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I am asked to find the Wronskian of $x$ and $|x|$ in $[-1,1]$.
But $|x|$ is not differentiable at $x=0$.
How do I calculate the Wronskian of such non-differentiable functions?
Could I do this: $W(x,|x|)|_{x\in[-1,0)}+W(x,|x|)|_{x\in(0,-1]}$ because $x$ and $|x|$ both are $0$ at $x=0$.

Manjoy Das
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  • Can you state the definition of the Wronskian that you were given? – Peter Jul 14 '22 at 20:24
  • @Pétur basically it is used to determine the linear dependency but i was just asked to find the value of wronskian for these functions with no further information available. i am also not sure whether the wronskians can be broken interval wise – Manjoy Das Jul 14 '22 at 20:48
  • @Pétur it is very clear that these two functions are linearly dependent if I use the formal definition using $c_i$'s. But I was not asked to check their dependency rather to find the value of their Wronskian. – Manjoy Das Jul 14 '22 at 20:51
  • The two functions are not linearly dependent on the interval $[-1,1]$, or any interval containing $0$. Can you see why? – Peter Jul 14 '22 at 20:53
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    Note: Wronskian zero does not imply linearly dependent, even for functions that are differentiable everywhere. The CONVERSE is true: if functions are linearly dependent, then the Wronskian is zero. – GEdgar Jul 14 '22 at 21:05
  • @Pétur why are they linearly independent? does 'not dependent' necessarily imply independence? – Manjoy Das Jul 14 '22 at 21:11
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    https://math.stackexchange.com/q/4490472/977780 – Sourav Ghosh Jul 15 '22 at 02:53
  • https://math.stackexchange.com/q/4490648/977780 – Sourav Ghosh Jul 15 '22 at 02:53
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    https://math.stackexchange.com/q/4490145/977780 – Sourav Ghosh Jul 15 '22 at 02:54
  • @ManjoyDas See all the MathSE post and then comeback here. It will be crystal clear. – Sourav Ghosh Jul 15 '22 at 02:55
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    @LostinSpace i am enlightened with all the discussions in your posts. I just want to clear one thing. Wronskian of the functions like $(x,|x|), (1,x^2)$, are continuous and changes sign in their suitable intervals. hence these Wronskian must have a zero value at some $x\in I$. what would you conclude from this result? are they linearly dependent or independent? – Manjoy Das Jul 15 '22 at 18:26
  • Wronskian non zero at least one point implies Linearly independent. Wronskian of a non differentiable functions is not defined. It will be helpful you comment on the respective post as it will be easy to answer in that context. – Sourav Ghosh Jul 15 '22 at 18:30
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    @LostinSpace দেশের মানুষকে কাছে পেয়ে ভালো লাগে। wanted to relate this question to the topic, that's why.. – Manjoy Das Jul 15 '22 at 18:58

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Wronskian is defined for differentiable functions. To define $W(f_1, ..., f_n)(x)$ on some interval $I$, functions $f_1, ...f_n$ must be at least $n-1$ times differentiable on that intreval.

Since $|x|$ is not differentiable on $[-1,1]$, so the Wronskian is not defined for $x$ and $|x|$ on that interval.

More precisely the Wronskian is not defined at $x=0$ and for $x \neq 0$ the Wronskian is

$$ W(x, |x|)(x) = \begin{vmatrix} x & |x| \\ 1 & \frac{|x|}{x}\\ \end{vmatrix} = |x| - |x| = 0. $$

Dan
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  • $|x|=x$ if $x\ge 0$. so if by definition i choose $c_1=1,c_2=-1$, the relation holds. so how they are linearly independent? – Manjoy Das Jul 14 '22 at 20:54
  • I guess just as there are weak derivatives, there can be weak Wronskians. – Galen Jul 14 '22 at 20:58
  • More precisely, $W(x, |x|)(x)$ is defined for all $x \neq 0$ and is undefined at $x=0$. Also, while it is true that $W$ not vanishing identically implies linear independence, $W = 0$ identically does not imply linear dependence, even if both functions are differentiable. Therefore, the last line of your answer ("falsely indicates ...") is misleading. – Peter Jul 14 '22 at 21:00
  • Assume that there exist numbers $a$ and $b$ such that $a x + b |x| = 0$ for all $x \in [-1,1]$. Try to calculate $a$ and $b$. – Dan Jul 14 '22 at 21:00
  • the definition says that if $W\ne 0$, then $f$'s are l.independent. hence the non-vanishing of the wronskian is a sufficient condition that the functions are l.independent – Manjoy Das Jul 14 '22 at 21:05
  • You're right. If $W(f,g)=0$ it doesn't imply that $f,g$ are linearly dependent. I edited the answer. – Dan Jul 14 '22 at 21:08