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Chapter 4

Wave equation I:Well-posedness ofCauchy problem

In this chapter, we prove that Cauchy problem for Wave equation is well-posed (see Ap-pendix A for a detailed account of well-posedness) by proving the existence of a solutionby explicitly exhibiting a formula, followed by uniqueness of solutions to Cauchy prob-lem. The third part of the Hadamard’s criterion of a well-posed problem, which is referredto as continuous dependence on parameters in general, is proved in the form of stabilityof solutions.

Cauchy problemGiven three functions f , φ, andψ from appropriate classes of functions, Cauchy problemfor wave equation is to find a solution of

□u ≡ ut t − c2�

ux1 x1+ ux2 x2

+ · · ·+ uxd xd

�= f (x , t ), x ∈Rd , t > 0. (4.1a)

u(x , 0) = φ(x), x ∈Rd , (4.1b)

ut (x , 0) =ψ(x), x ∈Rd . (4.1c)

where x denotes the vector (x1, x2, · · · , xd ) ∈Rd , and c > 0.

Definition 4.1 (Classical Solution).

(i) A function u ∈C 2�Rd × (0,∞)�∩C 1

�Rd × [0,∞)� is said to be a classical solution ofthe Cauchy problem (4.1) if u satisfies the wave equation (4.1a), and the initial conditions(4.1b) and (4.1c) are satisfied.

(ii) Let T > 0 be fixed. When the Cauchy problem (4.1) is posed for x ∈Rd , 0< t < T , thenthe corresponding notion of a classical solution of the Cauchy problem (4.1) is definedby replacing∞ with T in (i) above.

Owing to the linearity of the d’Alembertian operator □, we reduce the problem ofsolving (4.1) to solving two simpler problems: in one of them f = 0, and in the otherφ =ψ= 0.

Let v be a solution of the homogeneous wave equation satisfying the given initialconditions, i.e.,

□v ≡ vt t − c2�vx1 x1

+ vx2 x2+ · · ·+ vxd xd

�= 0, x ∈Rd , t > 0. (4.2a)

73

74 Chapter 4. Wave equation I

v(x , 0) = φ(x), x ∈Rd , (4.2b)

vt (x , 0) =ψ(x), x ∈Rd , (4.2c)

and v be a solution of the nonhomogeneous wave equation with zero initial conditions,i.e.,

□v ≡ vt t − c2�vx1 x1

+ vx2 x2+ · · ·+ vxd xd

�= f (x , t ), x ∈Rd , t > 0. (4.3a)

v(x , 0) = 0, x ∈Rd , (4.3b)

vt (x , 0) = 0, x ∈Rd . (4.3c)

Letting u := v + v, it is easy to see that u solves the Cauchy problem (4.1).We study Cauchy problem (4.1) in the space dimensions d = 1,2,3.The organization of this chapter is as follows: Existence of solutions to Cauchy prob-

lem is proved in Section 4.1, uniqueness of solutions is proved in Section 4.2, and stabil-ity of solutions is established in Section 4.3 and thereby completing the proof of well-posedness of Cauchy problem.

In Section 4.4, initial boundary value problems are considered for one dimensionalwave equation.

4.1 Existence of solutionsIn this section, we derive an expression for solution to the homogeneous Cauchy problem(4.2) in Subsection 4.1.1 (for d = 1), Subsection 4.1.2 (for d = 3), Subsection 4.1.3 (ford = 2). The nonhomogeneous Cauchy problem (4.3) will be solved in Subection 4.1.4using Characteristic coordinates for d = 1, and using Duhamel’s principle for a generald .

We also prove that the derived expression is indeed a classical solution, when the dataf ,φ,ψ are sufficiently smooth.

4.1.1 Case of one dimensional wave equation

Cauchy problem for one dimensional wave equation takes the form

ut t = c2ux x , x ∈R, t > 0, (4.4a)u(x, 0) = φ(x), x ∈R, (4.4b)

ut (x, 0) =ψ(x), x ∈R. (4.4c)

Main steps in solving the above Cauchy problem are

(i) The wave equation (4.4a) is transformed into its canonical form by changing theindependent variables from (x, t ) to the characteristic coordinates. The resultantequation is then solved.

(ii) The general solution of the wave equation (4.4a) is obtained by change of variablesin the solution of the transformed equation.

(iii) A solution of Cauchy problem is then obtained by imposing the initial conditions(4.4b) and (4.4c), which results in the d’Alembert’s formula.

Canonical form of Wave operator

In this section, we will deduce a canonical form of the wave equation (4.4a). To this end, letus analyze how the wave equation transforms when a new coordinate system is introduced

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4.1. Existence of solutions 75

onR×R i.e., the coordinate system (x, t ) undergoes a change of variables to (ξ ,η), where(ξ ,η) = (ξ (x, t ),η(x, t )).

We have the following identites:

w(ξ ,η) = u (x(ξ ,η), t (ξ ,η)) , and u (x, t ) = w (ξ (x, t ),η(x, t )) . (4.5)

We will compute ut t and ux x by choosing an appropriate identity from (4.5). Applyingone of the most fundamental and important results in differential calculus, namely, chainrule, to u (x, t ) = w (ξ (x, t ),η(x, t )) we get

ut (x, t ) = wξ (ξ (x, t ),η(x, t ))ξt (x, t )+wη (ξ (x, t ),η(x, t ))ηt (x, t )

ux (x, t ) = wξ (ξ (x, t ),η(x, t ))ξx (x, t )+wη (ξ (x, t ),η(x, t ))ηx (x, t )

Differentiating the above set of equations once more, we get

ut t (x, t ) = wξ ξ (ξ (x, t ),η(x, t )) (ξt (x, t ))2 + 2wξ η (ξ (x, t ),η(x, t ))ξt (x, t )ηt (x, t )+

wηη (ξ (x, t ),η(x, t )) (ηt (x, t ))2 ,

ux x (x, t ) = wξ ξ (ξ (x, t ),η(x, t )) (ξx (x, t ))2+ 2wξ η (ξ (x, t ),η(x, t ))ξx (x, t )ηx (x, t )+

wηη (ξ (x, t ),η(x, t )) (ηx (x, t ))2 .

Substituting the values of ut t (x, t ) and ux x (x, t ) into (4.4a), we get

□u(x, t ) =wξ ξ (ξ (x, t ),η(x, t ))�ξ 2

t − c2ξ 2x

�(x, t )+ 2wξ η (ξ (x, t ),η(x, t ))

�ξtηt − c2ξxηx

�(x, t )

+wηη (ξ (x, t ),η(x, t ))�η2

t − c2η2x

�(x, t ) (4.6)

To obtain a canonical form of Wave equation, which is an hyperbolic equation, werequire that the coefficients of wξ ξ and wηη be zero. That is,�

ξ 2t − c2ξ 2

x

�(x, t ) = 0,�η2

t − c2η2x

�(x, t ) = 0. (4.7)

Note that both ξ and η satisfy the same equation. Since we are interested in (ξ ,η)to be a change of coordinates from (x, t ), we must choose two functionally independent(Jacobian non-zero) solutions of (4.7). Let us consider the equation satisfied by ξ , whichmay be factored as

(ξt − cξx )(ξt + cξx ) = 0. (4.8)

The factorization (4.8) allows us to find ξ ,η satisfying

ξt + cξx = 0, and ηt − cηx = 0. (4.9)

From the last equation, we conclude that ξ and η are arbitrary differentiable functionsdepending solely on x − c t and x + c t respectively. For simplicity, we choose

ξ (x, t ) = x − c t , and η(x, t ) = x + c t . (4.10)

The new coordinate system found in this way is called system of characteristic coor-dinates, in view of the following definition

Definition 4.2 (characteristics). The two families of lines

x − c t = constant, x + c t = constant

September 3, 2015 Sivaji


are called the characteristics of the one dimensional wave equation.

Note that characteristics consist of two families of straight lines with slopes± 1c . When

c = 1, the two families of characterisitcs are orthogonal.Substituting the expressions for ξ (x, t ) and η(x, t ) from (4.10) into the equation (4.6)

yieldsut t (x, t )− c2ux x (x, t ) =−4c2wξ η (ξ (x, t ),η(x, t )) (4.11)

In other words,

ut t (x, t )− c2ux x (x, t ) =−4c2wξ η (ξ (x, t ),η(x, t )) . (4.12)

Thus the homogeneous wave equation (4.4a) reduces to

wξ η(ξ ,η) = 0 (4.13)

in the coordinates ξ and η.By integrating the equation (4.13) first w.r.t. η, and then w.r.t. ξ , we get

wξ (ξ ,η) = f (ξ )

w(ξ ,η) =∫ ξ

f (ξ )dξ +G(η)

Thus the general solution of (4.13) is of the form

w(ξ ,η) = F (ξ )+G(η).

Note that the function w ∈C 2(R×R) if and only if F ∈C 2(R) and G ∈C 2(R).Thus solution u of the Wave equation in (x, t ) coordinates is given by

u(x, t ) = F (x − c t )+G(x + c t ), (4.14)

where F ∈ C 2(R) and G ∈ C 2(R). If F and G are twice continuously differentiable,then the function u defined above is easily seen to be a classical solution; which we recordbelow in the form of a lemma whose proof is left to the reader.

Lemma 4.3 (Classical solution). Let F ∈ C 2(R) and G ∈ C 2(R). Then the function udefined by the formula (4.14) is a classical solution of the homogeneous wave equation (4.4a).

Remark 4.4 (Geometric interpretation: Wave propagation). To present a geometric in-terpretation, let us fix a time instant t = t0.

(i) The graph of F (x + c t0) is precisely the graph of F (x) shifted by c t0 units towardsthe left. Thus F (x + c t ) represents a backward moving wave with speed c .

(ii) The graph of G(x − c t0) is precisely the graph of G(x) shifted by c t0 units towardsthe right. Thus G(x − c t ) represents a forward moving wave with speed c .

Thus we conclude that any solution of the wave equation is a superposition of forward,and backward moving waves.

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Solution of Cauchy problem for homogeneous Wave equation: formula of d’Alembert

Recall from (4.14) that the general solution of the wave equation is given by

u(x, t ) = F (x − c t )+G(x + c t ).

Imposing the initial condition u(x, 0) = φ(x) yields the equation

F (x)+G(x) = φ(x).

Imposing the other initial condition ut (x, 0) =ψ(x) gives

−cF ′(x)+ cG′(x) =ψ(x).

Integrating the last equation over the interval [0, x] yields

−F (x)+G(x) =1c

∫ x0ψ(s )d s − F (0)+G(0)

Solving for F and G, we get

F (ξ ) =12φ(ξ )− 1

2c

∫ ξ0ψ(s )d s +

F (0)−G(0)2

(4.15)

G(η) =12φ(η)+

12c

∫ η0ψ(s )d s − F (0)−G(0)

2(4.16)

Substituting these values in the general solution, we get the following expression for u,

u(x, t ) =φ(x − c t )+φ(x + c t )

2+

12c

∫ x+c t

x−c tψ(s )d s , (4.17)

which is known as d’Alembert’s formula.We now have the following result whose proof is left as an exercise to the reader.

