Vanishing wronskian for linearly independent solutions

[issacnewton]

Hi I am trying to do this problem. Verify that \( y_1=x^3 \) and \(y_2=|x|^3 \) are linearly independent solutions of the diff. equation
\( x^2y''-4xy'+6y=0\) on the interval \((-\infty,\infty) \). Show that \( W(y_1,y_2)=0 \) for every real number x.

I could actually show the above by splitting the interval for \( x>0 \) ,\( x=0 \) and \( x<0 \). Now there is theorem about the wronskian. The set of
solutions \(y_1,y_2,\cdots,y_n \) is linearly independent on \( I\) iff \( W(y_1,y_2,\cdots,y_n)\neq 0 \) for every x in the interval. But in this problem,
\( W(y_1,y_2)=0 \) for every real number x. So the reason for this is that , we have \( a_2(x)=x^2 \) , which is zero when x=0 in the given interval. So
one of the conditions for the theorem is not satisfied here. Thats why we get weird behavior here. Is my reasoning correct ?

Thanks

 

[Ackbach]

I don't think you need to split the interval into three pieces. Just differentiate your functions there (note that $d|x|/dx=\text{sgn}(x)=x/|x|=|x|/x$) and plug them into the DE. As for linear independence, I think you might be able to do that from the definition.

 

[issacnewton]

ackbach, on second look at my work I see that I have not really done the problem correctly. I tried to use the approach suggested by you in the second line. But for \( y_2=|x|^3 \), when I plug this into the diff. equation, the second derivative of \(y_2 \) gives second derivative of \( |x| \), which doesn't cancel out, so I don't get zero . As for linear independence, consider the equation \( c_1 x^3 + c_2 |x|^3=0 \). How do I show that
\( c_1=c_2=0 \). ?

 

[Ackbach]

Hmm. Let me see here. We have the following:

\begin{align*}
y_{2}&=|x|^{3}\\
y_{2}'&=3|x|^{2}\,\frac{x}{|x|}=3x|x|,\quad\text{ for }x\not=0\\
y_{2}''&=3\left(|x|+x\,\frac{|x|}{x}\right)=6|x|.
\end{align*}

You can see here that I'm using whichever form of $d|x|/dx$ I want for convenience. Also note that the function is not differentiable at the origin. Plugging this into the DE yields
$$x^{2}(6|x|)-4x(3x|x|)+6|x|^{3}=-6x^{2}|x|+6|x|^{3}.$$
What can you say about this expression? Don't forget that $|x|\equiv\sqrt{x^{2}}$.

As for linear independence, plug in, say, $x=1$ to get one equation, and $x=-1$ to get another. Solve.

 

[issacnewton]

thanks for detailed explanation. About the constants c₁ and c₂ , I have a question. Why did you choose particular values of x to solve for them. We need to prove that \(c_1=c_2=0\) for all
x in \((-\infty,\infty) \).

 

[Ackbach]

thanks for detailed explanation. About the constants c₁ and c₂ , I have a question. Why did you choose particular values of x to solve for them. We need to prove that \(c_1=c_2=0\) for all
x in \((-\infty,\infty) \).

Hmm. Well, that's a good question. The definition of linearly dependent functions, according to MathWorld, is that two functions $f_{1}$ and $f_{2}$ are linearly dependent on $\mathbb{R}$ if there exists $c_{1}, c_{2}$, not both zero, such that $c_{1}f_{1}(x)+c_{2}f_{2}(x)=0$ for all $x\in\mathbb{R}$. We could write that in first-order logic as:

$((\exists c_{1})(\exists c_{2})(\forall x\in\mathbb{R})[((c_{1}\not=0)\lor(c_{2}\not=0)) \land (c_{1}f_{1}(x)+c_{2}f_{2}(x)=0)]\implies f_{1}\text{ and }f_{2}\text{ are linearly dependent.}$

The contrapositive gives us the definition of linearly independent functions:

$f_{1}\text{ and }f_{2}\text{ are linearly independent }\implies \lnot \left\{(\exists c_{1})(\exists c_{2})(\forall x\in\mathbb{R})[((c_{1}\not=0)\lor(c_{2}\not=0)) \land (c_{1}f_{1}(x)+c_{2}f_{2}(x)=0)]\right\}$

or

$f_{1}\text{ and }f_{2}\text{ are linearly independent }\implies (\forall c_{1})\lnot \left\{(\exists c_{2})(\forall x\in\mathbb{R})[((c_{1}\not=0)\lor(c_{2}\not=0)) \land (c_{1}f_{1}(x)+c_{2}f_{2}(x)=0)]\right\}$

or

$f_{1}\text{ and }f_{2}\text{ are linearly independent }\implies (\forall c_{1})(\forall c_{2})\lnot \left\{(\forall x\in\mathbb{R})[((c_{1}\not=0)\lor(c_{2}\not=0)) \land (c_{1}f_{1}(x)+c_{2}f_{2}(x)=0)]\right\}$

or

$f_{1}\text{ and }f_{2}\text{ are linearly independent }\implies (\forall c_{1})(\forall c_{2})(\exists x\in\mathbb{R})\lnot \left\{[((c_{1}\not=0)\lor(c_{2}\not=0)) \land (c_{1}f_{1}(x)+c_{2}f_{2}(x)=0)]\right\}$

or

$f_{1}\text{ and }f_{2}\text{ are linearly independent }\implies (\forall c_{1})(\forall c_{2})(\exists x\in\mathbb{R})[\lnot((c_{1}\not=0)\lor(c_{2}\not=0)) \lor (c_{1}f_{1}(x)+c_{2}f_{2}(x)\not=0)]$

or

$f_{1}\text{ and }f_{2}\text{ are linearly independent }\implies (\forall c_{1})(\forall c_{2})(\exists x\in\mathbb{R})[((c_{1}=0) \land (c_{2}=0)) \lor (c_{1}f_{1}(x)+c_{2}f_{2}(x)\not=0)].$

So what this is saying is that either $c_{1}=0$ and $c_{2}=0$, or the linear combination is not zero. Clearly, both cannot be true simultaneously. But you can see here, that because of the negation, the $\forall$ quantifier in the definition of linear dependence becomes the $\exists$ quantifier after negation for the definition of linear independence. So I don't need to prove this is true for all $x$.

When it comes to two functions, all you need to do is see if one of the functions is a constant multiple of the other. If they are, then they're linearly dependent. If they're not, then they're linearly independent.