Masujima M. Applied Mathematical Methods in Theoretical Physics

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9.6 Differential Equations, Integral Equations, and Extremization of Integrals 327

1 − 2e

Fig. 9.2 The shape of the trial function for Example 9.16.

 Example 9.16. Solve

y(x) =−λy(x), 0 < x < 1, (9.6.8)

with

y(0) = y(1) = 0. (9.6.9)

Solution. WechoosethetrialfunctiontobetheoneinFigure9.2.ThenI = 2 · h

ε,

J = h

(1 −

ε), and IJ = 2



ε(1 −

ε)



, which has a minimum value of

ε =

. This is compared with the exact value, λ = π

.Notethatλ = π

is a lower

bound for IJ.Ifwechoosethetrialfunctiontobe

y(x) = x(1 − x),

we get IJ = 10. This is accurate to almost one percent.

In order to obtain an accurate estimate of the eigenvalue, it is important to

choose a trial function that satisﬁes the boundary condition and looks qualitatively

like the expected solution. For instance, if we are calculating the lowest eigenvalue,

it would be unwise to use a trial function that has a zero inside the interval.

Let us now calculate the next eigenvalue. This is done by choosing the trial

function y(x) which is orthogonal to the exact lowest eigenfunction u

(x), i.e.,



y(x)u

(x)r(x)dx = 0, (9.6.10)

328 9 Calculus of Variations: Fundamentals

and ﬁnd the minimum value of IJ with respect to some parameters in the trial

function. Let us choose the trial function to be

y(x) = x(1 − x)(1 − ax),

and then the requirement that it is orthogonal to x(1 − x), instead of u

(x)whichis

unknown, gives a = 2. Note that this trial function has one zero inside the interval

[0, 1]. This looks qualitatively like the expected solution. For this trial function, we

have

I J = 42,

and this compares well with the exact value, 4π

 Example 9.17. Solve the Laplace equation

∇

φ = 0, (9.6.11)

with φ given at the boundary.

This problem is equivalent to extremizing

I =



(



∇φ)

dV. (9.6.12)

 Example 9.18. Solve the wave equation

∇

φ = k

φ, (9.6.13a)

with

φ = 0 (9.6.13b)

at the boundary.

This problem is equivalent to extremizing



(



∇φ)

dV







. (9.6.14)

 Example 9.19. Estimate the lowest frequency of a circular drum of radius R.

Solution.

≤



(



∇φ)

dV



dV. (9.6.15)

9.6 Differential Equations, Integral Equations, and Extremization of Integrals 329

Try a rotationally symmetric trial function,

φ(r) = 1 −

,0≤ r ≤ R. (9.6.16)

Then we get

≤



(φ

(r))

2πrdr



(φ(r))

2πrdr

. (9.6.17)

This is compared with the exact value,

5.7832

. (9.6.18)

Note that the numerator of the right-hand side of Eq. (9.6.18) is the square of the

smallest zero in magnitude of the 0th order Bessel function of the ﬁrst kind, J

(kR).

The homogeneous Fredholm integral equations of the second kind for the

localized, monochromatic, and highly directive classical current distributions in

two and three dimensions can be derived by maximizing the directivity D in the

far ﬁeld while constraining C = NT,whereN is the integral of the square of

the magnitude of the current density and T is proportional to the total radiated

power. The homogeneous Fredholm integral equations of the second kind and the

inhomogeneous Fredholm integral equations of the second kind are now derived

from the calculus of variations in general term.

