Lecture 10: Fourier Sine 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 Sine Series (Compiled 5 September 8 In the last lecture we reduced the problem of solving the initial-boundary value problem for the heat distribution along a conducting rod to solving two ODEs, one in space and one in time. The spatial ODE and boundary conditions lead to an eigenvalue problem, which identifies a discrete set of wavenumbers {λ n } and corresponding eigenfunctions {X n (} that satisfy both homogeneous boundary conditions and the spatial ODE. Because the heat equation is linear, the general solution of the heat equation is obtained by superimposing the product of the eigensolutions and the corresponding solution of the time ODE to obtain an infinite series. All that remains is that we determine the epansion coefficients b n for the terms of this series. These are obtained by letting t = in the general solution and equating the resulting series to the initial value function f(. This is known as a Fourier Series. This lecture deals with the procedure to determine the Fourier coefficients b n. Our approach is motivated by the process introduced in inear Algebra for projecting a vector onto a set of basis vectors. Key Concepts: Fourier Sine Series; Vector Projection; functions as infinite dimensional vectors; orthogonality; Fourier Coefficients.. Fourier Sine Series Observe that we have a new type of eigenvalue problem in which we seek a nontrivial solution to the following boundary value problem X = X = λ X or X + λ X = X( = = X(. (. Just as in the case with matrices we obtain a sequence of eigenvalues {λ n }. However, because of the infinite dimensional nature of this eigenvalue the problem there are an infinite number of eigenvalues: ( λ n = n =,,... (. and corresponding eigenfunctions {X n (} ( X n ( = sin λ n = sin or X n ( { sin ( π, sin ( π, sin ( } 3π,.... (.3 In order complete the solution of the heat equation we need to determine the coefficients b n such that ( u(, = f( = b n sin. (.4 Observations For symmetric matrices the eigenvalues are real - for the BVP the eigenvalues {λ n (} are also real. For symmetric matrices the eigenvectors form a basis - the eigenfunctions {X n (} are linearly independent. n=

2 .. Euler s Column: The Buckling oad for a Beam - perhaps the oldest eigenvalue problem Eigenvalue problems also arise independently without necessarily coming from a PDE problem. Consider a beam that is subjected to an aial load P applied to its endpoints. Our eperience tells us that as we increase P a critical load P c is reached at which the beam starts to buckle. Figure. Buckling of Euler s column The Bernoulli-Euler aw: A beam constructed from a material that is known to deform in such a way that the curvature κ is proportional to the bending moment καm. In particular, κ = y ( [ + (y ] = cm = M EI where E = Young s modulus and I = the moment of inertia of the beam. If the deflection of the beam is small (y << y << then we can make the approimation y = M EI The Bernoulli-Euler aw When subject to an aial load P as shown in figure.., the bending moment on the buckling beam is given by M( = P y( y = M EI = P EI y = k y k = P EI Thus determining the magnitude of P = EIk for which the beam will first buckle is reduced to solving the following eigenvalue problem: y + k y = y( = = y( y( = A cos k + B sin k } Eigenvalue Problem y( = A = y( = B sin k = k n = We have eigenvalues k n = ( and eigenfunctions y n( = sin. But n =,,.... kn = P ( n EI = n =,,.... Therefore the smallest buckling force P c = P = EIπ ( π Mode is sin. is known as the critical Euler load and the Euler Buckling

3 Fourier Series 3.. Finding the Fourier Coefficients How do we find the b n in the sine series epansion (.4 of f(? v 3 z v f. v y y f( f = (f(,..., f( N Figure. eft side: Epand f in terms of the basis vectors { v, v, v 3 } ; Right side: Approimation of a function f( by a vector f of sample points with values {f( k } Decomposition of vectors into components - projection: How do we epand a vector f in terms of linearly independent vectors v k? If v k v l, k l i.e. the v k are orthogonal Assume f = α v + α v + α 3 v 3 f v k = α v v k + α v v k + α 3 v 3 v k v v v v v v 3 α v v v v v v 3 α = v v 3 v v 3 v 3 v 3 α 3 f v f v f v 3 (.5 α k = f v k v k v k (.6 Functions as infinite dimensional vectors and projection: But functions are just infinite dimensional vectors: f [f, f,..., f N ] g [g, g,..., g N ] f g = f g + f g + + f N g N N = f( k g( k. k= = N (.7 The analogue of the dot product for functions is given by the so-called inner product: N f, g := f(g( d f( k g( k. (.8 k=

4 4 Back to finding b n : f( = ( b n sin n= ( kπ f, sin = ( kπ f( sin d = n= b n ( ( kπ sin sin d. (.9 Recall sin(a sin B = {cos(a B cos(a + B}. Therefore I nk = I nn = = ( ( kπ sin sin d cos(n k π cos(n + kπ d n k = [ ] sin(n kπ/ sin(n + kπ/ (n kπ/ (n + kπ/ = (. = / ( sin d = Therefore the Fourier Coefficients {b n } are given by: Eample. Therefore u(, t = 8 π b k = ( cos d ( kπ f( sin d. (. { < < f( = = ( < < b n = sin( d + ( sin( d = 8 sin(/ n = n π sin ( k= Observe as t u(, t (all the heat leaks out. u(, = 8 ( k π (k + sin [ (k + π ]. k= π 8 = (k + by letting = f( =. k= ( k (k + sin [ (k + π ] e (k+ π t. (.

5 Fourier Series 5 terms of the Fourier Series terms of the Fourier Series.5.5 f( f(.5.5 Eample. As t u(, t. f( = < < = b n = Therefore u(, t = π sin( d = cos( = ( n+ ( n+ sin(e (t. (.3 n n= u(, = ( n+ sin(. π n ( n= u, = π 4 = π = = π n= k= ( n+ n ( k (k + sin(/ k n sin ( (.4 terms of the Fourier Series terms of the Fourier Series.5.5 f( f(.5.5

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