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CHAPTER 4

The Wave Equation

The wave equation

utt = c2∆u,

is the prototype for second order hyperbolic PDE, modeling the propagation of sound waves,electromagnetic waves such as light, and waves in elastic solids. We show in detail how thewave equation describes the deformation of one-dimensional elastic solids, specifically thinrods and elastic strings.

Central to the study of the one-dimensional equation is d’Alembert’s solution, an explicitformula for solutions of initial value problems. The method of spherical means provides acorresponding explicit formula in two and three dimensions. In three dimensions, this formulaembodies Huygens principle of light propagation. From the wave equation we derive an energyprinciple in which the total energy (the sum of kinetic and potential energy) is conserved.

4.1. The Wave Equation in Elasticity

We introduce the wave equation with a simple derivation from one-dimensional elasticity.The derivation illustrates the basic notions of conservation laws and constitutive equationsintroduced in Chapter 2. Then we discuss a second application, to an elastic string vibratingin a plane. Conservation of momentum leads to a system of nonlinear PDE. Considering smallamplitude vibrations near a stationary string, we linearize the equations, thereby deriving twowave equations with different wave speeds. One equation represents longitudinal motion alongthe string, and the other represents transverse motion, the vibrations seen in a guitar or violinstring.

4.1.1. Longitudinal motion of a thin elastic rod. Consider a thin elastic rod under-going only longitudinal deformation (extension or compression), with no bending. We labellocations of cross sections in the rod by using points in a reference configuration, an interval,say 0 ≤ x ≤ 1. (See Fig. 4.1.) The physical configuration is also an interval 0 ≤ u ≤ L,depending on the deformation. The cross section labeled x in the reference configuration hascoordinate u(x, t) in the physical configuration at time t. It is convenient, but not essential,to think of the reference configuration as being the rod in equilibrium, with no forces actingon it. The function u is called the displacement; it is the unknown, or dependent variable. Weassume the density ρ (mass per unit volume in the reference configuration) is constant, andthat the cross sectional area A of the rod is constant along its length.

39

40 4. THE WAVE EQUATION

x u

Reference Configuration Physical Configuration

x u(x,t)

u( ,t)

0 1 0 L

(Lagrangian variables) (Eulerian variables)

Figure 4.1. Deformation u in a one-dimensional rod.

Consider forces on a cross section labeled x0 in the rod, at a specific time t. The part of the rodwith x > x0 exerts a force F (x0, t) on the part with x < x0, and the part with x < x0 exerts anequal and opposite force −F (x0, t) on the part with x > x0, so that across each cross-section,forces are balanced. However, the variation of these forces along the rod means that the netforce acting on a segment of rod may be non-zero, and induces a change in momentum.

In our formulation of the equation of motion, it is convenient to express the force distributionas a function of Lagrangian variable x rather than Eulerian variable u, even though we thinkof the force acting in the physical domain rather than the reference configuration. In fact, ifwe were to label forces in the physical domain as f(u, t), then F (x, t) = f(u(x, t), t). Moreover,it is convenient to consider the stress σ, which is force per unit area, rather than force F. Inthe present context, F = Aσ.

Since ut is the velocity of a point (i.e., cross-section) in the rod, the momentum density(meaning momentum per unit volume) is the quantity ρ ut. Now consider a short segment of

the rod a < x < b. The momentum of this section is∫ baρ ut(x, t)Adx. The balance law states

that the rate of change of momentum is equal to the net force:

d

dt

∫ b

a

ρ ut(x, t)Adx = F (b, t)− F (a, t).

Notice that if ut(x, t) is constant in x, then this is precisely Newton’s law

Mass× Acceleration = Force,

where Mass means the mass of the little section of rod.

Figure 4.2. Forces on a small section of the rod.

4.1. THE WAVE EQUATION IN ELASTICITY 41

As in Chapter 2, we can now derive a PDE from the balance law by writing both sides of theequation as integrals over a < x < b :∫ b

a

ρ utt(x, t)Adx =

∫ b

a

Fx(x, t) dx =

∫ b

a

σx(x, t)Adx.

Thus, provided utt, σx are continuous, we have the PDE

(1.1) ρutt = σx,

that expresses conservation of momentum.

To this equation we add a constitutive law, an equation that relates σ to u in a different way.In elasticity, this constitutive law states that the stress σ is a function of the strain. Thestrain is the deformation gradient; in the one-dimensional context of the rod, we have

strain = ux.

In engineering, it is common to define strain to be ux − 1, so that zero strain corresponds tono deformation: u(x, t) = x. In both cases, elasticity is expressed by a functional relationshipbetween σ and ux, written

σ = σ(ux).

Substituting into the PDE (1.1), we obtain

ρutt = σ(ux)x.

As we observed earlier, this equation is hyperbolic if σ′(ux) > 0, in which case, stress increaseswith strain, but is elliptic if σ′(ux) < 0. The hyperbolic case is more significant, especially forsmall deformations (more precisely, for small strains), but the elliptic case is also importantfor large deformations; it is associated with an effect called strain softening.

Perhaps the most important form of the constitutive law is Hooke’s Law, which states thatincreases in stress are proportional to increases in strain. This is expressed in the formula

(1.2) σ(ux) = k(ux − 1).

Note that this can also be stated as stress is proportional to strain if we define the strain tobe ux − 1.

