5 Eigenvalue problems

For example for the insulated wire, Dirichlet conditions correspond to applying a zero temperature at the ends, Neumann means insulating the ends, etc…. Other types of endpoint conditions also arise naturally, such as

These problems came up, for example, in the study of the heat equation $ut = kuxx$ when we were trying to solve the equation by the method of separation of variables. During the process we encountered a certain eigenvalue problem and found the eigenfunctions $Xn(x)$ . We then found the eigenfunction decomposition of the initial temperature $f(x) = u (x,0 )$ in terms of the eigenfunctions

Once we had this decomposition and once we found suitable $Tn(t)$ such that $Tn(0) = 1$ , we noted that a solution to the original problem could be written as

We will try to solve more general problems using this method. We will study second order linear equations of the form

Essentially any second order linear equation of the form $′′ ′ a(x)y + b (x)y + c(x)y + λd (x)y = 0$ can be written as (5.1) after multiplying by a proper factor.

Example 5.1.1 (Bessel):

( ) x2y′′ + xy′ + λx2 - n 2 y = 0.

Multiply both sides by $1 x$ to obtain

1 ( ( ) ) ( n 2) d ( dy ) n2 0 = -- x2y′′ + xy′ + λx2 - n2 y xy′′ + y′ + λx ---- y = --- x--- - --y + λxy. x x dx dx x

We can state the general Sturm-Liouville problem . We seek nontrivial solutions to

In particular, we seek $λ$ s that allow for nontrivial solutions. The $λ$ s for which there is no nontrivial solution are called the eigenvalues and the corresponding nontrivial solutions are called eigenfunctions. Obviously $α 1$ and $α2$ should not be both zero, same for $β1$ and $β2$ .

Theorem 5.1.1. Suppose $p (x)$ , $p′(x )$ , $q(x )$ and $r(x)$ are continuous on $[a,b]$ and suppose $p(x) > 0$ and $r(x) > 0$ for all $x$ in $[a,b]$ . Then the Sturm-Liouville problem (5.2) has an increasing sequence of eigenvalues

λ1 < λ2 < λ3 < ⋅⋅⋅

such that

lin→m∞λn = + ∞

and such that to each $λn$ there is (up to a constant multiple) a single eigenfunction $yn(x)$ .

Moreover, if $q(x) ≥ 0$ and $α1,α2,β1,β2 ≥ 0$ , then $λn ≥ 0$ for all $n$ .

Note: Be careful about the signs. Also be careful about the inequalities for $r$ and $p$ , they must be strict for all $x$ ! Problems satisfying the hypothesis of the theorem are called regular Sturm-Liouville problems and we will only consider such problems here. That is, a regular problem is one where $p (x)$ , $p′(x )$ , $q(x)$ and $r(x)$ are continuous, $p (x) > 0$ , $r(x) > 0$ , $q(x) ≥ 0$ , and $α 1,α 2,β1,β2 ≥ 0$ .

When zero is an eigenvalue, we will usually start labeling the eigenvalues at 0 rather than 1 for convenience.

Example 5.1.2: The problem $y′′ + λy$ , $0 < x < L$ , $y(0) = 0$ , and $y(L ) = 0$ is a regular Sturm-Liouville problem. $p(x) = 1$ , $q(x) = 0$ , $r(x) = 1$ , and we have $p (x) = 1 > 0$ and $r(x) = 1 > 0$ . The eigenvalues are $λ = n2π2 n L2$ and eigenfunctions are $y (x) = sin(nπx) n L$ . All eigenvalues are nonnegative as predicted by the theorem.

Exercise 5.1.1: Find eigenvalues and eigenfunctions for

′′ ′ ′ y + λy = 0, y (0) = 0, y (1) = 0.

Identify the $p,q, r,αj,βj$ . Can you use the theorem to make the search for eigenvalues easier?

