1. Spaces of Functions

Transcription

1 FUNCTION SPACES Analysis of systems of linear algebraic equations leads naturally to the notion of a linear space in which vectors (i.e., the unknowns in the problem) formation of linear combinations (i.e., the algebraic conditions imposed on the unkowns) are the principle ingredients. Considering linear systems in this setting of a linear space results in efficiency of expression economy of effort. We are motivated to try to discover similar structures for the analysis of partial differential equations. Here, of course, functions are going to play the role played by vectors in linear algebra the matrices will be replaced by partial differential operators. An additional difference is that the spaces of functions will be required to be closed under passage to the limit besides being closed under the operation of forming linear combinations. 1. Spaces of Functions A linear space X is a collection of objects x,y,z,... together with a set of scalars A,B,C... two operations: i) Addition x,y X, x y X ii) Scalar Multiplication x X, Ax X for all scalars A Examples of linear spaces are familiar from linear algebra algebra. These spaces are generally examples of finite dimensional linear spaces. For 1 p, let L p U denote the linear space of functions which are defined whose p-th power is integrable on an open set U R n. That is, the functions which are defined on U for which I fx p dx. 1.1 The open set U can be bounded or unbounded the functions clearly need not be continuous in order for (1.1) to be satisfied. In fact the set of singularities permitted for functions in L p U is quite large, as a result, the integral used must be more general than the Riemann integral. However, for what we propose to do, we will not require any detailed knowledge of a more general integration theory. The linear structure on L p U is defined by f,g L p U, A,B R, let Af Bgx Afx Bgx x U 1.2 That L p U is a linear space is the assertion of, Proposition 1.1 f,g L p U, A,B R, Af Bg L p U. This result follows from the inequality for each p, 1 p, A B p 2 p1 A p B p. A,B R. Note that the scalars here are taken to be the real numbers. A real valued function, N, defined on a linear space X is a norm if it has the following properties: 1. NAx A Nx A R, x X 2. Nx y Nx Ny x,y X 3. Nx 0 x X Nx 0 iff x 0 1

2 A norm is defined on L p U, 1 p, by f p U fx p dx 1/p, for f L p U. Then L p U becomes a normed linear space. That Nf f p satisfies 1 3 is clear but 2 requires proof. Using the inequality leads to ab ap p bq q a,b R, for 1 p 1 q 1, 1.3 Proposition 1.2 ( Holder Minkowski inequalities) (a) If 1 p 1 q 1, then f g 1 f p g q f L p U, g L q U (b) If 1 p, then f L p U f g p f p g p. We should be careful to notice that f g p 0 does not imply that fx gx at all x U. In fact, f g can differ at infinitely many points in U so long as the set of points contains no positive volume subsets (i.e., f g can differ on sets of measure zero ). This means that functions in L p U are defined only up to sets of measure zero that altering a function in L p U on such a set does not produce a different function in L p U. Then functions in L p I are, in fact, equivalence classes of pointwise defined functions where the equivalence relation is equality almost everywhere. When we say let f be an arbitrary function in L p U", what we really mean is let f be an arbitrary representative of one of the equivalence classes of functions in L p U". Convergence To say a sequence f n x in L p U converges of to a limit f in L p U means f n f p 0 as n. This type of convergence is weaker than uniform convergence on U neither implies nor is implied by pointwise convergence on U. Every sequence f n x in L p U that is convergent must be a Cauchy sequence; i.e., it must satisfy f n f m p 0 as m,n. In fact, it is true in any normed linear space that every convergent sequence must be a Cauchy sequence. On the other h, in an arbitrary normed linear space it is not necessarily the case that every Cauchy sequence converges to a limit that belongs to the space. A space with the property that every Cauchy sequence is convergent is said to be complete. The following result, known as the Riesz-Fischer Theorem in analysis, asserts that L p U is complete. The proof requires techniques of an integration theory more general than Riemann integration. Proposition 1.3 (Riesz-Fischer) For 1 p, L p U is complete for the norm f p. For each p, 1 p, L p U is a complete, normed linear space. Such spaces are called Banach spaces. We shall be most interested at this point in the special cases of L 1 U, the absolutely integrable functions L 2 U,the square integrable functions. The space L 2 U enjoys some properties not shared by L 1 U. 2

3 2. Inner Product Spaces An inner product space is a linear space X, on which there is defined a mapping which associates to every pair of elements x,y X, a scalar value which we denote by x,y X. This mapping must have the following properties 1 x,y X y,x X x,y X 2 Ax By,z X Ax,z X By,z X, x,y,z X, A,B R, x,x X 0 x X, x,x X 0 iff x 0 The mapping is called an inner product on X. The most familiar example of an inner product space is the space R n of n-tuples x x 1,...,x n, where the inner product is defined as x,y R n n i1 x i y i. 2.2 An inner product space has a norm, induced by the inner product. That is, x X x,x X 1/2 for x X 2.3 defines a norm on the linear space X. Recall that the norm defines a meaning for distance in the linear space X. In any inner product space, the following results are valid. Proposition 2.1 (Cauchy-Schwartz Triangle inequalities ) a) x,y X x X y X x,y X b) x y X x X y X, x,y X A Cauchy sequence in the inner product space X is a sequence of elements x m X with the property that x m x k X 0 as m,k. A sequence of elements x m X is said to be convergent if there exists an element x X such that x m x X 0 as m. The inner product space X is said to be complete if every Cauchy sequence in X is convergent. It is well known that the inner product space R n is complete (this is just a consequence of the fact that the real numbers have been constructed to be complete). We now consider two additional examples of inner product spaces. The first example is not a function space but will be important later. The Space of Square Summable Sequences Let 2 denote the linear space of infinite sequences of real numbers x x 1,x 2,... This space carries the same linear inner product structure as R n if we define Ax By Ax 1 By 1,Ax 2 By 2,... x,y 2 A,B R 2.4 x,y 2 1/2 i1 x i y i x 2 x,x Of course the inner product norm have to be restricted to those sequences for which the infinite sums are finite. Such sequences are said to be square summable sequences we use the notation 2 to indicate the linear space of all such sequences. It is straightforward to show that the dimension of the space 2 is infinite. In fact, by showing there is a basis for 2 which is in one to one correspondence with the natural numbers, it 3