Substituting (1.2) into (1.1), we obtain the one-dimensional wave equation

(1.3) utt = c2uxx,

in which c2 = kρ.

Remarks on Hooke’s Law.

1. The constant k > 0 is a constitutive parameter called the elastic modulus that dependson the elastic properties of the material; it can be measured in experiments. The sameexperiments assess the range of strains within which Hooke’s law is reasonable.

2. The parameter c has dimensions of a speed, i.e., L/T, where L and T are a typical referencelength (perhaps the length of the rod), and a typical time scale, respectively; correspondingly,


density (mass per unit volume) has dimensions M/L3, where M is the mass of the rod. Itfollows that k has dimensions LM/T 2, i.e., the dimensions of mass × acceleration, the samedimensions as force. Note that this is consistent with (1.2), since both u and x have dimensionsof length, so that ux is dimensionless.

3. Hooke’s law is familiar from elementary mechanics, or ODEs. It arises in relating theextension of a spring to the tension in the spring. To see the connection with the rod, considera uniform deformation given by u(x) = Lx. Then σ(ux) = k(L−1). But L−1 is the extension(if L > 1); the stress σ is constant and corresponds to the tension in the spring. Thus, thetension is proportional to the extension. Indeed, just as for springs, the constant k in Hooke’slaw can be found by performing simple extension experiments.

4.1.2. The Vibrating String. Consider a thin elastic string such as a guitar string orbungee cord, which we treat as a one-dimensional curve. For simplicity, we shall assume thatthe string moves only in two dimensions, and that the tension in the string is high enough thatwe can ignore gravity. Another scenario with no effect of gravity would be an experiment witha string constrained to a horizontal frictionless table. The effectively one-dimensional elasticbody is called a string when we assume that it can be bent with no resulting force. Then wesay there is no resistance to bending. Let’s consider the motion of a point on the string labeled

r

r0 L 1

2 rrr

x

x x

( x , t )=

Figure 4.3. The elastic string; one-dimensional string deforming in two dimensions.

by x ∈ [0, 1]. At each time t, this point will be located in the plane at (r1, r2) = r(x, t) ∈ R2,see Fig. 4.3. Then the tangent to the string is rx(x, t). Since there is no resistance to bending

and no gravity, the only force on the string is due to the tension T , which acts tangentiallyand is the only non-zero component of the stress. If the string has a uniform cross sectionalarea A and constant density ρ (gm/cm3), then the equations of motion (Newton’s second law,or conservation of momentum) are

ρArtt =

(AT

rx|rx|

)x

.

4.1. THE WAVE EQUATION IN ELASTICITY 43

Now we make the constitutive assumption that the tension T depends only on the strain |rx|and write T /ρ = T (|rx|). Thus, the string equations are

(1.4) rtt =

(T (|rx|)

rx|rx|

)x

, 0 < x < 1, t > 0.

Suitable boundary conditions, in which the string is fixed at two locations, are:

(1.5) r(0, t) = 0; r(1, t) = (L, 0).

With these boundary conditions, there is an equilibrium solution r0 = (xL, 0) in which thestring is stretched between the two fixed ends, as in a guitar string before it is plucked orstrummed. We shall assume that the tension at equilibrium is positive: T (L) > 0, and alsothat it is increasing with strain: T ′(L) > 0.

Now consider small deviations (u, v) from the equilibrium solution, and write r = (xL+u, v).We aim to find equations for the new variables u, v as functions of x, t. Of course, we cansimply substitute this expressi on into the string equations and get exact equations for u, v.However, we want to take advantage of the smallness of u, v. To do this, we substitute intothe PDE system (1.4) and then use a Taylor expansion about the equilibrium solution, whichis now u = v = 0.

When we substitute into (1.4), and expand each term as a Taylor series in u, v retaining onlyconstant and first order terms (linear in u, v), we get a lot of terms. For example, we need

|rx| =√

(L+ ux)2 + v2x = L

((1 +

uxL

)2

+v2x

L2

)1/2

= L+ ux + h.o.t.,

where h.o.t. represents the remainder in the Taylor series; specifically, h.o.t. = O(u2x + v2

x).Similarly,

T (|rx|) = T (L) + T ′(L)ux + h.o.t.;

rx|rx|

=

(L+ uxvx

)(L+ ux)

−1 + h.o.t. =

(1

vx/L

)+ h.o.t.

Thus,

(1.6) T (|rx|)rx|rx|

=

(T (L) + T ′(L)ux

T (L)vx/L

)+ h.o.t.

Finally, the equation (1.4) becomes a pair of wave equations if we retain only terms linear inu and v; that is, drop the h.o.t. terms in (1.6):

(1.7)utt = c2uxx

vtt = s2vxx,

In these linear wave equations, c =√T ′(L) is the longitudinal wave speed, and s =

√T (L)L

is the transverse wave speed. It can be argued that the longitudinal motion represented by


u is smaller than the transverse motion if the string is displaced laterally. Thus, a goodapproximation is to take the equation for v alone, and represent the string simply by thetransverse displacement v(x, t), 0 < x < 1.

This is a useful way to think of solutions of the wave equation; for each fixed time t the graphof v(x, t) represents the string. As time varies the graph evolves as a string in motion.

4.2. d’Alembert’s Solution

In 1747, d’Alembert1 published a paper on vibrating strings that included his famous solutionof the wave equation in one space variable x and time t:

(2.8) utt = c2uxx.

