Differential Operators and their Adjoint Operators
Differential Operators Linear functions from En to Em may be described, once bases have been selected in both spaces (ordinarily one uses the standard unit vectors), by means ofn×nmatricesA; we write Y = AX. This operation is linear becauseA(c1X1 + c2X2) = c1AX1 + c2AX2.
When m = n we often use the term linear transformation for this operation because it then transforms En into itself.
We will use the symbol P C2[a, b], or P C20[a, b] to stand for the set of piecewise continuous functions on the interval [a, b]; supplied with the inner product and norm
hf, gi = Z b a f(x)g(x)dx; kfk = sZ b a |f(x)| 2 dx.
This inner product and norm is ordinarily associated with the larger space L2[a, b], consisting of square Lebesgue 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 P C2[a, b] it is easily verified that a linear combination c1f + c2g continues to belong to that set, it follows that P C2[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 L2[a, b] it is
not complete; sequences of functions fk(x) ∈ P C2[a, b] which have the
Cauchy property limj,k→ ∞kfj −fkk = 0 do not necessarily converge
to a limit in P C2[a, b]; the limit does exist in the larger, complete space L[a, b] (which, being a complete inner product space is a Hilbert space) but it may not be piecewise continuous.
A linear transformation on P C2[a, b] is simply a linear function from that space into itself. It is very easy to cite examples. One may take a
fixed function, h(x), piecewise continuous on [a, b] and define H : P C2[a, b] → P C2[a, b]; H(f)(x) = h(x)f(x).
It is straightforward to see that for each f ∈ P C2[a, b] this “operation” H gives another function in that space and it is easy to verify the linearity of the operation sinceh(x)(c1f(x) +c2g(x)) = c1h(x)f(x) + c2h(x)g(x). Another example of a linear transformation on P C2[a, b] is given by
K(f)(x) = Z b
a k(x, y)f(y)dy, x ∈ [a, b],
where the “kernel” k(x, y) satisfies the conditions indicated in the sec-tion on linear transforms (in particular, k(x, y) continuous on [a, b]×
[a, b] would suffice).
There is little precision in the terminology for these linear transfor-mations; the terms linear function, linear transformation, linear opera-tor are all used with no clear distinction between them. In these notes we will reserve the word “operator” for use in the term linear
differ-ential operator (the term linear differential operator is standard but
“operator” by itself is often used in other contexts in the general liter-ature). For our purposes here an n-th order linear differential operator
on [a, b] takes the form
D = a0(x) dn dxn + a1(x) dn−1 dxn−1 + ... + an−1(x) d dx + an(x), where eachan−k(x) is a piecewise continuous function on [a, b] (we often
add further requirements). The “operation” ofDon a piecewisentimes continuously differentiable function y(x) defined forx ∈ [a, b] is simply
D(y)(x) = a0(x) dny dxn + a1(x) dn−1y dxn−1 + ... + an−1(x) dy dx + an(x)y(x). Most, but not all, of the examples we will cite correspond to ak(x) ≡
function, D(y)(x) and it is a simple matter to verify that the process is a linear one.
It is clear that the operatorD cannot be applied to an arbitrary func-tion y(x) ∈ P C2[a, b] because, in general, functions in that space may not have the required derivatives. Unlike linear transformations on fi-nite 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 P C2[a, b] = P C20[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 P C2n[a, b], which consists of functions y(x) for which the n-th derivativey(n)(x) is defined and piecewise continuous on [a, b]. This im-plies thaty(x) and the lower order derivativesy(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 restricted by imposing n boundary conditions at the end-points, a and b, 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 differen-tial operators. A function φ(x), not equivalent to the zero function, in the domain subspace S of the differential operator D is an eigenfunc-tion of D corresponding to the (possibly complex) eigenvalue λ of D
if
This means that φ(x) satisfies, for x ∈ [a, b], the differential equation a0(x) dnφ dxn + a1(x) dn−1φ dxn−1 + ... + an−1 dφ dx + an(x)φ(x) − λ φ(x) ≡ 0. 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
y(x) ∈ S and all z(x) ∈ S∗ we have
hDy, zi = hy, D∗zi,
then D∗ is the adjoint operator for D. If D = D∗ and S = S∗ we say
that D is self-adjoint.
Remark Strictly speaking we should say that D∗ is an adjoint oper-ator forD 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 defined by
(Dy)(x) = y0(x), x ∈ [0,1]; S = ny(x) ∈ P C21[a, b]|y(0) = 0}. For z(x) also in P C21[0,1] we compute
hDy, zi = Z 1 0 y 0 (x)z(x)dx = y(x)z(x)|10 − Z 1 0 y(x)z 0 (x)dx
= y(1)z(1) − Z 1
0 y(x)z
0
(x)dx.
