M344 - ADVANCED ENGINEERING MATHEMATICS
Lecture 16: More on the Wave Equation
The Damped Wave Equation
Consider the equation
utt + 2cut = β 2 uxx
where c is a small positive constant. The term 2cut represents a damping force
proportional to the velocity ut . For simplicity, we will assume that the length
of the string is L = π, and the constant β 2 = 1.
If we let u(x, t) = X(x)T (t), then
XT 00 + 2cXT 0
X 00 T
T 00 + 2cT 0
X 00
=
⇒
=
= −λ.
XT
XT
T
X
This gives us the two ordinary differential equations
X 00 + λX = 0, T 00 + 2cT 0 + λT = 0.
If we assume the two ends of the string are fixed, so u(0, t) = u(π, t) = 0
for all t > 0, then we again have the Sturm-Liouville problem
X 00 + λX = 0, X(0) = X(π) = 0
2 2
with eigenvalues λn = nππ2 = n2 and corresponding eigenfunctions Xn (x) =
bn sin(nx). The equation for Tn (t) is
Tn00 + 2cTn0 + n2 Tn = 0,
with characteristic polynomial r2 + 2cr + n2 = 0. We will assume that c < 1,
and then
√
√
−2c ± 4c2 − 4n2
r=
= −c ± c2 − n2 .
2
2
2
The discriminant c −n < 0 for any integer n ≥ 1, so the roots are complex
and the general solution is
√
√
Tn (t) = an e−ct cos( n2 − c2 t) + bn e−ct sin( n2 − c2 t).
This means that
u(x, t) =
∞
X
i
h
√
√
sin(nx)e−ct an cos( n2 − c2 t) + bn sin( n2 − c2 t) .
n=1
1
If the initial conditions are u(x, 0) = f (x), ut (x, 0) = g(x), then
u(x, 0) =
∞
X
an sin(nx) = f (x)
n=1
implies that an =
ut (x, t) =
∞
X
2
π
Rπ
0
f (x) sin(nx)dx. The partial of u with respect to t is
´i
h
³
√
√
−ct
2
2
2
2
sin(nx) −ce
an cos( n − c t) + bn sin( n − c t)
n=1
h
³
´i
√
√
√
+ sin(nx) e−ct n2 − c2 −an sin( n2 − c2 t) + bn cos( n2 − c2 t) ;
therefore,
ut (x, 0) =
∞
X
h
i
√
sin(nx) −can + n2 − c2 bn ≡ g(x).
n=1
This is a Fourier Sine Series for g(x) with coefficients
Z
√
2 π
2
2
g(x) sin(nx)dx;
−can + n − c bn =
π 0
therefore,
1
bn = √
n2 − c2
µ
2
can +
π
Z
¶
π
g(x) sin(nx)dx .
0
The solution of utt + 2cut = uxx , u(0, t) = u(π, t) = 0, for all t > 0 with
the given initial conditions is
∞
h
i
X
√
√
u(x, t) =
sin(nx)e−ct an cos( n2 − c2 t) + bn sin( n2 − c2 t) ,
2
an =
π
n=1
Z
π
0
1
f (x) sin(nx)dx, bn = √
2
n − c2
µ
2
can +
π
Z
¶
π
g(x) sin(nx)dx
0
Example 1 We will redo the “plucked” string problem from Lecture 15, letting
½ 2
x
0 ≤ x < π2
π
f (x) =
2
− π (x − π) π2 ≤ x ≤ π
and g(x) ≡ 0.
2
A graph of u(x, t) at t = 0, π8 , π4 , · · · π, and a 3-dimensional plot of u(x, t)
for 0 ≤ t ≤ 2π are shown in Figure 1.
It can be seen that with c equal to 0.2, the string moves down and back once
as t goes from 0 to approximately 2π. However the solution is no longer a sum
of exact harmonics; that is, the sines and cosines in successive terms do not
have frequencies that are exact integral multiples of a fundamental frequency.
In addition, the oscillations in the string damp out to zero as t → ∞.
1
0.8
t =πι/4
0.6
0.8
0.4
0
–0.2
0.4
t =πι/2
0.2
0.5
1
1.5
x
0
2
2.5
–0.4
0
3
–0.4
1
2
3
t 4
5
6
3
1
1.5
x
2
2.5
0
0.5
Figure 1: (a) Position of the string at Figure 1: (b) A 3-dimensional plot of
times t = 0, π8 , π4 · · · π
the function u(x, t) over 1 period
D’Alembert’s Solution of the Wave Equation
To solve the wave equation utt = β 2 uxx on the infinite line −∞ < x <
∞, t > 0, it is easier to make the change of independent variables
r = x + βt, s = x − βt,
and let u(x, t) = w(r, s) = w(r(x, t), s(x, t)).
To find uxx and utt in terms of the new variables r and s, we will need the
four derivatives
∂s
∂r
∂s
∂r
= 1,
= 1,
= β,
= −β.