Example 5.1.3: Find eigenvalues and eigenfunctions of the problem

y′′ + λy = 0, 0 < x < 1 ′ ′ hy(0) - y (0 ) = 0, y (1) = 0, h > 0.

These equations give a regular Sturm-Liouville problem.

Exercise 5.1.2: Identify $p,q,r,αj,βj$ in the example above.

First note that $λ ≥ 0$ by Theorem 5.1.1. Therefore, the general solution (without boundary conditions) is

√ -- √ -- y(x) = A cos λx + B sin λx if λ > 0, y(x) = Ax + B if λ = 0.

Let us see if $λ = 0$ is an eigenvalue: We must satisfy $0 = hB - A$ and $A = 0$ , hence $B = 0$ (as $h > 0$ ), therefore, 0 is not an eigenvalue (no eigenfunction).

Now let us try $λ > 0$ . We plug in the boundary conditions.

√ -- 0 = hA - λB, √ -- √-- √-- √-- 0 = -A λsin λ + B λ cos λ.

Note that if $A = 0$ , then $B = 0$ and vice-versa, hence both are nonzero. So $B = h√Aλ$ , and $0 = - A√ λ-sin √ λ + h√A√ λc os √ λ λ$ . As $A ⇔ 0$ we get

√ -- √-- √ -- 0 = - λsin λ + hcos λ,

-h-- √-- √--= tan λ. λ

Now use a computer to find $λn$ . There are tables available, though using a computer or a graphing calculator will probably be far more convenient nowadays. Easiest method is to plot the functions $h∕x$ and $tan x$ and see for which $x$ they intersect. There will be an infinite number of intersections. So denote by $--- √λ 1$ the first intersection, by $--- √λ 2$ the second intersection, etc…. For example, when $h = 1$ , we get that $λ ≈ 0.86 1$ , and $λ ≈ 3.4 3 2$ . A plot for $h = 1$ is given in Figure 5.1. The appropriate eigenfunction (let $A = 1$ for convenience, then $h √- B = ∕ λ$ ) is

∘--- --h-- ∘ --- yn(x) = cos λnx + √ λn-sin λnx.

Figure 5.1: Plot of $1x$ and $tanx$ .

5.1.2 Orthogonality

We have seen the notion of orthogonality before. For example, we have shown that $sin nx$ are orthogonal for distinct $n$ on $[0,π ]$ . For general Sturm-Liouville problems we will need a more general setup. Let $r(x)$ be a weight function (any function, though generally we will assume it is positive) on $[a,b ]$ . Then two functions $f (x )$ , $g(x)$ are said to be orthogonal with respect to the weight function $r(x)$ when

and then say $f$ and $g$ are orthogonal whenever $⟨f,g⟩ = 0$ . The results and concepts are again analogous to finite dimensional linear algebra.

The idea of the given inner product is that those $x$ where $r(x)$ is greater have more weight. Nontrivial (nonconstant) $r(x)$ arise naturally, for example from a change of variables. Hence, you could think of a change of variables such that $dξ = r(x) dx$ .

We have the following orthogonality property of eigenfunctions of a regular Sturm-Liouville problem.

Theorem 5.1.2. Suppose we have a regular Sturm-Liouville problem

( ) d-- p(x)dy- - q(x)y + λr(x)y = 0, dx dx α1y(a) - α2y′(a) = 0, ′ β1y(b ) + β2y (b) = 0.

Let $yj$ and $yk$ be two distinct eigenfunctions for two distinct eigenvalues $λj$ and $λk$ . Then

∫ b yj(x)yk(x)r(x) dx = 0, a

that is, $yj$ and $yk$ are orthogonal with respect to the weight function $r$ .

Proof is very similar to the analogous theorem from § 4.1. It can also be found in many books including, for example, Edwards and Penney [EP].

5.1.3 Fredholm alternative

We also have the Fredholm alternative theorem we talked about before for all regular Sturm-Liouville problems. We state it here for completeness.

