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Chapter 14

Waves on 2 and 3 dimensionaldomains

We now turn to the studying the initial boundary value problem for the waveequation in two and three dimensions. In this chapter we focus on the situationwhere the spatial domain ⌦ is a bounded region, and where we require that theDirichlet boundary condition hold at �, the boundary of ⌦.

Thus we seek functions u(t, x), defined for t 2 [0,T] and x 2 ⌦, such that23D-BVP

PDE @2u

@t

2 = �u t 2 [0,T], x 2 ⌦ (14.1a) 23D-wave

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

@t

(0, x) = v(x) x 2 ⌦. (14.2) 23D-IC

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), wesee that the functions A and must satisfy

dA

dt

= �A and � = �

for some constant �. Applying the Dirichlet boundary condition we furthermore seethat 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 (withDirichlet boundary conditions) must satisfy the following eigenvalue problem:23D-eigenvalue-problem

� = � on ⌦ (14.3a)

= 0 on �. (14.3b)

The following exercise shows that there are only non-trivial solutions to (14.3) inthe case that � < 0.

Exercise 14.1. Suppose that satisfies (14.3). Using the fact that� = Div(grad )and the divergence theorem, show that

�

Z

⌦ 2

dV =

Z

⌦ � dV = �

Z

⌦k grad k2 dV .

Explain why this implies that � 0.

Suppose, then, that � = 0. Explain why it must be that grad = 0 and thus that = 0.

Finally, conclude that we must have � < 0 in order to have a non-trivial solutionto (14.3) and that as a result we can set � = �!2 for some ! > 0.

If we make use of the inner product

hu, vi =Z

⌦u v dV (14.4) 23D-inner-product

we can also show that eigenfunctions with di�erent eigenvalues are orthogonal.

exercise:23D-orthogonal-eigenfunctions Exercise 14.2. Suppose � 1 = �1 1 and � 2 = �2 2, and that both 1, 2 satisfy


the Dirichlet boundary condition. Show that

�1h 1, 2i = �2h 1, 2i.

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-adjointoperator with an infinite list of eigenvalues that tend to negative infinity and with acorresponding list of eigenfunctions that form a complete orthogonal collection.

Theorem 14.3. (Hilbert-Schmidt Theorem) Suppose that ⌦ is a bounded domaintheorem:Hilbert-Schmidt

in Rn . 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, orthogonalcollection. In particular, if u is some function in L

2(⌦) then the sum

X

k

↵k k, where ↵k =hu, k ik k k2

, (14.5) 23D-Fourier-series

converges in norm to u.

Unfortunately, the proof of completeness of eigenfunctions is beyond the scope ofthis course. Most books on functional analysis and/or Hilbert spaces include thistheorem; you might search the name of the theorem, or search the term “spectraldecomposition.”

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 Ak (t) and Bk (t) suchthat

A

00k = �k Ak, Ak (0) = 1, A

0k (0) = 0,

B

00k = �k Bk, Bk (0) = 0, B

0k (0) = 1.

Exercise 14.4. Since each �k is negative, we know that we may write �k = �!2k

for some positive !k . Use this to find formulas for Ak and Bk .


Using the linearity of the wave equation, we see that the most general solution to(14.1) is

u(t, x) =X

k

(↵k Ak (t) k (x) + �k Bk (t) k (x)) .

Thus the initial condition (14.2) is satisfied if

s =X

k

↵k k and v =X

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 seethat the initial value problem can be solved using combinations of standing wavesolutions.

We conclude this discussion by emphasizing that, just as in the one-dimensionalsetting, understanding the standing wave solutions to the wave equation allows usto understand all solutions, as any solution can be expressed as the sum of standingwave solutions.

In the following sections, we analyze the eigenvalue problem (14.3), and the cor-responding initial value problem for the wave equation, in a number of specialcircumstances. Our primary goal in each section is to understand the shape of thestanding wave solutions. Thus we focus on the Dirichlet eigenvalue problem (14.3).

