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For the following system $$ x'=\left( \begin{array}{ccc} \frac{-1}{2} & 1 \\ -1 & \frac{-1}{2} \end{array} \right)x $$

To find a fundamental set of solutions, we assume that $$ x = Ee^{rt}$$

According to the solution, this set of linear algebraic equations is obtained [1]:

$$ x'=\left( \begin{array}{ccc} \frac{-1}{2} & 1 \\ -1 & \frac{-1}{2} \end{array} \right) \left( \begin{array}{ccc} E_1 \\ E_2 \end{array} \right) = \left( \begin{array}{ccc} 0 \\ 0 \end{array} \right)$$

For the eigenvalues and eigenvectors of A the characteristic equation is [2]

$$ \left| \begin{array}{ccc} \frac{-1}{2}-r & 1 \\ -1 & \frac{-1}{2}-r \end{array} \right| = r^2 + r+\frac{5}{4} $$

Therefore the eigenvalues are $\frac{-1}{2} + i$ and $\frac{-1}{2} - i$

I don't understand how eqn [1] and [2] are obtained.

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1 Answer 1

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We are given:

$$ x'=\left( \begin{array}{ccc} \frac{-1}{2} & 1 \\ -1 & \frac{-1}{2} \end{array} \right)x $$

To find the characteristic polynomial, we solve:

$$|A - \lambda I| = 0$$

So, for your matrix, we would write:

$$ \left| \begin{array}{ccc} \frac{-1}{2} - \lambda & 1 \\ -1 & \frac{-1}{2} - \lambda \end{array} \right|= 0 $$

Taking the determinant results in the equation you show (you are using $r$ instead of the more traditional $\lambda$).

Here are the steps of finding the determinant and solving for the roots (eigenvalues):

$$\displaystyle \lambda^2 + \lambda + \frac{5}{4} = 0 \rightarrow \lambda_{1,2} = -\frac{1}{2} \pm i$$

Now that we know the eigenvalues, there are various ways to find the exponential solution using eigenvectors or other methods.

Is that all clear?

In the end, we would end up with (you used $E$ as the initial condition, where I call that vector $x_0$):

$$\displaystyle x(t) = x_0 e^{A t} = x_0 e^{-\frac{t}{2}}\begin{bmatrix} \cos t & \sin t\\- \sin t & \cos t\end{bmatrix}$$

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  • $\begingroup$ nicely outlined and explained ;-) +1 $\endgroup$
    – amWhy
    Commented Apr 27, 2013 at 0:09
  • $\begingroup$ @amWhy: No feedback, not sure if it helped the OP. :-( $\endgroup$
    – Amzoti
    Commented Apr 27, 2013 at 0:32
  • $\begingroup$ I should read your answer. but let's start with a +. $\endgroup$
    – Mikasa
    Commented Apr 27, 2013 at 4:52
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    $\begingroup$ @BabakS.: Thank you my friend, it has been a wild day on MSE! REgards $\endgroup$
    – Amzoti
    Commented Apr 27, 2013 at 6:09
  • $\begingroup$ Don't worry, Amzoti! I appreciate the support! No harm done ;-) $\endgroup$
    – amWhy
    Commented Apr 30, 2013 at 3:14

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