I am trying to understand linear dependence and linear independence of real valued functions on a set. Say S. I want to know that using wronskain how can we say that a set S of functions is linearly dependent. I was thinking that if wronskain is zero everywhere on the domain then S is linearly dependent and if at least at one point of the domain wronskain is nonzero then S is linearly independent. Where is the problem actually ?
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$\begingroup$ No. Wikipedia give a counterexample of wronskian being 0 but the functions are not linear dependent. Gate: this $\endgroup$– xbhCommented Sep 18, 2018 at 16:23
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$\begingroup$ Thank you. I noticed that. Will you please tell me that in wiki it is written that " the Wronskian can be used to show that a set of differentiable functions is linearly independent on an interval by showing that it does not vanish identically. It may, however, vanish at isolated points" does it mean it can vanish at most at isolated points? If I say it vanishes everywhere except an isolated point then still set will be linearly independent. Am I doing right? $\endgroup$– prakash nainwalCommented Sep 19, 2018 at 2:26
1 Answer
Let us see the relation between the Wronskain and the linearly dependent and independent solutions. It may be clear your confusion.
Let $f$ and $g$ be two real valued differentiable functions on a set $S = [a,b]$ (say). If Wronskian $W(f,g)(t_{0})$ is nonzero for some $t_{0}$ in $[a,b]$, then $f$ and $g$ are linearly independent on $[a,b]$.
If $f$ and $g$ are linearly dependent then the Wronskian $W(f,g)(t_{0})$ is zero for all $t_{0}$ in [a,b] .
You can also verify this in
- "Differential Equations" by Shepley L. Ross ($3^{rd}$ Edition, Page 112, Theorem $4.4$).
- "An Introduction to Ordinary Differential Equations" by Ravi P. Agarwal, Donal O'Regan (Page $118$ (Theorem $17.1$))