Differential Eqtns - with a matrix

[flyingpig]

Homework Statement

http://img847.imageshack.us/img847/2259/unledgqc.th.png

Uploaded with ImageShack.us

The Attempt at a Solution

I am guessing both part a) and part b) implies that I must solve the differential equation first to answer those questions and I have

det(A-λI) = 0

$$\begin{bmatrix}<br /> -\lambda & 1\\ <br /> \frac{1}{4}& \frac{3}{4} - \lambda<br /> \end{bmatrix} = 0$$

Solving I got my λ = -1/4 or 1

So starting with λ = 1, I build my vector

[A - λI | 0]

After row reducing and I found my parametric eqtn to be

$$\vec{x} = x_2 \begin{bmatrix}<br /> 1\\ <br /> 1<br /> \end{bmatrix}$$

For λ = -1/4, I got

$$\vec{x} = x_2 \begin{bmatrix}<br /> -4\\ <br /> 1<br /> \end{bmatrix}$$

So my final vector should be (where I decided to let x₂ = 1 for convenience. Also does anyone know why I can't let x₂ be 0? I have a sneaking suspicion that it has to do with eigenvectors can't be 0...)

$$\vec{x(t)} = c_1\begin{bmatrix}1\\ 1\end{bmatrix}e^t + c_2\begin{bmatrix}4\\ 1\end{bmatrix}e^{\frac{-1t}{4}}$$

So now to answer the question

For part a) I am assuming they want to ask what happens t → ∞

In that case, one of them "diverges" and one of them "converges" to 0.

So (from my notes), this is a saddle point.

For part b)

$$\vec{x(0)} = c_1\begin{bmatrix}1\\ 1\end{bmatrix} + c_2\begin{bmatrix}4\\ 1\end{bmatrix} = \begin{bmatrix} -2\\ 1\end{bmatrix}$$

This is too trivial, I just have to solve for my constants and then plug it back into the neat equation.

Now my question is, for part a, am I right? What exactly happens if one of the vectors goes to negative infinity and one goes to positive infinity?

Also, there is only two vectors in this problem (2 x 2 matrix), what if this is bigger? Like if I ended up with three vectors, what happens if two goes to infinity and one goes to 0?

 

[flyingpig]

Am I right about the behavior?

 

[HallsofIvy]

You've done very well but it might help to have a formal way of thinking about problems like this.
You have X'= AX where A is a "diagonalizable" matrix. That means that there exist an invertible matrix,
P, such that $P^{-1}AP= D$ where D is a diagonal matrix having the eigenvalues of A on
its main diagonal. Since $P^{-1}$ is a constant matrix, $P^{-1}X'= (P^{-1}X)'$ so,.
multiply on both sides by $P^{-1}$,
$$(P^{-1}X)'= P^{-1}AX= P^{-1}A(PP^{-1})X= (P^{-1}AP)(P^{-1}X)= D(P^{-1}X)$$

Now let $Y= P^{-1}X$ so the equation becomes $Y'= DY$.

Here, that is
$$Y'= \begin{batrix}y_1' \\ y_2'}= \begin{bmatrix}-1/4 & 0 \\ 0 & 1\end{bmatrix}\begin{bmatrix}y_1 \\ y_2\end{bmatrix}$$

That gives the two uncoupled equations, $y_1'= -(1/4)y_1$ and $y_2'= y_2$.

The solutions to those are $y_1(t)= c_1e^{-(1/4)t}$ and $y_2(t)= c_2e^t$ just
as you have.

That gives
$$Y(t)= \begin{bmatrix}c_1e^{-(1/4)t} \\ c_2e^t\end{bmatrix}$$

Finally, since we defined $Y= P^{-1}X$, $X= PY$.

Yes, as you say, an eigenvector corresponding to eigenvalue -1/4 is <-4, 1> and an eigenvector
corresponding to eigenvalue 1 is <1, 1> so we have
$$P= \begin{bmatrix}-4 & 1 \\ 1 & 1\end{bmatrix}$$
and
$$X= \begin{bmatrix}-4 & 1 \\ 1 & 1\end{bmatrix}\begin{bmatrix}c_1e^{-(1/4)t} \\ c_2e^t\end{bmatrix}$$
$$X= \begin{bmatrix}-4c_1e^{-(1/4)t}+ c_2e^t \\ c_1e^{-(1/4)t}+ c_2e^t\end{bmatrix}$$

Now, as t goes to infinity, the negative exponential will go to 0 while the positive exponential gets
larger and larger. But for very large t, the solution will act like
$$X= \begin{bmatrix}c_2e^{t} \\ c_2e^t\end{bmatrix}$$
so that $x_1(t)\approx x_2(t)$.

 

[flyingpig]

What if there are three vectors? Like $$x_1 (t), x_2 (t), x_3 (t)$$?