d Alembert Solution

Chapter 7

Wave Equation

[Carrier & Pearson, “PDEs Theory and Technique”, Academic Press, NY, 1976, Chapter 3; Zauderer,“PDEs of Applied Mathematics”, John Wiley & Sons, NY, 1983, Chapter 6; See also Mint-U & Debnath,Chapter 4.]

In the following two to three weeks, we discuss primarily the solution and the properties of the solutionof the wave equation utt = c2uxx + h(x, t).

7.1 An Infinite Domain Problem

We start our discussion by solving the following problem.

Example 1. Solve the Cauchy problem

utt = c2uxx −∞ < x < ∞ 0 < t (7.1)

u(x, 0) = f(x) −∞ < x < ∞ (7.2)

ut(x, 0) = g(x) −∞ < x < ∞ (7.3)

[SOLUTION] Since the Cauchy data is not given on the characteristics, we can find the unique solutionby transforming the equation into the other canonical form.

Step 1. Introduce the new coordinates

ξ = x+ ct

η = x− ct

The canonical form of equation (7.1)uξη = 0

Step 2. Find the general solution from the canonical form as

u(ξ, η) = A(ξ) +B(η)

and write it in the original coordinate system

u(x, t) = A(x+ ct) +B(x− ct) (7.4)

63

64 CHAPTER 7. WAVE EQUATION

Step 3. Determine the arbitrary functions A and B in the general solution using the initial data

f(x) = u(x, 0) = A(x) +B(x)

g(x) = ut(x, 0) = cA′(x)− cB′(x)

or

A(x) +B(x) = f(x)

A(x)−B(x) =1

c

∫ x

0g(s)ds +A(0) −B(0)

Thus,

A(x) =1

2

[f(x) +

1

c

∫ x

0g(s)ds +A(0)−B(0)

]

B(x) =1

2

[f(x)− 1

c

∫ x

0g(s)ds −A(0) +B(0)

]

Step 4. Obtain the particular solution for the Cauchy problem as

u(x, t) =1

2[f(x+ ct) + f(x− ct)] +

1

2c

∫ x+ct

x−ctg(s)ds (7.5)

This solution is called the d’Alembert’s solution.

Remark 1. There are several interesting and important features associated with the d’Alembert’s solu-tion in particular, with the solution of wave equation in general:

(A) The general solution (7.4) represents two traveling waves, one with velocity c and the other withvelocity -c. In other words, information is propagated along two families of lines x ± ct = const, whichare the characteristics.

(B) From d’Alembert’s solution (7.5) of the Cauchy problem, we can observe that the solution at the pointP (x, t) is completely determined by the Cauchy data inside the domain [x− ct, x+ ct] on the x-axis. Asa matter of fact, the solution inside the triangular region enclosed by the two characteristics connecting(x, t), (x − ct, 0), and (x + ct, 0) is completely determined by the Cauchy data inside [x − ct, x + ct] onthe x-axis. This domain is called domain of dependence. On the other hand, for any point P0 on thex-axis, there are two characteristic lines coming out from this point, which form a triangular region. Anypoint inside this region is affected by the initial data at P0. Therefore, this region is called the domainof influence of this point P0. (See Fig. 1.)

In order to understand the solution behavior of the wave equation better, it is instructive to look at thesolution behavior of the following two initial data.

Example 2. Find and sketch the solution of the following problem1

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

1The initial data is discontinuous. The solution is also going to be discontinuous as one will see. Therefore, it won’tsatisfy the equation in the usual sense. One way to look at this problem is to view the discontinuous initial data as the limitof a sequence of continuous data, and the discontinuous solution is the limit of the corresponding sequence of the continuoussolutions. Or one can look at it from the view point of weak solution, see Zauderer, sec. 6.4.

7.1. AN INFINITE DOMAIN PROBLEM 65

x

t

C-: x-ct=const

(x-ct,0) (x+ct,0)

(x,t)

C+: x+ct=const

x

t

C+: x+ct=const C-: x-ct=const

(x,0)

(a) (b)

Figure 7.1: (a) Domain of determinacy (dependence): (i) the solution at (x, t) is determined by theinformation inside the triangular region; (b) domain of influence: the initial data at (x, 0) affect thesolution of any point inside the triangular region.

where

f(x) =

{1 |x| < �0 |x| > �

[SOLUTION] The d’Alembert’s solution is written

u(x, t) =1

2[f(x+ ct) + f(x− ct)]

which implies that the initial data propagates towards x = ±∞ with speed ±c. Therefore, the solutionis piecewise constant.

As shown in Fig. 2, the x− t plane can be divided into six regions, by four characteristics x± ct = ±�.We can work out the solution in each region as

t

(-l,0) (l,0)

x-ct=l

x-ct=-lx+ct=l

x+ct=-l

t*

(1)

(2)

(3) (4)

(5) (6)

x

Figure 7.2: Illustration of different regions for Examples 1 & 2.

Case 1. t < t∗ = �c :


In region 3 and 4, i.e., x < −ct− � or x > ct+ �, f(x+ ct) = f(x− ct) = 0 ⇒ u(x, t) = 12(0 + 0) = 0.

In region 5, i.e., −ct− � < x < ct− �, f(x+ ct) = 1, f(x− ct) = 0 ⇒ u(x, t) = 12(1 + 0) = 1.

In region 6, i.e., −ct+ � < x < ct+ �, f(x+ ct) = 0, f(x− ct) = 1 ⇒ u(x, t) = 12(0 + 1) = 1.

In region 1, i.e., ct− � < x < −ct+ �, f(x+ ct) = f(x− ct) = 1 ⇒ u(x, t) = 12(1 + 1) = 1.

