Orthogonal Function Expansion 正交函數展開 •Introduction of the Eigenfunction Expansion •Abstract Space •Function Sapce •Linear Operator and Orthogonal Function Introduction -The Eigenfunction Expansion Consider the equation : y d1 ( x) y d2 ( x) y 0, a x b With the b.c’s : y(a) y(b) 0 The g.s. is y(x) c1u1 (x) c2u 2 (x) where u1,u2 are linearly index. Fucs. And C1,C2 are arb consts. For b.c’s : c1u1 (a) c2u 2 (a) 0 c1u1 (b) c2u 2 (b) 0 The condition of nontrivial sol. of c1,c2 to be existed if : u1 ( a )u1 ( a ) u 2 (b)u 2 (b) 0 The Euler Column d 2v P 2 2 v 0 , where , dx2 EI The g.s. v( x) c1 sin x c2 cosx :v(0) v(l ) 0 c2 0 For the b.c.’s And c1 sin x 0 for nontrivial solution c1 0 sin l 0 i.e. l n , n 1,2,3,4..... So that we obtain the Eigen values n n , n 1,2,3...... l And the corresponding eigenfucs (nontrivial sols) are vn ( x) sin n x n ( x) According the analysis, we will have v( x) 0 unless the end force P such that: n 2 0 b a f ( x) g ( x) ? Function Space Abstract Space (N, )為一良序半群 (N,)為良序可交換單子 Topological Space Metric Space Normed Space Inner Product Space Rn Q Z N (Z ,)為一良序可交換群 (Z,)為良序可交換單子 (Z, ,)為一良序可交換單子環 (Q, ,)為一有序域 (R, ,)為一完備的有序域 連續的有序域 有序 : m n, m n, m n ml l n m n ml l n m n Hilbert Space Banach Space 由R→Rn(有序性喪失) 代數的原型: , ( - , ) 完備性:每一Cauchy系列均收斂 Topological Space Definition A topological space is a non-empty set E together with a family X (Ui i I ) of subsets of E satisfying the following axioms: (1) E X ,0 X (2) The union of any number of sets in X belongs to X i.e. J finite, J I U i X i j (3) The intersection of any finite number of sets in X belongs to X i.e. J I Ui X i j Metric Space Definition A metric space is a 2-tuple (X,d) where X is a set and d is a metric on X, that is, a function d : X × X → R, such that d(x, y) ≥ 0 (non-negativity) d(x, y) = 0 if and only if x = y d(x, y) = d(y, x) d(x, z) ≤ d(x, y) + d(y, z) (identity) (symmetry) (triangle inequality). Cauchy Sequence Definition: Complete space A sequence (Xn):in a metric space X=(X,d) is said to be Cauchy if for every there is an N=N(e) such that for m,n>N d ( xm , xn ) e x is called the limit of (Xn) and we write lim xn x ; or, simply, xn x n Definition: Completeness Any Cauchy Sequence in X is convergence d < xn, x >0 Ball and Sphere Definition: Given a point x0 X and a real number r>0, we define three of sets: (a) B( x0 ; r) x X d ( x, x0 ) r S ( x ; r) x X d ( x, x ) r (Open ball) ~ (b) B ( x0 ; r ) x X d ( x, x0 ) r (Closed ball) (c) (Sphere) 0 0 In all three case, x0 is called the center and r the radius. Furthermore, the definition immediately implies that ~ S ( x0 ; r) B( x0 ; r) B( x0 ; r) Definition (Open set and closed set): A subset M of a metric space X is said to be open if is contains a ball about each of its points. A subset K of X is said to be closed if its complement (in X ) is open, that is, Kc=X-K is open. A mapping from a normed space X into a normed space Y is called an operator. A mapping from X into the scalar filed R or C is called a functional. The set of all biunded linear operator from a given normed space X into a given normed space Y can be made into a normed space, which is denoted by B(x,y). Similarly, the set of all bounded linear functionals on X becomes a normed space, which is called the dual space X’ of X. Normed Space Definition of Normed Space Here a norm on a vector space X is a real-value function on X whose value at an x X is denoted by x x 0 x 0 x0 x y x y x y Here x and y are arbitrary vector in X and is any scalar Definition of Banach Space A Banach space is a complete normed space. Lemma (Translation invariance) A metric d induced by a norm on a normed space X satisfies (a) d(x+a,y+a)=d(x,y) (b) d (x,y) d ( x, y) For all x, y a X and every scalar