### Author Topic: Day Section, Question 1  (Read 10428 times)

#### Victor Ivrii ##### Day Section, Question 1
« on: January 31, 2013, 01:29:59 PM »
Determine the values of $\alpha$ , if any, for which all solutions of the following ODE tend to zero as $t\to\infty$ as well as all values of $\alpha$ , if any, for which all nonzero solutions become unbounded as $t\to\infty$
$$y'' - (2\alpha-1)y'+\alpha(\alpha-1)y=0.$$

#### Brian Bi ##### Re: Day Section, Question 1
« Reply #1 on: January 31, 2013, 04:26:03 PM »
The characteristic equation
\begin{equation}
r^2 - (2\alpha-1)r + \alpha(\alpha-1) = 0
\end{equation}
factors as $(r - \alpha)(r - (\alpha - 1))$, so the general solution to the ODE is given by
\begin{equation}
y = A e^{\alpha t} + B e^{(\alpha-1)t}
\end{equation}
where $A, B \in \mathbb{R}$.

We consider the following cases:
• $\alpha < 0$: Both exponentials will be decaying, so each solution tends to zero as $t \to \infty$.
• $\alpha = 0$ or $\alpha = 1$: Each $y = c$ for constant $c$ is a solution, so there exist solutions that neither tend to zero nor become unbounded as $t \to \infty$.
• $0 < \alpha < 1$: One exponential is growing and the other decaying, so there exist nonzero solutions that tend to zero as well as solutions that tend to infinity.
• $\alpha > 1$: Both exponentials will be growing. The larger of the two, $Ae^{\alpha t}$, dominates as $t \to \infty$, so $y$ is unbounded unless $A = 0$. If $A$ vanishes identically, then all nonzero solutions $Be^{(\alpha-1)t}$ again become unbounded.
We conclude that the answer is: (i) $\alpha < 0$, and (ii) $\alpha > 1$.

#### Zhuolin Liu

• Newbie
• • Posts: 1
• Karma: 0 ##### Re: Day Section, Question 1
« Reply #2 on: February 01, 2013, 04:02:18 PM »
I wonder, isn't that when 0<Î±<1, e(Î±-1)t tend to 0 as t tend to infinity, and eÎ±t tend to infinity? As a result, isn't that all nonzero solutions become unbounded as t tend to zero when Î±>0 instead of Î±>1?

#### Brian Bi ##### Re: Day Section, Question 1
« Reply #3 on: February 01, 2013, 05:43:49 PM »
I wonder, isn't that when 0<Î±<1, e(Î±-1)t tend to 0 as t tend to infinity, and eÎ±t tend to infinity? As a result, isn't that all nonzero solutions become unbounded as t tend to zero when Î±>0 instead of Î±>1?
The coefficient on $e^{\alpha t}$ might be zero, so you can have solutions that are just $A e^{(\alpha-1)t}$. These will decay to zero as $t \to \infty$.

#### Victor Ivrii ##### Re: Day Section, Question 1
« Reply #4 on: February 01, 2013, 05:51:45 PM »
We will discuss these things in details later but just simple observations:

If we have two characteristic roots $\lambda_2 >0>\lambda_1$ then almost all solutions (with $C_1\ne 0$ and $C_2\ne 0$) are unbounded as $t\to \pm \infty$, solutions $C_2e^{\lambda_2t}$ ($C_2\ne 0$) are unbounded as $t\to +\infty$ and tend to $0$ as $t\to-\infty$ and solutions $C_1e^{\lambda_1t}$ ($C_2\ne 0$) are unbounded as $t\to -\infty$ and tend to $0$ as $t\to +\infty$.

PS.  I prefer to write $+\infty$ rather than $\infty$ to avoid any ambiguity.

« Last Edit: February 01, 2013, 06:16:28 PM by Victor Ivrii »