Stone M., Goldbart P. Mathematics for Physics: A Guided Tour for Graduate Students

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2.4 Further exercises and problems 85

Exercise 2.24: Let the non-local functional S[f ] be defined by

S[f ]=

4π



∞

−∞



∞

−∞



f (x) − f (x



)

x − x





dxdx



Compute the functional derivative of S[f ] and verify that it is given by

δS

δf (x)





∞

−∞

f (x



)

x − x





See Exercise 6.10 for an occurence of this functional.

Linear ordinary differential equations

In this chapter we will discuss linear ordinary differential equations. We will not describe

tricks for solving any particular equation, but instead focus on those aspects of the general

theory that we will need later.

We will consider either homogeneous equations, Ly = 0 with

Ly ≡ p

(x)y

(n)

+ p

(x)y

(n−1)

+···+p

(x)y, (3.1)

or inhomogeneous equations Ly = f . In full,

(x)y

(n)

+ p

(x)y

(n−1)

+···+p

(x)y = f (x). (3.2)

We will begin with homogeneous equations.

3.1 Existence and uniqueness of solutions

The fundamental result in the theory of differential equations is the existence and

uniqueness theorem for systems of first-order equations.

3.1.1 Flows for first-order equations

Let x

, ..., x

be a system of coordinates in R

, and let X

, x

, ..., x

, t), i = 1, ..., n,

be the components of a t-dependent vector field. Consider the system of first-order

differential equations

= X

, x

, ..., x

, t),

= X

, x

, ..., x

, t),

= X

, x

, ..., x

, t). (3.3)

For a sufficiently smooth vector field (X

, X

, ..., X

) there is a unique solution x

(t)

for any initial condition x

(0) = x

. Rigorous proofs of this claim, including a statement

3.1 Existence and uniqueness of solutions 87

of exactly what “sufficiently smooth” means, can be found in any standard book on

differential equations. Here, we will simply assume the result. It is of course “physically”

plausible. Regard the X

as being the components of the velocity field in a fluid flow,

and the solution x

(t) as the trajectory of a particle carried by the flow. A particle initially

at x

(0) = x

certainly goes somewhere, and unless something seriously pathological is

happening, that “somewhere” will be unique.

Now introduce a single function y(t), and set

= y,

=˙y,

=¨y,

= y

(n−1)

, (3.4)

and, given smooth functions p

(t), ..., p

(t), with p

(t) nowhere vanishing, look at the

particular system of equations

= x

n−1

= x

=−



+ p

n−1

+···+p

. (3.5)

This system is equivalent to the single equation

(t)

+ p

(t)

n−1

+···+p

n−1

(t)

+ p

(t)y(t) = 0. (3.6)

Thus an n-th order ordinary differential equation (ODE) can be written as a first-order

equation in n dimensions, and we can exploit the uniqueness result cited above. We

conclude, provided p

never vanishes, that the differential equation Ly = 0 has a unique

solution, y(t), for each set of initial data (y(0), ˙y(0), ¨y(0), ..., y

(n−1)

(0)). Thus,

(i) If Ly = 0 and y(0) = 0, ˙y(0) = 0, ¨y(0) = 0, ..., y

(n−1)

(0) = 0, we deduce that

y ≡ 0.

(ii) If y

(t) and y

(t) obey the same equation Ly = 0, and have the same initial data,

then y

(t) = y

(t).

88 3 Linear ordinary differential equations

3.1.2 Linear independence

In this section we will assume that p

does not vanish in the region of x we are interested

in, and that all the p

remain finite and differentiable sufficiently many times for our

formulæ to make sense.

Consider an n-th order linear differential equation

(x)y

(n)

+ p

(x)y

(n−1)

+···+p

(x)y = 0. (3.7)

The set of solutions of this equation constitutes a vector space because if y

(x) and y

(x)

are solutions, then so is any linear combination λy

(x) + µy

(x). We will show that the

dimension of this vector space is n. To see that this is so, let y

(x) be a solution with

initial data

(0) = 1,



(0) = 0,

(n−1)

= 0, (3.8)

let y

(x) be a solution with

(0) = 0,



(0) = 1,

(n−1)

= 0, (3.9)

and so on, up to y

(x), which has

(0) = 0,



(0) = 0,

(n−1)

= 1. (3.10)

