CHAPTER WE
W.1
The Wave Equation
Let us next turn to the oscillatory problems described by the so-called wave equation
Lu ≡ uxx −
1
utt = 0
c2
where c is a non-vanishing function of time and space. If c is constant and f and g are any
twice continuously differentiable functions of a single argument then it follows by differentiation that
Lf (x − ct) = Lg(x + ct) = 0.
Since f (x − ct) describes a wave traveling to the right with speed c and g(x + ct) is a
wave moving left with speed c it is customary to call c the wave speed (even when c is not
constant). We shall solve the wave equation subject to given initial and boundary conditions
with an eigenfunction expansion.
W.1.1
The Solution Technique
To introduce the eigenfunction solution for the wave equation we shall consider the
following initial/boundary value problem
(1.1a)
Lu = uxx −
(1.1b)
u(0, t) = A(t),
1
utt = F (x, t)
c2
u(L, t) = B(t)
(1.1c)
u(x, 0) = u0 (x)
(1.1d)
ut (x, 0) = u1 (x)
where F , u0 and u1 are given smooth functions. In order to use an eigenfunction expansion
we need to zero out the boundary conditions. As before we shall choose
v(x, t) = A(t)
L−x
x
+ B(t) .
L
L
1
Then
w(x, t) = u(x, t) − v(x, t)
satisfies
µ
¶
1
1
Lw = wxx − 2 wtt = F (x, t) − vxx − 2 vtt ≡ G(x, t)
c
c
w(0, t) = w(L, t) = 0
w(x, 0) = u0 (x) − v(x, 0) ≡ w0 (x)
wt (x, 0) = u1 (x) − vt (x, 0) ≡ w1 (x)
where G(x, t), w0 (x) and w1 (x) are known data functions.
We now proceed as with the diffusion equation. The eigenvalue problem associated
with this equation is again
φ00 (x) = µφ(x)
φ(0) = φ(L) = 0
with eigenfunctions
φn (x) = sin λn x
with
λn =
nπ
,
L
µn = −λ2n .
The approximating problem is
Lw(x, t) = PN G(x, t) =
N
X
γn (t)φn (x)
n=1
w(0, t) = w(L, t) = 0
w(x, 0) =
N
X
α̂n φn (x)
n=1
wt (x, 0) =
N
X
β̂n φn (x)
n=1
where
γn (t) =
α̂n =
hG(x, t), φn i
,
hφn , φn i
hw0 (x), φn i
hφn , φn i
2
β̂n =
hw1 (x), φn i
hφn , φn i
This problem is solved by
(w.1.2)
wN (x, t) =
N
X
αn (t)φn (x)
n=1
when αn (t) is a solution of the problem
−λ2n αn (t) −
(w.1.3)
1 00
α (t) = γn (t)
c2 n
αn (0) = α̂n
αn0 (0) = β̂n .
Hence the difference to the diffusion solution is that αn (t) satisfies the second order equation
(w.1.2) rather than the first order equation (D.1.3).
The general structure of this solution is
αn (t) = c1 cos λn ct + c2 sin λn ct + αnp (t)
where the so-called particular integral αnp (t) is ANY solution of (w.1.3). Once it is known
the constants c1 and c2 can be found so that αn (t) satisfies the given initial conditions at
t = 0. We recall from the theory of ordinary differential equations and the discussion of the
method of variation of parameters that one can always find a particular of the form
αnp (t) = v1 (t) cos λn ct + v2 (t) sin λn ct
where v1 and v2 are chosen such that
µ
cos λn ct
sin λn ct
−λn c sin λn ct λn c cos λn ct
¶µ
v10
v20
However, when γn (t) has the special form
γn (t) = tk eωt
3
¶
=
µ
0
2
−c γn (t)
¶
.
for some integer k and some real or complex constant ω then the so-called method of
undetermined coefficients tends to be easier to apply. We guess a particular solution of the
form
αnp (t) =
"k+2
X
#
(Ci ti ) eωt
i=0
and compute the unknown coefficients Ci so that αnp (t) solves (w.1.3).
Once wN (x, t) has been found the approximate solution of the original problem is
uN (x, t) =
N
X
αn (t)φn (x) + v(x, t).
n=1
If the wave equation is to be solved subject to derivative boundary conditions then different
eigenfunctions apply but the general approach remains unchanged.
W.1.2
Applications
The simple problem of a vibrating string serves to illustrate the eigenfunction approach
for the solution of the wave equation.
