Waves on 2 and 3 dimensional domains

Chapter 14 Waves on 2 and 3 dimensional domains We now turn to the studying the initial boundary value problem for the wave equation in two and three dimensions. In this chapter we focus on the situation where the spatial domain is a bounded region, and where we require that the Dirichlet boundary condition hold at, the boundary of. 23D-BVP Thus we seek functions u(t, x), defined for t 2 [0, T] and x 2, such that @ 2 u PDE = u t 2 [0, T], x 2 (14.1a) 23D-wave @t2 BC u(t, x) = 0 t 2 [0, T], x 2 (14.1b) for some pre-specified functions s and v. There are two problems of interest. The first is to find standing wave solutions to (14.1). The second is to solve the initial value problem where we also require IC u(0, x) = s(x) @u (0, x) = v(x) x 2. (14.2) 23D-IC @t Just as in the one-dimensional setting, we begin by looking for standing wave 160

CHAPTER. WAVES ON 2 AND 3 DIMENSIONAL DOMAINS 161 solutions to (14.1a) of the form u(t, x) = A(t) (x). Inserting this in to (14.1a), we see that the functions A and must satisfy da dt = A and = 23D-eigenvalue-problem for some constant. Applying the Dirichlet boundary condition we furthermore see that we must have = 0 along the boundary. Thus we see that the spatial function that describes the shape of standing wave solutions to the wave equation (with Dirichlet boundary conditions) must satisfy the following eigenvalue problem: = on (14.3a) = 0 on. (14.3b) The following exercise shows that there are only non-trivial solutions to (14.3) in the case that <0. Exercise 14.1. Suppose that satisfies (14.3). Using the fact that = Div(grad ) and the divergence theorem, show that Z Z 2 dv = Explain why this implies that apple 0. dv = Z k grad k 2 dv. Suppose, then, that = 0. = 0. Explain why it must be that grad = 0 and thus that Finally, conclude that we must have <0 in order to have a non-trivial solution to (14.3) and that as a result we can set =! 2 for some!>0. If we make use of the inner product Z hu, vi = u v dv (14.4) 23D-inner-product we can also show that eigenfunctions with di erent eigenvalues are orthogonal. hogonal-eigenfunctions Exercise 14.2. Suppose 1 = 1 1 and 2 = 2 2, and that both 1, 2 satisfy

CHAPTER. WAVES ON 2 AND 3 DIMENSIONAL DOMAINS 162 the Dirichlet boundary condition. Show that 1h 1, 2 i = 2 h 1, 2 i. Use this to conclude that if 1, 2 then 1 and 2 are orthogonal. In fact, just as is the case for Sturm-Liouville problems, the Laplace operator, when subjected to Dirichlet boundary conditions, is a negative, self-adjoint operator with an infinite list of eigenvalues that tend to negative infinity and with a corresponding list of eigenfunctions that form a complete orthogonal collection. heorem:hilbert-schmidt Theorem 14.3. (Hilbert-Schmidt Theorem) Suppose that is a bounded domain in R n. Then there exists an infinite list of eigenvalues k and eigenfunctions k satisfying (14.3). The eigenvalues satisfy 0 > 1 > 2... and are such that k! 1 as k!1. Furthermore, the corresponding eigenfunctions k form a complete, orthogonal collection. In particular, if u is some function in L 2 ( ) then the sum converges in norm to u. X k k k, where k = hu, ki, (14.5) 23D-Fourier-series k k k2 Unfortunately, the proof of completeness of eigenfunctions is beyond the scope of this course. Most books on functional analysis and/or Hilbert spaces include this theorem; you might search the name of the theorem, or search the term spectral decomposition. We now show that Theorem 14.3 implies that we may solve the initial value problem (14.1) (14.2). For each eigenvalue k we can find functions A k (t) and B k (t) such that Ak 00 k A k, A k (0) = 1, Ak 0 (0) = 0, Bk 00 k B k, B k (0) = 0, Bk 0 (0) = 1. Exercise 14.4. Since each k is negative, we know that we may write k =! 2 k for some positive! k. Use this to find formulas for A k and B k.