4 follows that the dimension of this space is equal to the cardinality of the natural numbers. Proposition 2.2 Every Cauchy sequence in 2 is convergent. Proof- Let x n denote a Cauchy sequence in 2. Then x m x n 2 2 i x i m x i n 2 0 as m,n. This implies that for every 0, N such that x i m x i n 2 for all m,n N. Then for each i, x i m is a Cauchy sequence in R, since R is complete x i m X i as m, for some real number X i. It remains now to show that X X 1,X 2,... belongs to 2 that x n converges to X in 2. For m,n N, M i1 x m n 2 i x i im1 x i m x i n 2, for arbitrary M since each sum is nonnegative, M i1 x i m x i n 2. Now fix m let n tend to infinity. This leads to M i1 x m 2 i X i, m N every M 1. Since M is arbitrary, let M tend to infinity to conclude that for every 0 there is an N such that Then it follows that i1 x m 2 i X i, m N. X X m 2 0 as m, X 2 X X m 2 Xm 2 Xm 2 (so X 2 ). It should be noted that the set of vectors E m in 2 where E i m im, forms a basis for 2 in the sense that for any x 2, the sequence of elements m x m i1 x,e i 2 Ei is a Cauchy sequence in 2 converging to x as m. There is, in general, no finite combination of E s which equals x. The Function Space L 2 In the special case, p q 2, the Holder inequality looks like the Cauchy-Schwartz inequality. In fact, in this special case, L p U L 2 U is an inner product space. The inner product on L 2 U is defined by f,g 2 U fx gx dx, f 2 f, f 2 1/2 2.6 L 2 U is the only one of the L p U spaces that supports an inner product. Since it is an L p U space, it has all the relevant properties such as completeness. 4

5 Two elements in an inner product space are said to be orthogonal if their inner product is zero. In L 2 U this has no visualizable significance (e.g. it does not mean the graphs of two orthogonal functions are orthogonal trajectories). However, orthogonality in L 2 U does imply linear independence. Since it is possible to generate an infinite family of orthogonal functions in L 2 U, it follows that L 2 U is infinite dimensional. An infinite dimensional inner product space that is complete in the norm induced by the inner product is called a Hilbert space. Note that if U is bounded, so that U 1dx U, it follows that L 2 U is contained in L 1 U To see this suppose f L 2 U write f 1 U f dx U 1 f dx 1 2 f 2 When U is not bounded, then neither space is contained in the other. For example consider the functions fx 1/x if x 1 0 if x 1 gx 1/ x if 0 x otherwise. Then f belongs to L 2 R but does not belong to L 1 R, while g belongs to L 1 R but does not belong to L 2 R. These two spaces have many functions that are in both spaces but each contains some functions that are not in the other. 3. The Fourier Integral Transform For functions which are defined on all of R n we can define an alternative representation for the function. This alternative representation is called the Fourier transform of the function. The Fourier transform of the everywhere defined function fx is defined as follows T F f F 2 n R n fx ei x dx. Here, we will restrict our attention to the one dimensional case where we have T F f F 2 1 R fx e i x dx 3.1 The notations T F f F will be used interchangeably to denote the Fourier transform of the function f(x). Note that the transform does not exist for any function f(x) for which the improper integral (3.1) fails to converge. If the integral converges, it defines a possibly complex valued function of the real variable. Evidently, a sufficient condition for the Fourier transform to exist is that f is absolutely integrable, i.e., f L 1 R. Example 3.1 Some Fourier Transforms (a) Consider Then f L 1 R (b) Consider fx 1 if x 1 0 if x 1 F 2 1 fx R ei x dx e i x dx 1 fx e x. L 1 R. Then x ei i2 x1 x1 sin 5

6 F 2 1 fx R ei x dx 2 1 e x e i x dx 1 where we used the result ex1i 1i 0 ex1i 1i i 1 1i 1 lim x ex1i lim x e x1i 0. (c) The Gaussian function fx e x2 belongs to L 1 R But hence F 2 1 fx R ei x dx 2 1 e x 2 e i x dx 2 1 e x 2 ix 2 /4 2 /4 dx e 2 /4 1 e xi 2 dx 2 e xi 2 dx e z 2 dz F 1 4 e2 /4 T F e x2. e xi x dx 2 1 e xi x dx Each of the functions in this example belongs to L 1 R each transform is a continuous function of the transform variable. In addition, it appears that the transform tends to zero as. In fact, this is true in general. We will use the notation F C 0 R to indicate that F is continuous on R 1 F tends to zero as tends to infinity. Note that a function in C 0 R is not necessarily absolutely integrable. The function pairs fx, F listed above are examples of Fourier transform pairs. Theorem 3.1 If f L 1 R then F T F fx exists is a continuous function of R 1. Moreover, F C 0 R. The proof of this theorem requires the dominated convergence theorem for Lebesgue integrals it therefore omitted. We list now several useful properties of the Fourier transform. Theorem 3.2 If fx gx have Fourier transforms F, G, respectively, then 1. T F Afx Bgx AF BG A,B R 2. T F fbx 1 b F b 0 b A transformation with property 1 is said to be linear one with property 2 is said to be homogeneous. Problem 1 Use the definition of the Fourier transform to prove theorem 3.2 Problem 2 Use theorem 3.22 to show that (a) T F I A x sina where I A x 1 if x/a 1 0 if x/a 1 1 if x A 0 if x A 6

7 (b) T F e b x b 1 b 2 2 for b 0 (c) T F e bx2 1 4b e2 /4b for b 0 Theorem 3.3 If fx has Fourier transform F, then for all real values c the following transforms exist are related to F as indicated 1. T F fx c e ic F 2. T F e icx fx Fc Problem 3 Use the definition of the Fourier transform to prove theorem 3.3 Theorem 3.4 If both fx f x have Fourier transforms then the transform of the derivative, f x, is given in terms of F T F f by T F df/dx if. More generally, if fx all its derivatives up to order m have Fourier transforms, then T F d k f/dx k i k F for k 1,2,...,m Theorem 3.5 If both fx xfx have Fourier transforms then the transform of x fx, is given in terms of F T F f by T F xfx i d/df. More generally, if fx x k fx have Fourier transforms, for k 1,2,...,m then T F x k fx i d/d k F for k 1,2,...,m Problem 4 Use the definition of the Fourier transform to prove theorem 3.4 Problem 5 Use the definition of the Fourier transform to prove theorem 3.5 For functions fx gx defined on R, we formally define the convolution product of f g as f gx R fx y gy dy 3.2 Problem 6 Use the definition of the convolution product the change of variable, z x y, to show that f gx gfx. Theorem 3.6 If fx gx belong to L 1 R, then f g L 1 R T F f g 2F G. The L 1 R Inversion Theorem We have seen that for each f L 1 R, there is a Fourier transform F C 0 R. In some 7