Theorem 4.5 (Classical solution). Let the Cauchy data be such that φ ∈ C 2(R) and ψ ∈C 1(R). Then the function u defined by the formula (4.17) is a classical solution of the Cauchyproblem for homogeneous wave equation (4.4).

4.1.2 Case of three dimensional wave equation

Cauchy problem for three dimensional wave equation takes the form

ut t − c2�

ux1 x1+ ux2 x2

+ ux3 x3

�= 0, x = (x1, x2, x3) ∈R3, t > 0, (4.18a)

u(x , 0) = φ(x), x ∈R3, (4.18b)ut (x , 0) =ψ(x), x ∈R3. (4.18c)


(i) Note that the wave equation (4.18a) is already in a canonical form. In the case of onedimensional wave equation, the wave equation was reduced to a simpler canonicalform wξ η and a general solution was determined. As discussed in Remark 3.25,it is very cumbersome, and perhaps impossible to derive such a simpler a canonicalform as in the case of one dimensional wave equation. Thus following the procedureimplemented for wave equation in one space dimension is ruled out.



(ii) The Cauchy problem (4.18) will be reduced to an equivalent Cauchy problem for aone dimensional wave equation by following the method of spherical means. Formulaof d’Alembert will be used to solve the resultant one dimensional wave equation.

(iii) A solution of Cauchy problem (4.18) is then retrieved from the d’Alembert formulaso-obtained.

Spherical means and the Darboux formula

In this paragraph, we deal with functions of d real variables (not necessarily d = 3). LetB(x ;ρ) and S(x ;ρ) denote the open ball of radius ρ > 0 and sphere of radius ρ > 0, withcenter at x ∈ Rd respectively. That is, denoting the Euclidean norm of x ∈ Rd by ∥x∥,we have

B(x ;ρ) := {y : ∥x − y∥<ρ} and S(x ;ρ) := {y : ∥x − y∥= ρ}. (4.19)

Letωd denote the measure of the unit sphere S(0; 1) inRd , and dω denote the surfacemeasure on S(0; 1).

Definition 4.6 (Spherical means). Let g :Rd →R be a continuous function. The sphericalmean of g , denoted by M g (x ,ρ), is a function defined on Rd × (0,∞), given by

M g (x ,ρ) :=1

|S(x ;ρ)|∫

S(x ;ρ)g (y)dσ (4.20)

where dσ is the surface measure on the sphere S(x ;ρ), and |S(x ;ρ)| denotes the measure ofS(x ;ρ) w.r.t. dσ .

Remark 4.7 (on spherical means). The formula (4.20) for spherical means M g (x ,ρ)maybe written as

M g (x ,ρ) =1

|S(x ;ρ)|∫

S(x ;ρ)g (y)dσ =

1ρd−1ωd

∫∥x−y∥=ρ

g (y)dσ =1ωd

∫∥ν∥=1

g (x +ρν)dω,

(4.21)

since |S(x ;ρ)|= ρd−1ωd , and dσ = ρd−1dω.

Lemma 4.8. Let g : Rd → R be a continuous function. Let M g : Rd × (0,∞)→ R be asdefined by (4.20).

(i) The function M g can be extended to the domain Rd ×R such that ρ 7→M g (x ,ρ) is aneven function, for each fixed x .

(ii) If g ∈C k (Rd ), then so is the function x 7→M g (x ,ρ) for each ρ.(iii) The function g can be retrieved from M g (x ,ρ) in the following sense:

limρ→0

M g (x ,ρ) = g (x) for all x ∈Rd . (4.22)

Proof. Since the unit sphere is invariant under the mapping ν 7→ −ν, the last integral in(4.21) may be re-written as∫

∥ν∥=1g (x +ρν)dω =

∫∥ν∥=1

g (x −ρν)dω. (4.23)

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Thus we get

M g (x ,ρ) =1

ρd−1

∫∥ν∥=1

g (x +ρν)dω =1

ρd−1

∫∥ν∥=1

g (x −ρν)dω. (4.24)

The last equation (4.24) suggests that M g can be extended to the domainRd ×R such thatρ 7→M g (x ,ρ) is an even function, for each fixed x . Setting M g (x , 0) = g (x), the functionρ 7→M g (x ,ρ) is now meaningful for all ρ ∈R. This completes the proof of (i).

Assertion (ii) of lemma follows by differentiating under the integral sign. Assertion(iii) is an easy consequence of the continuity of g .

Lemma 4.9 (Darboux formula). Let g ∈C 2(Rd ). Then M g satisfies the partial differentialequation �

∂ 2

∂ ρ2+

d − 1ρ

∂

∂ ρ

�M g (x ,ρ) =∆M g (x ,ρ), (4.25)

where∆ stands for the Laplacian in the variables x = (x1, x2, · · · , xd ) ∈Rd .

Proof. Using integration by parts formula (see Appendix C), we get∫∥x−y∥<ρ

∆g (y)d y =∫∥x−y∥=ρ

∇g (y).ν(y)dσ(y) (4.26)

=ρd−1∫∥ν∥=1∇g (x +ρν).ν dω (4.27)

=ρd−1 ∂

∂ ρ

∫∥ν∥=1

g (x +ρν)dω. (4.28)

Also

∂

∂ ρM g (x ,ρ) =

1ρd−1ωd

∫∥x−y∥<ρ

∆g (y)d y (4.29)

=1

ρd−1ωd

∫ ρ0

r d−1∆

∫∥ν∥=1

g (x + r ν)dω d r. (4.30)

Multiply the equation (4.30) with ρd−1, and then differentiate w.r.t. ρ, to get

∂

∂ ρ

�ρd−1 ∂

∂ ρM g (x ,ρ)�=∆�ρd−1M g (x ,ρ)�

. (4.31)

Carrying out the differentiation on both sides of the equation (4.31), and then cancellingthe factor ρd−1 yields the Darboux formula (4.25).

Reduction of the Cauchy problem (4.18) to an equivalent Cauchy problem

Let u ∈C 2(Rd ×R) be a solution of (4.2). Then for each fixed t ∈R, we have

Mu (x , t ,ρ) =1ωd

∫∥ν∥=1

u(x +ρν , t )dω. (4.32)



The following equalities follow by applying Laplacian operator to the last equation:

∆Mu (x , t ,ρ) =1ωd

∫∥ν∥=1

∆u(x +ρν , t )dω

=1

c2ωd

∫∥ν∥=1

∂ 2

∂ t 2u(x +ρν , t )dω

=1c2

∂ 2

∂ t 2Mu (x , t ,ρ)

Thus we are led to define, for each fixed x ∈Rd ,

M (ρ, t ) :=1ωd

∫∥ν∥=1

u(x +ρν , t )dω. (4.34)

Lemma 4.10 (Equivalent Cauchy problems). The following statements concerning anu ∈C 2(Rd ×R) are equivalent.

1. The function u is a solution of the Cauchy problem (4.2).2. The function M (ρ, t ) solves the following Cauchy problem

∂ 2

∂ t 2M (ρ, t ) = c2�∂ 2

∂ ρ2+

d − 1ρ

∂

∂ ρ

�M (ρ, t ), ρ ∈R, t > 0, (4.35a)

M (ρ, 0) =Mφ(x ,ρ), ρ ∈R, (4.35b)

∂ M∂ t(ρ, 0) =Mψ(x ,ρ), ρ ∈R. (4.35c)

Proof. Statement 2 follows from Statement 1 by direct computation using the definitionof M (ρ, t ), computation in the equations (4.33), and Darboux formula, which is left as anexercise to the reader.

Statement 1 follows from Statement 2 in view of Lemma 4.8, the reader is urged to fillin the details of its proof.

Solution of Cauchy problem (4.18):Formulae of Poisson

From now on we consider the case d = 3, till the end of this subsection.Setting L(ρ, t ) := ρM (ρ, t ), we see that L satisfies the following Cauchy problem

∂ 2

∂ t 2L(ρ, t ) = c2 ∂

2

∂ ρ2(L(ρ, t )) , ρ ∈R, t > 0, (4.36a)

L(ρ, 0) = ρMφ(x ,ρ), ρ ∈R, (4.36b)

∂ L∂ t(ρ, 0) = ρMψ(x ,ρ), ρ ∈R. (4.36c)

Using the d’Alembert formula (4.17), we get

L(ρ, t ) = ρM (ρ, t ) =12

�(ρ− c t )Mφ(x ,ρ− c t )+ (ρ+ c t )Mφ(x ,ρ+ c t )

�+

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12c

∫ ρ+c t

ρ−c ts Mψ(x , s)d s . (4.37)

We cannot retrieve u from the equation (4.37) easily, by dividing both the sides withρ and then passing to the limit as ρ → 0. We work our way around this difficulty bydifferentiating w.r.t. ρ the equation (4.37), and use the resultant equation to pass to thelimit as ρ→ 0.

Differentiating the equation (4.37) w.r.t. ρ, we get

M (ρ, t )+ρ∂ M∂ ρ(ρ, t ) =

12

�Mφ(x ,ρ− c t )+Mφ(x ,ρ+ c t )

�+

12

�(ρ− c t )

∂

∂ ρMφ(x ,ρ− c t )+ (ρ+ c t )

∂

∂ ρMφ(x ,ρ+ c t )�

+12c

�(ρ+ c t )Mψ(x ,ρ+ c t )− (ρ− c t )Mψ(x ,ρ− c t )

�. (4.38)

Expanding the RHS of (4.38), we get

M (ρ, t )+ρ∂ M∂ ρ(ρ, t ) =

18π

�∫∥ν∥=1

φ(x +(ρ− c t )ν)dω+∫∥ν∥=1

φ(x +(ρ+ c t )ν)dω�

+1

8π

�(ρ− c t )∫∥ν∥=1∇φ(x +(ρ− c t )ν).ν dω+(ρ+ c t )

∫∥ν∥=1∇φ(x +(ρ+ c t )ν).ν dω

�+

18cπ

�(ρ+ c t )∫∥ν∥=1

ψ(x +(ρ+ c t )ν)dω− (ρ− c t )∫∥ν∥=1

ψ(x +(ρ− c t )ν)dω�

.(4.39)

Now taking the limit as ρ→ 0 in the equation (4.39), we get

u(x , t ) =1

8π

�∫∥ν∥=1

φ(x − c t ν)dω+∫∥ν∥=1

φ(x + c t ν)dω�

+1

8π

�−c t∫∥ν∥=1∇φ(x − c t ν).ν dω+ c t

∫∥ν∥=1∇φ(x + c t ν).ν dω

�+

18π

�t∫∥ν∥=1

ψ(x + c t ν)dω+ t∫∥ν∥=1

ψ(x − c t ν)dω�

. (4.40)

By Lemma 4.8, the equation (4.40) reduces to

u(x , t ) =1

4π

∫∥ν∥=1

φ(x + c t ν)dω+1

4πc t∫∥ν∥=1∇φ(x + c t ν).ν dω

+1

4πt∫∥ν∥=1

ψ(x + c t ν)dω. (4.41)

The last equation (4.41) may be written as

u(x , t ) =1

4π∂

∂ t

�t∫∥ν∥=1

φ(x + c t ν)dω�+

14π

t∫∥ν∥=1

ψ(x + c t ν)dω. (4.42)



In the formula (4.42), converting the integrals on the unit sphere to those on S(x , c t )yields us another formula for the solution of the Cauchy problem as

4πc2u(x , t ) =∂

∂ t

�1t

∫S(x ,c t )

φ(y)dσ�+

1t

∫S(x ,c t )

ψ(y)dσ (4.43)

On the other hand, the formula (4.41) gives us

4πc2u(x , t ) =1t 2

∫S(x ,c t ){tψ(y)+φ(y)+∇φ(y).(y − x)} dσ (4.44)

The formulae (4.41), (4.42), (4.43), (4.44) are known as Poisson’s formulae, which arealso called sometimes as Kirchoff’s formulae.

