4.5: Applications of Fourier Series
( \newcommand{\kernel}{\mathrm{null}\,}\)
Periodically Forced Oscillation
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).
The general solution consists of (???) consists of the complementary solution xc, which solves the associated homogeneous equation mx″+cx′+kx=0, and a particular solution of Equation (???) 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 xp 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)=Acos(ω0t)+Bsin(ω0t),
where ω0=√km. Any solution to mx″(t)+kx(t)=F(t) is of the form Acos(ω0t)+Bsin(ω0t)+xsp. 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)=c02+∞∑n=1cncos(nπLt)+dnsin(nπLt).
Then we write a proposed steady periodic solution x as
x(t)=a02+∞∑n=1ancos(nπLt)+bnsin(nπLt),
where an and bn are unknowns. We plug x into the differential equation and solve for an and bn in terms of cn and dn. This process is perhaps best understood by example.
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.
Solution
The equation is, therefore,
x″+2x=F(t),
where F(t) is the step function
F(t)={0if−1<t<0,1if0<t<1,
extended periodically. We write
F(t)=c02+∞∑n=1cncos(nπt)+dnsin(nπt).
We compute
cn=∫1−1F(t)cos(nπt)dt=∫10cos(nπt)dt=0 for n≥1,c0=∫1−1F(t)dt=∫10dt=1,dn=∫1−1F(t)sin(nπt)dt=∫10sin(nπt)dt=[−cos(nπt)nπ]1t=0=1−(−1)nπn={2πnif n odd,0if n even.
So
F(t)=12+∞∑n=1n odd2πnsin(nπt).
We want to try
x(t)=a02+∞∑n=1ancos(nπt)+bnsin(nπt).
Once we plug 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 bn=0 for n even. Hence we try
x(t)=a02+∞∑n=1n oddbnsin(nπt).
We plug into the differential equation and obtain
x″+2x=∞∑n=1n odd[−bnn2π2sin(nπt)]+a0+2∞∑n=1n odd[bnsin(nπt)]=a0+∞∑n=1n oddbn(2−n2π2)sin(nπt)=F(t)=12+∞∑n=1n odd2πnsin(nπt).
So a0=12, bn=0 for even n, and for odd n we get
bn=2πn(2−n2π2).
The steady periodic solution has the Fourier series
xsp(t)=14+∞∑n=1n odd2πn(2−n2π2)sin(nπt).
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 4.5.1 for the plot of this solution.

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=Acos(ω0t)+Bsin(ω0t),
where ω0=NπL for some positive integer N. In this case we have to modify our guess and try
x(t)=a02+t(aNcos(NπLt)+bNsin(NπLt))+∞∑n=1n≠Nancos(nπLt)+bnsin(nπLt).
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(aNcos(NπLt)+bNsin(NπLt)) 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.
Find the steady periodic solution to the equation
2x″+18π2x=F(t),
where
F(t)={−1if−1<t<0,1if0<t<1,
extended periodically. We note that
F(t)=∞∑n=1n odd4πnsin(nπt).
Compute the Fourier series of F to verify the above equation.
As √km=√18π22=3π, the solution to (???) is
x(t)=c1cos(3πt)+c2sin(3πt)+xp(t)
for some particular solution xp.
If we just try an xp given as a Fourier series with sin(nπt) as usual, the complementary equation, 2x″+18π2x=0, eats our 3rd harmonic. That is, the term with sin(3πt) is already in in our complementary solution. Therefore, we pull that term out and multiply it by t. We also add a cosine term to get everything right. That is, we try
xp(t)=a3tcos(3πt)+b3tsin(3πt)+∞∑n=1n oddn≠3bnsin(nπt).
Let us compute the second derivative.
x″p(t)=−6a3πsin(3πt)−9π2a3tcos(3πt)+6b3πcos(3πt)−9π2b3tsin(3πt)+∞∑n=1n oddn≠3(−n2π2bn)sin(nπt).
We now plug into the left hand side of the differential equation.
2x″p+18π2xp=−12a3πsin(3πt)−18π2a3tcos(3πt)+12b3πcos(3πt)−18π2b3tsin(3πt)−12a3πsin(3πt) +18π2a3tcos(3πt)+12b3πcos(3πt) +18π2b3tsin(3πt)+∞∑n=1n oddn≠3(−2n2π2bn+18π2bn)sin(nπt).
If we simplify we obtain
2x″p+18π2x=−12a3πsin(3πt)+12b3πcos(3πt)+∞∑n=1n oddn≠3(−2n2π2bn+18π2bn)sin(nπt.)
This series has to equal to the series for F(t). We equate the coefficients and solve for a3 and bn.
a3=4/(3π)−12π=−19π2,b3=0,bn=4nπ(18π2−2n2π2)=2π3n(9−n2) for n odd and n≠3.
That is,
xp(t)=−19π2tcos(3πt)+∞∑n=1n oddn≠32π3n(9−n2)sin(nπt.)
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. Basically what happens in practical resonance is that one of the coefficients in the series for x_{sp} can get very big. We will not go into details here.
- Jiří Lebl (Oklahoma State University).These pages were supported by NSF grants DMS-0900885 and DMS-1362337.