# Re(eigenvalue) inequality problem

1. Sep 28, 2005

### N2

Hello,
if a diff.eqn. has the characteristic equation $\lambda^2 + (3-K) \lambda + 1 = 0$
the eigenvalues solves to $\lambda=-3/2 + K/2 \pm 1/2 \sqrt{5-6K+K^2}$. No problem there. But when is the diff.eqn. asymp. stable, meaning $\Re(\lambda)<0$ ?

I can only get this far
$\Re(-3/2 + K/2 \pm 1/2*\sqrt{5-6K+K^2})<0$
$-3/2+1/2 \Re(K \pm \sqrt{5-6K+K^2})<0$

How can i find the values for K, where this inequality is true?

Thanks

2. Sep 29, 2005

### CarlB

My inclination would be to first consider the case when $$5-6K+K^2$$ is positive, and then consider the case when it is negative. If you separate them out, it shouldn't be too hard.

Carl

3. Sep 30, 2005

### HallsofIvy

Staff Emeritus
Solving K2- 6K+ 5> 0 tells us that there will be complex roots for K between 1 and 5 and real roots for K<= 1, >= 5.

I there are complex roots, the real part is just -3/2+ K/2. That is 0 for K= 3, negative for K< 3, positive for K> 3. The solution will be stable for 1< K< 3, unstable for 3< K< 5.

For K<= 1 or K>= 5, we need to look at all of $-3/2 + K/2 \pm 1/2 \sqrt{5-6K+K^2}$

The best way to determine where that is positive or negative is to set it equal to 0 and solve the resulting quadratic equation. Those will separate "< 0" from "> 0". Choose one value of K in each resulting interval to see whether this is positive or negative.