This finishes the derivation of a formula for the solution of Cauchy problem (4.18).The following results says that all these formulae represent a classical solution of the

Cauchy problem, if the Cauchy data is sufficiently smooth, the proof is left as an exerciseto the reader.

Theorem 4.11 (Classical solution). Let φ ∈ C 3(R3) and ψ ∈ C 2(R3). Then a classicalsolution of the Cauchy problem is given by

u(x , t ) =∂

∂ t

�1

4πc2 t

∫S(x ,c t )

φ(y)dσ�+

14πc2 t

∫S(x ,c t )

ψ(y)dσ . (4.45)

4.1.3 Case of two dimensional wave equation via Hadamard’s method ofdescent

Cauchy problem for two dimensional wave equation takes the form

ut t − c2�

ux1 x1+ ux2 x2

�= 0, x = (x1, x2) ∈R2, t > 0, (4.46a)



(i) All the hard work was already done in deriving a formula for solution of Cauchyproblem in three space dimensions. To find a solution in two space dimensions, wefollow Hadamard’s method of descent, which presumes that solutions of Cauchyproblem (4.46) are actually special solutions of the Cauchy problem (4.18) in threespace dimensions that do not depend on the x3 variable.

(ii) We then use one of the Poisson’s formulae (4.41)-(4.44), and reduce the integrals onthe spheres in three dimensions to those on disks in two dimensions. The resultantformulae yield a solution for the Cauchy problem (4.46) for two dimensional waveequation.

Let S denote the sphere with center at (x1, x2, 0) and radius c t , i.e.,

S = {y = (y1, y2, y3) ∈R3 : (x1− y1)2+(x2− y2)

2+ y23 = c2 t 2}. (4.47)

From the formula (4.45), we get

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u(x1, x2, t ) = u(x1, x2, 0, t )

=∂

∂ t

�1

4πc2 t

∫Sφ(y1, y2)dσ�+

14πc2 t

∫Sψ(y1, y2)dσ . (4.48)

Let us compute the integral∫

S φ(y1, y2)dσ , and computation of the other integral isexactly similar.

Let S+ denote the upper half of the sphere S. Since integrand is independent of the y3variable, we have ∫

Sφ(y1, y2)dσ = 2

∫S+φ(y1, y2)dσ . (4.49)

Note that the upper half of the sphere is indeed the graph of the functionh : D((x1, x2), c t )→R given by

h(y1, y2) =Æ

c2 t 2− (x1− y1)2− (x2− y2)2, (4.50)

where D((x1, x2), c t ) denotes the open disk with center at (x1, x2) having radius c t .Hence the surface measure dσ on S+ is thus given by

dσ =Æ

1+ ∥∇h(y1, y2)∥2 d y1d y2, (4.51)

which becomes

dσ =c tp

c2 t 2− (x1− y1)2− (x2− y2)2d y1d y2, (4.52)

Thus the formula (4.48) simplifies to

u(x1, x2, t ) =∂

∂ t

�1

2πc

∫D((x1,x2),c t )

φ(y1, y2)pc2 t 2− (x1− y1)2− (x2− y2)2

d y1d y2

�+

12πc

∫D((x1,x2),c t )

ψ(y1, y2)pc2 t 2− (x1− y1)2− (x2− y2)2

d y1d y2 (4.53)

Carrying out the differentiation in equation (4.53), we get

u(x , t ) =1

2πc t 2

∫D(x ,c t )

tφ(y)+ t∇φ(y).(y − x)+ t 2ψ(y)pc2 t 2−∥x − y∥2 d y. (4.54)

Re-writing the integral in the equation (4.54) on the unit disk, we get

u(x , t ) =1

2π

∫D(0,1)

φ(x + c t z )p1−∥z∥2 d z +

c t2π

∫D(0,1)

∇φ(x + c t z ).zp1−∥z∥2 d z

+t

2π

∫D(0,1)

ψ(x + c t z )p1−∥z∥2 d z . (4.55)

The following result says that the formula (4.53) represents a classical solution of theCauchy problem, if the Cauchy data is sufficiently smooth, the proof is left as an exerciseto the reader.



Theorem 4.12 (Classical solution). Let φ ∈ C 3(R2) and ψ ∈ C 2(R2). Then a classicalsolution of the Cauchy problem is given by

u(x1, x2, t ) =∂

∂ t

�1

2πc

∫D((x1,x2),c t )

φ(y1, y2)pc2 t 2− (x1− y1)2− (x2− y2)2

d y1d y2

�+

12πc

∫D((x1,x2),c t )

ψ(y1, y2)pc2 t 2− (x1− y1)2− (x2− y2)2

d y1d y2, (4.56)

where D((x1, x2), c t ) denotes the open disk with center at (x1, x2) having radius c t .

The formulae (4.56), (4.54), (4.55) are known as Poisson’s formulae, and sometimesalso as Kirchhoff’s formulae.

4.1.4 Nonhomogeneous problem

As discussed at the beginning of this section, a solution of (4.1) can be obtained by addinga solution of (4.2) to a solution of (4.3). In the previous subsections, we already solved theCauchy problem (4.2). In this paragraph we concentrate on finding a solution to (4.3),and then find a solution to (4.1).

We find a solution of (4.3) by a general principle called Duhamel’s principle for all di-mensions d . However note that wave equation in one space dimension is special whencompared to wave equation in more than one space dimensions, in the sense that a canoni-cal form exists which has a very simple form (in the characterisitcs coordinates) and henceis easier to solve. Hence we present a method based on characteristic coordinates and solve(4.3) for d = 1.

Existence of solutions via Characteristic Coordinates

The Cauchy problem (4.3) (for d = 1), in view of (4.12), is equivalent to the followingproblem in the characterisitcs coordinates given by (4.10):

wξ η(ξ ,η) =− 14c2

f�ξ +η

2,η− ξ

2c

�(4.57a)

w(ξ ,ξ ) = wξ (ξ ,ξ ) = wη(ξ ,ξ ) = 0 (4.57b)

While the equation (4.57a) follows from (4.12), the equations (4.57b) follow from

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

Integrating (4.57a) w.r.t. η, from η= ξ to η= η0, we get

wξ (ξ ,η0)−wξ (ξ ,ξ ) =− 14c2

∫ η0

ξ

f�ξ + s

2,

s − ξ2c

�d s , (4.58)

which, in view of the condition (4.57b), reduces to

wξ (ξ ,η0) =− 14c2

∫ η0

ξ

f�ξ + s

2,

s − ξ2c

�d s (4.59)

Integrating (4.59) w.r.t. ξ , from ξ = ξ0 to ξ = η0, we get

w(η0,η0)−w(ξ0,η0) =− 14c2

∫ η0

ξ0

∫ η0

zf� z + s

2,

s − z2c

�d s d z, (4.60)

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.. x.

t

.

ξ =x − c t

η=constan

t

.

η= x + c tξ = constant

.

η=η 0

.

ξ = ξ0

.x0− c t0

.x0+ c t0

.(ξ ,ξ )

.ξ = η

.

(x0, t0)

.

(ξ0,η0)

.

(ξ0,ξ0)

.(η0,η0)

Figure 4.1. Domain of integration in (x, t ) and (ξ ,η) coordinates.

which, in view of the condition (4.57b), reduces to

w(ξ0,η0) =1

4c2

∫ η0

ξ0

∫ η0

zf� z + s

2,

s − z2c

�d s d z. (4.61)

Let us now get an expression for u(x, t ) by making the following change of variables

s = x + c t , z = x − c t . (4.62)

Note that the absolute value of Jacobian of this transformation is 2c . Under this changeof variables, the domain of integration in (4.61)

D = {(s , z) : z ≤ s ≤ η0, ξ0 ≤ z ≤ η0} (4.63)

transforms into

{(x, t ) : x0− c(t0− t )≤ x ≤ x0+ c(t0− t ), 0≤ t ≤ t0}. (4.64)

Note that z ≤ s ≤ η0 implies x− c t ≤ x+ c t ≤ x0+ c t0, from which we get x0− c(t0−t )≤ x ≤ x0+ c(t0− t ). Also note that ξ0 ≤ z ≤ η0 implies x0− c t0 ≤ x − c t ≤ x0+ c t0,which in view of x − c t ≤ x + c t ≤ x0+ c t0, will yield the inequalities

x0− c t0 ≤ x − c t ≤ x + c t ≤ x0+ c t0.

From the above inequalities, we conclude that 0≤ 2c t ≤ 2c t0, from which we get 0≤ t ≤t0. Thus the equation (4.61) transforms to

u(x0, t0) =12c

∫ t0

∫ x0+c(t0−t )

x0−c(t0−t )f (x, t )d x d t =

12c

∫T

f (x, t )d x d t , (4.65)

where T is the characteristic triangle having vertices at (x0, t0), (x0 − c t0, 0), and (x0 +c t0, 0).



Solution via Duhamel’s principle

Duhamel principle gives us a way to solve nonhomogeneous problems corresponding toa linear differential operator, by superposition of solutions of a family of correspondinghomogeneous problems. In this subsection we motivate Duhamel principle using an ex-plicit example of an ordinary differential equation, and then compute solutions of Cauchyproblem for nonhomogeneous Wave equation given by (4.3) for d = 1,2,3.