Theorem 5.1.3 (Fredholm alternative). Suppose that we have a regular Sturm-Liouville problem. Then either

( ) d-- dy- dx p(x)dx - q(x)y + λr(x)y = 0, ′ α1y(a) - α2y (a) = 0, β1y(b ) + β2y′(b) = 0,

has a nonzero solution, or

( ) d-- dy- dx p (x )dx - q (x)y + λr(x)y = f (x), ′ α 1y(a) - α2y (a) = 0, β1y(b) + β 2y′(b) = 0,

has a unique solution for any $f(x)$ continuous on $[a,b]$ .

This theorem is used in much the same way as we did before in § 4.4. It is used when solving more general nonhomogeneous boundary value problems. The theorem does not help us solve the problem, but it tells us when does a solution exist and if it exists if it is unique, so that we know when to spend time looking for a solution. To solve the problem we decompose $f(x)$ and $y(x)$ in terms of the eigenfunctions of the homogeneous problem, and then solve for the coefficients of the series for $y(x)$ .

5.1.4 Eigenfunction series

What we want to do with the eigenfunctions once we have them is to compute the eigenfunction decomposition of an arbitrary function $f(x)$ . That is, we wish to write

where $yn(x)$ the eigenfunctions. We wish to find out if we can represent any function $f(x)$ in this way, and if so, we wish to calculate $cn$ (and of course we would want to know if the sum converges). OK, so imagine we could write $f (x)$ as (5.3). We will assume convergence and the ability to integrate the series term by term. Because of orthogonality we have

Note that $ym$ are known up to a constant multiple, so we could have picked a scalar multiple of an eigenfunction such that $⟨ym,ym⟩ = 1$ (if we had an arbitrary eigenfunction $~ym$ , divide it by $∘ -------- ⟨~ym,~ym⟩$ ). In the case that $⟨ym, ym⟩ = 1$ we would have the simpler form $cm = ⟨f,ym⟩$ as we essentially did for the Fourier series. The following theorem holds more generally, but the statement given is enough for our purposes.

Theorem 5.1.4. Suppose $f$ is a piecewise smooth continuous function on $[a,b]$ . If $y1,y2,...$ are the eigenfunctions of a regular Sturm-Liouville problem, then there exist real constants $c1,c2,...$ given by (5.4) such that (5.3) converges and holds for $a < x < b$ .

Example 5.1.4: Take the simple Sturm-Liouville problem

π y′′ + λy = 0, 0 < x < -, ( ) 2 y(0 ) = 0, y′ π- = 0. 2

The above is a regular problem and furthermore we actually know by Theorem 5.1.1 that $λ ≥ 0$ .

Suppose $λ = 0$ , then the general solution is $y(x) = Ax + B$ , we plug in the initial conditions to get $0 = y(0) = B$ , and $π 0 = y′(2) = A$ , hence $λ = 0$ is not an eigenvalue.

The general solution, therefore, is

√ -- √ -- y(x) = A cos λx + B sin λx.

Plugging in the boundary conditions we get $0 = y(0) = A$ and $√ -- √-- 0 = y′(π) = λB cos λ π 2 2$ . $B$ cannot be zero and hence $√--π cos λ 2 = 0$ . This means that $√--π λ 2$ must be an odd integral multiple of $π 2$ , i.e. $√ --- (2n - 1)π2 = λnπ2$ . Hence

λn = (2n - 1)2.

We can take $B = 1$ . And hence our eigenfunctions are

yn(x) = sin (2n - 1)x.

We finally compute

∫ π 2 ( )2 π f (x )sin (2n - 1)x dx = -. 0 4

So any piecewise smooth function on $[0,π2]$ can be written as

∞∑ f(x) = cnsin (2n - 1)x, n=1

where

∫ π ⟨f,yn⟩ 02f (x) sin (2n - 1)x dx 4∫ π2 cn = -------= -∫-π-(------------)2----= -- f (x) sin (2n - 1)x dx. ⟨yn,yn⟩ 02 sin (2n - 1)x dx π 0

Note that the series converges to an odd $2π$ -periodic (not $π$ -periodic!) extension of $f(x)$ .

