Edwards-Penney Fourier series, heat and wave equations, 2280 Chapter 9 Updated 2015 ================================================================ 9.1-6 f(t) = cos(2 Pi t), determine if periodic. If periodic, then find the smallest period. ================================================================ Periodic means f(t+P)=f(t) holds for all t, for some constant P>0. Trig courses advertise the period of cos(omega t) as 2 Pi /omega. Then f(t) is periodic with period P = 2 Pi/(2 Pi) = 1. ================================================================ 9.1-10 f(t) = cos^2(3t) ================================================================ Periodic means f(t+P)=f(t) holds for all t, for some constant P>0. The function h(t)=cos(3t) is periodic of period P = 2 Pi / 3. The function g(u)=u^2 is continuous and everywhere defined. According to the theory of periodic functions, f(t)=g(h(t)) is periodic of the same period as h(t), i.e., it has period P = 2 Pi/3. ================================================================ 9.1-13 f(t) = 0 on -Pi < t <= 0, f(t)=1 on 0 < t <= Pi. Let G(t) = 2 Pi periodic extension of f(t) to the whole real line, defined by G(t)=f(t-2 n Pi) on - Pi < t - 2 n Pi <= Pi. Sketch g(t) on several periods. Let F(t) be the Fourier series of f(t). Compute F(t) and compare to G(t). ================================================================ The sketch should be first done by hand. The method is to graph f(t) on -Pi < t <= Pi, then replicate this graphic left and right. There are no tools available in this section to compute the Fourier series F(t) indirectly. The only way provided to compute the Fourier series is by direct evaluation of the integrals in equations (14), (16), (17), called the Fourier Coefficient Formulas. (14) a0 = (1/Pi) integral(1,t=0..Pi) = 1, (16) an = (1/Pi) integral( 1*cos(nt),t=0..Pi) = 0 (17) bn = (1/Pi) integral( 1*sin(nt),t=0..Pi) = (1/Pi)(1/n)(1-cos(n Pi)) = (1/Pi)(1/n)(1-(-1)^n) = zero for n=2k, 2/(n Pi) for n=2k+1 The Fourier series F(t) is then the summation a0 + b1 sin(t) + b3 sin(3t) + b5 sin(5t) + ... with answer matching the back of book. ================================================================ 9.1-24 f(t) = | sin(t) | on |t| <= Pi Let G(t) = 2 Pi periodic extension of f(t) to the whole real line, defined by G(t)=f(t-2 n Pi) on - Pi < t - 2 n Pi <= Pi. Sketch g(t) on several periods. Let F(t) be the Fourier series of f(t). Compute F(t) and compare to G(t). ================================================================ The sketch should be first done by hand. The method is to graph f(t) on -Pi < t <= Pi, then replicate this graphic left and right. The only way provided to compute the Fourier series is by direct evaluation of the integrals in equations (14), (16), (17), called the Fourier Coefficient Formulas. (14) a0 = (1/Pi) integral(|sin(t)|,t=-Pi..Pi) = 1, (16) an = (1/Pi) integral( |sin(t)|*cos(nt),t=-Pi..Pi) = 0 (17) bn = (1/Pi) integral( |sin(t)|*sin(nt),t=-Pi..Pi) = 0 The Fourier series F(t) is then the summation a0 + a1 cos(t) + a2 cos(2t) + a3 cos(3t) + ... with answer matching the back of book. ================================================================ 9.2-8 f(t) = 0 on 0 < t < 1, f(t) = 1 on 1 < t < 2, f(t) = 0 on 2 < t < 3, f(0) = 0, f(1) = 1/2, f(2) = 1/2 G(t) has Period = 3, extends f(t) to the real line. F(t) = Fourier series of f(t). Function f(t) is designed to have the average value condition (13). Sketch the graph of f(t) on 0 < t < 3. Compute the Fourier series F(t). ================================================================ A more interesting problem is stated by Richard Haberman: Graph the Fourier series F(t) over 3 periods, using only the function f(t). No Fourier series is computed or used for the graphic. No computer graphics allowed. The graph of f(t) on 0 <= t <= 3 is zero except on 1 <= t <= 2. It has value one on 10 = 2/3 for n=0 bn = (1/L) integral( G(t)sin(n Pi t/L),|t|<=L) = (2/3) integral( f(t)sin(n Pi t/L), t=0 to t=2L) [this is (10b)] REASON: the integral of a P-periodic function has the same value over any interval of length P. = (2/3) integral( 1*sin(2n Pi t/3), t=1 to t=2) REASON: f(t)=0 except on 1<=t<=2. = (2/(n*Pi))*(-cos((2/3)*Pi*n)+cos((4/3)*Pi*n)) The answers are simplified by considering three separate cases: (1) n=3k; (2) n=3k+1; (3) n=3k+2 The simplified answers are (1) an=0, bn=0; (2) an=-sqrt(3)/(n Pi), bn=0; (3) an = sqrt(3)/(n Pi), bn=0. Then F(t) = 1/3 - (sqrt(3)/Pi)[cos(2 Pi t/3) - (1/2) cos(4 Pi t/3) + ...] ================================================================ 9.2-20 Derive the Fourier series equation on 0 < t < 2*Pi sum((1/n^2)*cos(n*t), n=1 to infinity) = (3*t^2-6*Pi*t+2*Pi^2)/12 ================================================================ Only the basic idea will be disclosed plus the intermediate formulas that are required. The idea is to write the equation backwards, and then view the identity as the sum of three series formulas, each of which is justified separately, by computing the corresponding Fourier series. (1) 3 t^3/12 = Fourier series of f(t)=t^2/4 on 0 < t < 2*Pi (2) -6*Pi*t/12 = Fourier series of f(t)=-Pi*t/2 on 0 < t < 2*Pi (3) 2*Pi^2/12 = Fourier series of f(t)=Pi^2/6 on 0 < t < 2*Pi The Fourier series on the right side of these equations are, respectively, (4) Pi^2/3 + sum(cos(n*t)/n^2,n=1 to infinity) + sum(-Pi*sin(n*t)/n, n=1 to infinity) (5) -Pi^2/2 + sum(Pi*sin(n*t)/n, n=1 to infinity) (6) Pi^2/6 The details expected include adding the three series termwise to obtain sum(cos(n*t)/n^2,n=1 to infinity). Derivation is expected for the three Fourier series expansions (4), (5), (6). ================================================================ 9.3-4 f(t) = triangle = t on 0 < t <= 1, = 2-t on 1 < t < 2, = 0 at t=0, = 0 at t=2. Fc(t) = Fourier cosine series of f(t). Fs(t) = Fourier sine series of f(t). Compute Fc, Fs. ================================================================ Fc Answers: a0 = 1 an = 0 for n odd, n=1,3,5,... an = 0 for n even, n=4,8,12... an = -16/(Pi^2*n^2) for n even, n=2,6,10,... Fs Answers: bn = 0 for n even, n=2,4,6,... bn = 8/(Pi^2*n^2) for n odd, n=1,5,9,... bn = -8/(Pi^2*n^2) for n odd, n=3,7,11,... Fourier series details. The trick is to split the integral into two intervals, e.g., an = integral(f(t)*sin(n*Pi*t/2),t=0 to t=2) = integral(t*sin(n*Pi*t/2),t=0 to 1) + integral((2-t)*sin(n*Pi*t/2),t=1 to 2) The integrals can be evaluated from tables or computer algebra systems. The final reduction via even and odd indices is a change of variable, like n=2k or n=2k+1. Subcases occur while simplifying values of sine/cosine. ================================================================ 9.3-17 Write integral calculus proofs for (a) and (b). (a) Let f(-t)=f(t) for all t. Show that integral(f(t), t=-a to t=0) = integral(f(t), t=0 to t=a) (b) Let f(-t)=-f(t) for all t. Show that integral(f(t), t=-a to t=0) = (-1)*integral(f(t), t=0 to t=a) ================================================================ To do (a), substitute t=-u into the left side integral. This changes the integration limits, u=a to u=0, but also there is an extra sign under the integrand, because of dt = -du. Because integral(F,u=a to u=b) = - integral(F, u=b to u=a) then properties of even/odd functions allow a re-write of the integrand, resulting in the integral on the right side. ================================================================ 9.3-20 Assume given the Fourier series expansion t^4/24 = Pi^2*t^2/12 + sum(2*(1-(-1)^n*cos(n*t))/n^4, n=1 to infinity) Substitute t=Pi and t=Pi/2 into this identity. First, justify why it is valid to do so. Second, prove the identities (a) sum(1/n^4,n=1 to infinity) = Pi^4/90, (b) sum((-1)^(n+1)/n^4,n=1 to infinity) = 7*Pi^4/720 ================================================================ The identity cos(n*Pi)=(-1)^n implies reductions of the formula obtained after substitution of t=Pi. Details for identity (a): The identity for t=Pi reduces to Pi^4/24 = sum(4/n^4, n odd, n=1 to infinity) Let S=sum(1/n^4,n=1 to infinity). By splitting off even and odd terms we get the identity S = sum(1/n^4, n odd, n=1 to infinity)+ sum(1/n^4, n even, n=1 to infinity) = (1/4)(Pi^4/24) + (1/2)^4 * sum(1/k^4,k=1 to infinity) = (1/4)(Pi^4/24) + (1/16)*S which implies S=(16/15)(1/4)(Pi^4/24) = Pi^4/90. Details