Abstract

The scattering of waves in an inhomogeneous infinite string with one change in the density of the string is well known. In this paper, we study the case where there are two discontinuities in the density of the string. It turns out that we can write the solution for this case as well. The form of the solution will be given in finite sums of reflected and transmitted waves over finite time. The finite sums become converging infinite series for infinite time.

1. Introduction

The scattering of waves in an infinite string due to a single change in the density of the string has been studied in graduate text books such as [1, 2]. In [1] the author notes that in practice a scattering of this sort occurs in physical systems such as submarine cables or telephone lines when a join in the system creates a transmitted as well as a reflected wave. Both of these waves can be computed in terms of the incoming signal. For practical purposes it is desirable to suppress the reflected waves by attaching, at the join, a damping mechanism or a point mass so that the reflected waves do not interfere with the incoming signals. All three cases of no suppression, suppression by damping, and suppression by point mass have been studied in [1] by assigning appropriate boundary conditions at the join. In this paper we study the case where the string has more than one change in its density and no suppression occurs at the joins. We write the form of the solution when there are only two changes in the density. We also show what the reflected and transmitted waves due to an incident wave on both sides of a join at an arbitrary point in the string look like. This will allow us to theoretically write the form of the reflected and transmitted waves for higher number of changes. But because of the repetitive bouncing back and forth of the waves between each interface writing a general formula for the solution of the wave problem with more than two discontinuities will be impossible. Just imagine that all the waves inside one pair of joins get transmitted to the neighboring pair and in turn bounce against the joins and create waves of their own. Writing a formula to account for all of these waves will require a horrendous amount of work if at all possible, and we are not going to tackle it here.

To fix the ideas, consider two semi-infinite strings with densities 𝜌1 and 𝜌2, 𝜌1𝜌2, joined together at the origin 𝑂. The speed of the waves traveling along the two pieces is 𝑐1 and 𝑐2, respectively. An incoming wave 𝑓(𝑥𝑐1𝑡), such that 𝑓(𝑠)=0,𝑠>0, will be scattered to partly reflected and partly outgoing at 𝑥=0. Suppose the resulting wave 𝑢(𝑥,𝑡) satisfies that𝑢𝑡𝑡𝑐2(𝑥)𝑢𝑥𝑥=0,(𝑥,𝑡)𝑅×(0,),(1.1) where𝑐𝑐(𝑥)=1𝑐,𝑥0,2,𝑥>0.(1.2) Under the geometric continuity (the string is continuous at 𝑥=0) and dynamical continuity (the transverse force is continuous at 𝑥=0) conditions,𝑢𝑢(0,𝑡)=𝑢(0+,𝑡),𝑡0,𝑥(0,𝑡)=𝑢𝑥(0+,𝑡),𝑡0,(1.3) and the initial conditions,𝑢𝑢(𝑥,0)=𝑓(𝑥),𝑥𝑅,𝑡(𝑥,0)=𝑐1𝑓(𝑥),𝑥𝑅,(1.4) the solution is well known to be (see [1, 2])𝑓𝑢(𝑥,𝑡)=𝑥𝑐1𝑡+𝑐2𝑐1𝑐2+𝑐1𝑓𝑥𝑐1𝑡,𝑥<0,2𝑐2𝑐2+𝑐1𝑓𝑐1𝑐2𝑥𝑐2𝑡,𝑥>0.(1.5) Now, consider the above setup extended to a string made of three pieces in the intervals (,0), (0,𝜎), and (𝜎,) for some constant 𝜎>0. Denote the densities and their corresponding wave speeds by 𝜌1, 𝜌2, 𝜌3 and 𝑐1, 𝑐2, 𝑐3, respectively. We will study the behavior of an incoming wave 𝑓(𝑥𝑐1𝑡), where 𝑓(𝑠)=0, 𝑠>0, at the interfaces and show how, under similar conditions as above, similar formulas for the scattered waves can be written. We note that each outgoing wave from an interface will be incident on the next, and the reflected waves at the next interface will be incident on the previous. Each incident wave will, in turn, be scattered into reflected and outgoing waves at the interfaces. We will write the solution to the wave problem as sums of these scattered waves.

2. The Problem

Consider the problem𝑢𝑡𝑡𝑐2(𝑥)𝑢𝑥𝑥𝑐=0,(𝑥,𝑡)𝑅×(0,),𝑐(𝑥)=1𝑐,𝑥0,2𝑐,0<𝑥𝜎,3,𝑥>𝜎.(2.1) Assume that a wave 𝑓(𝑥𝑐1𝑡)𝐶2(𝑅), where 𝑓(𝑠)0, 𝑠>0, is incoming. At 𝑥=0 the wave scatters as follows:𝑢1(𝑓𝑥,𝑡)=𝑥𝑐1𝑡+𝑐2𝑐1𝑐1+𝑐2𝑓𝑥𝑐1𝑡,𝑥<0,2𝑐2𝑐1+𝑐2𝑓𝑐1𝑐2𝑥𝑐2𝑡,0𝑥𝜎.(2.2) The transmitted wave (2𝑐2/(𝑐1+𝑐2))𝑓((𝑐1/𝑐2)(𝑥𝑐2𝑡)) becomes incident at the point 𝑥=𝜎. Let us denote𝑔𝑥𝑐2𝑡=2𝑐2𝑐1+𝑐2𝑓𝑐1𝑐2𝑥𝑐2𝑡,0𝑥<𝜎.(2.3) We will show here, for the sake of completeness, that 𝑔 also scatters at 𝑥=𝜎 and creates new waves.

