2 Wave equation in one dimension

Transcription

1 2 Wave equation in one dimension Here we will consider in detail the wave equation in one dimension (x), for a function H(x,t). H could be, for example, a component of an EM field, the pressure of air in a sound wave moving along the x axis, or it could be the displacement of a string, for a wave moving along a taught string. The one-dimensional wave equation is 2 H(x,t) x 2 = 1 c 2 2 H(x,t) 2 or equivalently 2 H(x,t) = c 2 2 H(x,t) 2 x 2 where c is the speed of the wave. Note that it is a linear homogeneous PDE. In the next section we consider d Alembert s very general solution of the wave function. After that we consider standing then travelling wave solutions, before finishing with the application of boundary conditions. 2.1 d Alembert s solution The 18th century mathematician and physicist Jean le Rond d Alembert showed that the solution, H, to the one-dimensional wave equation can always be written as the sum of two travelling waves. We call these two waves W F, and W B (F for forward and B for back). Both W F and W B depend only on a single variable, unlike H which depends on two: x and t. W F depends on the variable u = x ct, and W B depends on the variable v = x + ct. So d Alembert said that we could always write the solution of the wave PDE as H(x,t) = W F (x ct) + W B (x + ct) = W F (u) + W B (v) Note that this is not trivial because the left-hand side is a function of two variables, while each of the two functions on the right-hand side is just a function of one variable. This is a solution for the wave equation, whatever W F and W B are, i.e., these functions can be anything. As we will see W F and W B are usually taken to be sine or cosine waves, i.e., sin(x ct) or cos(x ct) for W F, or the closely related exponentials exp(ik(x ct)). But they can be any function: polynomials, Gaussians, etc. The basic physics behind d Alembert s solution is that solutions of the wave equation don t change shape as time advances, they just translate with unchanged shape (to the right or left, depending on the direction of the wave) Proof that d Alembert s W F (u) is a solution to the wave PDE To see that W F (u), with u = x ct, is a solution to the wave equation, we just substitute it into both sides of this PDE. Then by use of the chain rule of differentiation, we can show that for any function of u the LHS equals the RHS and it must be a solution. We start with the left-hand side LHS = 2 W F (x,t) x 2 = x x = ( WF (u) ) = x x x 7

2 because / x = 1. Continuing LHS = x Now for the right-hand side RHS = 1 c 2 2 W F (x,t) 2 = 1 c 2 = because / = c. Continuing RHS = 1 ( ) = 1 ( ) c c x = = 2 W F (u) 2 = 1 ( ) = 1 ( ) ( c) c 2 c 2 ( ) = 1 c ( ) ( c) = 2 W F (u) 2 So, both sides of the wave equation are just the second derivative with respect to u, and so are equal to each other, proving that any function W F (x ct) is a solution to the wave equation. The proof that W B (x + ct) is also a solution, is very similar. 2.2 Solution of the one-dimensional wave equation via separation of variables: the standing-wave solution A standard way of solving PDEs such as the wave equation, diffusion equation, Schrödinger s equation, etc, is to start by assuming that the solution, e.g., H(x,t), can be written as a product of functions, each of which is a function of only one of the variables. The idea is that H(x,t) is a function of x only, call it X(x), times a function of t only, call it T(t): H(x,t) = X(x)T(t) Note that T(t) is a function of t, not the temperature. At the moment, X and T are unknown functions. Now, by assuming that we can write the solution to the PDE as a product of a function of x times a function of t, we can obtain a solution of PDE but we will not obtain the general solution to the PDE. Typically (as we will see below) we use this method to obtain solutions, which are sine and cosine waves, and then we sum many of these waves (with different wavelengths) in a Fourier series to obtain the particular solution that we want, i.e., one that satisfies the boundary conditions on the solution. If we substitute H(x,t) = X(x)T(t) into the one-dimensional wave equation we get T(t) 2 X(x) x 2 = 1 c 2X(x) 2 T(t) 2 because we can take T out of the x differentiation as it is independent of x, and similarly we can take X out of the t differentiation. If we divide both sides by XT, we get 1 2 X(x) = T(t) X(x) x 2 c 2 T(t) 2 Now, we notice that the left-hand side is a function of x but not of t while the right-hand side is a function of t but not of x. So the LHS tells us that the equation cannot depend on t, and the RHS 8