Motivation for Duhamel principle: Method of variation of parameters for Ordinary differentialequations

Let µ> 0, y0, y1 ∈R. Consider the following initial value problem

d 2yd x2+µ2y = f (t ), (4.66a)

y(0) = y0, y ′(0) = y1. (4.66b)

We will solve the IVP (4.66) using method of variation of parameters. A general solu-tion of the homogeneous equation

y ′′+µ2y = 0 (4.67)

is C1 cosµt + C2 sinµt , where C1,C2 are arbitrary constants. Method of variation ofparameters attempts to find a solution of the nonhomogeneous ODE (4.66a) of the form

y(t ) =C1(t ) sinµt +C2(t )cosµt . (4.68)

Differentiating the equation (4.68), we get

y ′(t ) =µC1(t )cosµt +C ′1(t ) sinµt −µ2C2(t ) sinµt +C ′2(t )cosµt . (4.69)

Method of variaiton of parameters assumes that

C ′1(t ) sinµt +C ′2(t )cosµt = 0 (4.70)

In view of the equation (4.70), the expression for y ′(t ) in (4.69) reduces to

y ′(t ) =µC1(t )cosµt −µC2(t ) sinµt . (4.71)

Differentiating the equation (4.71), we get

y ′′(t ) =−µ2C1(t ) sinµt +µC ′1(t )cosµt −µ2C2(t )cosµt −µC ′2(t ) sinµt . (4.72)

From the equation (4.72), we get

y ′′(t )+µ2y(t ) =µC ′1(t )cosµt −µC ′2(t ) sinµt = f (t ). (4.73)

Solving for C ′1 and C ′2 from equations (4.70) and (4.73), we get

C ′1(t ) =1µ

f (t )cosµt , C ′2(t ) =− 1µ

f (t ) sinµt . (4.74)

Substituting expressions for C1(t ),C2(t ) in the formula for y(t ), and using the initial con-ditions (4.66b), we get

y(t ) = y1sinµtµ+ y0 cosµt +

1µ

∫ t0

sinµ(t −τ) f (τ)dτ. (4.75)

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Let us denote by S(t )y1, called source operator, the solution of the Cauchy problem

d 2yd x2+µ2y = 0,

y(0) = 0, y ′(0) = y1.

In terms of the source operator, the solution of (4.66) is given by

y(t ) = S ′(t )y0+ S(t )y1+∫ t

0S(t −τ) f (τ)dτ. (4.76)

Application to one dimensional wave equation

The Cauchy problem for d = 1 is

ut t − c2ux x = f (x, t ), x ∈R, t > 0, (4.77a)u(x, 0) = φ(x), ut (x, 0) =ψ(x) for x ∈R. (4.77b)

Let S(t )ψ denote the solution of the Cauchy problem

ut t − c2ux x = 0, x ∈R, t > 0,u(x, 0) = 0, ut (x, 0) =ψ(x) for x ∈R.

Taking cue from (4.76), we expect the following formula to yield a solution of theCauchy problem (4.77)

u(x, t ) =∂

∂ t(S(t )φ) (x)+ S(t )ψ(x)+

∫ t0(S(t −τ) fτ) (x)dτ, (4.78)

where fτ(x) := f (x,τ).We will now check that the last term on the right hand side of (4.78) solves the non-

homogeneous wave equation, as we will see later that first two terms therein correspondto solution of the homogeneous wave equation satisfying the given Cauchy data.

Indeed, differentiating the equation (4.78) w.r.t. t gives

∂

∂ t

�∫ t0(S(t −τ) fτ) (x)dτ

�= S(0) ft +∫ t

0

∂

∂ t(S(t −τ) fτ) (x)dτ

=∫ t

0

∂

∂ t(S(t −τ) fτ) (x)dτ.

Differentiating w.r.t. t once again, we get

∂ 2

∂ t 2

�∫ t0(S(t −τ) fτ) (x)dτ

�=�∂

∂ tS�(0) ft (x)+∫ t

0

∂ 2

∂ t 2(S(t −τ) fτ) (x)dτ

= f (x, t )+ c2∫ t

0

∂ 2

∂ x2(S(t −τ) fτ) (x)dτ

= f (x, t )+ c2 ∂2

∂ x2

�∫ t0(S(t −τ) fτ) (x)dτ

�.



Thus we have �∂ 2

∂ t 2− c2 ∂

2

∂ x2

��∫ t0(S(t −τ) fτ) (x)dτ

�= f (x, t ). (4.79)

Since S(t ) is explicitly known via d’Alembert formula, let us substitute in the above ex-pression and get an explicit formula for solution of the nonhomogeneous Cauchy prob-lem in terms of the given data f ,φ,ψ. Note that

(S(t )ψ) (x) =12c

∫ x+c t

x−c tψ(σ)dσ . (4.80)

Thus the formula (4.78) takes the form

u(x, t ) =∂

∂ t

�12c

∫ x+c t

x−c tφ(σ)dσ�+

12c

∫ x+c t

x−c tψ(σ)dσ +∫ t

0

12c

∫ x+c(t−τ)

x−c(t−τ)f (σ ,τ)dσ dτ

=φ(x − c t )+φ(x + c t )

2+

12c

∫ x+c t

x−c tψ(σ)dσ +

12c

∫ t0

∫ x+c(t−τ)

x−c(t−τ)f (σ ,τ)dσ dτ.

Thus we obtain a solution of the Cauchy problem

u(x, t ) =φ(x − c t )+φ(x + c t )

2+

12c

∫ x+c t

x−c tψ(σ)dσ +

12c

∫ t0

∫ x+c(t−τ)


(4.81)

Remark 4.13. Note that w(x, t ;τ) := S(t −τ) fτ solves the following problem

wt t − c2wx x = 0, x ∈R, t > τ (4.82a)

and satisfies the conditions

w(x,τ;τ) = 0, ut (x,τ;τ) = f (x,τ) for x ∈R. (4.82b)

Thus the integral ∫ t0(S(t −τ) fτ) (x)dτ

in the Duhamel’s formula has the form∫ t0

w(x, t ;τ)dτ

Application to two dimensional wave equation

In the case of two dimensional wave equation, the source operator S(t ) is given by

S(t )ψ=1

2πc

∫D(x ,c t )

ψ(y)pc2 t 2−∥x − y∥2 d y (4.83)

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Thus Duhamel principle gives the following solution of nonhomogeneous problemas

u(x , t ) =∫ t

0(S(t −τ) fτ) (x)dτ =

∫ t0

12πc

∫D(x ,c(t−τ))

f (y,τ)pc2 t 2−∥x − y∥2 d y dτ (4.84)

That is,

u(x , t ) =1

2πc

∫ t0

∫D(x ,c(t−τ))

f (y,τ)pc2 t 2−∥x − y∥2 d y dτ. (4.85)

Thus a solution of the Cauchy problem (4.1) for d = 2 is given by

u(x , t ) =∂

∂ t

�1

2πc

∫D(x ,c t )

φ(y)pc2 t 2−∥x − y∥2 d y

�+

12πc

∫D(x ,c t )

ψ(y)pc2 t 2−∥x − y∥2 d y

+1

2πc

∫ t0

∫D(x ,c(t−τ))

f (y,τ)pc2 t 2−∥x − y∥2 d y dτ, (4.86)

where D(x , c t ) denotes the open disk with center at x having radius c t .If we use the formula (4.55) for the solution of homogeneous wave equation, a solution

of the Cauchy problem (4.1) for d = 2 is

u(x , t ) =1

2π

∫D(0,1)

φ(x + c t z )p1−∥z∥2 d z +

c t2π

∫D(0,1)

∇φ(x + c t z ).zp1−∥z∥2 d z

+t

2π

∫D(0,1)

ψ(x + c t z )p1−∥z∥2 d z +

12πc

∫ t0

∫D(x ,c(t−τ))

f (y,τ)pc2 t 2−∥x − y∥2 d y dτ,

(4.87)

Application to three dimensional wave equation

In the case of three dimensional wave equation, the source operator S(t ) is given by

S(t )ψ=1

4πc2 t

∫S(x ,c t )

ψ(y)dσ (4.88)

Thus Duhamel principle gives the following solution of nonhomogeneous problemas

u(x , t ) =∫ t

0(S(t −τ) fτ) (x)dτ =

∫ t0

14πc2(t −τ)∫

S(x ,c(t−τ))f (y,τ)dσ(y)dτ (4.89)

That is,

u(x , t ) =1

4πc2

∫ t0

1(t −τ)∫

S(x ,c(t−τ))f (y,τ)dσ(y)dτ. (4.90)

Thus a solution of the Cauchy problem (4.1) for d = 3 is given by

u(x , t ) =1

4πc2 t 2

∫S(x ,c t ){tψ(y)+φ(y)+∇φ(y).(x − y)} dσ

+1

4πc2

∫ t0

1(t −τ)∫

S(x ,c(t−τ))f (y,τ)dσ(y)dτ. (4.91)



Since y ∈ S(x , c(t −τ))means that ∥y− x∥= c(t −τ), the last formula (4.91) may bere-written as

u(x , t ) =1

4πc2 t 2


+1

4πc

∫ t0

∫S(x ,c(t−τ))

f�

y, t − ∥y−x∥c

�∥y − x∥ dσ(y)dτ. (4.92)

Note that the last integral on the RHS of (4.92) is on the surface of the back-ward conewith vertex at (x , t ), and is the domain of dependence for the solution at (x , t ). The lastintegral reduces to an integral on B(x , c t ), and thus the formula (4.92) takes the form

u(x , t ) =1

4πc2 t 2


+1

4πc2

∫B(x ,c t )

f�

y, t − ∥y−x∥c

�∥y − x∥ d y. (4.93)

Note that the solution at the point (x , t ) requires the values of the source term f at earliertimes (retarded or delayed times) given by t− ∥y−x∥

c . Thus the last term on RHS of (4.93)is called retarded potential.

4.2 Uniqueness of solutionsIn this section, we will prove the uniqueness of solutions to Cauchy problem (4.1) ford = 1, d = 2, and d = 3. Uniqueness of solutions also follow from Causality prinicple,and also from energy considerations which will be discussed in Chapter 5.

Case of one dimensional Wave equation

We have in fact shown, while deriving d’Alembert formula, that any solution of the waveequation must be equal to the d’Alembert solution, thereby proving the uniquenesss ofsolutions.

Case of three dimensional Wave equation

If u and v are solutions of the Cauchy problem (4.1) for d = 3, then the spherical meanL(ρ, t ) := ρMw (ρ, t ), of the function w := u − v solves the Cauchy problem (4.36) withzero Cauchy data. Since (4.36) is a Cauchy problem for one dimensional wave equations,and by the uniqueness result that was established, it follows that Mw ≡ 0. By passing tothe limit as ρ→ 0, we get that the function w(x , t )≡ 0. Thus uniqueness of solutions forCauchy problem in the case of two dimensional wave equation is proved.

Case of two dimensional Wave equation

As was already observed, solutions of Cauchy problem for wave equation in two spacedimensions are also solutions of a Cauchy problem for wave equation in three space di-mensions. By uniqueness result in the case of three space dimensions, the correspondinguniqueness result follows for two space dimensions.