Problem 1: A uniform string of length L is held fixed at both ends. At time t = 0 it is
given an initial displacement and velocity. Find the motion of the string for t > 0.
Answer: Newton’s second law and some small amplitude assumptions are known to lead
to the wave equation
Lu ≡ uxx −
1
utt = 0
c2
for the displacement u(x, t) of the string at the point x and time t. Here c is a known
constant depending on the density of the string and the tension applied to it. Since the
ends are held fixed we have
u(0, t) = u(L, t) = 0.
The initial displacement from the equilibrium position and the initial velocity are
u(x, 0) = u0 (x)
ut (x, 0) = u1 (x)
where u0 and u1 are given functions. The eigenvalue problem associated with the spatial
part of the wave equation is
φ00 (x) = µφ(x)
4
φ(0) = φ(L) = 0.
Its solutions {µn , φn (x)} are
φn (x) = sin λn x with
λn =
nπ
L
and
µn = −λ2n .
The approximating problem is
Lu ≡ uxx −
1
utt = 0
c2
u(0, t) = u(L, t) = 0
u(x, 0) = PN u0 (x) =
N
X
α̂n φn (x)
n=1
ut (x, 0) = PN u1 (x) =
N
X
β̂n φn (x)
n=1
where
α̂n =
hu0 , φn i
hφn , φn i
β̂n =
hu1 , φn i
.
hφn , φn i
We shall take for granted that this problem has a unique solution u N (x, t) and show that it
can be written in the form
uN (x, t) =
N
X
αn (t)φn (x).
n=1
If we substitute this sum into the wave equation we obtain
N ·
X
n=1
−λ2n αn (t)
¸
1 00
− 2 αn (t) φn (x) = 0.
c
It follows that uN solves the wave equation and the initial and boundary conditions if α n (t)
is computed such that
λ2n αn (t) +
1 00
α (t) = 0
c2 n
αn (0) = α̂n
αn0 (0) = β̂n .
5
The solution αn (t) is readily found as
αn (t) = α̂n cos cλn t +
β̂n
sin cλn t.
cλ
We observe that each term of the sum defining uN (x, t) corresponds to a standing wave
with amplitude αn (t) and nodes at the zeros of φn (x) = sin nπ
L x. In other words, uN is the
superposition of standing waves.
As an illustration we show in Fig. 1 the standing wave α5 (t)φ5 (x) for t = .4 and t = .8,
as well as the the solution u10 (x, t) for t = 0, .4 and .8 for the following data: c = L = 1
and
u0 (x) =
½
u1 (x) = 0
3x,
x ∈ [0, 1/3]
3
(1
−
x),
x
∈ (1/3, 1].
2
so that
9 sin nπ
3
,
β̂n = 0.
α̂n =
(nπ)2
Since there is no exponential decay with time the number of terms N required in the
eigenfunction expansion is dictated at all times by the quality of the approximation of u 0
and u1 in terms of their orthogonal projections. It is important that initial and boundary
data are consistent to rule out any Gibbs phenomena because no smoothing appears as the
solution of the wave equation evolves with time. We also remark that the wave equation with
the above piecewise linear initial displacement does not, strictly speaking, have a solution
because u0 (x) is not differentiable at x = 1/3. However, PN u0 is infinitely differentiable
which guarantees the existence of a smooth solution uN (x, t).
Problem 2: A forced wave and resonance.
Suppose a uniform string of length L is held fixed at x = 0 and oscillated at x = L
according to
u(L, t) = A sin ωt.
Find the motion of the string for t > 0.
Answer: Our model is incomplete since the initial state of the string is not given. We
observe that the boundary data u(0, t) = 0 and u(L, t) = A sin ωt imply ut (0, t) = 0 and
ut (L, t) = Aω cos ωt. Hence we shall impose initial conditions
u(x, 0) = 0,
ut (x, 0) = Aω
6
x
L
which are consistent with the boundary data. In order to use an eigenfunction expansion
we need homogeneous boundary conditions. If we set
µ ¶
x
v(x, t) = A sin ωt
L
and
w(x, t) = u(x, t) − v(x, t)
then
µ ¶
x
Lw = Lu − Lv = vtt = −Aω sin ωt
L
2
where for ease of notation we have set c = 1. The boundary and initial conditions are
w(0, t) = w(L, t) = 0
w(x, 0) = u(x, 0) − v(x, 0) = 0
wt (x, 0) = ut (x, 0) − vt (x, 0) = 0.
