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General Solution using Normal Modes
Overview and Motivation : Last time we solved the initial-value problem (IVP) for
the 1D wave equation on a finite domain with "closed-closed" bc's using the general
form of the solution q ( x , t ) = f ( x + ct ) + g ( x − ct ). Today we solve the same problem
using the normal mode solutions for this system.
Key Mathematics : We utilize some integrations involving harmonic functions.
I. The Problem Defined
We are looking for the general solution to the wave equation
( ) ( )
2
2 2 2
2 , ,
x
qxt c t
qxt
on the finite domain 0 ≤ x ≤ L subject to the initial conditions
q^ (^ x , 0 )^ = a ( ) x and (^ x )^ b ( ) x t
q
∂
, 0 (2a), (2b)
and the boundary conditions
q ( 0 , t ) = 0 and q ( L , t ) = 0. (3a), (3b)
This time we write the solution q ( x , t )as linear combination
1
of the normal-mode
solutions qn ( x , t )
( ) ( ) ∑ (^ )
∞
=
1
n
q xt qn xt , (4)
where the normal modes can be expressed as
( ) (^ )^ (^ )(^ )
i t n
i t n n n
n n q xt k x Ae Be
ω − ω
, = sin +. (5)
Here k n = n π L is the wave vector and ω n = ckn is the angular frequency. As before,
let's make the normal-mode solutions explicitly real by setting
Bn = A n. Then Eq. (5)
can be written as
1 As written, Eq. (4) looks like a simple sum, not a linear combination, but as Eq. (5) shows, we have kept
undetermined amplitudes as part of our normal modes, so Eq. (4) may be justifiably thought of as a linear
combination of normal modes.
q ( (^) n ) ( x , t ) = sin( knx ) [Re ( an ) cos( ω n t ) −Im( an ) sin( ω (^) nt )], (6)
where we have defined a new amplitude an = An 2. Using Eq. (6) we may thus
express the general solution [via Eq. (4)] as
( ) ( ) [ ( ) ( ) ( ) ( )] ∑
∞
=
1
, sin Re cos Im sin
n
qxt kn x an ω nt an ω nt , (7)
As we shall shortly see, the amplitudes an are determined by the initial conditions.
Recall, however, that normal modes already satisfy the bc's, and so the general
solution as expressed by Eq. (7) automatically satisfies those bc's. We thus need not
consider the bc's any further.
II. The Initial Value Problem (Yet Again!)
Let's now apply the initial conditions and see what we get. From Eq. (7) we obtain
for the initial displacement
( ) ( ) ( ) ∑
∞
=
1
sin Re
n
L n
n ax x a
π
and for the initial velocity, we obtain after differentiating Eq. (7)
( ) ( ) ( ) ∑
∞
=
1
sin Im
n
L n
n bx n x a
π
In Eqs. (8) and (9) we have used k n = n π L. So what do we have here? Well, perhaps
not surprisingly, we have two equations for the amplitudes an in terms of the initial
conditions. However, unlike the (finite) N -oscillator case, the rhs's of Eqs. (8) and (9)
have an infinite number of amplitudes because the wave equation has an infinite
number of normal modes!
Aside: The N -oscillator problem
This looks pretty grim, but perhaps a look back at the N -oscillator case will give us
some insight into the current problem. For the N oscillator problem the equation
equivalent to Eq. (8) is the extension of Eq. (19) from the Lecture (6) notes to N
oscillators, which we can write as
( )
( ( ) ( ) ( ) ( ))
( )
( )
( )
( )
( ( ) ( ) ( ) ( ))
( )
( )
( )
( )
N
N
q
q
q
q
N
a
N
m
N
m
N
m
N
m
N
m N
m N
m N
m
N
N
m N
m N
m N
m
m
1
1
1
1
1 1 1 1
3
2
1
1 1 1 1
sin
sin 3
sin 2
sin 1
sin 1 sin 2 sin 3 sin
sin 1 sin 2 sin 3 sin
Re
π
π
π
π
π π π π
π π π π
M
K
M
K
The key point here is that multiplying Eq. (10) by the m th eigenvector allows us to
find Re( am ) in terms of the initial condition on the displacement of the system.