Exercise 5.1.3 (challenging): In the above example, the function is defined on $π 0 < x < 2$ , yet the series converges to an odd $2 π$ -periodic extension of $f(x)$ . find out how the extension is defined for $π< x < π 2$ .

5.1.5 Exercises

Exercise 5.1.4: Find eigenvalues and eigenfunctions of

′′ ′ y + λy = 0, y(0 ) - y(0) = 0, y(1) = 0.

Exercise 5.1.5: Expand the function $f (x) = x$ on $0 ≤ x ≤ 1$ using the eigenfunctions of the system

y′′ + λy = 0, y′(0) = 0, y(1) = 0.

Exercise 5.1.6: Suppose that you had a Sturm-Liouville problem on the interval $[0,1]$ and came up with $yn(x) = sin γnx$ , where $γ > 0$ is some constant. Decompose $f(x) = x$ , $0 < x < 1$ in terms of these eigenfunctions.

Exercise 5.1.7: Find eigenvalues and eigenfunctions of

y(4) + λy = 0, y(0) = 0, y′(0) = 0, y(1) = 0, y′(1 ) = 0.

This problem is not a Sturm-Liouville problem, but the idea is the same.

Exercise 5.1.8 (more challenging): Find eigenvalues and eigenfunctions for

d --(exy′) + λexy = 0, y(0) = 0, y(1 ) = 0. dx

Hint: First write the system as a constant coefficient system to find general solutions. Do note that Theorem 5.1.1 guarantees $λ ≥ 0$ .

5.2 Application of eigenfunction series

The eigenfunction series can arise even from higher order equations. Suppose we have an elastic beam (say made of steel). We will study the transversal vibrations of the beam. That is, assume the beam lies along the $x$ axis and let $y(x,t)$ measure the displacement of the point $x$ on the beam at time $t$ . See Figure 5.2.

Suppose the beam is of length 1 simply supported (hinged) at the ends. Suppose the beam is displaced by some function $f(x)$ at time $t = 0$ and then let go (initial velocity is 0). Then $y$ satisfies:

Again we try $y(x,t) = X (x )T (t)$ and plug in to get $a 4X (4)T + XT ′′ = 0$ or

We note that we want $T ′′ + λa4T = 0$ . Let us assume that $λ > 0$ . We can argue that we expect vibration and not exponential growth nor decay in the $t$ direction (there is no friction in our model for instance). Similarly $λ = 0$ will not occur.

Exercise 5.2.1: Try to justify $λ > 0$ just from the equations.

Write $4 ω = λ$ , so that we do not need to write the fourth root all the time. For $X$ we get the equation $(4) 4 X - ω X = 0$ . The general solution is

Now $0 = X (0 ) = A + B + D$ , $0 = X′′(0) = ω2(A + B - D )$ . Hence, $D = 0$ and $A + B = 0$ , or $B = -A$ . So we have

Now $0 = X (1 ) = A(eω - e-ω) + C sinω$ , and $0 = X′′(1) = Aω 2(eω - e-ω) - Cω 2sin ω$ . This means that $C sinω = 0$ and $ω -ω A (e - e ) = A2 sinh ω = 0$ . If $ω > 0$ , then $sinhω ⇔ 0$ and so $A = 0$ . This means that $C ⇔ 0$ else we do not have an eigenvalue. Also $ω$ must be an integer multiple of $π$ hence $ω = nπ$ and $n ≥ 1$ (as $ω > 0$ ). We can take $C = 1$ . So the eigenvalues are $λn = n 4π 4$ and the eigenfunctions are $sin nπx$ .

