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Week 5Linear operators and the Sturm–Liouville theory

1. Complex differential operators

2. Properties of self-adjoint operators

3. Sturm-Liouville theory

4. Non-homogeneous boundary-value problems

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۞ A function f assigns to a number x another number, f(x):

).(xfx

1. Complex differential operators

۞ An operator Ĥ assigns to a function f(x) another function, Ĥf(x):

).(ˆ)( xfHxf

Different notations can be used:

)].([)]([ˆ)(ˆ)(ˆ xfxfHxHffHxfH H

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Example 1:

1. Multiplication of a function by a given function, say x2, is an operator:

).()(ˆ 2 xfxxfH

2. Differentiation of a function is an operator:

.d

dˆx

ffD

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Example 1 (continued):

3. The Fourier and Laplace transformations,

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,de)()]([,de)()]([ ttftfxxfxf stikx LF

are integral operators.

Observe that the ‘resulting’ functions, F[f(x)] and L[f(t)], depend on k and s, whereas the ‘original’ functions depend

on x and t – i.e. integral operators change the function and its variable as well.

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۞ A linear operator is an operator such that

).(,ˆ)(ˆ xffHfH

1. All four operators in Example 1 are linear.

.d

dˆ,sinˆ,ˆ 22 fx

fxfQffPffH

,,ˆˆ)(ˆ212121 fffHfHffH (1)

Example 2:

2. The following operators are non-linear:

(2)

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۞ An operator of the form

,d

d)(ˆ

0

N

nn

n

n x

fxcfH

where the coefficients cn(x) are complex functions of a real

variable x, will be called a linear differential operator of order N.

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Comments:

The set of functions that are analytical (have infinitely

many derivatives) in a closed interval [a, b] (where the “closed” means “including the endpoints”) will be

denoted by C∞[a, b].

Unless specified otherwise, all operators considered below will satisfy the following property:

].,[ˆthen],,[if baCfHbaCf

This certainly holds for linear differential operators, but may not hold for integral operators (e.g. Laplace transformation) and some other types.

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We shall use the following inner product for complex functions,

.d)(*)(, b

a

xxgxfgf

Observe that g*(x) is ‘conjugated’ – hence,

.,, fggf

۞ Let S and S+ be subspaces of C∞ [a, b], and Ĥ and Ĥ+ be linear operators such that

.,*ˆ,ˆ, SgSffHggHf

Then Ĥ and Ĥ+ are said to be adjoint to each other, and so are S and S+.

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Example 3:

Let

,0)(],[)( afbaCxfS

Then

.d

dˆ,0)(],[)(x

DbgbaCxgS

To prove the above, consider

.0)(

,0)(

bgSg

afSf

.d

dˆx

D

(3)

“such that”

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Next, consider the l.-h.s. of the definition of adjoint operators,

b

aaxbx

b

a

xx

fggfgf

xx

gf

x

gfgHf

dd

d***

dd

*d

d

d,,

Then, taking into account (3), we obtain

,*

d

d,)(*00)(

d

d, f

xgagbfg

xf

as required.

D̂

D̂

.*

dd

*dsamesame

dd

d***

dd

*d

d

d,,

b

a

b

aaxbx

b

a

xx

fg

xx

fggfgf

xx

gf

x

gfgHf

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Example 4:

The proof is similar to the one in Example 3, but with two integration by parts instead of just one.

,0)()(],[)( bfafbaCxfS

Prove that

.,ˆˆ SSHH

Let

.d

dˆ2

2

xH

(4)

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۞ Let an operator Ĥ be defined in a space S, and Ĥ = Ĥ+, S = S+. Then Ĥ is said to be self-adjoint, or Hermitian in S.

Example 6:

Prove that the following operator:

,d

d)(

d

dˆx

xcx

H

where c(x) is a given real function, is self-adjoint in both spaces defined in Examples 4 and 5.

Example 5:

Prove that (4) is still valid if

.0)()(],[)( bfafbaCxfS

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۞ A linear integral operator is given by

,d)(),()(ˆ2

2

22212 b

a

xxfxxcxfH

where the kernel c(x1, x2) is a continuous complex function defined for

],,[],,[ 222111 baxbax

and f(x2) is a continuous function.

Observe that the ‘original’ function, f(x2), depends on x2,

while the ‘resulting’ one, Ĥf depends on x1 (which makes

integral operators different from differential operators).

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where x1 = k and x2 = x. The space in which this operator acts can be chosen as

Example 7:

The Fourier transform can be viewed as an integral operator with

,e),(,, 21

212121xixxxcbbaa

.as0],[)( xfCxfS

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A proof that one equals zero

Consider two equal non-zero numbers x and y, such that

x = y,hence,

x2 = x y,hence,

x2 − y2 = x y − y2.Divide by (x − y),

x + y = y.Since x = y, we see that

2 y = y.Thus, since y ≠ 0,

2 = 1, Subtract 1 from both sides,

1 = 0.