Notice that Eq. (10) is an equation for the initial displacement of the system in terms
of the set of coefficients Re( an ), while Eq. (13) is an equation for any coefficient
Re ( am )(labeled by m ) in terms of the initial displacement of the system. We can thus
think of Eq. (13) as the inversion of Eq. (10).
Now the notation used in Eqs. (10) – (13) is rather cumbersome. Fortunately there is
a more succinct way to express these equations. Looking at Eq. (10) we first note that
the j th element of that equation can be written as
( ) (^) ∑ ( ) ( )
−
N
n
N n
n qj j a
1
1
0 sin Re
π
We now notice that if we multiply this equation by ( j ) N
m 1
sin
π
and them sum the
equation on j ,
( ) ( ) ( ) ( ) ( (^) n )
N
j
N
n
N
n N
m
N
j
N j
m sin j q 0 sin j sin j Re a
1 1
1 1 1
∑ 1 ∑ ∑ = =
=
π π π
then this is indeed multiplication of Eq. (10) by the transpose of the m th eigenvector!
That is, Eq. (15) is the same equation as Eq. (11), only in a much more succinct form.
We now switch the order of the sums on the rhs
( ) ( ) ( ) ( ) ( ) ∑ ∑ ∑ = =
=
N
n
N
j
N
n N
m n
N
j
N j
m j q a j j
1 1
1 1 1
sin 1 0 Re sin sin
π π π
and notice that the sum on j on the rhs is simply the product of the (transpose of)
m th eigenvector with the n th eigenvector, which is nonzero only if they are the same
eigenvector. That is, the sum on j on the rhs is nonzero only if n = m. Thus Eq. (16)
simplifies to
=
=
N
j
N
m m
N
j
N j
m j q a j
1
1
2
1
1
sin 0 Re sin
π π
This is equivalent to Eq. (12), only in much more elegant notation. As with Eq. (12)
we now only have one amplitude, Re( am ), on the rhs. We can thus solve Eq. (12) for
this amplitude, obtaining
( )
=
=
N
j
N
m
N
j
N j
m
m
j
j q
a
1
1
2
1
1
sin
sin 0
Re
π
π
[which is equivalent to Eq. (13)].
Back to the wave-equation IVP
So how does this apply to the current problem? Well, the main difference between
the two problems is that x is a continuous variable while j is discrete. But if we
remember a bit of calculus, we should recall that an integral is really just a sum over a
continuous variable. We thus might expect the sums over j in Eq. (15) to be
replaced by integrals over x. Let's try it and wee what happens. Multiplying Eq. (8)
by ( L x )
m π
sin and integrating on x from 0 to L gives us
∞
=
L
n
L n
n L
m
L
L
m xax dx x x a dx
0 0 1
sin sin sin Re
π π π
and switching the order of integration and summation on the rhs produces
∫ ∑^ ∫
∞
=
0 1 0
sin Re sin sin
n
L
L
n L
m n
L
L
m xax dx a x x dx
π π π
normally use the variable n to label the amplitudes and write these last two equations
as
( ) ( ) ( ) ∫
L
L
n n xax dx L
a
0
sin
Re
π
and
( ) ( ) ( ) ∫
L
L
n
n
n
xbx dx L
a
0
sin
Im
π
ω
Summarizing, Eq. (7), which can be written as
( ) ( ) [ ( ) ( ) ( ) ( )] ∑
∞
=
1
, sin Re cos Im sin
n
L
n L n
n L n
n qxt x a c t a c t
π π π
along with Eqs. (26) and (27) are the complete solution to the initial value problem on
the finite domain 0 ≤ x ≤ L for the bc's q ( 0 , t ) = 0 and q ( L , t ) = 0.
III. An Example Revisited
Let's consider the problem that we solved in the Lecture 9 Notes. Looking at this
problem again will give us a good comparison of the two methods of solving the
initial value problem.
The problem in the Lecture 9 notes has the initial conditions
2
( )
( ) ( )
2 2 2 exp
exp
x L L
a x A (29a)
and
b ( ) x = 0. (29b)
(^2) OK, so we now have another amplitude A , but it is not the same as the amplitudes An ( n = 1 , 2 ,...) used
earlier in Eq. (5).
Recall that a ( ) x is a Gaussian peak that is (vertically) shifted so that the bc's are
satisfied. The following figure plots a ( x )for the same values of the width parameter
σ that we investigated in Lecture 9: σ = 0. 05 , σ = 0. 2 , and σ = 0. 5.