Now $T′′ + n4π4a4T = 0$ . The general solution is $T(t) = A sinn2π2a2t + B cos n2π2a2t$ . But $T ′(0 ) = 0$ and hence we must have $A = 0$ and we can take $B = 1$ to make $T (0 ) = 1$ for convenience. So our solutions are $Tn (t) = cosn2π2a2t$ .

As the eigenfunctions are just sines again, we can decompose the function $f(x)$ on $0 < x < 1$ using the sine series. We find numbers $bn$ such that for $0 < x < 1$ we have

The point is that $XnTn$ is a solution that satisfies all the homogeneous conditions (that is, all conditions except the initial position). And since and $Tn(0) = 1$ , we have

Note that the natural (circular) frequency of the system is $n 2π 2a 2$ . These frequencies are all integer multiples of the fundamental frequency $2 2 π a$ , so we will get a nice musical note. The exact frequencies and their amplitude are what we call the timbre of the note.

The timbre of a beam is different than for a vibrating string where we will get “more” of the smaller frequencies since we will get all integer multiples, $1,2,3,4,5,...$ For a steel beam we will get only the square multiples $1,4,9,16,2 5,...$ That is why when you hit a steel beam you hear a very pure sound. The sound of a xylophone or vibraphone is, therefore, very different from a guitar or piano.

Example 5.2.1: Let us assume that $f (x ) = x(x-1) 10$ . On $0 < x < 1$ we have (you know how to do this by now)

∞ ∑ ---4-- f(x) = 5 π3n3 sinnπx. nn=od1d

Hence, the solution to (5.5) with the given initial position $f (x)$ is

∑∞ y(x,t) = --4---(sin nπx)(cos n2π2a2t) . 5π3n3 nn=o1dd

5.2.1 Exercises

Exercise 5.2.2: Suppose you have a beam of length 5 with free ends. Let $y$ be the transverse deviation of the beam at position $x$ on the beam ( $0 < x < 5$ ). You know that the constants are such that this satisfies the equation $y + 4y = 0 tt xxxx$ . Suppose you know that the initial shape of the beam is the graph of $x(5 - x)$ , and the initial velocity is uniformly equal to 2 (same for each $x$ ) in the positive $y$ direction. Set up the equation together with the boundary and initial conditions. Just set up, do not solve.

Exercise 5.2.3: Suppose you have a beam of length 5 with one end free and one end fixed (the fixed end is at $x = 5$ ). Let $u$ be the longitudinal deviation of the beam at position $x$ on the beam ( $0 < x < 5$ ). You know that the constants are such that this satisfies the equation $utt = 4uxx$ . Suppose you know that the initial displacement of the beam is $x-5 50$ , and the initial velocity is $-(x-5) 100$ in the positive $u$ direction. Set up the equation together with the boundary and initial conditions. Just set up, do not solve.

Exercise 5.2.4: Suppose the beam is $L$ units long, everything else kept the same as in (5.5). What is the equation and the series solution.

Exercise 5.2.5: Suppose you have

4 a yxxxx + ytt = 0 (0 < x < 1,t > 0), y(0,t) = yxx(0,t) = 0, y(1,t) = yxx(1,t) = 0, y(x,0) = f (x), yt(x,0) = g(x).

That is, you have also an initial velocity. Find a series solution. Hint: Use the same idea as we did for the wave equation.

5.3 Steady periodic solutions

5.3.1 Forced vibrating string.

Suppose that we have a guitar string of length $L$ . We have studied the wave equation problem in this case, where $x$ was the position on the string, $t$ was time and $y$ was the displacement of the string. See Figure 5.3.

where $An$ and $Bn$ were determined by the initial conditions. The natural frequencies of the system are the (circular) frequencies $nπa L$ for integers $n ≥ 1$ .