Theorem 1. The incoming wave 𝑔 given by (2.3) scatters at 𝑥=𝜎, for 𝑡>𝜎/𝑐2, in the form 𝑢2𝑔(𝑥,𝑡)=𝑥𝑐2𝑡+𝑐3𝑐2𝑐2+𝑐3𝑔2𝜎𝑥+𝑐2𝑡,0<𝑥𝜎,2𝑐3𝑐2+𝑐3𝑔𝑐3𝑐2𝑐3𝑐𝜎+2𝑐3𝑥𝑐3𝑡,𝑥>𝜎.(2.4)

Proof. First note that 𝑔(𝑥𝑐2𝑡)=𝑔(𝑥𝜎) when 𝑡=𝜎/𝑐2, and 𝑔(𝑥𝜎)=0, 𝑥>𝜎, by the definition of 𝑔 in (2.3). Consider the wave function 𝑢(𝑥,𝑡), which we assume as 𝑢(𝑥,𝑡)=𝑢1(𝑥,𝑡) on (,0)×(0,𝜎/𝑐2]. The solution to the wave equation 𝑢𝑡𝑡𝑐2(𝑥)𝑢𝑥𝑥𝜎=0,(𝑥,𝑡)(0,)×𝑐2,,(2.5) where 𝑐𝑐(𝑥)=2𝑐,0<𝑥𝜎,3,𝑥>𝜎,(2.6) is of the form 𝑗𝑢(𝑥,𝑡)=𝑥𝑐2𝑡+𝑞𝑥+𝑐2𝑡𝑙,0<𝑥<𝜎,𝑥𝑐3𝑡+𝑚𝑥+𝑐3𝑡,0>𝜎.(2.7) We use the geometric and dynamical continuity conditions, 𝜎𝑢(𝜎,𝑡)=𝑢(𝜎+,𝑡),𝑡𝑐2,𝑢𝑥(𝜎,𝑡)=𝑢𝑥(𝜎𝜎+,𝑡),𝑡𝑐2,(2.8) and the initial conditions, 𝑢𝜎𝑥,𝑐2𝑢=𝑔(𝑥𝜎),𝑥(0,),𝑡𝜎𝑥,𝑐2=𝑐2𝑔(𝑥𝜎),𝑥(0,),(2.9) to find 𝑗, 𝑞, 𝑙, and 𝑚. From continuity of the wave function 𝑢(𝑥,𝑡) at 𝑥=𝜎, we have 𝑗𝜎𝑐2𝑡+𝑞𝜎+𝑐2𝑡=𝑙𝜎𝑐3𝑡+𝑚𝜎+𝑐3𝑡.(2.10) Using dynamical condition (2.8) at 𝑥=𝜎 and differentiating equation (2.10) with respect to 𝑡, we have the system 𝑗𝜎𝑐2𝑡+𝑞𝜎+𝑐2𝑡=𝑙𝜎𝑐3𝑡+𝑚𝜎+𝑐3𝑡𝑐2𝑗𝜎𝑐2𝑡+𝑐2𝑞𝜎+𝑐2𝑡=𝑐3𝑙𝜎𝑐3𝑡+𝑐3𝑚𝜎+𝑐3𝑡.(2.11) Solving the system (2.11) for 𝑞 and 𝑗 and integrating, we have 𝑞𝜎+𝑐2𝑡=𝑐3𝑐22𝑐3𝑙𝜎𝑐3𝑡+𝑐3+𝑐22𝑐3𝑚𝜎+𝑐3𝑡,(2.12)𝑗𝜎𝑐2𝑡=𝑐3+𝑐22𝑐3𝑙𝜎𝑐3𝑡+𝑐3𝑐22𝑐3𝑚𝜎+𝑐3𝑡.(2.13) From initial conditions (2.9) at 𝑡=𝜎/𝑐2, 𝑢𝜎𝑥,𝑐2𝑢=𝑔(𝑥𝜎)=𝑗(𝑥𝜎)+𝑞(𝑥+𝜎),𝑥<𝜎,𝑡𝜎𝑥,𝑐2=𝑐2𝑔(𝑥𝜎)=𝑐2𝑗(𝑥𝜎)+𝑐2𝑞(𝑥+𝜎),𝑥<𝜎.(2.14) The right-hand sides of (2.14) provide a system that can be solved for 𝑗 and 𝑞: 𝑗(𝑥𝜎)=𝑔(𝑥𝜎),𝑥<𝜎,(2.15)𝑞(𝑥+𝜎)=0,𝑥<𝜎.(2.16) At the time 𝑡=𝜎/𝑐2 and 𝑥>𝜎, we have 𝜎𝑔(𝑥𝜎)=𝑢𝑥,𝑐2𝑐=𝑙𝑥3𝑐2𝜎𝑐+𝑚𝑥+3𝑐2𝜎=0,𝑥>𝜎,(2.17)𝑐2𝑔(𝑥𝜎)=𝑢𝑡𝜎𝑥,𝑐2=𝑐3𝑙𝑐𝑥3𝑐2𝜎+𝑐3𝑚𝑐𝑥+3𝑐2𝜎=0,𝑥>𝜎,(2.18) Differentiating the right-hand side of (2.17) and solving the system obtained from (2.17) and (2.18), namely, 𝑙𝑐𝑥3𝑐2𝜎+𝑚𝑐𝑥+3𝑐2𝜎=0,𝑥>𝜎,𝑐3𝑙𝑐𝑥3𝑐2𝜎+𝑐3𝑚𝑐𝑥+3𝑐2𝜎=0,𝑥>𝜎,(2.19) we will have 𝑙𝑐𝑥3𝑐2𝜎𝑚𝑐=0,𝑥>𝜎,(2.20)𝑥+3𝑐2𝜎=0,𝑥>𝜎.(2.21) By (2.15), when the arguments of 𝑔 and 𝑗 are negative the functions are equal. This means that 𝑗𝑥𝑐2𝑡=𝑔𝑥𝑐2𝑡𝜎,𝑡>𝑐2.(2.22) Notice that by (2.21) if the argument of 𝑚 satisfies 𝑥+(𝑐3/𝑐2)𝜎>𝜎+(𝑐3/𝑐2)𝜎, then 𝑚=0. For 𝑡>𝜎/𝑐2 the argument of 𝑚 in (2.12) and (2.13) is larger than 𝜎+(𝑐3/𝑐2)𝜎. Therefore, 𝑚𝑥+𝑐3𝑡𝜎=0,𝑥𝜎,𝑡>𝑐2,(2.23) and (2.12) and (2.13) become, 𝑞𝜎+𝑐2𝑡=𝑐3𝑐22𝑐3𝑙𝜎𝑐3𝑡𝜎,𝑡>𝑐2,𝑗(2.24)𝜎𝑐2𝑡=𝑐3+𝑐22𝑐3𝑙𝜎𝑐3𝑡𝜎,𝑡>𝑐2.(2.25) From (2.22) and (2.25) we obtain 𝑔𝜎𝑐2𝑡=𝑐3+𝑐22𝑐3𝑙𝜎𝑐3𝑡𝜎,𝑡>𝑐2.(2.26) Equation (2.26) determines 𝑙 as follows: 𝑙𝜎𝑐3𝑡=2𝑐3𝑐3+𝑐2𝑔𝜎𝑐2𝑡𝜎,𝑡>𝑐2.(2.27) Denote the argument of 𝑙 in (2.27) by 𝜏, then 𝑙(𝜏)=2𝑐3𝑐3+𝑐2𝑔𝑐𝜎2𝑐3𝑐(𝜎𝜏),𝜏>3𝑐2𝑐2𝜎.(2.28) Due to (2.28), 𝑙𝑥𝑐3𝑡=2𝑐3𝑐3+𝑐2𝑔𝑐3𝑐2𝑐3𝑐𝜎+2𝑐3𝑥𝑐3𝑡𝜎,𝑡>𝑐2.(2.29) Finally, putting (2.23) and (2.27) in (2.12) determines 𝑞: 𝑞𝜎+𝑐2𝑡=𝑐3𝑐2𝑐3+𝑐2𝑔𝜎𝑐2𝑡𝜎,𝑡>𝑐2.(2.30) A similar change of independent variable in (2.30) yields 𝑞𝑥+𝑐2𝑡=𝑐3𝑐2𝑐3+𝑐2𝑔2𝜎𝑥+𝑐2𝑡𝜎,0<𝑥<𝜎,𝑡>𝑐2.(2.31) Plugging 𝑗, 𝑚, 𝑙, and 𝑞 determined by (2.22), (2.23), (2.29), and (2.31), respectively, in (2.7) yields 𝑢2 in (2.4).

