# Show that the next eigenfunction has a zero between 2 zeros

1. Dec 10, 2015

### Happiness

1. The problem statement, all variables and given/known data

2. Relevant equations
Wronskian theorem:

3. The attempt at a solution
I've gotten the relationship given by the question but I do not know how to continue.

Since $\psi_n(a)=\psi_n(b)=0$,
LHS $=\psi_n'(b)\,\psi_{n+1}(b)-\psi_n'(a)\,\psi_{n+1}(a)$

If LHS $=0$, RHS $=0$, then by the First Mean Value Theorem for Integrals, $\psi_{n+1}(c)=0$ for some $c\in(a,b)$.

But LHS is not necessarily $0$.

Attached below is the derivation of Wronskian theorem.

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Last edited: Dec 10, 2015
2. Dec 15, 2015

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Dec 16, 2015

### Samy_A

As a and b are consecutive zeros of $\psi_n$, $\psi_n$ won't change sign between a and b.
Assume, without loss of generality, that $\psi_n$ is positive between a and b.
What does that tell you about the signs of $\psi_n'(a)$ and $\psi_n'(b)$?

Now assume that $\psi_{n+1}$ has no zeros between a and b. It follows that $\psi_{n+1}$ won't change sign between a and b.
Now just check what the signs of the two expressions will be (remember that $E_{n+1}>E_n$).

Last edited: Dec 16, 2015