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$$\frac{\mathrm d^2y}{\mathrm dt^2} + p\frac{\mathrm dy}{\mathrm dt} + qy = 0$$

If the eigenvalues are complex, what conditions on $p$ and $q$ guarantee that solutions spiral around the origin in a clockwise direction ?

I found the eigenvalues to be: $\lambda = -\frac{p}{2} \pm \frac{\sqrt{p^2 - 4q}}{2}$

and not sure how we can answer this. All I note that if $p > 0$ and $p^2 - 4q < 0$, then we have a spiral sink. How does it help me if I can get conditions for spiral sink or source and match their directions (counterclockwise or clockwise) when the problem indicates we are able to do this without the use of technology sketching a phase portrait?

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So does this mean that we need p^2 - 4q < 0 ? This is when we get a spiral sink, and sinks are such that spiral around the origin in a clockwise direction, right? –  mary Jul 25 '11 at 5:34
    
@mary: Please edit in your question that you mean when $y$ results in a spiral around the origin in its phase portrait, as opposed to in the complex plane. Whether or not $y$'s image in $\mathbb{C}$ spirals clockwise or counterclockwise (or at all) into $0$ depends not just on the ODE's coefficients but on the initial conditions. –  anon Jul 25 '11 at 5:49
    
Ah - I eventually figured you were talking about the phase space, due to Robert's answer, but apparently my first impression was correct that you're talking about the complex plane. –  anon Jul 25 '11 at 13:08
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2 Answers

The general solution to this equation is $$y(t) = a_1 \exp x_1 t + a_2 \exp x_2 t$$ with amplitudes $a_1,a_2$ dependent on initial conditions and $x_1,x_2$ the roots (or eigenvalues as you have called them that depend on $p,q$). For your roots, the solution can be written in the form $$y(t) = e^{- (pt/2)}(a_1 e^{ist} + a_2 e^{-ist}) $$ where $s$ is the complex part assuming that it has a complex part or imaginary part of $\sqrt{p^2 - 4q}/2$. The two complex terms differ in sign. Whether the solution goes clockwise or counter clockwise depends on which amplitude $a_1,a_2$ is bigger, and that would be determined from the initial conditions, as @anon states above.

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Careful of potential degeneracy, e.g. if $a_1=\overline{a_2}$ or $a_1=-a_2$ then your solution $y(t)$ will be affixed to a linear subspace of $\mathbb{C}$ (though it will still oscillate about the origin), and depending on your definition, being stuck to a single line might not truly be "spiraling." –  anon Jul 25 '11 at 13:14
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Consider the situation when, say, $y > 0$ and $y' = 0$ (so on a graph where $y$ is the horizontal axis and $y'$ the vertical axis, this would be to the right of the origin). At such a point $y'' = - q y$. If $q > 0$, that is negative so $y'$ is decreasing, and your spiral is going clockwise. If $q < 0$, it is positive so $y'$ is increasing, and the spiral is going counterclockwise.

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