08 Properties of Solutions

7/30/2019 08 Properties of Solutions

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Qualitative Properties of solutions of a second orderhomogeneous Linear Differential equations.


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Throughout this chapter we shall be

looking at the second order homogeneouslinear differential equation

( ) ( ) 0 ....(1)y P x y Q x y


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We shall like to say something about

solutions of such an equation withoutactually solving it. For example, whether

such a solution oscillates (=has an

infinite number of zeros) or not (= hasonly a finite number of zeros).

If a functionf(x) vanishes at a pointx =x0(that is,iff(x0) = 0), then we sayx0 is a zero

of the functionf(x). For example, the zeros

off(x) = sinx arex = n, n = 0, 1, 2,


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Theorem 5(Sturms Separation Theorem)

Lety1(x) andy2(x) be two LI solutions of(1). Then the zeros ofy1(x) separate the

zeros ofy2(x) and vice-versa. That is

between two successive zeros ofy1(x), thereexists a zero ofy2(x) and between two

successive zeros ofy2(x), there exists a zero

ofy1(x).


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x1 x2x3

y1(x) y2(x)


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Exampley1(x) = cosx andy2(x) = sinx are

two LI solutions of the second orderhomogeneous l.d.e.

0y y Hence the zeros of cosx and sinx alternate.


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Normal Form

Obtain a suitable substitution for the dependent

variable which transforms the equation

into normal form, i.e. , from where the first

derivative is absent.

0 QyyPy


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Normal form of a second order

homogeneous l.d.e.

Consider a second order homogeneous l.d.e.

in thestandard form

( ) ( ) 0 ....(1)y P x y Q x y Let us put ,y u v u, v are functions ofx.

,y u v u v

2and y u v u v u v


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Substituting in (1), we get

( 2 )( ) ( ) ( ) 0

u v u v u vP x u v u v Q x u v

[2 ( ) ]

[ ( ) ( ) ] 0 .....(2)

u v u v P x v

u v P x v Q x v

i.e.

We shall choose v such that

2 ( ) 0v P x v


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Thus

2 ( ) 0v P x v

v

v

v

21 1

4 2P v P v

1

( )2P x dxe

1

2

P v

1 1

2 2

P v P v


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Hence Eqn (2) becomes

2 21 1 1[ ] 04 2 2

u v u P P P Q v

2

1 1[ ] 04 2u Q P P u

i.e. as v 0

Thus the given homogeneous l.d.e. (1) has

been transformed into the l.d.e.

( ) 0u q x u 21 1

( )4 2

q x Q P P where


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Thus any solutiony to the equation (1) is

of the form ,y u v

where u is a solutionof ( ) 0 (3)u q x u

and

1( )

2P x dx

v e

The equation (3) is called the normal formof the equation (1).

Since v(x) is never zero, the zeros of anysolution of (1) are the same as the zeros of

the corresponding solution of (3).


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Problem 1. Find the normal form of the d.e.

2 tan 5 0y x y y

Solution Here

12tan

2x dx

v e

2 tan , 5P x Q

secx

21 1( )4 2

q x Q P P 2 25 tan secx x

= 6


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Hence the normal form of the equation is

6 0u u Note: The general solution of the above

1 2cos 6 sin 6u c x c x Hence the general solution of the given

equation is

1 2[ cos 6 sin 6 ]secy c x c x x c1, c2 arbitrary constants.

equation is


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Problem 2. Find the normal form of the d.e.2

4

20

ay y y

x x

Solution Here

1(2/ )

2x dx

v e

2 42 / , /P x Q a x

1

x

21 1( )4 2

q x Q P P

22

4 2

1 1(4 / )

4

ax

x x

2

4

a

x


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Hence the normal form of the equation is

2

40

au u

x


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If q(x) in (1) is a negative function, then the

slution of this equation do not oscillate at all

Theorem B

If q(x) < 0, and if u(x) is a nontrivial solution of

( ) 0u q x u then u(x) has at most one zero.


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Theorem C

Let u(x) be any nontrivial solution of

0 uxqu

1

dxxq

then u(x) has infinitely many zeros on positive

x-axis.

where q(x) > 0 for all x > 0. If


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Problem: Find the normal form of following D.E.

and use it to show that non-trivial solution of D.E.

has infinitely many (+) ve zeros on the x-axis forx>0.

064

42 yxyxyx

(1)

0

642

2

yxxyxy


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2

2 6,4

xxxQ

xxP

Let y = uv be solution of (1)

dx

xPdx

eev

4

2

1

2

1

Choose

2

v x


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Then u is given by normal form

0 uxqu (3)

2

2

2

2

2

14

2

14

4

16

2

1

4

1

xxxx

xPxPxQxq

2

222

2 246 x

xxx

x


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22 ;03 xvuxufrom

is the normal form

0,02

xxxq

1 1

2dxxdxxqAlso

13

3

1x

many zeros on (+) ve x-axis. hasinfinitelyu x


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Problem: Find the normal form of following D.E.

and use it to show that non-trivial solution of

D.E. has not an infinitely many (+) ve zeros onthe x-axis for x>0.

22 3 2

4 4 1 1 0x y x y x y

Here2 2

2

( 1) 1( ) , ( )

4

xP x x Q x

x

21/ 2 / 4xdx x

v e e

2 2


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2 22

2 2

( 1) 1 1 1 1( ) 0, 0

4 24 2

xq x x x

x x

The normal form is

2

10

2u u

x

Also 2

1 1

1

1

2

1 1 12 2

q x dx dxx

x


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The d.e. (where p is a non-negative real

number)2 2 2( ) 0x y x y x p y

is known as Bessels d.e. of order p.