8 sense, knowledge of one of these functions is equivalent to knowledge of the other; i.e., they are two different representations for the same information. To make this assertion more precise, we have the next theorem. Theorem 3.7 If fx belongs to L 1 R,, in addition, the Fourier transform F is also in L 1 R, then x R F e i x d C 0 R f L1 0. The assertion of this theorem is that if f F both belong to L 1 R, then the inverse Fourier transform of F is defined by T F 1 F R F e i x d 3.3 T F 1 F fx where the equality is equality in the sense of L 1 R. Then 3.3 is known as the Fourier inversion formula. By comparing (3.3) with (3.1), we arrive at the following result which can be used to increase the number of transform pairs. Theorem 3.8 If fx belongs to L 1 R,, in addition, the Fourier transform F is also in L 1 R, then T F F 1 2 f. Problem 7 Use theorem 3.8 previous transform pairs to show that: a T F b T F sinax x 1 2 I Ax 1 2 I Ax 2b b 2 x 2 e b for b 0 c T F e x2 /4b b eb2 for b 0 Symmetry Symmetry in the graph of the absolutely integrable function fx implies certain related symmetries for the Fourier transform, F. Since, in general, fx may be complex valued, we will write it as the sum of real imaginary parts, each of which is then the sum of an odd an even function, Now so fx Re f E x Re f O x iim f E x Im f O x. e ix cosx i sinx F 1 0 Re fe x cosx Im f O x.sinxdx i Im fe x cosx Re f O x.sinxdx. 0 Clearly F can be complex valued even when fx is real valued, but in any case there are 8

9 certain correspondences between the evenness or oddness of f that of F: f even complex odd F even complex odd even real imaginary even real imaginary odd real imaginary odd imaginary real 4. The Fourier Transform in L 2 R Note that the functions f 1 x I A x, f 2 x e b x, f 3 x e bx2 all belong to L 1 R each has a Fourier transform in C 0 R by theorem 3.1. Note further that these transforms F 1 sina, F 2 b 1 F b /4b 4b e2 all belong to L 2 R, but only F 2 F 3 belong to L 1 R. Then only F 2 F 3 can be inverted using Theorem 3.7 to recover the functions f 2 f 3. So in what sense is it possible to say that the inverse transform of F 1 is f 1? If f L 1 R then its Fourier transform exists belongs to C 0 R. For an arbitrary function f L 2 R, it is not necessarily the case that f L 1 R it is not clear then that its Fourier transform exists. In order to extend the Fourier transform to L 2 R we will use an idea that is pervasive in the study of partial differential equations. We first introduce a special subspace of L 2 R where transforming inverting works more smoothly. This subspace has the additional property that the results which hold on the subspace can be extended to the whole space, L 2 R, by passing to the limit. We define this subspace as follows. The Space of Test Functions A function x defined on R is a test function if is infinitely differentiable vanishes on the complement of some closed bounded interval a,b; i.e., has compact support. We denote the linear space of test functions by C c R.It is obvious that C c R L 1 R C c R L 2 R but a more surprising fact is true. Theorem 4.1 For every f L p R, 1 p, such that n f p 0 as n. there exists a sequence n C c R We describe this by saying that C c R is a dense subspace of L p R. The proof of this theorem as well as additional information about test functions will be given later when we develop the theory of generalized functions. We now proceed to use the test functions to 9

10 extend the Fourier transform to L 2 R. Theorem 4.2 For every C c R, the Fourier transform,, exists, in addition, belongs to L 2 R. Moreover, It follows now from theorems that 4.1 1) for every f L 2 R, there exists a sequence n x C c R such that n f 2 0 as n. (hence n m 2 0 as m,n ) 2) since n C c R, the Fourier transforms, n, exist, in addition, n L 2 R. Moreover, n m n m as m,n 3) since L 2 R is complete, the sequence n is Cauchy, there exists a unique v L 2 R such that n v 2 0 as n 4) since n f, n v in L 2 R, n T F n, we define v T F f Note that we can also invert the transformation as follows, 1) for every F L 2 R, there exists a sequence n x C c R such that n F 2 0 as n. (hence n m 2 0 as m,n ) 2) since n C c R, the inverse Fourier transforms, n x R n e i x d exist, in addition, n L 2 R. Moreover, n m n m as m,n 3) since L 2 R is complete, the sequence n is Cauchy, there exists L 2 R such that n 2 0 as n 4) since n F, n in L 2 R, n T F 1 n, we define T F 1 F 5) since T F f F F 0, f T F f 2 2 0, we have f. We summarize these observations in the following theorem. Theorem 4.3 For every f L 2 R there exists a unique F L 2 R such that F T F f in the sense that if n x C c R is such that n f 2 0 as n, 10

11 then n T F n L 2 R is such that n F 2 0 as n, In addition T F 1 F f, where the equality is in the sense of L 2 R. Problem 8 Show that the definition of the L 2 R Fourier transform F does not depend on the choice of the sequence n Each of the Fourier transform properties detailed in theorems 3.2 to 3.6 holds for the L 2 R extension of the Fourier transform. In addition, we have Theorem 4.4 For f,g L 2 R the Fourier transforms, F,G satisfy f,g 2 R fx g x dx 2 R F G d 2F,G Here g, G are used to denote complex conjugates. Even though we are dealing exclusively with real valued functions f, g the Fourier transforms may be complex valued we have therefore stated the result for the complex form of the inner product on L 2 R. Note that when f g (4.2) reduces to (4.1). The result (4.1) is known as the Parseval relation (4.2) is called the Plancherel relation. Together they assert that the Fourier transform is a Hilbert space isometry, meaning that T F maps L 2 R onto itself in a one to one norm inner product preserving fashion. 5. Some Applications of the Fourier Transform Consider the functions gx 1 x if x 1 0 if x 1 px x if x 1 0 if x 1 Both g p belong to L 2 R, in fact g is continuous with compact support while p has compact support but is discontinuous at x 1. Neither function is differentiable in the classical sense. Since g p belong to L 2 R, they have Fourier transforms in L 2 R, given by G 2 2 1cos, P 2 sin1cos. 2 We observe that ig L 2 R, but ip L 2 R. Since ig T F g x we conclude that g has a derivative in the sense of L 2 R this derivative is given by fx T F 1 ig sgnx if x 1 0 if x 1. On the other h, ip L 2 R implies that p does not have a derivative in L 2 R (although we will see later that p has a derivative in the sense of generalized functions). Problem 9 Find the Fourier transforms use them to find the derivatives in the L 2 sense of 11

