Chapter 6 Series Solution of 1D Problems In the last chapter, we studied PDEs on unbounded domain and represented solution in terms of integral by the aid of Fourier transform method. To obtain series solutions of linear PDEs on bounded domains we need to represent functions as an innite series in terms of some functions called basis, similar to what we did in the last chapter of the rst volume of this book. In this chapter we learn how to derive dierent basis for boundary value problems in the form of eigenvalue problems generally called the Sturm-Liouville problem. The theory is a direct generalization of the Fourier series method and we strongly recommend the reader rst read the last chapter of the previous book on Fourier series method. 6.1 Inner product spaces We rst x the suitable space to work within. 6.1.1 Inner product Let T denote the interval (x0;x1). The set C(T ) denote the set of all real continuous functions dened on T and C r(T ) is the set of all real function on T such that f (k) C(T ) for k =0;:::;r. This set become a vector space with the familiar function addition and2scalar multiplication as (f + g)(x) := f(x) + g(x); (cf)(x) = cf(x); x T : 8 2 An extremely important operation in Cr(T ) is the inner product dened by the relation f ; g := f(x) g(x) dx; f ; g Cr(T ): (6.1) h i 8 2 ZT It is simply seen (and is left as an exercise to the reader) to verify the following familiar properties that hold also for dot product in Rn i. f ; f 0 for all f 0, and if f ; f = 0 then f 0, h i h i ii. f ; g = g; f (symmetric) h i h i iii. f ; g + h = f ; g + f ; h (additive) h i h i h i iv. cf ; g = c f ; g for all constant c (homogeneity) h i h i 1 2 Series Solution of 1D Problems Due to the above properties, it makes sense that we consider ; a natural generalization of the dot product in Rn. h i Denition 6.1. The vector space C r(T ) with the inner product ; is called an inner product space. h i Remember the notion of orthogonality in Rn. We say two non-zero vectors ~u; ~v are orthogonal if ~u:~v = 0. We keep this notion to dene the orthogonality in X(T ). In fact, we say two function f ; g Cr(T ) are orthogonal if f ; g =0. The orthogonality of two functions may seem somehow st2range, however, it is somehtimeis useful we consider functions as vectors in the vector space C r(T ). But, we know that a vector has two components: a) a direction, b) a magnitude or norm. Remember that the norm or magnitude of a vector ~u in Rn is dened by the relation ~u = p~u:~u. We keep this notion also for an inner product, that is, k k 1/2 f : = f ; f = f(x) 2dx : (6.2) k k h i j j T p Z ^ f Therefore, we can dene the direction of a nonzero function as f = f . In fact, we have k k innitely many orthogonal functions in C r(T ). For example, for T = ( L; L), it is seen that n n ¡ the set of all functions 1; cos x ; sin x 1 are orthogonal, that is, L L n=1 ¡ ¡ 0 n =/ m cos(nx/L); cos(mx/L) = sin(nx/L); sin(mx/L) = ; h i h i L n = m and also for all n; m, we have cos(nx/L); sin(mx/L) = 0: h i Therefore, we can consider Cr(T ) an innite dimensional vector space equipped with the inner product ; . h i Complex space.In the above formulation, we assumed that all functions in C r(T ) are real. We need sometimes to work with complex functions, and for this we should modify our denition of the inner product. Denition 6.2. If f ; g are two complex piecewise continuous functions dened on T, then their inner product f ; g is dened by the following relation hh ii f ; g = f(x)g(x) dx; hh ii ZT where g is the complex conjugate of g, that is, if g = g + ig then g= g ig . 1 2 1 ¡ 2 It is simply seen that the symmetric property of the real inner product ; is changed to h i f ; g = g; f : hh ii hh ii Accordingly, we have f ; cg = c f ; g . hh ii hh ii 6.1 Inner product spaces 3 Properties of norm.The norm of a function f as dened by the relation f = f ; f enjoys same properties as the norm of a vector in Rn, that is, k k h i p i. f 0 and f = 0 if and only if f = 0, k k k k k k ii. cf = c f for all c R, k k j j k k 2 iii. f + g f + g . k k k k k k The proof of the third property need the following important inequality. Lemma 6.1. (Cauchy-Schwarz) For any two functions f ; g C r(T ), the following inequality holds 2 f ; g f g : (6.3) jh ij k k k k The equality holds if and only if f ; g are linearly dependent. Proof. By the relation tf + g 0 for arbitrary t, we have k k 0 tf + g; tf + g = t2 f 2 + 2t f ; g + g 2: h i k k h i k k The above inequality holds for all t and thus f ; g 2 f g 0: h i ¡ k k k k f ; g If f ; g 2 = f g , and f =/ 0 (otherwise f ; g are linearly dependent), take t = h i , and h i k k k k ¡ f 2 thus tf + g = 0. This implies tf + g = 0 and thus f ; g are linearly dependent. k k k k Now, it is left an exercise to the reader to verify the third property of the norm. Remark 6.1. The dened inner product ; and its associated norm : are not unique. In fact, for any positive function (x) > 0, thhe ifollowing relation is also aknkinner product f ; g = (x) f(x) g(x) dx: h i ZT For the norm, we have more options. For example, the following relation is a norm called the innity norm of f f = max f(x) : k k1 x T j j 2 6.1.2 Convergence in inner product spaces n Remember that a set of vectors ~v1; :::; ~vn is called a basis for R if ~v1; :::; ~vn are linearly f n g independent and every vector in R is represented uniquely by a linear combination of ~v1;:::; ~vn. The story for functions is a little bit delicate and it leads to the concept of convergence. r An innite set of functions := 1; 2; C (T ) is called linearly independent if every nite subset of is a set of linefarly indepgendent functions (we discussed the concept of linear independence of functions in the rst volume of this book). The set is called a basis for C r(T ) if every function f C r(T ) can be represented by an innite series 2 f c + + c + : (6.4) 1 1 n n 4 Series Solution of 1D Problems Consider the relation (6.4). What is the exact meaning of this relation? We interpret this relation as follows. Let SN be the partial sum of the innite series in the right of (6.4), that is N SN = cn n: n=1 We say SN converges to f and write X lim SN = f ; N !1 and understand the convergence in the following sense 2 lim f(x) SN(x) dx = 0; N T j ¡ j or equivalently !1 Z lim f SN = 0: N k ¡ k !1 We call the above convergence, the convergence in norm. This is only one notion of the convergence. Another notion is the pointwise convergence, that is, for any x T , we have 2 lim SN(x) = f(x): N !1 As we will see below, these two notions are not equivalent. Example 6.1. Let (Sn) be the following sequence of functions 0 x = 0 1 Sn(x) =8 pn 0 < x < n : > <> 0 x 1 n > It is simply seen that the sequence conve:>rges to the zero function, f = 0 in [0; 1] pointwise. 1 In fact, Sn(0) = Sn(1) = 0 for all n. Also, for any x (0; 1), there exist N0 such that < x N0 and therefore S (x) = 0 for all n N . Therefore, fo2r any x [0; 1], we have n 0 2 lim Sn(x) = 0: n !1 On the other hand, Sn(x) does not converges in norm to the zero function. In fact, for all n 1 we have 1 1/n 0 S 2 = S (x) 2 = ndx = 1: k ¡ nk j n j Z0 Z0 However, in some cases, two convergences are the same. For example, consider the function x f(x) = e in the domain [0; 1]. Let Sn be the sequence of functions 1 1 S (x) = 1 + x + x2 + + xn: n 2! n! We show that Sn(x) converges to f(x) both pointwise and in norm to f(x).