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Using Everyday Examples in Engineering (E3)
Fourier Series/Partial Differential Equations: What is the Shape of a Vibrating String? 1
Bernd Schroeder
Louisiana Tech University
Photo Credit
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© Loren M. Winters, used with permission.
Adapted from http://www.wiley.com/WileyCDA/WileyTitle/productCd-0470447516.html by Bernd
Schroeder
© 2012 Bernd Schroeder. All rights reserved. Copies may be downloaded from
www.EngageEngineering.org. This material may be reproduced for educational purposes.
.
This material is based upon work supported by the National Science Foundation (NSF) under Grant No. 083306. Any
opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not
necessarily reflect the views of NSF.
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What is the shape of a vibrating string?
Adapted from
http://www.wiley.com/WileyCDA/WileyTitle/productCd-0470447516.html.
Where it fits. After Fourier series in a calculus class, as an extension/application. The Fourier coefficients could have been computed in earlier examples or exercises.
The separation of variables could also be used to motivate the idea of
representing functions with Fourier series. In this approach, the partial differential equation could be separated first, then Fourier series would be discussed, then the problem could be solved completely. (So the whole solution
would stretch over several classes.)
It helps if students have seen the solutions of linear ordinary differential
equations with constant coefficients, but it’s not mandatory. For most people
who see separation of variables for the first time, parts will feel like magic
or cheating. The procedure simply is intricate, even if experienced people
might not see the intricacy any more.
Another place where the topic fits is a module on partial differential
equations in a class on ordinary differential equations. It’s a perfect example
for separation of variables. Again, the topic could be used as an example
or as a bracket where the problem motivates the method and the solution is
given at the end.
Finally, of course, the topic fits into a class on partial differential equations.
Introductory advice to students. This computation is long and will
feel messy to sophomores who are accustomed to problems having “clean”
solutions like “2,” “π/2,” or “sin(25x).” It can help to note that the beauty
of our solution of the wave equation is not that the solution is “simple” like
solutions we have seen in calculus. For a problem like this, the beauty lies
in the fact that we can find a solution at all.
Model. The one-dimensional wave equation, which governs the motion
of a vibrating string under mild conditions, is
∂2
∂2
u(x,
t)
=
u(x, t).
∂t2
∂x2
We will solve this equation on the domain 0 ≤ x ≤ π and for t ≥ 0 with
∂
initial conditions u(x, 0) = f (x), u(x, 0) = 0 for 0 ≤ x ≤ π and boundary
∂t
condition u(0, t) = u(π, t) = 0 for t > 0.
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This means we consider a string that is fixed at both ends, which are at 0
and at π, and which is initially at rest. We will specify an initial shape f (x)
later.
(Deriving the wave equation would be another class period altogether.)
Solution, part 1. Separation of variables. To perform the solution
with separation of variables, we assume there are functions X(x) and T (t) so
that a solution of the form u(x, t) = X(x)T (t) exists. If u(x, t) = X(x)T (t),
then
∂
u(x, t)
∂t
∂2
u(x, t)
∂t2
∂
u(x, t)
∂x
∂2
u(x, t)
∂x2
∂
X(x)T (t) = X(x)T ′ (t)
∂t
∂2
=
X(x)T (t) = X(x)T ′′ (t)
∂t2
∂
=
X(x)T (t) = X ′ (x)T (t)
∂x
∂2
=
X(x)T (t) = X ′′ (x)T (t).
∂x2
=
Substituting into the equation gives
∂2
∂2
u(x,
t)
=
u(x, t)
∂t2
∂x2
X(x)T ′′ (t) = X ′′ (x)T (t).
We divide by X(x)T (t) and
T ′′ (t)
X ′′ (x)
=
=λ
T (t)
X(x)
separates into two ordinary differential equations
T ′′ (t) − λT (t) = 0
and
X ′′ (x) − λX(x) = 0.
Solution, part 2. Solving the space part. Both equations are of the
same form. We will first solve X ′′ − λX = 0. The solution for the other
equation is the same. Setting up X(x) = ecx gives the following.
X ′′ − λX = 0
c(2 ecx − )λecx = 0
c2 − λ ecx = 0
c2 = λ.
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Because we do not know the value of λ, we must analyze the solutions
for all possible values of λ.
For λ = 0, the solution is a constant function. This is neither very
interesting (we know that constant functions solve the wave equation) nor
helpful with this particular problem. (This is why constant solutions are
often ignored in separation
of variables.)