3 tells us that it cannot depend on x. Therefore it cannot depend on either x or t, and so must be a constant. We will call this constant k 2. As we will see the minus sign will give us the sine and cosine functions we want 2. The fact that k is squared is just to make some of the later equations look neater. So we have 1 2 X(x) = T(t) = k 2 X(x) x 2 c 2 T(t) 2 which gives us the two ODEs The solutions of these ODEs are 2 X(x) x 2 = k 2 X(x) and 2 T(t) 2 = k 2 c 2 T(t) X(x) = A cos(kx) + B sin(kx) and T(t) = C cos(kct) + D sin(kct) where A, B, C and D are constants. You can check this by substituting these solutions back into the ODEs and checking that the LHS equals the RHS. These ODEs were covered in the semester 1 ODE course, please refer to your notes or to a textbook if you are unsure about these solutions. You need to understand how to solve these simple ODEs if you are to understand how to solve the PDEs. We can multiply X and T together and get the solution for H or H(x,t) = X(x)T(t) = [A cos(kx) + B sin(kx)] [C cos(kct) + D sin(kct)] H(x,t) = AC cos(kx) cos(kct) + AD cos(kx) sin(kct) + BC sin(kx) cos(kct) + BD sin(kx) sin(kct) Defining the four new constants I = AC, J = AD, P = BC and Q = BD we have that the solution is H(x,t) = I cos(kx) cos(kct) + J cos(kx) sin(kct) + P sin(kx) cos(kct) + Q sin(kx) sin(kct) This solution has four terms: the four combinations of sines and cosines as functions of space and of time. It is a solution but not the most general solution. However, as the wave equation is a linear homogeneous PDE then the sum of any solutions of this PDE is also a solution to the PDE. In other words we can add lots of terms like cos(kx) cos(kct), with different k s, and the sum is also a solution, e.g., H(x,t) = I 1 cos(k 1 x) cos(k 1 ct) + I 2 cos(k 2 x) sin(k 2 ct) is a solution of the PDE, for any values of the constants I 1 and I 2, and of the wavevectors k 1 and k 2. This means that we can add lots of these terms together and vary the constants, and so get an expression which satisfies the boundary conditions. This is how we can use a Fourier series to obtain a particular solution of a PDE. To write a Fourier series general solution to the wave PDE we add subscript n s to the I, J, P, Q and k to indicate that we could have many terms like these with different values of I, k, etc. Then we have H(x,t) = [I n cos(k n x) cos(k n ct) + J n cos(k n x) sin(k n ct) + P n sin(k n x) cos(k n ct) + Q n sin(k n x) sin(k n ct)] n=1 2 We choose the sign as we wanted sine and cosine functions. If we picked a + sign, we would get exponentials, which are not useful for the problems we study here but might be in other problems. 9

4 Figure 2: A plot of a square wave H(x) of wavelength L = 1, plus two sine wave series approximations to it. The square wave is the black solid curve. A sine wave of the same wavelength is shown as a dot-dashed green (grey) curve. This is a sine series approximation to the square wave, truncated after the first term. A sum of up to the n = 9 terms is shown as the dashed red (grey) curve. We see that the dashed curve follows the square wave much better than the dot-dashed curve but it is still an approximation, only an infinite series perfectly follows the square wave. h x Note that for each solution all the constants are different, the I n, J n, P n, Q n and k n are all different. k n = 2π/λ n is the wavevector of the nth term in the sum, and so the nth term is 4 waves, each with the same wavelength, but each term has a different wavelength. The boundary conditions will determine the values of the constants I n, J n, P n and Q n, and of the k n and then once these values are determined we have the particular solution. 2.3 Travelling wave solutions The method of separation of variables gives us the solution in terms of standing waves, i.e., a sine (or cosine) of kx times a sine (or cosine) of kct, e.g., sin(kx) sin(kct). Any solution of the wave PDE can be expressed as a sum of standing wavs. However, any solution of the wave PDE can also be written in terms of a sum of travelling waves, i.e., functions like sin(k(x ct)). So, we can also write the solution as H(x,t) = [A n cos(k n (x ct)) + B n sin(k n (x ct)) + C n cos(k n (x + ct)) + D n sin(k n (x + ct))] n=1 The first two terms are sine and cosine waves moving to the right, with speed c, whereas the last two terms are sine and cosine waves moving to the left, with speed c. For example a simple sine wave with wavelength λ = 2π/k, amplitude B, moving with speed c is H(x,t) = B sin(k n (x ct)) This just gives a sine curve of course, but other shapes can be built up from a sum of sines (and/or 10

