# The Wave Equation

### - The Origin and Solutions of the Wave Equation

by Ronald Koster, version 1.0, 2005-03-25

Keywords: origin of the wave equation

## 1. Introduction

Waves are usually described by the so called Wave Equation: ∇2f - (1/c2)(∂2/∂t2)f = 0. Where does that equation come from? What kind of waves are described by it? Are there any other equations, ie. are there any wave functions not fitting the standard Wave equation?

## 2. Unattenuating one dimensional (1D) waves

One can define an unattenuating 1D wave f(x, t) with constant phase velocity c=dx/dt as follows:

f(x + dx, t + dt) = f(x, t)     (2.1)

⇔ f + fxdx + ftdt = f ⇒ fxc + ft = 0     (2.2)

(2.2) ⇒ fxxc + ftx = 0 ⇒ fxxc2 + ftxc = 0     (2.3)

(2.2) ⇒ fxtc + ftt = 0     (2.4)

Substracting (2.4) from (2.3) and using ftx = fxt gives:

fxxc2 - ftt = 0 ⇔ fxx - (1/c2)ftt = (∂2/∂x2 - (1/c2)∂2/∂t2)f = 0     (2.5)

This is the unattenuated 1D wave equation. A rather general solution (perhaps the general solution, but at least the one we are interested in) is:

f(x, t) = F(ct - x) + G(ct + x)     (2.6)

As can easily be verified as follows. Use U = ct - x and realize (using the chain rule):

Fxx = FUU and Ftt = c2FUU ⇒ Fxx - (1/c2)Ftt = FUU - (1/c2)c2FUU = 0     (2.7)

A similar deduction can be done for G and V = ct + x. Hence F and G are both solutions of (2.5). And since (2.5) describes a lineair operator their sum is a solution too.

F represents a wave traveling in the direction of increasing x at constant speed c. G represents a wave traveling in the opposite direction. Often G = 0.

## 3. Unattenuating three dimensional (3D) Waves

Now that we have found the 1D unattenuating wave equation we would like to find the 3D version. A possible candidate, purely for look alike reasons, could be:

fxx + fyy + fzz - (1/c2)ftt = 2f - (1/c2)(∂2/∂t2)f = ∂μ2f = 0.     (3.1)

with:

μ ≡ (∂/∂x1, ∂/∂x2, ∂/∂x3, ∂/∂x4) and x1 = x, x2 = y, x3 = z, x4 = jct. ⇒ ∂μ = (, (1/(jc))∂/∂t)     (3.2)

(using the chain rule: ∂/∂x4 = (∂/∂xt)(∂t/∂x4) = 1/(jc)(∂/∂t))

The operator μ2 is called the d'Alembertian.

Now the question is: Does (3.1) describe unattenuated 3D waves?

### 3.1 Test 1: Plane waves

One would expect plane waves to be a solution of (3.1). Based on the previous paragraph we know a plane wave traveling in the positive x direction has the form f(r, t) = g(ct - x). As can easily be verified by substitution g is a solution of (3.1) indeed.

In case the wave source is a harmonic oscillator one would expect:

f(r, t) = Aeejk(ct - x)     (3.3)

By definition the wave should be periodic in x with period λ, thus:

kλ = 2π ⇒ k = 2&pi/λ     (3.4)

Also by definition it should be periodic in t with period T, thus:

kcT = 2π     (3.5)

(3.4) and (3.5) ⇒ (2π/λ)cT = 2&pi ⇔ c = λ/T = λν     (3.6)

Which we know is true. Now let's consider a rotated frame u,v,w. Suppose a harmonic plane wave travels in the positive u direction, thus:

f(r, t) = Aeejk(ct - u)     (3.7)

Let n be the unit vector in the positive u direction, e1 the unit vector in the positive x direction, e2 the unit vector in the positive y direction, e3 the unit vector in the positive z direction, θ1 the angle between n and e1, θ2 the angle between n and e2, θ3 the angle between n and e3. It is easy to see the following is true:

u = xcosθ1 + ycosθ2 + zcosθ3 = xe1 ⋅ n + ye2 ⋅ n + ze3 ⋅ n = r ⋅ n     (3.8)

Notice by the way that:

n ⋅ ei = cosθi = (n1e1 + n2e2 + n3e3) ⋅ ei = ni     (3.9)

Substituting this into (3.7) gives:

f(r, t) = Aeejk(ct - n ⋅ r) = f(r, t) = Aeej(kct - k ⋅ r) with k = kn     (3.10)

So (3.9) describes a general harmical plane wave in the direction of k. Substitution it into (3.1) learns (3.9) is a solution of (3.1) (using k2 = k12 + k22 + k32).