12 fx 1x 2 if x if x 2 1, gx x 2 if x if x 2 1 Theorem 5.1 For f,g L 2 R, the following statements are equivalent (a) fx gx where the equality is in the L 2 R sense (b) if G (c) fx h fx D h f g 2 0 as h 0, where D h fx h for h 0 (d) there exists a sequence n x C c R such that n f 2 0 n g 2 0 as n, Proof-a b If f f L 2 then T F f if L 2 so clearly a b b c : If f L 2 then for h 0, theorem 3.3 implies, Since eih 1 ih T F D h fx T F fx h fx/h eih 1F h e ih 1 ih 1 as h 0 if : G h (L Hopital s rule), it follows that T F D h fx G h if : G as h 0.(convergence in L 2 ) Since the Fourier transform is an isometry on L 2 this implies D h fx T F 1 G h T F 1 G g f as h 0. c a :Note that for any test function, D h fx, 2 h 1 fs,she j 2 fx,x 2 fx,d h x 2 Now c) implies that as h tends to zero, D h fx, 2 g, 2 for any test function, hence we conclude g f L 2. fx,d h x 2 f, 2 d a : Suppose, as n, n f n g in L 2 For any test function, n, 0 n, 0 n as n Then g f L 2. g, 0 f, 0 That a) implies d) can be proved using the mollifier theorem. The examples have suggested that havingf F in L 2 implies that f is in L 2, which is to say that f has a derivative in the L 2 sense. While having a derivative in the L 2 sense is not the same as having a derivative in the classical point wise sense, it does imply some 12

13 additional regularity as the next proposition will show. Proposition 5.2 Suppose that f f L 2 R. Then f is continuous on R f 0 as x. Proof- For arbitrary x,y R, x y, we can use the Cauchy-Schwarz inequality to write fx fy y x f s ds y x 1 2 ds 1/2 y x f s 2 ds 1/2 x y 1/2 f 2, from which it is clear the f is continuous at each point. Next, since a b fx f x 2 dx 0, for all a,b we have since f is continuous, a b fx 2 dx a b f x 2 dx 2 a b fxf xdx, 2 a b fxf xdx fb 2 fa 2. Now since f f are both square integrable, lim b fx 2 b dx lim a,b a a,b f a x 2 dx 0 where a b are allowed to tend independently to plus infinity. Combining these last three results leads to the conclusion that fx tends to a constant as x. Since f is square integrable, this constant must be zero. Similar reasoning implies f tends to zero as x tends to minus infinity. The proposition asserts that having a derivative in the L 2 -sense does imply some additional smoothness for the function. Precisely, it says the following: i.e., f F in L 2 implies that f is in L 2 f f in L 2 implies that f is in C 0 We might expect this to generalize to, f p F in L 2 for p M implies that f p is in L 2 for p M f f p in L 2 for p M implies that f p is in C 0 for p? This precise assertion is given in the next proposition. Recall that if f L 1 R then F T F f C 0 R. If, in addition, F L 1 R, then T F 1 F C 0 R T F 1 Ff 1 0, which means T F 1 Fx fx belong to the same equivalence class in L 1 R. More generally, if p F L 1 R then T F 1 i p F C 0 R T F 1 i p F f p 1 0; i.e., f L 1 R has continuous derivatives up to order p if it is the case that p F L 1 R. Evidently, decay at infinity for the Fourier transform of a function in L 1 R implies smoothness for the function. Similarly, theorem 3.5 implies that if x k fx L 1 R, then the Fourier transform, F, must have continuous derivatives up to order k. We would like to consider this connection between decay at infinity for f or F smoothness for F or f, in the symmetric setting of L 2 R. 13

14 Theorem 5.3 (Sobolev Embedding Theorem in R 1 ) Suppose f L 2 R p F L 2 R for p M. Then f has derivatives in the L 2 R sense of all orders M, these L 2 derivatives, f j, belong to C 0 for j M1. Proof- Recall that if f L 1 R then F T F f C 0 R. If, in addition, F L 1 R, then T F 1 F C 0 R T F 1 Ff 1 0, which means T F 1 Fx fx belong to the same equivalence class in L 1 R.This will form the basis for our proof. Since we assumed that p F L 2 R for p M, it follows that f j T F 1 i p F, for p M, where this equality is in the L 2 R sense. Note that p F L 2 R for p M, implies 1 2 M/2 F L 2 R; i.e., R 1 2 M F 2 d C F 2 It remains to be discovered for which integers q it is true that q F L 1 R, since for these values of q it is then the case that T F 1 i q F C 0 R T F 1 i q F f q x for almost all x. Write R q F d R q R 2q 1 2 M 1 2 M/2 d C F R 2q 1 2 M d 1/2 1 2 M/2 F d R 1 2 M F 2 d 1/2 It only remains to be seen for which integers q it is true that IM,q R 2q 1 2 M d 1/2 1/2 is finite. It is easy to check that if M is 1 then we get q 0. If M is 2, then q 1, if M is 3 then q 2. Evidently, IM,q is finite if q M1. That is, if p F L 2 R for p M, then f q x C 0 R for q M1. We will now use this result to consider several example problems for PDE s. Example 5.1 Suppose 2 ux,y 0, x R, y 0 ux,0 fx x R, where f L 2 R. If U,y T F ux,y denotes the Fourier transform, in x, of u(x,y), then 2 U,y T F xx ux,y yy U,y T F yy ux,y, 2 U,y yy U,y 0, y 0 U,0 F. 14

15 It follows that U,y e y F, y 0. Now, two things are clear. First, so, by theorem 4.4, U,y U,0 2 e y F F 2 0, as y 0 ux,y ux,0 2 ux,y fx 2 0, as y 0. i.e., ux,y tends (in the L 2 sense) to fx as x,y tends to the boundary. Second, for all positive integers, p q T F x p y q ux,y i p y q U,y i p y q e y F i p q e y F. Now, because of the presence of e y this expression belongs to L 2 R for all positive y all positive integers p,q. It follows that ux,y has L 2 derivatives of all orders in both x y for all x all positive y. Then by theorem 5.2, ux,y has continuous derivatives of all orders as well. Then u is infinitely differentiable in x y for all x all positive y, given only that f is square integrable. To complete the solution, note that T F 1 e y 2y y 2 x 2 gx,y (see prob 7) T 1 F e y F 2 1 g,y f (Theorem 3.6). Then ux,y 2 1 g,y f y fz y 2 x z dz. 2 Problem 10 Show that in the special case fx I A x this solution becomes ux,y 1 arctan What would be the solution if fx I A x 2A? x A y arctan x A y. Example 5.2 Suppose t ux,t xx ux,t 0, x R, t 0, ux,0 fx, x R, where f L 2 R. If U,t T F ux,t denotes the Fourier transform, in x, of u(x,t), then 2 U,t T F xx ux,t t U,t T F t ux,t, t U,t 2 U,t 0, t 0 U,0 F. It follows that U,t e t2 F, t 0. As in the previous example, it is clear that U,t U,0 2 e t2 F F 2 0, as t 0 15

