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can give a particularly neat treatment which avoids the use of in¬nite sums.
(i) Explain why there exists a unique di¬erentiable function e : R ’ R
such that

e (x) = e(x) for all x ∈ R, e(0) = 0.

By di¬erentiating the function f de¬ned by f (x) = e(a ’ x)e(x), show that
e(a ’ x)e(x) = e(a) for all x, a ∈ R and deduce that e(x + y) = e(x)e(y)
318 A COMPANION TO ANALYSIS

for all x, y ∈ R. List all the properties of the exponential function that you
consider important and prove them.
(ii) Explain why there exist unique di¬erentiable functions s, c : R ’ R
such that

s (x) = c(x), c (x) = ’s(x) for all x ∈ R, s(0) = 0, c(0) = 1.

By di¬erentiating the function f de¬ned by f (x) = s(a’x)c(x)+c(a’x)s(x),
obtain an important addition formula for trigonometric functions. Obtain
at least one other such addition formula in a similar manner. List all the
properties of sin and cos that you consider important and prove them.
(iii) Write down a di¬erential equation for T (x) = tan x of the form

T (x) = g(T (x)).

Explain why, without using properties of tan, we know there exists a function
T with T (0) = 0 satisfying this di¬erential equation on some interval (’a, a)
with a > 0. State and prove, using a method similar to those used in parts (i)
and (ii), a formula for T (x + y) when x, y, x + y ∈ (’a, a).



Green™s function solutions ™
12.4
In this section we discuss how to solve the di¬erential equation for real-valued
functions on [0, 1] given as

y (t) + a(t)y (t) + b(t)y(t) = f (t)

subject to the conditions y(0) = y(1) = 0 by using the Green™s function. We
assume that a and b are continuous. Notice that we are dealing with a linear
di¬erential equation so that, if y1 and y2 are solutions and »1 + »2 = 1, then
»1 y1 + »2 y2 is also a solution. Notice also that the boundary conditions are
di¬erent from those we have dealt with so far. Instead of specifying y and y
at one point, we specify y at two points.
Exercise 12.4.1. (i) Check the statement about the solutions.
(ii) Explain why there is no loss in generality in considering the interval
[0, 1] rather than the interval [u, v].
Most of this section will be taken up with an informal discussion leading
to a solution (given in Theorem 12.4.6) that can be veri¬ed in a couple of
lines. However, the informal heuristics can be generalised to deal with many
interesting problems and the veri¬cation cannot.
319
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When a ball hits a bat its velocity changes very rapidly because the bat
exerts a very large force for a very short time. However, the position of
the ball hardly changes at all during the short time the bat and ball are in
contact. We try to model this by considering the system
y (t) + a(t)y (t) + b(t)y(t) = h· (t), y(0) = y(1) = 0
where
h· (t) ≥ 0 for all t,
for all t ∈ [s ’ ·, s + ·],
h· (t) = 0 /
s+·
h· (t) dt = 1
s’·

and · > 0, [s ’ ·, s + ·] ⊆ [0, 1]. We have
y (t) + a(t)y (t) + b(t)y(t) = 0 for t ¤ s ’ ·, y(0) = 0
y (t) + a(t)y (t) + b(t)y(t) = 0 for t ≥ s + ·, y(1) = 0
and
s+·
y (s + ·) ’ y (s ’ ·) = y (t) dt
s’·
s+·
h· (t) ’ a(t)y (t) ’ b(t)y(t) dt
=
s’·
s+· s+·
=1’ a(t)y (t) dt ’ b(t)y(t) dt.
s’· s’·

What happens as we make · small? Although y changes very rapidly we
would expect its value to remain bounded (the velocity of the ball changes
s+·
but remains bounded) so we would expect s’· a(t)y (t) dt to become very
small. We expect the value of y to change very little, so we certainly expect
s+·
b(t)y(t) dt to become very small.
s’·
If we now allow · to tend to zero, we are led to look at the system of
equations
y (t) + a(t)y (t) + b(t)y(t) = 0 for t < s, y(0) = 0
y (t) + a(t)y (t) + b(t)y(t) = 0 for t > s, y(1) = 0
y(s+) = y(s’) = y(s), y (s+) ’ y (s’) = 1.
Here, as usual, y(s+) = limt’s, t>s y(s) and y(s’) = limt’s, t<s y(s). The
statement y(s+) = y(s’) = y(s) thus means that y is continuous at s. We
write the system more brie¬‚y as
y (t) + a(t)y (t) + b(t)y(t) = δs (t), y(0) = y(1) = 0,
320 A COMPANION TO ANALYSIS