CHAPTER. WAVES ON 2 AND 3 DIMENSIONAL DOMAINS 163 Using the linearity of the wave equation, we see that the most general solution to (14.1) is X u(t, x) = ( k A k (t) k (x) + k B k (t) k (x)). Thus the initial condition (14.2) is satisfied if k X X s = k k and v = k k k k. (14.6) 23D-reduced-ic Since Theorem 14.3 implies that we may choose k and k to satisfy (14.6), we see that the initial value problem can be solved using combinations of standing wave solutions. We conclude this discussion by emphasizing that, just as in the one-dimensional setting, understanding the standing wave solutions to the wave equation allows us to understand all solutions, as any solution can be expressed as the sum of standing wave solutions. In the following sections, we analyze the eigenvalue problem (14.3), and the corresponding initial value problem for the wave equation, in a number of special circumstances. Our primary goal in each section is to understand the shape of the standing wave solutions. Thus we focus on the Dirichlet eigenvalue problem (14.3). 14.1 Rectangular domains Cartesian-eigenproblem As a first example, let us consider the case where the domain is a rectangle. In particular, we take = [0, L] [0, M] for some positive numbers L and M. Our goal is to find solutions to (14.3) with =! 2. Using Cartesian coordinates, we obtain the following problem: @ 2 @x 2 + @2 @y 2 =!2, (14.7a) Cartesian-eigeneqn (0, y) = 0 (L, y) = 0 (14.7b) Cartesian-BVx (x, 0) = 0 (x, M) = 0. (14.7c) Cartesian-BVy

CHAPTER. WAVES ON 2 AND 3 DIMENSIONAL DOMAINS 164 Our plan is to look for product solutions, meaning functions of the form (x, y) = X (x) Y (y) for some functions X and Y. This is motivated by our success in finding such solutions for the one-dimensional wave equation. (Hey! It worked once... why not try it again?) Plugging = XY in to (14.7a) yields X 00 (x) Y (y) + X (x) Y 00 (y) = X (x) Y (y). We re-arrange this in to the form X 00 (x) X (x) = Y 00 (y) Y (y). Since the left side is only a function of X, and the right side is only a function of Y, we conclude that each side is constant. (It might be helpful to review the discussion at the beginning of 4.1.) This means that there is some constant µ such that X 00 = µx and Y 00 = ( µ)y. Notice also that the boundary conditions (14.7b) and (14.7c) imply that X (0) = 0, X (L) = 0, Y (0) = 0, Y (M) = 0. In particular, we see that the functions X and Y each satisfy a Dirichlet eigenvalue problem. This is great we know exactly how to find the solutions! Let s begin with the equation for X, which is X 00 (x) = µx (x) X (0) = 0 X (L) = 0. (14.8) Cartesian-X-problem We know from 4.2 that we only obtain solutions when µ<0, in which case we set µ = 2. The general solution to the ODE in (14.8) is constructed from cos( x) and sin( x).

CHAPTER. WAVES ON 2 AND 3 DIMENSIONAL DOMAINS 165 The boundary condition X (0) = 0 rules out the cosine solution, while the boundary condition X (L) = 0 implies that must be an integer multiple of /L. Thus we obtain a list of solutions together with corresponding values X l (x) = sin l! L x, µ l =! 2 l. L We now turn to the equation for Y. Since we have a long list of possible values for µ, we in fact have a list of eigenvalue problems for Y a separate problem for each value of l! We need functions Y satisfying Y 00 (y) = ( µ l )Y (y), Y (0) = 0, Y (M) = 0. Using the same reasoning as we applied to the equation for X, we obtain a list of solutions m Y m (y) = sin M y for which we have µ l = m M 2. Combining the functions X l and Y m, we obtain a collection of functions lm(x, y) = sin l! m L x sin M y. that satisfy the eigenvalue problem (14.7) with eigenvalue Exercise 14.5. lm = 2 *,! 2 l m + L M 2+. -

