Wronskian Proof

**dogma** · May4-05, 11:47 AM

Hello out there.

I'm working on a proof by induction of the Wronskian and need a little boost to get going.

So, here goes:

If $LaTeX Code: y_1,...,y_n \\in C^n[a,b]$ , then their Wronskian is:

$LaTeX Code: Wr(y_1,...,y_n)=det\\left(\\begin{array}{ccc}y_1&\\cd ots&y_n\\\\y_1\\prime&\\cdots&y_n\\prime\\\\\\vdots&\\vdots &\\vdots\\\\y_1^{(n-1)}&\\cdots&y_n^{(n-1)}\\end{array}\\right)$

In general, a set of functions will be linearly independent IFF the Wronskian is not identically zero.

Prove this for n = 2, that is

are independent IFF

$LaTeX Code: \\left\\vert\\begin{array}{cc}f(x)&g(x)\\\\f^\\prime(x)& g^\\prime(x)\\end{array}\\right\\vert$ is not identically zero.

Okay, I think I understand how to prove this in one direction. That is, assuming the $LaTeX Code: Wr(f,g) \\neq 0$ and showing that

are independent.

But I'm a little stuck in the assumptions for proving the other direction.

This is what I have for the proof in one direction:

Assume $LaTeX Code: Wr(f(x),g(x)) \\neq 0$ .

Then there exists some

such that $LaTeX Code: Wr(f(x_o),g(x_o)) \\neq 0$ .

Assume:
$LaTeX Code: c_1 f(x) + c_2 g(x) \\equiv 0$
$LaTeX Code: c_1 f^\\prime(x) + c_2 g^\\prime(x) \\equiv 0$

then

$LaTeX Code: c_1 f^\\prime(x_o) + c_2 g^\\prime(x_o) = 0$

We have two equations in the unknowns

and

.

$LaTeX Code: \\left \\vert \\begin{array}{cc}f(x_o) & g(x_o) \\\\ f^ \\prime (x_o) & g ^\\prime (x_o) \\end{array} \\right \\vert \\neq 0$

Therefore there are unique solutions.

One such solution is:

This turns out to be the only solution.

Therefore

and

are independent.

I'm sure my wording is a little off. But I'm still confused about how I would prove it the other way.

Any help would be greatly appreciated.

Best,
dogma

**HallsofIvy** · May4-05, 05:37 PM

Looks good to me.

And the other way is easier!

Suppose f and g are NOT independent. Then there exist a, b such that af(x)+ bg(x)= 0 for all x. If b is not 0, then g(x)= (a/b)f(x) and the Wronskian if f(x)g'(x)- f'(x)g(x)= f(x)(b/a)f'(x)- f'(x)(b/a)f(x)= (b/a)(f(x)f'(x)- f'(x)f(x))= 0 for all x.

(If b is 0, then f(x) must be identically 0 so the Wronskian is 0(g'(x))- 0(g(x))= 0.)

**dogma** · May5-05, 05:52 AM

Thanks, Mr. Ivy!

That makes things a lot clearer! I can see the light.

Best,
dogma

~~kleinwolf~~ · May5-05, 12:01 PM

Ok, so Ivy proved the way you wanted : (Linearly dependent)->Wr(x)=0 forall x

The problem is that : this is equivalent to "exists t Wr(t)!=0->Lin independent" which is what dogma proved.

Since you said it's IFF, then remains to prove :

"Lin indep->exists t Wr(t)!=0" which is equivalent to "W(x)=0 \forall x->lin dep"

Since W(x)=0=f(x)g'(x)-f'(x)g(x).

If f(x)!=0 and g(x)!=0 for a certain domain of x...then clearly f(x)=Cg(x) from the diff. equ...(f'/f)=(g'/g)...but only on this domain

If exist a domain with f(y)=0, f'(y)=0 then g(x) can be anything

Hence the theorem is only in one direction since you can find f,g lin indep such that Wr(f,g)=0 forall x

For counterexample :

$LaTeX Code: f(x)=\\left\\{\\begin{array}{cc} 0 & |x|\\leq 1\\\\ (x+1)^2(x-1)^2 & else\\end{array}\\right.$

Then cleary f,g are in C^1(R) :

$LaTeX Code: fsingle-quote(x)=\\left\\{\\begin{array}{cc} 0 & |x|\\leq 1 \\\\4x(x-1)(x+1) & |x|\\geq 1\\end{array}\\right.$

You see that f'(\pm 1+)=0=f'(\pm 1-) so f'(x) is continuous on R.

If |x|<1 then W=0 clearly since f=0
if |x|>1 then W=0 clearly since f,g are dep on |x|>1

However f and g are lin indep on R, since if f-g=0 on |x|>1 then f-g!=0 in [-1;1]