 Example 9.20. Solve the homogeneous Fredholm integral equation of the second

kind,

φ(x) = λ



K(x, x



)φ(x



)dx



, (9.6.19)

with the square-integrable kernel K(x, x



), and the projection unity on ψ(x),



ψ(x)φ(x)dx = 1. (9.6.20)

This problem is equivalent to extremizing the integral

I =



ψ(x)K(x, x



)φ(x



)dxdx



, (9.6.21)

with respect to ψ(x), while keeping

J =



ψ(x)φ(x)dx = 1 ﬁxed. (9.6.22)

330 9 Calculus of Variations: Fundamentals

The extremization of Eq. (9.6.21) with respect to φ(x), while keeping J ﬁxed, results

in the homogeneous adjoint integral equation for ψ(x),

ψ(x) = λ



ψ(x



)K(x



, x)dx



. (9.6.23)

With the real and symmetric kernel , K(x, x



) = K(x



, x), the homogeneous integral

equations for φ(x)andψ (x), Eqs. (9.6.19) and (9.6.23), are identical and Eq. (9.6.20)

provides the normalization of φ(x)andψ(x)totheunity,respectively.

 Example 9.21. Solve the inhomogeneous Fredholm integral equation of the second

kind

φ(x) − λ



K(x, x



)φ(x



)dx



= f (x), (9.6.24)

with the square-integrable kernel K(x, x



), 0 <



< ∞.

This problem is equivalent to extremizing the integral

I =



[{

φ(x) − λ



K(x, x



)φ(x



)dx



}φ(x) + F(x)

φ(x)]dx, (9.6.25)

where F(x)isdeﬁnedby

F(x) =



f (x



)dx



. (9.6.26)

9.7

The Second Variation

The Euler equation is necessary for extremizing the integral, but it is not sufﬁcient.In

order to ﬁnd out whether the solution of the Euler equation actually extremizes the

integral, we must study the second variation . This is similar to the case of ﬁnding

an extremum of a function: To conﬁrm that the point at which the ﬁrst derivative

of a function vanishes is an extremum of the function, we must study the second

derivatives.

Consider the extremization of

I =



f (x, y, y



)dx, (9.7.1)

with

y(x

)andy(x

)speciﬁed. (9.7.2)

9.7 The Second Variation 331

Suppose y(x)issuchasolution.

Let us consider a weak variation,



y(x) → y(x) + εν(x),



(x) → y



(x) + εν



(x),

(9.7.3)

as opposed to a strong variation in which y



(x) is also varied, independent of εν



(x).

Then

I → I + εI

+···, (9.7.4)

where



[

∂f

∂y

ν(x) +

∂f

∂y



(x)]dx =



[

∂f

∂y

−

∂f

∂y



]ν(x)dx, (9.7.5)

and



[

∂

∂y

(x) + 2

∂

∂y∂y



ν(x)ν



(x) +

∂

∂y



(x)]dx. (9.7.6a)

If y = y(x) indeed minimizes or maximizes I,thenI

must be positive or negative

for all variations ν(x) vanishing at the end points, when the solution of the Euler

equation, y = y(x), is substituted into the integrand of I

The integrand of I

is a quadratic form of ν(x)andν



(x). Therefore, this integral

is always positive if

(

∂

∂y∂y



)

− (

∂

∂y

)(

∂

∂y



) < 0, (9.7.7a)

and

∂

∂y



> 0. (9.7.7b)

Thus, if the above conditions hold throughout

< x < x

is always positive and y(x) minimizes I. Similar considerations hold, of course,

for maximizations. The above conditions are, however, too crude, i.e., s tronger

than necessary. This is because ν(x)andν



(x) are not independent.

Let us ﬁrst state the necessary and sufﬁcient conditions for the solution to the

Euler equation to give the weak minimum:

332 9 Calculus of Variations: Fundamentals

Let

P(x) ≡ f

, Q(x) ≡ f



, R (x) ≡ f



, (9.7.8)

where P(x), Q(x), and R(x) are evaluated at the point which extremizes the integral

I deﬁned by

I ≡



f (x, y, y



)dx.