Bessels differential equation of

order p.


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We shall find its normal form.

Here2

21 , (1 )pP Qx x

Dividing byx2, the equation becomes2

21 (1 ) 0py y yx x

which is normal over the positive x-axis.


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Thus the normal form of the Bessels equation

is 2

2

1 4(1 ) 0

4

pu u

x

Hence the normal form of the Bessels

equation is ( ) 0u q x u where

21 1( )

4 2q x Q P P

2

2 2 2

1 11

4 2

p

x x x

2

2

1 414

p

x


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Problem 4, (Page 161): The hypothesis of

Theorem C is false for the Euler equation

2( / ) 0y k x y

but the conclusion is sometime true and

sometime false, depending on the magnitude

of the positive constant k. Show that every

nontrivial solution has infinite number of

positive zeros if k>1/4, and only finite numberif k 1/4 .


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Theorem (Sturms Comparison theorem)

( ) 0y q x y

( ) 0z r x z Assume that q(x) > r(x) for allx in the

interval [a, b] .Then between any two successive zeros of

z(x) in [a, b] , there is at least one zero ofy (x).

Lety (x) andz(x) be respectively non-

trivial solutions of


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Equivalently, we can say the following:

Ify (x) is a nontrivial solution of the biggerequation ( ) 0y q x y

and ify (x) has no zeros in the interval [a, b] ,then any nontrivial solutionz(x) of the

smaller equation ( ) 0z r x z

has at most one zero in the interval [a, b].


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Example. Let q(x) < 0 for allx in the interval

[a, b]. Then any nontrivial solutionz(x) of

the equation ( ) 0z q x z has at most one zero in the interval [a, b].

Solutiony(x) = 1 is a nontrivial solution of

of the bigger equation 0 0y y andy(x) = 1 has no zeros in the interval [a, b].

The result now follows from the previous

remark.


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Problem 1.(page 164) Show that every nontrivial

solution of

2(sin 1) 0y x y has an infinite number of positive zeros.

Solution Consider the equation

104z z


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Here2 1( ) sin 1 ( )

4

q x x r x

for all positivex.

Now ( ) sin 2

x

z x is a nontrivial solution of

10

4

z z

having infinite number of positive zeros,

namely, x = 2n, n = 1, 2, 3,


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Hence by Sturms Comparison theorem,

any non-trivial solution of

2(sin 1) 0y x y has at least one zero between 2n and

2(n+1) for n = 1, 2, 3,

i.e. has an infinite number of positive

zeros.


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Problem 2. Similarly we can show that any

non-trivial solution of

( ( ) 1) 0y f x y

has an infinite number of positive zeros.( wheref(x) 0 for all x 0 )


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Fact Any non-trivial solution of

0y y has an infinite number of positive zeros.

Proof Any solution of the above equationis of the form

1 2cos siny c x c x

sin( ),A x where 2 2

1 2 ,A c c 1 1

2

tanc

c


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And hence has an infinite number of

positive zeros of the form -+n , for all

large positive integers n.


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The zeros of Solutions of Bessels Equation

Bessels equation of orderp ( 0) is

2 2 2

( ) 0x y x y x p y

Its normal form is

2

21 4(1 ) 04

pu ux


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Case (i) 0 p < 1/2

Since2

2

1 4(1 ) 14

p

x

comparing the equation with 0v v we get from Sturm Comparison theorem

that every nontrivial solution of Bessels

equation has an infinite number of

positive zeros.


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Case (ii)p =1/2

Now the normal form of Bessels equation is

0u u

And hence every nontrivial solution of

Bessels equation has an infinite number

of positive zeros.


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Case (iii)p > 1/2

1 as x

Hence we can find anx0 > 0 such that

2

02

1 4 1(1 )

4 4

pfor all x x

x

2

2

1 4(1 )4

p

x


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We get from Sturm Comparison theorem that

every nontrivial solution of Bessels equation

has an infinite number of positive zeros.

Now ( ) sin

2

xv x

is a nontrivial solution of1 04

v v

having infinite number of positive zeros x0 ,namely, x = 2n, for all large n


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Problem 2 Page 164

Ify (x) is a nontrivial solution of ( ) 0y q x y show thaty (x) has an infinite number of

2( )k

q x xpositive zeros if for some

1

4

k

and only a finite number if 21

( )4

q xx


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Solution Let2

1( )

4q x

x

The bigger equation 21

04

y yx

has a nontrivial solution ( )y x xwhich has no positive zeros. Hence any

nontrivial solutionz(x) of the smallerequation has at most one

positive zero.

( ) 0y q x y


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Case (ii) Let2

( )k

q xx

for some1

4k

Thus 21 , 04

k for some

Look at the equation

2

2

(1/ 4 )0z z

x

Clearly sin( ln )z x x

is a nontrivial solution of the above equation

having an infinite number of positive zeros

namely/ , 1, 2,....nx e n


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Now2

2 2

(1/ 4 )( )

kq x

x x

Hence by Sturms Comparison theorem,

any nontrivial solution of the bigger

equation ( ) 0y q x y

also has an infinite number of positive zeros.

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08 Properties of Solutions