CHAPTER. WAVES ON 2 AND 3 DIMENSIONAL DOMAINS 166 1. For each function lm (x, y) find the functions A lm (t) and B lm (t) so that u lm (t, x, y) = A lm (t) lm (x, y) and v lm (t, x, y) = B lm (t0 lm (x, y) are standing wave solutions to the two-dimensional wave equation on the rectangle with Dirichlet boundary conditions, and are such that u lm (0, x, y) = lm (x, y) and @v lm @t (0, x, y) = lm (x, y). 2. What is the frequency at which the standing wave solutions u lm and v lm oscillate? Is it possible that two di erent u lm solutions oscillate at the same frequency? 3. What do the functions lm (x, y) look like? Draw a contour plot for generic values of l and m. Then describe the corresponding standing wave solutions u lm (t, x, y). Using the linearity of the wave equation, we see that X u(t, x, y) = ( lm u lm (t, x, y) + lm v lm (t, x, y)) l,m is also a solution for any constants lm and lm. We would like to choose the constants in order to ensure that the initial conditions u(0, x, y) = s(x, y) These conditions would be satisfied if and @u (0, x, y) = v(x, y). (14.9) Cartesian-IC @t X X s(x, y) = lm lm (x, y) and v(x, y) = l,m l,m lm lm (x, y). As the eigenfunctions lm form a complete orthogonal collection, we can compute the coe cients using (14.5). Exercise 14.6.

CHAPTER. WAVES ON 2 AND 3 DIMENSIONAL DOMAINS 167 1. Verify by direct computation that the functions lm form an orthogonal collection. 2. Compute k lm k 2. 3. Suppose s(x, y) = x(x L)y(y M) and v(x, y) = 0. Find the function u(t, x, y) that satisfies the IBVP given by (??). 14.2 Wave equation on the disk We now consider the case where the spatial domain is the unit disk. We make use of polar coordinates (r, ) and describe by 0 apple r apple 1 and apple apple. (14.10) polar-region Using (12.5), we see that that in polar coordinates the eigenvalue problem (14.3) becomes together with the boundary condition @ 2 @r 2 + 1 @ r @r + 1 @ 2 r 2 = (14.11) polar-eigenvalue-proble @ 2 (1, ) = 0. (14.12) polar-dirichlet While (14.12) does indeed describe the Dirichlet boundary condition for the unit disk, the parametrization of the disk using (14.10) has artificially introduced boundaries at r = 0, =, and =. Since = and = represent the same location on the disk, we require periodic boundary conditions (r, ) = (r, ) and @ @ (r, ) = @ (r, ). (14.13) polar-periodicbc @ We furthermore require that (r, ) be reasonable at r = 0. Just as in the previous section, we look for functions solutions that are in product form.

CHAPTER. WAVES ON 2 AND 3 DIMENSIONAL DOMAINS 168 cise:polar-reduced-ode Exercise 14.7. Suppose that (r, ) = R(r) ( ). Show that (14.11) becomes r 2 R 00 (r) + rr 0 (r) R(r) r 2 R(r) = 00 ( ) ( ). Conclude that each side must be equal to the same constant, which we call thus that µ, and r 2 R 00 (r) + rr 0 (r) + (µ r 2 )R(r) = 0, (14.14) first-bessel 00 ( ) = µ ( ). (14.15) Theta-ODE Finally, explain why must satisfy periodic boundary conditions, while R must satisfy R(1) = 0 and R(0) is reasonable. Exercise 14.7 reduces the problem of finding standing wave solutions to finding solutions to the ordinary di erential equations (14.14) and (14.15), subject to the appropriate boundary conditions. The problem for is a version of the periodic boundary value problem that we encountered in Chapter 4. Exercise 14.8. Show that (14.15) has solutions satisfying periodic boundary conditions only when µ is one of the numbers µ n = n 2 and that the corresponding solutions are cos(n ) and sin(n ). We now address the di erential equation (14.14). In fact, we must find a solution for each di erent value of µ. Our strategy is to try to express (14.14) as a Sturm- Liouville problem. Using the fact that µ = n 2, we re-write the equation as d 2 R dr 2 + 1 dr r dr n 2 R = R. (14.16) n-bessel r2