for identity (b): The identity cos(n*Pi)=(-1)^n and the identity cos((2*k+1)*Pi/2)=0 imply reductions of the formula obtained after substitution of t=Pi/2. Please provide details to the problem, following the argument above for part (a). ================================================================ 9.4-1 x'' + 5 x = F(t), F(t) = Fourier series of f(t), f(t) = 3 on 0 < t < Pi, f(t) = -3 on Pi < t < 2*Pi, Find the steady-state periodic solution x(t). ================================================================ Substitute into x'' + 5x = F(t) the formulas x(t) = sum(bn*sin(n*t),n=1 to infinity) F(t) = (12/Pi)*sum((1/n)*sin(n*t),n odd, n=1 to infinity) REF: Example 1, Section 9.1 Match coefficients of sine terms left and right of the equal sign to obtain bn=0 for n even and bn=12/(n*Pi*(5-n^2)) for n odd. This answer is the unique steady-state periodic solution x(t). ================================================================ 9.4-7 x'' + 9x = F(t), F(t) = Fourier series of f(t), f(t) = -1 on -Pi < t < 0, f(t) = 1 on 0 < t < Pi, Determine if pure resonance occurs. ================================================================ By Example 1, Section 9.1, we have the Fourier series formula F(t) = (4/Pi)*sum((1/n)*sin(n*t/3),n odd, n=1 to infinity) The series F(t) contain the term (4/3*Pi)*sin(3*t), which is a solution of the homogeneous problem x'' + 9x = 0. therefore, resonance occurs. ================================================================ 9.4-9 3x'' + 12x = F(t), F(t) = Fourier series of f(t), f(t) = -1 on -1 < t < 0, f(t) = 1 on 0 < t < 1, Determine if pure resonance occurs. ================================================================ By Example 1, Section 9.1, we have the Fourier series formula F(t) = (4/Pi)*sum((1/n)*sin(n*t/3),n odd, n=1 to infinity) The series F(t) does not contain a term with frequency sqrt(12/3)=2, therefore resonance does not occur. ================================================================ 9.4-13 x'' + 0.1 x' + 4 x = F(t), F(t) = Fourier series of f(t), f(t) = -3 on -Pi < t < 0, f(t) = 3 on 0 < t < Pi, Find the first three nonzero terms in the Fourier series for the unique steady-state periodic solution x(t). ================================================================ By Example 1, Section 9.1, we have the Fourier series formula F(t) = (4/Pi)*sum((1/n)*sin(n*t/3),n odd, n=1 to infinity) The unique steady-state periodic solution has the form x(t) = sum(bn*sin(n*t-alphan),n=1 to infinity) Substitute this formula into x'' + 0.1 x' + 4 x = F(t). The details proceed like Example 4, Section 9.4, which uses equations (14), (15), (16) in the textbook. The formula for F(t) implies bn=12/(n*Pi) for n odd and bn=0 for n even. The formula for alphan requires a calculator, with final answer equal to the textbook answer, back of book. ================================================================ 9.5-1 Solve the boundary value problem. u_t = 3 u_xx, u(0,t)=u(Pi,t)=0, u(x,0)=4*sin(2*x), on the set 00. ================================================================ The method expected is to follow Example 2, Section 9.5, using k=3, but replacing u(x,0)=u0 by u(x,0)=4*sin(2*x). Then the rod temperature is for L=Pi the series u(x,t) = sum(bn*exp(-3*n^2*Pi^2*t/L^2)*sin(n*Pi*x/L),n=1 to infinity) which simplifies to u(x,t) = sum(bn*exp(-3* n^2* t)*sin(n*x),n=1 to infinity) The Fourier coefficients of f(x)=4*sin(2*x) are b2=4 and otherwise bn=0. The reduced Fourier series is then u(x,t) = 4*exp(-3*2^2*t)*sin(2*x) ================================================================ 9.5-9 Solve the boundary value problem. 10u_t = u_xx, u(0,t)=u(5,t)=0, u(x,0)=25, on the set 00. ================================================================ Following Example 2, Section 9.5, we have for L=5 f(x) = u(x,0) = (100/Pi)*sum((1/n)*sin(n*Pi*x/5) Apply Theorem 1, equation (31), to obtain the solution u(x,t)= (100/Pi)*sum((1/n)*exp(-n^2*Pi^2*t/25)*sin(n*Pi*x/5),n=1 to infinity) ================================================================ 9.6-5 Solve the boundary value problem. u_tt = 25 u_xx, u(0,t)=u(3,t)=0, t>0, u(x,0)=(1/4)*sin(Pi*x), 0