# Spectral Approach to Derive the Representation Formulae for Solutions of the Wave Equation

Authors: Gusein Sh. Guseinov
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

Using spectral properties of the Laplace operator and some structural formula for rapidly decreasing functions of the Laplace operator, we offer a novel method to derive explicit formulae for solutions to the Cauchy problem for classical wave equation in arbitrary dimensions. Among them are the well-known d'Alembert, Poisson, and Kirchhoff representation formulae in low space dimensions.


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## Body

## 1. Introduction

The wave equation for a functionu(x1,…,xn,t)=u(x,t) of n space variables x1,…,xn and the time t is given by(1.1)∂2u∂t2=Δu,
where(1.2)Δ=∂2∂x12+⋯+∂2∂xn2
is the Laplacian. The wave equation is encountered often in applications. For n=1 the equation can represent sound waves in pipes or vibrations of strings, for n=2 waves on the surface of water, for n=3 waves in acoustics or optics. Therefore, formulae that give the solution of the Cauchy problem in explicit form are of great significance. In the Cauchy problem (initial value problem) one asks for a solution u(x,t) of (1.1) defined for x∈ℝn, t≥0 that satisfies (1.1) for x∈ℝn, t>0 and the initial conditions(1.3)u(x,0)=φ(x),∂u(x,0)∂t=ψ(x)(x∈Rn).Ifn=1 and φ∈C2(ℝ), ψ∈C1(ℝ), then the classical solution of problem (1.1), (1.3) is given by d’Alembert’s formula(1.4)u(x,t)=φ(x+t)+φ(x-t)2+12∫x-tx+tψ(y)dy.Ifn=2 and φ∈C3(ℝ2), ψ∈C2(ℝ2), then the solution of problem (1.1), (1.3) is given by Poisson’s formula(1.5)u(x,t)=12π∫|y-x|<tψ(y)dyt2-|y-x|2+∂∂t[12π∫|y-x|<tφ(y)dyt2-|y-x|2],
where x=(x1,x2), y=(y1,y2), and |y-x|2=(y1-x1)2+(y2-x2)2.Ifn=3 and φ∈C3(ℝ2), ψ∈C2(ℝ2), then the solution of problem (1.1), (1.3) is given by Kirchhoff’s formula(1.6)u(x,t)=14πt∫|y-x|=tψ(y)dSy+∂∂t[14πt∫|y-x|=tφ(y)dSy],
where x=(x1,x2,x3), y=(y1,y2,y3), |y-x|2=(y1-x1)2+(y2-x2)2+(y3-x3)2, and dSy is the surface element of the sphere {y∈ℝ3:|y-x|=t}.Passing to an arbitraryn let us denote by u(x,t)=Nφ(x,t) the solution of the problem(1.7)∂2u∂t2=Δu,x∈Rn,t>0,(1.8)u(x,0)=φ(x),∂u(x,0)∂t=0,x∈Rn.
It is easy to see that then the function(1.9)v(x,t)=∫0tu(x,τ)dτ
is the solution of the problem(1.10)∂2v∂t2=Δv,x∈Rn,t>0,(1.11)v(x,0)=0,∂v(x,0)∂t=φ(x),x∈Rn.
Indeed, integrating (1.7) we get(1.12)∫0t∂2u(x,τ)∂τ2dτ=∫0tΔu(x,τ)dτ=Δ∫0tu(x,τ)dτ=Δv(x,t).
Hence,(1.13)∂u(x,t)∂t-∂u(x,0)∂t=Δv(x,t)or∂u(x,t)∂t=Δv(x,t),
by the second condition in (1.8). On the other hand, from (1.9),(1.14)∂v(x,t)∂t=u(x,t),∂2v(x,t)∂t2=∂u(x,t)∂t.