The associated eigenvalue problem is the same as in Example l and the approximating
problem is
Lw ≡ wxx − wtt = sin ωt
N
X
γ̂n φn (x)
n=1
where
γ̂n = −
2 (−1)n Aω 2
A 2 hx, φn i
ω
=
,
L
hφn , φn i
L
λn
and
w(0, t) = w(L, t) = 0
w(x, 0) = 0
wt (x, 0) = 0.
This problem has the solution
wN (x, t) =
N
X
n=1
7
αn (t)φn (x)
where
−λ2n αn (t) − αn00 (t) = γ̂n sin ωt
αn (0) = 0
αn0 (0) = 0.
Let us assume at this stage that ω 6= λn for any n. Then αn (t) can be written in the form
αn (t) = c1 cos λn t + c2 sin λn t + αnp (t)
where αnp (t) is a particular integral. We can guess an αnp (t) of the form
αnp (t) = C sin ωt.
If we substitute αnp (t) into the differential equation we find that
C=
ω2
γ̂n
.
− λ2n
Determining c1 and c2 so that αn (t) satisfies the initial conditions we obtain
αn (t) =
γ̂n
[λn sin ωt − ω sin λn t]
λn (ω 2 − λ2n )
so that
N
X
γ̂n
Aω 2 (−1)n
[λn sin ωt − ω sin λn t] φn (x) + v(x, t).
uN (x, t) =
λ (ω 2 − λ2n ) λ2n (ω 2 − λ2n )
n=1 n
Let us now suppose that the string is driven at a frequency ω which is close to λ k for
some k, say
ω = λk + ²,
0 < ² ¿ 1.
Then
·
¸
γ̂k λk sin ωt − ω sin λk t
αk (t) =
λk
ω 2 − λ2k
can be rewritten as
αk (t) =
γ̂k [ω(sin ωt − sin λk t) + (λk − ω) sin ωt]
.
λk
ω 2 − λ2k
8
When we apply the identity
¶
µ
¶
x+y
x−y
cos
sin x − sin y = 2 sin
2
2
µ
we obtain
¶
µ
γ̂k ω
γ̂k sin ωt
2
²t
ω + λk
αk (t) =
t−
sin
cos
.
λk (ω + λk ) ²
2
2
λk (ω + λk )
The first term describes the contribution of a standing wave which oscillates with frequency
(ω+λk )
4π
∼
=
λk
2π
but whose amplitude rises and falls slowly in time with frequency
²
4π
which
gives the motion a so-called “beat.” Finally, we observe that if ² → 0, i.e. ω → λ k , then
l’Hospital’s rule applied to the first term of αk (t) yields
αk (t) =
γ̂k
[λk t cos λk t − sin λk t]
2λ2k
and after substituting for γ̂k ,
αk (t) = (−1)
k 2A
L
·
¸
t cos λk t sin λk t
φk (x).
−
2
2λk
Hence the amplitude of αk (t) grows linearly with time and will eventually dominate all other
terms in the eigenfunction expansion of uN (x, t). This phenomenon is called resonance and
the string is said to be driven at the resonant frequency λk .
Problem 3: Wave propagation in a resistive medium.
Suppose a string vibrates in a medium which resists the motion with a force proportional
to the velocity and the displacement of the string. Newton’s second law then leads to
Lu ≡ uxx − Butt − Cut − Du = 0
for the displacement u(x, t) from the equilibrium position where B, C and D are positive
constants. (We remark that the same equation is also known as the telephone equation and
describes the voltage or current of an electric signal travelling along a lossy transmission
line.) We shall impose the boundary and initial conditions of the last example
u(0, t) = 0,
u(x, 0) = 0,
u(L, t) = A sin ωt
ut (x, 0) =
9
Aω
x
L
and solve for the motion of the string.
Answer: If v(x, t) =
x
L
A sin ωt then w(x, t) = u(x, t) − v(x, t) satisfies
Lw = Lu − Lv = Bvtt + Cvt + Dv =
¤
Ax £
(−Bω 2 + D) sin ωt + Cω cos ωt
L
w(0, t) = w(L, t) = 0
w(x, 0) = wt (L, t) = 0.
The associated eigenvalue problem is again
φ00 (x) = µφ(x)
φ(0) = φ(L) = 0.
The eigenfunctions are
φn (x) = sin λn x
with λn =
nπ
,
L
µn = −λ2n .