Proof. d ( x a, y a) x a ( y a) x y d ( x, y) d (x,y) x y x y d ( x, y) Let X be the vector space of all ordered pairs x (1 , 2 ) y (1 ,2 ) of real numbers. Show norms on X are defined by x 1 1 2 2 12 x 2 (1 2 ) 2 x (1 2 ) p p p 1 p x max1 , 2 The sphere S (0; r) x X , x 1 In a normed space X is called the unit sphere. x 1 x 4 1 x 2 1 x 1 1 Unit Sphere in LP If a normed space X contains a sequence (en) with the property that every x X there is a uniquie sequence of scalars (an) such that x (1e1 .......nen ) 0 as Then (en) is called basis for X. series e k k k 1 Which has the sum x is then called the expansion of x x k ek k 1 n Inner Product Space A inner product space is a vector space X with an inner product define on X. Here, the inner product <x,y> is the mapping of into the scale filed, such that x y, z x, z y, z x, y x, y x, y y, x x, x 0 x, y 0 x 0 Hilbert Space A Hilbert space is a complete inner product space. x x, x d ( x, y) x y (Norm) x y, x y (Metric) Hence inner product spaces are normed spaces, and Hilber spaces are Banach spaces. Examples of finite-dimensional Hilbert spaces include 1. The real numbers x with (v, u ) the vector dot product of v and u 2. The complex numbers C x with (v, u ) the vector dot product of and the complex conjugate of u . v Euclidean space Rn The space Rn is a Hilbert space with inner product define by x, y 11 .... nn Where x (1 ) (1 ,.....n ) y (1 ) (1 ,.....n ) and 1 x x, x (1 ,..... n ) 2 2 1 1 2 d ( x, y) x y x y, x y [(1 1) ..... ( n n ) ] 2 2 2 1 2 Space L2[a,b] b x, y x(t ) y(t )dt a b x ( x(t ) dt) 2 1 2 a Hilbert sequence space l2 With the inner product x, y j j j 1 The norm 2 x x, x ( j ) 1 2 1 2 j 1 Space lp The space lp with p 0 is not inner product space, hence not a Hilbert space Orthonormal Sets and Sequences Orthogonality of elements plays a basis role in inner product and Hilbert spaces. The vectors form a basis for R3, so that every x R 3 has a unique representation. x 1e1 2e2 3e3 x, e1 1 1e1 2 2e2 3 3e3 1 Continuous functions Let X be the inner product space of all real-valued continuous functions on [0,2π] with inner product defined by 2 x, y x(t ) y(t )dt a An orthogonal sequence in X is (un), where un (i) cosnt n 0,1,....... Another orthogonal sequence in X is (vn), where vn (i) sin nt 0 2 u m ,u n cos m t cos ntdt 0 2 n 1,2,....... if mn if m n 1,2,....... if mn0 Hence an orthonormal sequence sequence is (en) e0 (t ) 1 2 en (t ) un (t ) cos nt un From (vn) we obtain the orthonormal sequence ( en ) where vn (t ) sin nt en (t ) vn ~ Homework 1. Does d(x,y)=(x-y)2 define a metric on the set of all real numbers? 2. Show that d ( x, y ) x y defines a metric on the set of all real numbers. 3. Let a, b R and ab Show that the open interval (a,b) is an incomplete subspace of R, whereas the closed interval [a,b] is complete. 4. Prove that the eigenfunction and eigenvalue are orthogonalization and real for the the Sturm-Lioville System. 5. Show the following when linear second-order difference equation is expressed in self-adjoint form: (a) The Wronskian is equal to constant divided by the initial coefficient p C p ( x) W [ y1 , y2 ] (b) A second solution is given by y2 ( x) Cy1 ( x) 6. For the very special case equation becomes x dt p[ y1 (t )]2 0 and q( x) 0, the self-adjoint eigenvalue d du ( x) [ p( x) ]0 dx dx Use this obtain a “second” solution of the following (a ) Legendre’s equation (b ) Laguerre’s equation (c ) Hermite’s equation Function Space (A) L2 [a,b] space: Space of real fucs. f(x) which is define on [a,b] and square integrable i.e . f 2 b ( f , f ) f 2 ( x)dx a In the language of vector space, we say that “any n linearly indep vectors form a basis in E ”space”. Similarly, in function space It is possible to