KINEMATIC DERIVATION OF THE HARMONIC WAVE EQUATION AND RELATED TOPICS

An extremely important type of wave in physics is the harmonic wave. This is a wave consisting of propagating simple harmonic oscillations or linear combinations thereof. Attach a weight to a spring and hang the spring so the weight is free to move. Then lift the weight straight up and release it; it will oscillate up and down in a close approximation to simple harmonic motion. Now imagine that you have a whole array of such mass-spring systems, all oscillating independently. If you could add all their oscillations together, you would get linear combination of simple harmonic motions. Whether a harmonic wave involves propagating oscillations of a solid, liquid, gas, or electromagnetic field, the oscillations will consist, in general, of a linear combination of simple harmonic motions as exemplified by the sum of the oscillations of the multiple mass-spring system.

The One-Dimensional Wave Equation

Mathematically, a harmonic wave can represented by a sinusoidal function of time and position or a linear superposition of such functions (that is, you just add them together as in the case of sum of the oscillations of the spring and weight system described above), since a single such sinusoidal function describes the propagation of a single simple harmonic oscillation. The argument of each sinusoidal function is of the form x ± vt, where x is the position along the wave where the oscillation is observed, t the time variable for the oscillation, and v the speed of the wave, which is the same for all frequencies.

The harmonic wave equation can be derived dynamically or kinematically. Dynamically, the derivation depends on the type of wave. For example, the wave equation for elastic media depends on the displacement of the medium through which the wave travels being proportional to the magnitude of the non-dissipative restoring force (or stress) that results from the displacement. Another example involves electromagnetic waves, which can be derived from electromagnetic theory. However, the kinematic derivation is not restricted to any particular type of medium. Kinematically, a wave is harmonic if it is described by a function of x ± vt, where the wave speed, v, does not depend on the shape of the wave. All harmonic waves are included in a kinematic derivation. A soliton also propagates without changing its shape, but there are restrictions on the form of the waves, whereas a harmonic wave can be of any reasonable (aka "differentiable") shape.

You can impose the requirement of constant shape for a wave moving in the positive direction by requiring the displacement of the medium through which the wave is propagating is same at a position x + Δx after the wave has moved a distance Δx as it was at the earlier position x. Since the wave is traveling at a speed v, it takes the wave a time Δt = Δx / v to travel the distance Δx. In equation form, the requirement that the displacement, call it y, at position x + Δx and time t +Δt is the same as that at position x and time t is

(1) y{x + Δx, t + Δt} = y{x,t},

where the wave speed is

(2) v = Δx / Δt.

The curly brackets indicate functionality; that is, at what position and time the displacement is evaluated. See Figure 1.

At this point you need to employ a bit of calculus. You may recall that an "expansion" of a function can be used to get the value of the function at x + Δx if you know its value at x. However, an expansion is only possible if the derivatives of the function evaluated at position x exist out the wazoo (that is, mathematically speaking, to "all orders", meaning you can differentiate it forever. Sinusoidal functions, for example, have this property.). The function f{x} can be expanded as

(3) f{x + Δx} = f{x} + (df{x}/dx)Δx + (d2f{x}/dx2)Δx2 + ... .

If Δx is extremely small, you can drop all terms with Δx raised to powers greater than one and write

(4) f{x + Δx} ≅ f{x} + (df{x}/dx)Δx

This expansion would be exact if f{x} were the equation of a straight line. Then df{x}/dx would be the slope and f{x} + (df{x}/dx)Δx would be the value of the line at x + Δx. Therefore this approximation is the same as if you were to choose a Δx small enough that the function is very close to a straight line over the range of Δx. See Figure 2.

Eq. (4) is what you are going to apply to Eq. (1) with one slight difference. The function f{x} in Eq. (4) is a function of one variable only. However, y{x,t} is a function of two variables. How do you handle that? If you have advanced in calculus to the point of partial derivatives, you may have an idea where to go from here. If not, just accept the fact that, if you have a function of more than one variable, you can always find out how the function changes with respect to one of the variables while the others are held constant by taking the partial derivative of the function. Practically speaking, this means you treat all the variables other than the one involved in the derivative as constants. As an example, say you have a function f{x,y} = xy2. The partial derivative with respect to x would just be

∂f/∂x = y2,

treating y as a constant. On the other hand, the "partial of f with respect to y" would be

∂f/∂y = 2xy,

where x is treated as a constant.