The approximating problem is
Lw =
where
N h
i
X
γ̂n sin ωt + δ̂n cos ωt φn (x)
n=1
A(D − Bω 2 ) hx, φn i
L
hφn , φn i
ACω hx, φn i
.
δ̂n =
L hφn , φn i
γ̂n =
The approximating problem has the solution
wN (x, t) =
N
X
αn (t)φn (x)
n=1
where αn (t) has to satisfy the initial value problem
−λ2n αn (t) − Bαn00 (t) − Cαn0 (t) − Dαn (t) = γ̂n sin ωt + δ̂n cos ωt
αn0 (0) = 0.
αn (0) = 0,
10
This is a constant coefficient second order equation with a source term. Its solution has the
form
αn (t) = c1 αn1 (t) + c2 αn2 (t) + αnp (t)
where αn1 (t) and αn2 (t) are complementary solutions of the homogeneous equation and
αnp (t) is any particular integral. For the complementary functions we try to fit an exponential function of the form
αc (t) = ert .
Substitution into the differential equation shows that r must be chosen such that
Br2 + Cr + (D + λ2n ) = 0.
This quadratic has the roots
r1,2 =
If r1 6= r2 we may take
−C ±
p
C 2 − 4B(D + λ2n )
.
2B
αn1 (t) = er1 t ,
αn2 (t) = er2 t .
The particular integral is best found with the method of undetermined coefficients. If we
substitute
αnp (t) = d1 sin ωt + d2 cos ωt
into the differential equation then we require
+ Bω 2 (d1 sin ωt + d2 cos ωt) − Cω(d1 cos ωt − d2 sin ωt)
− (D + λ2n )(d1 sin ωt + d2 cos ωt) = γ̂n sin ωt + δ̂n cos ωt.
Equating the coefficients of the trigonometric terms we find
µ
¶µ ¶ µ ¶
Bω 2 − (D + λ2n )
Cω
γ̂n
d1
=
.
−Cω
Bω 2 − (D + λ2n )
d2
δ̂n
For C > 0 this equation has a unique solution so that d1 and d2 may be assumed known.
Finally, we need to determine the coefficients c1 and c2 so that αn (t) satisfies the initial
conditions. We obtain the conditions
c1 + c2 = −d2
11
r1 c1 + r2 c2 = −ωd1 .
As long as r1 6= r2 this system has the unique solution
so that
αn (t) =
c1 =
ωd1 − r2 d2
r2 − r 1
c2 =
r1 d2 − ωd1
r2 − r 1
(ωd1 − r2 d2 )er1 t + (r1 d2 − ωd1 )er2 t
+ d1 sin ωt + d2 cos ωt.
r2 − r 1
It is possible that for some index k the two roots r1 and r2 are the same. Then the first
term for αk (t) is indeterminate and must be evaluated with l’Hospital’s rule. In analogy
to mechanical systems we may say that the nth mode of our approximate solution is overdamped, critically damped or underdamped if C 2 − 4B(D + λ2n ) is positive, zero or negative,
respectively.
The next example shows that special functions like Bessel functions also arise for certain
problems in cartesian coordinates.
Problem 4: A chain of length L with uniform density is suspended from a (frictionless)
hook and given an initial displacement and velocity which are assumed to lie in the same
plane. Describe the subsequent motion of the chain.
Answer. Let u(x, t) be the displacement from the equilibrium position where the coordinate
x is measured from the free end of the chain at x = 0 vertically upward to the fixed end
at x = L. Then the tension in the chain at x is proportional to the weight of the chain
below x. After scaling Newton’s second law leads to the following mathematical model for
the motion of the chain:
Lu ≡ (xux )x − utt = 0
u(L, t) = 0
u(x, 0) = u0 (x)
ut (x, 0) = u1 (x).
The associated eigenvalue problem is
(xφ0 (x))0 = µφ(x)
12
φ(L) = 0.
If we add the natural restriction that also |φ(0)| < ∞ then we again have a singular SturmLiouville problem. Its solution is not obvious. Fortunately, this eigenvalue problem can be
transformed to the problem solved in the last chapter when we discussed radial heat flow in
√
a disk. If we make the change of variable z = 2 x and define
Φ(z) = φ(z 2 /4) = φ(x)
then
dΦ
dφ dx
z dφ
=
=
dz
dx dz
2 dx
so that
dφ
2 dΦ
=
dx
z dz
d
dx
and
µ
dφ
x
dx
¶
2 d
=
z dz
µ
z 2 2 dΦ
4 z dz
¶
.