Comparing (1.13) and (1.14), we get (1.10). Besides,(1.15)v(x,0)=0,∂v(x,0)∂t=u(x,0)=φ(x)
so that initial conditions in (1.11) are also satisfied.Consequently, the solutionu(x,t) of problem (1.1), (1.3) is represented in the form(1.16)u(x,t)=Nφ(x,t)+∫0tNψ(x,τ)dτ.It follows that it is sufficient to know an explicit form of the solutionNφ(x,t) of problem (1.7), (1.8). It is known [1, 2] that(1.17)Nφ(x,t)=12m+1πm(∂∂t1t)m∫|y-x|=tφ(y)dSyifn=2m+1,(1.18)Nφ(x,t)=12mπm(∂∂t1t)m-1∂dt∫|y-x|<tφ(y)dyt2-|y-x|2ifn=2m,
where x=(x1,…,xn), y=(y1,…,yn), |y-x|2=(y1-x1)2+⋯+(yn-xn)2, and dSy is the surface element of the sphere {y∈ℝn:|y-x|=t}.In the present paper, we give a new proof of formulae (1.17), (1.18) for the solution of problem (1.7), (1.8). Our method of the proof is based on the spectral theory of the Laplace operator. We hope that such a method may be useful also in some other cases of the equation and space.The paper consists, besides this introductory section, of three sections. In Section2, we describe the structure of arbitrary rapidly decreasing function of the Laplace operator, showing that it is an integral operator and giving an explicit formula for its kernel. Next we use these results in Section 3 to derive the explicit representation formulae for the classical solution to the initial value problem for the wave equation in arbitrary dimensions. The final Section  is an appendix and contains some explanation of several points in the paper.


## 2. Structure of Arbitrary Function of the Laplace Operator

LetA be the self-adjoint positive operator obtained as the closure of the symmetric operator A′ determined in the Hilbert space L2(ℝn) by the differential expression(2.1)-Δ=-(∂2∂x12+⋅⋅⋅+∂2∂xn2),(x1,…,xn)∈Rn,
on the domain of definition D(A′)=C0∞(ℝn) that is the set of all infinitely differentiable functions on ℝn with compact support. Let Eμ denote the resolution of the identity (the spectral projection) for A:(2.2)Af=∫0∞μdEμf,f∈D(A).
Next, let g(t) be any infinitely differentiable even function on the axis -∞<t<∞ with compact support and(2.3)g̃(λ)=∫-∞∞g(t)eiλtdt
its Fourier transform. Note that the function g̃(λ) tends to zero as |λ|→∞  (λ∈ℝ) faster than any negative power of |λ|. Consider the operator g̃(A1/2) defined according to the general theory of self-adjoint operators (see [3]):(2.4)g̃(A1/2)f=∫0∞g̃(μ)dEμf,f∈L2(Rn).The following theorem describes the structure of the operatorg̃(A1/2) showing that it is an integral operator and giving an explicit formula for its kernel in terms of the function g(t).Theorem 2.1.
The operatorg̃(A1/2) is an integral operator
(2.5)g̃(A1/2)f(x)=∫RnK(x,y)f(y)dy,f∈L2(Rn).
Further, there is a smooth function k(t) defined on the interval 0≤t<∞ such that
(2.6)K(x,y)=k(|x-y|2).
The function k(t) depends on the function g(t) as follows. If one sets
(2.7)Q(t)=g(t),thatis,Q(t2)=g(t),0≤t<∞,
then
(2.8)k(t)={(-1)mπmQ(m)(t)ifn=2m+1,(-1)mπm∫t∞Q(m)(w)w-tdwifn=2m,
where Q(m)(t) denotes the mth order derivative of Q(t). Further, if suppg(t)⊂(-a,a), then suppk(t)⊂[0,a2). For any solution ψ(x,λ) of the equation
(2.9)-Δψ(x,λ)=λ2ψ(x,λ),
the equality
(2.10)∫Rnk(|x-y|2)ψ(y,λ)dy=g̃(λ)ψ(x,λ)
holds for λ∈ℝ.Proof.
