Page 206 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 12: Sturm Comparison and Separation Theorems

Notes
u 1 (x 1 )
exclude the possibility that W is ever zero. Suppose that W(x ) = 0. The vectors and
1 u 1 (x 1 )
u 2 (x 1 )
x
x
are then linearly dependent, and hence u (x ) = ku (x ) and u ( ) ku ( ) for some
u 2 (x 1 ) 1 1 2 1 1 2
constant k. The function u(x) = u (x) ku (x) is also a solution of (2) by linearity, and satisfies the
1 2
initial conditions u(x ) = 0, u’(x ) = 0. Since (2) has a unique solution, the obvious solution, u 0,
1 1
is the only solution. This means that u ku . Hence u and u are linearly dependent a
1 2 1 2
contradiction.
The non-singularity of the differential equation is crucial here. If we consider the equation
2
2
x y 2xy + 2y = 0, which has u (x) = x and u (x) = x as its linearly independent solutions, the
1 2
2
Wronskian is x , which vanishes at x = 0. This is because the coefficient of y also vanishes at
x = 0.
Self Assessment
2
3. Find the Wronskian of x, x on the interval ( 1, 1).
12.5 The Sturm Comparison Theorem

The theorem states that if f(x) and g(x) are nontrivial solutions of the differential equations
u + p(x)u = 0 ...(1)
and + q(x) = 0 ...(2)
and p(x) q(x), f(x) vanishes at least once between any two zeros of g(x) unless p q and f = g
where is a real number.
Proof: As p(s) q(x) for all values of x within the interval of interest. For example consider the
equation
2
2
w + a w = 0, a > 0 ...(3)
This equation has an oscillatory behaviour and the solution is of the form
w(x) = c sin ax + c cos ax ...(4)
1 2
2
since p(x) a > 0
then (1) will have an oscillatory solution and so will have zeros. As (1) is more oscillatory then
(2) it will have zeros also more frequently and hence in between zeros of (2) it have at least one
zero.

12.6 The Sturm Separation Theorem

If u (x) and u (x) are the linearly independent solutions of a non-singular homogeneous equation
1 2
(1), then the zeros of u (x) and u (x) occur alternately. In other words, successive zeros of u (x) are
1 2 1
separated by successive zeros of u (x) and vice versa.
2
Proof: Suppose that x and x are successive zeros of u (x); as the Wronskian W is given by
1 2 2
u 1 u 2
'
'
x
W ( ) u 1 ( ) ( ) u 2 ( ) ( )
x
x
u
x
u
x
1
2
u ' 1 u ' 2
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