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Let $y=\phi(x)$ and $y=\psi(x)$ be linearly independent solutions of the ODE $y^{\prime\prime} + p(x)y^{\prime} + q(x)y = 0$, where $p$ and $q$ are continuous on an open interval $I$.

I am looking at a problem that asks for a proof of the following statement:

Suppose that $x_{0}\in I$ is a zero of $\phi$, then $\phi$ cannot have a relative extremum value at $x_{0}$.

This problem also gives a hint by suggesting consideration of the Wronskian.

I know that $n$-th order linear homogeneous differential equation always has $n$ linearly independent solutions.
If $\phi$, $\psi$ are two linearly independent solutions of the ODE $y^{\prime\prime} + p(x)y^{\prime} + q(x)y = 0$, then $y = c_{1}\phi + c_{2}\psi$ is the general solution $\forall x \in I$ where $y=0$ only when $c_{1}=c_{2}=0$. However, if $\phi(x_{0})$ is a relative extremum, then $\phi^{\prime}(x_{0}) = \phi(x_{0}) = 0$ and the Wronskian, \begin{alignat*}{2} W(x_{0}) &= \left| \begin{matrix} \phi(x_{0}) & \psi(x_{0}) \\ \phi^{\prime}(x_{0}) & \psi^{\prime}(x_{0}) \\ \end{matrix} \right| &= \phi(x_{0})\psi^{\prime}(x_{0}) - \psi(x_{0})\phi^{\prime}(x_{0}) \end{alignat*} vanishes for this particular value.
I don't see why this is impossible.
I know that the Wronskian can be used to show that a set of differentiable functions is linearly independent on an interval by showing that it does not vanish identically, but my understanding is that the converse is not true without any extra conditions like analyticity.

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  • $\begingroup$ Furthermore, a property of Abel's theorem gives that the Wronskian is either always the zero function or always different from zero at every point $x \in I$. So, how do I know the Wronskian is always nonzero for this particular case? $\endgroup$ – jpb May 13 '14 at 2:08
  • $\begingroup$ If $\phi$ is a solution and $\phi(x_0)=\phi'(x_0)=0$ then, by uniqueness, $\phi$ is the zero function, contradicting the linear independence of $\{\phi, \psi\}.$ Am I missing something? $\endgroup$ – Gil Bor May 13 '14 at 5:55
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What you need is Abel's theorem and the fundamental uniqueness/existence theorem.

The Wronskian of a set of solutions for a second order ODE has a formula of the form: $W[\phi,\psi](x) = Ce^{f(x)}$ where $C$ is a constant and $f(x)$ is some function (follow the link for more details about the form of $f(x)$).

If $x=x_0$ was simultaneously a zero for $\phi(x)$ and a critical point, you correctly identified that you have $W[\phi,\psi](x_0)=0$. But then $Ce^{f(x_0)}=0$ so you must have $C=0$. This means $W[\phi,\psi](x)=0$ for all $x$ in your interval.

You are correct that the Wronskian vanishing is not enough to conclude linear dependence in general. However, it is enough for a set of solutions of a linear ODE.

Why? Let's stick to the second order case. Suppose $\phi$ and $\psi$ are solutions.

We know that if $\phi$ and $\psi$ are linearly dependent, then the system: $c_1\phi(x)+c_2\psi(x)=0$ and $c_1\phi'(x)+c_2\psi'(x)=0$ has a non-trivial solution and so the Wronskian is zero (for all $x$).

Conversely, if the Wronskian is zero for all $x$, it is zero for some particular $x_0$. Therefore, the system $c_1\phi(x_0)+c_2\psi(x_0)=0$ and $c_1\phi'(x_0)+c_2\psi'(x_0)=0$ has a non-trivial solution (some constants $c_1$ and $c_2$). Fix this non-trivial solution: $c_1,c_2$.

Now we have that the solution $c_1\phi+c_2\psi$ (linear combinations of solutions are solutions) solves the initial value problem with $y(x_0)=0$ and $y'(x_0)=0$. By uniqueness of solutions this must be the zero solution. Thus $c_1\phi(x)+c_2\psi(x)=0$ for all $x$ and so the functions are linearly dependent.

In general, a set of $n$ solutions of an $n$-th order homogeneous linear differential equation is linearly independent if and only if its Wronskian is never zero if and only if its Wronskian is non-zero at a point. [Assuming the linear DE is of the form $y^{(n)}+p_{n-1}(x)y^{(n-1)}+\cdots+p_0(x)y=0$ where the functions $p_0(x),\dots,p_{n-1}(x)$ are continuous on the interval in question.]

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Not sure how to do it, but here's my 2 cents. Suppose $\phi$ is a solution to the ODE. So we know

$$ \phi ''(x) + p(x) \phi '(x) + q(x) \phi(x) = 0 $$

Now suppose $x_0$ is such a point where $\phi(x_0)=0$. To show that it cannot have a relative extremum, we go for contradiction. Suppose it has one, this means $\phi'(x_0)=0$ and $\phi''(x_0)>0$ (or $\phi''=0$, we'll get there). But if we look at the ODE we see that

$$ \phi''(x_0) + p(x_0) *0 + q(x_0)*0 = \phi''(x_0) \neq 0$$

(Since $p$ and $q$ are continuous)So $\phi''(x_0)$ cannot be positive. Now we have to deal with the possibility that $\phi''(x_0) =0$. In this case, we check the wronskian. It's easy to see that

$$ W(x_0)=[ \phi, \psi ] =0 \quad \& \quad W'(x_0) = [ \phi, \psi]' =0 $$

So this point is also a critical point of the wronskian. I'm now unsure how to proceed.

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I'm going to assume that $x_0$ is in the interior of $I$(otherwise, if we consider the system $$y'' + y = 0$$ $I=[0,\pi]$ and $\phi(x) = sin(x)$ is a counterexample if $x_0 = 0$). If $\phi(x_0) = 0$ and $x_0$ is a relative extrema of $\phi$ then $\phi'(x_0) = 0$(because $x_0\in int I$). But the function $\eta(x) \equiv 0$ is also a solution of the equation and $\eta(x_0) = \eta'(x_0)= 0$ so, by existence and uniqueness theorem $$\phi(x) \equiv 0$$ but this contradict the fact that $\phi$ and $\psi$ are linearly independent $$1\cdot\phi(x) + 0\cdot\psi(x) \equiv 0$$


If you want to solve it using the Wronskian: If $W(x_0) = 0$ , this implies that the matrix $$\pmatrix{\phi(x_0) &\psi(x_0) \\ \phi'(x_0) &\psi'(x_0) } $$ is singular, so there exists $\alpha,\beta$ non trivial such that $$\alpha \phi(x_0) + \beta\psi(x_0) = 0$$ $$\alpha \phi'(x_0) + \beta\psi'(x_0) = 0$$ lets define $$\mu(x) := \alpha \phi(x) + \beta\psi(x)$$ Clearly $\mu$ solves the equation and $$\mu(x_0) = \alpha \phi(x_0) + \beta\psi(x_0) = 0$$ $$\mu'(x_0) = \alpha \phi'(x_0) + \beta\psi'(x_0) = 0$$ again by existence and uniqueness, $\mu \equiv 0$ so $$\alpha \phi + \beta\psi \equiv 0$$ that contradict the linear independece.

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