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#1 2007-03-24 08:32:16

rockysheedy
Member
Registered: 2006-06-11
Posts: 12

Differential Equation: Y1 and Y2 are Linearly dependant iff W(X) = 0

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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#2 2007-03-25 10:43:42

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: Differential Equation: Y1 and Y2 are Linearly dependant iff W(X) = 0

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, I’m not very sure about the second part of the proof because I didn’t 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! eek I am sure the first part is okay; it’s just the second part that might not be correct somewhere. Could someone please check the proof? dunno

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#3 2007-03-26 11:05:46

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: Differential Equation: Y1 and Y2 are Linearly dependant iff W(X) = 0

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? I’m afraid I don’t have a clue myself. dunno

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#4 2007-03-26 13:45:48

Zhylliolom
Real Member
Registered: 2005-09-05
Posts: 412

Re: Differential Equation: Y1 and Y2 are Linearly dependant iff W(X) = 0

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 smile) 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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