m²/s, find u at x = 0.5 m after 10 s.”
Chart
u(0.5) ≈ 37.27
This is an example. Type your own values to start a new problem.
u(
x (m)
,
t (s)
)
=
u(x, t) (°C)
u
=
A (°C)
sin
π
x
L
×
exp
−
t
τ
=
L
2
÷
(
k
×
(
n
π)²)
=
τ (s)
Formulas
τ
=
L
2
÷
(
k
×
(
n
× π)
2
)
10.1321
=
1
2
÷
(0.01
×
(1 × π)
2
)
u(
x
,
t
)
=
A
×
sin(
n
×
π
×
x
L
)
×
e
−
t
÷
τ
37.2708
=
100
×
sin(1
×
π
×
0.5
1
)
×
e
−10 ÷ 10.1321
Assumptions
•
uₜ = k·uₓₓ on 0 ≤ x ≤ L, with u = 0 at both ends for all time.
•
Separating variables, each sine mode A sin(nπx ÷ L) keeps its shape and decays as e−t ÷ τ.
•
τ falls as 1 ÷ n², so higher modes die n² times faster and the first mode is what lingers.
•
k = 0.01 m²/s is a classroom number; metals are near 10⁻⁴ m²/s, so a real rod cools over hours, not seconds.
Step-by-step
WE KNOW
A = 100 °C, L = 1 m, n = 1, k = 0.01 m²/s, x = 0.5 m, t = 10 s
FIND
decay time (τ), temperature at x and t (u(x, t))
Step 1 · Find decay time (τ)
Use τ = L² ÷ (k × (n × π)²)
Putting
the
mode
into
uₜ
=
k
·uₓₓ
gives
the
rate
k
(
nπ
L
)
2
;
τ
is
one
over
that
rate.
τ
=
1
2
÷
(0.01
×
(1 × π)
2
)
τ
=
1
2
0.01 × 9.8696
τ
=
1
2
0.098696
τ
=
1
0.098696
τ
=
10.13212
s
Step 2 · Find temperature at x and t (u(x, t))
Use u(x, t) = A × sin(n × π × x ÷ L) × e−t ÷ τ
The
starting
shape
at
x
,
shrunk
by
the
factor
e
−
t
÷
τ
the
mode
has
decayed
by.
u(
x
,
t
)
=
100
×
sin(1
×
π
×
0.5
1
)
×
e
−10 ÷ 10.13212
u(
x
,
t
)
=
100
×
sin(0.5π)
×
e
−0.987
u(
x
,
t
)
=
100
×
0.3727
u(
x
,
t
)
=
37.2708
°C
Check
Put the numbers back in. Both sides should match.
10.13212
=
1
2
÷
(0.01
×
(1 × π)
2
)
✓
37.2708
=
100
×
sin(1
×
π
×
0.5
1
)
×
e
−10 ÷ 10.13212
✓
Topic check
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