Differential Eqtns - with a matrix

[flyingpig]

Homework Statement

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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 &amp; 1\\ <br /> \frac{1}{4}&amp; \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&#039;= (P^{-1}X)&#039; so,.
multiply on both sides by P^{-1},
(P^{-1}X)&#039;= 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&#039;= DY.

Here, that is
Y&#039;= \begin{batrix}y_1&#039; \\ y_2&#039;}= \begin{bmatrix}-1/4 &amp; 0 \\ 0 &amp; 1\end{bmatrix}\begin{bmatrix}y_1 \\ y_2\end{bmatrix}

That gives the two uncoupled equations, y_1&#039;= -(1/4)y_1 and y_2&#039;= 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 &amp; 1 \\ 1 &amp; 1\end{bmatrix}
and
X= \begin{bmatrix}-4 &amp; 1 \\ 1 &amp; 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)?