Apply Eq. (4) to y{x + Δx, t + Δt} as follows. First find the approximate value of y{x + Δx, t + Δt} in terms of y{x, t + Δt}. This only involves an expansion in x. You get

(5) y{x + Δx, t + Δt} ≅ y{x, t + Δt} + (∂y{x, t + Δt}/∂x)Δx.

Now, expand y{x, t + Δt} and ∂y{x, t + Δt}/∂x from the values at y{x,t} and ∂y{x,t}/∂x. The latter can be expanded in terms of its derivatives just like y, because it is a function of x and t also. You get

(6) y{x, t + Δt} ≅ y{x,t} + (∂y{x,t}/∂t)Δt

and

(7) ∂y{x, t + Δt}/∂x = ∂y{x,t}/∂t + [∂(∂y{x,t}/∂x)∂t]Δt = ∂y{x,t}/∂t + (∂2y{x,t}/∂x∂t)Δt.

Note the partial derivative of y with respect to both x and t in the last term on the right in Eq. (7). This just means you took the derivative of y with respect to x holding t constant, then took the derivative with respect to t holding x constant. It does not matter which order you do this; that is,

2y/∂x∂t = ∂2y/∂t∂x.

Substituting the expansions (6) and (7) into Eq. (5) gives you

(8) y{x + Δx, t + Δt} ≅ y{x,t} + (∂y{x,t}/∂t)Δt + (∂y{x,t}/∂x)Δx + (∂2y{x,t}/∂x∂t)ΔxΔt.

If Δx and Δt are very small, the term at the far right in the above equation, called the "second-order term", will be small enough compared to the ones to the left that it can be ignored in the approximation. So, if we stick to very small "Δ"s, the expansion of y{x,t} becomes

(9) y{x + Δx, t + Δt} ≅ y{x,t} + (∂y{x,t}/∂t)Δt + (∂y{x,t}/∂x)Δx

This equation has a geometrical interpretation similar to that of Eq. (4). Instead of looking at a piece of a curve so small it looks like a straight line, in Eq. (9) you are looking at a piece of a "surface" (defined by the function of two variables, y{x,t}) so small that it looks like a plane. This is depicted in the following figure, where the blue shading is supposed to represent a piece of the surface y{x,t}. Eq. (9) is the approximate value of y at x + Δx and t + Δt in terms of its value at x and t, assuming the Δ values are small enough that the piece of the surface is close to a flat plane.

What you've done so far is just commonplace, boring old math. Now it's time to insert the requirement that the displacement due to the wave at a position x + Δx at a time t + Δx / v, where v is the wave speed, is the same as the displacement at position x at the earlier time t. This is Eq. (1) with Δt replaced by its equivalent, Δx / v.

(10) y{x + Δx, t + Δx / v} = y{x,t}.

Substituting Δx / v for Δt in (9) and cancelling out the equal terms by means of Eq. (10) gets you to the following equation.

(11) (∂y{x,t}/∂t)Δx / v + (∂y{x,t}/∂x)Δx ≅ 0.

Now you can cancel the "Δx"s, multiply through by v, et voila,

(12) ∂y{x,t}/∂t + v(∂y{x,t}/∂x) = 0.

You can replace "≅" by "=" because you consider the "Δx"s to get vanishingly small, thus eliminating all the higher order terms such as those in Eq. (3). Any differentiable function of the form f{x - vt} is a solution to this equation, as can easily be verified.

One slight problem, however. Equation 12 is the wave equation for harmonic waves moving in the positive x direction but not for those moving in the negative x direction. This is because I assumed a positively moving wave when I wrote Eq. (2). A wave moving in the negative direction would have a negative velocity v and you would have to rewrite Eq. (2) as

(13) v = -Δx / Δt

to ensure a positive time interval. Going through all the steps again with the negative sign results in the wave equation for a negatively propagating wave,

(14) ∂y{x,t}/∂t - v(∂y{x,t}/∂x) = 0.

