Differential Operators and their Adjoint Operators

Transcription

1 Differential Operators and their Adjoint Operators Differential Operators inear functions from E n to E m may be described, once bases have been selected in both spaces ordinarily one uses the standard unit vectors), by means of n n matrices A; we write Y = A X. This operation is linear because Ac 1 X 1 + c 2 X 2 )=c 1 A X 1 + c 2 A X 2. When m = n we often use the term linear transformation for this operation because it then transforms E n into itself. We will use the symbol PC 2 [a, b], or PC2 [a, b] to stand for the set of piecewise continuous functions on the interval [a, b]; supplied with the inner product and norm b f, g = fx)gx) dx; f = b fx) 2 dx. a This inner product and norm is ordinarily associated with the larger space 2 [a, b], consisting of square ebesgue integrable functions on the interval [a, b], but we don t want to get into the details of that at the present. Since for any two functions f and g in PC 2 [a, b] it is easily verified that a linear combination c 1 f + c 2 g continues to belong to that set, it follows that PC 2 [a, b] is a linear vector space; supplied with the inner product and norm we have just indicated it becomes an inner product space. Unlike the larger space 2 [a, b] it is not complete; sequences of functions f k x) PC 2 [a, b] which have the Cauchy property lim j,k f j f k = do not necessarily converge to a limit in PC 2 [a, b]; the limit does exist in the larger, complete space [ a, b] which, being a complete inner product space is a Hilbert space) but it may not be piecewise continuous. A linear transformation on PC 2 [a, b] is simply a linear function from that space into itself. It is very easy to cite examples. One may take a 1 a

2 fixed function, hx), piecewise continuous on [a, b] and define H : PC 2 [a, b] PC 2 [a, b]; Hf)x) =hx)fx). It is straightforward to see that for each f PC 2 [a, b] this operation H gives another function in that space and it is easy to verify the linearity of the operation since hx)c 1 fx)+c 2 gx)) = c 1 hx)fx)+ c 2 hx)gx). Another example of a linear transformation on PC 2 [a, b] is given by Kf)x) = b a kx, y)fy) dy, x [a, b], where the kernel kx, y) satisfies the conditions indicated in the section on linear transforms in particular, kx, y) continuous on [a, b] [a, b] would suffice). There is little precision in the terminology for these linear transformations; the terms linear function, linear transformation, linear operator are all used with no clear distinction between them. In these notes we will reserve the word operator for use in the term linear differential operator the term linear differential operator is standarut operator by itself is often used in other contexts in the general literature). For our purposes here an n -th order linear differential operator on [a, b] takes the form D = a x) dn dx + a 1x) dn 1 n dx a n 1x) d n 1 dx + a nx), where each a n k x) is a piecewise continuous function on [a, b] we often add further requirements). The operation of D on a piecewise n times continuously differentiable function yx) defined for x [a, b] is simply Dy)x) =a x) dn y dx + a 1x) dn 1 y n dx a n 1x) dy n 1 dx + a nx) yx). Most, but not all, of the examples we will cite correspond to a k x) a k, constant. As x is allowed to vary in [a, b] this process gives another 2

3 function, Dy)x) and it is a simple matter to verify that the process is a linear one. It is clear that the operator D cannot be applied to an arbitrary function yx) PC 2 [a, b] because, in general, functions in that space may not have the required derivatives. Unlike linear transformations on finite dimensional spaces, we here encounter a transformation, associated with the operator D, for which we have to specify a domain subspace; a subspace of the basic space in which we are operating, in this case PC 2 [a, b] = PC2 [a, b], consisting of those elements of the basic space to which the operation is to be applied. For the differential operator D as just defined, one begins by restricting the operation of D to the subspace PC2 n [a, b], which consists of functions yx) for which the n-th derivative y n) x) is defined and piecewise continuous on [a, b]. This implies that yx) and the lower order derivatives y k) x), k =1, 2,..., n 1 are all continuous and, in fact, y k) x) must have derivatives up to order n k. In addition, the domain subspace for D, which we will call S, needs to be further restrictey imposing n boundary conditions at the endpoints, a an, of the interval in question. In many cases n =2m is even and m boundary conditions are applied at each of these endpoints; exactly which boundary conditions depends on the specific operator and will become clearer as we discuss some of the examples to follow. Our main interest lies in eigenvalues and eigenfunctions of differential operators. A function φx), not equivalent to the zero function, in the domain subspace S of the differential operator D is an eigenfunction of D corresponding to the possibly complex) eigenvalue λ of D if Dφ)x) λφx), x [a, b]. 3

4 This means that φx) satisfies, for x [a, b], the differential equation a x) dn φ dx + a 1x) dn 1 φ n dx a dφ n 1 n 1 dx + a nx) φx) λφx). If we assume the eigenvalue λ is known, the general solution of this differential equation will involve n arbitrary constants. The purpose of the boundary conditions completing the definition of S, which we described earlier, is to determine these arbitrary constants. In general this specification of boundary conditions cannot be done in an entirely arbitrary way; it has to be done with due regard to the structure of the operator D. Definition 1 If D is a differential operator with domain subspace we will normally just say domain from this point on) S and for all yx) Sand all zx) S we have Dy, z = y, D z, then D is the adjoint operator for D. IfD = D and S = S we say that D is self-adjoint. Remark Strictly speaking we should say that D is an adjoint operator for D but it turns out that in virtually all cases of practical interest the adjoint operator D is unique if the domains S and S are chosen correctly. Example 1 Consider the differential operator definey Dy)x) =y x), x [, 1]; S = { yx) PC 1 2[a, b] y) = }. For zx) also in PC 1 2[, 1] we compute Dy, z = 1 x)zx) dx = yx)zx) 1 y 1 x) dx yx)z 4

