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•  » Wronksian determinant in 2nd order linear DE

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Fistfiz
2013-01-14 01:06:45

Thank you, this remark:
"Peano (1889) observed that if the functions are analytic, then the vanishing of the Wronskian in an interval implies that they are linearly dependent."
opened my eyes

scientia
2013-01-13 01:39:21

I see. Well, if
are differentiable, then linear independence implies
for all
. If they are not both differentiable, then it is possible that they are linearly independent yet
. See http://en.wikipedia.org/wiki/Wronskian# … dependence for an example.

Fistfiz
2013-01-13 01:06:52

Of course but what i meant was: can I avoid to include W(t0)!=0 for some t0 in my hypotesis? In other words, if I have two linearly independent solutions u and v, can I automatically say W(u,v)!=0 for all t?

scientia
2013-01-12 23:35:41

Yes, if you can show that
for some
then
for all
.

Fistfiz
2013-01-12 21:06:44

Hi guys,
I'm studying some demonstrations about 2nd order differential equations of the form:
y''+2by'+ay=f(t)
where a,b are constants.
Suppose that u,v are linearly independent solutions. Now, in several demonstrations, it's needed that the Wronksian determinant of u,v it's different from zero.
I see from Abel's identity (http://en.wikipedia.org/wiki/Abel's_identity) that if this is true for some t0 value, then it's true for all t. Provided this, can I always say that the Wronksian of u,v is always non-zero??