14.1 Rectangular domains

As a first example, let us consider the case where the domain is a rectangle. Inparticular, we take ⌦ = [0, L] ⇥ [0,M] for some positive numbers L and M . Ourgoal is to find solutions to (14.3) with � = �!2. Using Cartesian coordinates, weobtain the following problem:2D-Cartesian-eigenproblem

@2

@x

2 +@2

@y2 = �!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


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 successin finding such solutions for the one-dimensional wave equation. (Hey! It workedonce. . . 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 discussionat 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 eigenvalueproblem. 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 weset µ = �⌫2. The general solution to the ODE in (14.8) is constructed from

cos(⌫x) and sin(⌫x).


The boundary condition X (0) = 0 rules out the cosine solution, while the boundarycondition X (L) = 0 implies that ⌫ must be an integer multiple of ⇡/L. Thus weobtain a list of solutions

Xl (x) = sin ⇡l

L

x

!,

together with corresponding values

µl = � ⇡l

L

!2.

We now turn to the equation for Y . Since we have a long list of possible values forµ, we in fact have a list of eigenvalue problems for Y – a separate problem for eachvalue of l! We need functions Y satisfying

Y

00(y) = (� � µl )Y (y), Y (0) = 0, Y (M) = 0.

Using the same reasoning as we applied to the equation for X , we obtain a list ofsolutions

Ym (y) = sin✓⇡m

M

y◆

for which we have� � µl = �

✓⇡m

M

◆2.

Combining the functions Xl and Ym , we obtain a collection of functions

lm (x, y) = sin ⇡l

L

x

!sin

✓⇡m

M

y◆.

that satisfy the eigenvalue problem (14.7) with eigenvalue

�lm = �⇡2 *,

l

L

!2+

✓m

M

◆2+- .

Exercise 14.5.


1. For each function lm (x, y) find the functions Alm (t) and Blm (t) so that

ulm (t, x, y) = Alm (t) lm (x, y) and vlm (t, x, y) = Blm (t0 lm (x, y)

are standing wave solutions to the two-dimensional wave equation on therectangle with Dirichlet boundary conditions, and are such that

ulm (0, x, y) = lm (x, y) and@vlm@t

(0, x, y) = lm (x, y).

2. What is the frequency at which the standing wave solutions ulm and vlm

oscillate? Is it possible that two di�erent ulm solutions oscillate at the samefrequency?

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 solutionsulm (t, x, y).

Using the linearity of the wave equation, we see that

u(t, x, y) =X

l,m

(↵lmulm (t, x, y) + �lmvlm (t, x, y))

is also a solution for any constants ↵lm and �lm . We would like to choose theconstants in order to ensure that the initial conditions

u(0, x, y) = s(x, y) and@u

@t

(0, x, y) = v(x, y). (14.9) Cartesian-IC

These conditions would be satisfied if

s(x, y) =X

l,m

↵lm lm (x, y) and v(x, y) =X

l,m

�lm lm (x, y).

As the eigenfunctions lm form a complete orthogonal collection, we can computethe coe�cients using (14.5).

Exercise 14.6.


1. Verify by direct computation that the functions lm form an orthogonalcollection.

2. Compute k lm k2.

3. Suppose s(x, y) = x(x � L)y(y � M) and v(x, y) = 0. Find the functionu(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 makeuse of polar coordinates (r, ✓) and describe ⌦ by

0 r 1 and ⇡ ✓ ⇡. (14.10) polar-region

Using (12.5), we see that that in polar coordinates the eigenvalue problem (14.3)becomes

@2

@r

2 +1r

@

@r

+1r

2@2

@✓2 = � (14.11) polar-eigenvalue-problem

together with the boundary condition

(1, ✓) = 0. (14.12) polar-Dirichlet

While (14.12) does indeed describe the Dirichlet boundary condition for the unitdisk, the parametrization of the disk using (14.10) has artificially introduced“boundaries” at r = 0, ✓ = �⇡, and ✓ = ⇡. Since ✓ = �⇡ and ✓ = ⇡ repre-sent 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 productform.


exercise:polar-reduced-ode Exercise 14.7. Suppose that (r, ✓) = R(r)⇥(✓). Show that (14.11) becomes

r

2R

00(r) + r R

0(r) � �r

2R(r)

R(r)= �⇥

00(✓)⇥(✓)

.

