# If the Wronskian is zero at some point, does this imply linear dependency of functions?

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.

• Thanks a lot for clearing doubts. Now, the basic is very much clear to me. Commented Dec 23, 2014 at 9:45
• Can the above mentioned scenario be generalized for analytic functions?
– MAS
Commented 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):

• 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?
– EE18
Commented Jan 13, 2023 at 22:49
• Thank you for this description. It is very helpful! Commented Jun 18, 2023 at 3:44
• 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 $\mathbb R$? Commented Sep 5, 2023 at 17:21