CHAPTER. WAVES ON 2 AND 3 DIMENSIONAL DOMAINS 169 We see from Exercise 9.4 that in order for the left side to be a self-adjoint operator, we require a weight function w that satisfies 1 r = 1 d [1 w]. w dr It is easy to see that w = r is a function that satisfies this condition, and is also positive on the domain 0 < r < 1. Thus we may write (14.14) in the form d dr " r dr # dr n 2 R = rr, (14.17) SL-Bessel r which corresponds to the self-adjoint form (9.7) with p = r. The equation (14.17) also satisfies theconditions (9.8) on the domain 0 < r < 1, in particular the positivity condition. Finally, we see that by imposing the boundary conditions R(0) is finite and R(1) = 0, (14.18) SL-Bessel-BC then the condition (9.10) holds. Thus (14.18) is an admissible boundary condition for (14.17). We may now conclude from the Theorem 9.10 that for each value of n there exists an infinite sequence of eigenvalues 0 > n1 > n2 >... such that nk! 1 as k!1and and corresponding eigenfunctions R nk (r) comprising a complete orthogonal collection of functions. Here orthogonality is with respect to the inner product with weight w = r; hr nk, R nl i w = Z 1 0 8> 0 if k, l, R nk (r) R ml (r) rdr= < > : kr nk kw 2 if k = l. (14.19) Bessel-IP Using the functions R nk, we are able to construct solutions to the eigenvalue

CHAPTER. WAVES ON 2 AND 3 DIMENSIONAL DOMAINS 170 problem (14.11) satisfying the Dirichlet boundary condition (14.12), namely c nk (r, ) = R nk(r) cos(n ) and s nk (r, ) = R nk(r) sin(n ) (14.20) polar-eigenfunctions having eigenvalue negative nk. We know from the general discussion at the beginning of the chapter that the eigenfunctions (14.20) are orthogonal. In the following exercise, we see that this orthogonality comes from the orthogonality of the functions and R. Exercise 14.9. 1. Consider two of the cosine-type solutions, c nk and c. Show that ml h c nk, c ml i = Z! Z 1! cos(n ) cos(m ) d R nk (r) R ml (r) rdr. 0 (14.21) polar-ortho-coord 2. Explain why the integral in (14.21) is zero unless n = m. 3. Suppose that n = m. Explain why in this case the second integral in (14.21) is zero unless k = l. 4. Why is it that we never need to compute Z 1 0 R nk (r) R ml (r) rdr in the case that n, m? 5. Explain why similar reasoning applies to h c nk, s ml i and h s nk, s ml i. While the Sturm-Liouville theorem tells us that the functions R nk (r) exist, and are orthogonal, it does not tell us much else about these functions. In the remainder of this section, we learn more about these functions by constructing a power series expression for them. Since we know that each eigenvalue is negative, we set =! 2 for some!>0. It is convenient to make the change of variables x =!r. By the chain rule we

CHAPTER. WAVES ON 2 AND 3 DIMENSIONAL DOMAINS 171 compute dr dr = dr dx dx dr =! dr dx and thus d2 R dr 2 Making use of these, we see that (14.16) becomes =!2 d2 R dx 2. x 2 d2 R dx 2 + x dr dx + (x2 n 2 )R = 0, (14.22) Bessel which we want to solve for x > 0, and for which we want x =! to be a root. We furthermore require that R be finite at x = 0. Exercise 14.10. The Sturm-Liouville theorem tells us that there is, in fact, an increasing list of values for! that lead to a solution to our eigenvalue problem. This means that for each value of n, there should be an infinite number of roots x =! nk of the solution R(x) to (14.22). To see this, suppose that R(x) = 1 p x S(x) for some function S(x). expression in to (14.22) and show that S must satisfy Plug this d 2 S dx 2 = 1 + 1! 4n2 4x 2 S. (14.23) mod-bessel Conclude that for large values of x any solutions to (14.23) should be oscillating roughly in the way that cosine and sine do, and thus that S(x) must have an infinite number of roots. Further conclude that R(x) 1 p x (cosine or sine) as x gets large. We now investigate the behavior of solutions to (14.22) when x is small. Suppose R(x) is a solution to (14.22) and that R(x) = x p + R rem (x) for some power p and function R rem (x) that vanishes to higher order as x! 0 in the sense that R rem (x) R lim x!0 x p = 0, lim rem(x) 0 x!0 x p 1 = 0, etc. Plugging R(x) = x p + R rem (x) in to (14.22) yields (p 2 + n 2 )x p + x 2 R 00 rem(x) + xr 0 rem(x) + (x 2 n 2 )R rem (x) + x p+2 = 0.