We express I

in terms of P(x), Q(x), and R(x)as



[P(x)ν

(x) + 2Q(x)ν(x)ν



(x) + R(x)ν



(x)]dx. (9.7.6b)

Then we have the following conditions for the weak minimum:

Necessary condition Sufﬁcient condition

Legendre test R(x) ≥ 0 R(x) > 0

Jacobi test ξ ≥ x

ξ>x

,ornosuchξ exists,

(9.7.9)

where ξ is the conjugate point to be deﬁned later. Both the conditions must be

satisﬁed for sufﬁciency and both are necessary separately. Before we go on to prove

the above assertion, it is perhaps helpful to have an intuitive understanding of why

these two tests are relevant.

Let us consider the case in which R(x) is positive at x = a, and negative at x = b,

where a and b are both between x

and x

. Let us ﬁrst choose ν(x) to be the function

as in Figure 9.3.

Note that ν(x)isoftheorderofε,whileν



(x)isoftheorder

√

ε. If we choose ε to

be sufﬁciently small, I

is positive. Next, we consider the same variation located at

x = b. By the same consideration, I

is negative for this variation. Thus y(x)does

not extremize I. This shows that the Legendre test is a necessary condition.

The relevance of the Jacobi test is best illustrated by the problem of ﬁnding

the shortest path between two points on the surface of a sphere. The solution is

obtained by going along the great circle on which these two points lie. There are,

however, two paths connecting these two points on the great circle. One of them is

truly the shortest path, while the other is neither the shortest nor the longest. Take

the circle on the surface of the sphere which passes one of the two points. The

other point at which this circle and the great circle intersects lies on the one arc of

the great circle which is neither the shortest nor the longest arc. This point is the

conjugate point ξ of this problem.

We ﬁrst discuss the Legendre test.WeaddtoI

, (9.7.6b), the following term which

is zero,



(ν

(x)ω(x))dx =



(ν

(x)ω



(x) + 2ν(x)ν



(x)ω(x))dx. (9.7.10)

9.7 The Second Variation 333

− e

+ e

Fig. 9.3 The shape of ν(x) for the Legendre test.

Then we have I

to be





(P(x) + ω



(x))ν

(x) + 2(Q(x) + ω(x ))ν(x)ν



(x) + R(x)ν



(x)



dx.

(9.7.11)

We require the integrand to be a complete square, i.e.,

(Q(x) + ω(x ))

= (P(x) + ω



(x))R (x). (9.7.12)

Hence, if we can ﬁnd ω(x) satisfying Eq. (9.7.12), we will have



R(x )





(x) +

Q(x) + ω(x)

R(x )

ν(x)



dx. (9.7.13)

Now, it is not possible to have R(x) < 0 in any region between x

and x

for the

minimum. If R(x) < 0 in s ome regions, we can solve the differential equation



(x) =−P(x) +

(Q(x) + ω(x ))

R(x )

(9.7.14)

for ω(x) in such regions. Restricting the variation ν(x)suchthat



ν(x) ≡ 0, outside the region where R(x) < 0,

ν(x) = 0, inside the region where R (x) < 0,

334 9 Calculus of Variations: Fundamentals

we can have I

< 0. Thus it is necessary to have R(x) ≥ 0 for the entire region for

the minimum. If

P(x)R(x) − Q

(x) > 0, P(x) > 0, (9.7.15a)

then we have no need to go further to ﬁnd ω(x). In many cases, however, we have

P(x)R(x) − Q

(x) ≤ 0, (9.7.15b)

and we have to examine further. The differential equation (9.7.14) for ω(x)canbe

rewritten as

(Q(x) + ω(x))



=−P(x) + Q



(x) +

(Q(x) + ω(x ))

R(x )

, (9.7.16)

which is the Riccatti differential equation.MakingtheRiccatti substitution

Q(x) + ω(x) =−R(x)



(x)

u(x)

, (9.7.17)

we obtain the second-order linear ordinary differential equation for u(x),

[R(x)

u(x)] + (Q



(x) − P(x))u(x ) = 0. (9.7.18)

Expressing everything in Eq. (9.7.13) in terms of u(x), I

becomes



[R(x)(ν



(x)u(x) −u



(x)ν(x))

u

(x)]dx. (9.7.19)

If we can ﬁnd any u(x)suchthat

u(x) = 0, x

≤ x ≤ x

, (9.7.20)

we will have

> 0forR(x) > 0, (9.7.21)

which is the sufﬁcient condition for the minimum . This completes the derivation

of the Legendre test. We further clarify the Legendre test after the discussion of the

conjugate point ξ of the Jacobi test .