16 so, by theorem 4.4, ux,t ux,0 2 ux,t fx 2 0, as t 0. i.e., ux,t tends (in the L 2 sense) to fx as x,t tends to the initial line. Second, for all positive integers, p q T F x p t q ux,t i p t q U,t i p t q e t2 F i p 2 q e t2 F. Now, because of the presence of e t2 this expression belongs to L 2 R for all positive t all positive integers p,q. It follows that ux,t has L 2 derivatives of all orders in both x t for all x all positive t. Then by theorem 5.2, ux,t has continuous derivatives of all orders as well. Then u is infinitely differentiable in x t for all x all positive t, given only that f is square integrable. To complete the solution note that x2 T 1 F e t2 e 4t hx,t (see prob 7) t T 1 F e t2 F 2 1 h,t f (Theorem 3.6). Then ux,t 2 1 x y2 1 e 4t fydy 4t Problem 11 Show that in the special case fx I A x this solution becomes ux,t 1 A 4t A e x y2 4t dy where 1 2 erf 1 2 erfx 2 0 A x t e y 2 dy. erf 1 2 x A t, Example 5.3 Suppose tt ux,t a 2 xx ux,t 0, x R, t 0, ux,0 fx, x R, t ux,0 gx, x R, where f,g L 2 R. If then It follows that U,t T F ux,t denotes the Fourier transform, in x, of ux,t, 2 U,t T F xx ux,t tt U,t T F tt ux,t, tt U,t a 2 2 U,t 0, t 0 U,0 F t U,0 G. 16

17 In this case we have implying that Similarly, U,t cosat F sinat G a, t eiat F e iat F 1 2a eiat G i U,t U,0 2 cosatf F sinat G a cosatf F 2 sinat G a ux,t ux,0 2 ux,t fx 2 0, as t 0. 2 e iat G i 2 0, as t 0, t U,t t U,0 2 a sinatf cosat G G 2 a sinatf 2 cosat G G 2 0, as t 0, from which it follows, using theorem 4.4, that t ux,t t ux,0 2 t ux,t gx 2 0, as t 0. Then the initial conditions are satisfied if the limits are understood to mean L 2 limits. Note that by theorem 3.3, where Then T F 1 e iat F fx at T F 1 e iat G i G ih T F h x; i.e. h x gx. ux,t 1 2 fx at fx at 2a 1 hx at hx at 1 2 fx at fx at 2a 1 xat gs ds. xat hx at In contrast to the previous two examples, the transform of the solution in this case contains no smoothing operator in the form of a rapidly decreasing exponential function of. Here the modifier in the solution is an oscillatory function of t does not decrease at all as tends to inifinity. Here, in order to have T F x p t q ux,t i p t q U,t i p ia q U,t belong to L 2 R, it would be necessary to place conditions on the data; e.g., if 3 F 2 G both belong to L 2 R, then this is sufficient to imply ia 3 U,t belongs to L 2 R, which implies then that u(x,t) has continuous derivatives of up to order two with respect to both x t. If the data is not this smooth, then the PDE is not satisfied in the classical sense but only in some weaker sense. The smoothness of the solution in this case depends on the smoothness of the data. In each of these examples, the Fourier transform of the solution equals the Fourier transform of the data multiplied by a modifier that is a function of the transform variable,, 17

18 the nontransform variable, y or t. When the exponent in the modifier is real negative for all relevant values of the nontransform variable then the modifier decreases exponentially as tends to infinity, resulting in the extreme smoothness of the solutions to the equation, independent of the smoothness of the data. This is the case for the Laplace heat equations. If the exponent in the modifier is imaginary but linear in, then the modifier acts as a translation operator as prescribed in theorem 3.3. Then the solution is only as smooth as the data but the data is transmitted without distortion. This is the case for the transport wave equations (in the case of the transport equation with a lower order term, the lower order term introduces some distortion, see example 5.6 below). Thus, even without obtaining explicit formulas for the solutions in these problems, we can deduce qualitative properties of the solutions from looking at the transforms of the solutions. We will now consider additional examples see what we can determine about their solutions. Example 5.4 Suppose 2 ux,y,z 0, x,y R 2, z 0 ux,y,0 fx,y x,y R 2, where f L 2 R 2. If U 1, 2,z R R e ix 1iy 2ux,y,z dx dy T F ux,y,z denotes the Fourier transform, in (x,y) of ux,y,z, then U,z T F xx u yy u zz U,z T F zz ux,y,z, U,z zz U,z 0, z 0 U,0 F. It follows that U,z e z F, z 0, where It is clear that the qualitative behavior of the solution to Laplace s equation is essentially the same in 3 dimensions as it was in 2 dimensions. The formula representing the solution in terms of the data might be more complicated in the higher dimensional case but the smoothness limit at the boundary results are not changed. The heat equation shows a similar lack of variation in qualitative behavior as the number of dimensions increases. Example 5.5 Suppose where f L 2 R 2. If ux,y,t, then tt ux,y,t a 1 2 xx ux,y,t a 2 2 xx ux,y,t 0, x,y R 2, t 0, ux,y,0 fx,y, x,y R 2, t ux,y,0 0, x,y R 2, U 1, 2,t T F ux,y,t denotes the Fourier transform, in (x,y), of U,t T F xx ux,y,t yy ux,y,t tt U,t T F tt ux,y,t, 18

19 tt U,t a a U,t 0, t 0 U,0 F t U,0 0. It follows that U,t cos t a a F, t eit a a F e it a a F It is apparent from this that while the solution to the wave equation in 2 dimensions continues to have the same properties in terms of lack of smoothness limiting behavior as t tends to zero as we observed for the 1-dimensional solution, the modifier that multiplies the transform of the data is no longer a simple translation operator. The solution depends on the data in a much more complicated way than what we saw for the 1-dimensional wave equation. This is consistent with the remark that solutions to the wave equation are very sensitive to changes in the dimension. Example 5.6 Suppose t ux,t a x ux,t bux,t 0, x R, t 0, ux,0 fx, x R, where f L 2 R. If U,t T F ux,t denotes the Fourier transform, in x, of ux,t, then t U,t aibu,t 0, t 0 U,0 F It follows that U,t e iabt F e bt e iat F, t 0. ux,t e bt fx at, x, t 0. Clearly as in the case of the wave equation, there is L 2 convergence to the initial condition as t tends to zero there is no smoothing of the data since the modifier is a simple translation operator. The effect of the lower order term in the equation is simply to introduce exponential time growth or decay of the solution depending on the sign of the coefficient b. In more dimensions the solution becomes U,t e iabt F e bt e ita F e bt e ita e ita n n F 1,..., n ux,t e bt fx 1 a 1 t,...,x n a n t, x, t 0. a a 1,...,a n 1,..., n, Evidently the solutions to the transport equation do not have the same sensitivity to dimension as the solutions to the wave equation. Example 5.7 Suppose tt ux,t a 2 xx ux,t d 2 ux,t 0, x R, t 0, ux,0 fx, x R, t ux,0 0, x R, where f L 2 R. If U,t T F ux,t denotes the Fourier transform, in x, of ux,t, then, tt U,t a 2 2 d 2 U,t 0, t 0 U,0 F 19

