Fourier Analysis October 10, 2019 1 Fourier series as normal mode solutions to an eigenvalue problem We have seen how any bounded, piecewise continuous function may be written as a Fourier series. Now, we return to our picture of solutions to the wave equation to place this statement in terms of normal modes and eigenvectors. On an interval [0;L] we begin with the wave equation 1 @2q @2q − + = 0 c2 @t2 @x2 and substitute a single frequency mode for the time dependence, q (x; t) = 'n (x) sin (!nt + 'n) This gives 1 @2 @2 − (' (x) sin (! t + ' )) + (' (x) sin (! t + ' )) = 0 c2 @t2 n n n @x2 n n n !2 @2' (x) n ' (x) sin (! t + ' ) + n sin (! t + ' ) = 0 c2 n n n @x2 n n so cancelling the time dependence we have an eigenvalue equation, @2' (x) !2 n = − n ' (x) @x2 c2 n 2 As we found for coupled masses, the eigenvalues are minus the squares of the frequencies, −!n. This time, in place of an eigenvector, we have an eigenfunction, 'n (x). We easily see that orthonormal solutions to this eigenfunction equation are nπx nπx ' = a sin + b cos n n L n L Choosing an and bn to normalize to 1, L '2 (x) = 1 ˆ n 0 we have r 2 nπx r 2 nπx sin ; cos L L L L as the eigenfunctions, with normal mode frequencies coming from the eigenvalues as πnc ! = n L 1 A general solution to the wave equation is a linear superposition of the normal modes, 1 X nπx nπx q (x; t) = a sin + b cos sin (! t + ' ) (1) n L n L n n n=0 Thus, the solution to the wave equation may be described as linear combinations of a countably infinite sum of simple harmonic oscillations. 1.1 Importance for finding time evolution The normal mode form of solutions to a wide class of wave equations allows us to easily predict the time evo- lution of a solution from an initial spatial distribution. Consider, for example, the 1-dimensional Schrödinger equation, 2 @2 @ − ~ + V (x) = i (2) 2m @x2 ~ @t @2 This equation has only a first time derivative instead of @t2 as for string. This has its roots in the uncertainty principle. Because we cannot know both the initial time and initial position of a quantum particle exactly, it is impossible to specify the two initial conditions required for a second order equation. Even though the time derivative is only linear, this equation has wavelike solutions. To apply the normal mode approach to this more general equation, we again assume single-frequency solutions, setting − i Et (x; t) = 'E (x) e ~ for some constant E. Substituting into the Schrödinger equation, 2 2 ~ @ − i Et − i Et @ − i Et − ' (x) e ~ + V (x) ' (x) e ~ = i ' (x) e ~ 2m @x2 E E ~@t E 2 2 ~ @ 'E (x) − i Et − i Et − + V (x) ' (x) e ~ = E' (x) e ~ 2m @x2 E E and cancelling the time dependence, arrive at the stationary state Schrödinger equation, 2 d2' (x) − ~ E + V (x) ' (x) = E' (x) (3) 2m dx2 E E You may recognize this as an eigenvalue equation. On the left, we have a linear differential operator acting on ' (x), and on the right a constant times ' (x). Indeed, the operator on the right is often replaced by a matrix operator, depending on the problem. The solutions may no longer be simple sines and cosines, but we have a simpler equation to solve. Solutions to Eq.(3) may be either continuous, 'k (x) ; kreal, for unbound problems, or discrete, 'n (x) ; n = 1; 2;::: for bound states. Suppose we have a bound state so that the eigenvalues may be labeled by an integer, En. Then labelling the solutions as 'n (x) the initial wave function is given by an arbitrary superposition, 1 X (x; 0) = an'n (x) n=0 where the constants an determined by the initial conditions. The remarkable fact is that we can now immediately write the full time-dependent solution 1 i X − Ent (x; t) = an'n (x) e ~ n=0 since each normal mode oscillates with the single frequency En . ~ The method applies to a wide class of wave equations. 