Conclude that each side must be equal to the same constant, which we call �µ, andthus that

r

2R

00(r) + r R

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 mustsatisfy R(1) = 0 and R(0) is “reasonable.”

Exercise 14.7 reduces the problem of finding standing wave solutions to findingsolutions to the ordinary di�erential equations (14.14) and (14.15), subject to theappropriate boundary conditions.

The problem for ⇥ is a version of the periodic boundary value problem that weencountered in Chapter 4.

Exercise 14.8. Show that (14.15) has solutions satisfying periodic boundary con-ditions only when µ is one of the numbers

µn = �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 solutionfor 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

2R

dr

2 +1r

dR

dr

� n

2

r

2 R = �R. (14.16) n-Bessel


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

1r

=1w

d

dr

[1w] .

It is easy to see that w = r is a function that satisfies this condition, and is alsopositive on the domain 0 < r < 1. Thus we may write (14.14) in the form

d

dr

"r

dR

dr

#� n

2

r

R = �r R, (14.17) SL-Bessel

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 positivitycondition. 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 conditionfor (14.17).

We may now conclude from the Theorem 9.10 that for each value of n there existsan infinite sequence of eigenvalues

0 > �n1 > �n2 > . . .

such that �nk ! �1 as k ! 1 and and corresponding eigenfunctions Rnk (r)comprising a complete orthogonal collection of functions. Here orthogonality iswith respect to the inner product with weight w = r;

hRnk, Rnl iw =Z 1

0Rnk (r) Rml (r) r dr =

8>><>>:0 if k , l,

kRnk k2w if k = l .(14.19) Bessel-IP

Using the functions Rnk , we are able to construct solutions to the eigenvalue


problem (14.11) satisfying the Dirichlet boundary condition (14.12), namely

cnk (r, ✓) = Rnk (r) cos(n✓) and s

nk (r, ✓) = Rnk (r) sin(n✓) (14.20) polar-eigenfunctions

having eigenvalue negative �nk .

We know from the general discussion at the beginning of the chapter that theeigenfunctions (14.20) are orthogonal. In the following exercise, we see that thisorthogonality comes from the orthogonality of the functions ⇥ and R.

Exercise 14.9.

1. Consider two of the cosine-type solutions, cnk

and cml

. Show that

h cnk,

cml i =

Z ⇡

�⇡cos(n✓) cos(m✓) d✓

! Z 1

0Rnk (r) Rml (r) r dr

!.

(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

0Rnk (r) Rml (r) r dr

in the case that n , m?

5. Explain why similar reasoning applies to h cnk, s

mli and h s

nk, s

mli.

While the Sturm-Liouville theorem tells us that the functions Rnk (r) exist, and areorthogonal, it does not tell us much else about these functions. In the remainderof this section, we learn more about these functions by constructing a power seriesexpression 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


computedR

dr

=dR

dx

dx

dr

= !dR

dx

and thus d

2R

dr

2 = !2 d

2R

dx

2 .

Making use of these, we see that (14.16) becomes

x

2 d

2R

dx

2 + x

dR

dx

+ (x

2 � 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. Wefurthermore require that R be finite at x = 0.

Exercise 14.10. The Sturm-Liouville theorem tells us that there is, in fact, anincreasing 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 rootsx = !nk of the solution R(x) to (14.22).

To see this, suppose that R(x) = 1px

S(x) for some function S(x). Plug thisexpression in to (14.22) and show that S must satisfy

d

2S

dx

2 = � 1 + 1 � 4n

2

4x

2

!S. (14.23) mod-Bessel

Conclude that for large values of x any solutions to (14.23) should be oscillatingroughly in the way that cosine and sine do, and thus that S(x) must have an infinitenumber of roots.