Case 2. t > t∗ = �c :

In region 3 and 4, i.e., x < −ct− � or x > ct+ �, f(x+ ct) = f(x− ct) = 0 ⇒ u(x, t) = 12(0 + 0) = 0.

In region 5, i.e., −ct− � < x < −ct+ �, f(x+ ct) = 1, f(x− ct) = 0 ⇒ u(x, t) = 12 (1 + 0) = 1

2 .

In region 6, i.e., ct− � < x < ct+ �, f(x+ ct) = 0, f(x− ct) = 1 ⇒ u(x, t) = 12(0 + 1) = 1

2 .

In region 2, i.e., −ct+ � < x < ct− �, f(x+ ct) = f(x− ct) = 0 ⇒ u(x, t) = 12(0 + 0) = 0.

Fig. 3 presents the sketch of the solution at several different t.

x

u t=0

x

u 0<t<t*

x

u t>t*

(a) (b) (c)

Figure 7.3: Sketch of solutions for Example 1.

Example 3. Find and sketch the solution of the following problem

utt = c2uxx −∞ < x < ∞ t > 0u(x, 0) = 0 −∞ < x < ∞ut(x, 0) = g(x) −∞ < x < ∞

where

g(x) =

{1 |x| < �0 |x| > �

[SOLUTION] The d’Alembert’s solution is written

u(x, t) =1

2c

∫ x+ct

x−ctg(s)ds =

1

2c

∫ x+ct

0g(s)ds − 1

2c

∫ x−ct

0g(s)ds

As in the first example, this solution implies that the initial data propagates towards x = ±∞ with speed±c.

Since

G(y) =

∫ y

0g(s)ds =

⎧⎪⎨⎪⎩

−� y < −�y −� < y < �� y > �

7.1. AN INFINITE DOMAIN PROBLEM 67

we have

G(x+ ct) =

⎧⎪⎨⎪⎩

−� x < −�− ctx+ ct −�− ct < x < �− ct� x > �− ct

, G(x− ct) =

⎧⎪⎨⎪⎩

−� x < −�+ ctx− ct −�+ ct < x < �+ ct� x > �+ ct

Similarly as in Example 2, the x−t plane can be divided into six regions, by four characteristics x±ct = ±�(Fig. 2). We can work out the solution in each region as

Case 1. t < t∗ = �c :

In region 3, i.e., x < −ct− �, G(x+ ct) = G(x− ct) = −� ⇒ u(x, t) = 12c [−�− (−�)] = 0.

In region 4, i.e., x > ct+ �, G(x+ ct) = G(x− ct) = � ⇒ u(x, t) = 12c(�− �] = 0.

In region 5, i.e., −ct− � < x < ct− �, G(x+ ct) = x+ ct,G(x− ct) = −� ⇒ u(x, t) = 12c [x+ ct− (−�)] =

12c(x+ ct+ �).

In region 6, i.e., −ct+� < x < ct+�, G(x+ct) = x+ct,G(x−ct) = x−ct⇒ u(x, t) = 12c [x+ct−(x−ct)] = t.

In region 1, i.e., ct − � < x < −ct + �, G(x + ct) = �,G(x − ct) = x − ct ⇒ u(x, t) = 12c [� − (x − ct)] =

12c(−x+ ct+ �).

Case 2. t > t∗ = �c :

In region 3, i.e., x < −ct− �, G(x+ ct) = G(x− ct) = −� ⇒ u(x, t) = 12c [−�− (−�)] = 0.

In region 4, i.e., x > ct+ �, G(x+ ct) = G(x− ct) = � ⇒ u(x, t) = 12c(�− �] = 0.

In region 5, i.e., −ct− � < x < −ct+ �, G(x+ ct) = x+ ct,G(x− ct) = −� ⇒ u(x, t) = 12c [x+ ct− (−�)] =

12c(x+ ct+ �).

In region 6, i.e., −ct+ � < x < ct− �, G(x+ ct) = �,G(x − ct) = −� ⇒ u(x, t) = 12c [�− (−�)] = �

c .

In region 2, i.e., ct − � < x < −ct + �, G(x + ct) = �,G(x − ct) = x − ct ⇒ u(x, t) = 12c [� − (x − ct)] =

12c(−x+ ct+ �).

Fig. 4 presents the sketch of the solution at several different t.

-l

l

-l

l

G

y

u t=0

x

u t<t*

xx

u t>t*

(a) (b) (c) (d)

Figure 7.4: Sketch of solutions for Example 2.

Remark 1. There are other techniques and methods can be used to obtain the solution for the aboveproblem and more general problems. We will discuss those alternative techniques as we proceed.


7.2 Wave Equation on the Semi-infinite Domain

We now turn our attention to the solution of the wave equation on the semi-infinite domain. Let’s lookat an example first.

Example 3. Solveutt = c2uxx 0 < x < ∞ t > 0u(x, 0) = f(x) 0 < x < ∞ut(x, 0) = g(x) 0 < x < ∞u(0, t) = 0 t > 0

with f(0) = 0.

[SOLUTION] Several techniques can be used to obtain the solution of this problem. Here we use themethod of images.

Step 1. Extend both f and g to f and g, which are odd functions for all x

f(x) =

{f(x) x > 0−f(−x) x < 0

, g(x) =

{g(x) x > 0−g(−x) x < 0

Step 2. Similarly as the discussion of heat equation, we then solve the following problem

vtt = c2vxx −∞ < x < ∞ t > 0

v(x, 0) = f(x) −∞ < x < ∞vt(x, 0) = g(x) −∞ < x < ∞

and the solution can be obtained using the d’Alembert’s formula

v(x, t) =1

2

[f(x+ ct) + f(x− ct)

]+

1

2c

∫ x+ct

x−ctg(s)ds

It is clear that

v(0, t) =1

2

[f(ct) + f(−ct)

]+

1

2c

∫ ct

−ctg(s)ds = 0

due to the fact that both f and g are odd functions. Therefore

u(x, t) = v(x, t)

for x > 0.