By an argument similar to the one in Theorem 1, we can show the following. The reflected wave at 𝑥=𝜎, namely ((𝑐3𝑐2)/(𝑐2+𝑐3))𝑔(2𝜎(𝑥+𝑐2𝑡)) in (2.4), becomes incident at the point 𝑥=0. Let us denote𝑥+𝑐2𝑡=𝑐3𝑐2𝑐2+𝑐3𝑔2𝜎𝑥+𝑐2𝑡,0<𝑥<𝜎.(2.32) Then, the wave scatters at 𝑥=0, for 𝑡>2𝜎/𝑐2, as follows:𝑢3(𝑥,𝑡)=2𝑐1𝑐1+𝑐2𝑐2𝑐1𝑥+𝑐1𝑡,𝑥<0,x+𝑐2𝑡+𝑐1𝑐2𝑐1+𝑐2𝑥+𝑐2𝑡,0<𝑥<𝜎.(2.33) We note here that (𝑥+𝑐2𝑡) is the wave that moves to the left with (𝑠)=0, 𝑠<0. It is not difficult to check that 𝑢3 satisfies 𝑢3(𝑥,0)=(𝑥),𝑢3,𝑡(𝑥,0)=𝑐2(𝑥),𝑢3(0,𝑡)=𝑢3(0+,𝑡) and 𝑢3,𝑥(0,𝑡)=𝑢3,𝑥(0+𝑡).

