Chapter 21
Lecture 25 Solution by
separation of variables
Example 21.1
utt = c2 uxx
BC: u(0, t) = 0,
0 < x < L,
u(L, t) = 0
IC: u(x, 0) = f (x),
ut (x, 0) = g(x)
t>0
(21.1)
(21.2)
(21.3)
For a guitar string c =
T0
.
ρ0
Separate Variables u(x, t) = X(x)T (t)
T̈ (t)
X (x)
=
= −λ2
c2 T (t)
X(x)
(21.4)
T̈ (t) + λ2 c2 T (t) = 0 ⇒ T (t) = c1 cos(λct) + c2 sin(λct)
(21.5)
2
X +λ X =0
X(x) = A cos(λx) + B sin λx
⇒
X(0) = 0 = X(L)
X(0) = A = 0 X(L) = B sin λL = 0
nπ
n = 1, 2, . . .
λn =
L nπx .
Xn = sin
L
125
Lecture 25 Solution by separation of variables
Therefore
u(x, t) =
∞
An cos
n=1
u(x, 0) =
ut (x, t) =
∞
n=1
∞
An sin
−An
nπct
L
nπx L
nπc L
n=1
sin
nπx L
+ Bn sin
= f (x) ⇒ An =
sin
nπct
L
sin
2
L
L
∞
Bn
n=1
nπc L
sin
nπx L
sin
f (x) sin
+ Bn
cos
ut (x, 0) =
= g(x) ⇒ Bn
nπx (21.6)
L
nπx 0
nπx L
nπct
L
(21.7)
L
nπc L
nπx nπct
sin
L
L
nπc L
=
2
L
L
g(x) sin
(21.8)
nπx 0
L
dx .
(21.9)
Therefore
∞ nπx nπct
nπct
+ Bn sin
sin
. (21.10)
u(x, t) =
An cos
L
L
L
n=1
21.1
Notes
1. Period and Frequency:
cos
nπct
nπcT
(t + T ) = cos
provided
= 2π
L
L
L
nπc
(21.11)
2L 1
is the period (seconds per cycle) of mode n. fn =
thus Tn =
c n
1
c
=n
are the natural frequencies of vibration.
Tn
2L
2. Modes of Vibration: Standing waves of wavelength λn =
126
2L
.
n
21.2. NOW WE CAN USE THE TRIGONOMETRIC IDENTITIES
21.2
Now we can use the trigonometric identities
sin(A ± B) = sin A cos B ± cos A sin B;
cos(A ± B) = cos A cos B
∓ sin A sin B
(21.12)
to interpret the solution (21.10).
nπx nπ nπct
1
nπ
sin
=
sin
(x + ct) + sin
(x − ct) (21.13)
cos
L
L
2
L
L
nπx nπct
1
nπ
nπ
sin
=
cos
(x − ct) − cos
(x + ct) . (21.14)
sin
L
L
2
L
L
Now
∞
An cos
n=1
nπct
L
sin
nπx L
nπ 1
(x + ct)
An sin
2
L
n=1
nπ (x − ct)
(21.15)
+ sin
L
1
[f0 (x + ct) + f0 (x − ct)] (21.16)
2
∞
=
=
where f0 is the odd periodic extension of f .
∞
∞
nπx 1 nπct
nπ
nπ
sin
=
(x − ct) − cos
(x + ct) (21.17)
.
Bn sin
Bn cos
L
L
2
L
L
n=1
n=1
Let
2
bn =
L
Then Bn =
L
g(x) sin
0
L
nπc
g(s) ds =
∞
bn sin
n=1
nπx L
.
(21.18)
bn
∞
n=1
0
x
g(s) ds =
0
L
dx ⇒ g(x) =
x
1
c
nπx ∞
n=1
x
∞
nπx − cos nπs
L
0
nπ (21.19)
bn
cBn 1 − cos
=
L
L
n=1
Bn = −
∞
n=1
Bn cos
nπx L
.
(21.20)
127
Lecture 25 Solution by separation of variables
Therefore
∞
Bn sin
n=1
nπct
L
sin
nπx L
⎧ 0
⎫
x+ct
⎨
⎬
1
=
g0 (s) ds +
g0 (s) ds . (21.21)
⎭
2c ⎩
x−ct
0
Therefore
1
1
u(x, t) = [f0 (x + ct) + f0 (x − ct)] +
2
2c
x+ct
g0 (s) ds
(21.22)
x−ct
where f0 and g0 are the odd periodic extensions of f and g on [0, L] i.e.
f (x)
0 < x < L and f0 (x + 2L) = f0 (x0 )
f0 (x) =
(21.23)
−f (−x) −L < x < 0
g(x)
0 < x < L and g0 (x + 2L) = g0 (x)
(21.24)
.
g0 (x) =
−g(−x) −L < x < 0
Notes:
1. Equation (21.22) above shows that the Wave Equation Solution for
a string tied down at its ends is given by D’Alembert’s Solution (see
(19.29) of the notes) in which the initial displacement function is given
by the odd periodic extension f0 of the initial displacement of the
string, and the initial velocity function is given by the odd periodic
extension of g0 .
2. Information is carried along the characteristic curves x + ct = const
x − ct = const.
nπct
L
3. Observe that the time dependence of the solution involves sin
nπct
and cos
which do not decay with time. Thus the solutions
L
to the Wave Equation persist with time, whereas the solutions to the
Heat Equation typically decay exponentially with time.
128