First we consider the casen=1. In this case, the statements of the theorem take the following form: k(t)=Q(t)=g(t) for 0≤t<∞; the operator g̃(A1/2) is an integral operator of the form
(2.11)g̃(A1/2)f(x)=∫-∞∞g(x-y)f(y)dy,
and for any solution ψ(x,λ) of the equation
(2.12)-ψ′′(x,λ)=λ2ψ(x,λ),
the equality
(2.13)∫-∞∞g(x-y)ψ(y,λ)dy=g̃(λ)ψ(x,λ)
holds.
To prove the last statements note that, in the casen=1, the operator A is generated in the Hilbert space L2(-∞,∞) by the operation -d2/dx2 and the operator A1/2 by the operation id/dx. The resolvent Rμ=(A-μI)-1 of the operator A has the form
(2.14)Rμf(x)=i2μ∫-∞∞ei|x-y|μf(y)dy,
while the spectral projection Eμ of the operator A has the form (see [3, page 201])
(2.15)Eμf(x)=∫-∞∞sin⁡μ(x-y)π(x-y)f(y)dy,0≤μ<∞,Eμ=0forμ<0.
Therefore,
(2.16)g̃(A1/2)f(x)=∫0∞g̃(μ)dEμf(x)=∫0∞g̃(μ){∫-∞∞cos⁡μ(x-y)2πμf(y)dy}dμ=∫-∞∞{1π∫0∞g̃(λ)cos⁡λ(x-y)dλ}f(y)dy=∫-∞∞g(x-y)f(y)dy,
where we have used the inversion formula for the Fourier cosine transform. Therefore, (2.11) is proved. To prove (2.13) note that the general solution of (2.12) is
(2.17)ψ(x,λ)={c1cos⁡λx+c2sin⁡λxifλ≠0,c1+c2xifλ=0,
where c1 and c2 are arbitrary constants. Then, we have, for λ≠0,
(2.18)∫-∞∞g(x-y)ψ(y,λ)dy=c1∫-∞∞g(x-y)cos⁡λydy+c2∫-∞∞g(x-y)sin⁡λydy=c1∫-∞∞g(t)cos⁡λ(x-t)dt+c2∫-∞∞g(t)sin⁡λ(x-t)dt=c1∫-∞∞g(t)(cos⁡λxcos⁡λt+sin⁡λxsin⁡λt)dt+c2∫-∞∞g(t)(sin⁡λxcos⁡λt-sin⁡λtcos⁡λx)dt=c1cos⁡λx∫-∞∞g(t)cos⁡λtdt+c2sin⁡λx∫-∞∞g(t)cos⁡λtdt=(c1cos⁡λx+c2sin⁡λx)∫-∞∞g(t)cos⁡λtdt=ψ(x,λ)g̃(λ),
where we have used the fact that the function g(t) is even and therefore
(2.19)∫-∞∞g(t)sin⁡λtdt=0.
The same result can be obtained similarly for λ=0. Thus, (2.13) is also proved.
Now we consider the casen≥2. We shall use the integral representation
(2.20)Rμf(x)=∫Rnr(x,y;μ)f(y)dy
of the resolvent Rμ=(A-μI)-1 of the operator A. As is known [4, Section  13.7, Formula (13.7.2)],
(2.21)r(x,y;μ)=iμ(n-2)/42(n+2)/2π(n-2)/2|x-y|(n-2)/2H(n-2)/2(1)(|x-y|μ),
where Hν(1)(z) is the Hankel function of the first kind of order ν. Next, according to the general spectral theory of self-adjoint operators [3, page 150, Formula (11)], we have
(2.22)dEμf(x)=12πi(Rμ+i0-Rμ-i0)f(x)dμ.
Therefore, from (2.4) it follows that the representation (2.5) holds with
(2.23)K(x,y)=12πi∫0∞g̃(μ)[r(x,y;μ+i0)-r(x,y;μ-i0)]dμ.
Now the representation (2.6), which expresses that 𝒦(x,y) is a function of |x-y|2, follows from (2.23) by (2.21).