Step 3. Straightforward algebra yields

u(x, t) =

{12 [f(x+ ct) + f(x− ct)] + 1

2c

∫ x+ctx−ct g(s)ds x > ct

12 [f(x+ ct)− f(ct− x)] + 1

2c

∫ x+ctct−x g(s)ds 0 < x < ct

(7.6)

Remark. 1 The solution written in (7.6) can be better understood by examing the x− t plane (Fig. 5),which reveals some interesting features.

7.3. WAVE EQUATION ON A FINITE DOMAIN 69

x

t

(x,t)

(x-ct,0) (x+ct,0)

(x,t)

(ct-x,0)

I IIx=ct

Figure 7.5: Illustration of the “reflection” boundary.

The characteristics play an important role. The characteristic line x = ct separates the first quadrant,which is the region of our interest, into two parts, I (0 < x < ct) and II (x > ct). In region II, the presenceof the boundary x = 0 is not felt since any information from the boundary, traveling with a fixed speedc, has not reached this region yet. In region I, because of the homogeneous Dirichlet condition, x = 0is a reflection boundary. As shown in Fig. 5, the domain of dependence becomes the shaded area dueto the influence of the boundary. In other words, at any given t, the solution is the superposition of thepropagating initial data and their images propagating in the opposite direction.

Remark 2. The boundary behavior is associated with the condition imposed. For example, the effect ofthe Neumann condition is a “positive” image of the initial data on the side of x < 0. (Ex.?)

7.3 Wave Equation on a Finite Domain

We now consider the wave equation on the finite domain. Again, we start with an example.

Example 4. Solve

utt = c2uxx 0 < x < � t > 0u(x, 0) = f(x) 0 < x < �ut(x, 0) = g(x) 0 < x < �u(0, t) = u(�, t) = 0 t > 0

[SOLUTION] Method of images. We first extend both f and g to f and g, in a similar manner

f(x) =

{f(x) 0 < x < �−f(−x) −� < x < 0

, g(x) =

{g(x) 0 < x < �−g(−x) −� < x < 0

f(x+ 2�) = f(x), g(x+ 2�) = g(x)

We then solve the following problem

vtt = c2vxx −∞ < x < ∞ t > 0

v(x, 0) = f(x) −∞ < x < ∞vt(x, 0) = g(x) −∞ < x < ∞


and the solution can be obtained using the d’Alembert’s formula

v(x, t) =1

2

[f(x+ ct) + f(x− ct)

]+

1

2c

∫ x+ct

x−ctg(s)ds

It is clear that

v(0, t) =1

2

[f(ct) + f(−ct)

]+

1

2c

∫ ct

−ctg(s)ds = 0

v(�, t) =1

2

[f(�+ ct) + f(�− ct)

]+

1

2c

∫ �+ct

�−ctg(s)ds = 0

due to the fact that both f and g are anti-symmetric about x = 0 and x = �. Therefore

u(x, t) = v(x, t)

for 0 < x < �.

Remark 1. Recall that the same problem can be solved using the method of separation variables, andthe solution is obtained in the series form

u(x, t) =∞∑n=1

(Ancos

nπct

�+Bnsin

nπct

�

)sin

nπx

�

where

An =

∫ �0 sin

nπx� · f(x)dx∫ �

0 sin2 nπx

� dx=

2

�

∫ �

0sin

nπx

�· f(x)dx

Bn =

∫ �0 sin

nπx� · g(x)dx

nπc�

∫ �0 sin

2 nπx� dx

=2

nπc

∫ �

0sin

nπx

�· g(x)dx

Physically, this solution represents the superposition of standing waves, with the base tone and over tones.The solution obtained using the method of images represents, on the other hand, the superposition oftraveling waves and their reflections off both boundaries x = 0 and x = �. The connection between thistwo solutions can be shown in the following.

Straightforward algebra, using trigonometric identities, yields

u(x, t) =∞∑n=1

√A2

n +B2n

[sin

(nπ

�(x− ct+ αn)

)+ sin

(nπ

�(x+ ct− αn)

)]

where

sinαn =An√

A2n +B2

n

, cosαn =Bn√

A2n +B2

n

Therefore, the solution obtained using separation of variables can be interpreted as the superposition oftraveling waves as well2.

Remark 2. Other method, such as Laplace transform and method of characteristics can be used to solvethis problem as well as other problems. The use of Laplace transform for this problem is left as an home-work problem. In the following, we will illustrate the use of Laplace transform on the “Whip-cracking”problem while leaving a similar problem as another homework problem. The method of characteristicswill be discussed afterwards.

2To verify these two solutions are identical, one only needs to expand f and g into the Fouries sine series.

7.4. WAVE EQUATION WITH NON-HOMOGENEOUS BCS 71

7.4 Wave Equation with Non-homogeneous BCs

So far, we have discussed the homogeneous boundary condition. The non-homogeneous boundary condi-tions can be treated similarly. We now look at an example.