The first, and fundamental, step in deriving d’Alembert’s solution is to show that the generalsolution of (2.8) is

(2.9) u(x, t) = F (x− ct) +G(x+ ct),

where F and G are arbitrary C2 functions. The lines x− ct = const., x+ ct = const., whereF (x− ct), G(x+ ct) (respectively) are constant, are called characteristics.

Since c > 0 is constant, we can factor the partial differential operator ∂2t − c2∂2

x. and write thePDE as

(2.10) (∂t − c∂x) (∂t + c∂x)u = 0.

Then since (∂t+c∂x)F (x−ct) = 0, and (∂t−c∂x)G(x+ct) = 0, we see that (2.9) is a solution.

It will be useful when discussing solutions of the wave equation to interpret (2.9) as thesuperposition of two waves: The graph of F (x − ct) as a function of x for various times t isa wave traveling with speed c to the right, and G(x + ct) represents a wave moving to theleft with speed c. Thus, the wave equation (2.8) models waves of speed c moving in bothdirections, to the left and right, just as the linear transport equation models waves of speedc > 0 moving to the right only. We can add the two waves in (2.9) because the PDE (2.8) islinear and homogeneous.

To see that every solution can be represented in the form (2.9) for some choice of functionsF,G, we introduce characteristic variables suggested by the factorized equation (2.10):

ξ = x+ ct, η = x− ct

and write z(ξ, η) = u(x, t). Then

∂tu =∂z

∂ξ

∂ξ

∂t+∂z

∂η

∂η

∂t= c

∂z

∂ξ− c∂z

∂ηand c∂xu = c

∂z

∂ξ+ c

∂z

∂η.

1Jean le Rond d’Alembert (1717-1783)

4.2. D’ALEMBERT’S SOLUTION 45

Adding and subtracting as in (2.10), the equation becomes(−2c

∂

∂η

)(2c∂

∂ξ

)z = −4c2 ∂

∂η

∂

∂ξz = 0.

Integrating in η, we find∂z

∂ξ= g(ξ)

for some function g (which is constant with respect to η).

Thus, z(ξ, η) = G(ξ) + F (η), where G(ξ) =∫g(ξ) dξ, and F is another arbitrary function.

Back in the original variables, we arrive at

u(x, t) = z(x+ ct, x− ct) = F (x− ct) +G(x+ ct).

4.2.1. Initial Value Problem (Cauchy Problem). We use the general solution (2.9)of the wave equation to solve the Cauchy problem

(2.11)

utt = c2uxx, −∞ < x <∞, t > 0

u(x, 0) = φ(x), −∞ < x <∞

ut(x, 0) = ψ(x), −∞ < x <∞.

Just as for ODE, since the PDE is second order in t, to have a well-posed problem (cf. §2.1)we have to specify both the initial displacement u and the initial velocity ut.

Theorem 4.1. If φ is C2 and ψ is C1, then the unique C2 solution of (2.11) is given by

(2.12) u(x, t) =1

2(φ(x+ ct) + φ(x− ct)) +

1

2c

∫ x+ct

x−ctψ(y) dy

Proof: The general solution of the PDE is

(2.13) u(x, t) = F (x− ct) +G(x+ ct).

Then the initial conditions give

u(x, 0) = F (x) +G(x) = φ(x)

ut(x, 0) = −cF ′(x) + cG′(x) = ψ(x)

Integrating the second equation, −F (x) +G(x) = 1c

∫ x0ψ(y)dy + A.

Now we can solve for F and G, leading to (2.12). We leave it as an exercise to verify that theinitial conditions are satisfied.

Clearly from the formula (2.12), u(x, t) is a C2 function. Uniqueness is a consequence of thefact that F and G in the general solution are determined by the initial condition.


Formula (2.12) is known as d’Alembert’s solution. Note that since ψ specifies ut at t = 0,it is consistent that it should be integrated in a formula for u. In integrating ψ, we gain aderivative. Thus, we have the following regularity of the solution.

If φ is Cn, and ψ is Cn−1, for some n ≥ 2, then u is Cn.

In other words, the solution inherits regularity from the initial data. There is no gain or lossof regularity, a property typical of hyperbolic PDE. In fact, the formula (2.12) makes senseeven if φ or ψ are less regular than in the theorem; the corresponding function u(x, t) is knownas a weak solution, even though the derivatives of the solution seemingly required by the PDEmay not exist.

D’Alembert’s solution allows us to establish another part of well-posedness, namely continuousdependence on the data.

Proof of continuous dependence. Let u = u1, u = u2 be solutions of problem (2.11) withinitial data φk, ψk, k = 1, 2. that are bounded, and uniformly close in the sense of continuousfunctions:

|φ1(x)− φ2(x)| < ε; |ψ1(x)− ψ2(x)| < ε, −∞ < x <∞,

where ε > 0 is small. From (2.12),

|u1(x, t)− u2(x, t)| =

= |12(φ1 − φ2)(x+ ct) + 1

2(φ1 − φ2)(x− ct) + 1

2c

∫ x+ct

x−ct (ψ1 − ψ2)(y) dy|

≤ 12|(φ1 − φ2)(x+ ct)|+ 1

2|(φ1 − φ2)(x− ct)|+ 1

2c

∫ x+ct

x−ct |(ψ1 − ψ2)(y)| dy

≤ 12ε+ 1

2ε+ 1

2c2ct ε = (1 + t)ε.

