08 Properties of Solutions
-
Upload
shanmukha-sundaram -
Category
Documents
-
view
214 -
download
0
Transcript of 08 Properties of Solutions
-
7/30/2019 08 Properties of Solutions
1/46
Qualitative Properties of solutions of a second orderhomogeneous Linear Differential equations.
-
7/30/2019 08 Properties of Solutions
2/46
Throughout this chapter we shall be
looking at the second order homogeneouslinear differential equation
( ) ( ) 0 ....(1)y P x y Q x y
-
7/30/2019 08 Properties of Solutions
3/46
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,
-
7/30/2019 08 Properties of Solutions
4/46
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).
-
7/30/2019 08 Properties of Solutions
5/46
x1 x2x3
y1(x) y2(x)
-
7/30/2019 08 Properties of Solutions
6/46
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.
-
7/30/2019 08 Properties of Solutions
7/46
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
-
7/30/2019 08 Properties of Solutions
8/46
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
-
7/30/2019 08 Properties of Solutions
9/46
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
-
7/30/2019 08 Properties of Solutions
10/46
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
-
7/30/2019 08 Properties of Solutions
11/46
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
-
7/30/2019 08 Properties of Solutions
12/46
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).
-
7/30/2019 08 Properties of Solutions
13/46
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
-
7/30/2019 08 Properties of Solutions
14/46
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
-
7/30/2019 08 Properties of Solutions
15/46
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
-
7/30/2019 08 Properties of Solutions
16/46
Hence the normal form of the equation is
2
40
au u
x
-
7/30/2019 08 Properties of Solutions
17/46
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.
-
7/30/2019 08 Properties of Solutions
18/46
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
-
7/30/2019 08 Properties of Solutions
19/46
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
-
7/30/2019 08 Properties of Solutions
20/46
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
-
7/30/2019 08 Properties of Solutions
21/46
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
-
7/30/2019 08 Properties of Solutions
22/46
22 ;03 xvuxufrom
is the normal form
0,02
xxxq
1 1
2dxxdxxqAlso
13
3
1x
many zeros on (+) ve x-axis. hasinfinitelyu x
-
7/30/2019 08 Properties of Solutions
23/46
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
-
7/30/2019 08 Properties of Solutions
24/46
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
-
7/30/2019 08 Properties of Solutions
25/46
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.
-
7/30/2019 08 Properties of Solutions
26/46
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.
-
7/30/2019 08 Properties of Solutions
27/46
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
-
7/30/2019 08 Properties of Solutions
28/46
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 .
-
7/30/2019 08 Properties of Solutions
29/46
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
-
7/30/2019 08 Properties of Solutions
30/46
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].
-
7/30/2019 08 Properties of Solutions
31/46
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.
-
7/30/2019 08 Properties of Solutions
32/46
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
-
7/30/2019 08 Properties of Solutions
33/46
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,
-
7/30/2019 08 Properties of Solutions
34/46
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.
-
7/30/2019 08 Properties of Solutions
35/46
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 )
-
7/30/2019 08 Properties of Solutions
36/46
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
-
7/30/2019 08 Properties of Solutions
37/46
And hence has an infinite number of
positive zeros of the form -+n , for all
large positive integers n.
-
7/30/2019 08 Properties of Solutions
38/46
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
-
7/30/2019 08 Properties of Solutions
39/46
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.
-
7/30/2019 08 Properties of Solutions
40/46
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.
-
7/30/2019 08 Properties of Solutions
41/46
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
-
7/30/2019 08 Properties of Solutions
42/46
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
-
7/30/2019 08 Properties of Solutions
43/46
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
-
7/30/2019 08 Properties of Solutions
44/46
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
-
7/30/2019 08 Properties of Solutions
45/46
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
-
7/30/2019 08 Properties of Solutions
46/46
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.