# Disproving existence of sub-interval for which a solution f(t) of the ODE is equal to zero for all t in sub-interval.

Question:

Let $f(t)$ be a solution to the following ODE:

$y''+p(t)y'+q(t)y=0$

on interval I where $p$ and $q$ are continuous, and where $f(t)\neq0$ for some $t \in I$. Prove there is no sub-interval $J \subset I$ such that $f(t)=0$ for all $t \in J$.

Attempt at Solution:

We know this ODE will have a general solution of form:

$y(t)=c_1g(t)+c_2h(t)$.

where $g$ and $h$ are linearly independent. Now, since $f(t)$ is a solution of the ODE, it must be in the set generated by these fundamental solutions. Thus, $f(t)=c_1g(t)+c_2h(t)$ for some $c_1,c_2 \in R$. Also, as we know $f(t)$ is not the zero-function, then $(c_1,c_2) \neq (0,0)$.

Next, for sake of contradiction, assume that $f(t)=0$ for all $t \in J$. Then,

$c_1g(t)+c_2h(t)=0$ for all $t \in J$.

Well, this would imply that $g$ and $h$ are linearly dependent, which is a contradiction.

Therefore, there is no $J \subset I$ such that $f(t)=0$ for all $t \in J$.

My Issue/Question with this:

Is this actually a valid proof? My worry is with the following statement:

"$c_1g(t)+c_2h(t)=0$ for all $t \in J$.

Well, this would imply that $g$ and $h$ are linearly dependent, which is a contradiction."

Is this actually true? Or is it a faulty statement, as it implies that $g$ and $h$ are multiples of each other only for $t \in J$, as opposed to all $t$ in general?

Any help would be greatly appreciated. Thank you.

EDIT:

Based on the answer given below, I've written the following proof:

For a function $f(t)$ to be zero for all $t$ on some open interval $J$, it must be true that $f(t_0)=0$ and $f'(t_0)=0$ for some $t_0 \in J$.

Therefore, consider the initial value problem:

$y''+p(t)y'+q(t)y=0$

$y(t_0)=0$

$y'(t_0)=0$

We see that the zero-function is a solution to this problem. Since $p$ and $q$ are continuous, then by Existence and Uniqueness theorem, the solution must be unique. Thus, the zero-function is the ONLY solution. Since we know $f(t) \neq 0$ for some $t$, then we know $f(t)$ is not the zero-function, and is therefore not the solution to this initial value problem.

So, we see that there is no sub-interval $J$ for which $f(t)=0$ for all $t \in J$.

Hopefully...this proof looks better.

Your concern is valid, since it might still be that $$c_1 g(t) + c_2 h(t) \neq 0 \mbox{ for a }t\in I\setminus J.$$ A uniqueness statement (see e.g. https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem) could fill this gap. You can even shorten the prove by neglecting the idea of using a fundamental system and just rely on uniqueness of the initial value problem.
• You might do that, but in my opinion it is easier to proceed by contradiction and assume there exists such an intervall $J\subset I$, such that $f=0$ on $J$. Now fix a $t^*\in J$ such that $t^*$ lies in an open subset of $J$. Then you know what $f(t^*)$ and $f'(t^*)$ are and uniqueness gives you $f$ on the whole of $I$, which will be a contradiction. – humanStampedist Sep 22 '17 at 8:28