Wronskian and Linear Independence

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

Re: Wronskian and Linear Independence

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

Re: Wronskian and Linear Independence

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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.

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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

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 .

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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:

Attachment 28841

Re: Wronskian and Linear Independence

Quote:

Originally Posted by

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

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 $\displaystyle x|x|= x^2$ or all negative numbers in which $\displaystyle x|x|= -x^2$, in either case not independent of $\displaystyle x^2$.