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Wronskian Proof

  1. May 4, 2005 #1
    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 [tex]y_1,...,y_n \in C^n[a,b][/tex], then their Wronskian is:


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

    Prove this for n = 2, that is [tex]f(x), g(x)[/tex] are independent IFF

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

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

    Then there exists some [tex]x_o[/tex] such that [tex]Wr(f(x_o),g(x_o)) \neq 0[/tex].

    [tex]c_1 f(x) + c_2 g(x) \equiv 0[/tex]
    [tex]c_1 f^\prime(x) + c_2 g^\prime(x) \equiv 0[/tex]

    [tex]c_1 f(x_o) + c_2 g(x_o) = 0[/tex]
    [tex]c_1 f^\prime(x_o) + c_2 g^\prime(x_o) = 0[/tex]

    We have two equations in the unknowns [tex]c_1[/tex] and [tex] c_2[/tex].

    [tex]\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 [/tex]

    Therefore there are unique solutions.

    One such solution is: [tex]c_1 = c_2 = 0[/tex]

    This turns out to be the only solution.

    Therefore [tex]f(x)[/tex] and [tex]g(x)[/tex] 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.

  2. jcsd
  3. May 4, 2005 #2


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    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.)
  4. May 5, 2005 #3
    Thanks, Mr. Ivy!

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

  5. May 5, 2005 #4
    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 :

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

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

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

    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]
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