In this calculation, we have used the triangle inequality, and the integral estimate∣∣∫ f(x)dx

∣∣ ≤∫|f(x)|dx.

If follows that if |φ1 − φ2| and |ψ1 − ψ2| are uniformly small, then |u1(x, t)− u2(x, t)| is smallat each x, t <∞. Note that the estimate gets worse in time, so that to make u1−u2 uniformlysmall in x and t, we have to take a finite time interval, unless we include

∫∞−∞ |ψ1(y)−ψ2(y)| dy

in the smallness condition.

Example 1. Representing d’Alembert’s solution graphically.For simplicity, let’s take the initial velocity to be zero, ψ(x) ≡ 0, choose c = 2, and take thesupport of φ to be the interval [1, 3]. The support of a function φ is defined to be the closure

(including boundary points) of the set where φ is non-zero. Thus, spt φ ≡ {x : φ(x) 6= 0}.

The solution of (2.11) in example 1 is:

(2.14) u(x, t) =1

2(φ(x+ 2t) + φ(x− 2t)).


1

x0 1 2 3

φ

Figure 4.4. Initial displacement φ(x).

The initial data are piecewise linear, with changes in slope at x = 1, 2, 3. Correspondingly, thesolution (at a fixed t) will be piecewise linear, with changes in slope expected at values of xfor which

x± 2t = 1; x± 2t = 2; x± 2t = 3.

The graph of φ(x+ 2t) as a function of x has the shape of a triangle moving to the left withspeed 2, and φ(x− 2t) has the same shape and speed, but moving to the right. The solution(2.14) is the average of these two graphs.

We can represent the solution in the x-t plane, shown in Fig 4.5. The leading edge of theleft-moving triangular wave lies on the characteristic x+ 2t = 1; it gets to x = 0 when t = 1

2.

Starting at x = 1, and moving left with speed 2, the wave takes until t = 12

to get to 0.

x

t

1 3

u = 0

u = 0 u = 0

Figure 4.5. The x-t plane for Example 1. Note that the slope of the charac-teristics is the reciprocal of the wave speed.

Domain of dependence and region of influence.


The structure of the characteristics in the x-t plane in Fig. 4.5 suggests how the initial datapropagate and influence the solution. Likewise, we can consider the dependence on the initialdata of the solution at a point (x, t).

The backwards characteristics through a point a point (x0, t0) with t0 > 0 are the lines

x± ct = x0 ± ct0, 0 ≤ t ≤ t0.

The backwards characteristics intersect the x axis at x = x0± ct0; the solution of the Cauchyproblem at (x0, t0) depends only on the initial data in the interval between these points.The interval is referred to as the interval of dependence of the point (x0, t0). More commonterminology is to refer to the triangle with base given by the interval of dependence and sidesgiven by the backwards characteristics as the domain of dependence of the point (x0, t0).

The region of influence of a point (x0, t0) is the set bounded by the forward characteristics

x± ct = x0 ± ct0, t ≥ t0.

We can also speak of the region of influence of a subset of the x-t plane, but more commonlywe will refer to the region of influence of an initial interval a ≤ x0 ≤ b, with t = 0. Theregion of influence of the interval [a, b] on the x-axis is the set bounded by characteristicsx+ ct = a, x− ct = b.

x

t

x

t

(x ,t )0

0 00

0

( x ,t )0 0

x +ct0x - ct a b

Domain of Dependence of Region of In�uence of [a,b]

Figure 4.6. Domain of dependence and region of influence for the wave equa-tion.

These notions give us a graphical means of understanding how initial data propagates forwardin time, shown in Fig. 4.6. In particular, if the data have compact (i.e., bounded) supportin an interval [a, b], then the solution is necessarily zero outside the region of influence ofthe initial interval [a, b], as can be seen by drawing backwards characteristics from any pointoutside this region of influence. We say that initial disturbances (meaning where φ or ψ arenon-zero) propagate with finite speed c.


4.2.2. The wave equation on a semi-infinite domain. The Cauchy problem showshow initial disturbances propagate as waves in free space. To describe how these waves arereflected at a boundary, we formulate and solve an initial boundary value problem on thequarter-plane {(x, t) : x > 0, t > 0} with a single boundary.

Consider the initial boundary value problem

(2.15)

utt = c2uxx, x > 0, t > 0,

u(0, t) = 0, t > 0,

u(x, 0) = φ(x), x > 0,

ut(x, 0) = ψ(x), x > 0.

The boundary condition specifying u(0, t) means that the end of the string is held in place. Wecould instead specify the slope ux(0, t), which would mean the stress is specified. In particular,the boundary condition ux(0, t) = 0 is referred to as a stress-free boundary condition.

For x > ct, we have d’Alembert’s solution (2.12):

(2.16) u(x, t) =1

2[φ(x+ ct) + φ(x− ct)] +

1

2c

∫ x+ct

x−ctψ(y)dy,

with φ, ψ evaluated only at positive values of their arguments.

To obtain an expression for the solution in the region 0 < x < ct, we have to use the boundarycondition since formula (2.16) does not apply for x− ct < 0. Notice that characteristics withpositive speed c propagate into the domain from the boundary x = 0 as t increases, see Fig.4.7. These characteristics carry information from the boundary condition. Moreover, eachpoint in the quarter plane also has a characteristic moving left with speed c that originatedon the initial line x > 0, t = 0. Therefore, the solution for x < ct will involve both the initialcondition and the boundary condition.

