Ordinary Differential Equations/Linear Systems

A system of differential equations is a collection of two or more differential equations, which each ODE may depend upon the other unknown function.

For example consider the equations:

$\{\begin{array}{c}{x}^{\prime }(t)=2x(t)+y(t)\\ y^{\prime }(t)=3y(t)\end{array}$

In this case the equation for differential equation for $x^{\prime }(t)$ depends on both $x(t)$ and $y(t)$. In principle we could also allow $y^{\prime }(t)$ to depend on both $x$ and $y$, but it is not necessary.

Notice in some cases we find a solution for a system of ODE's. For example in the case above, because $y^{\prime }$ doesn't depend on $x$ we can solve the second equation (by separating variables or using an integrating factor) to get that $y=C_2{e}^{3t}$. Since there will be a second constant when we solve the first ODE, we choose to call the constant here $C_2$. Now we can plug this into the first equation to get that: $x^{\prime }=2x+C_2{e}^{3t}$. We can solve this equation by using an integrating factor to get that:

$\{\begin{array}{c}x(t)=C_1{e}^{2t}+C_2{e}^{3t}\\ y(t)=C_2{e}^{3t}\end{array}$

In other cases a clever change of variables allows one to separate the two ODE's. Consider the system

$\{\begin{array}{c}{x_1}^{\prime }(t)=4x_1(t)+2x_2(t)\\ {x_2}^{\prime }(t)=2x_1(t)+4x_2(t)\end{array}$.

If we let $y_1=x_1+x_2$ and $y_2=x_1-x_2$. Then we find that

$\{\begin{array}{c}{y_1}^{\prime }(t)=6y_1(t)\\ {y_2}^{\prime }(t)=2y_2(t)\end{array}$

and each of these are easy to solve: $y_1=C_1{e}^{6t}$ and $y_2=C_2{e}^{2t}$. And so we find $x_1=C_1{e}^{6t}+C_2{e}^{2t}$ and $x_1=C_1{e}^{6t}-C_2{e}^{2t}$. It turns out to be helpful with systems to work with vectors and matrices so if we introduce $\overrightarrow{x}(t)=\left(\begin{array}{c}{x}_1(t)\\ x_2(t)\end{array}\right).$ Then the above system can be re-written as:

$\frac{d}{dt}\overrightarrow{x}(t)=\left(\begin{array}{cc}4& 2\\ 2& 4\end{array}\right)\overrightarrow{x}(t).$

And we have solutions ${\overrightarrow{x}}_1(t)=C_1{e}^{6t}\left(\begin{array}{c}1\\ 1\end{array}\right)$ and ${\overrightarrow{x}}_2(t)=C_2{e}^{2t}\left(\begin{array}{c}1\\ -1\end{array}\right)$

Notice that the solutions we found were of the for $e^{\lambda t}\overrightarrow{\xi }$ for some constant vector $\overrightarrow{\xi }$. Using this as motivation we will investigate the question, when does $\overrightarrow{x}(t)=e^{\lambda t}\overrightarrow{\xi }$ solve the system:

$\frac{d}{dt}\overrightarrow{x}(t)=A\overrightarrow{x}(t).$

for some constant matrix $A$.

By substituting into the equation we see that:

$\begin{array}{cc}\frac{d}{dt}(e^{\lambda t}\overrightarrow{\xi })& =A(e^{\lambda t}\overrightarrow{\xi })\\ \lambda e^{\lambda t}\overrightarrow{\xi }& =e^{\lambda t}A\overrightarrow{\xi }\\ e^{\lambda t}(A-\lambda I)\overrightarrow{\xi }& =\overrightarrow{0}.\end{array}$

Since $e^{\lambda t}\ne 0$, the only way for the left hand side to be $\overrightarrow{0}$ is if $\lambda$ is an eigenvalue and $\overrightarrow{\xi }$ is a corresponding eigenvector.

This is not quite the end of the story. When the matrix is real we shall consider the following cases: