3 Wave components
Before diving into mechanical, electromagnetic, and quantum waves, it will helps us to understand the general solutions to the wave equation and the importance of boundary conditions to those solutions. And the waves in real-word applications often include multiple frequency components, which are the subject of Fourier synthesis. With the an understanding of basics of Fourier theory, we will be ready to deal with the important topics of wave packets and dispersion.
3.1 General solutions to the wave equation
In seeking a general solution to the classical one-dimensional wave equation
(3.1) ∂2𝑦/∂𝑥2 = 1/𝑣2 ∂2𝑦/∂𝑡2
we're likely to encounter an approach developed by the French mathematician Jean le Rond d'Alembert in the 18th century. To understand d'Alembert's solution, begin by defining two new variables involving both 𝑥 and 𝑡
𝜉 = 𝑥 - 𝑣𝑡, 𝜂 = 𝑥 + 𝑣𝑡
and consider how to write the wave equation using these variables. Since the wave equation involves second derivatives in both space and time, start with the chain rule:
∂𝑦/∂𝑥 = ∂𝑦/∂𝜉 ∂𝜉/∂𝑥 + ∂𝑦/∂𝜂 ∂𝜂/∂𝑥 = ∂𝑦/∂𝜉(1) + ∂𝑦/∂𝜂(1) = ∂𝑦/∂𝜉 + ∂𝑦/∂𝜂,
Taking a second derivative with respect to 𝑥 gives
∂2𝑦/∂𝑥2 = ∂/∂𝑥 (∂𝑦/∂𝜉 + ∂𝑦/∂𝜂) = ∂/∂𝜉 (∂𝑦/∂𝜉 + ∂𝑦/∂𝜂) ∂𝜉/∂𝑥 + ∂/∂𝜂 (∂𝑦/∂𝜉 + ∂𝑦/∂𝜂) ∂𝜂/∂𝑥
= (∂2𝑦/∂𝜉 2 + ∂2𝑦/∂𝜉∂𝜂)(1) + (∂2𝑦/∂𝜂∂𝜉 + ∂2𝑦/∂𝜂2)(1).
But, as long as the function 𝑦 has continuous second derivatives, the order of the differentiation in mixed partials is irrelevant, so
∂2𝑦/∂𝜉 ∂𝜂 = ∂2𝑦/∂𝜂 ∂𝜉, and
(3.2) ∂2𝑦/∂𝑥2 = ∂2𝑦/∂𝜉 2 + 2∂2𝑦/∂𝜉∂𝜂 + ∂2𝑦/∂𝜂2.
Applying the same procedure to the time derivative of 𝑦 gives
(3.3) ∂𝑦/∂𝑡 = ∂𝑦/∂𝜉 ∂𝜉/∂𝑡 + ∂𝑦/∂𝜂 ∂𝜂/∂𝑡.
In this case ∂𝜉/∂𝑡 = -𝑣 and ∂𝜂/∂𝑡 = +𝑣, so
∂𝑦/∂𝑡 = ∂𝑦/∂𝜉(-𝑣) + ∂𝑦/∂𝜂(𝑣) = -𝑣∂𝑦/∂𝜉 + 𝑣∂𝑦/∂𝜂,
Taking the second derivative with respect to 𝑡 gives
∂2𝑦/∂𝑡2 = ∂/∂𝑡 (-𝑣∂𝑦/∂𝜉 + 𝑣∂𝑦/∂𝜂) = ∂/∂𝜉 (-𝑣∂𝑦/∂𝜉 + 𝑣∂𝑦/∂𝜂) ∂𝜉/∂𝑡 + ∂/∂𝜂 (-𝑣∂𝑦/∂𝜉 + 𝑣∂𝑦/∂𝜂) ∂𝜂/∂𝑡
= (-𝑣∂2𝑦/∂𝜉 2 + 𝑣∂2𝑦/∂𝜉∂𝜂)(-𝑣) + (-𝑣∂2𝑦/∂𝜂∂𝜉 + 𝑣∂2𝑦/∂𝜂2)(𝑣).
Thus
(3.4) ∂2𝑦/∂𝑡2 = 𝑣2 ∂2𝑦/∂𝜉 2 -2𝑣2 ∂2𝑦/∂𝜉 ∂𝜂 + 𝑣2 ∂2𝑦/∂𝜂 2.
So the wave equation is
∂2𝑦/∂𝜉2 + 2∂2𝑦/∂𝜉 ∂𝜂 + ∂2𝑦/∂𝜂2 = 1/𝑣2(𝑣2∂2𝑦/∂𝜉 2 - 2𝑣2 ∂2𝑦/∂𝜉 ∂𝜂 + 𝑣2 ∂2𝑦/∂𝜂 2)
or
(∂2𝑦/∂𝜉2 - ∂2𝑦/∂𝜉 2) + (2∂2𝑦/∂𝜉 ∂𝜂 + 2∂2𝑦/∂𝜉 ∂𝜂) + (∂2𝑦/∂𝜂2 - ∂2𝑦/∂𝜂 2) = 0.
This shows that the change of variables from 𝑥 and 𝑡 to 𝜉 and 𝜂 has simplified the classical wave equation into this:
(3.5) ∂2𝑦/∂𝜉 ∂𝜂 = 0.
We may write this equation as
∂/∂𝜉 (∂𝑦/∂𝜂) = 0
If the partial derivative with respect to 𝜉 of ∂𝑦/∂𝜂 is zero, then ∂𝑦/∂𝜂 cannot depend on 𝜉. This means that ∂𝑦/∂𝜂 must be a function of 𝜂 alone, so we may write
(3.6) ∂𝑦/∂𝜂 = 𝐹(𝜂),
(3.7) 𝑦 = ∫(𝐹(𝜂)𝑑𝜂) + constant,
where "constant" means any function that doesn't depend on 𝜂 (so it could be any function 𝜉). So we may write
(3.8) 𝑦 = ∫(𝐹(𝜂)𝑑𝜂) + 𝑔(𝜉),
and letting the integral of 𝐹(𝜂) equal another function of 𝜂 called 𝑓(𝜂), we have
(3.9-10) 𝑦 = 𝑓(𝜂) + 𝑔(𝜉), or 𝑦 = 𝑓(𝑥 + 𝑣𝑡) + 𝑔(𝑥 - 𝑣𝑡).
This is the general solution to the classical one-dimensional wave equation, and it tells us that every wavefunction 𝑦(𝑥, 𝑡) can be interpreted as the sum of two waves propagating in opposite direction with the same speed.
Example 3.1 If the the functions 𝑓 and 𝑔 represent sine waves of amplitude of 𝐴, how does the wavefunction 𝑦(𝑥, 𝑡) behave?
𝑓(𝑥 + 𝑣𝑡) = 𝐴 sin(𝑘𝑥 + 𝜔𝑡), 𝑔(𝑥 - 𝑣𝑡) = 𝐴 sin(𝑘𝑥 - 𝜔𝑡)
𝑦 = 𝑓(𝑥 + 𝑣𝑡) + 𝑔(𝑥 - 𝑣𝑡) = 𝐴 sin(𝑘𝑥 + 𝜔𝑡) + 𝐴 sin(𝑘𝑥 - 𝜔𝑡).
But
sin(𝑘𝑥 + 𝜔𝑡) = sin(𝑘𝑥)cos(𝜔𝑡) + cos(𝑘𝑥)sin(𝜔𝑡), and sin(𝑘𝑥 - 𝜔𝑡) = sin(𝑘𝑥)cos(𝜔𝑡) - cos(𝑘𝑥)sin(𝜔𝑡), so
𝑦 = 𝐴[sin(𝑘𝑥)cos(𝜔𝑡) + cos(𝑘𝑥)sin(𝜔𝑡)] + 𝐴[sin(𝑘𝑥)cos(𝜔𝑡) - cos(𝑘𝑥)sin(𝜔𝑡)] = 2𝐴 sin(𝑘𝑥)cos(𝜔𝑡).
Notice that the 𝑘𝑥 and the 𝜔𝑡 appear in different terms. As we can see in Figure. 3.1, the wavefunction 𝑦(𝑥, 𝑡) is sinusoidal over space and oscillates in time. So, the resultant is a non-propagating wave called a "standing wave". As indicated in the figure, the locations of nulls are called "nodes" and the locations of the peaks and troughs are called the "anti-node" of the standing wave.
3.2 Boundary conditions
A boundary condition "ties down" a function or its derivatives to a specified value at a specified location in space or time.(1)
A subset of boundary conditions consist of the "initial conditions", which specify the value at the start time or at a lower spatial boundary of the region. For example, consider d'Alembert's general solution to the wave equation (Eq. (3.10)). When we are given the initial displacement as 𝐼(𝑥) = 𝑦(𝑥, 0) and the initial (transverse(2)) velocity as 𝑉(𝑥) = ∂𝑦(𝑥, 𝑡)/∂𝑡∣𝑡=0. Let's begin by setting 𝑡 = 0 in the general solution:
(3.11-12) 𝑦(𝑥, 𝑡)∣𝑡=0 = 𝑓(𝑥 + 𝑣𝑡)∣𝑡=0 + 𝑔(𝑥 - 𝑣𝑡)∣𝑡=0. or 𝑦(𝑥, 0) = 𝑓(𝑥) + 𝑔(𝑥) = 𝐼(𝑥),
where 𝐼(𝑥) is the initial displacement at each value of 𝑥. Using 𝜂 = 𝑥 + 𝑣𝑡 and 𝜉 = 𝑥 - 𝑣𝑡 and then taking derivative with respect to time and setting 𝑡 = 0 gives,
∂𝑦(𝑥, 𝑡)/∂𝑡∣𝑡=0 = ∂𝑓/∂𝜂 ∂𝜂/∂𝑡∣𝑡=0 + ∂𝑔/∂𝜉 ∂𝜉/∂𝑡∣𝑡=0.
We can write ∂𝜂/∂𝑡 = 𝑣 and ∂𝜉/∂𝑡 = -𝑣, as well as ∂𝑓/∂𝜂 = ∂𝑓/∂𝑥 and ∂𝑔/∂𝜉 = ∂𝑔/∂𝑥. Thus the initial condition for transverse velocity is
∂𝑦(𝑥, 𝑡)/∂𝑡∣𝑡=0 = ∂𝑓/∂𝑥) 𝑣 - ∂𝑔/∂𝑥 𝑣 = 𝑉(𝑥), ∂𝑓/∂𝑥 - ∂𝑔/∂𝑥 = 1/𝑣 𝑉(𝑥).
Integrating this equation over 𝑥 allows we to write the integral of 𝑉(𝑥) in terns of 𝑓(𝑥) and 𝑔(𝑥) as
(3.13) 𝑓(𝑥) - 𝑔(𝑥) = 1/𝑣 ∫0𝑥 𝑉(𝑥)𝑑𝑥,
where 𝑥 = 0 represents an arbitrary starting location that will turn out not to affect the solution. So we can isolate 𝑓(𝑥) by adding Eq. (3.13) to Eq. (3.12):
(3.14) 2𝑓(𝑥) = 𝐼(𝑥) + 1/𝑣 ∫0𝑥 𝑉(𝑥)𝑑𝑥, or 𝑓(𝑥) = 1/2 𝐼(𝑥) + 1/2𝑣 ∫0𝑥 𝑉(𝑥)𝑑𝑥.
Likewise, we can isolate 𝑔(𝑥) by subtracting Eq. (3.13) from Eq. (3.12):
(3.15) 2𝑔(𝑥) = 𝐼(𝑥) - 1/𝑣 ∫0𝑥 𝑉(𝑥)𝑑𝑥, or 𝑔(𝑥) = 1/2 𝐼(𝑥) - 1/2𝑣 ∫0𝑥 𝑉(𝑥)𝑑𝑥.
If we replace 𝑥 by 𝜂 = 𝑥 + 𝑣𝑡 in 𝑓(𝑥) and 𝑥 by 𝜉 = 𝑥 - 𝑣𝑡 in 𝑔(𝑥), we get
𝑓(𝜂) = 𝑓(𝑥 + 𝑣𝑡) = 1/2 𝐼(𝑥 + 𝑣𝑡) + 1/2𝑣 ∫0𝑥 + 𝑣𝑡 𝑉(𝑥 + 𝑣𝑡)𝑑𝑥, and
𝑔(𝜉) = 𝑔(𝑥 - 𝑣𝑡) = 1/2 𝐼(𝑥 - 𝑣𝑡) - 1/2𝑣 ∫0𝑥 - 𝑣𝑡 𝑉(𝑥 - 𝑣𝑡)𝑑𝑥.
𝑦(𝑥, 𝑡) = 𝑓(𝑥 + 𝑣𝑡) + 𝑔(𝑥 - 𝑣𝑡) = 1/2 𝐼(𝑥 + 𝑣𝑡) + 1/2 𝐼(𝑥 - 𝑣𝑡) + 1/2𝑣 ∫0𝑥 + 𝑣𝑡 𝑉(𝑥 + 𝑣𝑡)𝑑𝑥 + 1/2 𝐼(𝑥 - 𝑣𝑡) - 1/2𝑣 ∫0𝑥 - 𝑣𝑡 𝑉(𝑥 - 𝑣𝑡)𝑑𝑥,
which we can simplify by using the minus sign in front of the last term to switch the limit of integration and combining the integrals:
(3.16) 𝑦(𝑥, 𝑡) = 1/2 𝐼(𝑥 + 𝑣𝑡) + 1/2 𝐼(𝑥 - 𝑣𝑡) + 1/2𝑣 ∫𝑥 + 𝑣𝑡𝑥 + 𝑣𝑡 𝑉(𝑧)𝑑𝑧,
in which 𝑧 is a dummy variable. Equation (3.16) is a d'Alembert's general solution to the wave equation with initial conditions 𝐼(𝑥) and 𝑉(𝑥).
Example 3.2 Find 𝑦(𝑥, 𝑡) for a wave with initial displacement condition
⌈ 5[1 + 𝑥/(𝐿/2)] for -𝐿/2 < 𝑥 < 0,
𝑦(𝑥, 0) = 𝐼(𝑥) =┃5[1 - 𝑥/(𝐿/2)] for 0 < 𝑥 < 𝐿/2
⌊ 0 elsewhere
and initial transverse velocity condition
𝑦(𝑥, 𝑡)∣𝑡=0 = 0.
𝑦(𝑥, 𝑡) = 1/2 𝐼(𝑥 - 𝑣𝑡) + 1/2 𝐼(𝑥 + 𝑣𝑡) + 1/2𝑣 ∫𝑥 + 𝑣𝑡𝑥 + 𝑣𝑡 𝑉(𝑧)𝑑𝑧 = 1/2 [𝐼(𝑥 - 𝑣𝑡) + 𝐼(𝑥 + 𝑣𝑡)] + 0.
This is just the initial shape of wave (𝐼(𝑥)) scaled by 1/2 and propagating both in the negative and in the positive 𝑥-direction while maintaining its shape over time, as we can see in Fig. 3.3.
For the separation of variables technique for the one-dimensional wave equation, the solution 𝑦(𝑥, 𝑡) is assumed to be the product of a function 𝑋(𝑥) that depends only on 𝑥 and another function 𝑇(𝑡) that depends only on 𝑡. Thus 𝑦(𝑥, 𝑡) = 𝑋(𝑥) 𝑇(𝑡), and the classical wave equation becomes
(3.17) ∂2[𝑋(𝑥) 𝑇(𝑡)]/∂𝑥2 = 1/𝑣2 ∂2[𝑋(𝑥) 𝑇(𝑡)]/∂𝑡2.
But 𝑇(𝑡) has no 𝑥-dependence and 𝑋(𝑥) has no time dependence, so 𝑇 comes out in the left term and 𝑋 comes out of the second:
(3.18) 𝑇(𝑡) ∂2𝑋(𝑥)/∂𝑥2 = 1/𝑣2) 𝑋(𝑥) ∂2𝑇(𝑡)/∂𝑡2.
The next step is to divide both sides by the product of the term (𝑋(𝑥) 𝑇(𝑡)):
(3.19) 1/𝑋(𝑥) ∂2𝑋(𝑥)/∂𝑥2 = 1/𝑣2 1/𝑇(𝑡) ∂2𝑇(𝑡)/∂𝑡2.
Notice that the left side depends only on 𝑥 and the right side depends only on 𝑡. As described in Section 2.4, this means the both side must be constant. We can set that constant (called the "separation constant") equal to 𝛼 and write
1/𝑋 ∂2𝑋/∂𝑥2 = 𝛼, 1/𝑣2 1/𝑇 ∂2𝑇/∂𝑡2 = 𝛼 or
(3.20-21) ∂2𝑋/∂𝑥2 = 𝛼 𝑋, ∂2𝑇/∂𝑡2 = 𝛼 𝑣2𝑇.
They say that the second derivative of each function (𝑋 and 𝑇) is equal to a constant times that function. Harmonic functions (sines and cosines) fit this bill nicely. If set the constant 𝛼 to -𝑘2
(3.22-23) ∂2𝑋/∂𝑥2 = 𝛼 𝑋 = -𝑘2. ∂2𝑇/∂𝑡2 = 𝛼 𝑣2𝑇 = -𝑣2 𝑘2 𝑇,
for which the solutions include sin(𝑘𝑣𝑡) and cos(𝑘𝑣𝑡). This means that 𝑘𝑣 must represent angular frequency (𝜔), which is the case if 𝑘 represents wavenumber and 𝑣 does wave phase speed.
So separation of variables in the wave equation leads to the solutions in the form of 𝑦(𝑥, 𝑡) = 𝑋(𝑥) 𝑇(𝑡), where 𝑋(𝑥) and 𝑇(𝑡) are may be harmonic functions. Knowing boundary conditions we can select the appropriate functions and if we define our function 𝑋(𝑥) as a weighted combination of sine and cosine functions, so 𝑋(𝑥) = 𝐴 cos(𝑘𝑥) + 𝐵 sin(𝑘𝑥) as in Fig. 3.4. By combining functions such as sines and cosines with appropriate weighting factors, our solutions to the wave equation can satisfy a far greater range of conditions than can be achieved by sine or cosine alone.
The very useful ability to shift wavefunctions along the 𝑥- or 𝑡-axes can be achieved in other ways, using a phase constant (𝜙0) and writing the general solution as 𝐶 sin(𝜔𝑡 + 𝜙0).
Example 3.3 Find the displacement 𝑦(𝑥, 𝑡) produced by waves on a string fixed at both ends.
Since the strings fixed both ends, if we define one end to have value 𝑥 = 0 and the other end to have the value 𝑥 = 𝐿 (where 𝐿 is the length of the string), 𝑦(0, 𝑡) = 0 and 𝑦(𝐿, 𝑡) = 0. Separating 𝑦(𝑥, 𝑡) into the product of distance function 𝑋(𝑥) = 𝐴 cos(𝑘𝑥) + 𝐵 sin(𝑘𝑥) and time function 𝑇(𝑡) means that
𝑦(0, 𝑡) = 𝑋(0) 𝑇(𝑡) = [𝐴 cos(0) + 𝐵 sin(0)] 𝑇(𝑡) = 0, [(𝐴)(1) + (𝐵)(0)] 𝑇(𝑡) = 0.
Since this must be true at all time (𝑡), this means 𝐴 must equal zero. Applying the boundary condition at the the other end of the string (𝑥 = 𝐿) is also useful:
𝑦(𝐿, 𝑡) = 𝑋(𝐿) 𝑇(𝑡) = [𝐴 cos(𝑘𝐿) + 𝐵 sin(𝑘𝐿)] 𝑇(𝑡) = 0, [0 cos(𝑘𝐿) + 𝐵 sin(𝑘𝐿)] 𝑇(𝑡) = 0.
