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LECTURE 3 Orthogonal Functions 1. Orthogonal Bases The appropriate setting for our discussion of orthogonal functions is that of linear algebra. So let me recall some relevant facts about nite dimensional vector spaces. Every vector in Rn can be represented as a sum of the form n X v= (3.1) i vi ei =1 where ei is the unit vector along the ith coordinate axis. However, if ffi g is any set of n linearly independent vectors, then we also have a unique representation of v as v= (3.2) n X i ci fi : =1 It should be noted, however, that the vectors fi need not be orthogonal nor need they have unit length for the expansion (3.2) to work. If, however, (3.3) fi fj = ij ; that is, if the fi are orthonormal, then the coecients ci can easily be computed. For if we take the dot product of v with a basis vector fi we get 0n 1 n n X X X fi v = fi @ cj fj A = cj (fi fj ) = cj ij = ci; j that is to say, (3.4) =1 j =1 j =1 ci = fi v : 2. Bases of Orthogonal Functions The relevance of these remarks now comes from the observation that the set C[R; R] of continous real-valued functions on the real line is also a vector space; for the operations of addition and scalar multiplication of functions are well-dened: [f + g](x) = f(x) + g(x) [c f](x) = cf(x) 8 c 2 R: Note, however, that C[R; R] is an innite dimensional vector space. Indeed, the Taylor expansion (2.16) of can be thought of as an expansion of with respect to basis of monomial functions ffmn(x; t) = xm tng . Unfortunately, this basis is not orthonormal, at least not with respect to any obvious inner product. Fourier series are better examples, since they constitute expansions of functions with respect to an orthonormal basis of C[R; R]. 10 BASES OF ORTHOGONAL FUNCTIONS 2. 11 (Fourier Theorem) If f is a function such that f(x) and f 0 (x) that are piecewise continuous on the interval [0; L] R, then the series Theorem 3.1. 1 1 nx X nx a +X a cos + b sin n n 2 n L L n 0 =1 =1 where the coecients an, bn are determined by R 0 0 an = L L cos nx L f(x )dx R 0 0 bn = L L sin x L f(x )dx 2 0 0 2 2 0 0 converges pointwise to f(x) for all x 2 [0; L]. In view of the formulas 2 Z L sin mx sin nx x = m;n L L L Z 2 L sin mx cos nx x = 0 L L L Z L 2 cos mx cos nx x = m;n L L L , ; cos , nx the Fourier expansion of f amounts to a expansion of f with respect to the basis sin mx L L which is orthonormal with respect to the inner product ZL 2 f g = L f(x)g(x)dx : 0 0 0 0 Sturm-Liouville theory is a generalization of Fourier theory. It provides a means of constructing other sets of orthonomal bases for spaces of functions. Theorem 3.2. (Sturm-Liouville Theorem) Consider a boundary value problem of the form dy + (q(x) + r(x)) y = 0 p(x) dx (3.5) c y(a) + c y0 (a) = 0 d y(b) + d y0 (b) = 0 where p(x) and r(x) are smooth positive functions on the interval (a; b). Then d dx 1 2 1 2 (i) For all but a discrete set S of choices of , there are no solutions to (3.5). There exists a minimal 2 S . and one can arrange the set S of admissible so that S = f ; ; ; : : : ; g 0 with 1 2 < < < 0 One then has 1 2 nlim !1 n = +1 : (ii) To each eigenvalue n 2 S , there corresponds exactly one solution n (x) of (3.5). (iii) If n ; m 2 S and n (x), m (x) are the corresponding solutions, then n 6= m ) Zb a n (x)m (x)r(x)dx = 0 : 2. BASES OF ORTHOGONAL FUNCTIONS (iv) The set of functions 8 > < m (x) m = hR > b ( (x)) r(x)dsi = : a m 1 2 2 12 9 > = m = 0; 1; 2; > ; form a complete orthonormal basis for space of piecewise continuous functions on the interval [a; b]. That is to say, every piecewise continous function f : [a; b] ! R, can be expanded as f(x) (3.6) 1 X n nn (t) =1 where the coecients n are determined by (3.7) n = Zb a f(x)n (x)r(x)dx : The series (3.6) converges in the sense that (3.8) lim Zb N !1 a f(x) , N X n ! 