Author Topic: Reading Week Bonus problem 2  (Read 3317 times)

Victor Ivrii

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Reading Week Bonus problem 2
« on: February 16, 2013, 10:35:08 AM »
Using Reading Week Bonus problem 1 prove (in its framework)  that if $y(x_0)=y(x_1)=0$ and $y(x)>0$ on $(x_0,x_1)$, $x_0<x_1$ then $z(x)$ has a $0$ somewhere on $(x_0,x_1)$ unless $z(x_0)=y(x_1)=0$ and $Q(x)=q(x)$ on $(x_0,x_1)$.

Brian Bi

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Re: Reading Week Bonus problem 2
« Reply #1 on: February 16, 2013, 05:10:49 PM »
Assume that $z(x) > 0$ on $(x_0, x_1)$. We will show a contradiction.

Since $y(x) > 0$ and $z(x) > 0$ on $(x_0, x_1)$, recall the result of Bonus Problem 1:
\begin{equation}
W'(x) = (q(x) - Q(x))y(x)z(x)
\end{equation}
We know that $W'(x) \leq 0$ on $(x_0, x_1)$. By the mean value theorem, $W(x_0) \geq W(x_1)$. Consider what happens if we have $W(x_0) = W(x_1)$. Then if for some $x \in (x_0, x_1)$ we had $W(x) \neq W(x_0)$, then by the mean value theorem $W'$ would have to be positive on either $(x_0, x)$ or $(x, x_1)$, a contradiction. Conclude that if $W(x_0) = W(x_1)$, then $W$ is constant on $(x_0, x_1)$, implying that $q - Q$ vanishes identically on $(x_0, x_1)$. We proceed under the assumption that this is not the case, so that $W(x_0) > W(x_1)$.

Writing out the Wronskians explicitly gives
\begin{align}
\begin{vmatrix} y(x_0) & z(x_0) \\ y'(x_0) & z'(x_0) \end{vmatrix} &> \begin{vmatrix} y(x_1) & z(x_1) \\ y'(x_1) & z'(x_1)  \end{vmatrix} \\
\begin{vmatrix} 0 & z(x_0) \\ y'(x_0) & z'(x_0) \end{vmatrix} &> \begin{vmatrix} 0 & z(x_1) \\ y'(x_1) & z'(x_1) \end{vmatrix} \\
-z(x_0) y'(x_0) &> -z(x_1) y'(x_1) \label{eq:foo}
\end{align}
Now, because $y(x_0) = 0$ and $y(x_0) > 0$ on $(x_0, x_1)$, we must have $y'(x_0) \geq 0$. Likewise $y'(x_1) \leq 0$. We thus consider the following:
Case 1: If $z(x_0) > 0$ and $z(x_1) > 0$, then $-z(x_0) y'(x_0) \leq 0$ and $-z(x_1) y'(x_1) \geq 0$; together with $(\ref{eq:foo})$ we obtain the desired contradiction.
Case 2: If either $z(x_0) < 0$ or $z(x_1) < 0$, this contradicts the assumption that $z > 0$ on $(x_0, x_1)$ since $z$ is continuous.
Case 3: If $z(x_0) = 0$, then the RHS of $(\ref{eq:foo})$ must be negative, implying $z(x_1) < 0$. As in Case 2, find a contradiction.
Case 4: If $z(x_1) = 0$, then the LHS of $(\ref{eq:foo})$ must be positive, implying $z(x_0) < 0$. As in Case 2, find a contradiction.

So it is impossible for $z$ to be strictly positive on $(x_0, x_1)$; this implies that $z$ either has a zero, changes sign, or is strictly negative on $(x_0, x_1)$. The latter is impossible by an analysis analogous to the foregoing; and a sign change implies the presence of a zero anyway by the intermediate value theorem, so we are done.
« Last Edit: February 16, 2013, 05:17:37 PM by Brian Bi »

Victor Ivrii

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Re: Reading Week Bonus problem 2
« Reply #2 on: February 16, 2013, 05:32:30 PM »
Note: $y(x_0)=0\implies y'(x_0)\ne 0$ as otherwise $y\equiv 0$due to unicity theorem.