Consequences of Orthogonality

Transcription

1 Consequences of Orthogonality Philippe B. Laval KSU Today Philippe B. Laval (KSU) Consequences of Orthogonality Today 1 / 23

2 Introduction The three kind of examples we did above involved Dirichlet, Neumann and periodic boundary conditions. In all three cases, the boundary conditions were homogeneous. We reflect on these examples and put the techniques used into a more general mathematical framework. We will see that the nice properties integrals of sines and cosines have which allowed us to solve these problems more easily were not an accident or luck. In the process, we will introduce some important concepts from linear algebra and prove some important results about orthogonality and eigenvalues. We then apply these results to the last kind of boundary conditions we haven t studied yet: mixed or robin boundary conditions. We will also apply these results to boundary conditions which are not homogeneous. Philippe B. Laval (KSU) Consequences of Orthogonality Today 2 / 23

3 Linear Algebra Concepts There are many parallels between linear algebra and differential equations. They are based upon similar theoretical foundations we quickly review here. The underlying space under which both are studied is called a vector space. A vector space is not just a space containing the type of vectors studied in a multivariable class. It can contain many other objects such as matrices, functions,... A vector space is simply a space of elements which satisfies certain properties. As a refresher, we give the definition of a vector space. Philippe B. Laval (KSU) Consequences of Orthogonality Today 3 / 23

4 Linear Algebra Concepts: Vector Space Definition (Vector Space) A vector space is a non-empty set of objects V, called vectors, on which are defined two operations, called addition and scalar multiplication which must satisfy the ten axioms listed below. These axioms must be satisfied for every vector u, v, w in V and every scalar c and d. 1 u + v V, u + v = v + u, (u + v) + w = u + (v + w) 2 There exists a vector in V, called the zero vector and denoted 0 such that u + 0 = u 3 For every vector u in V, there exists a vector u also in V such that u + ( u) = 0 4 cu V, c (u + v) = cu + cv, (c + d) u = cu + du 5 c (du) = (cd) u, 1u = u Philippe B. Laval (KSU) Consequences of Orthogonality Today 4 / 23

5 Linear Algebra Concepts: Vector Space Remark When the above is satisfied, we say that (V, +,.) is a vector space. So it is important to understand that a vector space is a set of objects together with two operations acting on these objects. Remark How the two operations are defined depends on the set and the objects it contains. Remark The scalars which come into play in scalar multiplication can be either real numbers, in which case we have a real vector space, or complex numbers, in which case we have a complex vector space. Remark A vector is simply an element of a vector space. Philippe B. Laval (KSU) Consequences of Orthogonality Today 5 / 23

6 Linear Algebra Concepts: Inner Product Two important concepts associated with many vector spaces, are the concept of an inner product also known as a dot product and the concept of a norm. How the inner product is defined depends on the kind of vector space we have. For each example of vector spaces we give below, we will give the corresponding inner product. However, once we have an inner product, we can define the corresponding norm. The notation for the inner product is.,.. For example, if u and v are two vectors, then the dot or inner product between these two vectors is denoted u, v. When the vectors are elements of R 2, R 3, R n, we also use the notation u v for the inner product. Philippe B. Laval (KSU) Consequences of Orthogonality Today 6 / 23

7 Linear Algebra Concepts: Norm Definition Let V be a vector space and v V. The norm of v, denoted v is defined to be v = v v = v, v The notion of norm is very important, it allows us to have the concept of size or length of a vector. The notion of inner product allows us to determine when two vectors are orthogonal (perpendicular). Recall that u v u, v = 0 Philippe B. Laval (KSU) Consequences of Orthogonality Today 7 / 23

8 Linear Algebra Concepts Definition Let S = {v 1, v 2,..., v n } be a set of vectors in a vector space V. 1 We say that S is linearly independent if c 1 v 1 + c 2 v c n v n = 0 c 1 = c 2 =... = c n = 0 2 The expression c 1 v 1 + c 2 v c n v n is called a linear combination of the vectors v 1, v 2,..., v n. 3 If S in not linearly independent, then it is said to be linearly dependent. 4 We say that S spans V if every element in V can be written as a linear combination of elements of S. 5 A subset B of S is called a basis of V if B is linearly independent and spans V. In this case, the dimension of V is the number of elements in B. It can be proven that all the basis of a given vector space have the same number of elements. Philippe B. Laval (KSU) Consequences of Orthogonality Today 8 / 23

