Lecture 11: Fourier Cosine Series

Transcription

1 Introductory lecture notes on Partial Differential Equations - c Anthony Peirce Not to be copied, used, or revised without eplicit written permission from the copyright owner ecture : Fourier Cosine Series (Compiled 4 August 27 In this lecture we use separation of variables to solve the heat equation subject to Neumann boundary conditions In this case we reduce the problem to epanding the initial condition function f( in an infinite series of cosine functions - known as the Fourier Cosine Series Key Concepts: Heat Equation; Neumann Boundary Conditions; separation of variables; Fourier Cosine Series The heat equation subject to Homogenous Neumann Boundary Conditions We consider the heat equation subject to the following initial and boundary conditions: Insulator Insulation Insulator t Insultation u(,t = u t = α 2 u u(,t = u(, = f( Figure Consider a conducting bar with thermal conductivity α 2 that has an initial temperature distribution u(, = f( and whose endpoints are insulated Heat Equation : u t = α 2 u, < < ( u(, t u(, t Boundary Conditions : = = (2 Initial Condition : u(, = f( (3

2 2 Separation of Variables - Fourier sine Series: Consider the heat conduction in an insulated rod whose endpoints are insulate for all time and within which the initial temperature is given by f( as shown in figure Fourier s Guess: α 2 XT : Time equation Case I: Spatial equation assuming that λ : Now impose the boundary conditions: u(, t = X(T (t (4 u t = X( T (t = α 2 u = α 2 X (T (t X ( X( = T (t α 2 T (t = Constant = λ2 (5 T (t = α 2 λ 2 T (t ln T = α 2 λ 2 t + c T (t = De α2 λ 2t dt T = α2 λ 2 dt X ( + λ 2 X( = Guess X( = e r (r 2 + λ 2 e r = r = ±λi X = c e iλ + c 2 e iλ = A cos λ + B sin λ X = Aλ sin λ + Bλ cos λ = u(,t = X (T (t X ( = = u(,t = X (T (t X ( = Now substitute the solution from (8 and use the fact that we have assumed that λ = X ( = Aλ + Bλ B = = X ( = Aλsinλλ λ n = ( nπ n =, 2, Therefore for the case λ we have the countably infinite set of eigenvalues and eigenfunctions ( nπ ( nπ λ n = n =, 2, and X n ( = cos ( Case II: Spatial equation assuming that λ = : In this case the spatial ODE reduces to which has a general solution (6 (7 (8 (9 X ( = ( X( = A + B X ( = B (2 Now imposing the boundary conditions = X ( = B B = = X ( = B B = (3

3 The complete set of eigenvalues and eigenfunctions are thus: ( nπ λ n = n =,, 2, and Fourier Series 3 X ( =, X n ( = cos ( nπ, n =, 2, (4 Thus u n (, t = e α2 ( nπ 2t ( nπ cos n =,, 2, are all solutions of u t = α 2 u that satisfy the boundary conditions (2 (5 Since ( is linear, a linear combination of solutions is again a solution Thus the most general solution is for the form u(, t = A + ( nπ A n cos e α2 ( nπ 2t (6 n= What about the initial condition u(, = f(? If we let t = in (6, then to complete the solution process we are reduced to determining the coefficients A n in the series u(, = f( = A + ( nπ A n cos (7 As in the last lecture we use the inner product <, > to project f( onto the basis functions in the series: f( = A + ( nπ A n cos n= ( kπ f, cos = ( kπ f( cos d = A n= ( kπ cos d + Recall the identity cos(a cos B = {cos(a B + cos(a + B} Therefore 2 J nk = J nn = J = = 2 ( nπ ( kπ cos cos d cos(n k π n= + cos(n + kπ d A n n k (8 ( nπ ( kπ cos cos d(9 = [ ] sin(n kπ/ sin(n + kπ/ + 2 (n kπ/ (n + kπ/ = (2 = /2 ( cos 2 nπ d = 2 d = ( 2nπ + cos d

4 4 Substituting these integrals into (9 we obtain the following epressions for the Fourier Coefficients A k A = A k = 2 Finally the solution of the initial boundary value problem ( is u(, t = A + n= f(d (2 ( kπ f( cos d (22 ( nπ A n cos e α2 ( nπ 2t (23 where A n are defined in (2-(22 We observe that as t it follows that u(, t A, which is just the average value of the initial heat f( distributed in the bar as can be seen from (2 This is consistent with physical intuition It is sometimes convenient to re-define the Fourier coefficients as follows: a = 2A a k = A k, k =, 2, so that the a k assume the unified form a k = 2 ( kπ f( cos d k =,, 2, (24 In terms of the new coefficients a k defined in (24 the Fourier epansion for the initial condition function f( is of the form f( = a 2 + ( nπ a n cos n= while the solution of the heat equation ( is of the form n= (25 u(, t = a 2 + ( nπ a n cos e α2 ( nπ 2t (26 Eample Fourier Cosine Epansion: Determine the Fourier coefficients a k for the function and use the resulting Fourier Cosine epansion to prove the identity f( =, < < = (27 π 2 8 = (2k Solution: a = 2 a n = 2 [ ] d = 2 = 2 cos(nπd = 2 ( n n 2 π 2 = { 4 n 2 π 2, n odd, n even

5 substituting these epressions for the a n into (25, we obtain f( = = 2 4 π 2 Fourier Series 5 k= cos ((2k + π (28 (2k + 2 To obtain the required identity we set = in and rearrange terms The partial sums are shown in figure 2 2 terms of the Fourier Series 3 terms of the Fourier Series 5 terms of the Fourier Series f(= 6 4 f(= 6 4 f(= (a Sum till n = 2 terms (b Sum till n = 3 terms (c Sum till n = 5 terms Figure 2 These figures show the partial sums of the Fourier Cosine Series In figure 3 we plot the same graphs but on a larger domain than [, ] = [, ] 2 terms of the Fourier Series 3 terms of the Fourier Series 5 terms of the Fourier Series f(= 6 4 f(= 6 4 f(= (a Sum till n = 2 terms 2 2 (b Sum till n = 3 terms 2 2 (c Sum till n = 5 terms Figure 3 These figures show the partial sums of the Fourier Cosine Series

6 6 MATAB Code: %fourier cos eample clear;clf;d=;dt=; =-2:d:2;r=:d:;nterms=;ntime=; for nt=:ntime t = (nt-*dt; for n=:nterms K=::n; u(:,n+=5-4*(cos(pi*(2*k+'*'*(ep(-pi^2*(2*k+^2*t/(2*k+^2'/pi^2; plot(',u(:,n+,'r-',r',r','k-','linewidth',2;a=ais;a=[ 2];ais(a; tit=[num2str(n+,' terms of the Fourier Series '];title(tit;label('';ylabel('u(,t, f(=';pause( end if mod(nt,5==,pause(2;end end