CHAPTER. WAVES ON 2 AND 3 DIMENSIONAL DOMAINS 172 Dividing by x p and taking the limit as x! 0 we conclude that p 2 n 2 = 0 and thus p = ±n. Since we require R(x) to be bounded at x = 0, we conclude that R(x) = x n +... when x is small. The di erential equation (14.22) is actually a famous one it is called Bessel s equation, and the solutions to it are called Bessel functions of order n and are given the symbol J n (x). Most mathematical handbooks devote several pages to these functions, which can actually be defined for any value of n. Here, however, we focus on the case when n is a positive integer. We begin by seeking a power series expression for the solution J n (x) to (14.22) of the form 1X J n (x) = x n a k x k ; (14.24) Bessel-anzatz the reason for the factor of x n in front of the sum is that we expect the behavior of J n to be x n as x! 0. Exercise 14.11. k=0 1. Plug (14.24) in to (14.22) and show that, after re-indexing, the result is a 1 (2n + 1)x + 1X [a k (2n + k)k + a k 2 ] x k+n = 0. k=2 2. Use the previous step to conclude that a 1 = 0 and that the coe cients satisfy the recurrence relation a k = a k 2 (k + 2n)k. Conclude from this that, in fact, all of the odd-indexed coe 3. Since the odd-indexed coe cients are zero, we can write (14.24) as J n (x) = x n 1X a 2l x 2l. l=0 cients are zero.

CHAPTER. WAVES ON 2 AND 3 DIMENSIONAL DOMAINS 173 Show, using the recurrence relation from the previous step, that a 2l = a 0( 1) l n! (l + n)! l! 4. Show how to choose a 0 in order to achieve J n (x) = x n 1X l=0 1 2 2l. ( 1) l x 2l. (l + n)! l! 2 5. For n = 1, 2, 3,...10 use technology to plot the approximations on the domain 0 apple x apple 30. X100 x n ( 1) l x 2l (l + n)! l! 2 l=0 (a) Do the solutions oscillate? (b) What happens to the roots as n increases? r 2 (c) Compare the plots with the function. What do you observe? x As you see in the previous exercise (and as predicted by the Sturm-Liouville theorem), the Bessel function J n (x) has an infinite list of roots; we denote these roots by! n1,! n2,! n3,... so that! nk is the kth root of the nth order Bessel function J n (x). Many software packages can easily find (or have as built-in functions) the roots! nk. We now return to the problem that motivated all this Bessel stu finding our standing wave eigenfunctions. Recall that R(r) = R(! x) where! was a root of the function that solved (14.22). Since we now know that the solutions to (14.22) are the Bessel functions, and have a list of all their roots, we know that R nk (r) = J n (! nk r).

CHAPTER. WAVES ON 2 AND 3 DIMENSIONAL DOMAINS 174 Using this, we see that our standing wave eigenfunctions are c nk (r, ) = J n(! nk r) cos(n ) and s nk (r, ) = J n(! nk r) sin(n ). Exercise 14.12. 1. What does a typical plot of J n (! nk r) look like? What does a typical plot of cos(nr) look like? 2. Use the previous part to figure out what the contour plots of c (r, ) are. nk What does the number n indicate? What does the number k indicate? 3. Get a piece of technology to either plot the function c (r, ), or plot its nk contours, for various values of n and k. Describe the results. 14.3 Exploration: Wave equation on the ball This section is written so that it can be used as a take-home exam and/or a longer homework assignment. In this exploration you consider the wave equation on the unit ball subject to Dirichlet boundary conditions. Be warned: There are a lot of functions and change of variables running around this exploration it is imperative that you keep your work organized! Working in spherical coordinates (r,, ), the unit ball is described by 0 apple r apple 1, apple apple, and 0 apple apple. We require periodic boundary conditions at = ±, and that functions are finite at r = 0 and at = 0,. 1. Express the eigenvalue problem = in spherical coordinates. 2. Assume that the eigenfunction takes the product form (r,, ) = R(r)Y (, ), and recall that we can write =! 2 for some positive constant!. Show