Example 5. Solve the “Whip-cracking” problem

utt = c2uxx 0 < x < ∞ t > 0u(x, 0) = 0 0 < x < ∞ut(x, 0) = 0 0 < x < ∞u(0, t) = µ(t) t > 0|u(∞, s)| < ∞ t > 0

[SOLUTION] Using Laplace transform in t. Let U = L[u], µ = L[µ]. The ODE for U in x can be obtainedas

c2Uxx − s2U = 0 0 < x < ∞U(0, s) = µ(s)|U(∞, s)| < ∞

The solution of this problem can be obtained as

U(x, s) = µ(s)e−scx

Using L[f(t− a)H(t− a)] = e−asL[f ], we then obtain

u(x, t) = µ

(t− x

c

)H

(t− x

c

)

Remark 1. The solution form clearly shows the importance of the characteristic line and the characterof the solution. Once again, since the information propagates with finite velocity ±c, the boundary datawill not reach any point in the region where x > ct, where the solution is uniquely determined by theinitial data on x = 0. In the region 0 < x < ct, the solution is influenced both by the boundary data µ(t)and the initial data. Since the initial data is zero, the solution is, therefore, uniquely determined by theboundary data3.

Remark 2. The solution of this problem can be obtained using the same approach we used to obtainthe d’Alembert’s solution on the infinite domain. Here is the outline.

Step 1. We obtain the general solution in the form

u(x, t) = A(x+ ct) +B(x− ct)

Step 2. Since the argument of A is always positive while the argument of B is positive in the regionx > ct and negative in the region 0 < x < ct, we need to determine A(x) for x > 0, and B(x) for bothx < 0 and x > 0. Using the initial conditions, we have

A(x) = −B(x) = constant

3If the data µ(t) is imposed on a moving boundary, we have a more complicated and more interesting situation. The abilityof finding a unique solution depends on how fast the boundary moves. Mathematically, this problem can be generalizedinto the following problem: Given two monotonic curves, under what circumstances do we need to impose both u and un

on both curves, impose u and un on one of the curve with only u or un on the other one, or impose only one of the twoconditions on both curves? Here un is the normal derivative with respect to the curve. Part of the answer comes from ahomework problem. The so-called Gousat problem helps to address the rest of the equation, which will be discussed later.


for x > 0, therefore the solution in the region x > ct is

u(x, t) = 0

Using the boundary condition, we have

B(x) = µ

(−x

c

)+B(−x)

for x < 0, therefore the solution in the region 0 < x < ct is

u(x, t) = µ

(−x

c

)

Note that the above procedure works because we only need to know A(x) for x > 0. Note also that forthe initial value problem A(x) is the function of all x. That’s why we need two initial condition on theentire x-axis.

Remark 3. Laplace transform method as well as other techniques can be used to solve the wave equationon the finite domain with non-homogeneous Dirichlet boundary condition. One example is the followingproblem

utt = c2uxx 0 < x < � t > 0u(x, 0) = 0 0 < x < �ut(x, 0) = 0 0 < x < �u(0, t) = µ1(t) t > 0u(�, t) = µ2(t) t > 0

This is a homework problem with µ2(t) = 0. Here is the outline.

Step 1. Let U = L[u], µ1 = L[µ1], µ2 = L[µ2]. The equation for U is

c2Uxx − s2U = 0 0 < x < �U(0, s) = µ1(s)U(�, s) = µ2(s)

Step 2. Obtain the solution as

U(x, s) =1

sinhsc �

(µ1(s)sinh

s

c(�− x) + µ2(s)sinh

s

cx

)

= µ1(s)

[ ∞∑n=0

(−1)ne−sc(2n�+x) +

∞∑n=1

(−1)ne−sc(2n�−x)

]

+ µ2(s)

[ ∞∑n=1

(−1)ne−sc[(2n+1)�+x] −

∞∑n=1

(−1)ne−sc[(2n+1)�−x]

]

The solution u(x, t) can be obtained using the inverse Laplace transform

u(x, t) =∞∑n=0

(−1)nµ1

(t− 2n�+ x

c

)H

(t− 2n�+ x

c

)+

∞∑n=1

(−1)nµ1

(t− 2n�− x

c

)H

(t− 2n�− x

c

)

+∞∑n=1

(−1)nµ1

(t− (2n − 1)�+ x

c

)H

(t− (2n − 1)�+ x

c

)

−∞∑n=0

(−1)nµ1

(t− (2n + 1)�− x

c

)H

(t− (2n + 1)�− x

c

)

7.5. NON-HOMOGENEOUS WAVE EQUATION 73

7.5 Non-homogeneous Wave Equation

The wave equation with a non-homogeneous source term

utt = c2uxx + h

can be solved with several different approaches. For Cauchy problem, we have already found the solutionin an earlier homework problem when discussing the classification of second-order PDEs. The method weused there is a special case of the so-called Riemann method to be discussed later. Here we introduce ageneral principle, called Duhamel’s principle, which homogenize the equation and transform the originalproblem into a Cauchy problem of homogeneous wave equation.

Duhamel’s Principle4 Let u(x, t) be the solution of the following Cauchy problem

utt = c2uxx + h(x, t) −∞ < x < ∞, 0 < tu(x, 0) = ut(x, 0) = 0 −∞ < x < ∞

and v(x, t, τ) be the solution of

vtt = c2vxx −∞ < x < ∞, 0 < t, τv(x, τ, τ) = 0 −∞ < x < ∞, 0 < τvt(x, τ, τ) = h(x, τ) −∞ < x < ∞, 0 < τ

Then

u(x, t) =

∫ t

0v(x, t, τ)dτ

Remark 1. This result can be verified directly by differentiation.

