# Differential Equations: 'Fun Problem' involving eigenvalues and predicting limits.

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.

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

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 $b$.