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.


2 Answers 2


"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.

  • 1
    $\begingroup$ Thanks a lot for clearing doubts. Now, the basic is very much clear to me. $\endgroup$
    – monalisa
    Dec 23, 2014 at 9:45
  • $\begingroup$ Can the above mentioned scenario be generalized for analytic functions? $\endgroup$
    – MAS
    Feb 4, 2022 at 7:55

Suppose $f, g$ two real valued differentiable functions defined on an interval $I$. $W(f, g) =\begin{vmatrix}f&g\\f'&g'\end{vmatrix}=fg'-gf'$

Supposed $f, g$ are Linearly dependent on $I$.Then $\exists \lambda\in\Bbb{R}$ such that $f=\lambda g$ on $I$.

Then $W(f, g) =W(f, g) =\begin{vmatrix}\lambda g&g\\\lambda g'&g'\end{vmatrix}=0$

$\color{red}{\text{$\{f, g\}$ Linearly dependent on $I\implies W(f, g) =0$ on $I$}}$ $$[W(f, g) (x) =0\quad \forall x\in I]$$

But converse is not true in general. The celebrated example $f(x) =x^2 $ and $g(x) =x|x|$ on an interval $I$ containing $0$ is Linearly independent but $W(f, g) =0$ on $I$.

If $f, g$ are two solutions of $y"+p(x) y'+q(x) y=0\tag 1$ on $I$ where $p, q\in C(I) $ then by Abel's identity we have

$$W(f, g) (x) =W(f, g) (x_o) e^{-\int_{x_0}^{x} p(t) dt}$$

Then $W(f, g) (x_0) \neq 0$ for some $x_0\in I$ implies $W(f, g) \neq 0$ on $I$

Moreover $W(f,g)$ different from zero with the same sign at every point ${\displaystyle x} \in {\displaystyle I}$

$\color{red}{\text{ $f, g$ are solutions of $(1) $ then $W(f, g)(x) =0\quad \forall x\in I \iff \{f,g\}$ Linearly dependent}}$

But it's not difficult to find two functions $f, g$ such that $W(f, g)(x_0) =0$ for some $x_0\in I$ but $W(f, g) $ not identically $0$ on $I$.

Points to be considered :

  1. $W(f, g) (x_0) \neq 0\quad $ for some $x_0\in I$ implies Linearly independent.

  2. $f, g$ can't be solutions of $(1) $.

Example : $f=1$ and $g=x^2$ on $(-1, 1) $ Then $W(f, g) =2x $

It's clear that $W(f, g) (x_0) =0\iff x_0=0$

For further study (you might be interested in):

  1. Does there exists two functions $f, g\in C^1(I)$ for which $W(f, g) (x) >0$ for some $x$ and $W(f, g) (x) <0$ for some $x$?

  2. Does there exists two differentiable functions $f, g$ on $I$ such that $W(f, g) (x) >0$ on $A$ and $W(f, g) <0$ on $I\setminus A$?

  3. Does the Wronskian of three or more linearly independent functions change its sign?

  4. Do Wronskians have the intermediate value property?

  • $\begingroup$ Thank you for this useful answer. To be clear, we have in general that the vanishing of the Wronskian is a necessary but not sufficient condition for the linear dependence of some set of functions. But (as you add), if those functions are solutions to a linear ODE (you give of second order -- is it true for linear ODEs of all orders?) then the vanishing of the Wronskian is sufficient and necessary for linear dependence of the set of solutions. Is this correct? $\endgroup$
    – EE18
    Jan 13 at 22:49
  • 1
    $\begingroup$ Thank you for this description. It is very helpful! $\endgroup$
    – Charith
    Jun 18 at 3:44
  • $\begingroup$ What about the case of independent solutions $x^3$ and $x^2|x|$ for the ODE $x^2 \frac{d^2y}{dx^2}-4x\frac{dy}{dx}+6y=0$ on $\mathbb R$? $\endgroup$
    – Messi Lio
    Sep 5 at 17:21

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