Now it is darned inconvenient to have separate equations for the positive and negative directions. For one thing, you can't consider standing waves to be wave-equation solutions, because you need to combine both positively and negatively propagating waves to construct a standing wave. And it only gets worse when you generalize to waves moving in two and three dimensions. Somehow you have to combine the two wave equations, (12) and (14) to get a single equation that includes waves propagating in either direction. One way is to muliply the two together. This option will be discussed later.

Another way to combine the two equations is suggested by rewriting (12) as

(15) [(∂/∂t) + v(∂/∂x)]y{x,t} = 0.

Written this way the stuff inside the square brackets can be considered an operator, operating on y. All you've done is factor out the derivative operations to construct the operator in square brackets, which is a linear combination of partial derivative operations. The same can be done for Eq. (14),

(16) [(∂/∂t) - v(∂/∂x)]y{x,t} = 0.

The operators in square brackets in Eqs. (15) and (16) commute because the partial derivatives do, and one will yield zero when operating on y{x,t}. Therefore, why not operate on y with both operators, first one and then the other? (The order you do this does not matter, again because partial derivatives commute.) The result will be zero, because, whichever wave you choose - one propagating in the positive direction or one propagating in the negative direction - one of the operations will give zero. Finally, you use the math identity (a + b)(a - b) = a2 - b2, to combine the operators to give

(17) ∂2y/∂t2 - v22y/∂x2 = 0.

This is the full-blown one-dimensional harmonic wave equation, good for waves traveling in both the positive and negative directions. Functions of the form

(18) y{x,t} = Acos[k(x ± vt) + φ],

where A is the wave amplitude, k is a constant (called the "wave number" and equal to 2π divided by the wavelength), and φ is an arbitrary (phase) angle, are solutions of the equation with a single frequency (= kv / 2π) and wavelength (= 2π / k). In the terminology of light waves, these waves might be called "monochromatic". However the derivation of the equation does not depend on the shape of the wave (so long as the shape can be differentiated). The general solution to this equation is any doubly differentiable function f{x ± vt}, that is, any harmonic wave is a solution to the equation.

Now examine the option mentioned above, that is, multiply Eqs. (12) and (14) together. The equation you get is

(19) (∂y/∂t)2 - v2(∂y/∂x)2 = 0.

It is not hard to show that this equation has the same solutions as Eq. (17), that is, any function singly differentiable of the form f{x ± vt) is a solution of Eq. (19). Therefore, it is just as viable a wave equation as the canonical form, Eq. (17). Furthermore, functions that are only differentiable once, not twice, qualify, mathematically, as solutions, in contrast to the double-differentiation requirement for Eq. (17). However, to be a physically acceptable solution, the function must be doubly differentiable, as will be shown presently.

The next topic to cover is the physical interpretations of Eqs. (17) and (19). The first term in Eq. (17) is readily recognized as the oscillatory acceleration of an element of the medium at a certain position x through which the wave is traveling. For example, if the medium is an elastic string, the term describes the up-and-down acceleration of a point at position x along the string. Recall that Newton's second law of motion says the acceleration of a mass times that mass equals the (net) force acting on the mass. Therefore, if you multiply the mass per unit length times the first term in Eq. (17), you get the net force per unit length acting on the medium. Apply this idea by multiplying Eq. (17) through by the mass per unit length, call it "μ", and moving the second term to the right side. You get

(20) μ∂2y/∂t2 = μv22y/∂x2.

The term on the right therefore has to also be the net force per unit length, and what you have here is just a form of Newton's second law: mass times acceleration equals net force - per unit length in this case. Call the net force per unit length "f". The equation for f is

(21) f = μv22y/∂x2.

For an elastic string, it is known that the wave speed equals the square root of the tension in the string, F, divided by the mass per unit length of the string, μ, or,

(22) v = (F / μ)1/2.