9 Linear Algebra Concepts Example A typical vector space studied in linear algebra is R n. Consider the vectors u = (u 1, u 2,..., u n ) and v = (v 1, v 2,..., v n ) from R n. n The inner product is u, v = u 1 v 1 + u 2 v u n v n = u i v i. The norm of a vector is u = u1 2 + u n u2 n = u and v are orthogonal, denoted u v when u, v = 0 that is when u 1 v 1 + u 2 v u n v n = 0. Let e i denote the vector in R n with 1 in the i th entry and 0 everywhere else. Then B = {e 1, e 2,..., e n } is a basis for R n. It is easy to see that it is linearly independent and that it spans R n. Since B has n elements, R n has dimension n. i=1 i=1 u 2 i Philippe B. Laval (KSU) Consequences of Orthogonality Today 9 / 23

10 Linear Algebra Concepts Example Consider the functions y (x) : [α, β] R, and let V = {y (x) : ay + by + cy = 0, where a, b, c R and a 0}. It can be shown V satisfies the properties of a vector space. The inner product is u (x), v (x) = β α u (x) v (x) dx for any two functions u (x) and v (x) in V. The norm of a function is then u (x) = ( ) 1 β u (x), u (x) = α u (x) u (x) dx 2. u (x) and v (x) are orthogonal, denoted u v if u (x), v (x) = β α u (x) v (x) dx = 0. Let B = {y 1 (x), y 2 (x)} be a fundamental solution set for ay + by + cy = 0 then B is a basis for V. Since B has two elements, V has dimension 2. Philippe B. Laval (KSU) Consequences of Orthogonality Today 10 / 23

11 Orthogonality and Eigenvalues We now focus on functions which are orthogonal. We look at families of functions. Definition A family of functions {f n (x) : [a, b] R} n=1 none of which is identically 0, is said to be an orthogonal family on [a, b] if f i, f j = 0, i j that is if b a f i (x) f j (x) dx = 0 We have already encountered several such families. Philippe B. Laval (KSU) Consequences of Orthogonality Today 11 / 23

12 Orthogonality and Eigenvalues Example { sin nπ } L x is an orthogonal family for 0 < x < L. It forms the basis of n=1 the Fourier sine series on 0 < x < L. Example { cos nπ } { L x or equivalently 1, cos nπ } n=0 L x is an orthogonal family for n=1 0 < x < L. It forms the basis of the Fourier cosine series on 0 < x < L. Example { 1, sin nπ L x, cos nπ } L x is an orthogonal family for L < x < L. It n=1 forms the basis of the Fourier series on L < x < L. Philippe B. Laval (KSU) Consequences of Orthogonality Today 12 / 23

13 Orthogonality and Eigenvalues Orthogonality played an important part in determining the coeffi cients in the Fourier series expansion of the initial condition when we were solving the IBVP in the preceding two sections. This was not an accident as we now show. Consider X + λx = 0 with Dirichlet, Neumann, periodic or mixed (Robin) boundary conditions. Suppose that λ 1 is an eigenvalue with corresponding eigenfunction X 1 and λ 2 is another eigenvalue with corresponding eigenfunction X 2. Then we have Using the identity X 1 = λ 1 X 1, X 2 = λ 2 X 2, λ 1 λ 2 (1) ( X 1 X 2 + X 1 X 2) = X 1 X 2 + X 1 X 2 (2) and integrating each side over a < x < b, we have Philippe B. Laval (KSU) Consequences of Orthogonality Today 13 / 23

14 Orthogonality and Eigenvalues or b a ( X 1 X 2 + X 1 X 2 ) b dx = a ( X 1 X 2 + X 1 X 2 ) dx (3) = ( X 1X 2 + X 1 X 2 ) b a b (λ 1 λ 2 ) X 1 X 2 dx a = X 1 (b) X 2 (b) + X 1 (b) X 2 (b) + X 1 (a) X 2 (a) X 1 (a) X 2 (a) We now see what the above equation becomes for each boundary condition. (4) Philippe B. Laval (KSU) Consequences of Orthogonality Today 14 / 23

15 Orthogonality and Eigenvalues Dirichlet: In this case, both eigenfunctions X 1 and X 2 satisfy X (a) = X (b) = 0 and the right-hand side of equation 4 is 0. Neumann: In this case, both eigenfunctions X 1 and X 2 satisfy X (a) = X (b) = 0 and the right-hand side of equation 4 is 0. Mixed or Robin: In this case, both eigenfunctions X 1 and X 2 satisfy αx (a) + βx (a) = 0 and αx (b) + βx (b) = 0 and the right-hand side of equation 4 is 0 (see homework). Periodic: In this case, both eigenfunctions X 1 and X 2 satisfy X (a) = X (b) and X (a) = X (b) and the right-hand side of equation 4 is 0 (see homework). Philippe B. Laval (KSU) Consequences of Orthogonality Today 15 / 23