The easy part of this particular problem is solving for Im( an ). Using Eq. (27) we
immediately see that Im( an ) = 0. Similarly (but not as simply!), using Eq. (26) we see
that Re( an )is given by
( ) ( )
( ) ( ) ( ) ∫
−
L x L L L
n n x dx L
A
a
0
2 2 22 sin exp exp
Re σ σ
π
Unfortunately, the integral in Eq. (30) has no analytic solution.
3
Fortunately, a
program such as Mathcad can numerically solve the integral. Unfortunately, as Eq.
(28) indicates, we need an infinite number of an 's! Fortunately, in most cases we only
need to use a finite number of the an 's in order to get a very good approximation to
the exact solution. That is, in practice we typically use a truncated version of Eq. (28),
which we can write as
( ) ( ) [ ( ) ( ) ( ) ( )] ∑ =
M
n
L
n L n
n L n
n M
qxt x a c t a c t
1
, sin Re cos Im sin
π π π
where q ( x , t ) M is the M -term approximation to q ( x , t ). For the example at hand
Im ( an ) = 0 so we have
(^3) At least as far as I know! Actually, it is not too difficult to show that = 0 an if n (^) is even. But that still
leaves the odd values of n to deal with.
0 0.2 0.4 0.6 0.
0
1 sigma = 0.
sigma = 0.
sigma = 0.
sigma = 0.
sigma = 0.
sigma = 0.
x
a(x)
where we have defined a ( ) xM (^) = q ( x , 0 ) M. That M ≈ 29 does a good job for σ = 0. 05 is
illustrated in the following figure, which plots ax ( x ) M for several values of M. Notice
that as M increases ax ( ) xM more faithfully represents the function a ( ) x.
The next 2 figures show the same sort of thing for σ = 0. 2 and σ = 0. 5 , but in a more
direct fashion. This figures on the left plots the difference a ( x ) − a ( ) x 5 for σ = 0. 2 ,
while the figure on the right plots a ( x ) − a ( ) x 3 for a 0 (^) = 0. 5. In both cases, the error is
< 0. 06 for all values of x. Given the overall size of a ( x ), these also seems like
reasonable approximations.
0 0.2 0.4 0.6 0.
0
1
2
M = 3
M = 9
M = 19
M = 29
M = 3
M = 9
M = 19
M = 29
sigma = 0.
POSITION
a(x)M
0 0.5 1
0
sigma = 0.
POSITION
a(x) - a(x)
0 0.5 1
0
sigma = 0.
POSITION
a(x) - a(x)
The following two graphs are also somewhat illuminating. They plot the initial
condition a ( ) x along with the individual terms on the rhs of Eq. (27), ( L ) ( n )
n sin x Re a
π
for n = 1 ,..., M. From the graph for σ = 0. 5 it is not hard to imagine that only the n = 1
and n = 3 are needed to accurately describe a ( x ).
OK, so what about the time dependence? Now that we know how many normal
modes we need, we can simply use Eq. (26) [where Re( an )is calculated with Eq. (24)]
with the appropriate value of M. On the class web site there are videos of the
resulting wave motion for all three values of σ discussed here. Both the σ = 0. 2 and
σ = 0. 5 videos also show the motion of the individual normal modes that are used to
produce the approximation q ( x , t ) M.
Exercises
- 10.1 Show that Re( a (^) n ) cos( ω (^) n t ) − Im( an ) sin( ω nt )can be expressed more succinctly as
i t n
n ae
ω
Re. Thus Eq. (6) can be alternatively expressed as ( ) ( ) ( ) ( )
i t n n n
n q xt k x ae
ω
, = sin Re.
* 10.2 In the notes it is stated that ( ) ( )
L
L
n L
m x x dx
0
sin sin
π π
is nonzero only if m = n
(where m and n are both integers). Using the trig identities for cos( x + y )and
cos ( x − y ), do this integral and show that this statement is indeed true.
0 0.5 1
0
1
a(x)
n = 1
n = 3
n = 5
a(x)
n = 1
n = 3
n = 5
sigma = 0.
POSITION
a(x), 3 Normal Modes
0 0.5 1
0
a(x)
n = 1
n = 3
a(x)
n = 1
n = 3
sigma = 0.
POSITION
a(x), 2 Normal Modes