Had we not started with (3.7) but with:

f(r, t) = g(ct - u)     (3.11)

We would have found:

f(r, t) = g(ct - r ⋅ n)     (3.12)

For (using U = ct - r ⋅ n, the chain rule, n ⋅ ei = cosθi = ni and n12 + n22 + n32 = 1):

t2g = c2gUU andx2g = n12gUU andy2g = n22gUU andz2g = n32gUU ⇒ ∂μ2g = 0     (3.13)

Notice (3.10) is a special case of (3.12). So far we have not considered the wave traveling in the opposite direction. It can easily be verified (using V = ct + r ⋅ n) that the following equation gives the general plane wave solution:

f(r, t) = g(ct - r ⋅ n) + h(ct + r ⋅ n)    (3.14)

Conclusion: Unattenuated plane waves (harmonic or not) are a solution of (3.1).

### 3.2 Test 2: Spherical waves

Consider a wave of the form:

f(r, t) = g(r)ejωt     (3.15)

Is this a valid solution of (3.1)? Let's substitute it into (3.1) and use ∇2 in polar coordinates:

2 = (1/r2)∂r(r2r) + (1/(r2sinθ))∂θ(sinθ∂θ) + (1/(r2sin2θ))∂φ2     (3.16)

(3.15) ⇒ ∂μ2f = (2g)ejωt + (ω2/c2)ejωt = 0 ⇔ 2g + k2g = 0 (using k = &omega/c)     (3.17)

2g = (1/r2)(d/dr)(r2gr) = (1/r2)(2rgr + r2grr) = (2/r)gr + grr     (3.18)

Furthermore realize that:

(1/r)(d2/dr2(rg)) = (2/r)gr + grr     (3.19)

For:

(rg)r = g + rgr ⇒ (rg)rr = gr + gr + rgrr = 2gr + rgrr     (3.20)

Hence (3.17) and (3.18) and (3.19) give:

2g + k2g = (1/r)(rg)rr + k2g = 0 ⇔ (rg)rr + k2(rg) = 0     (3.21)

The general solution of (3.21) is:

rg = ae-jkr + bejkr     (3.22)

With a and b being complex constants. Which can be verified by substituting it in (3.21). Hence the solution of (3.15) is:

(3.1) and (3.15) ⇒ f(r, t) = (a/r)ejk(ct - r) + (b/r)ejk(ct + r)     (3.23)

Which in the absence of an outwards traveling wave simplifies to (using A = |a| and φ = arga):

f(r, t) = (Ae/r)ejk(ct - r)     (3.24)

Notice these kind of waves are not really unattenuated, since they fall of (are attenuated) by 1/r. Also notice that waves of the form:

f(r, t) = (Ae/rn)ejk(ct - r) and n ≠ 1     (3.25)

are not solutions of the wave equation (3.1). Though we consider these froms to be wave too...

Example
Consider the situation in which the source is harmonically emitting a conserved property E (for example energy). Then it should be possible to describe the flux Φ of that quantity as:

Φ = |f|2     (3.26)

with f given by (3.24). For then the amount of E passing through the sperical surface at r is then given by:

E = 4πr2|f|2 = constant     (3.27)

Ergo, no energy gets lost nor is created between the surfaces at r = r1 and r = r2 with r2 > r1.

### 3.3 Conclusion

Equation (3.1) has solutions we identify as waves indeed. Thus it is a 3D wave equation. Despite not all wave forms are a solution of it, some very common wave forms are. Since other wave forms exist, it is not the only possible wave equation. However, it turns out to be a very common one, appearing in many physics theories. It is therefore commonly know as the (unattenuated) 3D wave equation.