2 1.2 Time evolution of wave solutions As an example of finding the time evolution of an initial solution, suppose we deform a guitar string of unit length with into a square initial pulse and release it from rest. We have already seen that the initial square wave 1 a < x < b f (x) = 0 elsewhere may be written as the Fourier series 1 1 2 X 1 f (x) = (b − a) + (− (cos kπb − cos kπa) sin kπx + (sin kπb − sin kπa) cos kπx) 2 π k k=1 1 1 P1 1 (We further reduced this to f (x) = 2 (b − a)+ π k=1 k ((sin kπ (x − a)) − sin kπ (x − b)), but here we need the pure sin kπx and cos kπx terms.) th To find the time evolution, we need only multiply the k mode of the initial series by sin (!kt + 'k), then choose the phases 'k to match the initial conditions. Therefore, the displacement of the guitar string is given by 1 1 2 X 1 1 q (x; t) = (b − a) + − (cos kπb − cos kπa) sin kπx + (sin kπb − sin kπa) cos kπx sin (! t + ' ) 2 π k k k k k=1 For initial conditions, we require q (x; 0) = f (x). Therefore sin (!kt + 'k)jt=0 = sin 'k = 1 π π and we choose the phases to be 'k = 2 . Then sin !kt + 2 = cos !kt satisfies both the initial position and velocity conditions, since the time derivative, !k sin !kt, vanishes at t = 0. Next, we write q (x; t) in terms of right and left moving waves. To do this, recall the addition formulas 1 sin a cos b = (sin (a + b) + sin (a − b)) 2 1 cos a cos b = (cos (a − b) + cos (a + b)) 2 Using these we have 1 1 2 X 1 q (x; t) = (b − a) + (− (cos kπb − cos kπa) sin kπx cos ! t + (sin kπb − sin kπa) cos kπx cos ! t) 2 π k k k k=1 1 1 1 X 1 = (b − a) + (− (cos kπb − cos kπa) (sin kπ (x + ct) + sin kπ (x − ct))) 2 π k k=1 1 1 X 1 + ((sin kπb − sin kπa) (cos (kπ (x + ct)) + cos kπ (x − ct))) π k k=1 Collecting the right and left moving pieces, we have 1 1 1 X 1 q (x; t) = (b − a) + (− (cos kπb − cos kπa) sin kπ (x + ct) + (sin kπb − sin kπa) cos (kπ (x + ct))) 4 π k k=1 1 1 1 X 1 + (b − a) + (− (cos kπb − cos kπa) sin kπ (x − ct) + (sin kπb − sin kπa) cos kπ (x − ct)) 4 π k k=1 Each part reproduces the original wave with half the amplitude. A half-height square wave moves off to the left, and an idential half-height square wave moves off to the right. 3 Exercise: Suppose we pluck a guitar string of length L and fixed ends by raising the center to form a triangle, then releasing it from rest. We have already seen that the initial triangle wave 8 L x 0 < x < 2 < L f (x) = L − x 2 < x < L : 0 elsewhere may be represented by a Fourier series, 1 m 4L X (−1) (2m + 1) πx f (x) = sin π2 2 L m=0 (2m + 1) 1. Find the time evolution of the guitar string if we release the it at rest from its stretched triangular position. 2. Write the solution as a sum of right moving and left moving waves. 3. Describe the resulting waves. 1.3 Equivalence to the integrated general solution We would like to show that the general normal mode superpostion, Eq.(1), may be written as a sum of right- and left-moving functions. Begin with the general Fourier series, 1 X q (x; t) = (an sin knx + bn cos knx) sin (!nt + 'n) n=0 1 X = (an sin knx sin (!nt + 'n) + bn cos knx sin (!nt + 'n)) n=0 nπx nπcx where kn = L and !n = L . Using sin (a + b) = cos a sin b + cos b sin a to separate the initial phases and regrouping, the first (an) sum becomes 1 1 X X an sin knx (cos !nt sin 'n + cos 'n sin !nt) = (an sin 'n (sin knx cos !nt) + an cos 'n (sin knx sin !nt)) n=0 n=0 with a similar result for the bn sum. Now, grouping the x and t dependent terms, we use the sum and difference formulas to write each of the new products as a sum and difference, for example 1 nπx nπx sin k x cos ! t = sin + ! t − sin − ! t n n 2 L n L n All the terms in the sum may be rewritten in this way, leading to 1 X nπx nπx q (x; t) = a sin sin (! t + ' ) + b cos sin (! t + ' ) n L n n n L n n n=0 1 1 X nπx nπx = a sin ' sin + ! t − sin − ! t 2 n n L n L n n=0 1 1 X nπx nπx + a cos ' − cos + ! t + cos − ! t 2 n n L n L n n=0 1 1 X nπx nπx + b sin ' cos + ! t + cos − ! t 2 n n L n L n n=0 1 1 X nπx nπx + b cos ' sin + ! t + sin − ! t 2 n n L n L n n=0 4 Collecting the right- and left-moving modes, we have 1 1 X nπx nπx q (x; t) = (a sin ' + b cos ' ) sin + ! t + (b sin ' − a cos ' ) cos + ! t 2 n n n n L n n n n n L n n=0 1 1 X nπx nπx + (b cos ' − a sin ' ) sin − ! t + (b sin ' + a cos ' ) cos − ! t 2 n n n n L n n n n n L n n=0 Now define An = an sin 'n + bn cos 'n Bn = bn sin 'n − an cos 'n Cn = bn cos 'n − an sin 'n Dn = bn sin 'n + an cos 'n and notice that these four linear combinations of an; bn and sin 'n; cos 'n are independent.