But these are free vibrations. What if there is an external force acting on the string. Let us assume say air vibrations (noise), for example a second string. Or perhaps a jet engine. For simplicity, assume nice pure sound and assume the force is uniform at every position on the string. Let us say $F (t) = F 0cosωt$ as force per unit mass. Then our wave equation becomes (remember acceleration is force times mass)

That is, the string is initially at rest. First we find a particular solution $yp$ of (5.7) that satisfies $y(0,t) = y(L,t) = 0$ . We define the functions $f$ and $g$ as

We then find solution $yc$ of (5.6). If we add the two solutions, we find that $y = yc + yp$ solves (5.7) with the initial conditions.

Exercise 5.3.1: Check that $y = y + y c p$ solves (5.7) and the side conditions (5.8).

So the big issue here is to find the particular solution $yp$ . We look at the equation and we make an educated guess

or $- ω 2X = a2X′′ + F 0$ after canceling the cosine. We know how to find a general solution to this equation (it is an nonhomogeneous constant coefficient equation) and we get that the general solution is

Exercise 5.3.2: Check that $yp$ works.

Now we get to the point that we skipped. Suppose that $sin ωL = 0 a$ . What this means is that $ω$ is equal to one of the natural frequencies of the system, i.e. a multiple of $πa L$ . We notice that if $ω$ is not equal to a multiple of the base frequency, but is very close, then the coefficient $B$ in (5.9) seems to become very large. But let us not jump to conclusions just yet. When $ω = nπa- L$ for $n$ even, then $cos ωL = 1 a$ and hence we really get that $B = 0$ . So resonance occurs only when both $cos ωL = - 1 a$ and $sin ωL = 0 a$ . That is when $ω = nπa L$ for odd $n$ .

We could again solve for the resonance solution if we wanted to, but it is, in the right sense, the limit of the solutions as $ω$ gets close to a resonance frequency. In real life, pure resonance never occurs anyway.

The above calculation explains why a string will begin to vibrate if the identical string is plucked close by. In the absence of friction this vibration would get louder and louder as time goes on. On the other hand, you are unlikely to get large vibration if the forcing frequency is not close to a resonance frequency even if you have a jet engine running close to the string. That is, the amplitude will not keep increasing unless you tune to just the right frequency.

Similar resonance phenomena occur when you break a wine glass using human voice (yes this is possible, but not easy ) if you happen to hit just the right frequency. Remember a glass has much purer sound, i.e. it is more like a vibraphone, so there are far fewer resonance frequencies to hit.

When the forcing function is more complicated, you decompose it in terms of the Fourier series and apply the above result. You may also need to solve the above problem if the forcing function is a sine rather than a cosine, but if you think about it, the solution is almost the same.

Example 5.3.1: Let us do the computation for specific values. Suppose $F 0 = 1$ and $ω = 1$ and $L = 1$ and $a = 1$ . Then

( ) cos1---1- yp(x,t) = co sx - sin 1 sin x - 1 cost.

Call $cos1-1 B = sin1$ for simplicity.

Then plug in $t = 0$ to get

f(x) = -yp(x,0) = - cos x + Bsinx + 1,

and after differentiating in $t$ we see that $∂yp g(x) = - ∂t (x,0) = 0$ .

Hence to find $yc$ we need to solve the problem

ytt = yxx, y(0, t) = 0, y(1,t) = 0, y(x, 0) = - cosx + B sin x + 1, yt(x,0) = 0.

Note that the formula that we use to define $y(x, 0)$ is not odd, hence it is not a simple matter of plugging in to apply the D’Alembert formula directly! You must define $F$ to be the odd, 2-periodic extension of $y(x,0)$ . Then our solution would look like

F (x + t) + F (x - t) ( co s1 - 1 ) y(x,t) =------------------+ cosx - ---------sinx - 1 co st. 2 sin 1

(5.10)

Figure 5.4: Plot of $F(x+t)+F(x-t) ( cos1-1 ) y(x,t) = 2 + cosx - sin1 sin x - 1 cost$ .

