Applications of Fourier series
5.2 Partial differential equations
5.2.2 The wave equation
We now consider the example of a vibrating string of length L and having fixed ends.
Just as in the heat conduction problem, we will not discuss the physical arguments needed to derive the wave equation.
The equation describing the vertical displacement u(x, t) of a vibrating string is EXAMPLE 5.9
utt = a2ux x for 0< x < L, t > 0. (5.17) Here a is a constant which is related to the tension in the string. This equation is called the wave equation. Since the ends of the string are fixed, one has the following Wave equation
boundary conditions:
u(0, t) = 0, u(L, t) = 0 for t≥ 0. (5.18)
FIGURE 5.5
Temperature distribution in a rod with insulated ends.
0
t u
x L u(x, 0)
FIGURE 5.6
Displacement of a vibrating string.
In figure 5.6 we see the displacement of the string at time t= 0. We thus have the initial condition
u(x, 0) = f (x) for 0≤ x ≤ L. (5.19)
Moreover, it is given that at t= 0 the string has no initial velocity. So as a second initial condition we have
ut(x, 0) = 0 for 0≤ x ≤ L. (5.20)
Analogous to the heat conduction problem in a rod we will derive a formal solution in three steps.
Separation of variables Step 1
We first determine fundamental solutions of the form u(x, t) = X(x)T (t) that sat-isfy the partial differential equation and the linear homogeneous conditions. Substi-tution of u(x, t) = X(x)T (t) into (5.17) gives XT= a2XT . Dividing by a2X T leads to
T
a2T = X
X = c,
where c is the constant of separation. From this we obtain the two ordinary differ-ential equations
X− cX = 0, T− ca2T = 0. (5.21)
Conditions (5.18) and (5.20) are linear homogeneous conditions. Substitution of u(x, t) = X (x)T (t) in condition (5.18), and using that u(x, t) should not be the trivial solution, leads to the conditions X(0) = 0 and X (L) = 0. Subsequently substituting u(x, t) = X (x)T (t) in condition (5.20) gives the relation T(0)X (x) = 0 for 0≤ x ≤ L. Hence T(0) = 0.
Calculating eigenvalues and eigenfunctions Step 2
For the function X(x) we derived in step 1 the following differential equation with boundary conditions:
X− cX = 0 for 0< x < L, X(0) = 0, X(L) = 0.
So here we again find the eigenvalues c = −(nπ/L)2for n = 1, 2, . . . and the corresponding eigenfunctions
Xn(x) = sin(nπx/L) for n= 1, 2, 3, . . ..
In order to find the fundamental solutions we still have to determine T(t). From the eigenvalues we have found, we obtain from (5.21) the differential equation
T+ n2π2a2 L2 T = 0.
Its general solution is
T(t) = α cos(nπat/L) + β sin(nπat/L), which has as derivative
T(t) = nπα
L (−α sin(nπat/L) + β cos(nπat/L)) .
Substitution of the condition T(0) = 0 gives β = 0 and so we obtain for the eigenvalue−(nπ/L)2the following fundamental solution:
Tn(t) = cos(nπat/L) for n= 1, 2, 3, . . ..
We have thus found the following fundamental solutions:
un(x, t) = Tn(t)Xn(x) = cos(nπat/L) sin(nπx/L) for n= 1, 2, 3, . . ..
Superposition Step 3
Superposition of the fundamental solutions gives the series u(x, t) =∞
n=1
Ancos(nπat/L) sin(nπx/L). (5.22)
By substituting the remaining initial condition (5.19), a Fourier series arises:
u(x, 0) = f (x) =∞ n=1
Ansin(nπx/L) for 0≤ x ≤ L.
For the Fourier coefficients Anone thus has An= 2
L
L 0
f(x) sin(nπx/L) dx.
The series (5.22) together with these coefficients give the formal solution of our problem. One can indeed show that the formal solution thus obtained is the unique
solution, provided that f(x) is piecewise smooth.
In many cases the assumption of an initial velocity ut(x, 0) = 0 is artificial. This assumption made it possible for us to find a simple solution for Tn(t). For a string which is struck from its resting position, one takes as initial conditions u(x, 0) = 0 and ut(x, 0) = g(x). When both the initial displacement and the initial velocity are unequal to 0, then we are dealing with a problem with two inhomogeneous con-ditions. As a consequence, the functions Tn(t) will contain sine as well as cosine terms. We must then determine the coefficients in two distinct Fourier series. For detailed results we refer to Fourier series, transforms and boundary value problems by J. Hanna and J.H. Rowland, pages 228 – 233. In the same book, pages 219 – 227, one can also find the derivation of the wave equation and the heat equation, as well as the verification that the formal solutions constructed above are indeed solutions, which moreover are unique.
