Toronto Math Forum
MAT2442013S => MAT244 MathTests => Quiz 2 => Topic started by: Victor Ivrii 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 $t\to\infty$
$$
y''  (2\alpha1)y'+\alpha(\alpha1)y=0.
$$

The characteristic equation
\begin{equation}
r^2  (2\alpha1)r + \alpha(\alpha1) = 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^{(\alpha1)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 $Be^{(\alpha1)t}$ again become unbounded.
We conclude that the answer is: (i) $\alpha < 0$, and (ii) $\alpha > 1$.

I wonder, isn't that when 0<Î±<1, e^{(Î±1)t} tend to 0 as t tend to infinity, and e^{Î±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?

I wonder, isn't that when 0<Î±<1, e^{(Î±1)t} tend to 0 as t tend to infinity, and e^{Î±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 $A e^{(\alpha1)t}$. These will decay to zero as $t \to \infty$.

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.