In this manner, the forms of outgoing waves through the interfaces 𝑥=0 and 𝑥=𝜎 and the ones bouncing back and forth between the two are determined. In order to write the solution to the problem𝑢𝑡𝑡𝑐2(𝑥)𝑢𝑥𝑥=0,(𝑥,𝑡)(𝑥,𝑡)𝑅×(0,),(2.34) where,𝑐𝑐(𝑥)=1𝑐,𝑥0,2𝑐,0<𝑥𝜎,3,𝑥>𝜎,(2.35) subject to𝑢𝑢𝑢(0,𝑡)=𝑢(0+,𝑡),𝑡0,(𝑥,0)=𝑓(𝑥),𝑓(𝑠)0,𝑠>0,𝑡(𝑥,0)=𝑐1𝑓𝑢(𝑥),𝑥𝑅,𝑥(0,𝑡)=𝑢𝑥(𝑢0+,𝑡),𝑡0,𝑢(𝜎,𝑡)=𝑢(𝜎+,𝑡),𝑡0,𝑥(𝜎,𝑡)=𝑢𝑥(𝜎+,𝑡),𝑡0,(2.36) for 𝜎, 𝑐1, 𝑐2, 𝑐3 positive constants, and the incoming wave 𝑓(𝑥𝑐1𝑡) for some 𝑓𝐶2(𝑅), we need sums of the waves in each interval (,0), (0,𝜎), and (𝜎,). One way to write such solution is through the use of composition of the arguments of the scattered waves. For this purpose we introduce the following functions:𝐿𝑐(𝜂)=3𝑐2𝑐3𝑐𝜎+2𝑐3𝑐𝜂,𝑅(𝜂)=2𝑐1𝜂,𝑀(𝜂)=2𝜎𝜂,𝑆(𝜂)=𝜂.(2.37) Let us write the functions 𝑔 and introduced in (2.3) and (2.32), respectively, in terms of the variable 𝜂:𝑔(𝜂)=2𝑐2𝑐1+𝑐2𝑓𝑐1𝑐2𝜂𝑐,(2.38)(𝜂)=3𝑐2𝑐3+𝑐2𝑔(2𝜎𝜂).(2.39) From 𝑀 in (2.37) and (2.38)-(2.39), can be written in terms of 𝑓 as follows:𝑐(𝜂)=3𝑐2𝑐3+𝑐22𝑐2𝑐1+𝑐2𝑓𝑐1𝑐2𝑀(𝜂).(2.40) In terms of the functions in (2.37), (2.38) and (2.40), (2.4) and (2.33) in the time intervals [0,2𝜎/𝑐2) and [2𝜎/𝑐2,3𝜎/𝑐2) will, respectively, become𝑢2(𝑥,𝑡)=2𝑐2𝑐1+𝑐2𝑓𝑐1𝑐2𝑥𝑐2𝑡+𝑐3𝑐2𝑐3+𝑐22𝑐2𝑐1+𝑐2𝑓𝑐1𝑐2𝑀𝑥+𝑐2𝑡,0<𝑥<𝜎,2𝑐3𝑐2+𝑐32𝑐2𝑐1+𝑐2𝑓𝑐1𝑐2𝐿𝑥𝑐3𝑡𝑢,𝑥>𝜎,(2.41)3(𝑥,𝑡)=2𝑐1𝑐1+𝑐2𝑐3𝑐2𝑐3+𝑐22𝑐2𝑐1+𝑐2𝑓𝑐1𝑐2𝑀𝑅𝑥+𝑐1𝑡𝑐,𝑥<0,3𝑐2𝑐3+𝑐22𝑐2𝑐1+𝑐2𝑓𝑐1𝑐2𝑀𝑥+𝑐2𝑡+𝑐1𝑐2𝑐1+𝑐2𝑐3𝑐2𝑐3+𝑐22𝑐2𝑐1+𝑐2𝑐𝑓1𝑐2𝑀𝑆𝑥𝑐2𝑡,0<𝑥<𝜎.(2.42) Now, we begin to write the solution of the wave problem (2.34)-(2.36) in each interval (,0), (0,𝜎), and (𝜎,) using (2.2), (2.41), and (2.42). Consider the following interpretations of (2.2), (2.41), and (2.42). First we have the following wave on the interval (,0):𝑓𝑥𝑐1𝑡.(2.43) When this wave hits the join at 0 from the left, by (2.2) the reflected wave gets a coefficient of the form (𝑐2𝑐1)/(𝑐2+𝑐1) and an argument change to 𝑆(𝑥+𝑐1𝑡), that is, the reflected wave corresponding to (2.43) is𝑐2𝑐1𝑐2+𝑐1𝑓𝑆𝑥+𝑐1𝑡,𝑥<0,(2.44)𝑆  is as given in (2.37). The transmitted wave due to (2.43) on the other hand gets a coefficient of 2𝑐2/(𝑐1+𝑐2) and argument change to (𝑐1/𝑐2)(𝑥𝑐2𝑡), that is, the transmitted wave traveling to the right at 0 is2𝑐2𝑐1+𝑐2𝑓𝑐1𝑐2𝑥𝑐2𝑡,0<𝑥<𝜎.(2.45) Now let us see what changes the signal (2.45) undergo when it hits the join at 𝜎 on the left. From (2.41) part of the signal gets reflected to the left acquiring the coefficient (𝑐3𝑐2)/(𝑐3+𝑐2) and argument change to 𝑀(𝑥+𝑐2𝑡). The transmitted wave traveling right acquires the coefficient 2𝑐3/(𝑐2+𝑐3) and the argument change 𝐿(𝑥𝑐3𝑡). Lastly, by (2.42) a signal that hits the join at 0 on the right its transmitted part acquires the coefficient 2𝑐1/(𝑐1+𝑐2) and the change of argument 𝑅(𝑥+𝑐1𝑡). Its reflected part acquires the coefficient (𝑐1𝑐2)/(𝑐1+𝑐2) and the argument change of 𝑆(𝑥𝑐2𝑡).

