Wronskian and linear independence

These are from "An introduction to Linear Analysis" - Kreider, Kuller, Ostberg, Perkins:

1. "Theorem 3-4.

The functions

y_1, \cdots, y_n

are linearly independent in ..

C(I)

, whenever their Wronskian is *not* identically zero on

I

"

Fair enough. This is fine. I know the proof. (Here

I

is an interval of

\mathbb{R}

.)

Now we have:

2. "Theorem 3-6

A set of solutions of a normal nth-order homogeneous linear DE is linearly independent in

C(I)

, and hence is a basis for the solution space of the equation, if and only if its Wronskian *never* vanishes on

I

"

Now perhaps I'm just being thick, but these seem to be in contradiction to each other. It's been so long since I studied this stuff that I can't remember the exact result I learnt, and the reasoning in KKOP amounts to a single line when presenting this result, and I don't follow it.

Can anyone clarify this? The only way I can interpret this meaningfully is that Theorem 3-6 is saying that a set of l.i. functions will only satisfy the *additional* requirement that they be a set of solutions to a n-th order DE if their Wronskian never vanishes.

(edited 9 years ago)

Reply 1

Smaug123

15

Original post by atsruser

Can anyone clarify this? The only way I can interpret this meaningfully is that Theorem 3-6 is saying that a set of l.i. functions will only satisfy the *additional* requirement that they be a set of solutions to a n-th order DE if their Wronskian never vanishes.

Yep, that seems right to me. I don't really see what further clarification is needed :P

Reply 2

atsruser

OP

11

Original post by Smaug123

Yep, that seems right to me. I don't really see what further clarification is needed :P

To my rapidly fading mental faculties, the result seemed to be slightly surprising (probably because I've forgotten it) - it's saying that linear independence of a set of functions isn't enough to allow it to be the basis of the solution space of a linear DE, no? That strikes me as somewhat counter-intuitive.

[edit:]

(edited 9 years ago)

Reply 3

Smaug123

15

Original post by atsruser

To my rapidly fading mental faculties, the result seemed to be slightly surprising (probably because I've forgotten it) - it's saying that linear independence of a set of functions isn't enough to allow it to be the basis of the solution space of a linear DE, no? That strikes me as somewhat counter-intuitive.

[edit:]

Well,

x \mapsto x^2

isn't the basis of a solution space to any linear ODE, is it, despite being linearly independent? No homogeneous ODE exists whose solutions are

y(x) = A x^2

. Unless I'm being really silly?

Reply 4

atsruser

OP

11

Original post by Smaug123

Well,

x \mapsto x^2

isn't the basis of a solution space to any linear ODE, is it, despite being linearly independent? No homogeneous ODE exists whose solutions are

y(x) = A x^2

. Unless I'm being really silly?

I think that's right though I'm not actually sure how to prove it (I note that

\{x,x^2\}

is a basis for the solns of

\frac{d^3y}{dx^3}=0

and that seems the only possible DE it *could* work for - I think).

But then how do you find the Wronskian of a single function? I've only ever seen it defined for > 2. By determinant theory, I would have thought that

W[x^2] = x^2

which is never zero over, say,

[1,2]

so by that argument it *would* be the basis for an homogeneous ODE, if I'm reading Theorem 3-6 right.

Reply 5

Smaug123

15

Original post by atsruser

I think that's right though I'm not actually sure how to prove it (I note that

\{x,x^2\}

is a basis for the solns of

\frac{d^3y}{dx^3}=0

and that seems the only possible DE it *could* work for - I think).

But then how do you find the Wronskian of a single function? I've only ever seen it defined for > 2. By determinant theory, I would have thought that