CHAPTER. WAVES ON 2 AND 3 DIMENSIONAL DOMAINS 175 that this implies that 1 sin 2 for some constant µ. r 2 d2 R dr + 2r dr2 dr + (!2 r 2 + µ)r = 0 (14.25) 3D-R @ 2 Y @ 2 + 1 sin and @ @ " sin @Y @ # = µy (14.26) 3D-Y 3. It is possible to show that µ<0 using an integration-by-parts argument. To do this, multiply the Y equation by Y and then integrate over the sphere using the spherical area element da = sin d d. Integrate by parts in order to show that we must have µ<0 in order to have a nontrivial solution Y. 4. We now suppose that the function Y takes product form. Assume that Y (, ) = ( ) ( ) and show that the functions and must satisfy for some constant. sin d d " sin d 2 = (14.27) 3D-Theta d 2 d # = (µ sin 2 ) (14.28) 3D-Phi d Take a minute to review the logic here: The plan is to first solve (14.27), which involves understanding what are the possible values of. With this information in hand, you will turn to (14.28). The functions and are to be used in constructing the functions Y. Finally, you will address (14.25). 5. Explain why the function must satisfy periodic boundary conditions on the interval apple apple, and thus that (14.27) only has solutions when = m 2 for some positive integer m. What are the corresponding functions? 6. Now you get to address (14.28). Make the change of variables x = cos in order to obtain d dx " (1 x 2 ) d # dx m 2 1 x 2 = µ, (14.29) ass-legendre-ode

CHAPTER. WAVES ON 2 AND 3 DIMENSIONAL DOMAINS 176 which holds for 1 < x < 1. Explain why, for each fixed integer m, the equation (14.29) satisfies the hypotheses of the Sturm-Liouville theorem with respect to the inner product h f, gi = Z 1 1 f (x) g(x) dx. Conclude that for each fixed m there exists a sequence of negative eigenvalues and corresponding orthogonal eigenfunctions satisfying (14.29). The eigenfunctions are called associated Legendre functions and are given the symbol P l,m (x); the corresponding eigenvalues are µ l,m = l(l + 1). Verify that P 0,0 (x) = 1 and that P 1,0 (x) = x P 2,0 (x) = 3 1 2 x2 p p 2 P 1,1 (x) = 1 x 2 P 2,1 (x) = 3x 1 x 2 P 2,2 (x) = 3(1 x 2 ) P 3,2 (x) = 15x(1 x 2 ) satisfy (14.29). (It turns out that one must have l m in order to have nonzero P l,m.) 7. Conclude from the above that for each m and l m there exists l,m( ) = P l,m (cos ) satisfying (14.28) with = m 2 and µ = l(l + 1). Use this to construct functions Y l,m (, ) satisfying (14.26). These functions are called the spherical harmonics. List the first few spherical harmonics. Draw, as best you can, the contour plots of the first few harmonics on the sphere. 8. Finally, we are ready to address (14.25). Set µ = l(l + 1) and make the change of variable x =!r in order to obtain x 2 d2 R dr + 2x dx2 dx + (x2 l(l + 1))R = 0. Then make the change of variables R(x) = 1 p x S(x). Show that S(x) must satisfy x 2 d2 S dx 2 + x ds dx + x2 " l(l + 1) + 1 #! S = 0, (14.30) half-bessel 2

CHAPTER. WAVES ON 2 AND 3 DIMENSIONAL DOMAINS 177 which is Bessel s equation with order l(l + 1) + 1 2. The solutions are called Bessel functions and are given the symbol J l (l+1)+ 1 (x). 2 9. It is straightforward to find power series expressions for J l (l+1)+ 1 2 (x), following the procedure used in the previous section. At this stage, however, you should simply verify the following: r 2 J 1/2 (x) = x sin(x) r 2 3 x 2 J 5/2 (x) = sin(x) x x 2! 3 x cos(x) 10. Each Bessel function J l (l+1)+ 1 has an infinite number of roots! l,k. Use these 2 roots to construct solutions R l,k (r) satisfying (14.25) with µ = l(l + 1) and! =! l,k. Sketch R l,k (r) for the first few l, k. 11. Conclude this whole discussion by constructing standing wave forms l,m,k (r,, ) using the functions R l,k and Y l,m obtained above. Give the explicit form for the first few wave shapes. How should one understand these shapes geometrically and physically? 12. Look up the the orbital table in the Wikipedia article Atomic Orbital https://en.wikipedia.org/wiki/atomic_orbital#orbitals_table. What do you notice?