The solution can be found from the general solution (2.13), obtaining expressions for thefunctions F,G from the initial and boundary conditions, much as was done in the proof ofTheorem 4.1. The result of this calculation is the formula

(2.17) u(x, t) =1

2(φ(x+ ct)− φ(ct− x)) +

1

2c

∫ x+ct

ct−xψ(y)dy, 0 < x < ct.

Observe that this formula satisfies the boundary condition u(0, t) = 0. Formula (2.17) canbe interpreted as the solution of the Cauchy problem with initial data on the entire real lineobtained by extending both φ and ψ to be odd functions, so that φ(−x) = −φ(x), ψ(−x) =−ψ(x), x > 0. Then formula (2.17) uses the oddness of the extension to express the solutionentirely in terms of the given data on the positive x-axis.

Moreover, there is another interpretation of this construction. The solution of the Cauchyproblem with odd initial data involves waves moving left and right in the upper half plane.Those with x < 0 and moving right have the property that along the line x = 0 (the t-axis)


they exactly cancel waves moving left. This cancellation explains how the boundary conditionu(0, t) = 0 is satisfied.

Example 2Let’s suppose suppφ ⊂ [a, b] and suppψ ⊂ [a, b]. The solution is represented in the x-tplane in Fig. 4.7, where we have drawn forward characteristics from the boundary of thesupport of the initial data, including their reflections from the boundary x = 0. The reflectedcharacteristics record the switch from x − ct to ct − x in the solution; equally, we can thinkof the reflected characteristics as originating from the x-axis, and carrying information fromthe extended initial data. Finally, the reflected characteristics carry information from theboundary (specifically, the boundary condition) into the interior of the domain. This lastpoint of view is helpful when considering non-zero boundary data. In Fig. 4.7, the domain isdivided into sectors in which we can write the solution in more detail. For example, the solutionis zero in three of the regions. In each of the other regions, we use backwards characteristicsto see which part of the support of the initial data is used in calculating the solution.

x

t

a b

u = 0

u = 0 u = 0

IV

III

I

V

II

x-ct ct-x x+ct

(x,t)

Δ

x = ct

Figure 4.7. Characteristics for the quarter-plane problem.

For (x, t) in the triangular region labeled ∆, the backward characteristics hit the x-axis atx− ct and x + ct, both of which lie in the interval [a, b] Thus, there is no simplification, andu is given by d’Alembert’s formula (2.16).

In region I, x− ct < a, and ct−x < a, while a < x+ ct < b, so that both formulae (2.16,2.17)reduce to

u(x, t) =1

2φ(x+ ct) +

1

2c

∫ x+ct

a

ψ(y)dy.


this is a wave traveling to left, a function of x+ ct.

In region II, u(x, t) is likewise a function of x− ct; the graph is a wave traveling to the right.

In region III, part of the wave is reflected by the boundary and interacts with the wave travelingtowards the boundary. In fact there is just enough cancellation so that the boundary conditionis satisfied at x = 0. The reflected wave emerges as a function of (x− t) in region IV. Thus inregion III, where both waves are present, and a < ct− x < x+ ct < b, u(x, t) is given by thefull formula (2.17).

In region IV, after the left moving wave has been fully reflected, and is now moving to theright, we have x+ ct > b, and thus

u(x, t) = −1

2φ(ct− x) +

1

2c

∫ b

t−xψ(y) dy.

In region V, x − ct and ct − x are less than a, while x + ct > b. This case is shown in Fig.4.7. Therefore, there is no contribution from the initial displacement φ, and the contributionfrom the initial velocity is constant:

u(x, t) =1

2c

∫ b

a

ψ(y) dy.

Similarly, in the region between IV and the t axis, we find u ≡ 0. We see this by noting thatboth x+ ct and ct− x are larger than b. Equivalently, since x− ct < −b < b < x+ ct, we findthat u is the integral of the odd extension of ψ from −b to b, and hence is zero.

It is instructive to sketch the solution carefully for various values of t, say for the exampleearlier, in which φ is triangular, and ψ is zero.

Example 3Consider the initial boundary value problem with non-zero boundary condition:

(2.18)

utt = c2uxx, x > 0, t > 0,

u(0, t) = h(t), t > 0,

u(x, 0) = φ(x), x > 0,

ut(x, 0) = ψ(x), x > 0.

The solution is similar to that in (2.16), except there is an additional term that propagates theboundary data into the domain x > 0. To derive the additional term, first observe that if φ andψ are zero then the string is initially horizontal and at rest. The displacement h(t), specifiedat the boundary, propagates into the interior, inducing motion of the string. Consequently,this disturbance propagates as a wave u(x, t) = F (x− ct) with speed c. Setting x = 0 in thissolution, we match the boundary condition: F (−ct) = h(t). Consequently, F (ξ) = h(−ξ/c),for ξ < 0, and the solution

u(x, t) = h

(ct− xc

)


follows immediately. The full solution is obtained by simply superimposing the solution withh ≡ 0. Note that right-moving characteristics carry only half the information of the solutionfor the Cauchy problem, but carry all the boundary information, since the boundary datatravels only to the right in the physical domain x > 0.

4.3. The Energy E(t) and Uniqueness of Solutions

In this section, we define an energy function for the wave equation, show that energy isconserved for the Cauchy problem (2.11), and use this property to establish uniqueness ofsolutions of the Cauchy problem.

