Let $u$ and $v$ be two solutions of $y''+P(x)y'+Q(x)y=0$,Let $W(u,v)$ denote the wronskian of $u$ and $v$ then $W(u,v)$ vanishes at a point $x_0\in[a,b]\implies u$ and $v$ are linearly dependent

$W(u,v)(x_0)=0 $ for some $x_0\in [a,b] \implies W(x)=0~,~~ \forall x\implies W(x)$ is identically zero on $[a,b]\implies u$ and $v$ are linearly dependent

Where I'm commiting mistake?

  • $\begingroup$ Probably a duplicate of Proof that ODE solutions with Wronskian identically zero are linearly dependent. $\endgroup$ – user539887 Apr 11 '18 at 21:23
  • $\begingroup$ @user539887:It is not duplicate of this,my intention is two about the fuctions whose wronskian is zero at any particular point of the domain but the functions are LI?Read warning just before example 2 here tutorial.math.lamar.edu/Classes/DE/Wronskian.aspx $\endgroup$ – Styles Apr 11 '18 at 21:29
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    $\begingroup$ For any two functions it can happen that their Wronskian is constantly equal to zero but they are linearly independent (see my remark below Emilio Novati's answer). But this is impossible for solutions of a linear homogeneous second order ODE. $\endgroup$ – user539887 Apr 12 '18 at 8:07


The theorem is:

If the Wronskian $W(u,v)(x_0)$ is nonzero for some $x_0 \in [a,b]$ then $u$ and $v$ are linearly independent on $[a,b]$.

The contraposition is:

If $u$ and $v$ are linearly dependent then the Wronskian is zero for all $x \in [a,b]$.

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  • $\begingroup$ :The theorem says that if wronskian is nonzero at some point of the domain of Differential equation the solutions are Linearly independent.It does not say anything about the case if wronskian is zero at some point.Right??Can you give me an example of two linearly independent functions whose wronskian is zero at some particular point of the domain?? $\endgroup$ – Styles Apr 11 '18 at 21:38
  • $\begingroup$ @PKStyles See, e.g., here, or here. $\endgroup$ – user539887 Apr 12 '18 at 8:04
  • $\begingroup$ The point here is that when we say that a linear combination of two functions $\alpha u + \beta v=0$ for $\alpha, \beta$ not zerol ( i.e. the functions are linearly dependent), the $0$ at RHS is the zero function of the vector space, i.e. the function that is null fon any value in the domain. $\endgroup$ – Emilio Novati Apr 12 '18 at 8:14

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