We claim that the functions y

(x) are linearly independent. Suppose, to the contrary, that

there are constants λ

, ..., λ

such that

0 = λ

(x) + λ

(x) +···+λ

(x). (3.11)

3.1 Existence and uniqueness of solutions 89

Then

0 = λ

(0) + λ

(0) +···+λ

(0) ⇒ λ

= 0. (3.12)

Differentiating once and setting x = 0 gives

0 = λ



(0) + λ



(0) +···+λ



(0) ⇒ λ

= 0. (3.13)

We continue in this manner all the way to

0 = λ

(n−1)

(0) + λ

(n−1)

(0) +···+λ

(n−1)

(0) ⇒ λ

= 0. (3.14)

All the λ

are zero! There is therefore no non-trivial linear relation between the y

(x),

and they are indeed linearly independent.

The solutions y

(x) also span the solution space, because the unique solution with

initial data y(0) = a

, y



(0) = a

, ..., y

(n−1)

(0) = a

can be written in terms of them as

y (x) = a

(x) + a

(x) +···+a

(x). (3.15)

Our chosen set of n solutions is therefore a basis for the solution space of the differential

equation. The dimension of the solution space is therefore n, as claimed.

3.1.3 The Wronskian

If we manage to find a different set of n solutions, how will we know whether they are

also linearly independent? The essential tool is the Wronskian:

W (y

, ..., y

; x)

def

... y



... y



(n−1)

... y

(n−1)

. (3.16)

Recall that the derivative of a determinant

D =

... a

(3.17)

90 3 Linear ordinary differential equations

may be evaluated by differentiating row-by-row:



... a



... a



... a



... a

+···

···+

... a



... a



Applying this to the derivative of the Wronskian, we find

... y



... y



(n)

... y

(n)

. (3.18)

Only the term where the very last row is being differentiated survives. All the other row

derivatives give zero because they lead to a determinant with two identical rows. Now,

if the y

are all solutions of

(n)

+ p

(n−1)

+···+p

y = 0, (3.19)

we can substitute

(n)

=−



(n−1)

+ p

(n−2)

+···+p

, (3.20)

use the row-by-row linearity of determinants,

λa

+ µb

λa

+ µb

... λa

+ µb

... c

= λ

... a

... c

+ µ

... b

... c

, (3.21)

3.1 Existence and uniqueness of solutions 91

and find, again because most terms have two identical rows, that only the terms with p

survive. The end result is

=−





W . (3.22)

Solving this first-order equation gives

W (y

; x) = W (y

; x

) exp



−





(ξ)



dξ



. (3.23)

Since the exponential function itself never vanishes, W (x) either vanishes at all x,or

never. This is Liouville’s theorem, and (3.23) is called Liouville’s formula.

Now suppose that y

, ..., y

are a set of C

functions of x, not necessarily solutions

of an ODE. Suppose further that there are constants λ

, not all zero, such that

(x) + λ

(x) +···+λ

(x) ≡ 0 (3.24)

(i.e. the functions are linearly dependent). Then the set of equations

(x) + λ

(x) +···+λ

(x) = 0,



(x) + λ



(x) +···+λ



(x) = 0,

(n−1)

(x) + λ

(n−1)

(x) +···+λ

(n−1)

(x) = 0 (3.25)

has a non-trivial solution λ

, λ

, ..., λ

, and so the determinant of the coefficients,

W =

... y



... y



(n−1)

... y

(n−1)

, (3.26)

must vanish. Thus

Linear dependence ⇒ W ≡ 0.

There is a partial converse of this result: suppose that y

, ..., y

are solutions to an n-th

order ODE and W (y

; x) = 0atx = x

. Then there must exist a set of λ

, not all zero,

such that

Y (x) = λ

(x) + λ

(x) +···+λ

(x) (3.27)

92 3 Linear ordinary differential equations

has 0 = Y (x

) = Y



) = ··· = Y

(n−1)

). This is because the system of linear

equations determining the λ

has the Wronskian as its determinant. Now the function

Y (x) is a solution of the ODE and has vanishing initial data. It is therefore identically

zero. We conclude that

ODE and W = 0 ⇒ Linear dependence.

If there is no ODE, the Wronskian may vanish without the functions being linearly

dependent. As an example, consider

(x) =



0, x ≤ 0,

exp{−1/x

}, x > 0,

(x) =



exp{−1/x

}, x ≤ 0,

0, x > 0.