5 1 = y1)z1) x) dx. yx)z Clearly, then, if we take D z)x) = z x) and take S to consist of functions zx) PC2[, 1 1] such that z1) = we will have 1 Dy, z = x)zx) dx = x) dx = y, D z y yx)z and we see that D as defined on S is the adjoint operator for D. Example 2 et D be the first order) differentiation operator of Example 1 but, in specifying S, we replace the boundary condition used there by the periodic condition y) = y1). Essentially the same calculations as carried out in Example 1 then show that D = D with S = S. In this situation, where the adjoint of D is D with the same domain, we say that D is skew-adjoint or anti-hermitian.) The eigenvalue - eigenfunction equation is dφ dx = λφ for which the general solution is φx, λ) =ce λx, c being an arbitrary constant. Using the periodic boundary condition which is part of the definition of S = S we have 1 ce λ = c = e λ 1 = ce λ. This is satisfied for c just in case λ = 2kπi for some integer k, <k<. These eigenvalues are clearly complex, as are the eigenfunctions, which we may take to be φ k x) =e 2kπix, <k<. Example 3 We consider the constant coefficient, second order, differential operator D y)x) = 2 y x 2 cyx), c real, 5

6 defined on the domain S = { yx) PC 2 2 [,]) a y)+b y ) =, a y)+b y ) =} with a,b,a,b all real, a,b not both zero, a,b not both zero. We claim that D is self-adjoint. This assertion rests on the calculation, for y, z S, Dy, z = = y x)zx) = y x)zx) yx)z x) ) y x) cyx) ) zx) dx y x)z x) +cyx)zx) ) dx + = y x)zx) yx)z x) ) yx)z x) czx)) dx + y, Dz. Clearly we have the self-adjointness property just in case y x)zx) yx)z x) ) =. Suppose, first of all, that a. Then, still for y, z S, we compute y )z) y)z ) =y ) z)+ b ) z ) a b y )+y) z ) =. a A similar result is obtained at x =ifa. If a = then b and the boundary condition at x = gives y ) =, z ) =, in which case we again have y )z) y)z ) =. A similar result is obtained at x =ifa =. The self-adjointness of D is proved. Our final example for this section shows how we deal with a second order differential operator with coefficients depending on x. 6 )

7 Example 4 et D be the differential operator definey Dyx) =c x) d2 y dx + c 1x) dy 2 dx + c 2x) yx), where the real valued functions c k x) PC 2 k, k =, 1, 2. We will assume c x), x [a, b]. We take the domain of D to be the subspace of PC 2 [a, b] consisting of functions yx) which also lie in PC 2 [a, b] and satisfy boundary conditions, with real coefficients, d a y a) +e a ya) =, y b) +e b yb) =. We assume that at least one of d a,e a and at least one of,e b are different from zero. Assuming that zx) PC 2 [a, b], we use integration by parts to compute b Dy, z = c x) d2 y a dx + c 1x) dy 2 dx + c 2x) yx) zx) dx b = c x) y x)zx) yx) c x)z x) + c 1 x) c x) ) zx) )) b + a yx) c x) z x)+ 2c x) c 1 x) ) z x)+ c x) c 1x)+c 2 x) ) zx) ) dx. The last integral shows that the adjoint differential operator shoule D zx) =c x) d2 z 2c dx x) c 2x)+ 1 x) ) dz dx x)+ c x) c 1x)+c 2 x) ) zx). If the boundary terms at x = b can be rewritten as c b) y b)+ e ) b yb) zb) c b)e b yb)zb) yb) c b)z b) + c 1 b) c b) ) zb) ). Applying the boundary condition satisfiey yx) atx = b, the boundary term at x = b vanishes just in case = c b)e b yb)zb) yb) c b)z b) + c 1 b) c b) ) zb) ) 7 a

8 = yb) c b)z b) + c 1 b) c b)+ c b)e b which is true if zx) satisfies the boundary condition c b)z b) + c 1 b) c b)+ c b)e b If we assume d a we similarly obtain c a)z a) + c 1 a) c a)+ c a)e a d a zb), zb) =. za) =. The adjoint domain D consists of functions in PC 2 [a, b] satisfying these boundary conditions. The cases wherein d a = and/or = and hence e a and/or e b are left as exercises. QuickCheck Exercises 1. Find the adjoint operator for Dyx) = d ) dx 1 + x) dy dx, the domain of D consisting of functions in PC 2 [, 1] for which y) = y1) =. 2. Find the adjoint operator for Dyx) =y x), the domain consisting of functions in PC 3 [, 1] for which y) = y ) =, y 1) =. 3. Show that c is an eigenvalue of D as defined in Example 3 if a = a =. 4. Compute the eigenvalues of D in Example 3 if b = b =. What are the corresponding eigenfunctions? 5. How are the boundary conditions characterizing the domain of D modified in the case of the operator D of Example 4 if one or both of d a, is equal to zero? 8