Once again this must be true over all time, this can mean that either 𝐵 = 0 or sin(𝑘𝐿) = 0. But we will have more fun if we consider the case for which 𝐵 is non-zero and sin(𝑘𝐿) is zero. So in this case
sin(𝑘𝐿) = sin(2π𝐿/𝜆) = 0, 2π𝐿/𝜆 = 𝑛π, 𝜆 = 2𝐿/𝑛,
where 𝑛 can be any positive integer. This means that the wavelength (𝜆) takes on values of 2𝐿 (if 𝑛 = 1), 𝐿 (if 𝑛 = 2), 2𝐿/3 (if 𝑛 = 3), and so on and the cases looks like Fig. 3.5. So the general solution will be some weighted combination of these waveforms for 𝑋(0) = 0 and 𝑋(𝐿) = 0:
𝑋(𝑥) = 𝐵1 sin(π𝑥/𝐿) + 𝐵2 sin(2π𝑥/𝐿) + 𝐵3 sin(3π𝑥/𝐿) + ⋅ ⋅ ⋅ = ∑𝑛=1∞𝐵𝑛 sin(𝑛π𝑥/𝐿),
where the weighting coefficients 𝐵𝑛 determine exactly how much of each of these waveforms is present. And size of those coefficient depends on how we excite the string.
Separating the space and time components as described above and writing 𝑇(𝑡) as a weighted combination of sines and cosines gives
𝑇(𝑡) = 𝐶 cos(𝑘𝑣𝑡) + 𝐷 sin(𝑘𝑣𝑡) = 𝐶 cos[(2π/𝜆)𝑣𝑡] + 𝐷 sin[(2π/𝜆)𝑣𝑡]
The analysis of 𝑋(𝑥) has told us that 𝜆 = 2𝐿/𝑛 for a string fixed at both ends, so 𝑇(𝑡) becomes
𝑇(𝑡) = 𝐶 cos[(2π/𝜆)𝑣𝑡] + 𝐷 sin[(2π/𝜆)𝑣𝑡] = 𝐶 cos[(2π/2𝐿/𝑛)𝑣𝑡] + 𝐷 sin [(2π/2𝐿/𝑛)𝑣𝑡] = 𝐶 cos[(𝑛π/𝐿)𝑣𝑡] + 𝐷 sin[(𝑛π/𝐿)𝑣𝑡].
We can apply the boundary condition of zero displacement at time 𝑡 = 0:
𝑦(𝑥, 0) = 𝑋(𝑥) 𝑇(0) = 𝑋(𝑥) [𝐶 cos(0) + 𝐷 sin(0)] = 0, 𝑋(𝑥) [(𝐶)(1) + (𝐷)(0)] = 0, then 𝐶 = 0.
Thus the general solution for 𝑇(𝑡) is the sum of the sine terms for each value of 𝑛:
𝑇(𝑡) = ∑𝑛=1∞𝐷𝑛 sin[(𝑛π/𝐿)𝑣𝑡].
Combining the expression for 𝑋(𝑥) and 𝑇(𝑡) and absorbing the 𝐷𝑛 into 𝐵𝑛 makes the solution for displacement
(3.24) 𝑦(𝑥, 𝑡) = 𝑋(𝑥) 𝑇(𝑡) = ∑𝑛=1∞𝐵𝑛 sin (𝑛π𝑥/𝐿) sin (𝑛π𝑣𝑡/𝐿).
We can see this for the 𝑛 = 3 case in Fig. 3.8, for which which the spacial sine term result in 1.5 cycles over the distance from 𝑥 = 0 to 𝑥 = 𝐿. But we haven't discussed how the value of weighting coefficients such as 𝐵𝑛 in Eq. (3.24) can be determined from boundary conditions. You understand that process we need to study Fourier theory.
(1) "Dirichlet" boundary conditions specify the value of the function itself, "Neumann" boundary conditions specify the value of the function's derivative, and "Cauchy" (or "mixed") boundary conditions combine both Dirichlet and Neumann boundary conditions.
(2) The transverse velocity is the velocity of particles of the medium moving perpendicular to the direction of the wave's propagation.