where δc may be considered as ˜a unit impulse at c™ or ˜the idealisation of
h· (t) for small ·™ or a ˜delta function at s™ or a ˜Dirac point mass at s™ (this
links up with Exercise 9.4.11 on Riemann-Stieljes integration).
By the previous section, we know that there exists a unique, twice di¬er-
entiable, y1 : [0, 1] ’ R such that

y1 (t) + a(t)y1 (t) + b(t)y1 (t) = 0, y1 (0) = 0, y1 (0) = 1,

and a unique, twice di¬erentiable, y2 : [0, 1] ’ R such that

y2 (t) + a(t)y2 (t) + b(t)y2 (t) = 0, y2 (1) = 0, y2 (1) = 1.

We make the following

key assumption: y1 (1) = 0

(so that y2 cannot be a scalar multiple of y1 ).
If y is a solution of , the uniqueness results of the previous section
tell us that

y(t) = Ay1 (t) for 0 ¤ t < s, y(t) = By2 (t) for s < t ¤ 1

for appropriate constants A and B. Since y(s+) = y(s’) = y(s), we can ¬nd
a constant C such that A = Cy2 (s), B = Cy1 (s) and so

y(t) = Cy1 (t)y2 (s) for 0 ¤ t < s, y(t) = Cy2 (t)y1 (s) for s < t ¤ 1.

The condition y (s+) ’ y (s’) = 1 gives us

C y1 (s)y2 (s) ’ y1 (s)y2 (s) = 1

and so, setting W (s) = y1 (s)y2 (s) ’ y1 (s)y2 (s), and assuming, without proof
for the moment, that W (s) = 0, we have

y(t) = y1 (t)y2 (s)W (s)’1 for 0 ¤ t ¤ s, y(t) = y2 (t)y1 (s)W (s)’1 for s ¤ t ¤ 1.

Although we shall continue with our informal argument afterwards, we
take time out to establish that W is never zero.

De¬nition 12.4.2. If u1 and u2 are two solutions of

y (t) + a(t)y (t) + b(t)y(t) = 0,

we de¬ne the associated Wronskian W by W (t) = u1 (t)u2 (t) ’ u1 (t)u2 (t).
321
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Lemma 12.4.3. (i) If W is as in the preceding de¬nition, then W (t) =
’a(t)W (t) for all t ∈ [0, 1].
(ii) If W is as in the preceding de¬nition, then
t
W (t) = Ae’ a(x) dx
0



for all t ∈ [0, 1] and some constant A.
(iii) If y1 and y2 are as in the discussion above and the key assumption
holds, then
W (s) = y1 (s)y2 (s) ’ y1 (s)y2 (s) = 0
for all s ∈ [0, 1].
Proof. (i) Just observe that
W (t) = u1 (t)u2 (t) + u1 (t)u2 (t) ’ u1 (t)u2 (t) ’ u1 (t)u2 (t) = u1 (t)u2 (t) ’ u1 (t)u2 (t)
= u2 (t)(’a(t)u1 (t) ’ b(t)u1 (t)) ’ u1 (t)(’a(t)u2 (t) ’ b(t)u2 (t)) = ’a(t)W (t).
(ii) We solve the di¬erential equation formally, obtaining
W (t)
= ’a(t),
W (t)
whence
t
log W (t) = ’ a(x) dx + log A
0
t
and so W (t) = Ae’ 0 a(x) dx for some constant A.
We verify directly that this is indeed a solution. The uniqueness results
of the previous section (note that W (0) = A), show that it is the unique
solution.
(iii) Observe that W (1) = y1 (1)y2 (1) ’ y1 (1)y2 (1) = ’y1 (1) = 0 by the
key assumption. Since W does not vanish at 1, part (ii) shows that it
vanishes nowhere.
Exercise 12.4.4. Prove part (ii) of Lemma 12.4.3 by considering the deriva-
t
tive of the function f given by f (t) = W (t) exp( 0 a(x) dx).
A more general view of the Wronskian is given by Exercise K.272.
We write G(s, t) = y(s), where y is the solution we obtained to , that
is, we set
G(s, t) = y1 (t)y2 (s)W (s)’1 for 0 ¤ t ¤ s,
G(s, t) = y2 (t)y1 (s)W (s)’1 for s ¤ t ¤ 1.
322 A COMPANION TO ANALYSIS