√
For λ ̸= 0, let µ := |λ|. For λ > 0 the general solution of the differential
equation X ′′ − µ2 X = 0 is X(x) = k1 eµx + k2 e−µx and for λ < 0 the general
solution of the differential equation X ′′ + µ2 X = 0 is X(x) = k1 cos(µx) +
k2 sin(µx). Although we introduce more notation here,√this is a small price
to pay. We would rather not carry a term of the form |λ| through the rest
of the computation.
The initial condition states that the function u(x, 0) = X(x)T (0) is not
constant. Thus X(x) is not constant and hence λ ̸= 0. The boundary
conditions 0 = u(0, t) = X(0)T (t) and 0 = u(π, t) = X(π)T (t) dictate that
X(0) = X(π) = 0 [because T (t) must be nonzero for some time t].
In case λ > 0 we would have X(x) = k1 eµx + k2 e−µx and hence
0 = X(0) = k1 + k2
0 = X(π) = k1 eµπ + k2 e−µπ
(
)
0 = k1 eµπ − e−µπ ,
which forces k1 = k2 = 0 and then X(x) = 0, which cannot be.
Thus we must have that λ = −µ2 < 0 and X(x) = k1 cos(µx)+k2 sin(µx).
We can quickly see that 0 = X(0) = k1 shows that k1 is zero. Hence k2
is not equal to zero. The only way we can have 0 = X(π) = k2 sin(µπ) is if
µ is a positive integer n.
Hence
X(x) = c sin(nx)
for some integer n and some real number c.
Solution, part 3. Solving the time part and combining the solutions. From T ′′ − λT = 0 and λ = −n2 , we obtain T ′′ + n2 T = 0. The
general solution of this equation is T (t) = c1 cos(nt) + c2 sin(nt). Moreover, T ′ (0) = 0 because the string is initially at rest. But 0 = T ′ (0) =
−c1 n sin(n·0)+c2 n cos(n·0) = c2 n gives us that c2 = 0. Hence T = c1 cos(nt).
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But then any solution u(x, t) = X(x)T (t) must be a multiple of the function
un (x, t) = sin(nx) cos(nt).
Each un is equal to sin(nx) at t = 0 and sums of constant multiples of
functions un will solve the wave equation. Hence any function (any initial
shape) that can be represented as a Fourier series with sine terms only can
be approximated with the functions un (x, 0).
Solution, part 4. Enter
Fourier
series. Now let’s consider the specific
π π initial shape f (x) = − x − for 0 ≤ x ≤ π. This would be the initial
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2
shape of a plucked string, see Figure 1.
Figure 1: (Exaggerated) shape of a plucked string.
This function is not of the form c sin(nx), so we must represent it using functions
To do this, we compute the Fourier coefficients
 of this form.
 π − x − π for 0 ≤ x ≤ π,
2
Basically, by extending f in this
of f (x) = 2
 x + π − π for −π ≤ x ≤ 0.
2
2
fashion, we make sure that the function that we are interested in (the part
that exists on [0, π]) can be represented with sine functions (see Figure 2).
Because f is odd, all products f (x) cos(nx) are odd functions. Because
the integral of an odd function over [−π, π] is zero, we only need to compute
the Fourier sine coefficients of f .
For the Fourier sine coefficients, we obtain the following.
bn =
1∫π
f (x) sin(nx) dx
π −π
The product f (x) sin(nx) is even, which means that the
integral over [−π, π] is twice the integral over [0, π].
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Figure 2: An extension of the original shape that can be represented with
sine functions.
(
)
(
)
π
2 ∫ π π =
− x − sin(nx) dx
π 0 2
2
Now we use another symmetry property. On [0, π], the
function f is symmetric with respect to the vertical line
π
x = . Therefore the integral over [0, π] will vanish for
2
even n (because the product is “odd with respect to π/2”)
and
the
] integral over [0, π] will be twice the integral over
[
π
for odd n (because the product is “even with re0,
2
spect to π/2”).
b2k = 0
4
b2k+1 =
π
4
=
π
4
=
π
4
=
π
∫
π
2
0
∫
π
2
0
∫
π
2
π π
− x − sin ((2k + 1)x) dx
2
2
(
(
))
π
π
−
− x sin ((2k + 1)x) dx
2
2
x sin ((2k + 1)x) dx
0
[
]π
2
1
1
−x
cos ((2k + 1)x) +
sin
((2k
+
1)x)
2k + 1
(2k + 1)2
0
(
)
(
)
π
4
π
2
1
cos (2k + 1)
+
sin (2k + 1)
= −
2k + 1
2
π (2k + 1)2
2
k
4 (−1)
=
.
