set5 - Lecture 13 Qualitative Behaviour of Solutions to...

Lecture 13
Qualitative Behaviour of Solutions to ODEs (cont’d)
Applications of the Sturm Comparison Theorem
The most natural class of DEs with which we can perform a Sturm-type comparison are those
with constant coefficients, i.e.,
y
′′
+ k
2
y = 0 , k ∈ R , (1)
with solutions of the form
y ( x ) = A cos( kx − φ ) . (2)
The period of these solutions is T = 2 π/k . Consequently, the distance between consecutive
zeros of these solutions is π/k .
For example, suppose that for the following DE in normal form,
u
′′
+ q ( x ) u = 0 , (3)
we have been able to establish that
q ( x ) > k
2
(4)
on some sufficiently large domain, e.g, [ x
0
, ∞ ). Now let u ( x ) be a solution to (3). Then by the
Sturm Comparison Theorem (SCT), for any solution y in (2), a zero of u ( x ) must lie between
two consecutive zeros of y .
At first, one might think that it is sufficient that two consecutive zeros of u can be no further
apart that 2 π/k , as shown in the left figure below. But the SCT must hold for all solutions of
the form (2) which are obtained by continuously varying the phase φ . (Note that we do not
pay attention to the amplitudes since they do not affect the positioning of the zeros.) If we
slide the y solution in the above figure enough, then we produce the unacceptable situation on
the right. We cannot have two consecutive zeros of y enclosing a region that is devoid of a zero
of u .
1

x
π/k
y ( x ) = A cos k ( x − φ
1
)
π/k
u ( x )
u ( x )
y ( x ) = A cos k ( x − φ
2
)
x
Now imagine sliding the y wave continuously to the right: Between any two consecutive
zeros of the moving y wave - a kind of moving grating - we must always be able to find a zero
of v ( x ). This leads to the following consequence:
Every interval [ a, b ] of length π/k must contain at least one zero of u .
And the consequence of this consequence is that the distances between consecutive zeros of v
must be less than or equal to π/k .
Application to Bessel’s Equation
Recall Bessel’s equation of order p ≥ 0:
x
2
y
′′
+ xy
′
+ ( x
2
− p
2
) y = 0 (5)
The substitution
y ( x ) =
1
√
x
u ( x ) (6)
yields the following normal or reduced form equation for u(x):
u
′′
+
parenleftBigg
1 +
1 − 4 p 2
4 x
2
parenrightBigg
u = 0 . (7)
Recall that all zeros of u are zeros of y and vice versa – it is much easier to study properties of
zeros, etc., from the u equation. There are three cases to consider:
Case No. 1: p = 1 / 2. Then q ( x ) = 1 so that the general solution of (7), written in phase-
shifted form is
u ( x ) = A cos( x − φ ) . (8)
2

Thus the general solution of Bessel’s DE for p = 1 / 2 is
y ( x ) =
A
√
x
cos( x − φ ) . (9)
The distance between successive zeros of y ( x ) is π .
Case No. 2: 0 ≤ p < 1 / 2. Then q ( x ) > 1. From the SCT, every interval of length π must
contain a zero of u or y .
However, we can say even more: As x → ∞ , we see that q ( x ) → 1, so we expect that the
distances between consecutive zeros of u ( x ), hence y ( x ), will approach π . In fact, let’s assume
that x > 0 is sufficiently large so that, for a small ǫ > 0,
1 < 1 +
1 − 4 p 2
4 x
2
< 1 + ǫ. (10)
• The leftmost term, “1”, corresponds to the DE y ′′ + y = 0 , whose solutions have
consecutive zeros that are π apart.