(3.28)

We have W (y

, y

; x) ≡ 0, but y

, y

are not proportional to one another, and so not lin-

early dependent. (Note that y

1,2

are smooth functions. In particular they have derivatives

of all orders at x = 0.)

Given n linearly independent smooth functions y

, can we always find an n-th order

differential equation that has them as its solutions? The answer had better be “no”, or

there would be a contradiction between the preceding theorem and the counter-example

to its extension. If the functions do satisfy a common equation, however, we can use a

Wronskian to construct it: let

Ly = p

(x)y

(n)

+ p

(x)y

(n−1)

+···+p

(x)y (3.29)

be the differential polynomial in y(x ) that results from expanding

D(y) =

(n)

(n−1)

... y

(n)

(n−1)

... y

(n)

(n−1)

... y

. (3.30)

Whenever y coincides with any of the y

, the determinant will have two identical rows,

and so Ly = 0. The y

are indeed n solutions of Ly = 0. As we have noted, this

construction cannot always work. To see what can go wrong, observe that it gives

(x) =

(n−1)

(n−2)

... y

(n−1)

(n−2)

... y

(n−1)

(n−2)

... y

= W (y; x). (3.31)

3.2 Normal form 93

If this Wronskian is zero, then our construction fails to deliver an n-th order equation.

Indeed, taking y

and y

to be the functions in the example above yields an equation in

which all three coeffecients p

, p

are identically zero.

3.2 Normal form

In elementary algebra a polynomial equation

+ a

n−1

+···a

= 0, (3.32)

with a

= 0, is said to be in normal form if a

= 0. We can always put such an equation

in normal form by defining a new variable ˜x with x =˜x − a

(na

)

−1

By analogy, an n-th order linear ODE with no y

(n−1)

term is also said to be in normal

form. We can put an ODE in normal form by the substitution y = w ˜y, for a suitable

function w(x). Let

(n)

+ p

(n−1)

+···+p

y = 0. (3.33)

Set y = w ˜y. Using Leibniz’ rule, we expand out

(w ˜y)

(n)

= w ˜y

(n)

+ nw



˜y

(n−1)

n(n − 1)



˜y

(n−2)

+···+w

(n)

˜y. (3.34)

The differential equation becomes, therefore,

(wp

)˜y

(n)

+ (p

w + p



)˜y

(n−1)

+···=0. (3.35)

We see that if we chose w to be a solution of

w + p



= 0, (3.36)

for example

w(x) = exp



−





(ξ)



dξ



, (3.37)

then ˜y obeys the equation

(wp

)˜y

(n)

+˜p

˜y

(n−2)

+···=0, (3.38)

with no second-highest derivative.

Example: For a second-order equation,



+ p



+ p

y = 0, (3.39)

94 3 Linear ordinary differential equations

we set y(x) = v(x) exp{−



(ξ)dξ} and find that v obeys



+ v = 0, (3.40)

where

 = p

−



−

. (3.41)

Reducing an equation to normal form gives us the best chance of solving it by inspec-

tion. For physicists, another advantage is that a second-order equation in normal form

can be thought of as a Schrödinger equation,

−

+ (V (x) − E)ψ = 0, (3.42)

and we can gain insight into the properties of the solution by bringing our physics

intuition and experience to bear.

3.3 Inhomogeneous equations

A linear inhomogeneous equation is one with a source term:

(x)y

(n)

+ p

(x)y

(n−1)

+···+p

(x)y = f (x). (3.43)

It is called “inhomogeneous” because the source term f (x) does not contain y, and so is

different from the rest. We will devote an entire chapter to the solution of such equations

by the method of Green functions. Here, we simply review some elementary material.

3.3.1 Particular integral and complementary function

One method of dealing with inhomogeneous problems, one that is especially effective

when the equation has constant coefficients, is simply to try and guess a solution to

(3.43). If you are successful, the guessed solution y

is then called a particular integral.

We may add any solution y

of the homogeneous equation

(x)y

(n)

+ p

(x)y

(n−1)

+···+p

(x)y = 0 (3.44)

to y

and it will still be a solution of the inhomogeneous problem. We use this freedom

to satisfy the boundary or initial conditions. The added solution, y

, is called the

complementary function.

Example: Charging capacitor. The capacitor in the circuit in Figure 3.1 is initially

uncharged. The switch is closed at t = 0.