Therefore, the net oscillatory force per unit length along a string is given by

(23) f = F∂2y/∂x2.

If the wave is a harmonic wave of the form of Eq. (18), the net force per unit length varies as

(24) -k2FAcos[k(x ± vt) + φ] = -k2Fy{x,t}.

Making an analogy with a mass-spring system, where the restoring force is the negative of the spring constant (aka spring stiffness factor) times the displacement, you see that the spring constant equals the wave number squared times the tension, that is, k2F. This means the shorter the wavelength and the greater the tension, the stiffer the "spring" acting to restore the element of string to its equilibrium position.

What about Eq. (19)? The first term in this equation is the square of the speed of an element of the medium in oscillatory motion. If you multiply this by one-half times mass per unit length, the term becomes the kinetic energy per unit length of the medium. For the wave to not change shape in the absence of some means of amplifying it, the medium must not dissipate the wave's energy. In other words, the medium must be conservative. For a conservative medium, the kinetic plus potential energy per unit length must be constant. Also, the potential energy must be a function of position only, which means it is a function of the displacement y. Multiplying Eq. (19) through by (1/2)μ you get

(25) (1/2)μ(∂y/∂t)2 - (1/2)μv2(∂y/∂x)2 = 0.

Say the total energy per unit length = e. Add this to both sides to get

(26) (1/2)μ(∂y/∂t)2 - (1/2)μv2(∂y/∂x)2 + e = e.

This shows that the term

-(1/2)μv2(∂y/∂x)2

equals the potential energy minus the total energy. However, since the system is conservative, the zero of the potential energy can be re-assigned arbitrarily. Therefore, define the zero of potential energy to be such that the total energy is zero. (This means the potential energy is always negative and is equal to the negative of the kinetic energy.) If you redefine the potential energy this way, then the potential energy per unit length (call it "u"), is

(27) u{x,t} = -(1/2)μv2(∂y/∂x)2.

The relationship between force and potential energy is

(28) f = -∂u/∂y,

because y is the displacement caused by the force. It is easy to show that the partial derivative of u in Eq. (27) with respect to y gives Eq. (21).

The above analysis seems to imply that Newton's laws of motion depend on the homogeneity and isotropy of space and time. Without this property of space and time there would be no guarantee that a wave could propagate through an isotropic, homogeneous medium without changing shape. It is common knowledge among physicists that Newton's first law of motion (due to Galileo) is the basis of the conservation of momentum and needs a homogeneous, isotropic space for it to be realized. The fact that the wave equation can be derived from kinematic considerations alone implies that the conservation of momentum and energy depend on the homogeneity and isotropy of space and time. This is not new but does confirm previous concepts.

The Three-Dimensional Wave Equation

The three dimensional equations analogous to Eqs. (12) and (14) are

(29) ∂ψ/∂t ± v·∇ψ = 0,

where you have to use something other than "y" for the wave displacement if y is used as one of the three position variables. (I use x, y, and z here.) I have chosen to use ψ to stand for the wave displacement. You need some knowledge of vector algebra to understand this equation, where v is the velocity vector, the symbol "" is the gradient operator,

(30) = i∂/∂x + j∂/∂y + k∂/∂z,

and "·" denotes a vector multiplication operation known as the "scalar product" or "dot product". The dot product between two vectors A and B is defined as

(31) A·B = ABcosθ,

where A and B are the magnitudes of the vectors and θ is the angle between them. Finally, i, j and k are unit vectors in the x, y, and z directions, respectively.

You can combine Eqs. (29) in the same way as before to get two wave equations, but the algebra is a bit more difficult. It is easier to get the counterpart to Eq. (14), so I'll do this one first. The assumption is made that the wave propagates in the same direction as the gradient of ψ, that is,

(32) v || ψ.