20 t U,0 0. It follows that U,t costq F, t 0, 1 2 eitq e itq F, where Q a 2 d/ 2. Note that if d 0 then Q a the exponent in the modifier is linear in. Then the modifier is just a simple shifting operator leading to The d Alembert solution of the wave equation where the speed of propagation is equal to a Q. On the other h, if d is not zero then the exponent in the modifier is not linear in, the modifier is not a shifting operator. Instead, expitq acts like a shift operator in which the propagation speed, Q, is a function of. Propagation where waves with different frequencies propagate at different speeds is sometimes referred to as dispersive wave propagation. Each of the following examples exhibits a similar type of modifier, consequently describes some kind of dispersive propagation. i) Dispersion equation t ux,t a 2 xxx ux,t 0, x R, t 0, ux,0 fx, x R, U,t e ita2 3 F, ii) Schrodinger equation t ux,t i xx ux,t 0, x R, t 0, ux,0 fx, x R, U,t e it2 F, iii) Vibrating beam equation tt ux,t xxxx ux,t 0, x R, t 0, ux,0 fx, x R, t ux,0 gx, x R, U,t Cos 2 t F sin2 t G e it2 e it2 F e it2 e it2 G. In each of these cases, it is evident that the solutions will tend to the initial condition in the L 2 sense as t tends to zero that there is no smoothing of the data. It is also evident that the modifier acts as some version of the shifting operator that appears in the solution of the simple wave equation but that the shifting action varies with the transform variable. Exactly how this will act on the data when the Fourier transform is inverted is not clear, beyond the fact that the solution will involve a convolution integral operator of the form ux,t R Kx y,t fy dy. Problem 12 Analyze the following problem as to assumption of the initial condition, 20

21 smoothness of the solution whether the propagation is more wave-like or diffusion-like. t ux,t xxxx ux,t 0, x R, t 0 ux,0 fx, x R. Example 5.8 The following example illustrates some of the tricks that are often employed in order to find the inverse transform using only the short table of transforms that can be developed without contour integrations. Suppose Apply Fourier transform in x: 2 ux,y 0 x R, 0 y 1, ux,0 0, x R, y ux,1 gx L 2 R, x R. Then, solving the ode, we get But so Now so sinh y cosh 2 U,y U,y 0, U,0 0, U,1 G. U,y G sinh y cosh e y e y 1 e 1y e 1y e e 1e 2 e 2n1y e 2n1y, n0 U,y G e b e a e 2n1y e 2n1y n0 b a e s ds 2n1y U,y G e s ds 2n1y n0 Recall T F 1 e s 2s s 2 x 2 T 1 F G e s 2 1 2s gz dz s 2 2 x z so 2n1y ux,y 2n1y 1 n n1y 2n1y n0 2s gz dz s 2 2 x z ds 2s s 2 x z 2 ds gz dz Then, since we get 2n1y 2n1y 2s s 2 x z 2 ds log 2n1y 2 x z 2 2n1y 2 x z 2, 21

22 ux,y 2 1 2n1y 2 x z 2 log n0 2n1y 2 x z 2 gz dz where gx z,y gz dz gx z,y 2 1 log n0 2n1y 2 x z 2 2n1y 2 x z Orthogonal Families Generalized Fourier Series Using a basis for representing arbitrary vectors in R n has numerous advantages in dealing with problems in linear algebra. For most computational purposes, it is convenient if the basis is an orthonormal basis, meaning that the vectors in the basis are mutually orthogonal unit vector. If u 1,..., u n is such an orthonormal basis then an arbitrary v R n can be uniquely expressed as v n j1 v u j u j. It is our aim now to develop such representations in infinite dimensional function spaces. Let U R n denote a bounded open connected set in R n consider functions f,g L 2 U. The functions are said to be orthogonal if f,g 2 U fx gx dx 0. A countable family g 1 x,g 2 x,..., of functions in L 2 U is said to be an orthogonal family in L 2 U if g i,g j 2 0 if i j. The orthogonal family g j x is said to be an orthonormal family if, in addition to be mutually orthogonal, the functions satisfy g j 2 1 for every j; e.g., The family of functions G n x sinnx, n 1,2,... is an orthogonal family in L 2 0,1, the family g n x 2 sinnx, n 1,2,... is an orthonormal family in L 2 0,1. Suppose g 1 x,g 2 x,..., is an orthonormal family of functions in L 2 U, for an arbitrary f L 2 U, form the infinite sum n1 f,g n 2 g n x. It is not clear that this sum is convergent in L 2 U, even if it is convergent, it is not evident that it is convergent to f. In any case, we refer to the sum as the generalized Fourier series for f we refer to the coefficients f n f,g n 2 as the generalized Fourier coefficients for f. Proposition 6.1 Suppose g 1 x,g 2 x,..., is an orthonormal family of functions in L 2 U, for an arbitrary f L 2 U, let f n f,g n 2. Then for any integer N, any choice of constants a 1,a 2,...a N we have i) fx N n1 a n g n x ii) fx N n1 a n g n x Proof- For N a fixed positive integer, let 2 2 f 2 N n1 2 fx N f 2 n1 n g n x f,g n 2 2 N n1 f n a n

23 Then But S N x N n1 a n g n x. U fx S N x 2 dx U fx 2 dx 2 U fxs N xdx U S N x 2 dx. fxs N U xdx U fx N n1 S N U x 2 N dx U m1 a n g n x dx N n1 a m g m x N n1 a n g n x dx a n U fx g n xdx N n1 a n f n N m1 a m N n1 a n g n U xg m xdx N 2 n1 a n since g n,g m 2 mn Then U fx S N x 2 dx U fx 2 dx 2 N n1 a n f n N 2 n1 a n U fx 2 dx N n1 f 2 n N n1 U fx 2 dx N n1 f 2 n N n1 f n a n 2. f 2 n 2 N n1 a n f n N 2 n1 a n This is the result (i). Since N n1 f n a n 2 0, we get the result (ii). This last result asserts that among all linear combinations of the functions g 1,..., g N, the one that is closest to f in the L 2 U norm, is the combination with a n f n f,g n 2, n 1,...,N. Theorem 6.2 Suppose g 1 x,g 2 x,..., is an orthonormal family of functions in L 2 U, for an arbitrary f L 2 U, let f n f,g n 2. Then 1 (Bessel s inequality) n1 f 2 n f 2 2 U fx 2 dx. 2 (Riemann-Lebesgue lemma) f n 0 as n Proof-(1) Choosing a n f n in the sum S N x, using the results of the previous proposition, i.e., 0 f S N 2 dx U f 2 dx U N n1 f 2 n, N n1 f 2 n f 2 dx U f 2 2. Since this result holds for all positive integers N, the right side of the estimate does not depend on N, we are entitled to let N tend to infinity to obtain (1). Then the n-th term test implies the result (2). In the special case that the orthogonal family is 1,cosx,sinx,cos2x,sin2x,... in L 2 0,2 then the (2) takes the form 0 2 fx sinnx dx 0, 0 2 fx cosnx dx 0, as n. This is what is often referred to as the Riemann-Lebesgue lemma. The Bessel s inequality implies that for an arbitrary f L 2 U, the sequence f n f,g n 2 of generalized Fourier coefficients belongs to 2. Then it is also evident that the sequence of partial sums S N x, is a Cauchy sequence therefore converges in L 2 U to 23

