# Thread: Wronskian and Linear Independence

1. ## Wronskian and Linear Independence

Suppose $y_1$ and $y_2$ are solutions of $y''+p(x)y'+q(x)y=r(x)$, with p, q, and r being continuous functions of x. If $y_1$ and $y_2$ are linearly independent on an interval, is it possible to still get a Wronskian of 0 (id est $x^2$ and $x|x|$)?

2. ## Re: Wronskian and Linear Independence

My guess is that the answer is no, but I would like someone to confirm this.

3. ## Re: Wronskian and Linear Independence

Originally Posted by Phantasma
My guess is that the answer is no, but I would like someone to confirm this.
At first I though you were correct, it will not vanish as that would show that they are in fact linearly dependent.

A common misconception is that W = 0 everywhere implies linear dependence, but Peano (1889) pointed out that the functions x2 and |x|x have continuous derivatives and their Wronskian vanishes everywhere, yet they are not linearly dependent in any neighborhood of 0.
-wikipedia Wronskian - Wikipedia, the free encyclopedia

4. ## Re: Wronskian and Linear Independence

Wronskian=0 implies that the functions are dependent if they are analytic on some intervals. The abs(x) is not analytic on the neighborhood of zero. So x^2 and |x|x are linearly dependent on any open interval not containing zero.So even though y1 & y2 seems linearly independent, they are actually not on any open interval not containing zero .

5. ## Re: Wronskian and Linear Independence

Hi,
The attachment does not answer your question, and you may will know everything I say. But it does give a partial solution:

6. ## Re: Wronskian and Linear Independence

Originally Posted by Phantasma
Suppose $y_1$ and $y_2$ are solutions of $y''+p(x)y'+q(x)y=r(x)$, with p, q, and r being continuous functions of x. If $y_1$ and $y_2$ are linearly independent on an interval, is it possible to still get a Wronskian of 0 (id est $x^2$ and $x|x|$)?
The problem with your example is that x|x|, not being differentiable on any interval containing 0, cannot be a solution to such an equation. And if we are restricted to an interval that does NOT contain 0, we have all positive numbers, in which $x|x|= x^2$ or all negative numbers in which $x|x|= -x^2$, in either case not independent of $x^2$.