This should be true for isotropic media. In this case, Eqs. (29) can be written as

(33) ∂ψ/∂t ± v|ψ| = 0,

where |ψ| is the magnitude of the gradient of ψ. This is possible because θ = 0° (See Eq. (31).) Using the definition of the gradient operator and the dot product, Eqs. (33) can be written in the expanded form

(34) ∂ψ/∂t ± v[(∂ψ/∂x)2 + (∂ψ/∂y)2 + (∂ψ/∂z)2]1/2 = 0.

(The expansion follows from the dot product, where

(35) |A| = (A·A)1/2 = (Ax2 + Ay2 + Az2)1/2.)

Multiplying the two equations in (34) together gives the 3-D version of Eq. (14).

(36) (∂ψ/∂t)2 - v2[(∂ψ/∂x)2 + (∂ψ/∂y)2 + (∂ψ/∂z)2] = 0.

Like its 1-D counterpart, this equation is related to the energy carried by the wave. The solutions to this equation are of the form

(37) ψ{k·r ± |k|vt},

where k points in the direction of propagation of the wave, r is the position vector of the wave displacement, t is the time, and v is the speed of the wave. The only requirement for this function to be a solution of the equation is that its first derivatives exist. However, as in the case for the one-dimensional wave equation, the solution will not be physically acceptable if the second derivatives don't exist.

To get the canonical three-dimensional wave equation, you define two operators by analogy to what was done in the one-dimensional case,

(38) (∂/∂t ± v·∇).

You operate on a function of the form of Eq. (37) with each operator in turn (jumping to the correct conclusion that this function is a general solution of the canonical wave equation). Again, it doesn't matter in which order this is done, and the result will be zero for the same reason as before. You will come up with

(39) [∂2/∂t2 - (v·∇)(v·∇)]ψ = 0.

The second term on the left-hand side of the equation can be evaluated using the vector identity (valid if A and B are parallel)

(40) (A·B)(A·B) = A2(Bx2 + By2 + Bz2).

Applying this to Eq. (39) yields

(41) ∂2ψ/∂t2 - v2[2ψ/∂x2 + ∂2ψ/∂y2 + ∂2ψ/∂z2] = 0.

And, like in the one-dimensional case, this equation is an expression of Newton's second law of motion. Any function of the form of Eq. (37) is a solution so long as its second derivatives exist.

For those more familiar with vector calculus, you can use vector identities as follows to get the second term on the left in Eq. (41) from the second term in Eq. (39). First handle the term

v·∇ψ = ∇·(vψ) - ψ∇·v = ∇·(vψ),

where the divergence of v vanishes since v is a constant. Now look at the next operation.

(∇·(vψ)) = ∇ × (∇ × vψ) + v2ψ.

In this equation, the first term on the right hand side is zero, because

∇ × vψ = ψ × v + ψ∇ × v = 0.

The last term on the right is zero due to v being a constant. The first term vanishes due to ψ and v being parallel. Putting this all together (and taking the final dot product) gives

(v·∇)(v·∇)ψ = v22ψ = v2[2ψ/∂x2 + ∂2ψ/∂y2 + ∂2ψ/∂z2].

The Continuity Equation

Eq. (29) was written with a plus/minus sign because v is the same velocity in each equation, thus making it necessary to use a positive sign for a wave moving in the +v direction and a negative sign for a wave moving in the -v direction. However, if you let v "take its sign" - that is let v contain the positive or negative sign - then you only have to write one equation to stand for both equations in Eq. (29). That is,

(42) ∂ψ/∂t + v·∇ψ = 0,

Because the divergence of v (a constant) is zero, a vector identity allows you to write (42) as

(43) ∂ψ/∂t + ∇·(vψ) = 0.

This can be recognized as a continuity (or conservation) equation. For example, for electric charge, the continuity equation is

(44) ∂ρ/∂t + ∇·j = 0,

where ρ is the electric charge per unit volume and j is the charge current density. This equation says that the time rate of change of the charge in a given volume is equal to the current flowing into or out of that volume, so it expresses the conservation of electric charge.

Looking at Eq. (43) in light of Eq. (44), you see that, for a wave traveling without changing its shape, there is a sort of "conservation" of wave displacement. The change in the displacement in a given region has to be equal to the amount of displacement that either enters or leaves that region. With that interpretation, the term vψ is the displacement current "density". I will use this result below in the discussion of electromagnetic waves.

