Consider the linear system
$\dfrac{dx}{dt} = 2x$
$\dfrac{dy}{dt} = 3x + y$
The eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = 2$
The eigenvectors are $\vec{V_1} = \left[\begin{array}{c}
0\\
1\end{array}\right]$ and $\vec{V_2} = \left[\begin{array}{c}
1\\
3\end{array}\right]$
So this system has a source at the origin. I understand all that. But how do I know what the solutions inbetween the eigenvectors will look like?
The general solution is of the form
$y(t) = k_1e^{-t}\left[\begin{array}{c}
0\\
1\end{array}\right] + k_2e^{-2t}\left[\begin{array}{c}
1\\
3\end{array}\right]$.
So it seems to me that in the long term, solutions should look like whichever of the sumands in the general solution is dominant. Since the second sumand has a factor of $e^{-2t}$, which goes to zero much faster than $e^{-t}$, I would think thast in the long term, all solutions should look like $\vec{V_1}$. But, the displayed phase portrait in my textbook shows the exact opposite, that all solutions in the long term looke like $\vec{V_2}$. Why is this?
Best Answer
This is what the vector field looks like in between the eigenvectors. Red dot is the origin.
The eigenvalues are positive, so your solution cannot have decay like that for any initial value (see note) aside from $(0,0)$ which is an equilibrium point. The correct solution should be:
$y(t) = k_1e^{t}\left[\begin{array}{c} 0\\ 1\end{array}\right] + k_2e^{2t}\left[\begin{array}{c} 1\\ 3\end{array}\right]$.
And as you have already indicated, there is a source. Source causes vector field to expand in all directions.
So in the long run the solution would look like the part of the solution with the $\exp(2t)$ term. This explains the slight "spin" of the vector field as shown above. There is such a "spin" effect because the vector field is growing fastest along the eigenvector with the largest exponential term. That is sometimes called the "fast solution" of your system.
Note: In general, given a system of differential equation:
$\dot x(t) = Ax(t)$, $x(0) = x_o$, $A$ has distinct eigenvalues
Then the solution to this equation is given as:
$x(t) = \sum\limits_{i=1}^n c_ie^{\lambda_it}v_i$
where $c_i$ are the coefficient corresponding to initial condition $x(0)$, $v_i$ is the $i$th eigenvector, and $\lambda_i$ is the $i$th eigenvalue.