choose a set of basis function such that any function, satisfying Appropriate condition can be expressed as a linear combination to a basis in L2[a,b] Certainly, any such set of fucs. Must have infinitely many numbers; that is, such a L2[a,b] comprises infinitely many dimensions. (B) Schwarz Inequality: Given f(x), g(x) in L2[a,b], Define ( f , g ) b a f ( x) g ( x)dx, then, ( f , g ) 2 ( f , f )(g , g ) Proof: ( f , g ) ( f ag, f ag) a 2 ( g, g ) 2 ( f , g ) ( f , f ) 0 ( f , g ) 2 ( f , f )(g , g ) 0 ( f , g ) 2 ( f , f )(g , g ) (C) Linear Dependence, Independence: Criterion: A set of fucs. 1 ( x),......n ( x) In L2 [a,b] is linear dep.(indep.) if its If its Gramian (G) vanishes (does not vanish), where (1 , 1 )(1 , 2 ).........(n , n ) G (2 , 1 )(2 , 2 ).........(2 , n ) ................ (n , 1 )....................(n , n ) The proof is the same as in linear vector space. (D) The orthogonal System A set of real fucs. 1 ( x),......n ( x) …….is called an orthogonal set of fucs. In L2[a,b] if these fucs. are define in L2[a,b] if all the integral (m ( x),n ( x)) exist and are zero for all pairs of distinct Properties of Complete System Theorem: Let f(x), F(x) be defined on L2 [a,b] for which n f ( x) lim ckk . n k 1 n F ( x) lim Ckk n Then we have b a k 1 f ( x) F ( x)dx ck Ck k 1 Proof: Since f+F, and f-F are square integer able, from the completeness relation [ f F ] dx (c b 2 a k 1 k [ f F ] dx (c 2 a k 1 Ck ) 2 b 2 k Ck ) 4 f ( x) F ( x)dx 4 ck Ck b a k 1 Theorem: Every square integer able fnc. f(x) is uniquely determined (except for its value at a finite number of points) by its Fourier series. Proof: Suppose there are two fucs. f(x),g(x) having the identical Fourier series representation n lim [ f ( x) ckk ( x)]2 dx 0 b i.e. n a k 1 n lim [ g ( x) ckk ( x)]2 dx 0 b n a k 1 Then using ( ) 2( ) we find 2 2 2 2 0 [ g ( x) f ( x)] dx [(g ( x) c k k ) ( c k k ) f ( x)] dx b b 2 a a 2 2 2 [(g ( x) c k k )] 2 ( c k k ) f ( x)] dx 0 [(g ( x) f ( x)] 0 b a 2 b a b a g(x)=f(x) at the pts of continuity of the integrand g(x) and f(x) coincide everywhere, except possibly at a finite number of pts. of discontinuity Theorem: An continuous fuc. f(x) which is orthogonal to all the fucs. of the complete system must be identically zero. Proof: Since… n lim [ f ( x) ckk ( x)]2 dx 0 b n a k 1 n And let g ( x ) ck k ( x ) We can prove f(x)=g(x) at every point. k 1 n f ( x ) ck k ( x ) k 1 Theorem: The fourier series of every square integer able fuc. f(x) can be integrated term by term. In other words, if f ( x) ~ c11 ( x) ...... cnn ( x) ...... Then x2 x2 x2 x1 x1 f ( x)dx c1 1 ( x)dx .... cn n ( x)dx ....... x1 Where x1,x2 are any points on the inteval [a,b] Proof: Since… Assume x2>x1 x2 x1 n fdx ck k dx k 1 x2 x2 x1 x1 Take n limn n b n k 1 a k 1 f ckk dx f ckk dx 2 n f ckk dx 1dx b a b a k 1 x2 x1 n fdx ck k dx 0 k 1 x2 x1 The Sturm-Liouville Problem Self-adjoint Operator d2 d2 Lu( x) p0 ( x) 2 u ( x) P1 ( x) 2 u ( x) p2 ( x)u ( x) dx dx For a linear operator L the analog of a quadratic form for a matrix is the integral u L u u Lu u( x) Lu( x)dx up0u p1u p2udx b b a a Because of the analogy with the transposed matrix, it is convenient to define the linear operator 2 d d b u L u u Lu [u ( x)( p1 p0 )u ( x)]x a 2 [ pou ] [ p1u ] p2u udx a dx dx b Comparing the integrands u( p0 p1)u 2u( p0 p1 )u 0 p0 ( x) P1 ( x) d2 d d 2u du Lu 2 [ p0u ] [ p1u ] p2u p0 2 (2 p0 p1 ) ( p0 p1 p2 )u dx dx dx dx As the adjoint operator L. The necessary and sufficient condition that L L Lu Lu d du ( x) [ p( x) ] q( x)u ( x) dx dx The operator L is said to be self-adjoint. The Sturm-Liouville Boundary Value Problem A differential equation defined on the interval a x b having the form of d dy p ( x ) [q( x) w( x)] y 0 