The function G : [0, 1]2 ’ R is called a Green™s function.
We return to our informal argument. Since G(s, t) is the solution of

y (t) + a(t)y (t) + b(t)y(t) = δs (t), y(0) = y(1) = 0,
m
it follows, by linearity, that y(t) = »j G(sj , t) is the solution of
j=1

m
y (t) + a(t)y (t) + b(t)y(t) = »j δsj (t), y(0) = y(1) = 0.
j=1

N
In particular, if f : [0, 1] ’ R, then yN (t) = N ’1 f (j/N )G(j/N, t) is
j=1
the solution of
N
y (t) + a(t)y (t) + b(t)y(t) = N ’1 f (j/N )δj/N (t), y(0) = y(1) = 0.
j=1


Now imagine yourself pushing a large object. You could either give a con-
tinuous push, applying a force of magnitude f (t) or give a sequence of sharp
taps N ’1 N f (j/N )δj/N (t). As you make the interval between the taps
j=1
ever smaller (reducing the magnitude of each individual tap proportionally)
the two ways of pushing the object become more and more alike and
N
N ’1 f (j/N )δj/N ’ f in some way which we cannot precisely de¬ne.
j=1


It is therefore plausible that, as N ’ ∞,

yN ’ y— in some way to be precisely determined later,

where y— is the solution of

y— (t) + a(t)y— (t) + b(t)y— (t) = f (t), y— (0) = y— (1) = 0.

It also seems very likely that
N 1
’1
f (j/N )G(j/N, t) ’
yN (t) = N f (s)G(s, t) dt
0
j=1


(We could prove this rigorously, but a chain is as strong as its weakest link
and the real di¬culties lie elsewhere.)
We are therefore led to the following plausible statement.
323
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Plausible statement 12.4.5. The equation

y (t) + a(t)y (t) + b(t)y(t) = f (t), y(0) = y(1) = 0

has the solution
1
y(t) = f (s)G(s, t) ds.
0

We summarise the results of our informal argument in the next theorem
which we then prove formally.

Theorem 12.4.6. Suppose that f, a, b : [0, 1] ’ R are continuous. By
Exercise 12.3.11, there exists a unique twice di¬erentiable y1 : [0, 1] ’ R
such that

y1 (t) + a(t)y1 (t) + b(t)y1 (t) = 0, y1 (0) = 0, y1 (0) = 1

and a unique twice di¬erentiable y2 : [0, 1] ’ R such that

y2 (t) + a(t)y2 (t) + b(t)y2 (t) = 0, y2 (1) = 0, y2 (1) = 1.

We make the following

key assumption: y1 (1) = 0.

If we set W (t) = y2 (t)y1 (t) ’ y2 (t)y1 (t), then W is never zero and we may
de¬ne G : [0, 1]2 ’ R by

G(s, t) = y1 (t)y2 (s)W (s)’1 for 0 ¤ t ¤ s,
G(s, t) = y2 (t)y1 (s)W (s)’1 for s ¤ t ¤ 1.

With this notation, G is continuous so we can de¬ne
1
y(t) = G(s, t)f (s) ds
0

The function y, so de¬ned, is twice di¬erentiable and satis¬es

y (t) + a(t)y (t) + b(t)y(t) = f (t)

together with the conditions y(0) = y(1) = 0. Moreover this solution is
unique.
324 A COMPANION TO ANALYSIS