It is not hard to compute specific values for an odd extension of a function and hence (5.10) is a wonderful solution to the problem. For example it is very easy to have a computer do it, unlike a series solution. A plot is given in Figure 5.4.

5.3.2 Underground temperature oscillations

Let $u(x,t)$ be the temperature at a certain location at depth $x$ underground at time $t$ . See Figure 5.5.

The temperature $u$ satisfies the heat equation $ut = kuxx$ , where $k$ is the diffusivity of the soil. We know the temperature at the surface $u(0, t)$ from weather records. Let us assume for simplicity that

For some base temperature $T0$ , then $t = 0$ is midsummer (could put negative sign above to make it midwinter). $A0$ is picked properly to make this the typical variation for the year. That is, the hottest temperature is $T0 + A0$ and the coldest is $T 0 - A0$ . For simplicity, we will assume that $T0 = 0$ . $ω$ is picked depending on the units of $t$ , such that when $t = 1year$ then $ωt = 2π$ .

It seems reasonable that the temperature at depth $x$ will also oscillate with the same frequency. And this in fact will be the steady periodic solution, independent of the initial conditions. So we are looking for a solution of the form

We will employ the complex exponential here to make calculations simpler. Suppose we have a complex valued function

We will look for an $h$ such that $R eh = u$ . To find an $h$ , whose real part satisfies (5.11), we look for an $h$ such that

Exercise 5.3.3: Suppose $h$ satisfies (5.12). Use Euler’s formula for the complex exponential to check that $u = Re h$ satisfies (5.11).

where $∘ -- α = ± iω k$ . Note that $√ - ± i = ± 1√+i 2$ so you could simplify to $∘ -- α = ±(1 + i) -ω 2k$ . Hence the general solution is

We assume that an $X(x)$ that solves the problem must be bounded as $x → ∞$ since $u(x,t)$ should be bounded (we are not worrying about the earth core!). If you use Euler’s formula to expand the complex exponentials, you will note that the second term will be unbounded (if $B ⇔ 0$ ), while the first term is always bounded. Hence $B = 0$ .

Exercise 5.3.4: Use Euler’s formula to show that $√ -- e(1+i) ω2kx$ will be unbounded as $x → ∞$ , while $-(1+i)√-ωx e 2k$ will be bounded as $x → ∞$ .

Furthermore, $X (0 ) = A 0$ since $h (0,t) = A eiωt 0$ . Thus $A = A 0$ . This means that

Notice the phase is different at different depths. At depth $x$ the phase is delayed by $-- x ∘ ω- 2k$ . For example in $cgs$ units (centimeters, grams, seconds) we have $k = 0.005$ (typical value for soil), $----2π----- ---2π-- -7 ω = secondsin ayear = 31,557,341 ≈ 1.99 × 10$ . Then if we compute where the phase shift $∘ ω- x 2k = π$ we find the depth in centimeters where the seasons are reversed. That is, we get the depth at which summer is the coldest and winter is the warmest. We get approximately 700 centimeters which is approximately 23 feet below ground.

Be careful not to jump to conclusions. The temperature swings decay rapidly as you dig deeper. The amplitude of the temperature swings is $-√-ωx A 0e 2k$ . This decays very quickly as $x$ grows. Let us again take typical parameters as above. We also will assume that our surface temperature swing is $± 15∘$ Celsius, that is, $A0 = 15$ . Then the maximum temperature variation at 700 centimeters is only $± 0.66∘$ Celsius.

You need not dig very deep to get an effective “refrigerator.” That is why wines are kept in a cellar; you need consistent temperature. The temperature differential could also be used for energy. A home could be heated or cooled by taking advantage of the above fact. Even without the earth core you could heat a home in the winter and cool it in the summer. There is also the earth core, so temperature presumably gets higher the deeper you dig. We did not take that into account above.