Electromagnetic Waves

The electromagnetic wave equation from Maxwell's electromagnetic theory for a wave moving through a vacuum is exactly the same as the one discussed so far, with one exception. There are two vector components to an electromagnetic wave, not just the one component of displacement as, for example, in the case of a wave on a string. One is the electric field E and the other the magnetic field B. The equation for each is the same as Eq. (41) with the scalar (or one-dimensional vector) ψ replaced with E and B. The two equations are

(45) ∂2E/∂t2 - c2[2E/∂x2 + ∂2E/∂y2 + ∂2E/∂z2] = 0,

and

(46) ∂2B/∂t2 - c2[2B/∂x2 + ∂2B/∂y2 + ∂2B/∂z2] = 0,

where the speed v is replaced by the speed of light, traditionally symbolized by "c".

You lose no generality for the following work if you take the electric-field wave to be moving in the positive or negative x direction with the field E vibrating in some direction, arbitrary except that, for a plane wave in vacuum, it must vibrate in a direction perpendicular to the propagation direction, x. Hence, you can assume it is vibrating in the y direction, or the z direction, or some direction in between. If you take the electric field to lie along an axis to make it a one-dimensional vector and call it "E" without the bold type, then Eq. (42) becomes

(47) ∂2E/∂t2 - c22E/∂x2 = 0,

with an identical equation for the magnetic field B. (By electromagnetic theory these fields have to be at right angles for a wave traveling in a vacuum.)

What I intend to do is to apply what was covered earlier to the wave equation for the electric field, Eq. (44). A similar result would apply to the wave equation of the magnetic field. I will specialize to the case of a "plane wave". Perhaps the best way to describe what is meant by a plane wave is to give some examples. Imagine you are flying over the ocean and you see straight rows of ocean swells, forming what looks like a moving furrowed field. The crests and troughs form straight lines and the profile of the waves is sinusoidal. These constitute two-dimensional plane waves. Now imagine that you have a long tube extended in a straight line. You place a tuning fork of a particular frequency at one open end of the tube. The sound wave traveling down the tube will be a good approximation to a three-dimensional plane wave. The wave fronts of a plane wave in three dimensions are flat planes, hence the name of the wave.

A three-dimensional electric field plane wave moving in the positive x direction is given by the equation

(48) E = Eocos(kx - ωt + φ).

The constant φ is the phase angle, which is arbitrary and can be assigned a value of zero. Eo is the amplitude of the wave, k is the wave number and equal to 2π over the wavelength (k = 2π/λ), and ω is the angular frequency, equal to 2π times the ordinary frequency, f (ω = 2πf). This wave is a solution of Eq. (44), since it can be written as a function of "x - ct" as follows.

(49) E = Eocos[k(x - ct)],

where c = ω / k = λf = the speed of light (about 3.0 × 108 m/s).

The electric field wave equation can be written in the alternate form of Eq. (19).

(50) (∂E/∂t)2 - c2(∂E/∂x)2 = 0.

From electromagnetic theory, the energy density of the wave, call it u{x,t}, is given by the formula (in scientific units, SI, of joule per cubic meter)

(52) u{x,t} = εoE{x,t}2.

This equation includes the energy in both the electric and magnetic fields. The energy density is a function of position and time because the electric and magnetic fields are. εo is a constant that depends on the units used. In SI,

(52) ε0 = 8.854 × 10-12 C2/N·m2 (coulomb squared per newton-meter squared).

For the solution given in Eq. (46),

(53) u{x,t} = εoEo2cos2(kx - ωt).

Note that the energy density goes from zero to a maximum of εoEo2. What happens to the energy contained in the fields when u goes to zero? By electromagnetic theory, this is the energy propagated forward by the wave according to conservation of energy. Note that the total energy density is given by Eq. (50) when the cosine term equals one, namely,

(54) umax = εoEo2.

