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4.5 Applications of Fourier series

Note: 2 lectures, §9.4 in [EP], not in [BD]

4.5.1 Periodically forced oscillation

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Let us return to the forced oscillations. Consider a mass-spring system as before, where we have a mass m on a spring with spring constant k , with damping c , and a force F (t) applied to the mass. Suppose the forcing function F (t) is 2L -periodic for some L > 0 . We have already seen this problem in chapter 2 with a simple F(t) . The equation that governs this particular setup is

mx ′′(t) + cx ′(t) + kx(t) = F (t).
(4.9)

The general solution consists of the complementary solution xc , which solves the associated homogeneous equation mx ′′ + cx′ + kx = 0 , and a particular solution of (4.9) we call xp . For c > 0 , the complementary solution xc will decay as time goes by. Therefore, we are mostly interested in a particular solution x p that does not decay and is periodic with the same period as F (t) . We call this particular solution the steady periodic solution and we write it as xsp as before. What will be new in this section is that we consider an arbitrary forcing function F (t) instead of a simple cosine.

For simplicity, let us suppose that c = 0 . The problem with c > 0 is very similar. The equation

mx ′′ + kx = 0

has the general solution

x(t) = A cos(ω 0t) + B sin(ω 0t),

where  ∘ -- ω = k- 0 m . Any solution to mx ′′(t) + kx(t) = F (t) is of the form A cos(ω t) + B sin(ω t) + x 0 0 sp . The steady periodic solution xsp has the same period as F (t) .

In the spirit of the last section and the idea of undetermined coefficients we first write

 ∑∞ ( ) ( ) F(t) = c0+ cn cos nπ-t + dnsin nπ-t . 2 n=1 L L

Then we write a proposed steady periodic solution x as

 a0 ∑∞ (nπ ) (nπ ) x(t) = ---+ an cos ---t + bnsin ---t , 2 n=1 L L

where an and bn are unknowns. We plug x into the differential equation and solve for an and bn in terms of c n and d n . This process is perhaps best understood by example.

Example 4.5.1: Suppose that k = 2 , and m = 1 . The units are again the mks units (meters-kilograms-seconds). There is a jetpack strapped to the mass, which fires with a force of 1 newton for 1 second and then is off for 1 second, and so on. We want to find the steady periodic solution.

The equation is, therefore,

x′′ + 2x = F (t),

where F (t) is the step function

 ( ||{0 if −1 < t < 0, F(t) = ||(1 if 0 < t < 1,

extended periodically. We write

 c ∑∞ F (t) = -0+ cncos(nπt) + dnsin(nπt). 2 n=1

We compute

 ∫ 1 ∫ 1 cn = F(t)cos(nπt) dt = co s(nπt) dt = 0 for n ≥ 1, −1 0 ∫ 1 ∫ 1 c 0 = F(t) dt = dt = 1, ∫ −1 0 1 dn = F(t)sin(n πt) dt ∫ −11 = 0 sin(nπt) dt [ ]1 = −-cos(nπt) nπ t=0 n (| 2 1-−-(−1)- |{πn if n odd, = πn = ||(0 if n even.
So
 ∑∞ 1- 2-- F(t) = 2 + πn sin(nπt). nn=od1d

We want to try

 a0- ∑∞ x(t) = 2 + an cos(nπt) + bnsin(nπt). n=1

Once we plug x into the differential equation x′′ + 2x = F (t) , it is clear that an = 0 for n ≥ 1 as there are no corresponding terms in the series for F(t) . Similarly b = 0 n for n even. Hence we try

 a ∑∞ x(t) =--0+ bnsin(nπt). 2 n=1 nodd

We plug into the differential equation and obtain

 ∑∞ [ ] ∞∑ [ ] x′′ + 2x = − b n2π2sin (n πt) + a + 2 b sin(n πt) n=1 n 0 n=1 n nodd n odd ∞∑ = a0 + bn(2 − n2π2)sin(nπt) n=1 nodd 1 ∑∞ 2 = F(t) = -+ ---sin(nπt). 2 n=1πn n odd

So a0 = 1 2 , bn = 0 for even n , and for odd n we get

 2 bn = --------2-2-. πn(2 − n π )

The steady periodic solution has the Fourier series

 1 ∑∞ 2 xsp(t) = --+ -------2-2--sin(nπt). 4 n=1 πn(2 − n π ) n odd

We know this is the steady periodic solution as it contains no terms of the complementary solution and it is periodic with the same period as F(t) itself. See Figure 4.12 for the plot of this solution.


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Figure 4.12: Plot of the steady periodic solution xsp of Example 4.5.1.


4.5.2 Resonance

Just like when the forcing function was a simple cosine, resonance could still happen. Let us assume c = 0 and we will discuss only pure resonance. Again, take the equation

mx′′(t) + kx(t) = F (t).

When we expand F(t) and find that some of its terms coincide with the complementary solution to  ′′ mx + kx = 0 , we cannot use those terms in the guess. Just like before, they will disappear when we plug into the left hand side and we will get a contradictory equation (such as 0 = 1 ). That is, suppose

xc = A cos(ω0t) + B sin (ω 0t),

where  Nπ ω 0 = L for some positive integer N . In this case we have to modify our guess and try

 ( ( ) ( )) ∑∞ ( ) ( ) x(t) = a0+ t aN cos N-π t + bN sin N-π t + ancos nπt + bn sin nπt . 2 L L n=1 L L n⇔N

In other words, we multiply the offending term by t . From then on, we proceed as before.