Note that the signals inside the interval (0,𝜎) keep scattering, but the ones that are transmitted outside this interval continue traveling to the right or left forever. So, now let us look at a few more wave signals that are produced inside (0,𝜎). Look at the wave ((𝑐1𝑐2)/(𝑐1+𝑐2))((𝑐3𝑐2)/(𝑐3+𝑐2))((2𝑐2)/(𝑐1+𝑐2))𝑓((𝑐1/𝑐2)𝑀𝑆(𝑥𝑐2𝑡)) in (2.42). As it travels to the right it reaches the join 𝜎 as an incident wave on the left. By the argument above its reflected component picks up a coefficient of (𝑐3𝑐2)/(𝑐3+𝑐2) and an argument change of 𝑀𝑥+𝑐2𝑡. The transmitted component picks up a coefficient of 2𝑐3/(𝑐2+𝑐3) and an argument change 𝐿(𝑥𝑐3𝑡). These two wave are, respectively,𝑐3𝑐2𝑐3+𝑐2𝑐1𝑐2𝑐1+𝑐2𝑐3𝑐2𝑐3+𝑐22𝑐2𝑐1+𝑐2𝑐𝑓1𝑐2𝑀𝑆𝑀𝑥+𝑐2𝑡,(2.46)2𝑐3𝑐2+𝑐3𝑐1𝑐2𝑐1+𝑐2𝑐3𝑐2𝑐3+𝑐22𝑐2𝑐1+𝑐2𝑐𝑓1𝑐2𝑀𝑆𝐿𝑥𝑐3𝑡.(2.47) Now the wave (2.46) is incident on the join 0 from the right. The argument above shows that the transmitted part must pick up the coefficient 2𝑐1/(𝑐1+𝑐2) and the change of argument 𝑅(𝑥+𝑐1𝑡), and the reflected part picks up the coefficient (𝑐1𝑐2)/(𝑐1+𝑐2) and the argument change of 𝑆(𝑥𝑐2𝑡). These two new waves are given below, respectively,2𝑐1𝑐1+𝑐2𝑐3𝑐2𝑐3+𝑐2𝑐1𝑐2𝑐1+𝑐2𝑐3𝑐2𝑐3+𝑐22𝑐2𝑐1+𝑐2𝑐𝑓1𝑐2𝑀𝑆𝑀𝑅𝑥+𝑐1𝑡𝑐,(2.48)1𝑐2𝑐1+𝑐2𝑐3𝑐2𝑐3+𝑐2𝑐1𝑐2𝑐1+𝑐2𝑐3𝑐2𝑐3+𝑐22𝑐2𝑐1+𝑐2𝑐𝑓1𝑐2𝑀𝑆𝑀𝑆𝑥𝑐2𝑡.(2.49) The waves (2.47) and (2.48) will be moving to the right and left away from the interval (0,𝜎) indefinitely. But the wave (2.49) will be incident on the join 𝜎 on the left, and the process of reflection and transmission repeats as before. It is not too difficult now to decipher the general pattern of the reflected and transmitted waves in (0,𝜎) and the outgoing waves outside of this interval. Here they are. The waves outgoing in 𝑥<0 are of the form𝑐2𝑐1𝑐1+𝑐2𝑓𝑥𝑐1𝑡,4𝑐1𝑐2𝑐1+𝑐22𝑐3𝑐2𝑐3+𝑐2(𝑘1)/2𝑐1𝑐2𝑐1+𝑐2(𝑘3)/2𝑐𝑓1𝑐2(𝑀𝑆)(𝑘3)/2𝑀𝑅𝑥+𝑐1𝑡,𝑥<0,0<𝑡<(2𝑖+1)𝜎𝑐2,𝑘=3,5,7,.(2.50) The waves in (0,𝜎) moving left are 𝑐1𝑐2𝑐1+𝑐2(𝑗2)/2𝑐3𝑐2𝑐3+𝑐2𝑗/22𝑐2𝑐1+𝑐2𝑐𝑓1𝑐2(𝑀𝑆)(𝑗2)/2𝑀𝑥+𝑐2𝑡,0<𝑥<𝜎,0<𝑡<2𝑖𝜎𝑐2,𝑗=2,4,6,.(2.51) The waves in (0,𝜎) moving right are2𝑐2𝑐1+𝑐2𝑓𝑐1𝑐2𝑥𝑐2𝑡,𝑐3𝑐2𝑐3+𝑐2(𝑘1)/2𝑐1𝑐2𝑐1+𝑐2(𝑘1)/22𝑐2𝑐1+𝑐2𝑐𝑓1𝑐2(𝑀𝑆)(𝑘1)/2𝑥𝑐2𝑡,0<𝑥<𝜎,𝑘=3,5,7,.(2.52) Finally the ones moving right for 𝑥>𝜎 are4𝑐2𝑐3𝑐1+𝑐2𝑐2+c3𝑐1𝑐2𝑐1+𝑐2𝑐3𝑐2𝑐3+𝑐2(𝑗2)/2𝑐𝑓1𝑐2(𝑀𝑆)(𝑗2)/2𝐿𝑥𝑐3𝑡,𝑥>𝜎,0<𝑡<2𝑖𝜎𝑐2,𝑗=2,4,6,.(2.53) Now, we recall that 𝑢1 given in (2.2) is only good for the time interval (0,𝜎/𝑐2), and after 𝑡=𝜎/𝑐2, the transmitted part hits the join at 𝜎 and splits. Therefore, in this time interval we have the wave 𝑣1 defined as𝑣1𝑓(𝑥,𝑡)=𝑥𝑐1𝑡+𝑐2𝑐1𝑐2+𝑐1𝑓𝑥𝑐1𝑡,𝑥<0,2𝑐2𝑐1+𝑐2𝑓𝑐1𝑐2𝑥𝑐2𝑡,0<𝑥<𝜎,0,𝑥>𝜎.(2.54) We note that 𝑣1 is the original incoming signal 𝑓(𝑥𝑐1𝑡) incident on 0 plus the reflected and transmitted waves. The transmitted wave has not reached the join 𝜎 yet, and so there is no wave beyond the point 𝑥=𝜎. Therefore 𝑣1 in (2.54) satisfies the wave problem (2.34)–(2.36) in this time interval. We now bring in another wave 𝑣2 in the interval (𝜎/𝑐2,2𝜎/𝑐2). We define it by𝑣2𝑐(𝑥,𝑡)=0,𝑥<0,3𝑐2𝑐3+𝑐22𝑐2𝑐1+𝑐2𝑓𝑐1𝑐2𝑀𝑥+𝑐2𝑡,0<𝑥<𝜎,2𝑐3𝑐2+𝑐32𝑐2𝑐1+𝑐2𝑓𝑐1𝑐2𝐿𝑥𝑐3𝑡,𝑥>𝜎.(2.55) The function 𝑣2 in (2.55) represents the reflected and transmitted waves at 𝜎, caused by the incident signal (2𝑐2/(𝑐1+𝑐2))𝑓((𝑐1/𝑐2)(𝑥𝑐2𝑡)). We see this incident wave in the definition of 𝑣1 in (2.54). We argue that the sum of the wave in (2.54) and (2.55), 𝑣1+𝑣2, over the interval (0,2𝜎/𝑐2) satisfies the wave problem (2.34)–(2.36). The fact that they satisfy the wave (2.34) over each space interval is clear. The boundary conditions at 0 follow from the fact that 𝑣1 is the solution to the wave problem (1.1)–(1.4) and the function ((𝑐3𝑐2)/(𝑐3+𝑐2))(2𝑐2/(𝑐1+𝑐2))𝑓((𝑐1/𝑐2)𝑀(𝑥+𝑐2𝑡)) = ((𝑐3𝑐2)/(𝑐3+𝑐2))(2𝑐2/(𝑐1+𝑐2))𝑓((𝑐1/𝑐2)(2𝜎(𝑥+𝑐2𝑡)))=0 for 𝑥=0,𝑡<2𝜎/𝑐2. The boundary conditions at 𝑥=𝜎 are satisfied, because the sum of the parts for 0<𝑥<𝜎 and 𝑥>𝜎 is the the solution of the wave problem (2.5), (2.8)–(2.9), with 𝑔 given by (2.3) there. The initial conditions are satisfied because 𝑣1 satisfies them, and 𝑣2(𝑥,0)0 due to the argument of 𝑓 staying positive when 𝑡=0.

