Linearly independent functions with identically zero Wronskian

[zenterix]

Homework Statement

Consider the two functions $f(x)=x^3$ and $g(x)=x^2|x|$ on the interval $[-1,1]$.

(a) Show that their Wronskian $W(g,f)$ vanishes identically.

(b) Show that $f$ and $g$ are not linearly dependent.

(c) Do (a) and (b) contradict Lemma 2? If not, why not?

Relevant Equations

Here is Lemma 2

If $y_1(x)$ and $y_2(x)$ are two solutions of the homogeneous differential equation

$$y''+P(x)y'+Q(x)y=R(x)\tag{1}$$

on $[a,b]$, then they are linearly dependent on this interval $\iff$ their Wronskian is identically zero.

(By the way, I think it is implicit that $P$ and $Q$ are continuous on $[a,b]$, right?)

My question will be about item (c).

Part (a)

Note that for $x\geq 0$ we have $f(x)=g(x)$.

For $x<0$ we have $f(x)=-g(x)$.

Since $f$ is a constant times $g$ then one column of the matrix in the Wronskian is a constant times the other column. Thus, the Wronskian is zero, Note that there are two separate calculations, one for $x\geq 0$ and one for $x<0$, but in each case the justification for the Wronskian being zero is the same.

Part (b)

There is no linear combination of $f$ and $g$ that is identically zero on the entire $[-1,1]$. There are two different linear combinations that are zero on each of $[-1,0]$ and $[0,1]$.

Thus, $f$ and $g$ are linearly independent.

Part (c)

I've looked at the proof of Lemma 2 a few times to see where it is required that $y_1$ and $y_2$ be solutions to some homogeneous differential equation.

It doesn't seem (according to my guess) to be the case that if we just pick two functions then they will be solutions to some differential equation.

However, in any case, I did not see this assumption being explicitly used in the proof. (see next post)

That being said, if $P(x)$ and $Q(x)$ continuous on $[-1,1]$ then there should be exactly one solution with a given value and slope at a point.

The fact that $f$ and $g$ do have the same values and slopes at any point $x\geq 0$ means that they should be the same solution on the entire interval. But they aren't for $x<0$.

Therefore, can I conclude that these two functions cannot both be solutions to some equation like (1)?

If I can conclude this then it seems one of the assumptions of Lemma 2 is not satisfied (even though I can't really see yet how this assumption is used in the proof of that Lemma) and so it need not be the case that the zero Wronskian means the functions are linearly independent.

 

[zenterix]

Here is the proof of Lemma 2 in the textbook I am reading (Simmons - Differential Equations)

Now I can see the use of the assumption that $y_1$ and $y_2$ are solutions to a homogeneous 2nd order linear differential equation. It is in the proof of the converse of the biconditional.

We have a subinterval $[c,d]$ of $[a,b]$ and $y_1$ that is non-zero on this subinterval.

Because the Wronskian is zero on $[a,b]$ we conclude that $y_1=ky_2$ on $[c,d]$ for some constant $k$ and all $x$ in $[c,d]$.

Since $y_1$ and $ky_2$ are the same on $[c,d]$ then they have equal values and derivatives on points in this subinterval.

And here is where we use the assumption that $y_1$ and $y_2$ are solutions to $y''+P(x)y'+Q(x)y=0$ with continuous $P$ and $Q$.

Since $y_1$ and $ky_2$ have same value and slope at a point (actually on an entire subinterval) then they must be the same solution on the entire interval $[a,b]$.