Differential Equations - Systems: Real Eigenvalues - i

It’s now time to start solving systems of differential equations. We’ve seen that solutions to the system,

$${\overrightarrow{x}}^{\prime }=A\overrightarrow{x}$$

will be of the form

$$\overrightarrow{x}=\overrightarrow{\eta }{e}^{\lambda t}$$

where $\lambda$ and $\overrightarrow{\eta }$are eigenvalues and eigenvectors of the matrix $A$. We will be working with $2\times 2$ systems so this means that we are going to be looking for two solutions, ${\overrightarrow{x}}_1\left(t\right)$ and ${\overrightarrow{x}}_2\left(t\right)$, where the determinant of the matrix,

$$X=\left({\overrightarrow{x}}_1{\overrightarrow{x}}_2\right)$$

is nonzero.

We are going to start by looking at the case where our two eigenvalues, ${\lambda }_1$ and ${\lambda }_2$ are real and distinct. In other words, they will be real, simple eigenvalues. Recall as well that the eigenvectors for simple eigenvalues are linearly independent. This means that the solutions we get from these will also be linearly independent. If the solutions are linearly independent the matrix $X$ must be non-singular and hence these two solutions will be a fundamental set of solutions. The general solution in this case will then be,

$$\overrightarrow{x}\left(t\right)=c_1{e}^{{\lambda }_{1}t}{\overrightarrow{\eta }}^{\left(1\right)}+c_2{e}^{{\lambda }_{2}t}{\overrightarrow{\eta }}^{\left(2\right)}$$

Note that each of our examples will actually be broken into two examples. The first example will be solving the system and the second example will be sketching the phase portrait for the system. Phase portraits are not always taught in a differential equations course and so we’ll strip those out of the solution process so that if you haven’t covered them in your class you can ignore the phase portrait example for the system.

Example 1 Solve the following IVP.$${\overrightarrow{x}}^{\prime }=\left(\begin{array}{cc}1& 2\\ 3& 2\end{array}\right)\overrightarrow{x},\overrightarrow{x}\left(0\right)=\left(\begin{array}{c}0\\ -4\end{array}\right)$$

So, the first thing that we need to do is find the eigenvalues for the matrix.
$$\begin{array}{cc}det\left(A-\lambda I\right)& =\left|\begin{array}{cc}1-\lambda & 2\\ 3& 2-\lambda \end{array}\right|\\ & ={\lambda }^2-3\lambda -4\\ & =\left(\lambda +1\right)\left(\lambda -4\right)\Rightarrow {\lambda }_1=-1,{\lambda }_2=4\end{array}$$Now let’s find the eigenvectors for each of these.
${\lambda }_1=-1$ :
We’ll need to solve,
$$\left(\begin{array}{cc}2& 2\\ 3& 3\end{array}\right)\left(\begin{array}{c}{\eta }_1\\ {\eta }_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)\Rightarrow 2{\eta }_1+2{\eta }_2=0\Rightarrow {\eta }_1=-{\eta }_2$$The eigenvector in this case is,
$$\overrightarrow{\eta }=\left(\begin{array}{c}-{\eta }_2\\ {\eta }_2\end{array}\right)\Rightarrow {\overrightarrow{\eta }}^{\left(1\right)}=\left(\begin{array}{c}-1\\ 1\end{array}\right),{\eta }_2=1$$${\lambda }_2=4$ :
We’ll need to solve,
$$\left(\begin{array}{cc}-3& 2\\ 3& -2\end{array}\right)\left(\begin{array}{c}{\eta }_1\\ {\eta }_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)\Rightarrow -3{\eta }_1+2{\eta }_2=0\Rightarrow {\eta }_1=\frac{2}{3}{\eta }_2$$The eigenvector in this case is,
$$\overrightarrow{\eta }=\left(\begin{array}{c}\frac{2}{3}{\eta }_2\\ {\eta }_2\end{array}\right)\Rightarrow {\overrightarrow{\eta }}^{\left(2\right)}=\left(\begin{array}{c}2\\ 3\end{array}\right),{\eta }_2=3$$Then general solution is then,
$$\overrightarrow{x}\left(t\right)=c_1{e}^{-t}\left(\begin{array}{c}-1\\ 1\end{array}\right)+c_2{e}^{4t}\left(\begin{array}{c}2\\ 3\end{array}\right)$$Now, we need to find the constants. To do this we simply need to apply the initial conditions.
$$\left(\begin{array}{c}0\\ -4\end{array}\right)=\overrightarrow{x}\left(0\right)=c_1\left(\begin{array}{c}-1\\ 1\end{array}\right)+c_2\left(\begin{array}{c}2\\ 3\end{array}\right)$$All we need to do now is multiply the constants through and we then get two equations (one for each row) that we can solve for the constants. This gives,
$$\begin{array}{c}-c_1+2c_2=0\\ c_1+3c_2=-4\end{array}\}\Rightarrow c_1=-\frac{8}{5},c_2=-\frac{4}{5}$$The solution is then,
$$\overrightarrow{x}\left(t\right)=-\frac{8}{5}{e}^{-t}\left(\begin{array}{c}-1\\ 1\end{array}\right)-\frac{4}{5}{e}^{4t}\left(\begin{array}{c}2\\ 3\end{array}\right)$$