Use the solution for v, we obtain the d’Alembert’s solution

u(x, t) =1

2c

∫ t

0

∫ x+c(t−τ)

x−c(t−τ)h(z, τ)dzdτ

Remark 2. For the initial-boundary-value problem on semi-finite and finite domain, same result exists.For example, if u(x, t) is the solution of the following Cauchy problem

utt = c2uxx + h(x, t) 0 < x < �, 0 < tu(0, t) = u(�, �) = 0 0 < tu(x, 0) = ut(x, 0) = 0 0 < x < �

and v(x, t, τ) be the solution of

vtt = c2vxx 0 < x < �, 0 < t, τv(0, t, τ) = 0 0 < x < �, 0 < τv(�, t, τ) = 0 0 < x < �, 0 < τv(x, τ, τ) = 0 0 < x < �, 0 < τvt(x, τ, τ) = h(x, τ) 0 < x < �, 0 < τ

Then

u(x, t) =

∫ t

0v(x, t, τ)dτ

4see Zauderer, p. 160.


7.6 Method of Characteristics

As we have seen so far, characteristics play an important role in the discussion of the wave equation.In this section, we look further at the characteristics by introducing the so-called Riemann invariants.These invariants can then be used to obtain the solution of the wave equation, which is called the methodof the characteristics.

Riemann Invariant. Consider the solution of the wave equation utt = c2uxx. We introduce twoquantities

R± = cux ± ut

and calculate the total differential of R± as

d(R±) = (

R±)x dx+

(R±)

t dt = (cuxx ± utx) dx+ (cutx ± utt) dt

Along two characteristics, C± : x± ct = const, we have

dx± cdt = 0

ordx = ∓cdt

Therefore, assuming the continuity of the second derivatives, the total differential can be written as

d(R+) = (

−c2uxx − cuxt + cutx + utt)dt =

(−c2uxx + utt

)dt = 0

along C+, and

d(R−) = (

c2uxx − cuxt + cutx − utt)dt =

(c2uxx − utt

)dt = 0

along C−, respectively. Thus, we conclude that R± are constants, or invariants along C±.

Method of Characteristics. The Riemann invariants can be used to obtain the solution of the waveequation, which is called the method of characteristics. We will illustrate this method in the followingseveral examples.

Example 6. Use the method of characteristics to solve the wave equation on infinite domain

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

[SOLUTION] Let P0 (x0, t0) denote a point on the x− t plane. Let (R±)0 denote the Riemann invariantsat this point. We note that there are two characteristics x ± ct = const intersecting with the x-axis atP1 (x0 + ct0, 0) and P2 (x0 − ct0, 0), where the Riemann invariants are denotes by (R+)1 and (R−)2,respectively. (Fig. 6.)

R+ and R− are constants along C+ and C−, i.e.,

(cux + ut)0 = (cux + ut)

1

(cux − ut)0 = (cux − ut)

2

7.6. METHOD OF CHARACTERISTICS 75

x

t

(x -ct ,0) (x +ct ,0)

C+: x+ct=x + ct C-: x-ct =x -ct

(x ,t )P

P P

0 0 0

0 0

000

0

12

0

0

Figure 7.6: Illustration of Characteristics paths for Example 6.

Using the Cauchy data at t = 0, we have

u1x = f ′(x0 + ct0), u1t = g(x0 + ct0), u2x = f ′(x0 − ct0), u2t = g(x0 − ct0)

Therefore,

cu0x + u0t = cf ′(x0 + ct0) + g(x0 + ct0)

cu0x − u0t = cf ′(x0 − ct0)− g(x0 − ct0)

Thus,

ut(x0, t0) = u0t =c

2

[f ′(x0 + ct0)− f ′(x0 − ct0)

]+

1

2[g(x0 + ct0) + g(x0 − ct0)]

And the solution can be obtained as

u(x0, t0) =

∫ t

0ut(x0, t)dt

=c

2

∫ t0

0

[f ′(x0 + ct)− f ′(x0 − ct)

]dt+

1

2

∫ t0

0g(x0 + ct) + g(x0 − ct)dt

=1

2[f(x0 + ct0) + f(x0 − ct0)] +

1

2

∫ x0+ct0

x0−ct0g(s)ds

which is the d’Alembert’s solution.

Example 7. Use the method of characteristics to solve the wave equation on a finite domain

utt = c2uxx −� < x < � t > 0u(x, 0) = f(x) −� < x < �ut(x, 0) = g(x) −� < x < �u(0, x) = 0 t > 0u(�, x) = 0 t > 0

[SOLUTION] Similarly, let P0 (x0, t0) denote a point in the x− t-plane. Let P0P1P3P5 and P0P2P4P6 betwo characteristics path traced back to the x-axis from P0 to two points P5 and P6 on the x-axis. (Fig. 7)


x

t P (x ,t )0 0 0

P P

P

P

P

P

1

2

3

4

5 60 l

Figure 7.7: Illustration of Characteristics paths for Example 7.

Use the boundary condition, we have u1t = u2t = u3t = u4t = 0. Therefore,

cu0x + u0t = cu1x + u1t = cu1x = cu1x − u1t

= cu3x − u3t = cu3x = cu3x + u3t

= cu5x + u5t = cf ′(x5) + g(x5)

cu0x − u0t = cu2x − u2t = cu2x = cu2x + u2t

= cu4x + u4t = cu4x = cu4x − u4t

= cu6x − u6t = cf ′(x6)− g(x6)

In principle, we can determine u0x and u0t , and integrate either one of them to obtain the solution.

Remark 1. These problems have been solved earlier using other techniques. We use them here simplyto demonstrate the method of characteristics. For these particular problems, there is no advantage usingthe method of characteristics. However, this method can be applied to more general problems while theother method fail.

Remark 2. The quantities R± are invariants along the characteristics C±, respectively if u is the solutionof the homogeneous wave equation. For non-homogeneous wave equation, they are not invariants, butmethod of the characteristics is still applicable with proper modifications. Here is an example.