24 some limit, Sx. However, it is not necessarily the case that S f. An orthonormal family with the property that for every f L 2 U, the generalized Fourier series converges to f in L 2 U is said to be a complete orthonormal family. Theorem 6.3 Suppose g 1 x,g 2 x,..., is an orthonormal family of functions in L 2 U, for an arbitrary f L 2 U, let f n f,g n 2. Then the following assertions are all equivalent: 1. g 1 x,g 2 x,..., is a complete orthonormal family 2. n1 f 2 2 n f 2 3. f n 0 n, if only if f 0 4. f,g L 2 U, f,g 2 n1 f n g n 5. f S N 2 0 as N As a result of this theorem, it follows that L 2 U is isometrically isomorphic to 2. Theorem 6.4 Suppose g 1 x,g 2 x,..., is a complete orthonormal of functions in L 2 U. Then for an every f L 2 U, f n f,g n 2 belongs to 2 with f n 2 f 2. Conversely, for every f n 2, the sequence S N N n1 f n g n, converges to a unique limit, f, in L 2 U f n 2 f 2. Just as the Fourier transform provided an alternative but equivalent representation for functions in L 2 R n, the sequence of generalized Fourier coefficients provides an alternative but equivalent representation for functions in L 2 U. The functions in the complete orthonormal family are the elements of a countable basis for L 2 U, every function in L 2 U can be expressed uniquely as a (possibly infinite) linear combination of these elements. While it is relatively easy to find examples of orthonormal families of functions in L 2 U, it is not clear how to determine whether or not a family is complete. We shall see now one source of complete orthonormal families. Sturm-Liouville Systems Consider the problem of finding all scalars for which there exist nontrivial solutions to the following boundary value problem d/dxpx du/dx qxux rxux, a x b, C 1 ua C 2 u a 0, C 3 ub C 4 u b 0, 6.1 Clearly the trivial solution satisfies 6.1 for all choices of the parameter. Any value which leads to a nontrivial solution will be called an eigenvalue of the problem the corresponding nontrivial solution will be called an eigenfunction of the problem, corresponding to the eigenvalue. Note that if u ux is an eigenfunction corresponding to the eigenvalue, the for every nonzero constant k, the function kux is also an eigenfunction for the same eigenvalue. A problem of the form (6.1) is called a Sturm-Liouville problem. Theorem 6.5 Suppose the coefficients in the Sturm-Liouville problem (6.1) satisfy 24

25 Then px, p x, qx rx are all continuous on [a,b] px 0, rx 0 for all x a,b C 1 2 C 2 2 0, C 3 2 C 4 2 0, i) the Sturm-Liouville problem (6.1) has countably many real eigenvalues n ii) for each eigenvalue there is a single independent eigenfunction, eigenfunctions corresponding to distinct eigenvalues satisfy a b ux,j ux, k rx dx 0 j k. iii) The family of normalized eigenfunctions ux, n are a complete orthonormal family in L 2 a,b for the weighted inner product f,g 2 a b fx gx rx dx. The proof of the Sturm-Liouville theorem is beyond the scope of this course. Instead we will list several examples of S-L problems the associated eigenvalues eigenfunctions. In each of the following examples we have px rx 1, qx 0, on a,b 0,1. Then since rx 1, the weighted inner product of the theorem reduces to the usual L 2 inner product. Example 6.1- Choose C 1 C 3 1, C 2 C 4 0, Then u"x ux 0 x 1, u0 u1 0. In this case, we find n n 2 2, n 1,2,... u n x sinnx Example 6.2- Choose C 1 C 3 0, C 2 C 4 1, Then u"x ux 0 x 1, u 0 u 1 0. In this case, we find n n 2 2, n 0,1,2,... u n x cosnx Example 6.3- Choose C 1 C 4 1, C 2 C 3 0, Then u"x ux 0 x 1, u0 u 1 0. In this case, we find n n , n 1,2,... 25

26 u n x sin n 1 2 x Example 6.4- Choose C 1 C 4 0, C 2 C 3 1, Then u"x ux 0 x 1, u 0 u1 0. In this case, we find n n , n 1,2,... u n x cos n 1 2 x Problem 12 Show that if Lux d/dxpx du/dx qxux, then for smooth functions ux, vx Lu,v 2 u,lv 2 pxu v v u a b Show that if ux, vx both belong to D 1 u C 2 a,b : ua ub 0 then this reduces to Lu,v 2 u,lv 2 ; i.e., L is self adjoint. Show the result continues to hold if D 1 is replaced by any one of the following domains for L : Problem 13 D 2 u C 2 a,b : u a u b 0 D 3 u C 2 a,b : ua u b 0 D 4 u C 2 a,b : u a ub 0. Show that if is an eigenvalue of problem (6.1), then is necessarily real. Problem 14 Show that if, are distinct eigenvalues of problem (6.1), then associated eigenfunctions must satisfy a b rx ux, ux, dx 0. Example 6.5- Consider the problem, u"x ux 1 x 1, u1 u1 u 1 u 1. The boundary conditions in this problem are called periodic boundary conditions are not of the type indicated in problem (6.1). Nevertheless, in this slightly modified S-L problem there are an infinite set of eigenvalues n n 2 2, n 1,2,... but now there are two independent eigenfunctions for each eigenvalue, except the eigenvalue zero for which there is just one eigenfunction. We have 0 0 u 0 x 1 n n 2 2 u n x cosnx, v n x sinnx, 26