∂x
∂x
∂t
∂t
Then, using the Chain Rule for differentiation of a function of a function
ut ≡
∂u
∂w ∂r ∂w ∂s
=
+
= β(wr − ws )
∂t
∂r ∂t
∂s ∂t
3
and
∂
∂r
∂s
∂r
∂s
(β(wr − ws )) = β(wrr
+ wrs − wsr
− wss )
∂t
∂t
∂t
∂t
∂t
2
= β(βwrr − βwrs − βwsr + βwss ) = β (wrr − 2wrs + wss ).
utt =
Differentiating with respect to x,
ux ≡
∂w ∂r ∂w ∂s
∂u
=
+
= wr + ws
∂x
∂r ∂x
∂s ∂x
and
uxx =
∂
∂r
∂s
∂r
∂s
(wr + ws ) = wrr
+ wrs
+ wsr
+ wss
= wrr + 2wrs + wss .
∂x
∂x
∂x
∂x
∂x
If the formulas for utt and uxx are substituted into the wave equation
utt = β 2 uxx ,
β 2 (wrr − 2wrs + wss ) = β 2 (wrr + 2wrs + wss )
and cancelling like terms on each side gives 4β 2 wrs = 0. Since β is a non-zero
constant, this implies that
wrs (r, s) ≡ 0.
This is easily solved by integrating twice, first with respect to s:
wr (r, s) = p(r)
where p is an arbitrary function of r (since for any such function
second integration with respect to r gives:
Z
w(r, s) = p(r)dr + G(s),
∂p
∂s
= 0). A
where G is an arbitrary function of s. Now we can write w(r, s) = F (r)+G(s),
and therefore
u(x, t) = F (x + βt) + G(x − βt)
where F and G can be any functions of a single variable. This is the general
solution of the wave equation on the infinite line.
Now suppose we are given initial conditions u(x, 0) = f (x), ut (x, 0) =
g(x). The partial of u with respect to t is
ut (x, t) =
∂
(F (x + βt) + G(x − βt)) = F 0 (x + βt) · β + G0 (x − βt) · (−β).
∂t
4
Remember that F and G are functions of one variable. At t = 0 we have
u(x, 0) = F (x) + G(x) ≡ f (x)
(1)
The initial velocity condition gives
ut (x, 0) = βF 0 (x) − βG0 (x) = β
d
(F (x) − G(x)) ≡ g(x).
dx
Integrating the last equality,
1
F (x) − G(x) =
β
Z
x
g(τ )dτ + C.
(2)
0
If we add equations (1) and (2) we get
µ
¶
Z
Z
1
1 x
1 x
2F (x) = f (x) +
g(τ )dτ + C ⇒ F (x) =
f (x) +
g(τ )dτ + C .
β 0
2
β 0
Subtracting equations (1) and (2) gives
µ
¶
Z
Z
1
1 x
1 x
g(τ )dτ − C ⇒ G(x) =
f (x) −
g(τ )dτ − C .
2G(x) = f (x) −
β 0
2
β 0
Using these formulas for F (x) and G(x), the function u can be written as
u(x, t) = F (x + βt) + G(x − βt)
¶
¶
µ
µ
Z
Z
1
1
1 x+βt
1 x−βt
=
g(τ )dτ + C +
g(τ )dτ − C ,
f (x + βt) +
f (x − βt) −
2
β 0
2
β 0
and the exact analytic solution of the heat equation is
1
1
u(x, t) = (f (x + βt) + f (x − βt)) +
2
2β
Z
x+βt
g(τ )dτ.
x−βt
Example 2 To study the behavior of a wave on an infinite line, we will assume
the initial position is given by
½
1 −1 ≤ x ≤ 1
u(x, 0) ≡ f (x) =
0 otherwise
and the initial velocity g(x) is identically zero. Let the constant velocity β = 1.
5
1
0.8
1
0.8
0.6
0.4
0.2
0
0
0.5
0.6
0.4
0.2
–4
–2
0
2
x
4
Figure 1: Graph of u at time t = 0.5
–4
1
1.5
t
2
2.5
2
3
–2
0
x
4
Figure 2: 3-dimensional plot of u
In Figure (2), the position of the wave at time t = 0.5 is shown on the left,
and the 3-dimensional plot shows the wave moving out along the infinite line.
(x−t̄)
At any time t̄ the function u(x, t̄) is equal to f (x+t̄)+f
. This can be graphed
2
by moving the graph of f t̄ units to the right and t̄ units to the left, adding the
two functions and dividing by two. When t > 1 the two graphs are completely
separated, and the graph of u consists of two copies of f (t)/2 moving to the
right and left with speed determined by β.
Practice Problems:
1. * Find the solution of utt + 0.4ut = uxx if u(x, 0) ≡ 0 and

0 ≤ x < 0.4π
 0
0.2 0.4π ≤ x ≤ 0.6π
ut (x, 0) =

0
0.6π < x ≤ π
This is the damped version of the piano string, struck at time t = 0.
2. * Solve the equation utt = β 2 uxx on the infinite line, given the initial
condition u(x, 0) = sin(x), ut (x, 0) ≡ 0. Describe the behavior of u(x, t)
for t > 0. This is a model of a standing wave with no boundaries. How
does the value of β affect the wave?
3. * Use MAPLE to make a 3-dimensional plot of the solution of Problem
#2, for −20 ≤ x ≤ 20, 0 ≤ t ≤ 6π. Do the plot for β = 1.0 and for
β = 0.5.
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