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Theorem:-
Let y1 and y2 be two solutions of the 2nd order Homogenous Linear differential equation y" + P(x)y' + Q(x)y = 0 on [a,b]. Show that y1 and y2 are Linearly dependant if and only if Wronskian is identically zero [a, b].
Proof:- How to prove this theorem. Please post the proof if you have any for this theorem.
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One way round is easy. Suppose y[sub]1[/sub] and y[sub]2[/sub] are linearly dependent. Then there are constants R and S not both zero such that
Differentiating,
Hence
Since R and S are not both zero, the matrix on the left must be singular. Therefore the Wronskian of y[sub]1[/sub] and y[sub]2[/sub]
The converse is the tricky part. Assume that the Wronskian is 0.
Now, if y[sub]1[/sub] or y[sub]2[/sub] is identically 0 in [a.b], they would obviously be linearly dependent. So assume that neither of them is 0 in [a.b].
Since y[sub]1[/sub] ≠ 0, C ≠ 0.
∴ y[sub]1[/sub] and y[sub]2[/sub] are linearly dependent.
Unfortunately, Im not very sure about the second part of the proof because I didnt seem to make use of the fact that y[sub]1[/sub] and y[sub]2[/sub] are solutions of the second-order homogenous linear differential equation! I am sure the first part is okay; its just the second part that might not be correct somewhere. Could someone please check the proof?
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I think I see where the flaw in the second part of my proof lies. It was in going from
to
I had assumed that y[sub]1[/sub]⁄y[sub]2[/sub] was continuous. This might not be true if y[sub]2[/sub] = 0 in some parts of [a,b]. (Note: y[sub]2[/sub] was assumed to be not identically zero in [a,b] but it could be zero for some values in [a,b].)
For example, suppose y[sub]1[/sub] = x and y[sub]2[/sub] = |x|, and [a,b] = [−1,1]. Then y[sub]1[/sub]⁄y[sub]2[/sub] = 1 if x > 0 and y[sub]1[/sub]⁄y[sub]2[/sub] = −1 if x < 0, so
but
This should be where the fact that y[sub]1[/sub]⁄y[sub]2[/sub] are solutions to the second-order homogeneous linear differential equation comes in. Can anybody show that if y[sub]1[/sub]⁄y[sub]2[/sub] are solutions to such a differential equation, then y[sub]1[/sub]⁄y[sub]2[/sub] would be continuous? Im afraid I dont have a clue myself.
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By hypothesis, we have a point x = x[sub]0[/sub] ∈ [a, b] such that
We want to show linear dependence, so consider the system
Since
it follows by a theorem (stated without proof here ) that there is an infinite number of pairs of nontrivial solutions of this system for c[sub]1[/sub] and c[sub]2[/sub]. Consider one such nontrivial pair and write the linear combination as
y[sub]3[/sub](x) is also a solution to our differential equation. We also have that
So y[sub]3[/sub](x) is a solution of
satisfying y[sub]3[/sub](x[sub]0[/sub]) = 0, y'[sub]3[/sub](x[sub]0[/sub]) = 0.
But y(x) = 0 is also a solution satisfying y(x[sub]0[/sub]) = 0 and y'(x[sub]0[/sub]) = 0. By the uniqueness of the solutions of our differential equation, y[sub]3[/sub](x) and y(x) = 0 are identical. Then
But by the construction of y[sub]3[/sub](x), c[sub]1[/sub] and c[sub]2[/sub] are not both zero. Then this equation has a nontrivial solution, and therefore y[sub]1[/sub](x) and y[sub]2[/sub](x) are linearly dependent.
Last edited by Zhylliolom (2007-03-26 13:46:13)
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