16 for any pair of function f and g which satisfy 6. Philippe B. Laval (KSU) Consequences of Orthogonality Today 16 / 23 Orthogonality and Eigenvalues Hence, in all four cases we see that equation 4 becomes (λ 1 λ 2 ) b a X 1 X 2 dx = 0 (5) since λ 1 λ 2, we must have b a X 1X 2 dx = 0 which says that X 1 and X 2 are orthogonal for a < x < b. Definition Boundary conditions of the form α 1 X (a) + β 1 X (b) + γ 1 X (a) + δ 1 X (b) = 0 (6) α 2 X (a) + β 2 X (b) + γ 2 X (a) + δ 2 X (b) = 0 are called symmetric boundary conditions if ( f (x) g (x) f (x) g (x) ) b a = 0 (7)

17 Orthogonality and Eigenvalues Remark Before we continue, we make a few remarks. 1 The four boundary conditions we have studied so far are of the form described in equation 6 as long as they are homogeneous. 2 The four boundary conditions we have studied so far are symmetric for pairs of eigenfunctions as the computations above show, as long as they are homogeneous. This leads us to the following important theorem. Philippe B. Laval (KSU) Consequences of Orthogonality Today 17 / 23

18 Orthogonality and Eigenvalues Theorem Consider X + λx = 0 with any type of symmetric boundary conditions. 1 The sequence of eigenfunctions {X n (x)} forms an orthogonal family. 2 Suppose that b a f 2 (x) dx <. If f is expanded in an infinite series of these eigenfunctions f (x) = C n X n (x), a < x < b (8) n=1 then the coeffi cients for n = 1, 2,... are given by C n = f, X b n X n, X n = a f (x) X n (x) dx b a X n 2 (x) dx = b a f (x) X n (x) dx x n (x) 2 (9) Philippe B. Laval (KSU) Consequences of Orthogonality Today 18 / 23

19 Orthogonality and Eigenvalues: We make some important remarks before continuing. The first part of the theorem shows us that the nice formulas we used to find the coeffi cients in the two sections above was not just a calculus identity involving sine and cosines. It is in fact an inherent characteristic of eigenvalue problems with symmetric boundary conditions. In particular, this means that we don t have to verify the orthogonality condition for pairs of eigenvalues in an eigenvalue problem with symmetric boundary conditions. Theorem 11 guarantees they always happen. The second part of the theorem tells us that the technique we used to find the coeffi cients of the Fourier sine or cosine series apply to general eigenfunction expansions. Philippe B. Laval (KSU) Consequences of Orthogonality Today 19 / 23

20 Orthogonality and Eigenvalues In the work we did, we assumed there was such an expansion. In the next chapters, we will discuss when we have such an expansion. In all the work we did so far, we kind of assumed that λ was real. We are now in a position to prove it. Theorem Consider X + λx = 0 with any type of symmetric boundary conditions. All the eigenvalues are real and the corresponding eigenfunctions can be chosen to be real-valued functions. Again, let us emphasize that the four boundary conditions we have studied (Dirichlet, Neumann, mixed, and periodic) fall in this category as long as they are homogeneous. Philippe B. Laval (KSU) Consequences of Orthogonality Today 20 / 23

21 Orthogonality and Eigenvalues Looking back at the various examples we did, we noted that there was never a nontrivial solution for negative eigenvalues. We can now prove that this was not an accident. Theorem Consider X + λx = 0 with any type of symmetric boundary conditions. If the eigenfunctions satisfy then there are no negative eigenvalues. X (x) X (x) b a 0 (10) It is important to note that each of the boundary conditions we used in the previous two sections (Dirichlet, Neumann, and periodic) satisfy equation 10 as long as they are homogeneous. We can use this theorem to quickly rule out negative eigenvalues. We will see in the next section that mixed boundary conditions do not always satisfy equation 10. Philippe B. Laval (KSU) Consequences of Orthogonality Today 21 / 23

22 Summary We list the important results we found in this section. When solving the BVP X + λx = 0, the following is true: 1 The four boundary conditions (Dirichlet, Neumann, mixed, and periodic) are symmetric. 2 For symmetric boundary conditions, the eigenfunctions form an orthogonal family. Hence, for the four boundary conditions (Dirichlet, Neumann, mixed, and periodic), the eigenfunctions form an orthogonal family. 3 For symmetric boundary conditions, the eigenvalues are real. Hence, for the four boundary conditions (Dirichlet, Neumann, mixed, and periodic), the eigenvalues are real. 4 For symmetric boundary conditions for which the eigenfunctions also satisfy X (x) X (x) b a 0, there are no negative eigenvalues. This extra condition is always satisfied for Dirichlet, Neumann and periodic boundary conditions. It is not always satisfied for mixed boundary conditions. Hence, for mixed boundary conditions, the case λ < 0 will have to be checked. Philippe B. Laval (KSU) Consequences of Orthogonality Today 22 / 23

23 Exercises See the problems at the end of my notes on consequences of orthogonality. Philippe B. Laval (KSU) Consequences of Orthogonality Today 23 / 23