Now we consider wave 𝑣3 over the interval (2𝜎/𝑐2,3𝜎/𝑐2). It is defined by𝑣3(𝑥,𝑡)=4𝑐1𝑐2𝑐1+𝑐22𝑐3𝑐2𝑐3+𝑐2𝑐𝑓1𝑐2𝑀𝑅𝑥+𝑐1𝑡,𝑥<0,2𝑐2𝑐1+𝑐2𝑐3𝑐2𝑐3+𝑐2𝑐1𝑐2𝑐1+𝑐2𝑓𝑐1𝑐2(𝑀𝑆)𝑥𝑐2𝑡,0<𝑥<𝜎,0,𝑥>𝜎.(2.56) This is the wave due to the scattering of ((𝑐3𝑐2)/(𝑐3+𝑐2))(2𝑐2/(𝑐1+𝑐2))𝑓((𝑐1/𝑐2)𝑀(𝑥+𝑐2𝑡)), at the join 0. This incident wave can be seen in the definition of 𝑣2 in (2.55). We claim that the sum 𝑣1+𝑣2+𝑣3, where 𝑣1,𝑣2,𝑣3 are given by (2.54)–(2.56), satisfies the wave problem (2.34)–(2.36) over the interval (0,3𝜎/𝑐2). We note that 𝑣3 when added to the incident wave ((𝑐3𝑐2)/(𝑐3+𝑐2))(2𝑐2/(𝑐1+𝑐2))𝑓((𝑐1/𝑐2)𝑀(𝑥+𝑐2𝑡)) is the wave 𝑢3 that was given in (2.33) except for the notation. This can be seen by noting that in (2.33) is defined in terms of 𝑔 in (2.32) and 𝑔 is given in terms of 𝑓 in (2.3). We constructed 𝑢3 so that it satisfies the boundary conditions 𝑢(0,𝑡)=𝑢(0+,𝑡),𝑢𝑥(0,𝑡),𝑢𝑥(0+,𝑡) in (2.36), as well as the wave equation (2.34) for 𝑥<𝜎. The sum 𝑣1+𝑣2+𝑣3 contains 𝑢3 for 𝑥<𝜎. In light of the previous argument about 𝑣1+𝑣2 and the way 𝑣3 contributes to the sum, it is not difficult to see why 𝑣1+𝑣2+𝑣3 satisfies the boundary conditions in (2.36) and the wave equation in (2.34) in the interval (0,3𝜎/𝑐2). The fact that at time 0<𝑡<𝜎/𝑐2 all terms involving 𝑓 except 𝑓(𝑥𝑐1𝑡) are zero in 𝑣1+𝑣2+𝑣3 shows that the initial conditions of (2.30) are also satisfied.