Sivaji IIT Bombay

4.3. Stability of solutions and Well-posedness of Cauchy Problem 91

4.3 Stability of solutions and Well-posedness of CauchyProblemIn this section we prove stability (in an appropriate sense) of solutions to Cauchy problemfor d = 1, d = 2, and d = 3, thereby establishing the well-posedness of the Cauchyproblem. Indeed, we state and prove stability of solutions as part of the well-posednesstheorems in each of the dimensions.

4.3.1 Case of one dimensional Wave equation

We are going to prove that Cauchy problem is well-posed for wave equation in one spacedimension.

Theorem 4.14 (Well-posedness). Let T > 0. For f ∈C (R× [0,T ]) such that fx ∈C (R×[0,T ]), and for Cauchy data φ ∈C 2(R), and ψ ∈C 1(R), the Cauchy problem

ut t − c2ux x = f (x, t ), x ∈R, t ∈ (0,T ), (4.94a)u(x, 0) = φ(x), x ∈R, (4.94b)

ut (x, 0) =ψ(x), x ∈R. (4.94c)

is well-posed in the following sense:

(i) There exists a classical solution to the Cauchy problem (4.94). That is, there exists anu ∈ C 2 (R× (0,T )) ∩ C 1 (R× [0,T ]) such that u solves the wave equation (4.94a),and satisfies the Cauchy data (4.94b) and (4.94c).

(ii) There exists at most one classical solution to the Cauchy problem (4.94).

(iii) Given ε > 0, there exists a δ > 0 such that for every pair of data triples ( f1,φ1,ψ1) and( f2,φ2,ψ2) having the smoothness as stated above, satisfying

|φ1(x)−φ2(x)|<δ, |ψ1(x)−ψ2(x)|<δ, ∀x ∈R, (4.95a)

and| f1(x, t )− f2(x, t )|<δ ∀(x, t ) ∈R× [0,T ], (4.95b)

the corresponding solutions u1 and u2 of the Cauchy problem (4.94) satisfy the estimate

|u1(x, t )− u2(x, t )|< ε, ∀(x, t ) ∈R× [0,T ] (4.96)

Proof.Proof of (i): We derived d’Alembert formula for the solution of Cauchy problem (4.94),and let us recall the formula (4.81) here.

u(x, t ) =φ(x − c t )+φ(x + c t )

2+

12c

∫ x+c t

x−c tψ(σ)dσ +

12c

∫ t0

∫ x+c(t−τ)


(4.97)

The above formula gives us a classical solution can be verfied by using the smoothness ofthe data triple ( f ,φ,ψ), and is left as an exercise.



Proof of (ii): We established uniqueness of solutions to the Cauchy problem in the Sec-tion 4.2 already.

Proof of (iii): Let u1 and u2 be solutions of the Cauchy problem with the data ( f1,φ1,ψ1)and ( f2,φ2,ψ2) respectively. By d’Alembert’s formula, we have

(u1− u2)(x, t ) =φ1(x + c t )−φ2(x + c t )

2+φ1(x − c t )−φ2(x − c t )

2+

12c

∫ x+c t

x−c t(ψ1−ψ2)(s)d s

+12c

∫ t0

∫ x+c(t−τ)

x−c(t−τ){ f1(σ ,τ)− f2(σ ,τ)} dσ dτ.

If |φ1(x)−φ2(x)|<δ, |ψ1(x)−ψ2(x)|<δ, | f1(x, t )− f2(x, t )|<δ, then we get

|u1(x, t )− u2(x, t )| ≤ 12(δ +δ)+

12c

2c tδ +12c

2c t 2δ ≤ (1+T +T 2)δ. (4.98)

Now we will choose δ such that δ < ε1+T+T 2 . This finishes the proof of the theorem.

Remark 4.15. Note that the well-posedness of Cauchy problem is established by consid-ering distances of data and solutions in the ‘uniform sense’ (see (iii) of Theorem 4.14). Onecan also consider questions of stability w.r.t. other notions of distance between functions,in which case the same Cauchy problem could be either well-posed or ill-posed. In anycase, in any result of well-posedness one must clearly mention what is really being proved.

4.3.2 Case of two dimensional Wave equation

We are going to prove that Cauchy problem is well-posed for wave equation in two spacedimensions.

Theorem 4.16 (Well-posedness). Let T > 0. For f ∈ C (R2 × [0,T ]) such that ∇x f ∈C (R2× [0,T ]), and for Cauchy data φ ∈C 3(R2), and ψ ∈C 2(R2), the Cauchy problem

ut t − c2�

ux1 x1+ ux2 x2

�= f (x , t ), x = (x1, x2) ∈R2, t > 0, (4.99a)



(i) There exists a classical solution to the Cauchy problem (4.99). That is, there exists anu ∈ C 2�R2× (0,T )�∩C 1�R2× [0,T ]�

such that u solves the wave equation (4.99a),and satisfies the Cauchy data (4.99b) and (4.99c).



|φ1(x)−φ2(x)|<δ, ∥∇φ1(x)−∇φ2(x)∥<δ, |ψ1(x)−ψ2(x)|<δ, ∀x ∈R2,(4.100a)

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4.3. Stability of solutions and Well-posedness of Cauchy Problem 93

and| f1(x , t )− f2(x , t )|<δ ∀(x , t ) ∈R2× [0,T ], (4.100b)


|u1(x , t )− u2(x , t )|< ε, ∀(x , t ) ∈R2× [0,T ] (4.101)

Proof.Proof of (i): We derived d’Alembert formula for the solution of Cauchy problem (4.99),and let us recall the formula (4.87) here.

u(x , t ) =1

2π

∫D(0,1)

φ(x + c t z )p1−∥z∥2 d z +

c t2π

∫D(0,1)

∇φ(x + c t z ).zp1−∥z∥2 d z

+t

2π

∫D(0,1)

ψ(x + c t z )p1−∥z∥2 d z +

12πc

∫ t0

∫D(x ,c(t−τ))

f (y,τ)pc2 t 2−∥x − y∥2 d y dτ,

(4.87)

where D(x , r ) denotes the open disk with center at x having radius r .The above formula gives us a classical solution can be verfied by using the smoothness

of the data triple ( f ,φ,ψ), and is left as an exercise.


Proof of (iii): Let u1 and u2 be solutions of the Cauchy problem with the data ( f1,φ1,ψ1)and ( f2,φ2,ψ2) respectively. Using the formula (4.87) we write expressions for u1 and u2,and then subtract one from another to get

(u1− u2)(x , t ) =1

2π

∫D(0,1)

(φ1−φ2)(x + c t z )p1−∥z∥2 d z +

c t2π

∫D(0,1)

(∇φ1−∇φ2)(x + c t z ).zp1−∥z∥2 d z

+t

2π

∫D(0,1)

(ψ1−ψ2)(x + c t z )p1−∥z∥2 d z

+1

2πc

∫ t0

∫D(x ,c(t−τ))

( f1− f2)(y,τ)pc2 t 2−∥x − y∥2 d y dτ,

(4.102)

There are four terms on the RHS of the equation (4.102), and let A1,A2,A3,A4 denote thefirst, second, third, and fourth terms. On applying triangle equality to RHS of (4.102),we get

|(u1− u2)(x , t )| ≤ |A1|+ |A2|+ |A3|+ |A4| (4.103)

Using (4.100a) (δ to be determined shortly), we get

|A1|+ |A2|+ |A3| ≤ δ

2π(1+ cT +T )∫

D(0,1)

d zp1−∥z∥2 . (4.104)



Using polar coordinates, we compute the integral as∫D(0,1)

d zp1−∥z∥2 =∫ 2π

0

∫ 10

r d rp1− r 2

= 2π.

Thus the estimate (4.104) reduces to

|A1|+ |A2|+ |A3| ≤ δ(1+ cT +T ). (4.105)

Proceeding in a similar manner, we get

|A4| ≤ 2πcδT 2. (4.106)

Combining all the estimates, we get

|u1(x , t )− u2(x , t )| ≤ δ(1+(c + 1)T + 2πcT 2) (4.107)

For a given ε > 0, we will choose δ such that δ < ε1+(c+1)T+2πcT 2 . This finishes the proof

of the theorem.

4.3.3 Case of three dimensional Wave equation

We are going to prove that Cauchy problem is well-posed for wave equation in three spacedimensions.

Theorem 4.17 (Well-posedness). Let T > 0. For f ∈ C (R3 × [0,T ]) such that ∇x f ∈C (R3× [0,T ]), and for Cauchy data φ ∈C 3(R3), and ψ ∈C 2(R3), the Cauchy problem

ut t − c2�

ux1 x1+ ux2 x2

+ ux3 x3

�= f (x , t ), x = (x1, x2, x3) ∈R3, t > 0, (4.108a)



(i) There exists a classical solution to the Cauchy problem (4.108). That is, there exists anu ∈C 2�R2× (0,T )�∩C 1�R2× [0,T ]�

such that u solves the wave equation (4.108a),and satisfies the Cauchy data (4.108b) and (4.108c).



|φ1(x)−φ2(x)|<δ, ∥∇φ1(x)−∇φ2(x)∥<δ, |ψ1(x)−ψ2(x)|<δ, ∀x ∈R3,(4.109a)

and| f1(x , t )− f2(x , t )|<δ ∀(x , t ) ∈R3× [0,T ], (4.109b)


|u1(x , t )− u2(x , t )|< ε, ∀(x , t ) ∈R3× [0,T ] (4.110)

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4.4. Wave equation on subdomains 95

Proof.Proof of (i): Let us recall the formula (4.91) for a solution of the Cauchy problem (4.108)here.

u(x , t ) =1

4πc2 t 2


+1

4πc2

∫ t0

1(t −τ)∫

S(x ,c(t−τ))f (y,τ)dσ(y)dτ.

The above formula gives us a classical solution can be verfied by using the smoothnessof the data triple ( f ,φ,ψ), and is left as an exercise.


Proof of (iii): Proof of this stability estimate is similar to that of wave equation in twospace dimensions, which is left as an exercise to the reader.

4.4 Wave equation on subdomainsIn this section we study wave equation in bounded domains. We mainly concentrateon d = 1, in which case we have explicit formulae for solutions of initial-boundary valueproblems for wave equation. In Subection 4.4.1, we consider initial-boundary value prob-lems (IBVP) for Wave equation posed on an interval (0, l ), and the case of IBVP posed onsemi-infinite interval (0,∞) is similar, which is left as an exercise to the reader. Solution ofIBVP posed on a general bounded interval (a, b ) is similar to those posed on (0, l ), whichis also left as an exercise.