Proof. We have already established the contents of the ¬rst paragraph. The
proof of the continuity of G is left to the reader (see Exercise 12.4.7 (iii) for a
general result from which this follows). To show that y is di¬erentiable and
satis¬es we observe that

t 1
y(t) = G(s, t)f (s) ds + G(s, t)f (s) ds
0 t
t 1
’1
y1 (t)y2 (s)W (s)’1 f (s) ds
= y2 (t)y1 (s)W (s) f (s) ds +
0 t
t 1
’1
y2 (s)W (s)’1 f (s) ds
= y2 (t) y1 (s)W (s) f (s) ds + y1 (t)
0 t



[Experience shows that people get into frightful muddles over this calculation.
At each stage you must ask yourself ˜is s bigger than t or vice versa and what
does this mean for the de¬nition of G(s, t)?™]
We now use the product rule and the fundamental theorem of the calculus
to show that y is twice di¬erentiable with

t
y1 (s)W (s)’1 f (s) ds + y2 (t)y1 (t)W (t)’1 f (t)
y (t) = y2 (t)
0
1
y2 (s)W (s)’1 f (s) ds ’ y1 (t)y2 (t)W (t)’1 f (t)
+ y1 (t)
t
t 1
’1
y2 (s)W (s)’1 f (s) ds
= y2 (t) y1 (s)W (s) f (s) ds + y1 (t)
0 t


and

t
y1 (s)W (s)’1 f (s) ds + y2 (t)y1 (t)W (t)’1 f (t)
y (t) = y2 (t)
0
1
y2 (s)W (s)’1 f (s) ds ’ y1 (t)y2 (t)W (t)’1 f (t)
+ y1 (t)
t
t 1
’1
y2 (s)W (s)’1 f (s) ds
= y2 (t) y1 (s)W (s) f (s) ds + y1 (t)
0 t
’1
+ (y2 (t)y1 (t) ’ y1 (t)y2 (t))W (t) f (t)
t 1
’1
y2 (s)W (s)’1 f (s) ds + f (t),
= y2 (t) y1 (s)W (s) f (s) ds + y1 (t)
0 t
325
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so that

y (t) + a(t)y (t) + b(t)y(t)
1
y2 (s)W (s)’1 f (s) ds
= (y1 (t) + a(t)y1 (t) + b(t)y1 (t))
t
t
y1 (s)W (s)’1 f (s) ds + f (t)
+ (y2 (t) + a(t)y2 (t) + b(t)y2 (t))
0
1 t
’1
y1 (s)W (s)’1 f (s) ds + f (t) = f (t),
=0— y2 (s)W (s) f (s) ds + 0 —
t 0

as required.
To prove uniqueness, suppose that u1 and u2 are solutions of satisfying
u1 (0) = u2 (0) = u1 (1) = u2 (1) = 0. If we set u = u1 ’ u2 then, by linearity,

u (t) + a(t)u (t) + b(t)u(t) = 0, u(0) = 0, u(1) = 0.

Suppose u (0) = ». Then u and »y1 both satisfy

w (t) + a(t)w (t) + b(t)w(t) = 0, w(0) = 0, w (0) = »

so, by the uniqueness results of the previous section, u = »y1 . But y1 (1) = 0
and u(1) = 0, so » = 0 and u = 0. Thus u1 = u2 as required.
Exercise 12.4.7. Let A and B be subsets of some metric space and consider
a function f : A ∪ B ’ R.
(i) Show that, if f is not continuous at x, then at least one of the following
two statements must be true
(±) We can ¬nd an ∈ A with an ’ x and f (an ) f (x).
(β) We can ¬nd bn ∈ B with bn ’ x and f (bn ) f (x).
(ii) Give an example where f |A and f |B are continuous, but f is not.
(iii) Show that, if A and B are closed, then the continuity of f |A and f |B
implies the continuity of f .
[For further remarks along these lines see Exercise K.178.]
The proof of Theorem 12.4.6 is rather disappointing, since it uses only
rather elementary results and gives no hint as to how proceed in more general
circumstances. (In Exercise K.278, I outline proof which involves slightly
less calculation and slightly harder theorems but it still does not get to the
heart of the matter.) However the general idea of using ˜delta functions™ or
˜impulses™ to study ordinary and, particularly, partial di¬erential equations
is very important in both pure and applied mathematics. (The acoustics
of concert halls are tested by letting o¬ starting pistols and recording the
results.)
326 A COMPANION TO ANALYSIS

From the point of view of the pure mathematician, one of the chief ad-

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