9
$\begingroup$

I am wondering the intuition in regard to the following; (let $w$ represent the wronskian function).

Please correct me If I am mistaken, but I will write what I do know and what I am confused about.

Suppose we consider a set of $n$ differentiable functions, say, $\{f_1,…f_n\}$ on some open interval $I=(\alpha,\beta)$

Then why is it that

if $$w(f_1,…,f_n)(x) \neq 0$$ for $\mathbf{some}$ $x \in I$, then they are linearly independent on the interval.

But, if $$w(f_1,…,f_n)(x)=0$$ for even one $x \in I$, then they are linearly dependent in I?

Is this because if it is zero for one x that we can find, this implies that it will be zero for all $x \in I$? how can we conclude this?

in regard to its relationship with differential equations, ( which is why I am currently learning this), I understand that we would require $w \neq 0$ to have a unique solution, and I also understand why have a row that is a multiple of another would gives $w=0$ from simple determinant rules.

But I am having trouble tying it all together.

Thank you

$\endgroup$
1

1 Answer 1

2
$\begingroup$

Intuition:

Because Abel's Formula. it states that: $$W[y_1,\ldots,y_n]=Ce^h(x)$$ Since $e^h(x)\neq 0$ the only posibility to $W=0$ is when $C=0$, so $w=0$ always. The linear independence comes for the determinant of Wronskian, take a look of http://en.wikipedia.org/wiki/Abel%27s_identity.

$\endgroup$
1
  • $\begingroup$ Thanks, this is the type of thing I was looking for $\endgroup$
    – Quality
    Commented May 11, 2015 at 22:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .