Wronskian & Linear independence of functions

Question

We usually study wronskian concerned with solutions of ODE, but using this idea to check linear dependence of any two differentiable functions seems Okay because it is the determinant of a Matrix having functions & their derivatives.If two rows/columns are Linearly dependent, then Determinant is zero. The Wronskian for two real-valued differentiable functions f and g is defined by $W(f(t),g(t))$ = $\begin{vmatrix} f(t) & g(t)\\ f'(t) & g'(t)\\ \end{vmatrix}$ = $f(t)g'(t)-f(t)g(t)$. If f and g are linearly dependent on some interval $I$, then, WLOG we may assume $f = cg$ for some scalar $c$, so that the Wronskian becomes

$W(f(t),g(t))$ = $\begin{vmatrix} cg(t) & g(t)\\ cg'(t) & g'(t)\\ \end{vmatrix}$ = $c.g(t)g'(t)-c.g'(t)g(t)=0$

on $I$. In other words, if $f$ and $g$ are linearly dependent on $I$, the Wronskian $W(f(t),g(t))$ is

identically $0$ on $I$.

Now consider this question

Let $f, g : [- 1, 1] \to \mathbb{R}, f(x) = x ^ 3$ and $g(x) = x ^ 2|x|$. And options are

(a) $f$ and $g$ are linearly independent on $[- 1, 1]$

(b) $f(x) g' (x) - f'(x)g(x)$ is NOT identically zero on $[- 1, 1]$

(c) There exist continuous function $p(x)$ and $q(x)$ such that $f$ and $g$ satisfy $y'' + py' + qy = 0$ on $[- 1, 1].$

I had a discussion with my friend over this question, I said that the functions are Linearly Independent (abbrivated as LI) on $[-1,1]$ because we cannot write these functions as a scalar multiple of one another, now if they are LI their wronskian should not be zero. But my friend says define $$g(x) = \begin{cases} x³ & x≥0\\-x³ & x<0 \end{cases}$$ Now wronskian is for $x ≥0$ , $\begin{vmatrix} x³ & x³\\ 3x² & 3x²\\ \end{vmatrix}$ = $0$ and for $x<0$, $\begin{vmatrix} x³ & -x³\\ 3x² & -3x²\\ \end{vmatrix}$ = $0$ ,So it is zero on $[-1,1]$ So they are LI. Then, I said that you are calculating wronskian of a piecewise function over two different regions, then how can you combine the two individual results for the whole domain because LI and LD is domain based property. If we are going to check LI and LD on individual points in $\mathbb{R}$ then every function is LD because every point in $\mathbb{R}$ can be written as a scalar multiple of one another. I added, In Option B, they aren't asking to calculate wronskian, they are just asking that term to be NOT zero.

Option C I'm not sure about this but my view is that a piecewise function cannot be a solution to a differential equation.

Kindly help, what is correct explanation for this question.

Answer 1

The important thing to note is that the correct implication is $$f,g \text{ linearly dependent} \Longrightarrow W(f,g)=0$$ or equivalently $$W(f,g)\neq 0 \text{ somewhere} \Longrightarrow f,g \text{ linearly independent}.$$ The other implication does not hold true.

As for your question: Let $f(x) = x^3 =x^2x$, $g(x) = x^2|x|$. Then $$f'(x) = 3x^2,\; g'(x) = 3x|x|$$

(a): Suppose they were linearly dependent, i.e. $f(x) = λg(x)$ for a $λ$ and all $x$. Then in particular $1 = f(1) = λg(1) = λ$ and $-1 = f(-1) = λg(-1)=λ$, so $λ=1=-1$, which is absurd.

(b): We calculate $$f(x)g'(x) - f'(x)g(x) = x^3\cdot 3x|x| - 3x^2\cdot x^2|x| = 0$$ so the Wronskian is in fact identically $0$.

(c): We have $f''(x) = 6x$, $g''(x) = 6|x|$. So we want to find $p,q$ such that $$\begin{align*} 6x + p(x)\cdot 3x^2 + q(x)\cdot x^3 &= x(6 + p(x)\cdot 3x + q(x)\cdot x^2) = 0\\ 6|x| + p(x)\cdot 3x|x| + q(x)\cdot x^2|x| &= |x|(6 + p(x)\cdot 3x + q(x)\cdot x^2) = 0 \end{align*} $$ or, in other words, we want $F(x) := 6+3xp(x) + x^2q(x) = 0$ for all $x\in[-1;1]\setminus\{0\}$ (because for $x=0$, the factor $x$ or $|x|$ above takes care of the resulting $0$). But if $p,q$ are continuous, then the entire expression $F$ is continuous. Moreover $F(0) = 6$ and by assumption, $F(1) = 0$, so by the intermediate value theorem, there is some $x_0\in (0;1)$ such that $0<F(x_0)<6$. But then $x_0F(x_0) \neq 0$, which is a contradiction to the above equations. Thus the statement (c) is also false.

Edit: I forgot to adress one of your questions directly. The Wronskian also takes into account the derivatives of the functions, which contain information not only about the point you're looking at, but also how the functions behave close by. Thus, you can in fact check the Wronskian at every point and get a conclusion. But of course as you said, if you only look at the functions pointwise, then at every point you can find some scalar (which depends on the point!) which scales one to the other; but this is not linear (in)dependence!