Further conclude that R(x) ⇠ 1px

(cosine or sine) as x gets large.

We now investigate the behavior of solutions to (14.22) when x is small. SupposeR(x) is a solution to (14.22) and that R(x) = x

p + Rrem(x) for some power p andfunction Rrem(x) that vanishes to higher order as x ! 0 in the sense that

limx!0

Rrem(x)x

p= 0, lim

x!0

R

0rem(x)x

p�1 = 0, etc.

Plugging R(x) = x

p + Rrem(x) in to (14.22) yields

(p

2 + n

2)x

p + x

2R

00rem(x) + xR

0rem(x) + (x

2 � n

2)Rrem(x) + x

p+2 = 0.


Dividing by x

p and taking the limit as x ! 0 we conclude that p

2 � n

2 = 0 andthus p = ±n. Since we require R(x) to be bounded at x = 0, we conclude thatR(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 aregiven the symbol Jn (x). Most mathematical handbooks devote several pages tothese 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 Jn (x) to (14.22) ofthe form

Jn (x) = x

n1X

k=0ak 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 ofJn to be x

n as x ! 0.

Exercise 14.11.

1. Plug (14.24) in to (14.22) and show that, after re-indexing, the result is

a1(2n + 1)x +

1X

k=2[ak (2n + k)k + ak�2] x

k+n = 0.

2. Use the previous step to conclude that a1 = 0 and that the coe�cients satisfythe recurrence relation

ak =ak�2

(k + 2n)k

.

Conclude from this that, in fact, all of the odd-indexed coe�cients are zero.

3. Since the odd-indexed coe�cients are zero, we can write (14.24) as

Jn (x) = x

n1X

l=0a2l x

2l .


Show, using the recurrence relation from the previous step, that

a2l =a0(�1)ln!(l + n)! l!

122l .

4. Show how to choose a0 in order to achieve

Jn (x) = x

n1X

l=0

(�1)l

(l + n)! l!

✓x

2

◆2l.

5. For n = 1, 2, 3, . . . 10 use technology to plot the approximations

x

n100X

l=0

(�1)l

(l + n)! l!

✓x

2

◆2l

on the domain 0 x 30.

(a) Do the solutions oscillate?

(b) What happens to the roots as n increases?

(c) Compare the plots with the functionr

2⇡x

. What do you observe?

As you see in the previous exercise (and as predicted by the Sturm-Liouville theo-rem), the Bessel function Jn (x) has an infinite list of roots; we denote these rootsby !n1,!n2,!n3, . . . so that !nk is the kth root of the nth order Bessel functionJn (x). Many software packages can easily find (or have as built-in functions) theroots !nk .

We now return to the problem that motivated all this Bessel stu� – finding ourstanding wave eigenfunctions. Recall that R(r) = R(!x) where! was a root of thefunction that solved (14.22). Since we now know that the solutions to (14.22) arethe Bessel functions, and have a list of all their roots, we know that

Rnk (r) = Jn (!nkr).


Using this, we see that our standing wave eigenfunctions are

cnk (r, ✓) = Jn (!nkr) cos(n✓) and s

nk (r, ✓) = Jn (!nkr) sin(n✓).

Exercise 14.12.

1. What does a typical plot of Jn (!nkr) look like? What does a typical plot ofcos(nr) look like?

2. Use the previous part to figure out what the contour plots of cnk

(r, ✓) are.What does the number n indicate? What does the number k indicate?

3. Get a piece of technology to either plot the function cnk

(r, ✓), or plot itscontours, 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 longerhomework assignment.

In this exploration you consider the wave equation on the unit ball subject toDirichlet boundary conditions. Be warned: There are a lot of functions and changeof variables running around this exploration – it is imperative that you keep yourwork organized!

Working in spherical coordinates (r, ✓, �), the unit ball is described by

0 r 1, �⇡ ✓ ⇡, and 0 � ⇡.