27 Alternatively, we can express the eigenfunctions as follows, 0 0 u 0 x 1 n n 2 2 u n x e inx, v n x e inx, In the first case, the Fourier expansion would have the form where fx 1 a 2 0 n1 a n cosnx b n sinnx a n fx cosnx dx, bn 1 0 in the second case it has the form 1 fx sinnx dx, fx n c n e inx, c n fx e inx dx. 0 In either case we have S N f 2 0, as N. 7. Applications to Partial Differential Equations Consider the following problem for Laplace s equation 2 ux,y Fx,y 0 x,y 1 ux,0 ux,1 0 0 x 1, 7.1 u0,y u1,y 0, 0 y 1, recall the eigenfunctions from example 6.1, u n x sinnx, n 1,2... We note that the functions satisfy U mn x,y u m x u n y 2 U mn x,y u" m x u n y u m x u" n y m u m x u n y n u m x u n y m n u m x u n y. Evidently, the functions U mn x,y are eigenfunctions for the problem (7.1), with eigenvalues mn m n m 2 n 2 2, m,n 1,2,... It can be shown that since u n x is a complete orthogonal family in L 2 0,1 then it is necessarily the case that the orthogonal family U mn x,y u m x u n y is a complete orthogonal family in L 2 0,1 0,1. We can use this family of eigenfunctions to construct a Hilbert scale for L 2 0,1 0,1 L 2 U. We define H s U f L 2 U : 1 mn s/2 f mn 2, s 0. Now this scale can be used to asses the regularity of the solution to problem (7.1). First we show how this solution can be constructed. Since the family U mn x,y u m x u n y is a complete orthogonal family in L 2 U, any function in L 2 U can be expressed in terms of this family. In particular, Fx,y m,n1 F mn U mn m1 n1 F mn u m x u n y ux,y m,n1 C mn U mn m1 n1 C mn u m x u n y. 27

28 Since F is known, we can compute the coefficients F mn from F,U mn 2 U F, U mn, d d U j1 k1 F jk u j u k U mn, d d m1 1 1 n1 F jk uj u m 0 d uk u n d 0 m1 n1 F jk u j,u m 2 u k,u n 2 F mn u m 2 2 u n 2 2 4F mn. The coefficients, C mn, in the expansion for the solution, ux,y, are computed from the equation by noting that 2 ux,y ux,y 2 m1 n1 C mn U mn x,y m1 n1 2 U mn x,y C mn m1 n1 mn U mn x,y C mn, 2 ux,y Fx,y m1 n1 mn C mn F mn U mn x,y 0, from which it follows that i.e., C mn F mn mn, m,n 1,2,... ux,y m1 n1 F mn mn u m x u n y. Now observe that if F L 2 U H 0 U, then F mn 2 1 mn C mn 1 mn F mn mn ~F mn 2. This implies that u H 2 U if F H 0 U. More generally, if F H m U then 1 mn m/2 F mn 2 1 mn m2/2 C mn 1 mn m/2 1 mn F mn mn This implies that u H m2 U if scale from the data. Now consider the related problem ~1 mn m/2 F mn 2. F H m U; i.e, the solution is 2 steps up the regularity 2 ux,y 0 0 x,y 1 ux,0 f 0 x, ux,1 f 1 x 0 x 1, 7.2 u0,y g 0 y u1,y g 1 y, 0 y 1. For convenience, suppose f 0 x g 0 y 0, where 2 vx,y 0 0 x,y 1 note then that ux,y vx,y wx,y 28

29 vx,0 0, vx,1 f 1 x 0 x 1, 7.3a v0,y 0, v1,y 0, 0 y 1, 2 wx,y 0 0 x,y 1 wx,0 0, wx,1 0, 0 x 1, 7.3b w0,y 0, w1,y g 1 y, 0 y 1. Instead of exping v w in terms of the eigenfunctions U mn (this is called the full eigenfunction expansion ) we will exp v in terms of the eigenfunctions u m y we will exp w in terms of the eigenfunctions u n x. The reason for this will become clear in a moment. We write vx,y m1 v m y u m x m1 v m y sinmx wx,y n1 w n x u n y n1 w n x sinny. Substituting these expressions into the equations 7.3a,b for v w leads to m1 v" m y m 2 v m y sinmx 0 n1 w" n x n 2 w n x sinny 0. Since the eigenfunctions are orthogonal, they are certainly linearly independent then it follows from these last 2 equations that v" m y m 2 v m y 0, m 1,2, a w" n x n 2 w n x 0, n 1,2, b Substituting the eigenfunction expansions into the boundary conditions, leads to v m 0 0 v m 1 f 1,u m 2 f m 1, m 1,2, a w n 0 0 w n 1 g 1,u n 2 g n 1, n 1,2, b The general solutions for 7.4a,b are given by v m y Ae my Be my, w n x Ce nx De nx then the boundary conditions 7.5a,b lead to the results v m y f 1 m e my e my e m e m, w n x g 1 n e nx e nx e n e n, ux,y m m1 f e my e my 1 e m e m sinmx n n1 g e nx e nx 1 e n e n sinny. Note first that which means that e my e my e m e m e m1y for 0 y 1, m 1,2,... v m y f 1 m e my e my e m e m Ce my f 1 m with y 1y 0 for 0 y 1. 29

30 But then the decaying exponential factor ensures that m pq v m y 2 for all positive integers p,q all y, 0 y 1, which implies that x p y q vx,y L 2 U for all positive integers p,q. A similar argument shows that x p y q wx,y L 2 U for all positive integers p,q, together, these results show the smoothness we have come to expect of solutions to Laplace s equation on the interior of the domain. Note that this smoothness is independent of the data smoothness since we assumed only L 2 regularity for the data. Note finally that hence f 1 m e my e my e m e m f 1 m as y 1 v m y f 1 m 2 0 as y 1. Then the isometric isomorphism between L 2 0,1 2 ensures that Similarly, thus vx,y f 1 x 2 0 as y 1. wx,y g 1 y 2 0 as x 1, ux,y f 1 x in L 2 0,1 as y 1 ux,y g 1 y in L 2 0,1 as x 1 i.e., ux,y satisfies the boundary conditions in the L 2 sense. Exping the solution in terms of just one or the other of the 2 families of eigenfunctions in the problem was necessary in the case of inhomogeneous boundary data since the 2-d eigenfunctions U mn x,y vanish at all points of the boundary. Exping the solution in terms of one of the 1-d families of eigenfunctions is referred to as a partial eigenfunction expansion. Consider now the following initial boundary value problems t ux,t K xx ux,t 0 x 1, t 0, ux,0 fx, 0 x 1, 7.6 u0,t u1,t 0 t 0, tt wx,t a 2 xx wx,t 0 x 1, t 0, wx,0 fx, 0 x 1, 7.7 t wx,0 0, 0 x 1, w0,t w1,t 0 t 0. If we were to assume a solution of the form this into the pde, we would find T t KTt X "x Xx. ux,t Xx Tt for (7.6), then on substituting Since the two sides of this equation depend on different variables, it follows that both sides 30