Of course, the solution will not be a Fourier series (it will not even be periodic) since it contains these terms multiplied by t . Further, the terms  ( ( ) ( )) t aN cos Nπt + bN sin Nπt L L will eventually dominate and lead to wild oscillations. As before, this behavior is called pure resonance or just resonance.

Note that there now may be infinitely many resonance frequencies to hit. That is, as we change the frequency of F (we change L ), different terms from the Fourier series of F may interfere with the complementary solution and will cause resonance. However, we should note that since everything is an approximation and in particular c is never actually zero but something very close to zero, only the first few resonance frequencies will matter.

Example 4.5.2: We want to solve the equation

2x ′′ + 18 π2x = F (t),

where

 (| F (t) = |{ −1 if −1 < t < 0, ||( 1 if 0 < t < 1,

extended periodically. We note that

 ∞ ∑ -4- F (t) = πn sin(nπt). nn=od1d

Exercise 4.5.1: Compute the Fourier series of F to verify the above equation.

The solution must look like

x(t) = c1 cos(3 πt) + c2sin (3πt) + xp(t)

for some particular solution xp .

We note that if we just tried a Fourier series with sin(n πt) as usual, we would get duplication when n = 3 . Therefore, we pull out that term and multiply by t . We also have to add a cosine term to get everything right. That is, we must try

 ∑∞ xp(t) = a3tcos(3πt) + b3tsin(3πt) + bnsin(nπt). nn= o1dd n⇔3

Let us compute the second derivative.

x′′p(t) = −6a 3πsin(3πt) − 9π2a3tcos(3πt) + 6b3π cos(3πt) − 9π2b3tsin(3πt) ∑∞ + (−n 2π 2b ) sin(nπt). n nn= o1dd n⇔3
We now plug into the left hand side of the differential equation.
 ′′ 2 2 2 2xp + 1 8π x = − 1 2a3πsin(3πt) − 18π a3tco s(3πt) + 1 2b3πcos(3πt) − 1 8π b3tsin
If we simplify we obtain
 ′′ 2 ∑∞ 2 2 2 2xp + 18π x = −12a 3πsin(3πt) + 12b3π cos(3 πt) + (− 2n π bn + 18π bn)sin(nπt). nn= o1dd n⇔3

This series has to equal to the series for F (t) . We equate the coefficients and solve for a3 and bn .

 4∕(3π) − 1 a3 = -------= --2, − 12π 9π b3 = 0, 4 2 bn = ------2-----2-2-= -3-------2- for n odd and n ⇔ 3. nπ(18π − 2n π ) π n(9 − n )

That is,

 ∑∞ xp(t) =-−1-tcos(3πt) + -----2-----sin(nπt). 9 π2 n=1 π3n(9 − n2) n on⇔d3d

When c > 0 , you will not have to worry about pure resonance. That is, there will never be any conflicts and you do not need to multiply any terms by t . There is a corresponding concept of practical resonance and it is very similar to the ideas we already explored in chapter 2. We will not go into details here.

4.5.3 Exercises

Exercise 4.5.2: Let  ∑ F (t) = 12 + ∞n=1 1n2-cos(nπt) . Find the steady periodic solution to x′′ + 2x = F (t) . Express your solution as a Fourier series.

Exercise 4.5.3: Let F (t) = ∑∞ 1-sin(nπt) n=1 n3 . Find the steady periodic solution to  ′′ ′ x + x + x = F (t) . Express your solution as a Fourier series.

Exercise 4.5.4: Let  ∑ F(t) = ∞n=1-12 co s(nπt) n . Find the steady periodic solution to x ′′ + 4x = F (t) . Express your solution as a Fourier series.

Exercise 4.5.5: Let F(t) = t for − 1 < t < 1 and extended periodically. Find the steady periodic solution to x′′ + x = F (t) . Express your solution as a series.

Exercise 4.5.6: Let F(t) = t for − 1 < t < 1 and extended periodically. Find the steady periodic solution to x′′ + π2x = F (t) . Express your solution as a series.

Exercise 4.5.101: Let F (t) = sin(2πt) + 0.1co s(10 πt) . Find the steady periodic solution to  √ -- x′′ + 2x = F (t) . Express your solution as a Fourier series.

Exercise 4.5.102: Let  ∑ ∞ F(t) = n=1e−nco s(2nt) . Find the steady periodic solution to x′′ + 3x = F (t) . Express your solution as a Fourier series.

Exercise 4.5.103: Let F (t) = |t| for − 1 ≤ t ≤ 1 extended periodically. Find the steady periodic solution to  ′′ √-- x + 3x = F (t) . Express your solution as a series.

Exercise 4.5.104: Let F (t) = |t| for − 1 ≤ t ≤ 1 extended periodically. Find the steady periodic solution to x′′ + π2x = F (t) . Express your solution as a series.