What's next is to solve for the eigenvalues λ using the quadratic formula in terms of α. As with every quadratic equation, there are 3 possibilities:

I. two distinct real solutions, λ_{1} and λ_{2}

II. one real solution λ

III. complex conjugate solutions, λ = a±ib

As to the solution for the system, these correspond to the following cases:

I. two independent real exponential functions, e^{λ1t} and e^{λ2t}

II. two functions of the form e^{λt} and t*e^{λt}

III. two independent trig-exponential functions, e^{at}sin(bt) and e^{at}cos(bt)

The phase portraits are qualitatively different for these three cases. The critical (or degenerate) case is case II. This happens when the discriminant of the quadratic formula (the term under the root sign) equals 0.

In your example, the characteristic equation is

(α-λ)(-4-λ)+10=0, or

λ²+(4-α)λ+(10-4α)=0

Use the quadratic formula and get the eigenvalues

λ=1/2( α-4±√(α²+8α-24) ),

so the discriminant is α^{2}+8α-24. Set this discriminant to zero and use the quadratic formula again to find the critical values of α:

α^{2}+8α-24=0 ⇒ α=-4±2√10, about -10.3 and 2.32

If α lies between these two values, i.e.

-4-2√10 < α < -4+2√10,

the discriminant is negative, so you are in case III, complex-conjugate eigenvalues giving rise to trig-exponential functions.

Now what if α lies outside these two critical values, i.e.

α ≤ -4-2√10 or α ≥ -4+2√10 ?

Then you get two real eigenvalues (case I: real exponential functions) EXCEPT at the boundaries,

α = -4-2√10 and α = -4+2√10,

when there is only one eigenvalue, namely

λ=-4+√10 and λ=-4-√10

(case II).

They claim there is a third critical value at α=5/2, corresponding to λ=0 and λ=-3/2. I don't see anything special about this value though.

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