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For $t<0$, we have $t|t|= -t^2$. For $t>0$, we have $t|t|=t^2$.

Hence the Wronskian $$W[y_1,y_2](t) = y_1 y_2' -y_1' y_2 \\ = t^2 \cdot (-2t) - (2t)\cdot(-t^2) = 0 \quad t<0 \\ t^2\cdot (2t)-(2t)\cdot t^2 = 0 \quad t>0\ . $$ In either case, the Wronskian is zero. How are these functions then linearly independent on [-1,1]?

Context: this is problem 2.1.11 of Braun's Differential Equations and Their Applications, 4th edition.

(a) Show that $y_1$ and $y_2$ are linearly dependent on the interval $0\leq t \leq 1$.

(b) Show that $y_1$ and $y_2$ are linearly independent on the interval $-1\leq t \leq 1$.

Mussé Redi
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2 Answers2

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You don't need the Wronskian to know these two functions are linearly independent on the interval $[-1,1]$ because two functions are linearly dependent just when one is a constant multiple of the other. That's clearly not the case here. These function are dependent on $[0,1]$ since they are identical. They are also dependent on $[-1,0]$.

At https://math.libretexts.org/Bookshelves/Analysis/Supplemental_Modules_(Analysis)/Ordinary_Differential_Equations/3%3A_Second_Order_Linear_Differential_Equations/3.6%3A_Linear_Independence_and_the_Wronskian you can read that

Let $f$ and $g$ be differentiable on $[a,b]$. If Wronskian $W(f,g)(t_0)$ is nonzero for some $t_0$ in $ [a,b]$ then $f$ and $g$ are linearly independent on $[a,b]$.

Note the implication there: if the Wronskian is nonzero then the functions are independent. You are invoking the converse. But nowhere does it say that an identically zero Wronskian implies dependence.

Your example shows that the converse is in fact false. You can find it in the wikipedia page that @ChristianBlatter notes in his comment.

Ethan Bolker
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  • You mean to say, if the Wronskian is nonzero than they are independent. I couldn't suggest an edit to your question since it's only two characters. Could you edit your answer. – Mussé Redi Nov 19 '20 at 16:28
  • @MusséRedi Thanks. I saw that and fixed it before I saw your comment. – Ethan Bolker Nov 19 '20 at 16:30
  • No, it's not fixed yet. I'm referring to the following sentence. Note the implication there: if the Wronskian is nonzero then the functions are dependent. – Mussé Redi Nov 19 '20 at 16:31
  • Btw, I accepted the other question since (I understand it now) it shows directly why their independent for $-1\leq t \leq 1$. – Mussé Redi Nov 19 '20 at 16:33
  • On why they're dependent on $[-1,0]$. Doesn't @hitechphysics directly show why they're independent on $[-1,1]$? I find it tricky to see the difference between these two domains. – Mussé Redi Nov 19 '20 at 16:36
  • I fixed the second typo thanks. I'm glad the other answer helped you. I think you can learn something from this one too, that's more general than an argument with Wronskians. It's easy to think "if" means "if and only if" so you should be careful when reading theorems. – Ethan Bolker Nov 19 '20 at 16:36
  • My final comment. They are dependent on $[-1,0]$ because there one is $-1$ times the other. They are dependent on $[0,1]$ because they are equal there. Since $-1 \ne 1$ they are independent on the whole interval. – Ethan Bolker Nov 19 '20 at 17:07
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The Wronskian test seems inconclusive in this context. As you have pointed out, $t\lvert t \rvert = -t^{2}$ for $t < 0 $ and $t \lvert t \rvert = t^{2} $ for $t \ge 0$ ( "$\ge$" since the absolute value has $0$ inclusive) so this translates to the system of equations \begin{align*} c_{1} t^{2} + c_{2}t^{2} &= 0 \quad ,t \ge 0 \\ c_{1}t^{2} - c_{2}t^{2} &= 0 \quad ,t <0 \end{align*} Solving this system of equation yields $$ \frac{t^{2}}{t^{2}}\begin{pmatrix} 1 & 1 & | \ \ \ 0 \\ 1 & -1 &| \ \ \ 0 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & |\ \ \ 0 \\ 0 & 1 & |\ \ \ 0 \end{pmatrix} \implies c_{1} = c_{2} \equiv 0,$$ the system is linearly independent for arbitrary $t$ so it is linearly independent for $t \in (0,1)$.

hexaquark
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  • How did you obtain the system of equations? Where does the sign come from? How does this relate to the Wronskian, or dependence or indepence of the solutions? – Mussé Redi Nov 12 '20 at 18:24