Fourier series A periodic function f(x) with period T = 2π can be represented by a Fourier series: sinnx dx = sinnxsinmx dx =

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1 Periodic functions and boundary conditions Afunctionisperiodic,withperiodT,ifitrepeatsitselfexactlyafteranintervaloflengthT. i.e. y(x = y(x+ T for any x. Evidently, the derivatives of y(x are also periodic (unless the function cannot be differentiated otherwise just differentiate y(x = y(x+t. ODEs can be solved over an interval of length T subject to periodic boundary conditions, which state that the function and its derivatives at one end of the interval equal their values at the other end of the interval. The solution is then a periodic function with period T. For example, for a second-order ODE to be solved over the interval < x < T, two conditions are needed to uniquely specify the solution. The corresponding periodic boundary conditions are y( = y(t and y ( = y (T. e.g. y +y = ( 4π (πx, y( = y(, y ( = y ( (so T =. The homogeneous and particular solutions are y h = Ax+Bsinx y p = (πx. The general solution is therefore y = Ax+Bsinx+(πx. Applying the boundary conditions: A+ = A+Bsin+ and B = B Asin, which can be solved to find that A = B =. Thus y = (πx. This solution can be found more straightforwardly by recognizing that (πx is periodic with period T =, as needed, but the two homogeneous solutions x and sinx, though periodic, have the wrong period (π and must therefore be eliminated. Fourier series A periodic function f(x with period T = π can be represented by a Fourier series: f(x = a + (a n nx+b n sinnx ( for some constant a and a set of (constant coefficients {a n,b n }, n =,,... Helpful integrals: for any integers n and m, nx dx = nxmx dx = which follow on using our handy trig formulae. sinnx dx = sinnxsinmx dx = nxsinmx dx = {, if n m π, if n = m Using the preceding integrals, one can determine the constant and coefficients of the Fourier series by multiplying ( by one of, mx or sinmx, and then integrating x from to π. We obtain a = π f(xdx, a m = π f(xmx dx, b m = π f(xsinmx dx e.g. The sawtooth function, f(x = π x for < x < π, with f(x = f(x+π = f(x π furnishing the function for points outside this interval. We find πa m = πa = x dx+ +x dx =, x mx dx+ +x mx dx = n [ ( n ] (integrating by parts

2 and Hence πb m = π x = n π [ ( n ]nx = 4 π x sinmx dx+ +x sinmx dx =. m= [(m x] (m = 4 π Application to an ODE: If f(x is the preceding sawtooth, solve y y = f(x, y( = y(π y ( = y (π. (x+ 3 3x+ 5 5x+.... We use the fact that f(x can be represented as the Fourier series above (with a = b n =. Hence, y y = a n nx, a n = n π [ ( n ]. The homogeneous solutions are y h = Ae x +Be x. The inhomogeneous term is a series of ines, so we pose the trial particular solution, y p = d n nx. Plugging nto the ODE gives (+n d n nx = a n nx, then matching up each of the ines gives d n = a n /(+n. The general solution is therefore y = Ae x +Be x a n nx + n +. However, the periodic boundary conditions demand that y(x be periodic with period π, whereas the homogeneous solution is never periodic, and so A = B =. Alternatively, one can substitute the general solution into the boundary conditions, and solve the resulting pair of algebraic equations to find these values of A and B explicitly. Thus, y = 4 [(m x] π [(m +](m. m= The square wave: Find the Fourier series for { k for x > f(x = k for x <, We find and Hence πa = k f(x = 4k π dx k dx =, πb n = k m= πa n = k f(x = f(x+π. nx dx k nx dx = sinnx dx k sinnx dx = k n [ ( n ]. sin[(m x] (m = 4k π (sinx+ 3 sin3x+ 5 sin5x+.... Note that the sawtooth is an even function (i.e. f(x = f( x and has b n =, whereas the square wave is an odd function (i.e. f(x = f( x and has a = a n =. These are, in fact, general properties of even and odd functions (which reduces by at least one half the degree of effort required to compute their Fourier series!.

3 Figure : The sawtooth and square wave. The functions are shown in (thick red. The blue lines show the Fourier series truncated at m = (dashed, (dotted, 3 (dash-dotted and 8 (solid. Note the Gibbs phenomenon for the square wave (residual oscillations in the truncated Fourier series. Funks with Jumps If f(x has a discontinuity at x = a, the Fourier series conveges to f(a + f(a+, where f(a is the limit of f(x as x approaches a from the left, and f(a + is the limit of f(x as x approaches a from the right. Thus, if f(x is defined to be anything other than f(a + f(a+ at the jump, the Fourier series will not converge to f(x at x = a. At any jumps of f(x, the Fourier series displays persistent ringing if truncated at a finite number of terms because smooth functions (i.e. sines and ines are being used to represent something that is discontinuous. This is Gibbs phenomenon. Even and odd functions A function is even if f(x = f( x; it is odd if f(x = f( x. In view of these properties (Even Functiondx = (Even Functiondx (Odd Functiondx =. The ine function is even (( x = x, and the sine function is odd ( sin( x = sinx. 3

4 Products of even function remain even; i.e. if f(x and g(x are both even, then f(xg(x is even. Products of odd function are also even: if f(x and g(x are both odd, then f(xg(x is even. The product of an even and an odd function is odd: if f(x is even and g(x is odd, then f(xg(x is odd. Thus, (Even Functionnx dx = (Even Function nx dx (Even Functionsinnx dx =, and (Odd Functionnx dx = (Odd Functionsinnx dx = (Odd Function sin nx dx. This means that for an even function, the coefficients of the sine terms of the Fourier series must vanish (b n =, and f(x = a + a n nx with a = π f(xdx a n = π f(xnx dx. This is called a Fourier ine series. Similarly, for an odd function, the constant and the coefficients of the ine terms of the Fourier series must vanish (a = a n =, and f(x = b n nx with b n = π This is called a Fourier sine series. with Fourier series for arbitrary period f(xsinnx dx. A periodic function with period T = can be represented by the Fourier series, a = f(x = a + f(xdx, a n = [ a n f(x +b n sin the proof follows as before, but using the alternative handy integrals dx = sin dx = sin Again, even functions have the Fourier ine series, f(x = a + a n, a = and odd functions have the Fourier sine series, f(x = ] dx, b n = f(xsin dx; dx = sin b n sin, b n = sin dx = {, if n m dx =, if n = m f(xdx, a n = f(xsin f(x dx. dx, 4

5 Figure : The even and odd periodic extensions of the function f(x = x for < x < π. Extensions of functions If a function f(x is defined over the interval < x <, then it can be extended to < x < as an even function by defining f(x = f( x for x <. Similarly, the function can be extended to < x < as an odd function by demanding that f( x = f(x for x <. Both of these new functions can then be made periodic with period by also demanding that f(x = f(x+ = f(x if x lies outside of the interval (,. The first is the even periodic extension of f(x and is described by a Fourier ine series. The second is the odd periodic extension of f(x and has a Fourier sine series. The odd periodic extension of f(x necessarily satisfies f( +f( + = and f( +f( + =. In view of the periodicity condition f(x = f(x±, the latter further implies that f( +f( + =. The Fourier series respresentation of the odd periodic extension of the function f(x will therefore vanish at x = and. This is a handy result when we start solving PDEs. 5