Let’s assume that u = u(x, t) is a smooth solution of the Cauchy problem and the derivativesut(x, t), ux(x, t) are square integrable (i.e., in L2(R)) for each t ≥ 0. Then the total energydefined by

E(t) =

∫ ∞−∞

(1

2u2t +

c2

2u2x

)dx

is finite.2

Note that E(t) is the sum of kinetic and potential energy. The potential energy PE(t) =∫∞−∞

c2

2u2x dx is the energy stored in the string due to tension, and the kinetic energy KE(t) =∫∞

−∞12u2t dx is akin to the quantity 1

2mv2 in classical mechanics of a rigid body with mass m

and velocity v.

To see how E(t) is connected to the one-dimensional wave equation (2.8) we multiply thePDE by ut and integrate by parts.∫ ∞

−∞ut utt dx =

∫ ∞−∞

c2 uxx ut dx.

Thus,

∫ ∞−∞

∂

∂t

u2t

2dx = c2 ux ut

∣∣∣∣+∞x=−∞

− c2

∫ ∞−∞

ux uxt dx

= −c2

∫ ∞−∞

∂

∂t

u2x

2dx.

That is, E ′(t) =d

dt

∫ ∞−∞

(1

2u2t +

c2

2u2x

)dx = 0

Therefore, we have conservation of total energy: E(t) = constant, from which we deduce

E(t) = E(0) =1

2

∫ ∞−∞

(ψ(x)2 + c2 φ′(x)2

)dx, t > 0.

2Since the constant density ρ has been absorbed into c2, the physical energy is actually ρE(t).

4.4. DUHAMEL’S PRINCIPLE FOR THE INHOMOGENEOUS WAVE EQUATION. 53

This identity is an important tool for existence and regularity of solutions, but also for unique-ness of solutions, as we now discuss.

Uniqueness of Solutions

Consider the Cauchy problem

(3.19)utt = c2uxx + f(x, t), −∞ < x <∞, t > 0

u(x, 0) = φ(x), ut(x, 0) = ψ(x), −∞ < x <∞,

in which the inhomogeneity f(x, t) is a specified function representing a time dependent dis-tribution of force along the one-dimensional elastic body. For example, if the PDE representssmall transverse vertical vibrations of an elastic string, then f(x, t) = −g could be the forcedistribution due to gravity. (Note that the density ρ has been absorbed into c2.)

We can use the energy calculation to prove uniqueness of C2 solutions of (3.19). Consider twoC2 solutions u1, u2 with the same data φ, ψ, f. To prove uniqueness, we show u1 = u2.

Define u(x, t) = u1(x, t) − u2(x, t). Then u satisfies the homogeneous version of (3.19), withzero initial data:

utt = c2uxx

u(x, 0) = 0, ut(x, 0) = 0.

Since E(t) = E(0) = 0 for this problem, we have∫ ∞−∞

(1

2u2t +

c2

2u2x

)dx = 0.

Therefore,

ut = 0 and ux = 0.

Thus, u is constant in x and t. But u(x, 0) = 0, so the constant is zero. Hence u = u1−u2 ≡ 0.

4.4. Duhamel’s Principle for the Inhomogeneous Wave Equation.

Duhamel’s Principle is used to solve inhomogeneous initial boundary value problems, whenwe have the solution of the homogeneous problem in hand.

Consider the initial value problem

utt = c2uxx + f(x, t), −∞ < x <∞, t > 0

u(x, 0) = 0, ut(x, 0) = 0 −∞ < x <∞.(4.20)


To solve (4.20) using Duhamel’s Principle, let u(x, t; s) be the solution (for each s > 0) of

utt = c2uxx, t > s, −∞ < x <∞u(x, s; s) = 0ut(x, s; s) = f(x, s)

(4.21)

To understand why u might be helpful, consider the special case in which f(x, t) = F (t) isindependent of x, and we seek a solution v(t) independent of x. Then

v′′(t) = F (t), v(0) = 0 = v′(0).

Integrating twice and reversing the order of integration, we see that

v(t) =

∫ t

0

(t− s)F (s) ds.

But for each s > 0, w(s, t) = (t− s)F (s) solves the x-independent version of (4.21), namely

wtt = 0, w(s, s) = 0, wt(s, s) = F (s),

and v(t) =∫ t

0w(s, t) ds.

This calculation suggests that u(x, t) =

∫ t

0

u(x, t, s)ds solves (4.20). To complete the solution,

it remains to find u. But u satisfies a Cauchy problem for the homogenous wave equation,with initial condition at time t = s. We can adapt d’Alembert’s solution by translating t bys in d’Alembert’s formula, with φ(x) = 0; ψ(x) = f(x, s). This gives a formula for u :

u(x, t; s) =1

2c

∫ x+c(t−s)

x−c(t−s)f(y, s) dy

Now let u(x, t) =

∫ t

0

u(x, t, s)ds. That is, u(x, t) =1

2c

∫ t

0

∫ x+c(t−s)

x−c(t−s)f(y, s) dy ds.

The double integral is an integration of f over the triangular domain of dependence of (x, t)shown in Fig. 4.6 .

Claim: u(x, t) satisfies (4.20).