The ordinary differential equation (9.7.18) for u(x) is related to the Euler equation

through the inﬁnitesimal variation of the initial condition. In the Euler equation,

(x, y, y



) −



(x, y, y



) = 0 (9.7.22)

9.7 The Second Variation 335

we make the following inﬁnitesimal variation:

y(x) → y(x) + εu(x), ε = positive inﬁnitesimal. (9.7.23)

Writing out this variation explicitly,

(x, y + εu , y



+εu



) −



(x, y + εu , y



+ εu



) = 0,

we have, with the use of the Euler equation,

u + f



−





u + f





= 0,

or,

Pu + Qu



−

(Qu + Ru



) = 0, (9.7.24)

i.e.,

[R(x)

u(x)] + (Q



(x) − P(x))u(x) = 0,

which is nothing but the ordinary differential equation (9.7.18). The solution to the

differential equation (9.7.18) corresponds to the inﬁnitesimal change of the initial

condition. We further note that the differential equation (9.7.18) is of the self-adjoint

form.ThusitsWronskianW(u

(x), u

(x)) is given by

W(x) ≡ u

(x)u



(x) − u



(x)u

(x) = CR(x), (9.7.25)

where u

(x)andu

(x) are the linearly independent solutions of Eq. (9.7.18) and C

in Eq. (9.7.25) is some constant.

We now discuss the conjugate point ξ and the Jacobi test . We claim that the

sufﬁcient condition for the minimum is that R(x) > 0on(x

, x

) and that the

conjugate point ξ lies outside (x

, x

), i.e., both the Legendre test and the Jacobi

test are satisﬁed. Suppose that

R(x ) > 0on(x

, x

), (9.7.26)

which implies that the Wronskian has the same sign on (x

, x

). Suppose that u

(x)

vanishes only at x = ξ

and x = ξ

(ξ

) = u

(ξ

) = 0, x

<ξ

< x

. (9.7.27)

We claim that u

(x) must vanish at least once between ξ

and ξ

.TheWronskian

W evaluated at x = ξ

and x = ξ

are given, respectively, by

W(ξ

) =−u



(ξ

), (9.7.28a)

336 9 Calculus of Variations: Fundamentals

and

W(ξ

) =−u



(ξ

). (9.7.28b)

By the continuity of u

(x)on(x

, x

), u



(ξ

) has the opposite sign to u



(ξ

), i.e.,



(ξ



(ξ

) < 0. (9.7.29)

But, we have

W(ξ

)W(ξ

) = u



(ξ



(ξ

) > 0, (9.7.30)

from which we conclude that u

(ξ

) has the opposite sign to u

(ξ

), i.e.,

(ξ

) < 0. (9.7.31)

Hence u

(x) must vanish at least once between ξ

and ξ

Since u(x) provide the inﬁnitesimal variation of the initial condition, we choose

(x)andu

(x)tobe

(x) ≡

∂

∂α

y(x, α, β), u

(x) ≡

∂

∂β

y(x, α, β), (9.7.32)

where y(x, α, β) is the solution of the Euler equation,

−



= 0, (9.7.33a)

with the initial conditions,

y(x

) = α, y(x

) = β, (9.7.33b)

and u

(x)(i = 1, 2) satisfy the differential equation (9.7.18). We now claim that the

sufﬁcient condition for the weak minimum is that

R(x ) > 0andξ>x

. (9.7.34)

We construct U(x)by

U(x) ≡ u

(x)u

) − u

(x), (9.7.35)

which vanishes at x = x

. We deﬁne the conjugate point ξ as the solution of the

equation,

U(ξ ) = 0, ξ = x

. (9.7.36)