If one waits other 𝜎/𝑐2 units of time, another scattering happens at 𝜎 from the wave (2𝑐2/(𝑐1+𝑐2))((𝑐3𝑐2)/(𝑐3+𝑐2))((𝑐1𝑐2)/(𝑐1+𝑐2))𝑓((𝑐1/𝑐2)(𝑀𝑆)(𝑥𝑐2𝑡)) and then again at 0 from the reflected wave of (2𝑐2/(𝑐1+𝑐2))((𝑐3𝑐2)/(𝑐3+𝑐2))((𝑐1𝑐2)/(𝑐1+𝑐2))𝑓((𝑐1/𝑐2)(𝑀𝑆)(𝑥𝑐2𝑡)). This process continues forever, and new waves appear every 𝜎/𝑐2 units of time. The functions 𝑣4,𝑣5, can be defined as before such that their sum 𝑣𝑖,𝑖=1,2,3 will satisfy the wave problem (2.34)–(2.36) over longer and longer time intervals. In the statement below we write the solution in the form of a finite series whose upper limit depends on the number of 𝜎/𝑐2 elapsed. In doing so we use the general forms of the waves in (2.50)–(2.53).

We summarize these results in the following theorem.

Theorem 2. Let 𝑓(𝑥𝑐1𝑡),𝑓(𝑠)=0, 𝑠>0, 𝑓𝐶2(𝑅) be an incoming wave. Then the solution to the problem (2.34)–(2.36) in the time interval 0𝑡<𝑖𝜎/𝑐2, 𝑖=1,2,3, is given by 𝑓𝑢(𝑥,𝑡)=𝑥𝑐1𝑡+𝑐2𝑐1𝑐2+𝑐1𝑓𝑥𝑐1𝑡,𝑥<0,2𝑐2𝑐1+𝑐2𝑓𝑐1𝑐2𝑥𝑐2𝑡,0<𝑥<𝜎,0,𝑥>𝜎+Σ𝑖𝑗even=2𝑐0,𝑥<0,1𝑐2𝑐1+𝑐2(𝑗2)/2𝑐3𝑐2𝑐3+𝑐2𝑗/22𝑐2𝑐1+𝑐2𝑐𝑓1𝑐2(𝑀𝑆)(𝑗2)/2𝑀𝑥+𝑐2𝑡,0<𝑥<𝜎,4𝑐2𝑐3𝑐1+𝑐2𝑐2+𝑐3𝑐1𝑐2𝑐1+𝑐2𝑐3𝑐2𝑐3+𝑐2(𝑗2)/2𝑐𝑓1𝑐2(𝑀𝑆)(𝑗2)/2𝐿𝑥𝑐3𝑡,𝑥>𝜎+Σ𝑖𝑘odd=34𝑐1𝑐2𝑐1+𝑐22𝑐3𝑐2𝑐3+𝑐2(𝑘1)/2𝑐1𝑐2𝑐1+𝑐2(𝑘3)/2𝑐𝑓1𝑐2(𝑀𝑆)(𝑘3)/2𝑀𝑅𝑥+𝑐1𝑡,𝑥<0,2𝑐2𝑐1+𝑐2𝑐3𝑐2𝑐3+𝑐2𝑐1𝑐2𝑐1+𝑐2(𝑘1)/2𝑐𝑓1𝑐2(𝑀𝑆)(𝑘1)/2𝑥𝑐2𝑡,0<𝑥<𝜎,0,𝑥>𝜎.(2.57) Furthermore, assume that 𝑓 and its derivatives of up to the order two are bounded in 𝑅. Then, the solution to (2.34)–(2.36) over the time interval [0,) is the limit as 𝑖 of the above finite time solution.

