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Many Coupled Oscillators

A VIBRATINGSTRING

Say we havenparticles with the same massmequally spaced on a string having tensiont. Letykdenote the vertical displacement if the kthmass. Assume the ends of the string are fixed; this is the same as having additional particles at the ends, but with zero displacement:y0=0 andyn+1=0. Letfkbe the angle the segment of the string between the k thand k+1stparticle makes with the horizontal. Then Newton"s second law of motion applied to the k thmass asserts that m¨yk=tsinfktsinfk1;k=1;:::;n:(1) If the particles have horizontal separationh, then tanfk= (yk+1yk)=h. For the case of small vibrations we assume thatfk0; then sinfktanfk= (yk+1yk)=hso we can rewrite (1) as

¨yk=p2(yk+12yk+yk1);k=1;:::;n;(2)

wherep2=t=mh. This is a system of second order linear constant coefficient differential equations with the boundary conditionsy0(t) =0 andyn+1(t) =0. As usual, one seeks special solutions of the formyk(t) =vkeat. Substituting this into (2) we find a

2vk=p2(vk+12vk+vk1);k=1;:::;n;

that is,a2is an eigenvalue of the matrixp2(T2I), where T=0 B

BBBBBBBB@0 1 0 0:::0 0 0

1 0 1 0:::0 0 0

0 1 0 1:::0 0 0........................

0 0 0 0:::0 1 0

0 0 0 0:::1 0 1

0 0 0 0:::0 1 01

C

CCCCCCCCA:(3)

From the work in the next section (see (9)), we conclude that a

2k=2p2(1coskpn+1) =4p2sin2kp2(n+1);k=1;:::;n;

so a k=2ipsinkp2(n+1);k=1;:::;n: The corresponding eigenvectorsVkare the same as forT(see (10)). Thus the special solutions are Y k(t) =Vke2iptsinkp2(n+1);k=1;:::;n; whereY(t) = (y1(t);:::;yn(t)). 1

A SPECIALTRIDIAGONALMATRIX

We investigate the simplennreal tridiagonal matrix: M=0 B

BBBBBBBB@a b0 0:::0 0 0

b a b0:::0 0 0

0b a b:::0 0 0........................

0 0 0 0:::a b0

0 0 0 0:::b a b

0 0 0 0:::0b a1

C

CCCCCCCCA:=aI+bT;

whereTis defined by (3). This matrix arises in many applications, such asncoupled har- monic oscillators (see the previous section) and solving the Laplace equation numerically. ClearlyMandThave the same eigenvectors and their respective eigenvalues are related byμ=a+bl. Thus, to understandMit is sufficient to work with the simpler matrixT.

EIGENVALUES ANDEIGENVECTORS OFT

Usually one first finds the eigenvalues and then the eigenvectors of a matrix. ForT, it is a bit simpler first to find the eigenvectors. Letlbe an eigenvalue (necessarily real) and V= (v1;v2;:::;vn)be a corresponding eigenvector. It will be convenient to writel=2c. Then

0= (TlI)V=0

B

BBBBBBBB@2c1 0 0:::0 0 0

12c1 0:::0 0 0

0 12c1:::0 0 0........................

0 0 0 0:::2c1 0

0 0 0 0:::12c1

0 0 0 0:::0 12c1

C

CCCCCCCCA0

B

BBBBBBBB@v

1 v 2 v 3... v n2 v n1 v n1 C

CCCCCCCCA

0 B

BBBBBBBB@2cv1+v2

v

12cv2+v3...

v k12cvk+vk+1... v n22cvn1+vn v n12cvn1 C

CCCCCCCCA(4)

Except for the first and last equation, these have the form v k12cvk+vk+1=0:(5) 2 We can also bring the first and last equations into this same form by introducing new arti- ficial variablesv0andvn+1, setting their values as zero:v0=0,vn+1=0. The result (5) is asecond order linear difference equation with constant coefficients along with theboundary conditions v0=0, andvn+1=0. As usual for such equations one seeks a solution with the formvk=rk. Equation (5) then gives 12cr+r2=0 whose roots are r =cpc 21

Note also

2c=r+r1andr+r=1:(6)

Case 1:c6=1. In this case the two rootsrare distinct. Letr:=r+=c+pc 21.

