We know that

  1. if for functions $f$ and $g$, the Wronskian $W(f,g)(x_0)$ is nonzero for some $x_0$ in [a,b] then f and g are linearly independent on [a,b].
  2. If f and g are linearly dependent then the Wronskian is zero for all $x_0$ in [a,b].

My doubt is : If for some $x$ $W(f,g)(x)$ is zero, can we conclude that wronskian is identically zero as we know that wronskian is zero or never zero.

In one problem Wronskian $W$ was coming as $-x^2$ on $(\infty,-\infty)$. Since $W$ is $0$ for $x=0$ can we say wronskian is identically zero OR using point 1 we may conclude that we are getting more than one point where wronskian is not zero and hence functions are linearly independent.


"Identically zero" means "equal to zero for all values of $x$".

The function $-x^2$ is not identically zero, because there are values of $x$ (such as $1,2,3,\dots$) for which it's nonzero.

Since the Wronskian of linearly dependent functions is identically zero, the functions whose Wronskian is $-x^2$ are not linearly dependent.

As an aside: there is a scenario in which $W$ is either always zero or never zero: it happens when the two functions are solutions of the ODE of the form $y''+p(x)y'+q(x)y=0$. For such solutions, the Wronskian satisfies the identity $W(t)=W(s)\exp\left(-\int_s^t p(x)\,ds\right)$ which implies that if $W$ is zero at some point, it is zero everywhere.

  • $\begingroup$ Thanks a lot for clearing doubts. Now, the basic is very much clear to me. $\endgroup$ – monalisa Dec 23 '14 at 9:45

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