18.1 Straight line solutions

Consider this following linear system of equations:

In Section 15 we found the two straight line solutions:

• Solution 1: \(\displaystyle \vec{s}_{1}(t) = \begin{pmatrix} 0 \\ e^{t} \end{pmatrix}= \begin{pmatrix} 0 \\ e^{t} \end{pmatrix} =e^{t} \begin{pmatrix} 0 \\ 1 \end{pmatrix}\).
• Solution 2: \(\displaystyle \vec{s}_{2}(t) = \begin{pmatrix} e^{2t} \\ e^{2t} \end{pmatrix}= \begin{pmatrix} e^{2t} \\ e^{2t} \end{pmatrix} = e^{2t} \begin{pmatrix} 1 \\ 1\end{pmatrix}\).

Let’s verify that Solution 2 is indeed a solution to this linear system. First we will take the derivative of Solution 2:

\[\begin{equation} \frac{d}{dt} \left( \vec{s}_{2}(t) \right) = \frac{d}{dt} \left( e^{2t} \begin{pmatrix} 1 \\ 1\end{pmatrix} \right) = 2 e^{2t} \begin{pmatrix} 1 \\ 1\end{pmatrix} \end{equation}\]

Let’s compare this solution to the right hand side of the differential equation: \[\begin{equation} \begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix} \left( e^{2t} \begin{pmatrix} 1 \\ 1\end{pmatrix} \right) = \left( e^{2t} \begin{pmatrix} 2 \\ 2 \end{pmatrix} \right) = 2 e^{2t} \begin{pmatrix} 1 \\ 1\end{pmatrix} \end{equation}\]

So, indeed Solution 2 is a solution to the differential equation. However something interesting is occurring. Notice how \(\displaystyle \frac{d}{dt} \left( \vec{s}_{2}(t) \right)\) equals \(2 \vec{s}_{2}(t)\), so if \(\displaystyle \frac{d}{dt} \left( \vec{s}_{2}(t) \right) = \begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix} \vec{s}_{2}(t)\), then the following equality has to hold:

\[\begin{equation} \begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix} \vec{s}_{2}(t) = 2 \vec{s}_{2}(t) \end{equation}\]

In fact, for any linear system \(\vec{y'}=A\vec{y}\), straight line solutions have the property that \(\displaystyle A \vec{v} = \lambda \vec{v}\). In the example we just computed \(\displaystyle \vec{v} = \begin{pmatrix} 1 \\ 1\end{pmatrix}\), and \(\lambda = 2\).

Some apply some terminology here. For these special straight line solutions, we give a particular name to \(\vec{v}\) - we call it the eigenvector. The name we give to \(\lambda\) is the eigenvalue. (Eigen means own in German - get it?)

Notice how \(s_{2}(t)\) was expressed as \(e^{2t}\begin{pmatrix} 1 \\ 1\end{pmatrix}\). In fact, any straight line solution has the form \(\displaystyle \vec{s} = e^{\lambda t} \vec{v}\), where \(\lambda\) and \(\vec{v}\) are the eigenvalue and eigvenvector.

So how do we determine an eigenvalue or eigenvector? A straight line solution satisfies the equation \(A\vec{v} = \lambda{v}\). Re-arranging this equation we can express this as \((A - \lambda I) \vec{v} = \vec{0}\), where \(\vec{0}\) is a vector of all zeros and \(I\) is called the identity matrix, or a square matrix with ones along the diagonal and zero everywhere else. The goal is to find a \(\lambda\) and \(\vec{v}\) consistent with this equation.

The eigenvalues of \(\lambda\) that solve \(( A \vec{v} - \lambda \vec{v} ) = 0\) can be found by solving \(\det (A - \lambda I ) =0\), where \(\det(M)\) is the determinant. Once the eigenvalues are found, we then determine the eigenvectors.

Time out. I recognize that we are starting to get deeper into linear algebra. We will just borrow some key results that we will need - so hopefully that will give you a leg up when you study linear algebra - it is a great topic! Let’s get to work.