To prove (2.10) we use (2.23). By virtue of (2.23),
(2.24)∫Rnk(|x-y|2)ψ(y,λ)dy=∫RnK(x,y)ψ(y,λ)dy=lim⁡ε→+0∫Rn{12πi∫0∞g̃(μ)[r(x,y;μ+iε)-r(x,y;μ-iε)]dμ12πi∫0∞}ψ(y,λ)dy=ψ(x,λ)lim⁡ε→+0επ∫0∞g̃(μ)(μ-λ2)2+ε2dμ=ψ(x,λ)g̃(λ),
see Appendix. Here we have used the fact that from (2.9) it follows that
(2.25)(-Δ-z)ψ(x,λ)=(λ2-z)ψ(x,λ),
that is,
(2.26)ψ(x,λ)=(λ2-z)(-Δ-z)-1ψ(x,λ),
and therefore
(2.27)∫Rnr(x,y;z)ψ(y,λ)dy=1λ2-zψ(x,λ).
Finally, to deduce the explicit formulae (2.7), (2.8), we take ψ(x,λ)=eiλx1 in (2.10). Then, putting x̃=(x2,…,xn), we can write
(2.28)∫Rnk(|x1-y1|2+|x̃-ỹ|2)eiλy1dy1dỹ=g̃(λ)eiλx1.
If we set
(2.29)(x1-y1)2=w,
then the left-hand side of (2.28) equals
(2.30)∫-∞∞{∫Rn-1k(w+|x̃-ỹ|2)dỹ}eiλy1dy1.
On the other hand,
(2.31)∫Rn-1k(w+|x̃-ỹ|2)dỹ=∫0∞{∫|x̃-ỹ|=rk(w+|x̃-ỹ|2)dS}dr=∫0∞k(w+r2){∫|x̃-ỹ|=rdS}dr=σn-1∫0∞rn-2k(w+r2)dr=12σn-1∫w∞(t-w)(n-3)/2k(t)dt,
where
(2.32)σn=2πn/2Γ(n/2)
is the surface area of the (n-1)-dimensional unit sphere (Γ is the gamma function) and dS denotes the surface element of the sphere {ỹ∈ℝn-1:|x̃-ỹ|=r}. Therefore, setting
(2.33)Q(w)=12σn-1∫w∞(t-w)(n-3)/2k(t)dt,
we get that (2.28) takes the form
(2.34)∫-∞∞Q(w)eiλy1dy1=g̃(λ)eiλx1.
Substituting here the expression of w given in (2.29) and making then the change of variables x1-y1=t, we obtain
(2.35)∫-∞∞Q(t2)eiλtdt=g̃(λ)=∫-∞∞g(t)eiλtdt.
Hence (2.7) follows. Further, it is not difficult to check that the formula (2.33) for n≥2 is equivalent to (2.8), see Appendix.
Sinceg(t) is smooth and has a compact support, it follows from (2.7), (2.8) that the function k(t) also is smooth and has a compact support; more precisely, if suppg(t)⊂(-a,a), then suppk(t)⊂[0,a2). This implies, in particular, convergence of the integral in (2.10) for each fixed x. The theorem is proved.


## 3. Derivation of Formulae (1.17), (1.18)

Consider the Cauchy problem (1.7), (1.8):(3.1)∂2u∂t2=Δu,x∈Rn,t>0,(3.2)u(x,0)=φ(x),∂u(x,0)∂t=0,x∈Rn,
where u=u(x,t), t≥0, x=(x1,…,xn)∈ℝn, φ(x)∈C0∞(ℝn).Forν=(ν1,…,νn), x=(x1,…,xn)∈ℝn, let us set(3.3)|ν|2=ν12+⋯+νn2,(ν,x)=ν1x1+⋯+νnxn.
Since (3.4)-Δei(ν,x)=|ν|2ei(ν,x),
applying (2.9), (2.10), we get(3.5)∫Rnk(|x-y|2)ei(ν,y)dy=g̃(|ν|)ei(ν,x)(ν∈Rn).
Hence, by the inverse Fourier transform formula,(3.6)k(|x-y|2)=1(2π)n∫Rng̃(|ν|)ei(ν,x)e-i(ν,y)dν.