5 cosines). We can consider a simple example: the square wave, defined by { +1 0 < x < L/2 H(x,t = 0) = 1 L/2 < x < L This can be written as the sum of sine waves H(x,t = 0) = n=2,6,10,14,... 8 ( nπx ) nπ sin L Note that for the square wave, the sum only includes every 4th term, the n = 3, 4 and 5 terms are zero, the n = 6 term is non-zero, the n = 7, 8and9 terms are zero, and so on. Starting with this square wave at t = 0, a square wave moving with speed c is then obtained by just subtracting ct from x in the argument of the sine waves, i.e., The first few terms are H(x,t) = H(x,t) = 4 π sin ( 2π(x ct) L n=2,6,10,14, Applying boundary conditions 8 nπ sin ) + 4 ( ) 6π(x ct) 3π sin L ( ) nπ(x ct) L + 4 5π sin ( ) 10π(x ct) + L Above we started with the wave at time t = 0, and if we know the direction the wave is travelling in, then we can just subtract ct from x in the sine functions, to get the solution at any time t. When we did this we were effectively specifying boundary conditions: these were the initial condition, which is the wave at t = 0, plus a direction. Here we will look at two more examples of the application of boundary conditions to solutions of the wave PDE. One to a standing wave solution and another to a travelling wave solution. First the standing wave solution Simple example boundary conditions applied to a standing wave solution In general the solution will be an infinite sum of waves (also called Fourier modes), each with a different k, i.e., with a different wavelength. But here we will consider a simple example, where there is only one standing wave, a wavelength λ, with a corresponding wavevector k = 2π/λ. In this case the solution is just H(x,t) = I cos(kx) cos(kct) + J cos(kx) sin(kct) + P sin(kx) cos(kct) + Q sin(kx) sin(kct) As a simple example of how boundary conditions are applied, we consider boundary conditions that are initial conditions. The first initial condition is that at time t = 0, the function H(x,t = 0) is given by H(x,t = 0) = 10 cos(kx) and the second initial condition is that the time derivative is zero, i.e., the string is stationary at t = 0 = 0 11

6 where the subscript t = 0 on the brackets indicates that we evaluate the time derivative at t = 0. Now, having written down the boundary conditions that the particular solution must satisfy, we must apply these boundary conditions. We start 3 by applying the boundary condition on H at t = 0 H(x,t = 0) = 10 cos(kx) = I cos(kx) cos(0) + J cos(kx) sin(0) + P sin(kx) cos(0) + Q sin(kx) sin(0) and as sin(0) = 0 and cos(0) = 1, this simplifies to 10 cos(kx) = I cos(kx) + P sin(kx) So I = 10 and P = 0. Putting these values in our expression for H(x,t) we get H(x,t) = 10 cos(kx) cos(kct) + J cos(kx) sin(kct) + Q sin(kx) sin(kct) Now we apply the boundary condition on the initial time derivative. First we take the time derivative of H H(x, t) = 10kc cos(kx) sin(kct) + Jkc cos(kx) cos(kct) + Qkc sin(kx) cos(kct) At t = 0, this time derivative equals zero = 0 = 10kc cos(kx) sin(0) + Jkc cos(kx) cos(0) + Qkc sin(kx) cos(0) 0 = Jkc cos(kx) + Qkc sin(kx) so both J and Q are zero, J = 0, and Q = 0. So the particular solution that satisfies both boundary conditions is H(x,t) = 10 cos(kx) cos(kct) Simple example boundary conditions applied to a travelling wave solution We will now look at exactly the same boundary conditions but with a travelling wave. The general travelling wave solution for one fixed value of k is H(x,t) = A cos(k(x ct)) + B sin(k(x ct)) + C cos(k(x + ct)) + D sin(k(x + ct)) As before, the first initial condition is that at time t = 0, the function H(x,t = 0) is given by so we have H(x,t = 0) = 10 cos(kx) H(x,t = 0) = 10 cos(kx) = A cos(kx) + B sin(kx) + C cos(kx) + D sin(kx) H(x,t = 0) = (A + C) cos(kx) + (B + D) sin(kx) To satisfy this boundary condition, A + C = 10 and B + D = 0. 3 We start with this boundary condition as it simplifies the solution by eliminating some terms. Note that in principle we can apply the boundary conditions in any order, but that in practice some orders are a lot easier than others to do. 12

7 The second initial condition is that the time derivative is zero at t = 0. The time derivative of the solution is H(x, t) = Akc sin(k(x ct)) Bkc cos(k(x ct)) Ckc sin(k(x + ct)) + Dkc cos(k(x + ct)) At t = 0 this derivative is zero = Akc sin(kx) BkcB cos(kx) Ckc sin(kx) + Dkc cos(kx) = kc(a C) sin(kx) + kc(d B) cos(kx) = 0 so to satisfy this boundary condition A C = 0 and D B = 0. By combining the equations for A and C from the two boundary conditions we obtain two simultaneous linear equations for A and C A + C = 10 A C = 0 which have the solution A = C = 5. For B and D we have two simultaneous linear equations B + D = 0 D B = 0 which have the solution B = D = 0. So the particular travelling wave solution that satisfies these boundary conditions is H(x,t) = 5 cos(k(x ct)) + 5 cos(k(x + ct)) These are two cosine travelling waves, one travelling left to right and the other right to left. These two travelling waves sum to give the standing wave we found in the previous section, i.e., the two solutions we found, one in terms of a standing wave, and the other in terms of two travelling waves, are completely equivalent. We can choose to use one or the other, whichever we find more convenient for the problem we are studying. 13