We require periodic boundary conditions at ✓ = ±⇡, and that functions are finite atr = 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


that this implies that

r

2 d

2R

dr

2 + 2r

dR

dr

+ (!2r

2 + µ)R = 0 (14.25) 3D-R

and1

sin2 �

@2Y

@✓2 +1

sin �@

@�

"sin �@Y

@�

#= µY (14.26) 3D-Y

for some constant µ.

3. It is possible to show that µ < 0 using an integration-by-parts argument. Todo this, multiply the Y equation by Y and then integrate over the sphere usingthe spherical area element dA = sin � d✓ d�. Integrate by parts in order toshow 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 thatY (✓, �) = ⇥(✓)�(�) and show that the functions ⇥ and � must satisfy

d

2⇥

d✓2 = ⌫⇥ (14.27) 3D-Theta

sin �d

d�

"sin �

d�

d�

#= (µ sin2 � � ⌫)� (14.28) 3D-Phi

for some constant ⌫.

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 thisinformation in hand, you will turn to (14.28). The functions � and ⇥ are tobe used in constructing the functions Y . Finally, you will address (14.25).

5. Explain why the function⇥must satisfy periodic boundary conditions on theinterval �⇡ ✓ ⇡, 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 � inorder to obtain

d

dx

"(1 � x

2)d�

dx

#� m

2

1 � x

2� = µ�, (14.29) ass-Legendre-ode


which holds for �1 < x < 1. Explain why, for each fixed integer m, theequation (14.29) satisfies the hypotheses of the Sturm-Liouville theoremwith respect to the inner product

h f , gi =Z 1

�1f (x) g(x) dx.

Conclude that for each fixed m there exists a sequence of negative eigenvaluesand corresponding orthogonal eigenfunctions satisfying (14.29).

The eigenfunctions are called associated Legendre functions and are giventhe symbol Pl,m (x); the corresponding eigenvalues are µl,m = �l (l + 1).Verify that P0,0(x) = 1 and that

P1,0(x) = x P2,0(x) =32

x

2 � 12

P1,1(x) =p

1 � x

2P2,1(x) = 3x

p1 � x

2

P2,2(x) = 3(1 � x

2) P3,2(x) = 15x(1 � x

2)

satisfy (14.29). (It turns out that one must have l � m in order to havenonzero Pl,m .)

7. Conclude from the above that for each m and l � m there exists �l,m (�) =Pl,m (cos �) satisfying (14.28) with ⌫ = �m

2 and µ = �l (l + 1). Use this toconstruct functions Yl,m (✓, �) satisfying (14.26). These functions are calledthe spherical harmonics. List the first few spherical harmonics. Draw, asbest 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 thechange of variable x = !r in order to obtain

x

2 d

2R

dx

2 + 2x

dR

dx

+ (x

2 � l (l + 1))R = 0.

Then make the change of variables R(x) = 1px

S(x). Show that S(x) mustsatisfy

x

2 d

2S

dx

2 + x

dS

dx

+

x

2 �"l (l + 1) +

12

#!S = 0, (14.30) half-Bessel


which is Bessel’s equation with order l (l + 1) + 12 . The solutions are called

Bessel functions and are given the symbol Jl (l+1)+ 12(x).

9. It is straightforward to find power series expressions for Jl (l+1)+ 12(x), follow-

ing the procedure used in the previous section. At this stage, however, youshould simply verify the following:

J1/2(x) =

r2⇡x

sin(x)

J5/2(x) =

r2⇡x

3 � x

2

x

2 sin(x) � 3x

cos(x)!

10. Each Bessel function Jl (l+1)+ 12

has an infinite number of roots!l,k . Use theseroots to construct solutions Rl,k (r) satisfying (14.25) with µ = �l (l + 1) and! = !l,k . Sketch Rl,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 Rl,k and Yl,m obtained above. Give the explicit form forthe first few wave shapes. How should one understand these shapes geomet-rically 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 younotice?

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