Multiplying both sides of the last equality by φ(y) and then integrating on y∈ℝn, we get(3.7)∫Rnk(|x-y|2)φ(y)dy=1(2π)n∫Rng̃(|ν|)ei(ν,x)[∫Rnφ(y)e-i(ν,y)dy]dν.
Substituting here for g̃(|ν|) its expression(3.8)g̃(|ν|)=2∫0∞g(t)cos⁡(|ν|t)dt
and setting(3.9)u(x,t)=1(2π)n∫Rn(cos⁡|ν|t)ei(ν,x)[∫Rnφ(y)e-i(ν,y)dy]dν,
we obtain(3.10)∫Rnk(|x-y|2)φ(y)dy=2∫0∞g(t)u(x,t)dt.
Obviously, the function u(x,t) defined by (3.9) is the solution of problem (3.1), (3.2). Next we will transform the left-hand side of (3.10) using Theorem 2.1.First we consider the casen=1. In this case, (3.10) takes the form(3.11)∫-∞∞k(|x-y|2)φ(y)dy=2∫0∞g(t)u(x,t)dt
and from (2.7), (2.8) we have(3.12)k(t2)=Q(t2)=g(t).
Therefore, making the change of variables y-x=t and taking into account the evenness of the function g(t), we can write(3.13)∫-∞∞k(|x-y|2)φ(y)dy=∫-∞∞k(t2)φ(x+t)dt=∫-∞∞g(t)φ(x+t)dt=∫0∞g(t)[φ(x+t)+φ(x-t)]dt.
Substituting this in the left-hand side of (3.11), we obtain(3.14)∫0∞g(t)[φ(x+t)+φ(x-t)]dt=2∫0∞g(t)u(x,t)dt.
Hence, by the arbitrariness of the smooth even function g(t) with compact support, we get(3.15)u(x,t)=φ(x+t)+φ(x-t)2.Further assume thatn≥2. Making the change of variables(3.16)y-x=tω,0≤t<∞,|ω|=1,ω=(ω1,…,ωn),dy=tn-1dtdSω,
where dSω is the surface element of the unit sphere {ω∈ℝn:|ω|=1}, we get(3.17)∫Rnk(|x-y|2)φ(y)dy=∫0∞tn-1k(t2){∫|ω|=1φ(x+tω)dSω}dt.
Further, making in the right-hand side of (3.17) the change of variables(3.18)x+tω=y,dSy=tn-1dSω,
where dSy is the surface element of the sphere {y∈ℝn:|y-x|=t}, we have(3.19)tn-1∫|ω|=1φ(x+tω)dSω=∫|y-x|=tφ(y)dSy=:Pφ(x,t).
Therefore,(3.20)∫Rnk(|x-y|2)φ(y)dy=∫0∞k(t2)Pφ(x,t)dt,
and (3.10) becomes(3.21)∫0∞k(t2)Pφ(x,t)dt=2∫0∞g(t)u(x,t)dt.
Consider the cases of odd and even n separately.Letn=2m+1(m∈ℕ). Then, by (2.8) we have(3.22)k(t2)=(-1)mπmQ(m)(t2)
and it follows from (2.7) (by successive differentiation) that(3.23)Q(m)(t2)=(12t∂∂t)mg(t).
Therefore,(3.24)k(t2)=(-1)m2mπm(1t∂∂t)mg(t),
and (3.21) takes the form(3.25)(-1)m2mπm∫0∞{(1t∂∂t)mg(t)}Pφ(x,t)dt=2∫0∞g(t)u(x,t)dt.
Further, integrating m times by parts, we get(3.26)∫0∞{(1t∂∂t)mg(t)}Pφ(x,t)dt=R(x,t)|t=0t=∞+(-1)m∫0∞g(t)(∂∂t1t)mPφ(x,t)dt,
where
(3.27)R(x,t)=∑k=1m(-1)k-1t{(1t∂∂t)m-kg(t)}(∂∂t1t)k-1Pφ(x,t)=∑k=1m(-1)k-1t{(1t∂∂t)m-kg(t)}(∂∂t1t)k-1t2m∫|ω|=1φ(x+tω)dSω.