dx dx and the boundary conditions a1 y ( a ) a2 y ' ( a ) 0 b1 y (b) b2 y ' (b) 0 is called as Sturm-Liouville boundary value problem or Sturm-Liouville system, where , ; the weighting function r(x)>0 are given functions; a1 , a2 , b1 , b2 are given constants; and the eigenvalue is an unspecified parameter. The Regular Sturm-Liouville Equation It is a special kind of boundary value problem which consists of a second-order homogeneous linear differential equation and linear homogeneous boundary conditions of the form d dy p ( x ) q( x) w( x)y 0 dx dx d dy L( y ) p ( x) q ( x) y ( x) dx dx where the p, q and r are real and continuous functions such that p has a continuous derivative, and p(x) > 0, r(x) > 0 for all x on a real interval a x b; and is a parameter independent of x. L is the linear homogeneous differential operator defined by L(y) = [p(x)y´]´+q(x)y.And two supplementary boundary conditions A1y(a)+A2y´(a) = 0 B1y(b)+B2y´(b) = 0 . where A1 , A2 , B1 and B2 are real constants such that A1 and A2 not both zero and B1 and B2 are not both zero. Definition 1.1 : Consider the Sturm-Liouville problem consisting of the differ entail equation and supplementary conditions. The value of the parameter in for which there exists nontrivial solution of the problem is called the eigenvalue of the problem. The corresponding nontrivial solution is called the eigenfunction of the problem. The Sturm-Liouville problem is also called an eigenvalue problem. The Nonhomogeneous Sturm-Liouville Problems Consider boundary value problem consisting of the nonhomogeneous differential equation L[y] = - [p(x)y´]´+q(x)y = w(x)y+f(x), where is a given constant and f is a given function on a a x b and the boundary conditions A1y(a)+A2y´(a)=0 B1y(b)+B2y´(b)=0 . And as in regular Sturm-Liouville problems we assume that p, p, q, and r are continuous on a x b and p(x) > 0, r(x) > 0 there.We solve the problem by making use of the eigenfunctions of the corresponding homogeneous problem consisting of the differential equation The Bessel's Differential Equation In the Sturm-Liouville Boundary Value Problem, there is an important special case called Bessel's Differential Equation which arises in numerous problems, especially in polar and cylindrical coordinates. Bessel's Differential Equation is defined as: x 2 y xy ( x 2 n2 ) y 0 where is a non-negative real number. The solutions of this equation are called Bessel Functions of order n . Although the order n can be any real number, . the scope of this section is limited to non-negative integers, i.e., , unless specified otherwise. Since Bessel's differential equation is a second order ordinary differential equation, two sets of functions, the Bessel function of the first kind Jn(x) and the Bessel function of the second kind (also known as the Weber Function) Yn(x) , are needed to form the general solution: y( x) c1J n( x) c2Yn ( x) Five Approaches The Bessel functions are introduced here by means of a generating function. Other approaches are possible. Listing the various possibilities, we have 1. Gram-Schmidt Orthogonalization b a i 2 wdx Ni 2 We now demand that each solution i be multiplied by Ni b a 1 i 2 ( x)w( x)dx 1 b ( x) ( x)w( x)dx a i j ij The presence of the new un(x) will guarantee linear independence. We star with n=0, letting Then normalize 0 ( x) u0 ( x) 0 ( x) 0 ( x) [ 0 wdx] 2 1 2 . Fro n=1, let 1 ( x) u1 ( x) a100 ( x) This demand of orthogonality leads to 2 wdx u wdx a 10 0 wdx 0 10 10 As 0 is normalized to unity, we have a10 u1 0 wdx Fixing the value of a10. Normalizing, we have 1 ( x) 1 ( x) 1 2 ( 1 wdx) 2 We demand that 1 ( x) be orthogonal to 0 ( x) 1 ( x) Where 1 ( x) ( 1 ( x) w( x)dx) 2 1 2 1 ( x) ui ai 00 ai11 ........ aii1i 1 The coefficients aij are given by aij ui j wdx If some order normalization is selected b a [ j ( x)]2 w( x)dx N j 2 The equation can be replaced by i ( x) N i And aij becomes aij i ( x) ( 1 wdx) 2 1 2 u wdx i i Nj 2 p j ui ( x) { ui (t ) j (t ) w(t )dt } j ( x) i 1 i ( x) {1 Pj }ui ( x) j 1 Orthogonal polynomial Generated by Gram-Schmidt Orthogonalization of un ( x) X n , n 0,1,2........ 2. Series solution of Bessel’s differential equation x 2 y xy ( x 2 n2 ) y 0 Using y’ for dy/dx and y for d2y/dx2 . Again, assuming a solution of the form y( x) a x k 0 Inserting these coefficients in our assumed series solution, we have 2 2 n ! x n ! x y( x) a0 x n [1 2 4 .......] 2 1!(n 1) 2 2!(n 2)! Inserting these coefficients in our assumed series solution, we have y( x) a0 2 n! (1) j n j 0 1 x ( ) n 2 j j!(n j )! 2 With the result that….. J n ( x) (1)n J n( x) 3. Generating function g ( x, t ) e x 1 ( )( t ) 2 t Expanding this function in a Laurent series, we obtain e x 1 ( )( t ) 2 t n J ( x ) t n n It is instructive to compare. The coefficient of tn, Jn(x), is defined to be Bessel function of the first kind of integral order n. Expanding the exponential, we have a product of Maclaurin series in xt/2 and –x/2t, respectively. xt 2 e e x 2t x t x s t s ( ) (1) ( ) r! s 0 2 s! r 0 2 r r s For a given s we get tn(n>=0) from r=n+s; s x n s t n s s x s t ( ) (1) ( ) 2 (n s)! 2 s! The coefficient tn is then (1) s x n 2 s xn x n2 J n ( x) ( ) n n 2 ...... 2 n! 2 (n 1)! s 0 s!(n s )! 2 Bessel function J0(x), J1(x) and J2(x) 4. Contour integral: Some writers prefer to start with contour integral definitions of the Hankel function, and develop the Bessel function Jv(x) from the Hankel functions. The integral representation g ( x, t ) e x 1 ( )( t ) 2 t may easily be established as a Cauchy integral for v=n, that is , an integer. [Recognizing that the numerator is the generating function and integrating around the origin] 1 ( x )( t 1 ) dt J v ( x) e 2 t v 1 2i t Cut line (Schlsefli integral) 1 ei ( x 2 )( t 1t ) dt ( x) e i 0 t v 1 Hv (1) Hv ( 2) 1 0 ( x 2 )( t 1t ) dt ( x ) i e i e t v 1 1 (1) ( 2) J v ( x) [ H v ( x) H v ( x)] 2 1 (1) ( 2) N v ( x) [ H v ( x) H v ( x)] 2i 5. Direct solution of physical problems, Fraunhofer diffraction with a circular aperture illusterates this. Incidentally, can be treated by series expansion if desired. Feynman develop Bessel function from a consideration of cavity resonators. In the theory of diffraction through a circular aperture we encounter the integral ~ a 0 2 0 eibr cos drdr The parameter B us given by b 2 sin a ~ 2 J 0 (br)rdr 0 Feynman develop Bessel function from a consideration of cavity resonators. (Homework 1) ~ 2ab a 2a J ( ab ) ~ J ( sin ) 1 1 2 b sin The intensity of the light in the diffraction pattern is proportional to Ф2 and 2a ) sin ] J [( 1 2 ~{ }2 sin 2a sin 3.8317 ..... Fro green light 5.5 105 cm Hence, if a=0.5 cm sin 6.7 10 5(radian) 14s. of arc. Homework (2) Using only the generating function e ( x 1 )( t ) 2 t n J ( x ) t n n Explicit series form Jn(x), shoe that Jn(x) has odd or even parity according to whether n is odd or even, this J n ( x) (1)n J n ( x) (3) Show by direct differentiation that (1) s x 2s J v ( x) ( ) s ! ( s )! 2 s 0 Satisfies the two recurrence relations 2 J v ( x) x J v 1 ( x) J v 1 ( x) 2 J v ( x) J v 1 ( x) J v 1 ( x) And bessel‘s differential equation x 2 J v( x) xJ v ( x) ( x 2 2 ) J v ( x) 0 (4) Show that e ( x 1 )( t ) 2 t n I ( x ) t n n Thus generating modified Bessel function In(x) (5) The chebyshev polynomials (typeII) are generated, 1 n U ( x ) t n 1 2 xt t 2 n 0 Using the techniques for transforming series, develop a series representation of Un(x) PS:請參考補充講義
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