Example 8. Solve the following problem by method of characteristics

utt = c2uxx + h −∞ < x < ∞, 0 < tu(x, 0) = f(x) −∞ < x < ∞ut(x, t) = g(x) −∞ < x < ∞

[SOLUTION] Introduce R± = cux ± ut. The total differential of R± along C± are

dR+ = (−c2uxx + utt)dt = hdt on C+

dR− = (−c2uxx + utt)dt = hdt on C−

Therefore,R+(x, t)−R+(x1, t1) =

∫ tt1h(x1 + c(t1 − t), t)dt

R−(x, t)−R−(x2, t2) =∫ tt2h(x2 − c(t2 − t), t)dt

7.7. GOURSAT PROBLEM 77

In general, these two relations can be used, similarly as in the homogeneous case, to obtain ux and ut atany point (x, t) provided the information at (x1, t1) and (x2, t2) are know. Here x1+ct1 = x+ct, x2−ct2.u at (x, t) can be obtained by integrate either ux or ut.

In particular, for the present example, let t1 = t2 = 0, we find x1 = x+ ct and x2 = x− ct, and

R+(x1, t1) = R+(x+ ct, 0) = cux(x+ ct, 0) + ut(x+ ct, 0) = cf ′(x+ ct) + g(x+ ct)R−(x2, t2) = R+(x− ct, 0) = cux(x− ct, 0)− ut(x− ct, 0) = cf ′(x− ct)− g(x− ct)

Thereforecux(x, t) + ut(x, t) = cf ′(x+ ct) + g(x+ ct) +

∫ t0 h(x+ c(t− t), t)dt

cux(x, t)− ut(x, t) = cf ′(x− ct)− g(x− ct) +∫ t0 h(x− c(t− t), t)dt

and

ut(x, t) =c

2

[f ′(x+ ct)− f ′(x− ct)

]+1

2[g(x+ ct) + g(x− ct)]+

1

2

∫ t

0[h(x+ c(t− t), t)− h(x− c(t− t), t)] dt

Integrate it to recove the d’Alembert’s formula.

Remark 3. So far we have discussed the initial value or initial-boundary value problems. From ourdiscussion before, we also know that the initial data imposed on the characteristics is not sufficient forobtaining a unique solution, for hyperbolic equation in general. However, there are situations that weneed to solve for problems with initial data imposed on one or two of the characteristics. These are theso-called Goursat problems to be discussed in the following.

7.7 Goursat Problem

The simplest Goursat problem is the following characteristic initial value problem

utt = c2uxx −ct < x < ct t > 0u(x, t) = f(x) x = ct t > 0u(x, t) = g(x) x = −ct t > 0

with f(0) = g(0).

[SOLUTION] Using the general solution of the form

u(x, t) = A(x+ ct) +B(x− ct)

and the initial conditions

f(ct) = u(ct, t) = A(2ct) +B(0)

g(−ct) = u(−ct, t) = A(0) +B(−2ct)

we obtainA(x) = f

(x2

)−B(0) x > 0B(x) = g

(x2

)−A(0) x < 0


Therefore, the solution is obtained as

u(x, t) = f

(x+ ct

2

)+ g

(x− ct

2

)

for t > 0 and −ct < x < ct.

Remark 1. Note that if the initial data is only specified for t > 0, which is a sensible restriction for mostphysical problems, we can only determine the solution u(x, t) in the region between two characteristicsx = ±ct for t > 0. A more general problem is the following problem

utt = c2uxx vt < x < ct t > 0u(x, t) = f(x) x = ct t > 0u(x, t) = g(x) x = vt t > 0

with f(0) = g(0), and v �= c.

[SOLUTION] For v = −c, we recover to the previous problem. A similar discussion reveals that we canonly determine the solution if −c ≤ v < c. A more general discussion can be found in problem # 5 ofhomework # 5.

Remark 2. This situation can be rescued by imposing the initial data for all t as shown in the followingexample 5

utt = c2uxx xγ(t) < x < ctu(x, t) = f(x) x = ctu(x, t) = g(x) x = xγ(t)

where xγ(t) is a strictly monotonic function with xγ(0) = 0, and f(0) = g(0).

[SOLUTION] The solution can be obtained using the same approach used earlier. First we write downthe general solution as

u(x, t) = A(x+ ct) +B(x− ct)

Use the initial conditions

f(ct) = u(ct, t) = A(2ct) +B(0)

g(xγ(t)) = u(xγ(t), t) = A(xγ(t) + ct) +B(xγ(t)− ct)

we obtain

A(x) = f

(x

2

)−B(0)

Let y = xγ(t)− ct, we have t = h(y) due the fact that xγ(t) is a monotonic function. Therefore,

B(y) = g(xγ(h(y))) − f

(xγ(h(y))

2

)+B(0)

Therefore, the solution is obtained as

u(x, t) = f

(x+ ct

2

)− f

(xγ(h(x − ct))

2

)+ g(xγ(h(x− ct)))

5Myint-U & Debnath, Example 4.9.1, p. 83.

7.8. RIEMANN METHOD 79

Remark 1. We have only solved the Goursat problems for the simple wave equation. There are Goursatproblems for the general hyperbolic equations. We will discuss that in the next section in conjunctionwith our discussion of the Riemann method.

7.8 Riemann Method

The Riemann method is a general technique used for solving the hyperbolic equations. Often, it isapplicable to the problems when the other methods fail to generate a solution. We start our discussionusing the following Cauchy problem6

Lu := uxy + aux + buy + cu = h

with u and un = ∂u/∂n given on Γ, which is not a characteristic and intersects at most once with eachcharacteristic (Sec 4.8, Myint-U & Debnath).