Now, let’s take a look at the phase portrait for the system.

Example 2 Sketch the phase portrait for the following system.$${\overrightarrow{x}}^{\prime }=\left(\begin{array}{cc}1& 2\\ 3& 2\end{array}\right)\overrightarrow{x}$$

From the last example we know that the eigenvalues and eigenvectors for this system are,
$$\begin{array}{cccc}{\lambda }_1& =-1& {\overrightarrow{\eta }}^{\left(1\right)}& =\left(\begin{array}{c}-1\\ 1\end{array}\right)\\ {\lambda }_2& =4& {\overrightarrow{\eta }}^{\left(2\right)}& =\left(\begin{array}{c}2\\ 3\end{array}\right)\end{array}$$It turns out that this is all the information that we will need to sketch the direction field. We will relate things back to our solution however so that we can see that things are going correctly.
We’ll start by sketching lines that follow the direction of the two eigenvectors. This gives,

Now, from the first example our general solution is
$$\overrightarrow{x}\left(t\right)=c_1{e}^{-t}\left(\begin{array}{c}-1\\ 1\end{array}\right)+c_2{e}^{4t}\left(\begin{array}{c}2\\ 3\end{array}\right)$$If we have $c_2=0$ then the solution is an exponential times a vector and all that the exponential does is affect the magnitude of the vector and the constant $c_1$ will affect both the sign and the magnitude of the vector. In other words, the trajectory in this case will be a straight line that is parallel to the vector, ${\overrightarrow{\eta }}^{\left(1\right)}$. Also notice that as $t$ increases the exponential will get smaller and smaller and hence the trajectory will be moving in towards the origin. If $c_1>0$ the trajectory will be in Quadrant II and if $c_1<0$ the trajectory will be in Quadrant IV.
So, the line in the graph above marked with ${\overrightarrow{\eta }}^{\left(1\right)}$ will be a sketch of the trajectory corresponding to $c_2=0$ and this trajectory will approach the origin as $t$ increases.
If we now turn things around and look at the solution corresponding to having $c_1=0$ we will have a trajectory that is parallel to ${\overrightarrow{\eta }}^{\left(2\right)}$. Also, since the exponential will increase as $t$ increases and so in this case the trajectory will now move away from the origin as $t$ increases. We will denote this with arrows on the lines in the graph above.

Notice that we could have gotten this information with actually going to the solution. All we really need to do is look at the eigenvalues. Eigenvalues that are negative will correspond to solutions that will move towards the origin as $t$ increases in a direction that is parallel to its eigenvector. Likewise, eigenvalues that are positive move away from the origin as $t$ increases in a direction that will be parallel to its eigenvector.
If both constants are in the solution we will have a combination of these behaviors. For large negative $t$’s the solution will be dominated by the portion that has the negative eigenvalue since in these cases the exponent will be large and positive. Trajectories for large negative $t$’s will be parallel to ${\overrightarrow{\eta }}^{\left(1\right)}$ and moving in the same direction.
Solutions for large positive $t$’s will be dominated by the portion with the positive eigenvalue. Trajectories in this case will be parallel to ${\overrightarrow{\eta }}^{\left(2\right)}$ and moving in the same direction.
In general, it looks like trajectories will start “near” ${\overrightarrow{\eta }}^{\left(1\right)}$, move in towards the origin and then as they get closer to the origin they will start moving towards ${\overrightarrow{\eta }}^{\left(2\right)}$ and then continue up along this vector. Sketching some of these in will give the following phase portrait. Here is a sketch of this with the trajectories corresponding to the eigenvectors marked in blue.

In this case the equilibrium solution $\left(0,0\right)$ is called a saddle point and is unstable. In this case unstable means that solutions move away from it as $t$ increases.

So, we’ve solved a system in matrix form, but remember that we started out without the systems in matrix form. Now let’s take a quick look at an example of a system that isn’t in matrix form initially.