Wronskian Proof

1. May 4, 2005

dogma

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

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

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

Prove this for n = 2, that is $$f(x), g(x)$$ are independent IFF

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

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

Then there exists some $$x_o$$ such that $$Wr(f(x_o),g(x_o)) \neq 0$$.

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

then
$$c_1 f(x_o) + c_2 g(x_o) = 0$$
$$c_1 f^\prime(x_o) + c_2 g^\prime(x_o) = 0$$

We have two equations in the unknowns $$c_1$$ and $$c_2$$.

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

Therefore there are unique solutions.

One such solution is: $$c_1 = c_2 = 0$$

This turns out to be the only solution.

Therefore $$f(x)$$ and $$g(x)$$ 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

2. May 4, 2005

HallsofIvy

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.)

3. May 5, 2005

dogma

Thanks, Mr. Ivy!

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

Best,
dogma

4. May 5, 2005

kleinwolf

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 :

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

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

$$f'(x)=\left\{\begin{array}{cc} 0 & |x|\leq 1 \\4x(x-1)(x+1) & |x|\geq 1\end{array}\right.$$

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]