[SOLUTION] Recall that a simpler problem was solved earlier. Here we take a similar approach, butmuch more general. First, we note that the adjoint operator L∗ of L on a given function v is

L∗v := vxy − (av)x − (bv)y + cv

It is straightforward to show thatvLu− uL∗v = Ux + Vy (7.7)

where U = auv − uvy and V = buv + vux.

x0

y

D

Γ

P

P

01

2

0P(x ,y )0

Figure 7.8: Illustration of the Riemann method.

Let P0 (x0, y0) be a point off the initial Γ and P1 and P2 are two points on Γ such that P0P1 and P0P2

are two characteristics. Let D be the region enclosed by P0P1P2, denoted by C. (Fig. 8.) We integrate(7.7) on D ∫ ∫

DvLu− uL∗v =

∫ ∫DUx + Vy =

∫CUdy − V dx

where have used Green’s theorem. Since ???CHECK∫CUdy − V dx =

(∫ P2

P1

+

∫ P0

P2

+

∫ P1

P0

)Udy − V dx

6Without losing generality, we write the hyperbolic equation in the canonical form.


=

∫ P2

P1

Udy − V dx−∫ P0

P1

V dx+

∫ P2

P0

Udx

=

∫ P2

P1

(auv − uvy)dy − (buv + vux)dx−∫ P0

P1

(buvvux)dx+

∫ P2

P0

auv − uvydx

=

∫ P2

P1

(auv − uvy)dy − (buv + vux)dx−∫ P0

P1

(buv − uvx)dx+

∫ P2

P0

auv − uvydx− uv|P1P0

Introduce the Riemann function v(x0, y0, x, y) as the solution of the following Goursat problem

L∗v = 0av − vy = 0, x = x0bv − vx = 0, y = y0v = 1 x = x0, y = y0

We have ∫ ∫DvLu− uL∗vdxdy =

∫ ∫Dhdxdy

∫CUdy − V dx =

∫ P2

P1

(auv − uvy)dy − (buv + vux)dx− uv|P1+ u(x0, y0)

Therefore,

u(x0, y0) =

∫ ∫Dhdxdy −

∫ P2

P1

(auv − uvy)dy − (buv + vux)dx+ uv|P1(7.8)

Similarly, we have

u(x0, y0) =

∫ ∫Dhdxdy −

∫ P2

P1

(auv + vuy)dy − (buv − uvx)dx+ uv|P2(7.9)

using

uv|P2− uv|P1

=

∫ P2

P1

(uv)xdx+ (uv)ydy

Add (7.8) and (7.9) yields

u(x0, y0) =

∫ ∫Dhdxdy −

∫ P2

P1

uv(ady − bdx)− 1

2

∫ P2

P1

u(vxdx− vydy)

+1

2

∫ P2

P1

v(uxdx− uydy) +1

2

(uv|P1

+ uv|P2

)(7.10)

Remark 1. All the three forms of the solution can be used for the Cauchy problem. There are advantagesto using specific one if u and ux or u and uy are given as the initial data.

Example 9. Use Riemann method to solve the wave equation on infinite domain

utt = c2uxx + h −∞ < x < ∞, t > 0u(x, 0) = f(x) −∞ < x < ∞ut(x, 0) = g(x) −∞ < x < ∞

[SOLUTION] Step 1. Transform the equation into the canonical form. Let ξ = x − ct, η = x + ct,or x = (ξ + η)/2, t = (η − ξ)/2c. Then, ut = c(uη − uξ), utt = c2(uηη − 2uηξ + uξξ, ux = uη + uξ,uxx = uηη + 2uηξ + uξξ. The canonical form is

uξη = h

7.9. FUNDAMENTAL SOLUTION 81

where h = −h/4c2. The Cauchy data are imposed on η = ξ as

u(ξ, ξ) = f(ξ), uη(ξ, ξ) − uξ(ξ, ξ) =1

cg(ξ)

Step 2. Use Riemann method to solve the Cauchy problem in the new coordinates. First we solve forthe Riemann function v from

vξη = 0vξ(ξ, η0) = 0vη(ξ0, η) = 0v(ξ0, η0) = 1

It is straightforward to show that v ≡ 1.

Step 3. Using (7.10), the solution can be written as

u(x0, y0) =

∫ ∫Dhdxdt −

∫ P2

P1

uv(adt− bdx)− 1

2

∫ P2

P1

u(vxdx− vtdt)

+1

2

∫ P2

P1

v(uxdx− utdt) +1

2

(uv|P1

+ uv|P2

)

=1

2[f(ξ0) + f(η0)] +

1

4

∫ P2

P1

[f ′(ξ) +

g(ξ)

c

]dξ

−1

4

∫ P2

P1

[f ′(η) +

g(η)

c

]dη +

∫ ∫Dhdξdη

=1

2[f(ξ0) + f(η0)] +

1

2c

∫ η0

ξ0g(s)ds +

∫ ∫Dhdξdη

=1

2[f(x0 − ct0) + f(x0 + ct0)] +

1

2c

∫ x0+ct0

x0−ct0g(s)ds +

1

2c

∫ ∫Dhdxdt

7.9 Fundamental Solution

Similarly as for the heat equation, the fundamental solution of the wave equation is the solution with aconcentrated source.