Proof: Let (x0, t0) be fixed with t0 > 0. The proof involves integrating the PDE (4.20) overthe domain of dependence ∆ = {(x, t) : x0 − c(t − t0) < x < x0 + c(t − t0), 0 < t < t0}and using Green’s Theorem in the plane (see the Appendix). With the exact differentialdu = uxdx+ utdt, the integral on the boundary is reduced to u(x0, t0).

It is straightforward to use this procedure to solve the more general initial value problem inwhich the initial data can be non-zero:

utt = c2uxx + f(x, t), −∞ < x <∞ t > 0

u(x, 0) = φ(x), ut(x, 0) = ψ(x), −∞ < x <∞.

4.5. THE WAVE EQUATION ON R2 AND R3 55

We solve this problem by considering each of f , φ, ψ separately, setting the others to zero;the solution is then the sum of the corresponding solutions:

u(x, t) =1

2c

∫ t

0

∫ x+c(t−s)

x−c(t−s)f(y, s)dyds+

1

2(φ(x+ ct) + φ(x− ct)) +

1

2c

∫ x+ct

x−ctψ(y)dy.

4.5. The Wave Equation on R2 and R3

The method of spherical means uses the rotational and translational invariance of the waveequation to find solutions that are the analog in Rn, n ≥ 2 of d’Alembert’s solution in onedimension. The resulting formulae allow us to understand the Huygens principle of wavepropagation. Huygens originally expressed his principle geometrically using spheres centeredat points of a wavefront, with radius given by an incremental time, arguing that the intensitieswould cancel except along the expanding surface formed as the envelope of the overlappingspheres.

Here, we show briefly how to derive an explicit formula for the solution of initial value problemsin R2 and R3. It is in fact easier to start with R3. Suppose u(x, t) is a solution of the waveequation

utt = c2∆u, x ∈ R3, t ≥ 0,

with Cauchy data u(x, 0) = φ(x), ut(x, 0) = ψ(x). For r > 0, define the spherical meansv(r, t) = u(r, t) = −

∫S(0,r)

u(x, t) dS, where S(x, r) denotes the sphere with center x, radius r.

(See Appendix A for the integral average notation −∫

.) If we can find v(r, t), then we recoveru(0, t) = limr→0 v(r, t) But this will give a formula for u(x, t) for any x, t, by centering thespheres at a general point x ∈ R3 instead of at x = 0. Here are the main steps in constructingthe formula.

1. Observe that v(r, t) is a rotationally invariant solution of the wave equation:

vtt = c2(vrr +2

rvr),

with initial data given by the averages of the data for u : v(r, 0) = φ(r); vt(r, 0) = ψ(r).Notice that v(r, t) is symmetric about the origin, but we could equally well have centered thespheres at any point x in space, taking integral averages over the resulting spheres S(x, r),leading to the same Cauchy problem for v.

2. Let w(r, t) = rv(r, t). Then w satisfies the one-dimensional wave equation

wtt = c2wrr, r > 0, t > 0,

with initial data w(r, 0) = rφ(r); wt(r, 0) = rψ(r), and boundary condition w(0, t) = 0. Con-sequently, d’Alembert’s solution can be used to solve this quarter-plane problem, effectivelyusing the even extensions of φ(r);ψ(r), or equivalently the odd extensions of rφ(r); rψ(r).


3. Since we are interested only in u(0, t) = limr→0 v(r, t) = limr→0w(r, t)/r, we write thesolution with r < ct :

(5.22) w(r, t) =1

2c

∫ ct+r

ct−rsψ(s) ds+

∂

∂t

1

2c

∫ ct+r

ct−rsφ(s) ds.

Notice that we write the final φ term as the derivative of an integral, which turns out to bemore convenient than the equivalent form used in subsection 4.2.1.

4. Now we compute the limit as r → 0 :

u(0, t) = limr→0

w(r, t)/r =∂w

∂r(0, t),

since w(0, t) = 0. Computing this derivative using (5.22) above, and including the formulaefor the integral averages, we find

(5.23) u(0, t) = t−∫S(0,ct)

ψ(y) dS +∂

∂t

{t−∫S(0,ct)

φ(y) dS

}.

5. Finally, we observe that by translation invariance, the corresponding formula applies bycentering the spheres at any point x ∈ R3.

(5.24) u(x, t) = t−∫S(x,ct)

ψ(y) dS +∂

∂t

{t−∫S(x,ct)

φ(y) dS

}.

Solution in the plane (n = 2)The formula (5.24) can be adapted to find the solution of the wave equation in two spacedimensions. The idea is to consider solutions v(x1, x2, t) = u(x1, x2, x3, t) that are indepen-dent of x3 Then v satisfies the wave equation in two dimensions, and in three dimensions.Since initial data φ, ψ are in two dimensions, we consider them as functions of x1, x2, x3 butindependent of x3. Then we calculate the integrals in (5.24) by representing the spheres overthe projections onto the x1 − x2 plane, which are disks. This process gives the formula

(5.25) v(x, t) =1

2πc

∫B(x,ct)

ψ(y)

[c2t2 − |x− y|2]1/2dy +

∂

∂t

1

2πc

∫B(x,ct)

φ(y)

[c2t2 − |x− y|2]1/2dy.

Everything here is to be interpreted in two dimensions. Thus, x = (x1, x2),y = (y1, y2),B(x, ct) is the two-dimensional disk centered at x, and with radius ct, and dy = dy1dy2.