Proof. Note that for 𝑖=1,𝑢(𝑥,𝑡)=𝑣1(𝑥,𝑡) as described in (2.54), for 𝑖=2,𝑢(𝑥,𝑡)=𝑣1+𝑣2, where 𝑣2 is given in (2.55), and for 𝑖=3,𝑢(𝑥,𝑡)=𝑣1+𝑣2+𝑣3, where 𝑣3 is defined in (2.56). We showed that in all these cases these sums satisfy the wave problem (2.34)–(2.36). For 𝑖>3, we will be adding reflected and transmitted waves at each interface 0 or 𝜎 which, when added to their corresponding incident waves, satisfy the boundary conditions in (2.36). We also argued that the initial conditions are satisfied because for 𝑡 close to 0 all terms involving 𝑓 except 𝑓(𝑥𝑐𝑡) will be zero. So 𝑢 solves the wave problem in (0,𝑖𝜎/𝑐2).
It remains to show the limiting case when 𝑖. For that, let us just focus on one piece of the function 𝑢 in the interval (,0):𝑢(𝑥,𝑡)=𝑓𝑥𝑐1𝑡+𝑐2𝑐1𝑐2+𝑐1𝑓𝑥𝑐1𝑡+𝑘odd=34𝑐1𝑐2𝑐1+𝑐22𝑐3𝑐2𝑐3+𝑐2(𝑘1)/2𝑐1𝑐2𝑐1+𝑐2(𝑘3)/2𝑐𝑓1𝑐2(𝑀𝑆)(𝑘3)/2𝑀𝑅𝑥+𝑐1𝑡.(2.58) First notice that by definitions in (2.37) the argument of 𝑓 in the summation (2.58) can be simplified to 4𝑐1𝑐2𝑐1+𝑐22𝑐3𝑐2𝑐3+𝑐2(𝑘1)/2𝑐1𝑐2𝑐1+𝑐2(𝑘3)/2𝑐𝑓1𝑐2(𝑀𝑆)(𝑘3)/2𝑀𝑅𝑥+𝑐1𝑡=4𝑐1𝑐2𝑐1+𝑐22𝑐3𝑐2𝑐3+𝑐2(𝑘1)/2𝑐1𝑐2𝑐1+𝑐2(𝑘3)/2𝑐𝑓1𝑐2𝑐(𝑘1)𝜎2𝑐1𝑥+𝑐1𝑡.(2.59) The coefficients of 𝑓 satisfy |(𝑐3𝑐2)/(𝑐3+𝑐2)|<1,|(c1𝑐2)/(𝑐1+𝑐2)|<1. Let ||||𝑐𝑟=max3𝑐2𝑐3+𝑐2||||,||||𝑐1𝑐2𝑐1+𝑐2||||,(2.60) then 𝑟<1. Now with 𝑓 being bounded, the partial sums of the series (2.58) are bounded above by the partial sums of a convergent geometric series in powers of 𝑟. Therefore it is absolutely and uniformly convergent. On the other hand, the term-by-term differentiation of (2.58) in terms of 𝑡 results in an extra coefficient 𝑐1 and no extra coefficient in terms of 𝑥, as can be seen by (2.59). Since 𝑓 and 𝑓 are also bounded, the resulting series also converge absolutely and uniformly to the derivatives of the limit of the series (2.58). A similar argument can be applied to the series for 𝑢 in the intervals (0,𝜎) and (𝜎,).

We have shown the form of the scattering of an incident wave on the right- and left-hand side of the join 0. But we have only shown the scattering of an incident wave on the left of 𝜎. If one wanted to study a higher number of interfaces, then it would be necessary to know what happens when an incoming wave hits an interface at an arbitrary point 𝜎0 on the right. Here, and for the completeness of the argument, we point out that the transmitted and reflected waves at 𝑥=𝜎 from an incoming wave 𝑘(𝑥+𝑐3) from the right would be as follows:𝑢4(𝑥,𝑡)=2𝑐2𝑐2+𝑐3𝑘𝑐2𝑐3𝑐2𝑐𝜎+3𝑐2𝑥+𝑐2𝑡𝑘,0<𝑥<𝜎,𝑥+𝑐3𝑡+𝑐2𝑐3𝑐2+𝑐3𝑘2𝜎𝑥𝑐3𝑡,𝑥>𝜎.(2.61) Theoretically, this should enable us to extend the result of Theorem 2 to a higher number of discontinuities in the density of the string. The difficulty would be the ability to keep track of all incoming and outgoing waves as well as the ones that bounce back and forth between the interfaces. Since there are so many such waves, with even three interfaces, a general solution for more than two interfaces is impossible to write down. However, for a specific, relatively short length of time the solution can be found, when the discontinuities are few, say three. In this case the form of the solution will also depend on the distance between the discontinuities. Then, one can ask up to how many joins or for what length of time will the scattered waves be tractable. Another interesting question would lie in the area of inverse scattering. Since by our experience the solution contains the location 𝜎 of a discontinuity, in the absence of such knowledge will it be possible to find the coordinate of the interface from the form of the scattered waves.