Remark: We know there are at most N distinct eigenvalues of VN . The index n does not start at zero because this would imply θ = 0, but θ = 0—due to the presence of sin θ in the denominator of DN —is not a zero of the determinant and hence does not correspond to an eigenvalue of VN . We conclude there are N distinct eigenvalues of VN . These eigenfrequencies are also called normal modes or characteristic oscillations.

6.4The Eigenvectors

6.4.1Constructing the Eigenvectors fN

We now find the eigenvector fn corresponding to eigenvalue λn. That is, we want a column vector fN that satisfies

the above equation in component form is

−fn,j−1 + 2fn,j − fn,j+1 = 2(1 − cos θn)fn,j

with

fn,0 = fn,N +1 = 0.

This is a constant coe cient di erence equation in the j index . Assume, therefore, a solution

of the form

fn,j = eijϕ .

The recursion relation becomes with this guess

−2 cos ϕ + 2 = 2(1 − cos θn),

i.e.

ϕ = ±θn.

The fn,j will be linear combinations of e±ijθN ,

fn,j = c1 sin(jθn) + c2 cos(jθn).

We require fn,0 = fn,N +1 = 0 which implies c2 = 0.

To summarize,

The general solution to (6.4) is

n=1

Thus every oscillation of the weighted string is a sum of characteristic oscillations.

6.4.2Orthogonality of Eigenvectors

The set of eigenvectors {fN}Nn=1 forms a basis for RN since the matrix VN is symmetric. (Another reason they form a basis is that the eigenvalues of VN are distinct.) We claim the eigenvectors have the additional (nice) property that they are orthogonal, i.e.

fN · fM = 0, n 6= m,

where · denotes the vector dot product. The orthogonality is a direct result of the fact that VN is a symmetric matrix. Another way to prove this is to use (6.9) to compute

j=1

To see that this is zero for n 6= m, we leave as an exercise to prove the trigonometric identity

where δn,m is the Kronecker delta function. (One way to prove this identity is first to use the formula sin θ = (eiθ − e−iθ )/2i to rewrite the above sum as a sum of exponentials. The resulting sums will be finite geometric series.) From this identity we also get that the length of each vector, kfnk, is

6.5Determination of constants an and bn

Given the initial vectors u(0) and u˙(0), we now show how to determine the constants an

and bn. At t = 0,

N

X

u(0) = anfN.

n=1

Dotting the vector fP into both sides of this equation and using the orthogonality of the eigenvectors, we see that

Di erentiating u(t) with respect to t and then setting t = 0, we have

X

u˙(0) = ωnbnfN.

n=1

Likewise dotting fP into both sides of this equation results in

If we assume the weighted string starts in an initial state where all the initial velocities are zero,

u˙ j (0) = 0,

then the solution u(t) has components

n=1

where the constants an are given by (6.12) in terms of the initial displacements uj (0). The special solutions obtained by setting all the an except for one to zero, are called the normal modes of oscillation for the weighted string. They are most interesting to graph as a function both in space (the j index) and in time (the t variable). In figures we show a “snapshot” of various normal mode solutions at various times t.