Thus our eigenvalue problem is
(zΦ0 (z))0 = µzΦ(z)
|Φ(0)| < ∞
and
√
Φ(2 L) = 0.
This problem is identical to the eigenvalue problem (6.1) and has the solutions
{µn , J(λn z)}
√
where µn = −λ2n and where λn 2 L = zn . As before, zn is the nth zero of the first order
Bessel function J0 (z). In summary, the eigenvalue problem associated with the hanging
chain has the solution {µn , φn (x)} with
√
φn (x) = J0 (2λn x),
zn
λn = √ ,
2 L
µn = −λ2n .
We also know from our discussion of problem 8 that
hJ0 (λm z), J0 (λn z)i =
Z
2L
J0 (λm z)J0 (λn z)z dz =
0
Z
L
φm (x)φn (x)dx = 0
0
so that the eigenfunctions {φn (x)} are orthogonal in L2 (0, L). We note that orthogonality
with respect to the weight function w(x) = 1 is predicted by the Sturm-Liouville theorem
13
of Chapter xxx if the eigenvalue problem were a regular problem given on an interval [², L]
with ² > 0. The approximate solution is written as
uN (x, t) =
N
X
αn (t)φn (x).
n=1
Substitution into the wave equation leads to the initial value problems
−λ2n φn (t) − αn00 (t) = 0
αn (0) =
hu0 (x), φn (x)i
,
hφn , φn i
αn0 (0) =
and the approximate solution
uN (x, t) =
N ·
X
n=1
hu1 (x), φn (x)i
hφn , φn i
¸
√
αn0 (0)
sin λn t J0 (2λn x).
αn (0) cos λn t +
λn
Problem 5. At time t = 0 a spherically symmetric pressure wave is created in a rigid
sphere of radius R. Find the subsequent pressure distribution in the sphere.
Answer. Since there is no angular dependence the mathematical model for the pressure
u(r, t) in the sphere is
urr +
2
1
ur − 2 utt = 0.
r
c
We shall assume that the initial state can be described by
u(r, 0) = u0 (r)
ut (r, 0) = u1 (r).
At all times we have to satisfy the symmetry condition
ur (0, t) = 0.
And since the sphere is rigid there is no pressure loss through the shell so we require
ur (R, t) = 0.
The eigenvalue problem associated with this model is
φ00 (r) +
2 0
φ (r) = µφ(r)
r
14
φ0 (0) = φ0 (R) = 0.
To insure a finite pressure we shall require that |φ(0)| < ∞. We encountered an almost
identical problem in our discussion of heat flow in a sphere and already know that bounded
solutions for non-zero λ have to have the form
φ(r) =
sin λr
.
r
The boundary condition at r = R requires that
φ0 (R) =
Rλ cos λR − sin λR
= 0.
2
Hence the eigenvalues λn are roots of
f (λ) ≡ Rλ cos λR − sin λR = 0.
The existence and distributions of the roots of f were discussed in connection with onedimensional heat transfer in a rod with convective cooling at one end. There are countably
many values λn , and
λn →
³π
2
´
+ nπ /R
as n → ∞
because the cosine term will dominate for large n. In addition we observe that for the
boundary conditions of this application λ0 = µ0 = 0 is an eigenvalue with eigenfunction
φ0 (x) ≡ 1.
It follows from the general theory that distinct eigenfunctions are orthogonal in L 2 (0, R, r 2 ).
This orthogonality can also be established by simple integration. We see that for λ m 6= λn
R
sin λm r sin λn r 2
r dr
r
r
0
1
= 2
[λm cos λm R sin λn R − λn cos λn R sin λm R] = 0
λn − λ2m
hφm (r), φn (r)i =
Z
in view of f (λm ) = f (λn ) = 0. It is straightforward to verify that this conclusion remains
valid if m = 0 and n 6= 0. It follows that
un (r, t) =
N
X
n=1
15
αn (t)φn (r)
where
α000 (t) = 0
and for n > 0
−(cλn )2 αn (t) − αn00 (t) = 0
αn (0) =
hu0 (r), φn i
,
hφn , φn i
αn0 (0) =
hu1 (r), φn i
.
hφn , φn i
Note that α0 (0) and α0 (0) are the average values for the initial pressure and velocity over
the sphere. The equations for αn (t) are readily integrated. We obtain
uN (x, t) = α0 (0) +
α00 (0)t
+
N
X
λn
n=1
·
¸
αn0 (0)
sin λn r
αn (0) cos cλn t +
.
sin cλn t
cλn
λn r
For example, let us suppose that R = 1 and
u0 (r) =
½ sin(10πr)
0
0 < r < 1/10
1/10 ≤ r < 1
r
and
u1 (r) = 0.
