Systems of ODE's - Complex Eigenvalues

  • #1

Homework Statement


Find the general solution of the given system.

The given matrix is X' = (1st row (1,-1,2) 2nd row (-1,1,0) 3rd row (-1,0,1))X

2. The attempt at a solution

The eigenvalue determinant = (1st row (1-λ,-1,2) second row (-1,1-λ,0) 3rd row (-1,0,1-λ)

Solving the determinant gives -(λ-1)(λ^2-2λ+2)

So λ1,2 = 1+i and 1-i and λ3 = 1

Now my question is how do I solve the 3x3 matrix with copmlex numbers in it? I used gaussian elimination to solve for the real value λ3.

(1st row (-i,-1,2) 2nd row (-1,-i,0) 3rd row (-1,0,-i)) multiplied by the column (k1,k2,k3) = 0
 
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Answers and Replies

  • #2
gabbagabbahey
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Solving the determinant gives -(λ-1)(λ^2-2λ+2)

So λ1,2 = -1+i and -1-i and λ3 = 1

Your two complex eigenvalues have a sign error. The roots of [itex]\lambda^2-2\lambda+2=0[/itex] are [itex]\lambda=1\pm i[/itex].

Now my question is how do I solve the 3x3 matrix with copmlex numbers in it? I used gaussian elimination to solve for the real value λ3.

(1st row (2-i,-1,2) 2nd row (-1,2-i,0) 3rd row (-1,0,2-i)) multiplied by the column (k1,k2,k3) = 0

You can still use Gaussian elimination to solve for the eigenvectors.
 
  • #3
Your two complex eigenvalues have a sign error. The roots of [itex]\lambda^2-2\lambda+2=0[/itex] are [itex]\lambda=1\pm i[/itex].



You can still use Gaussian elimination to solve for the eigenvectors.

Can someone show me how?
 
  • #4
gabbagabbahey
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You do it exactly the same way you did it for your real eigenvalue....try it out and post your work if you get stuck.
 
  • #5
You do it exactly the same way you did it for your real eigenvalue....try it out and post your work if you get stuck.

Is multiplying a row by a complex number a valid operation?
 
  • #6
gabbagabbahey
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Yes. It's exactly the same as multiplying both sides of an equation by a complex number.
 
  • #7
The corrected eigenvalue matrix is
[-i,-1,2]
[-1,-i,0]
[-1,0,-i]

using the operations 1) i*R1, 2) R1+R2, 3) R1+R3, 4) (-1/2)R2+R3, 5) -2*R2+R1 gives

[1,-i,0]
[0,-i, i]
[0,0,0]

which says k2=k3; picking k1=1 and k2=k3= i gives one solution,k=
[1]

which is split into the real and imaginary parts B1 and B2, respectively

[1]
[0]
[0]

[0]
[1]
[1]

The general solution is c1[B1*cos(t)-B2sin(t)]e^t + c2[B2*cos(t)-B1sin(t)]e^t +c3
([0])e^t
([1])
([2])
 
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