Because of the way the formula (5.25) in two dimensions is related to the formula (5.24) inthree dimensions, it is not surprising that the integrals are over the disks B(x, ct). However,this has a profound consequence, because now the solution depends on values of φ and ψinside the disk B(x, ct), in contrast to the three-dimensional case, in which the dependenceis only on values of the data on the expanding sphere S(x, ct). Thus, the Huygens principledoes not hold in two dimensions. For example, waves generated by dropping a stone into theflat surface of a body of water generates not just a circular expanding wave, but also lots ofconcentric ripples behind the leading wave.


Problems

1. Consider the initial value problem

utt = uxx, −∞ < x <∞, t > 0,

u(x, 0) = φ(x), −∞ < x <∞,

ut(x, 0) = ψ(x), −∞ < x <∞.

Let φ(x) be the function with graph shown in Fig. 4.4, and let ψ(x) ≡ 0. In the xt−planerepresentation of the solution in Fig. 4.5, we find u ≡ 0 in the middle section with t > 1

2.

Show that if we keep the same φ but make ψ non-zero, with supp ψ = [1, 3], then u will stillbe constant in this middle section. Find a condition on ψ that is necessary and sufficient tomake this constant zero.

2. Consider C3 solutions of the wave equation

(5.26) utt = c2uxx

For c = 1, define the energy density e = 12(u2

t +u2x), and let p = utux (the momentum density).

(i) Show that et = px, pt = ex.

(ii) Conclude that both e and p satisfy the wave equation.

3. Suppose u(x, t) satisfies the wave equation (5.26). Show that:

(i) For each y ∈ R, the function u(x− y, t) also satisfies (5.26).

(ii) Both ux and ut satisfy (5.26).

(iii) For any a > 0, the function u(ax, at) satisfies (5.26). Note that the restriction a > 0 isnot necessary.

4. (i) Let u(x, t) be a solution of the wave equation (5.26) with c = 1, valid for all x, t. Provethat for all x, t, h, k,

u(x+ h, t+ k) + u(x− h, t− k) = u(x+ k, t+ h) + u(x− k, t− h).

(ii) Write a corresponding identity if u satisfies (5.26) with c = 2.

5. Consider the quarter-plane problem

utt = 4uxx, x > 0, t > 0,

u(0, t) = 0, t > 0,

u(x, 0) = φ(x), x > 0,

ut(x, 0) = ψ(x), x > 0.

Let φ(x) be the function with graph shown in Fig 4.4, and let ψ(x) ≡ 0. Sketch the solutionu(x, t) as a function of x for t = 1

4, 3

8, 1

2, 1, 2..


6. Consider the quarter-plane problem with a homogeneous Neumann boundary condition:

utt = uxx

ux(0, t) = 0, t > 0

u(x, 0) = φ(x), x > 0

ut(x, 0) = ψ(x), x > 0

Suppose sptφ = [1, 2] = sptψ.

(i) Solve for u(x, t), x ≥ 0, t > 0.

(ii) Where can you guarantee u = 0 in the first quadrant of the xt−plane?

(iii) Consider φ ≡ 0; write a formula for u.

(iv) If 0 is in the support of φ or ψ (e.g., if limx→0+ φ(x) 6= 0), write conditions that guaranteeu is (a) continuous and (b) C1. Explain your answers in terms of the behavior of the dataaround the boundary of the domain. (Any compatibility condition will be effectively at theorigin, but you will need to match u, ux and ut across x = t. )

7. Consider problem 6, but in the more general case in which ux(0, t) = h(t) is not identicallyzero. Here, the general solution can be employed with the boundary condition to find F (ξ)for ξ < 0 in terms of G(−ξ) and h. Using this approach, derive the solution

u(x, t) = −∫ t−x

0

h(y) dy +1

2(φ(x+ t) + φ(t− x)) +

1

2

∫ x+t

t−xψ(y)dy

for x < t. Derive a suitable compatibility condition at the origin that ensures the solution iscontinuous when the data are continuous. What about the first derivatives across x = t?

8. Consider the wave equation that includes frictional damping:

(5.27) utt + µut = c2uxx,

in which µ > 0 is a damping constant. Show that if u(x, t) is a C2 solution with ux → 0 asx→∞, then the total energy E(t) =

∫∞−∞

12(u2

t + c2u2x) dx is a decreasing function.

Incidentally, can you devise a C2 function f(x) with the property f(x) approaches a constantas x→ ±∞, but f ′(x) does not approach zero?

9. Consider the quarter plane problem (2.15).

(i) Formulate the mechanical energy E(t) for solutions, and show that it is conserved. Specifyany assumptions you need on the initial data.

(ii) For the non-zero boundary conditions of Example 3, (2.18), evaluate E ′(t) in terms of thedata φ, ψ, h.


10. Let f(x, t) be a continuous function, and let ∆(x, t) denote the domain of dependenceof the point (x, t) for equation (5.26). Use the Fundamental Theorem of Calculus to showdirectly that u(x, t) = 1

2c

∫ ∫∆(x,t)

f(y, τ) dydτ satisfies

utt = c2uxx + f(x, t), u(x, 0) = 0 = ut(x, 0).

11. Consider the wave equation in three dimensions, with initial conditions in which φ(x) =f(|x|) is rotationally symmetric, the function f satisfies f(r) = 0, r ≥ ε, and ψ ≡ 0. Showthat the solution u(x, t) is (i) rotationally symmetric, and (ii) zero outside a circular stripcentered at the origin and having width ε.

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