Then
α0 (0) =
3
100π
(10π−λ )
αn (0) = 2
n
sin
10
(10π−λn )
(10π+λ )
−
n
sin
10
(10π+λn )
2 − sin(2λn )
and
αn0 (0) = 0.
The next example is reminiscent of constrained Hilbert space minimization problems
discussed, for example, in [ ]. The problem will be stated as follows:
Determine the “smallest” force F (x, t) such that the wave u(x, t) described by
uxx − utt = F (x, t)
u(0, t) = u(L, t) = 0
16
u(x, 0) = ut (x, 0) = 0
satisfies the final condition
u(x, T ) = uf (x)
ut (x, T ) = 0
where uf (x) is a given function and T is a given final time.
We shall ignore the deep mathematical questions of whether and in what sense this
problem does indeed have a solution and concentrate instead on showing that we can actually
solve the approximate problem formulated for functions of x which belong to the subspace
MN = span{φn (x)}N
n=1
where as before φn (x) is the nth eigenfunction of
φ00 (x) = µφ(x)
φ(0) = φ(L) = 0.
To make the problem tractable we shall agree that the size of the force will be measured in
the least squares sense
kF k =
ÃZ
T
0
Z
L
F (x, t)2 dx dt
0
!1/2
.
The approximation to the above problem can then be formulated as:
Example 6: Find
F̂N (x, t) =
N
X
γ̂n (t)φn (x)
n=1
such that
kF̂N k ≤ kFN k
for all FN (·, t) ∈ M for which the solution uN of
Lu = uxx − utt = FN (x, t)
u(0, t) = u(L, t) = 0
17
u(x, 0) = ut (x, 0) = 0
satisfies
u(x, T ) = PN uf (x)
ut (x, T ) = 0.
Answer: If
N
X
γn (t)φn (x)
N
X
αn (t)φn (x)
FN (x, t) =
n=1
and
uN (x, t) =
n=1
then as in Example 2 it follows that
−λ2n αn (t) − αn00 (t) = γn (t)
αn (0) = αn0 (0) = 0
where λn =
nπ
L
, n = 1, 2, . . . , N . The variation of parameters solution for this problem can
be readily verified to be
αn (t) = −
Z
t
0
1
γn (s) sin λn (t − s)ds.
λn
uN (x, T ) will satisfy the final condition if γn (t) is chosen such that
αn (T ) =
huf (x), φn i
hφn , φn i
αn0 (T ) = 0.
Hence γn (t) must be chosen such that
(w.2.1)
(w.2.2)
Z
T
0
γn (s) sin λn (T − s)ds = −λn
Z
huf , φn i
hφn , φn i
T
0
γn (s) cos λn (T − s)ds = 0.
18
Finally, we observe that
N Z
LX T
γn (t)2 dt
kFN k =
2 n=1 0
RT
so that kFN k will be minimized whenever 0 γn (t)2 dt is minimized for each n. But it is
2
known that the minimum norm solution in L2 [0, T ] of the two constraint equations (w.2.1,2)
must be of the form
γ̂n (t) = c1 sin λn (T − t) + c2 cos λn (T − t).
Substitution into (w.2.1,2) and integration with respect to t show that c 1 and c2 must satisfy
Ã
!
!µ ¶ Ã
huf ,φn i
sin2 λn T
sin 2λn T
T
−λ
−
c
n
1
hφn ,φn i
2
4λn
2λn
=
sin2 λn T
T
c2
+ sin 2λn T
0
2λn
2
4λn
We observe that the determinant of the coefficient matrix is
¤
1 £
2
2
(T
λ
)
−
sin
λ
T
n
n
4λ2n
and hence never zero for T > 0. Thus each γ̂n (t) is uniquely defined and the approximating
problem is solved. Whether
FN (x, t) =
N
X
γ̂n (t)φn (x)
n=1
remains meaningful as N → ∞ depends strongly on uf (x). It can be shown by actually
solving the linear system for c1 and c2 that
c1 ∼
2λn huf , φn i
T
and |c1 | ¿ |c2 |
as n → ∞.
