Please help me with this math problem?

Question

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?)

Answer 1

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, e^λ₁t and e^λ₂t

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

III. two independent trig-exponential functions, e^atsin(bt) and e^atcos(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.

Answer 2

From what you got you can obtain:

 

(λ+4)(λ−α)+10=0; or

λ²+(4-α)λ+10-4α=0;

 

Using quadratic formula, we obtain:

 

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 e^kt 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;