Since g(t) is identically zero for large values of t, we have from (3.27) that R(x,∞)=0. Also, it follows directly from (3.27) that R(x,0)=0. Therefore, (3.25) becomes(3.28)12mπm∫0∞g(t)(∂∂t1t)mPφ(x,t)dt=2∫0∞g(t)u(x,t)dt.
Since in (3.28) g(t) is arbitrary smooth even function with compact support, we obtain that(3.29)u(x,t)=12m+1πm(∂∂t1t)mPφ(x,t).
This coincides with (1.17) by (3.19).Now let us consider the casen=2m(m∈ℕ). In this case, by (2.8) we have(3.30)k(r2)=(-1)mπm∫r2∞Q(m)(w)w-r2dw=(-1)mπm∫r∞Q(m)(t2)2tt2-r2dt,
and therefore(3.31)∫0∞k(r2)Pφ(x,r)dr=(-1)mπm∫0∞{∫r∞Q(m)(t2)2tt2-r2dt}Pφ(x,r)dr=(-1)mπm∫0∞{∫r∞Q(m)(t2)2tt2-r2dt}{∫|y-x|=rφ(y)dSy}dr=(-1)mπm∫0∞{∫r∞Q(m)(t2)2t[∫|y-x|=rφ(y)dSyt2-|y-x|2]dt}dr=(-1)mπm∫0∞Q(m)(t2)2t{∫0t[∫|y-x|=rφ(y)dSyt2-|y-x|2]dr}dt.
Hence, setting(3.32)Hφ(x,t):=∫0t[∫|y-x|=rφ(y)dSyt2-|y-x|2]dr=∫|y-x|<tφ(y)dyt2-|y-x|2,
we get(3.33)∫0∞k(r2)Pφ(x,r)dr=(-1)mπm∫0∞Q(m)(t2)2tHφ(x,t)dt.
Substituting this in the left-hand side of (3.21) (beforehand replacing t by r in the left side of (3.21)), we obtain(3.34)(-1)mπm∫0∞Q(m)(t2)2tHφ(x,t)dt=2∫0∞g(t)u(x,t)dt
or, using (3.23),(3.35)(-1)m2m-1πm∫0∞{(1t∂∂t)mg(t)}tHφ(x,t)dt=2∫0∞g(t)u(x,t)dt.
Further, integrating m times by parts, we get(3.36)∫0∞{(1t∂∂t)mg(t)}tHφ(x,t)dt=L(x,t)|t=0t=∞+(-1)m∫0∞g(t)(∂∂t1t)mtHφ(x,t)dt,
where(3.37)L(x,t)=∑k=1m(-1)k-1{(1t∂∂t)m-kg(t)}(∂∂t1t)k-1tHφ(x,t).
Since g(t) is identically zero for large values of t, we have from (3.37) that L(x,∞)=0. Also, using the expression of Hφ(x,t),(3.38)Hφ(x,t)=∫0t[∫|y-x|=rφ(y)dSyt2-|y-x|2]dr=∫0tr2m-1[∫|ω|=1φ(x+rω)t2-r2dSω]dr=∫0tr2m-1t2-r2[∫|ω|=1φ(x+rω)dSω]dr=∫0t(t2-ξ2)2m-2[∫|ω|=1φ(x+t2-ξ2ω)dSω]dξ,
we can check directly from (3.37) that L(x,0)=0. Therefore, (3.35) becomes(3.39)12m-1πm∫0∞g(t)(∂∂t1t)mtHφ(x,t)dt=2∫0∞g(t)u(x,t)dt.
Since in (3.39) g(t) is arbitrary smooth even function with compact support, we obtain that(3.40)u(x,t)=12mπm(∂∂t1t)mtHφ(x,t)=12mπm(∂∂t1t)m-1∂∂tHφ(x,t).
This coincides with (1.18) by (3.32).



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