Boundedness of |u00f | and the consistency condition uf (0) = uf (L) = 0 guarantee that
c1 = O(λ−2
n ) so that FN will converge absolutely as N → ∞.
The last example in this chapter is chosen to illustrate that the eigenfunction approach
depends only on the solvability of the eigenvalue problem, not on the order of the differential
operators.
Example 7: Find the natural frequencies of a cantilevered uniform beam of length L.
Answer: The mathematical model for the displacement u(x, t) of the beam is [ ]
1 ∂2u
∂4u
+ 2 2 =0
∂x4
c ∂t
19
u(0, t) = ux (0, t) = 0
uxx (L, t) = uxxx (L, t) = 0.
The associated eigenvalue problem is
φ(iv) (x) = µφ(x)
φ(0) = φ0 (0) = 0
φ00 (L) = φ000 (L) = 0.
It is straightforward to show as in Chapter 2 that the eigenvalue must be positive. For
notational convenience we shall write
µ = λ4
for some positive λ. We observe that the function
φ(x) = erx
will solve the differential equation if
r 4 = λ4 .
The four roots of the positive number λ4 are
r1 = λ,
r2 = −λ,
r3 = iλ and r4 = −iλ.
A general solution of the equation is then
φ(x) = c1 sinh x + c2 cosh λx + c3 sin λx + c4 cos λx.
It is straightforward to verify (but a little laborious to compute) that there are countably
many functions
(w.2.3)
φn (x) = [cosh λn L + cos λn L][sinh λn x − sin λn x]
− [sinh λn L + sin λn L][cosh λn x − cos λn x]
which satisfy the boundary conditions provided
cosh λn L cos λn L = −1.
20
If we write
f (x) ≡ cos x +
1
cosh x
then f behaves essentially like cos x and has two roots in every interval (nπ − π/2, nπ + π/2)
for n = 1, 3, 5, . . . The first five numerical roots of f (x) = 0 are tabulated below.
i
xi
1
1.87510
2
4.69409
3
7.85476
10.9955 ∼
= 3π − π/2
4
14.1372 ∼
= 3π + π/2.
5
Since cosh x grows exponentially all subsequent roots are numerically the roots of cos x. The
eigenvalues and eigenfunctions for the vibrating beam then are
µn = λ4n ,
φn (x)
as given by (w.2.3),
where λn = xn /L.
An oscillatory solution of the beam equation is obtained when we write
un (x, t) = αn (t)φn (x)
and compute αn (t) such that
λ4n αn (t) +
1 00
α (t) = 0.
c2 n
It follows that
αn (t) = An cos(cλ2n t + Bn )
where the amplitude An and the phase Bn are the constants of integration. Each un (x, t)
describes a standing wave. A snapshot of the first two modes φ1 (x) and φ2 (x) is shown in
Fig. ....
The motion of a vibrating beam subject to initial conditions and a forcing function
is found in the usual way by projecting the data into the span of the eigenfunctions and
solving the approximating problem in terms of an eigenfunction expansion.
21
W.2
Theory
W.2.1 Convergence of uN (x, t) to the analytic solution
Bounds for the solution of the wave equation are obtained from so-called energy estimates
which we shall introduce for the following model problem
(w.2.1)
wxx − wtt = F (x, t)
w(0, t) = w(L, t) = 0
w(x, 0) = w0 (x)
wt (x, t) = w1 (x).
Theorem W.2.1. Assume that the data of problem (w.2.1) are sufficiently smooth so that
it has a smooth solution w(x, t) on D = {(x, t) : 0 < x < L, 0 < t < T } for some T > 0.
Then for t ≤ T
Z
L
0
[wx2 (x, t)
+
wt2 (x, t)]dx
<e
t
Z
L
0
[w12 (x)
+
w002 (x)]dx
+
Z tZ
0
L
et−s F 2 (x, t)dx dt
0
Proof. We multiply the wave equation by wt
wt wxx − wt wtt = wt F (x, t)
and use the smoothness of w to rewrite this expression in the form
(w.2.2)
(wt wx )x − (wxt wx ) − wt wtt = wt F (x, t).
Since w(0, t) = w(L, t) = 0 it follows that wt (0, t) = wt (L, t) = 0. The integral of (w.2.2)
with respect to x can then be written in the form
(w.2.3)
1 d
2 dt
Z
L
0
(wt2
+
wx2 )dx
=−
Z
L
wt F (x, t)dx.