Well, not really. Electromagnetic theory doesn't really say what the energy density is, just that the loss or gain of energy density in a given volume has to be balanced by the flow of energy out of or into that volume such that the conservation of energy is observed. (This is known as Poynting's theorem.) But, for convenience, for the time being I will accept Eq. (51) as the total energy density of an electromagnetic wave.

You can turn the electric field wave equation (47) into an energy density wave equation by multiplying through by εo / ω2. This changes the units of each term to joule per cubic meter. You get

(55) (εo / ω2)(∂E/∂t)2 - (εoc2 / ω2)(∂E/∂x)2 = 0.

Or, since ω / c = k,

(56) (εo / ω2)(∂E/∂t)2 - (εo / k2)(∂E/∂x)2 = 0.

Following the same arguments leading to Eq. (26), you can add the "total" energy density to both sides of the equal sign and write Eq. (53) as

(57) (εo / ω2)(∂E/∂t)2 - (εo / k2)(∂E/∂x)2 + ε0E02 = ε0E02.

You might be tempted to treat the first term of (54) as analogous to the kinetic energy density as in Eq. (26). Then the second and third terms on the left of the equal sign constitute the analogy of the potential energy density. The sum of the terms on the left equals the total energy density on the right of the equal sign. However, once again you have the arbitrariness of the zero level of the potential energy. After all, you can add any energy density to both the left and right side of the equal sign, and the equation is still valid. If you take the total energy to be zero as explained in the text surrounding Eq. (26), you come up with Eq. (53).

From the Continuity Equation for Displacement to Poynting's Theorem

Recall the continuity equation for displacement derived above.

(43) ∂ψ/∂t + ∇·(vψ) = 0.

For a plane electromagnetic wave with an electric field "displacement" E, the equation becomes

(58) ∂E/∂t + ∇·(cE) = 0,

where c is the velocity of light, that is, it has the magnitude of the speed of light, c, and the direction in which the light is going. By a vector identity used before, this equation can be written as

(59) ∂E/∂t + c·∇E + E∇·c = 0,

where the third term on the left is zero due to c being a constant. Multiply both sides of Eq. (59) by the electric field E and by 1/2.

(60) (1/2)E∂E/∂t + (1/2)Ec·∇E = 0,

This equation is the same as

(61) ∂E2 + c·∇E2 = 0.

Applying the same vector identity as before but in the reverse direction you get

(62) ∂E2 + ∇·(cE2) = 0.

This is actually a form of Poynting's theorem for an electromagnetic wave in vacuum. To see this recall from electromagnetic theory that such a wave travels in the direction of E×B. Since this is the direction of c, c can be written as

(63) c = (E×B / |E×B|)c.

Replace c by this expression and note that for an EM wave in vacuum, |E×B| = EB = E2 / c. Also multiply the equation through by the permittivity of free space, εo, noting that the energy density of the wave is given by Eq. (48). With these changes you get

(64) ∂u/∂t + εoc2∇·(E×B) = 0.

Since Poynting's vector is

(65) S = ε0c2(E×B),

You get Poynting's theorem for an EM wave in vacuum in its usual form from Eq. ((64) as

(66) ∂u/∂t + ∇·S = 0.

As an aside, you can go back to Eq. (56),

(56) (εo / ω2)(∂E/∂t)2 - (εo / k2)(∂E/∂x)2 = 0,

replace E with (u / ε0)1/2, perform some differention and come up with

(67) (∂u/∂t)2 - c2(∂u/∂x)2 = 0,

showing that the energy density of an electromagnetic wave also obeys the wave equation.

What you have seen so far implies that the amplitude squared of a wave is always proportional to the energy or energy density. I don't think the developments given above prove this, but note that for a wave that propagates without changing its shape, there is always a conservation equation in the form of Eq. (42). Since waves carry energy, this strongly implies this relationship between amplitude and energy holds. In fact, you can expand this concept to a wave packet where the individual frequencies maintain their shape as they propagate. Although the wave packet as a whole may change its shape due to the different frequencies having different speeds, each harmonic wave would conserve its energy, conserving the total energy of the wave packet.