Sincer=cpc

21=1=r, we deduce that the general solution of (4) is

v k=Ark+Brk;k=2;:::;n1 (7) for some constantsAandBwhich.

The first boundary condition,v0=0, givesA+B=0, so

v k=A(rkrk);k=1;:::;n1:(8) Since for a non-trivial solution we needA6=0, the second boundary condition,vn+1=0, implies r n+1r(n+1)=0;sor2(n+1)=1: In particular,jrj=1. Using (6), this gives 2jcj jrj+jrj1=2. Thusjcj 1. In fact, jcj<1 because we are assuming thatc6=1. Case 2:c=1. Thenr=cand the general solution of (4) is now v k= (A+Bk)ck: The boundary conditionv0=0 implies thatA=0. The other boundary condition then gives 0=vn+1=B(n+1)cn+1. This is satisfied only in the trivial caseB=0. Conse- quently the equations (4) have no non-trivial solution forc=1. It remains to rewrite our results in a simpler way. We are in Case 1 sojrj=1. Thus r=eiq,c=cosq, and 1=r2(n+1)=e2i(n+1)q. Consequently 2(n+1)q=2kpfor some

1kn(we excludek=0 andk=n+1 because we know thatc6=1, sor6=1).

Normalizing the eigenvectorsVby the choiceA=1=2i, we summarize as follows: 3

Theorem 1The nn matrix T has the eigenvalues

l k=2c=2cosq=2coskpn+1;1kn(9) and corresponding eigenvectors V k= (sinkpn+1;sin2kpn+1;:::;sinnkpn+1):(10) REMARK1. Ifn=2k+1 is odd, then the middle eigenvalue is zero because(k+1)p=(n+

1) = (k+1)p=2(k+1) =p=2.

REMARK2. Since 2ab=a2+b2(ab)2a2+b2with equality only ifa=b, we see that for anyx2Rn with equality only ifx=0. Similarlyhx;Txi 2kxk2. Thus, the eigenvalues ofTare in the interval2These are theGershgorin disks. Theorem 2 (Gershgorin)Each eigenvalues of A lies in at least one of these Gershgorin discs. Proof:SayAx=lxand sayjxij=maxjjxjj. Theithcomponent ofAx=lxis (laii)xi=å j6=ia ijxj so j(laii)xij å j6=ijaijjjxjj Rijxij:

That is,jlaiij Ri, as claimed.

By Gershgorin"s theorem, we observed immediately that all of the eigenvalues ofT satisfyjlj 2. 4

DETERMINANT OFTlI

We use recursion onn, the size of thennmatrixT. It will be convenient to build on (4) and letDn=det(TlI). As before, letl=2c. Then, expanding by minors using the first column of (4) we obtain the formula D n=2cDn1Dn2n=3;4;::::(11) SinceD1=2candD2=4c21, we can use (11) to defineD0:=1. The relation (11) is, except for the sign ofc, is identical to (5). The solution forc6=1 is thus D k=Ask+Bsk;k=0;1;:::;(12) where

2c=s+s1ands=c+pc

21:(13)

This time we determine the constantsA,Bfrom theinitial conditions D0=1 andD1=

2c. The result is

D k=8 :12 pc

21(sk+1s(k+1))ifc6=1;

(c)k(k+1)ifc=1:(14)

For many purposes it is useful to rewrite this.

Case 1:jcj<1. Thens=c+ip1c2hasjsj=1 sos=eiaandc=cosafor some

0 D k=sin(k+1)asina:(15) Case 2:c>1. Writec=coshbfor someb>0. Sinceebeb=2c=s+s1, write s=eb. Then from (14), D k= (1)ksinh(k+1)bsinhb;(16) where we chose the sign in pc

21=sinhbso thatD0=1.

Case 3:c<1. Writec=coshbfor someb>0. Sinceet+et=2c=s+s1, writes=eb. Then from (14), D k=sinh(k+1)bsinhb;(17) where we chose the sign in pc

21= +sinhtso thatD0=1.

Note that ast!0 in (15)-(17), that is, asc! 1. these formulas agree with the case c=1 in (14). 5quotesdbs_dbs20.pdfusesText_26

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