Clearly, then, if we take (D∗z)(x) = −z0(x) and take S∗ to consist of functions z(x) ∈ P C21[0,1] such that z(1) = 0 we will have
hDy, zi = Z 1 0 y 0 (x)z(x)dx = − Z 1 0 y(x)z 0 (x)dx= hy, D∗zi
and we see that D∗ as defined on S∗ is the adjoint operator for D. Example 2 Let 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(0) = y(1). 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 = λ φ
for which the general solution is φ(x, λ) = c eλ x, c being an arbitrary constant. Using the periodic boundary condition which is part of the definition of S∗ = S we have
c eλ·0 = c = eλ·1 = c eλ.
This is satisfied for c 6= 0 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 φk(x) = e2kπix, −∞ < k < ∞.
Example 3 We consider the constant coefficient, second order, dif-ferential operator
(Dy) (x) = ∂ 2y
defined on the domain S = ny(x) ∈ P C22([0, L])|a0y(0)+b0y 0 (0) = 0, aLy(L)+bLy 0 (L) = 0}
with a0, b0, aL, bL all real, a0, b0 not both zero, aL, bL not both zero.
We claim that D is self-adjoint. This assertion rests on the calcula-tion, for y, z ∈ S, hDy, zi = Z L 0 y00(x)−c y(x)z(x)dx = y0(x)z(x) L 0 − Z L 0 y0(x)z0(x) + c y(x)z(x)dx = y0(x)z(x) − y(x)z0(x) L 0 + Z L 0 y(x)(z 00 (x) − c z(x))dx = y0(x)z(x) − y(x)z0(x) L 0 + hy, Dzi. Clearly we have the self-adjointness property just in case
y0(x)z(x) − y(x)z0(x) L 0 = 0.
Suppose, first of all, that aL 6= 0. Then, still for y, z ∈ S, we compute
y0(L)z(L) − y(L)z0(L) = y0(L) z(L) + bL aL z0(L) ! − bL aL y0(L) +y(L) ! z0(L) = 0.
A similar result is obtained at x = 0 if a0 6= 0. If aL = 0 then bL 6= 0
and the boundary condition at x = L gives y0(L) = 0, z0(L) = 0, in which case we again have y0(L)z(L) − y(L)z0(L) = 0. A similar result is obtained at x = 0 if a0 = 0. 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.
Example 4 Let D be the differential operator defined by Dy(x) = c0(x)d 2y dx2 + c1(x) dy dx + c2(x)y(x),
where the real valued functions ck(x) ∈ P C2−k, k = 0,1,2. We
will assume c0(x) 6= 0, x ∈ [a, b]. We take the domain of D to be the subspace of P C2[a, b] consisting of functions y(x) which also lie in P C2[a, b] and satisfy boundary conditions, with real coefficients,
day
0
(a) + eay(a) = 0, dby
0
(b) + eby(b) = 0.
We assume that at least one of da, ea and at least one of db, eb are
different from zero. Assuming thatz(x) ∈ P C2[a, b], we use integration by parts to compute hDy, zi = Z b a c0(x) d2y dx2 + c1(x) dy dx + c2(x)y(x) z(x)dx = c0(x)y 0 (x)z(x)−y(x)c0(x)z 0 (x) + c1(x)−c 0 0(x) z(x) b a + Z b a y(x) c0(x)z00(x) +2c00(x)−c1(x)z0(x) +c000(x)−c01(x) +c2(x)z(x)dx. The last integral shows that the adjoint differential operator should be
D∗z(x) = c0(x) d 2z dx2(x)+ 2c00(x)−c1(x) dz dx(x)+ c000(x)−c01(x) +c2(x)z(x). If db 6= 0 the boundary terms at x = b can be rewritten as
c0(b) y0(b) + eb db y(b) ! z(b) − c0(b)eb db y(b)z(b) −y(b)c0(b)z 0 (b) + c1(b)−c 0 0(b) z(b).
Applying the boundary condition satisfied byy(x) atx = b, the bound-ary term at x = b vanishes just in case
0 = −c0(b)eb
db
= −y(b) c0(b)z0 (b) + c1(b)−c00(b) + c0(b)eb db z(b) ,
which is true if z(x) satisfies the boundary condition c0(b)z 0 (b) + c1(b)−c00(b) + c0(b)eb db z(b) = 0.
If we assume da 6= 0 we similarly obtain
c0(a)z0(a) +
c1(a)−c00(a) + c0(a)ea da
z(a) = 0.
The adjoint domainD∗consists of functions inP C2[a, b] satisfying these boundary conditions. The cases wherein da = 0 and/or db = 0 (and
hence ea 6= 0 and/or eb 6= 0 are left as exercises.
QuickCheck Exercises
1. Find the adjoint operator for Dy(x) = dxd (1 +x)dxdy, the domain of D consisting of functions in P C2[0,1] for which y(0) = y(1) = 0. 2. Find the adjoint operator for Dy(x) = y000(x), the domain consist-ing of functions in P C3[0,1] for which y(0) = y00(0) = 0, y0(1) = 0. 3. Show that −c is an eigenvalue of D as defined in Example 3 if a0 = aL = 0.
4. Compute the eigenvalues of D in Example 3 if b0 = bL = 0.
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 da, db is equal to zero?