4.4.1 Wave equation in one space dimension: Case of a bounded interval

Let us consider the following initial-boundary value problem (IBVP)

ut t − c2ux x = 0 for 0< x < l , t > 0, (4.111a)u(x, 0) = φ(x) for 0≤ x ≤ l , (4.111b)

∂ u∂ t(x, 0) =ψ(x) for 0≤ x ≤ l , (4.111c)

u(0, t ) = 0 for t ≥ 0, (4.111d)u(l , t ) = 0 for t ≥ 0. (4.111e)

Note that we are dealing with homogeneous Dirichlet boundary conditions where thevalues of the unknown function u are prescribed to be equal to zero, on the boundary forall positive times. We may consider other boundary conditions (homogeneous and non-homogeneous) such as Neumann conditions (where ∂ u

∂ x (0, t ) and ∂ u∂ x (l , t ) are prescribed),

Robin conditions (where αu(0, t ) +β ∂ u∂ x (0, t ) and αu(l , t ) +β ∂ u

∂ x (l , t ) are prescribed),or a mix of any of these three boundary conditions in place of Dirichlet conditions. Weaddress some of these problems in exercises.



We derive an expression for the solution of IBVP (4.111) using two different methods.The first of these methods is called solution via first principles, and the other method ismethod of separation of variables.

Solution via first principles

Step 1: Solution in the region 0,0Recall that the general solution of the homogeneous wave equation is given by

u(x, t ) = F (x − c t )+G(x + c t ), (4.112)

where F ,G ∈C 2(R). The boundary condition u(x, 0) = φ(x) yields the relation

F (x)+G(x) = φ(x) for 0≤ x ≤ l .

The boundary condition ut (x, 0) =ψ(x) gives

−cF ′(x)+ cG′(x) =ψ(x) for 0≤ x ≤ l .

Integrating the last equation over the interval [0, x] gives

−F (x)+G(x) =1c

∫ x0ψ(s)d s − F (0)+G(0)

Let us recall from equation (4.15) that solving for F and G, we get

F (ξ ) =12φ(ξ )− 1

2c

∫ ξ0ψ(s)d s +

F (0)−G(0)2

for 0≤ ξ ≤ l (4.113)

G(η) =12φ(η)+

12c

∫ η0ψ(s)d s − F (0)−G(0)

2for 0≤ η≤ l (4.114)

Since the constant terms in (4.113) cancel each other in the expression for u(x, t ), we dropthem from now onwards, and we write

F (ξ ) =12φ(ξ )− 1

2c

∫ ξ0ψ(s)d s for 0≤ ξ ≤ l (4.115)

G(η) =12φ(η)+

12c

∫ η0ψ(s )d s for 0≤ η≤ l (4.116)

Substituting these values in the general solution, we get

u(x, t ) =φ(x − c t )+φ(x + c t )

2+

12c

∫ x+c t

x−c tψ(s )d s (4.117)

for (x, t ) such that 0 ≤ x − c t ≤ l and 0 ≤ x + c t ≤ l . Thus the solution is completelydetermined in the triangular region�

(x, t ) ∈ [0, l ]× [0,∞) : 0≤ x − c t ≤ l , 0≤ x + c t ≤ l, (4.118)

which is the triangular region denoted by 0,0 in Figure 4.2.Note that the boundary conditions (4.111d) and (4.111e) do not play any role in de-

termining (and thus does not influence) the solution within the region 0,0.

Sivaji IIT Bombay


..x − c t =

0

.

x + c t = l

.

x − c t =−l

.

x + c t = 2l

.

x − c t =−2l

.

x + c t = 3l

.0,0

.

1,1

.

2,2

.

1,0

.

0,1

.

2,1

.

1,2

.

3,2

.

2,3

.0.

l.

lc

.

lc

.

2lc

.

2lc

.

3lc

.

3lc

Figure 4.2. Diamond picture.

Step 2: Solution in the region 1,0Note that in Step 1, we could obtain the solution in the region denoted by 0,0. In

order to find the solution at other points of the domain [0, l ]×[0,∞), we need to extendthe functions F ,G outside the interval [0, l ] in which they were already determined by(4.115). The boundary conditions (4.111d) and (4.111e) impose conditions on such exten-sions by intertwining the values of the extended functions (which we still denote usingthe same F ,G) as seen below.

Using the boundary condition (4.111d), in the general form of the solution (4.112),we get

F (−c t )+G(c t ) = 0 for t ≥ 0, (4.119)

Denoting ζ = −c t , the equation (4.119) gives a way to extend the function F (ζ ) fornegative arguments ζ , namely, by setting

F (ζ ) =−G(−ζ ). (4.120)

For the moment, the formula (4.120) is meaningful for −l ≤ ζ < 0. This is because Gis defined only on the interval [0, l ], and whenever −l ≤ ζ < 0 holds, the −ζ satisfies0<−ζ ≤ l , and thus G(−ζ ) is defined.

F (ζ ) =−G(−ζ ) =−12φ(−ζ )− 1

2c

∫ −ζ0

ψ(s )d s for − l ≤ ζ < 0. (4.121)

Thus, in the triangular region 1,0 in Figure 4.2 which is described by�(x, t ) ∈ [0, l ]× [0,∞) : −l ≤ x − c t ≤ 0, 0≤ x + c t ≤ l

, (4.122)



we have the following expression for the solution u(x, t ) of the IBVP

u(x, t ) = F (x − c t )+G(x + c t ) =−G(c t − x)+G(x + c t ),

which takes the following form in terms of the given data:

u(x, t ) =−φ(c t − x)+φ(x + c t )

2+

12c

∫ x+c t

c t−xψ(s)d s . (4.123)

The formula (4.123) takes the d’Alembert form, in terms of suitably extended initialdata in the following sense:

Let φo and ψo denote the extensions of the functions φ and ψ respectively to theinterval [−l , 0) as odd functions w.r.t. 0. That is,

φ0(x) =−φ(−x) for − l ≤ x < 0, (4.124a)ψ0(x) =−ψ(−x) for − l ≤ x < 0. (4.124b)

In terms of the extended functions φo and ψo , the formula (4.121) takes the form

F (ζ ) =12φ0(ζ )− 1

2c

∫ ζ0ψ0(s )d s for − l ≤ ζ ≤ 0, (4.125)

and thus the solution takes the following d’Alembert form

u(x, t ) =φo(x − c t )+φo(x + c t )

2+

12c

∫ x+c t

x−c tψo(s )d s . (4.126)

Step 3: Solution in the region 0,1, region 1,1 and region 1,2Let us use the second boundary condition (4.111e), in the general form of the solution

(4.112), to getF (l − c t )+G(l + c t ) = 0 for t ≥ 0. (4.127)

Denoting ζ = l + c t , the equation (4.127) gives a way to extend the function G(ζ ) forζ > l , namely, by setting

G(ζ ) =−F (2l − ζ ). (4.128)

For the moment, the formula (4.128) is meaningful for l < ζ ≤ 3l . This is because F isknown only on the interval [−l , l ], and whenever l < ζ < 3l holds, the 2l − ζ satisfies−l ≤ 2l − ζ ≤ l , and thus F (2l − ζ ) is defined.

G(ζ ) =−F (2l − ζ ) = −

12φ(2l − ζ )+ 1

2c

∫ 2l−ζ0 ψ(s )d s for l < ζ ≤ 2l .

G(ζ − 2l ) for 2l ≤ ζ ≤ 3l .(4.129)

Thus the function G is now known on the interval [0,3l ], and F is known on theinterval [−l , l ]. Using this information, the solution of the IBVP will be determined inthree regions.

Solution in the region 0,1The region 0,1 is given by�

(x, t ) ∈ [0, l ]× [0,∞) : 0≤ x − c t ≤ l , l ≤ x + c t ≤ 2l, (4.130)

Sivaji IIT Bombay


and the solution in this region is given by

u(x, t ) = F (x − c t )+G(x + c t ) = F (x − c t )− F (2l − x − c t ),

which in terms of the initial data take the following form

u(x, t ) =φ(x − c t )−φ(2l − x − c t )

2+

12c

∫ 2l−x−c t

x−c tψ(s)d s . (4.131)

As before, the formula (4.131) may easily seen to be in d’Alembert form (4.126), providedwe extend the functions φ0 and ψ0 to whole of the real line as 2l -periodic functions.


(x, t ) ∈ [0, l ]× [0,∞) : −l ≤ x − c t ≤ 0, l ≤ x + c t ≤ 2l, (4.132)




u(x, t ) =−φ(c t − x)−φ(2l − x − c t )

2+

12c

∫ 2l−x−c t

c t−xψ(s )d s . (4.133)


(x, t ) ∈ [0, l ]× [0,∞) : −l ≤ x − c t ≤ 0, 2l ≤ x + c t ≤ 3l, (4.134)




u(x, t ) =−φ(c t − x)+φ(x + c t − 2l )

2+

12c

∫ x+c t−2l

c t−xψ(s )d s . (4.135)

Step 4: Solution in other regions By continuing the above procedure, we can determinethe solution in the region [0, l ]× [0,∞). As we saw in the previous steps, the main ideais to incrementally extend the functions F ,G to have domains equal to R, which wereinitially determined on [0, l ], using the boundary conditions via equations (4.120) and(4.128).

Theorem 4.18 (Existence and uniqueness theorem). The IBVP (4.111) has a uniqueclassical solution when the initial data φ ∈ C 2(0, l ), ψ ∈ C 1(0, l ), and satisfy the followingcompatibility conditions:

φ(0) = φ(l ) =ψ(0) =ψ(l ) = 0, (4.136a)φ′+(0), φ′−(l ), φ′′+(0), φ′′−(l ), ψ′+(0), ψ′−(l ) exist, (4.136b)

φ′′+(0) = φ′′−(l ) = 0. (4.136c)



Proof. Note that in the region 0,0 the expression for u given by the d’Alembert formula(4.117), and thus it is the unique classical solution to the IBVP in this region due to thesmoothness assumptions on φ,ψ.

The solution in the region 1,0 is given by the formula (4.126), which is once again inthe d’Alembert form in terms of the extended functionsφo andψo . Thus ifφo ∈C 2(−l , l )and ψo ∈ C 1(−l , l ) then the formula (4.126) gives the unique classical solution in theregion 1,0. Let us find out the necessary and sufficient conditions to have φo ∈C 2(−l , l ).They are

(i) φo is a continuous function on (−l , l ) if and only if φ(0) = 0.(ii) The function φo is differentiable at all points of the interval (−l , l ) except possibly

at zero. Since we assumed that

φ′+(0) = limx→0+

φ(x)−φ(0)x

exists,

we have

φ′−(0) = lims→0−

φ(s )−φ(0)s

= limx→0+

φ(x)−φ(0)x

,

where the last equality holds by a change of variable s = −x and since φ is an oddfunction about x = 0. Thus the function φo is differentiable on the interval (−l , l ),and the derivative is also continuous.

(iii) By a similar analysis, using the remaining compatibility conditions, it also followsthat φo has a continous second derivative on the interval (−l , l ).