EXERCISES
A thin rod of length L with insulated sides has its ends kept at 0◦C. The initial 5.7
temperature is u(x, 0) =
x for 0≤ x ≤ L/2, L− x for L/2 ≤ x ≤ L.
Show that the temperature u(x, t) is given by the series
u(x, t) = 4L π2
∞ n=0
(−1)n
(2n + 1)2e−(2n+1)2π2kt/L2sin((2n + 1)πx/L).
Both ends and the sides of a thin rod of length L are insulated. The initial tem-5.8
perature of the rod is u(x, 0) = 3 cos(8πx/L). Write down the heat equation for this situation and determine the initial and boundary conditions. Next determine the temperature u(x, t).
For a thin rod of length L the end at x= L is kept at 0◦C, while the end at x= 0 5.9
is insulated (as well as the sides). The initial temperature of the rod is u(x, 0) = 7 cos(5πx/2L). Write down the heat equation for this situation and determine the initial and boundary conditions. Next determine the temperature u(x, t).
Determine the solution of the following initial and boundary value problem:
5.10
ut = ux x for 0< x < 2, t > 0, ux(0, t) = 0, u(2, t) = 0 for t ≥ 0,
u(x, 0) =
1 for 0< x < 1, 2− x for 1 ≤ x < 2.
A thin rod of length L has initial temperature u(x, 0) = f (x). The end at x = 0 is 5.11
kept at 0◦C and the end at x= L is insulated (as well as the sides). Write down the heat equation for this situation and determine the initial and boundary conditions.
Next determine the temperature u(x, t).
Determine the displacement u(x, t) of a string of length L, with fixed ends and 5.12
initial displacement u(x, 0) = 0.05 sin(4πx/L). At time t = 0 the string has no initial velocity.
A string is attached at the points x= 0 and x = L and has as initial displacement 5.13
f(x) =
0.02x for 0< x < L/2, 0.02(L − x) for L/2 ≤ x < L.
At time t= 0 the string has no initial velocity. Write down the corresponding initial and boundary value problem and determine the solution. One could call this the problem of the ‘plucked string’: the initial position is unequal to 0 and the string is pulled at the point x= L/2, while the initial velocity is equal to 0.
A string is attached at the points x = 0 and x = 2 and has as initial displacement 5.14
u(x, 0) = 0. The initial velocity is
ut(x, 0) = g(x) =
0.05x for 0< x < 1, 0.05(2 − x) for 1 < x < 2.
Write down the corresponding initial and boundary value problem and determine the solution. This problem could be called the problem of the ‘struck string’: the initial position is equal to 0, while the initial velocity is unequal to 0, and the string is struck at the midpoint.
Determine the solution of the following initial and boundary value problem, where 5.15
k is a constant:
ut = a2ux x for 0< x < π, t > 0, ux(0, t) = 0, ux(π, t) = 0 for t > 0,
ut(x, 0) = 0, u(x, 0) = kx for 0 < x < π.
S U M M A R Y
In this chapter Fourier series were first applied to determine the response of anLTC -system to a periodic input. Here the frequency response, introduced in chapter 1, played a central role. It determines the response to a time-harmonic input. Since the input can be represented as a superposition of time-harmonic signals, using Fourier series, one can easily determine the line spectrum of the output by applying the superposition rule. This line spectrum is obtained by multiplying the line spectrum of the input with the values of the frequency response at the integer multiples of the fundamental frequency of the input:
yn= H(nω0)un.
Here ynis the line spectrum of the output y(t), un the line spectrum of the input u(t), H(ω) the frequency response, and ω0the fundamental frequency of the input.
For real inputs and real systems the properties of the time-harmonic signals are taken over by the sinusoidal signals.
Systems occurring in practice are often described by differential equations of the form
am dmy
dtm + am−1dm−1y
dtm−1 + · · · + a0y= bn dnu
dtn + bn−1dn−1u
dtn−1 + · · · + b0u.