2 nn (x) dx = 0 : =1 Remarks: (i) Existence of Eigenvalues and Eigenfunctions. Statement (i) is made a plausible by considering the simple example y00 + y = 0 y(0) = 0 y(L) = 0 2 9 = ; ) = n L , y(x) = A sin nx L and indeed in applications before one can make use of the expansion (3.6), one has to rst nd the eigenvalues i and so part (i) may be regarded as proved constructively. (However, an abstract proof also exists.) (ii) Uniqueness of Eigenfunctions. Statement (ii) follows from the existence and uniqueness theorem for second order linear ODE's. (iii) Orthogonality Property of Eigenfunctions. Let 1 and 2 be solutions of (3.5) for = and = , respectively. Assume = 6 . Then 1 2 1 2 d ,p(x)0 + (q(x) + r(x)) = 0 1 1 dx d ,p(x)0 + (q(x) + r(x)) = 0 2 2 dx 1 2 Multiplying the rst equation by 2 and the second by 1 and subtracting the two equations yields d ,p(x)0 , d ,p(x)0 ( , ) r(x)1 2 = 2 dx 1 dx 1 2 1 2 3. EXAMPLES 13 Integrating both sides of the expression above between x = a and x = b yields Zb Zb d ,p(x)0 (x) dx ( , ) 1 (x)2 (x)r(x)dx = 2 (x) dx 1 a a Zb d ,p(x)0 (x) dx , 1 (x) dx 2 a 1 2 Zb , b 0 = (x) (x)p(x) a , 0 (x)0 (x)p(x)dx a Zb , b , (x)0 (x)p(x) a + 0 (x)0 (x)p(x)dx a , = r(b) 0 (b) (b) , (b)0 (b) , ,r(a) 0 (a) (a) , (a)0 (a) 2 1 1 1 2 1 2 1 1 2 1 2 2 2 1 2 Now, in order for the boundary conditions at x = a c 1 (a) + c 01 (a) = 0 c 2 (a) + c 02 (a) = 0 to have solutions, with c ; c not both zero, we must have 01 (a)2 (a) , 1 (a)02 (a) = 0 : This can be seen as follows: Recall from linear algebra that if M is a n n matrix and v is a n-dimensional vector, then M v = 0 has non-trivial solutions if and only if det M = 0. The statement above then follows by considering c 01 (a) M = 1 (a) ; v = : 0 c 2 (a) 2 (a) 1 1 2 1 2 2 1 2 Similarly, in order for the boundary conditions at x = b to be satised for c ;c not both zero we must have 01 (b)2 (b) , 1 (b)02 (b) = 0 : Thus, we have 3 ( , ) 1 So, 2 Zb a if 6= . 1 Zb a 4 1 (x)2 (x)r(x)dx = 0 1 (x)2 (x)r(x)dx = 0 2 (iv) Completeness of Eigenfunctions The hard thing to understand is the remarkable completeness property expressed in statement (iv). The proof of this statement is not terribly dicult - however, it does require a moderate digression into the Calculus of Variations. At the end of the course, time-permitting, we will develop the Calculus of Variations, and then prove (iv) as a sample application. 3. Examples Example 3.3. Fourier Sine Series: y00 + y = 0 y(0) = 0 y(L) = 0 3. EXAMPLES 14 This is a Sturm-Liouville type problem with p(x) = r(x) = 1, q(x) = 0. The general solution of the ODE is given by p p y(x) = A sin x + B cos x however such solutions will satisfy the boundary conditions y(0) = y() = 0 if and only if B = 0 and p n = L ; n = 1; 2; 3; : : : : The Sturm-Liouville inner product is (f; g) = and the functions ZL 0 f(x)g(x)dx r yn = L2 sin nx L constitute a complete orthonormal basis for the set of piecewise continuous functions on the interval (0; L). Example 3.4. Fourier Cosine Series: y00 + y = 0 y0 (0) = 0 y0 (L) = 0 This is a Sturm-Liouville type problem with p(x) = r(x) = 1, q(x) = 0. The general solution of the ODE is given by p p y(x) = A sin x + B cos x however such solutions will satisfy the boundary conditions y(0) = y() = 0 if and only if A = 0 and p 2n = L ; n = 1; 2; 3; : :: : The Sturm-Liouville inner product is (f; g) = and the functions ZL 0 f(x)g(x)dx r yn = L2 cos 2nx L constitute a complete orthonormal basis for the set of piecewise continuous functions on the interval (0; L). Example 3.5. Bessel Functions d xdJn;a , n J + ,x + a J = 0 n;a dx dx x n;a 2 2 Example 3.6. (3.9) Example 3.7. (3.10) Legendre Functions d dx ,1 , x dLl + (1 + l(l + 1)) L = 0 l dx Hermite Functions d 2 ,x2 dH + 2e,x2 H = 0 e dx dx In each of the last three examples, just as in the case of the rst two examples, there exist dierent sets of orthogonal functions depending on the boundary condtions imposed.