0
Using the algebraic-geometric inequality 2wt F (x, t) ≤ wt2 + F 2 (x, t) and defining the “energy” integral
E(t) =
Z
L
0
[wt2 (x, t) + wx2 (x, t)]dx
22
we obtain from (w.2.3) the inequality
d
E(t) ≤ E(t) + kF (·, t)k2
dt
(w.2.4)
where kF (·, t)k2 =
RL
0
F 2 (x, t)dx. Gronwall’s inequality applies to (w.2.4) and yields
Z t
t
E(t) ≤ E(0)e +
et−s kF (·, s)k2 ds
0
which was to be shown.
We note from (w.2.3) that if F = 0 then for all t
E(t) = E(0).
We also observe that the energy estimate allows the pointwise estimate
¯
¯Z x
¯ √
¯
√ p
wx (r, t)dr ¯¯ ≤ x kwx (·, t)k < x E(t)
|w(x, t)| = ¯¯
0
as well as the mean square estimate
kw(·, t)k <
Lp
L
kwx (·, t)k <
E(t).
π
π
These estimates are immediately applicable to the error
eN (x, t) = w(x, t) − wN (x, t)
where w solves (w.2.1) and wN is the computed approximation obtained by projecting
F (·, t), u0 and u1 into the span{sin λn x}N
n=1 with λn =
nπ
L .
Since eN (x, t) satisfies (w.2.1) with the substitutions
F ← F − PN F,
w 0 ← w0 − PN w 0 ,
w 1 ← w1 − PN w 1
we see from Theorem W.2.1 that
£
¤
kex (·, t)k2 + ket (·, t)k2 ≤ et kw1 − PN w1 k2 + k(w0 − PN w0 )0 k2
Z tZ L
et−s kF (·, s) − PN F (·, s)k2 ds.
+
0
0
This estimate implies that if PN F (·, t) converges in the mean square sense to F (·, t) uniformly with respect to t and PN w00 → w00 then the computed solution converges pointwise
and in the mean square sense to the true solution. But in contrast to the diffusion setting
the error will not decay with time. If it should happen that F ∈ span{sin λ n x} for all t then
the energy of the error remains constant and equal to the initial energy. If the source term
F − PN F does not vanish then the energy could conceivably grow exponentially with time.
23
W.2.2 Eigenfunction expansions and Duhamel’s principle
The influence of v(x, t) chosen to zero out non-homogeneous boundary conditions imposed
on the wave equation can be analyzed as in the case of the diffusion equation and will not be
repeated here. However, it may be instructive to show that the eigenfunction approach leads
to the same equations as Duhamel’s principle for the wave equation with time dependent
data so that we only provide an alternative view but not a different computational method.
Consider the problem
(2.2.1)
wxx − wtt = F (x, t)
w(0, t) = w(L, t) = 0
w(x, 0) = 0
wt (x, 0) = 0.
Duhamel’s principle yields the solution
w(x, t) =
Z
t
W (x, t, s)ds
0
where
Wxx (x, t, s) − Wtt (x, t, s) = 0
W (0, t, s) = W (L, t, s) = 0
W (x, s, s) = 0
Wt (x, s, s) = −F (x, s).
The absence of a source term makes the calculation of an eigenfunction solution of W (x, t, s)
straightforward.
An eigenfunction solution obtained directly from (2.2.1) is found in the usual way in
the form
wN (x, t) =
N
X
αn (t)φn (x)
n=1
where αn (t) solves the initial value problem
−λ2n αn (t) − αn00 (t) = γn (t)
24
αn (0) = αn0 (0) = 0
with
γn (t) =
hF (x, t), φn i
.
hφn , φn i
The variation of parameters solution for this problem is
1
αn (t) = −
λn
Z
t
0
sin λn (t − s)γn (s)ds.
If we set
Wn (x, t, s) = −
1
sin λn (t − s)γn (s)φn (x)
λn
then the eigenfunction solution is given by
wN (x, t) =
N
X
αn (t)φn (x) =
Z tX
N
Wn (x, t, s).
0 n=1
n=1
By inspection
Wnxx − Wntt = 0
Wn (x, s, s) = 0
Wnt (x, s, s) = −γn (s)φn (x) = −
so that
N
X
n=1
hF (x, s), φn i
φn (x)
hφn , φn i
Wn (x, s, s) = −Pn F (x, s).
Hence both methods yield the same solution and require the evaluation of identical integrals.
25