In order to determine a periodic solution y(t) for a given periodic signal u(t), it is important to know whether or not there are any periodic eigenfunctions. These are periodic solutions of the homogeneous differential equation arising from the dif-ferential equation above by taking the right-hand side equal to 0. Periodic eigen-functions correspond to zeros s = iω of the characteristic polynomial A(s) = amsm + am−1sm−1+ · · · + a0, and these lie on the imaginary axis. When the period of a periodic input coincides with the period of a periodic eigenfunction, then resonance may occur, that is, for a given u(t) the differential equation does not have periodic solutions, but instead unbounded solutions.
When the differential equation describes a stable system, then all zeros of A(s) lie in the left-half plane and the frequency response is then for allω equal to H(ω) = B(iω)
A(iω),
with B(s) = bnsn+bn−1sn−1+· · ·+b0. For real systems this means that there are no sinusoidal eigenfunctions, that is, no sinusoidal signals with an eigenfrequency.
Secondly, Fourier series were applied in solving the one-dimensional heat equation
ut = kux x,
and the one-dimensional wave equation utt = a2ux x.
Using the method of separation of variables, and solving an eigenvalue problem, one can obtain a collection of fundamental solutions satisfying the partial differen-tial equation under consideration, as well as the corresponding linear homogeneous conditions, but not yet the remaining inhomogeneous condition(s). By superposi-tion of the fundamental solusuperposi-tions one can usually construct a formal solusuperposi-tion which also satisfies the inhomogeneous condition(s). In most cases the formal solution is the solution of the problem being posed. In the superposition of the fundamental so-lutions lies the application of Fourier series. The fundamental soso-lutions describe, in relation to one or several variables, sinusoidal functions with frequencies which are an integer multiple of a fundamental frequency. This fundamental frequency already emerges when one calculates the eigenvalues. The superposition is then a Fourier series whose coefficients can be determined by using the remaining inhomogeneous condition(s).
S E L F T E S T
For the frequency response of anLTC-system one has 5.16
H(ω) = (1 − e−2iω)2.
a Is the response to a real periodic input real again? Justify your answer.
b Calculate the response to the input u(t) = sin(ω0t).
c What is the response to a periodic input with period 1?
For anLTC-system the relation between an input u(t) and the corresponding output 5.17
y(t) is described by the differential equation y+ 4y+ 4y = u. Let u(t) be the periodic input with period 2π, given on the interval (−π, π) by
u(t) =
πt + t2 for−π < t < 0, πt − t2 for 0< t < π.
Determine the first harmonic of the output y(t).
For anLTC-system the relation between an input u(t) and the output y(t) is de-5.18
scribed by the differential equation y+ y+ 4y+ 4y = u+ u.
a Does the differential equation determine the periodic response to a periodic in-put uniquely? Justify your answer.
b Let u(t) = cos 3t and y(t) the corresponding output. Calculate the power of the output.
A thin rod of length L has constant positive initial temperature u(x, 0) = u0for 5.19
0< x < L. The ends are kept at 0◦C. The so-called heat-flux through a cross-section of the rod at position x0(0< x0 < L) and at time t > 0 is by definition equal to−K ux(x0, t). Show that the heat-flux at the midpoint of the rod (x0= L/2) equals 0.
Consider a thin rod for which one has the following equations:
5.20
ut = kux x for 0< x < L, t > 0, u(0, t) = 0, u(L, t) = 0 for t ≥ 0,
u(x, 0) =
a for 0≤ x ≤ L/2, 0 for L/2 < x ≤ L, where a is a constant.
a Determine the solution u(x, t).
b Two identical iron rods, each 20 cm in length, have their ends put against each other. Both of the remaining ends, at x= 0 and at x = 40 cm, are kept at 0◦C. The left rod has a temperature of 100◦C and the right rod a temperature of 0◦C. Cal-culate for k = 0.15 cm2s−1the temperature at the boundary layer of the two rods, 10 minutes after the rods made contact, and show that this value is approximately 36◦C.
c Calculate approximately how many hours it will take to reach a temperature of 36◦C at the boundary layer, when the rods are not made of iron, but concrete (k= 0.005 cm2s−1).
Given is the following initial and boundary value problem:
5.21
utt = a2ux x for 0< x < L, t > 0, u(0, t) = 0, u(L, t) = 0 for t > 0,
u(x, 0) = sin(πx/L) for 0< x < L, ut(x, 0) = 7 sin(3πx/L) for 0< x < L.
Show that the first two steps of the method described in section 5.2 lead to the collection of fundamental solutions
un(x, t) = (Ansin(nπat/L) + Bncos(nπat/L)) sin(nπx/L),
and subsequently determine the formal solution which is adjusted to the given initial displacement and initial velocity.