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Find the critical value or values of α where the qualitative nature of the phase portrait for x'=(alpha, -1, 10, -4)x changes. (this is 2x2 matrix, alpha and -1 on the left, 10 and -4 on the right, I got (α-λ)(-4-λ)+10=0, but what's next and how to get the answer?)

Sun,

the eigenvalues you should get from the quadratic formula are

λ=1/2( a-4±√(a²+8a-24) )

Set the term under the radical equal to zero and get your first two critical values,

α=-4±2√10,

then set the entire λ equal to zero, to get one more critical value, α=5/2.

Sun:

The critical value α=5/2 correspond to the situation of det |A|=-4α+10=0; In this case there are infinitely many critical points, whereas if det |A|≠0, the only critical point is x=(0;0). It has nothing to do with the fact that eigenvalue is zero.

I am surprised though that they did not distinguish cases of purely imaginary eigenvalues from complex ones.
It's impossible to have purely imaginary eigenvalues in this problem. If α=4, λ is real.

### 2 Answers by Expert Tutors

Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
5.0 5.0 (3 lesson ratings) (3)
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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, λ1 and λ2
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, eλ1t and eλ2t
II. two functions of the form eλt and t*eλt
III. two independent trig-exponential functions, eatsin(bt) and eatcos(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 α2+8α-24. Set this discriminant to zero and use the quadratic formula again to find the critical values of α:

α2+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.
Kirill Z. | Physics, math tutor with great knowledge and teaching skillsPhysics, math tutor with great knowledge...
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From what you got you can obtain:

(λ+4)(λ−α)+10=0; or
λ2+(4-α)λ+10-4α=0;

λ1=½[(α-4)-√(α2-8α-24+16α)]=½[(α-4)-√(α2+8α-24)]
λ2=½[(α-4)+√(α2-8α-24+16α)]=½[(α-4)+√(α2+8α-24)]

Everything depends on the sign of the expression under the square root.

Let us solve α2+8α-24=0;

α1=-4-√(16+24)=-4-2√10;
α2=-4+√(16+24)=-4+2√10;

Now if α1<α<α2, then α2+8α-24<0; two eigenvalues are complex. If α=4, two eigenvalues are purely imaginary. In the case of complex eigenvalues, solution has the exponent with real and imaginary part, which reduces to the product of sine or cosine function and ekt type of function. So we will have a wave with exponentially increasing or decreasing amplitude. It increases exponentially if α>4, decreases if α<4, or remains constant if α=4.

If α<α1 or α>α2, then two eigenvalues are real.
In this particular example, since the expression under square root is less than (α-4), we have either two positive or two negative roots. In the first case we have two exponentially growing solutions, in the second case--two exponentially decreasing solutions.

So, critical values of α are:

α=4;
α=-4-2√10;
α=-4+2√10.

If det |A|=0, where A is your original matrix, the system has infinitely many stationary points, which clearly changes a phase portrait. This happens when -4α+10=0 or α=5/2;