1
$\begingroup$

My ODE Professor gave us a 'fun' problem and I haven't made much progress on it. We are going over eigenvalues and oscillation currently.

Here are both problems:


a) For the following system of ODE's, use the eigenvalue method to find the condition (in terms of relative size $a$ and $b$) such that $\lim_{t\to\infty} R(t)=\infty$ and $\lim_{t\to\infty} J(t)=\infty$ (as $t$ approaches infinity, both solutions $R(t)$ and $J(t)$ approach positive infinity) for some initial conditions.

$\begin{cases} R' =-aR+bJ \\ J'= bR -aJ \end{cases}$

$(a>0,b>0)$

b) For the following system, predict the long term behaviors of the solutions $R(t) and J(t)$ (e.g. approaching infinity, approaching zero, or oscillating), depending on the signs of $a$ and $b$.

$\begin{cases} R'=aJ \\ J'=bR \end{cases}$

$(a\neq 0,b \neq 0)$


What I have done so far for a.) is translated the system of equations into matrix form, and it looks something like this:

$\begin{bmatrix} R'\\ J' \end{bmatrix}$ $=$ $\begin{bmatrix} -a & b\\ b & -a \end{bmatrix}$ $\begin{bmatrix} R\\ J \end{bmatrix}$

After attempting to take the eigenvalues using the determinant to find the characteristic polynomial, I have found the following polynomial:

$det= \lambda^2 +2a\lambda +a^2 -b^2 = 0$

Using the quadratic formula, I have: $\lambda = -a + b$, or $-a-b$

Producing eigenvectors from this is where I am running into trouble.

I have read part $b$ and I have predicted that if $a$ and $b$ are both the same sign, then $R(t)$ and $J(t)$ will approach infinity, but if $a$ and $b$ have different signs, than $R(t)$ and $J(t)$ will approach $0$, and thus they will oscillate. I have no justification for this, other than it seems to make sense based on the way linear systems of differential equations work.

Any assistance would be greatly appreciated on these.

$\endgroup$
  • 2
    $\begingroup$ Where do you get $2b$ from? the determinant is $a^2-b^2$. Eigenvalues should be $-a\pm b$. $\endgroup$ – Lutz Lehmann Oct 29 '15 at 8:35
  • 1
    $\begingroup$ For the b) part, try differentiating one of the equations, and see if you can use the other one! $\endgroup$ – krvolok Oct 29 '15 at 8:51
  • $\begingroup$ Ah my mistake. After using the quadratic formula, I have reached the same eigenvalues that you have, LutzL, Thanks. $-a+b$ and $-a-b$ $\endgroup$ – klorzan Oct 29 '15 at 9:09
1
$\begingroup$

So, after reviewing this further, it seems that there are some things to note about the behavior of these differential equations with respect to the signs of the eigenvalues.

enter image description here

So the first equation's behavior is determined by the sign of $b$.

The second equation's behavior is determined by the signs of both $a$ and $b$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.