(iv) Finally it follows that φo when extended to R by 2l -periodicity, belongs to C 2(R)by using compatibility conditions at x = l . By a similar reasoning, ψo ∈C 1(R).

Since φo ∈ C 2(R), and ψo ∈ C 1(R), it follows by induction that the solution u as con-structed earlier (in various regions) is indeed a classical solution. The proof will be com-plete if we verify that u is twice continuously differentiable at all points on the boundariesof different regions, whose proof is left to the reader.

Example 4.19. Let us find the value of u( 12 , 32 )where u solves the following initial-boundary

value problem (IBVP)

ut t − ux x = 0 for 0< x < 1, t > 0, (4.137a)u(x, 0) = 0 for 0≤ x ≤ 1, (4.137b)

∂ u∂ t(x, 0) = x(1− x) for 0≤ x ≤ 1, (4.137c)

u(0, t ) = 0 for t ≥ 0, (4.137d)u(1, t ) = 0 for t ≥ 0. (4.137e)

From (4.120), (4.128), and (4.115), we get

u�

12

,32

�= F (−1)+G(2) =−G(1)− F (0) =−1

2

∫ 10

s(1− s )d s − 0=−16

. (4.138)

Sivaji IIT Bombay


Remark 4.20 (interpretation of the solution using Reflections of waves). Have a lookat the formulae (4.118), (4.122), (4.130), (4.132), and (4.134). Each of them has the form

u(x, t ) =(−1)pφ(L)+ (−1)qφ(R)

2+

12c

∫ RLψ(s)d s , (4.139)

Let us start at the point (x, t ) and follow the left characteristic till we hit the left boundary,and then follow the right characteristic till we hit right boundary, and once again followthe left characteristic through that point, and continue this procedure till we hit a pointL (lying in the interval [0, l ]) on the x-axis. Let p denote the number of turns that wetook till reaching L. Similarly by following the right characteristic reach a point R, andlet q denote the number of turns that we took till reaching a point R (lying in the interval[0, l ] on the x-axis.

Method of separation of variables

We were considering the following IBVP at the beginning of this section, and we discussedhow to solve the same starting from first principles.

ut t − c2ux x = 0 for 0< x < l , t > 0, (4.140a)u(x, 0) = φ(x) for 0≤ x ≤ l , (4.140b)

∂ u∂ t(x, 0) =ψ(x) for 0≤ x ≤ l , (4.140c)

u(0, t ) = 0 for t ≥ 0, (4.140d)u(l , t ) = 0 for t ≥ 0. (4.140e)

In this subsection, we use a general method called separation of variables method tosolve the IBVP (4.140). Solution of the IBVP (4.140) can be found by finding solutionsu1 (with φ ≡ 0), and u2 (with ψ ≡ 0) of the IBVP (4.140). Since the equation (4.140a) islinear, by superposition principle u = u1+u2 solves the given IBVP (4.140). Thus we willassume that ψ≡ 0 for the rest of this discussion.

Method of separation of variables assumes a solution of (4.140a) is of the form

u(x, t ) =X (x)T (t ) for x ∈ (0, l ), t > 0. (4.141)

Substituting the expression for u from (4.141) in the equation (4.140a) gives

X (x)T ′′(t )− c2X ′′(x)T (t ) = 0. (4.142)

On dividing both sides of the equation (4.142) with X (x)T (t ) and re-arranging termsyields

T ′′(t )T (t )

= c2 X ′′(x)X (x)

. (4.143)

Note that the LHS of the equation (4.143) is a function of t only, while the RHS is afunction of x only. Thus both of them must be equal to a constant, and let us denote itby λ. Thus we have

T ′′(t )T (t )

= c2 X ′′(x)X (x)

= λ. (4.144)



Thus we get two ODEs from (4.144) which are given by

X ′′− λc2

X = 0, T ′′−λT = 0. (4.145)

Using the boundary condition (4.140d), we get

u(0, t ) =X (0)T (t ) = 0 for all t > 0. (4.146)

Since we do not want to find a trivial solution u ≡ 0, we cannot admit T (t )≡= 0. Thuswe get X (0) = 0.

Similarly using the boundary condition (4.140e), we get

u(l , t ) =X (l )T (t ) = 0 for all t > 0. (4.147)

Since we do not want to find a trivial solution u ≡ 0, we cannot admit T (t )≡= 0. Thuswe get X (l ) = 0.

Thus we are led to the boundary value problem for X given by

X ′′− λc2

X = 0, (4.148a)

X (0) =X (l ) = 0. (4.148b)

We are interested in finding non-trivial solutions to the boundary value problem (4.148),which depends on the value of λ. The λs for which the BVP admits a non-trivial solutionis called an eigenvalue and corresponding non-trivial solutions are called eigenfunctions.Let us find the eigenvalues, and corresponding eigenfunctions now.

(i) (λ = 0) General solution of the ODE (4.148a) is given by X (x) = ax + b . Apply-ing the boundary conditions (4.148b), we get a = b = 0. Thus λ = 0 is not aneigenvalue.

(ii) (λ > 0) When λ > 0, we may write λ = µ2 where µ > 0. The ODE (4.148a) thenbecomes

X ′′− µ2

c2X = 0,

whose general solution is given by X (x) = aeµc x + b e−

µc x . Applying the boundary

conditions (4.148b), we get a = b = 0. Thus λ > 0 is not an eigenvalue.(iii) (λ < 0) When λ < 0, we may write λ=−µ2 where µ> 0. The ODE (4.148a) then

becomes

X ′′+ µ2

c2X = 0,

whose general solution is given by X (x) = a cos�µ

c x�+ b sin�µ

c x�. Applying the

boundary conditions (4.148b), we get

a = 0, (4.149a)

a cos�µ

cl�+ b sin�µ

cl�= 0. (4.149b)

Sivaji IIT Bombay


Since we are interested in non-trivial solutions to the BVP (4.148), at least one of theconstants a, b should be non-zero. However, we already have a = 0 from (4.149).Thus in order to have b = 0, we must have

sin�µ

cl�= 0. (4.150)

Solving the equation (4.150), we get µn =cnπ

l for each n ∈ N. Thus we have thefollowing eigenvalue and corresponding eigenfunction, indexed by n ∈N:

λn =− c2n2π2

l 2, Xn(x) = sin�nπ

lx�

. (4.151)

We now solve the ODE for T with λ= λn for each n ∈N. The initial condition (4.140c)gives rise to the condition T ′(0) = 0. Let us solve the ODE T ′′ − λnT = 0 with thecondition T ′(0) = 0. We get the following solutions Tn indexed by N ∈N:

Tn(t ) = b cos�nπc

lt�

.

Since the equation is linear, and each of the functions Xn(x)Tn(t ) is a solution, we proposea formal solution using ‘superposition principle’ by the formula

u(x, t ) =∞∑

n=1

bn sin�nπ

lx�

cos�nπc

lt�

(4.152)

We use the initial condition (4.140b) to determine the coefficients bn . Indeed the functionu defined by (4.152) satisfies the initial condition (4.140b) if and only if

φ(x) = u(x, 0) =∞∑

n=1

bn sin�nπ

lx�

. (4.153)

Thus bn are the fourier sine coefficients of the function φ. Extend the function φ to theinterval [−l , l ] as an odd function around x = 0, and let the extended function be stilldenoted by φ. Then the fourier series of φ takes the form

φ(x) =∞∑

n=1

bn sin�nπ

lx�

, (4.154)

where bn is given by

bn =2l

∫ l0φ(x) sin�nπ

lx�

d x. (4.155)

Thus the formal solution of the IBVP (4.140) is given by

u(x, t )≈∞∑

n=1

�2l


lx�

d x

�sin�nπ

lx�

cos�nπc

lt�

. (4.156)

Theorem 4.21. Let φ ∈ C 4[0, l ] such that φ(0) = φ(l ) = φ′′(0) = φ′′(l ) = 0. Then thefunction defined by (4.156) is a solution of the IBVP (4.140).

Proof. In order to establish that the formal solution given by (4.156) is indeed a solutionto the IBVP (4.140), we need to show that



(i) the series in (4.156) is twice continuously differentiable w.r.t. both the variables xand t , thus making sure that ut t and ux x are meaningful. Then we need to showthat equation (4.140a) is satisfied,

(ii) the function u(x, t ) given by (4.156) is continuous upto the boundary of (0,π)×(0,T ) and that the initial-boundary conditions (4.140b) - (4.140e) are satisfied.

Step 1: Proof of (i) By formal differentiation of the equation (4.156), using the nota-tion introduced in (4.155), we get

ut (x, t ) =−πcl

∞∑n=1

nbn sin�nπ

lx�

sin�nπc

lt�

(4.157)

ut t (x, t ) =−�πc

l

�2 ∞∑n=1

n2bn sin�nπ

lx�

cos�nπc

lt�

(4.158)

ux (x, t ) =πcl

∞∑n=1

nbn cos�nπ

lx�

cos�nπc

lt�

(4.159)

ux x (x, t ) =−�πc

l

�2 ∞∑n=1

n2bn sin�nπ

lx�

cos�nπc

lt�

. (4.160)

From (4.157), it is clear that ut t = ux x . Thus it remains to justify that the functiondefined by (4.156) is twice differentiable w.r.t. x and t variables. By a theorem of realanalysis Theorem D.1, it is enough to show that the series in (4.157) are uniformly con-vergent, and the series in (4.156) converges at some point. Clearly all the series in (4.157)are uniformly convergent (by comparison test) if the following condition is satisfied: thereexists a constant C > 0 such that for each n ∈N the following inequalities hold.

n2|bn | ≤ 1n2

(4.161)

Let us show that the conditions (4.161) are satisfied.

bn =2l


lx�

d x

=− 2nπ

∫ l0φ(x)

dd x

�cos�nπ

lx��

d x

=2

nπ

∫ l0φ′(x)cos�nπ

lx�

d x − φ(x)cos�nπ

lx��l

0

=2

nπ

∫ l0φ′(x)cos�nπ

lx�

d x

=2l

n2π2

∫ l0φ′(x) d

d x

�sin�nπ

lx��

d x

=− 2ln2π2

∫ l0φ′′(x) sin�nπ

lx�

d x + φ′(x) sin�nπ

lx��l

0

=− 2ln2π2

∫ l0φ′′(x) sin�nπ

lx�

d x.

Sivaji IIT Bombay


Continuing the above computations two more times, we get

bn =2l 3

n4π4

∫ l0φ(i v)(x) sin�nπ

lx�

d x.

The last equation gives the following estimate:

|bn | ≤M2l 4

n4π4,

where M =maxx∈[0,l ] |φ(i v)(x)|, from which the estimate (4.161) follows.

