# Non-vanishing Solution in Differencial Equation

• Sep 1st 2008, 03:34 PM
ThePerfectHacker
Non-vanishing Solution in Differencial Equation
Say we wish to solve $\displaystyle y' = y$ on $\displaystyle (-\infty,\infty)$*.

1)$\displaystyle y$ is non-vanishing i.e. $\displaystyle y(\xi) \not = 0$ for any $\displaystyle \xi \in \mathbb{R}$.
Then we can write $\displaystyle y'/y = 1$ and thus $\displaystyle (\log |y|)' = 1$.
This leads to $\displaystyle \log |y| = x + k_1 \implies y = ke^x$ with $\displaystyle k\not = 0$

2)$\displaystyle y$ is the zero function but that is definitely a solution.

3)$\displaystyle y$ vanishes at some point i.e. $\displaystyle y(\xi) = 0$ for some $\displaystyle \xi \in \mathbb{R}$.
This is were the difficultly is. If it vanishes at $\displaystyle \xi$ and in the open interval $\displaystyle I$ containing $\displaystyle \xi$ then I think this leads to $\displaystyle y$ being the zero function**. However, if $\displaystyle y$ does not vanish on any open interval containing it then it means there is a point $\displaystyle \xi_1$ such that $\displaystyle y(\xi_1) \not = 0$. But by continuity it means there is an interval $\displaystyle J$ containing $\displaystyle \xi_1$ such that $\displaystyle y$ is non-vanishing on $\displaystyle J$. By applying #1 to this restricted function, $\displaystyle y|J$, we find that $\displaystyle y|J$ must be of the form $\displaystyle ke^x,k\not = 0$. And this perhaps shows that $\displaystyle y$ is $\displaystyle ke^x$ everywhere.*** Therefore $\displaystyle y$ is non-vanishing everywhere. A contradiction if $\displaystyle y$ is a non-zero function. Thus, this case does not exist for non-zero functions.

How do we complete the proof?
And is there a general theorem that tells us we can divide the functions we are solving for without worrying it can vanish.

*)This can be solved without seperation of variables i.e. with integrating fractor.
But with this something interesting happens.

**)Though I did not find the complete proof. I guess the proof comes down to a analogue of the identity theorem from complex analysis applied to real functions. This is because the solution to the differencial equation is $\displaystyle \mathcal{C}^{\infty}$ and we can possibly even proof its analyticity. With analyticity we can have an analogue of the identity theorem.

***)Again by identity theorem for real function - by using power series.
• Oct 18th 2008, 04:55 AM
Rebesques
This is a homogeneous linear equation, so by the uniqueness theorem the only solution that can vanish at any point is y=0.