Example 10. Solve the following problem

Ftt = c2Fxx + δ(x − z)δ(t − τ) −∞ < x, z < ∞ t, τ > 0F (x, z, 0, τ) = Ft(x, z, 0, τ) = 0 −∞ < x, z < ∞ τ > 0

[SOLUTION] Using the d’Alembert’s solution, we obtain

F (x, z, t, τ) =1

2c

∫ t

0

∫ x+c(t−τ)

x−c(t−τ)δ(x− z)δ(t − τ)dxdt

=

{12c |x− z| < c(t− τ)0 |x− z| > c(t− τ)

7.10 Discontinuity of the Solution

As evident in the fundamental solution and few examples earlier, the discontinuity in the solutions of thewave equation is persistent. Since the solution does not satisfy the wave equation in the usual sense, we


can view the solution as the limit of a sequence of smooth solutions, or in the weak sense. The discussionof weak (or general) solutions can be found in Sec. 6.4 in Zauderer.

While these discontinuities are interesting, there is another kind of discontinuity which is only associatedwith the characteristic curves. Suppose that u, ut, ux are continuous, and uxx, utt, uxt are continuouseverywhere except across a curve Γ. We now show that this is only possible if Γ is a characteristics.

Consider a second-order partial differential equation in general form

Auxx + 2Buxt + Cutt = H

where

H = G−Dux − Eut − Fu

Let [f ]Γ = f(Γ+) − f(Γ−) denote the jump of the values of f across a curve Γ, wheref(Γ±) are thelimits of f from each side of the curve. Let the parametric form of the curve be x = xΓ(s), t = tΓ(s) withparameter s. The directional derivatives along Γ are

duxds

= uxxx′Γ + uxtt

′Γ

dutds

= utxx′Γ + uttt

′Γ

With u, its first derivatives, and all the coefficients continuous and the second-derivatives discontinuous,we have

A[uxx]Γ + 2B[uxt]Γ + C[utt]Γ[H]Γ = 0

Furthermore,

[duxds

]Γ = [dutds

]Γ = 0, [uxt]Γ = [utx]Γ

Therefore

[uxt]Γ = −[uxx]Γx′Γt′Γ

= −[uxx]Γdx

dt

[utt]Γ = −[utx]Γx′Γt′Γ

= [utt]Γ

(dx

dt

)2

Thus the jump in second-derivatives are not independent. With these relationships, we obtain

[uxx]Γ

[A− 2B

dx

dt+C

(dx

dt

)2]= 0

If [uxx]Γ �= 0, we must have

A− 2Bdx

dt+ C

(dx

dt

)2

= 0

In other words, Γ must be a characteristic.

7.11. MOVING CONCENTRATED FORCE 83

7.11 Moving Concentrated Force

The fundamental solution can be viewed as the solution of the wave equation with a concentrated forcingterm in both time and space. In the following, we examine a different case where a concentrated forcingterm in space is moving.

Example 11. Solve the following problem

utt = c2uxx − F0δ(t− x

v

)vt < x < ∞, 0 < t

u(x, 0) = ut(x, 0) = 0 0 < x < ∞u(0, t) = 0, |u(∞, t)| < ∞ 0 < t

[SOLUTION] Method I. Laplace Transform. Let U = L[u] in t. Then

s2U = c2Uxx − F0e− sx

v vt < x < ∞U(0, s) = 0, |U(∞, s)| < ∞

The general solution of the equation is

U(x, s) = A(s)e−sxc +B(s)e

sxc −

{v2F0

s2(v2−c2)e−

sxv v �= c

xF02cs e

− sxc v = c

Apply the boundary conditions yields B = 0 and

A(s) =

{v2F0

s2(v2−c2) v �= c

0 v = c

which leads to

U(x, s) = −{

v2F0s2(v2−c2)

(e−

sxv − e−

sxc

)v �= c

xF02cs e

− sxc v = c

Apply the inverse transform yields

u(x, t) = −{

v2F0(v2−c2)

[(t− x

v

)H

(t− x

v

)− (t− x

c

)H

(t− x

c

)]v �= c

xF02c H

(t− x

c

)v = c

Method II. Apply d’Alembert’s formula directly. (Exercise?)

7.12 Higher-dimensional Wave Equation

The wave equation in two or higher space dimension is written as

utt = c2∇2u+ h

where ∇2 =∑n

i=1(∂2/∂x2i ) and n is the dimension. Similar discussions as those in 1D can be carried out

for this equation. Here we only look at two examples briefly. Both problems have radial symmetry andcan be reduced to 1D problems.

Example 12. Solve the following problemutt = c2∇2u


with spherical symmetry, i.e.,

∇2 =1

r2∂

∂r

(r2

∂

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

r2 sin2 θ

∂2

∂φ2=

1

r2∂

∂r

(r2

∂

∂r

)

andu(r, 0) = f(r), ut(r, 0) = g(r)

[SOLUTION] Let v = ru, and the equation for v is

vtt = c2vrr

The solution of this equation can be obtained using the d’Alembert’s form, which yields the final solution

u(r, t) =

{12r [(r + ct)f(r + ct) + (r − ct)f(r − ct) + 1

c

∫ r+ctr−ct sg(s)ds] r ≥ ct

12r [(r + ct)f(r + ct) + (ct− r)f(ct− r) + 1

c

∫ r+ctct−r sg(s)ds] r ≤ ct

Example 13. Solve the following problemutt = c2∇2u

with cylindrical symmetry, i.e.,

∇2 =1

r

∂

∂r

(r∂

∂r

)+

1

r2∂2

∂θ2+

∂2

∂z2=

1

r

∂

∂r

(r∂

∂r

)

andu(r, 0) = f(r), ut(r, 0) = g(r)

[SOLUTION] Carrier & Pearson, p. 48, sec. 3.7; Myint-U & Debnath, pp.87-92, secs. 4.10, 4.11.

Remark 1. It is also quite common that the wave phenomenon is governed by first-order PDEs, whichcan be studied using the method of characteristics. For more details, see standard texts. (e.g., Carrier& Pearson).

d Alembert Solution

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