Step 2: Proof of (ii) Due to assumptions on φ, the fourier sine series of φ convergesto φ, and hence the initial conditions (4.140b) are satisfied. Other conditions (4.140c) -(4.140e) are checked by substituting appropriately the values of x and t in (4.156).

Remark 4.22. (i) Note that the conditions on the initial data φ in the above theoremare very restrictive. Luckily they are only sufficient conditions, and one can findweaker conditions on φ under which the above theorem is valid. Note that theoremcannot be applied in the case of an initially plucked string, which corresponds to φthat is piecewise linear (graph of φ is of triangle shape).

(ii) When φ is only piecewise continuous, the formal solution may be interepreted as aweak solution. For more details on weak solutions to wave equation, see Section 5.7(in Chapter 5).



ExercisesGeneral

4.1. Any solution of ξt − cξx = 0 is a function of x+ c t . Any solution of ξt + cξx = 0is a function of x − c t .

4.2. Let u be a twice continuously differentiable function that satisfies the wave equa-tion for x ∈ R and t ∈ R. That is, ut t − c2ux x = 0. Prove that u is constanton a member of the family of characteristics x − c t = constant if and only if it isconstant along each member of this family.

4.3. Let u be a classical solution to Wave equation ut t −ux x = 0 for (x, t ) ∈R× (0,∞).Show that

(i) for each fixed y ∈R, the function w(x, t ) := u(x − y, t ) is also a solution.

(ii) for each k ∈N, the function w(x, t ) := ∂ k u∂ xk (x, t ) is also a solution.

(iii) for each a > 0, the function w(x, t ) = u(ax,at ) is also a solution.

Cauchy problem in full space

4.4. Show that the Cauchy problem for wave equation for x ∈R and t > 0 is ill-posed,unlike the corresponding problem posed for x ∈R and 0≤ t ≤ T for a fixed T > 0.

4.5. Let T > 0. Show that function u given b d’Alembert formula is also a solution forx ∈ R and T < t ≤ 0 and show that the Cauchy problem is well-posed for x ∈ Rand T ≤ t ≤ 0.

4.6. If the Cauchy dataφ,ψ are even (resp. odd or periodic) functions, then the solutionof Cauchy problem for the homogeneous wave equation is also an even (resp. oddor periodic) function.

4.7. Isn’t it a surprise that we proved the existence of solutions to Cauchy problem ford = 1 followed by d = 3, and then for d = 2 via method of descent? Where doesthe method of spherical means hits the roadblock when d = 2?

4.8. Consider the Cauchy problem

ut t − ux x = 0, x ∈R, t > 0,

u(x, 0) = f (x) =

0 if −∞< x <−1,

x + 1 if − 1≤ x ≤ 0,1− x if 0≤ x ≤ 1,

0 if 1< x <∞.

,

ut (x, 0) = g (x) =

0 if −∞< x <−1,1 if − 1≤ x ≤ 1,0 if 1< x <∞.

.

(i) Evaluate u at the point (1, 12 ).

(ii) On what domain does d’Alembert’s formula gives a classical solution to theabove Cauchy problem? Discuss the smoothness of the function given byd’Alembert’s formula.

4.9. [32] Let u be a solution of the Cauchy problem

ut t − 9ux x = 0, x ∈R, t > 0,

Sivaji IIT Bombay

Exercises 107

u(x, 0) = f (x) =§

1 if |x| ≤ 2,0 if |x|> 2. ,

ut (x, 0) = g (x) =§

1 if |x| ≤ 2,0 if |x|> 2. .

(i) Evaluate u at the point (0, 16 ).

(ii) Discuss the large time behaviour of the solution, i.e., on limt→∞ u(x0, t ) foreach fixed x0.

(iii) What is the largest subdomain ofR×(0,∞) on which u is a classical solutionto the Cauchy problem?

4.10. [32] Solve the following Cauchy problem for a nonhomogeneous wave equation

ut t − ux x = t 7, x ∈R, t > 0,u(x, 0) = f (x) = 2x + sin x for x ∈R,

ut (x, 0) = g (x) = 0 for x ∈R.

4.11. [23] Let φ ∈ C 2(R) and ψ ∈ C 1(R) be such that φ ≡ ψ ≡ 0 outside an interval[a, b ]. Let u(x, t ) be the solution of the Cauchy problem (4.4).

(i) Show that for each fixed t > 0, the function x 7→ u(x, t ) is identically zerooutside an interval.

(ii) Show that the functions F ,G in the decomposition (4.14) for u can be ofcompact support only when

∫Rψ(s)d s = 0.

4.12. [39] Find all the spherical solutions of the three-dimensional wave equation.4.13. [39] Solve the three-dimensional wave equation inR3\{0}×(0,∞)with zero initial

conditions and with the limiting condition

limr→0

4πr 2ur (r, t ) = g (t ).

Assume that g (0) = g ′(0) = g ′′(0) = 0.

Wave equation on proper subdomains

4.14. [23] Let φ,ψ, h ∈C 2 ([0,∞)). Solve the following initial boundary value problem(IBVP).

ut t − ux x = 0, 0< x <∞, t > 0,u(x, 0) = φ(x), 0≤ x <∞,

ut (x, 0) =ψ(x), 0≤ x <∞,u(0, t ) = h(t ), t > 0.

from the first principles. Derive the compatibility condition on φ,ψ, h so thatthe solution derived above is indeed a classical solution. State and prove a relevantwell-posedness result for the IBVP.

4.15. Let φ,ψ, h ∈ C 2 ([0,∞)). Solve the following initial boundary value problemIBVP.

ut t − ux x = 0, 0< x <∞, t > 0,



u(x, 0) = φ(x), 0≤ x <∞,ut (x, 0) =ψ(x), 0≤ x <∞,∂ u∂ x(0, t ) = h(t ), t > 0

from the first principles. Derive the compatibility condition on φ,ψ, h so thatthe solution derived above is indeed a classical solution. State and prove a relevantwell-posedness result for the IBVP.

4.16. [23] Let α ∈ R be such that α+ c = 0. Let u be the solution of the homogeneouswave equation posed in the domain (x, t ) ∈ (0,∞)× (0,∞) such that

u(x, 0) = φ(x), ut (x, 0) =ψ(x) for x ≥ 0ut (0, t ) = αux (0, t ) for t ≥ 0,

where φ andψ belong to the class C 2(0,∞)∩C [0,∞) and φ(0) =ψ(0) = 0. Showthat generally no solution exists when α+ c = 0. (Hint: Use the decomposition(4.14)).

4.17. Let u be the solution to the IBVP

ut t − 4ux x = 0, 0< x <∞, t > 0,u(x, 0) = x2, 0≤ x <∞,

ut (x, 0) = 6x, 0≤ x <∞.u(0, t ) = t 2, t > 0,

Evaluate u(3,2) and u(2,3).4.18. Let φ,ψ ∈C 2 ([0, l ]). Solve the following initial boundary value problem IBVP.

ut t − ux x = 0, 0< x < l , t > 0,u(x, 0) = φ(x), 0≤ x ≤ l ,

ut (x, 0) =ψ(x), 0≤ x ≤ l ,∂ u∂ x(0, t ) = 0, t > 0,

∂ u∂ x(l , t ) = 0, t > 0

from the first principles. Derive the compatibility condition on φ,ψ so that thesolution derived above is indeed a classical solution. State and prove a relevantwell-posedness result for the IBVP.

4.19. Let φ,ψ ∈C 2 ([0, l ]). Solve the following initial boundary value problem IBVP.

ut t − ux x = 0, 0< x < l , t > 0,u(x, 0) = φ(x), 0≤ x ≤ l ,

ut (x, 0) =ψ(x), 0≤ x ≤ l ,∂ u∂ x(0, t ) = 0, t > 0,

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

from the first principles. Derive the compatibility condition on φ,ψ so that thesolution derived above is indeed a classical solution. State and prove a relevantwell-posedness result for the IBVP.

Sivaji IIT Bombay

Exercises 109

4.20. [41] Solve the following initial boundary value problem IBVP.

ut t − ux x = 0, 0< x <π

2, t > 0,

u(x, 0) = sin x, 0≤ x ≤ π2

,

ut (x, 0) = 0, 0≤ x ≤ π2

,

u(0, t ) = 0, t > 0,∂ u∂ x

�π2

, t�= 0, t > 0.

4.21. In the context of the IBVP (4.111), What is the domain of dependence for a point(x0, t0) ∈ (0, l )× (0,∞)? What is the range of influence of a subinterval [a, b ] of(0, l )?

4.22. Using separation of variables method solve the following IBVP for the dampedwave equation:

ux x + 2ut − ut t = 0, 0< x <π, t > 0,u(x, 0) = φ(x), 0≤ x ≤π,

ut (x, 0) = 0, 0≤ x ≤π,u(0, t ) = 0, t > 0,u(π, t ) = 0, t > 0.

4.23. Using separation of variables method solve the following IBVP for the dampedwave equation:

ut t + ut = ux x , 0< x <π, t > 0, (4.164a)u(0, t ) = u(π, t ) = 0, 0≤ t (4.164b)u(x, 0) = sin(x), 0≤ x ≤π (4.164c)

ut (x, 0) = 0, 0≤ x ≤π (4.164d)

(Ans: u(x, t ) = e−t/2�cos(p

32 t )+ 1p

3sin(p

32 t )�

sin x)

4.24. [41] Let c > 0 be such that c2 < 1, and φ ∈ C 2 ([0, l ]). Solve the following initialboundary value problem IBVP.

ut t − c2ux x = 0, 0< x < 1, t > x,u(x, x) = φ(x), 0≤ x ≤ 1,

ut (x, x) = 0, 0≤ x ≤ 1,u(0, t ) = 0, t > 0,u(l , t ) = 0, t > 0.

4.25. For each C ∈R, verify that

uC (x, t ) =C sinπx sinπt

is a solution to the following boundary value problem for the wave equation:

ut t = ux x , 0< x < 1, 0< t < 1,u(x, 0) = 0, u(x, 1) = 0 for 0≤ x ≤ 1,

u(0, t ) = u(1, t ) = 0 for 0≤ t ≤ 1.



What are the maximum and minimum values of the solution uC in the domain[0,1]× [0,1]? Where are they attained? What is your conclusion regarding a max-imum principle for Wave equation?

4.26. [19, 32]

(i) Solve the Darboux problem:

ut t − ux x = 0 , t > max{x,−x}, t ≥ 0, (4.165a)

u(x, t ) =§ϕ(t ) if x = t , t ≥ 0,ψ(t ) if x =−t , t ≥ 0, (4.165b)

where ϕ,ψ ∈C 2([0,∞)) satisfies ϕ(0) =ψ(0).(ii) Find the domain of dependence for any point (x0, t0) with t0 > |x0|.

(iii) Explain what is meant by saying that above problem is well-posed. Is theabove problem well-posed? On what domain is it well-posed? State and provea precise statement concerning this.

Sivaji IIT Bombay

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