π (2k + 1)2
Solution, part 5. Constructing the solution. With these coefficients
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and the un , we should be able to construct a function
u(x, t) =
∞
∑
bn un (x, t)
n=1
that is equal to the triangular shape at t = 0 (because at t = 0 it’s the
function’s Fourier series), that has time derivative zero at t = 0 (because
the un do), that is zero at x = 0 and at x = π (because the un satisfy
this condition) and that is a solution of the wave equation (because the un
are). We are omitting some technical concerns (related to the problems with
derivatives of Fourier series). As the final answer, we can record that the
function
u(x, t) :=
∞
∑
4
(−1)k
sin ((2k + 1)x) cos ((2k + 1)t)
2
k=0 π (2k + 1)
is a solution of the partial differential equation with the given initial and
boundary values.
Checking the solution. Now the fun starts, because we can compare
the predicted shapes with high-speed camera images.
Figure 3: Some predicted shapes of a plucked string according to the onedimensional wave equation.
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Figure 3 shows the snapshots of the solution for t = 0, , 1, . Our com2
2
puted solution predicts that the string will flatten in the middle and only
the flat part in the middle of the string will move towards the equilibrium
position. In particular, a particle located at horizontal position x on the
sloping sides will not move until the “traveling corner” of the flat part has
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reached x. At that time, the string will flatten out at x, and the particle will
start traveling towards the equilibrium position.
Figure 3 omits the continuation of the motion below the x-axis, but nothing new happens there. The center part of the string remains flat and pushes
back out until the mirror image of the original triangle is produced below the
axis. Then the reverse motion starts until we are back at the original shape.
In the model, this motion continues indefinitely, because we have assumed
that there is no friction and hence no energy loss. The graphs in Figure 3
were produced with a sum in which k ran from 0 to 34. This explains why the
corner at the top of the triangle still is slightly rounded. Fourier polynomials
are differentiable, so they do not have sharp corners. At the same time, with
a sufficiently high order Fourier polynomial, the top of the triangle will look
very much like a sharp corner.
The above finally gives the solution to a mathematical problem that is
supposed to describe a plucked string. But as we look at Figure 3, the result
seems counterintuitive. We don’t see oscillating strings retain corners in their
oscillations.
It would be a little too fast to dismiss the model, though. The wave
equation we set up assumes that there is no friction, that the oscillations are
small and that the string’s particles move vertically. A real string experiences
friction and the particles of the string do not move exactly vertically. But
for a short period of time, the effect of friction as well as of the problems
with the other modeling assumptions should be negligible. Unfortunately,
our eyes are not fast enough to observe the extremely short time in which
the first oscillation takes place.
Figure 4: Two-flash photograph of a plucked string, showing the initial shape
c Loren M. Winters, used
and one snapshot of the moving string. Image ⃝
with permission.
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Figure 5: Multiflash photograph of a plucked string, showing multiple snapshots as the string goes through a half cycle from being released on the
c Loren
bottom to almost reproducing its original shape at the top. Image ⃝
M. Winters, used with permission.
A high speed camera can see what the eye cannot. Consider Figure 4,
which shows the initial shape of a plucked string and, via double exposure,
the shape of the string shortly after it was released. Figure 4 shows what the
model predicted! Now consider Figure 5, which shows multiple exposures of
the early motion of a plucked string. This time the string is released from
below the equilibrium position, but that would not change the prediction of
the model (except that everything is flipped over the x-axis). We can see
that until the string reaches the equilibrium, the prediction of our model is
quite accurate. Even after that, the string retains a shape that is made up of
several straight lines. So our model is, as expected, accurate for a little while
(here about one quarter of a period). Once the effects of friction (against
the surrounding air molecules) as well as the violations of the other modeling
assumptions manifest themselves, the model’s prediction becomes less and
less accurate (even though it’s still pretty good). As long as we accept that
our modeling assumptions limit the model’s range of validity, we can say that
the results have been experimentally verified.
Overall, we have here a complicated problem with a surprising answer
that turns out to be not just mathematically correct, but experimentally
verifiable.
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