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$f, g\in C^1(I) $ where $I$ is an open interval and $f, g$ both are real valued.

Let $W(f,g)(x) =\begin{vmatrix}f(x) &g(x) \\f'(x)&g'(x)\end{vmatrix}$ denote the Wronskian of $f, g$ at $x\in I$

$\textbf{Question}$ Does there exists such $f, g$ for which $W(f, g) (x) >0$ for some $x$ and $W(f, g) (x) <0$ for some $x$ ?

$W(f, g) (x) \neq 0$ for some $x\in I$ implies $\{f, g\}$ linearly independent.

If two functions are solutions of a differential equation $y"+p(x) y'+q(x) y=0$ 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 ${\displaystyle x} \in {\displaystyle I}$

Hence we have to find two functions $f, g$ with the properties:

  1. $f, g$ must have to be linearly independent.

  2. $f, g\in C^1(I) $

  3. $f, g$ can't be the solution of $2$nd order homogenous linear ODE.

  4. $W(f, g) $ attains both positive and negative values on $I$.

Let $0\in I$ be an open interval and $f, g\in C^1(I) $ defined by $f(x) =x^2$ and $g(x) =x|x| $

Then $f,g$ satisfy $1, 2,3 $ but not $4$ as $W(f, g) (x) =0$ on $I$.

Sourav Ghosh
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  • Remember that $W(f,g)(x) = f^x(x) \cdot \frac{d}{dx} \frac{g(x)}{f(x)}$. This expression changes sign at each strict local extremum of the quotient $\frac{g}{f}$. – Hans Engler Jul 10 '22 at 19:49
  • You can find a second order linear ODE for any pair of independent functions, $0=L[y]=W[f,g,y]$. Obviously, this has to be singular at one or more points to allow the sign change in $W[f,g]$. – Lutz Lehmann Jul 10 '22 at 20:06
  • why can't $f$ and $g$ be solutions of a 2nd order ODE? – Manjoy Das Jul 15 '22 at 19:01
  • If $f,g$ both are solutions of second order ode , then they satisfy Abel's identity. Then Wronskian is identically $0$ or never $0$ – Sourav Ghosh Jul 15 '22 at 19:17

1 Answers1

4

Yes, that is possible: For $f, g\in C^1(I)$ with $g(x) \ne 0$ on $I$ has $$ W(g, fg)(x) = g(x)^2 \cdot W(1, f)(x) = g(x)^2 f'(x) $$ the same sign as $f'(x)$, that allows the easy construction of examples such that the Wronskian takes both positive and negative values.

Some concrete examples are

  • $W(1, x^2) = 2x$,
  • $W(e^x, e^x \sin(x)) = e^{2x} \cos(x)$.
Martin R
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  • Excellent. I have another question . If a set of points given, can I construct two functions whose Wronskian is positive on that set and negative in it's complement? – Sourav Ghosh Jul 11 '22 at 06:41
  • If $A \subsetneq \Bbb R$ is open then $w(x) = dist(x, A^C)$ is continuous, and positive exactly on $A$. Then $f(x) = \int^x_{x_0} w(t) dt$ satisfies $W(1, f) = f' = w$. – Martin R Jul 11 '22 at 06:52
  • To find some explicit example suppose $I=(0, 1) $ is given and $S\subset I$ be the all cantor points. – Sourav Ghosh Jul 11 '22 at 07:28
  • @LostinSpace: I am not sure if I understand what you mean. Note that $w = W(f, g)$ is continuous, so that $A = { x \mid w(x) > 0 }$ is always an open set. Conversely (as I argued above), if an open set $A$ is given, there is a function $f$ such that $w=W(1, f)$ is positive exactly on $A$. – Martin R Jul 11 '22 at 07:31
  • That means it is impossible to find two functions $f, g$ such that the set of all points where $W(f, g) >0$ is the Cantor set as cantor set isn't open. – Sourav Ghosh Jul 11 '22 at 07:35
  • @LostinSpace: Definitely impossible with $f, g \in C^1(I)$. If one relaxes the condition to “$f,g$ differentiable on $I$” then I do not know the answer, that might be related to your older question https://math.stackexchange.com/q/4472636/42969. – Martin R Jul 11 '22 at 07:39
  • Nice. It is on of my fav. unanswered question. I am working on the solution. – Sourav Ghosh Jul 11 '22 at 07:43
  • Last question. If $x\to W(f, g) (x) $ is continuous and $W(f, g) >0$ for some $x$ and $W(f, g) <0$ for some $x$ then by Darboux property $W(f, g)(x) =0$ for some $x$ . If $f, g$ are real analytic then $W(f, g) =0$ on $I$. Is it true? According to your example no. – Sourav Ghosh Jul 11 '22 at 07:46
  • @LostinSpace: The two concrete examples in my answer show that this is not true. – Martin R Jul 11 '22 at 07:49
  • So Real analyticity isn't enough. They are "not the solution of same differential equal " . This is very much required. Is there any alternative properties of $f, g$ for which the above criteria can be relaxed? – Sourav Ghosh Jul 11 '22 at 07:52
  • Let us continue this discussion in chat. – Martin R Jul 11 '22 at 07:52