2 Systems of first order differential equations

Systems of first order ordinary differential equations arise in many areas of mathematics and engineering, for example in control theory and in the analysis of electrical circuits. In each case the basic unknowns are each a function of the time variable $t$ . A number of techniques have been developed to solve such systems of equations; for example the Laplace transform. Here we shall use eigenvalues and eigenvectors to obtain the solution. Our first step will be to recast the system of ordinary differential equations in the matrix form $Ẋ=AX$ where $A$ is an $n\times n$ coefficient matrix of constants, $X$ is the $n\times 1$ column vector of unknown functions and $Ẋ$ is the $n\times 1$ column vector containing the derivatives of the unknowns.. The main step will be to use the modal matrix of $A$ to diagonalise the system of differential equations. This process will transform $Ẋ=AX$ into the form $Ẏ=DY$ where $D$ is a diagonal matrix . We shall find that this new diagonal system of differential equations can be easily solved. This special solution will allow us to obtain the solution of the original system.

Task!

Obtain the solutions of the pair of first order differential equations

$\left.\begin{array}{c}ẋ=-2x\hfill \\ ẏ=-5y\hfill \end{array}\right\}$ $\left(1\right)$

given the initial conditions

$\begin{array}{ccc}x\left(0\right)=3\hfill & \hfill & \text{ i.e.}x=3\text{ at}t=0\hfill \\ y\left(0\right)=2\hfill & \hfill & \text{ i.e.}y=2\text{ at}t=0\hfill \end{array}$

(The notation is that $ẋ\equiv \frac{dx}{dt},ẏ\equiv \frac{dy}{dt}$  )

[Hint: Recall, from your study of differential equations, that the general solution of the differential equation $\frac{dy}{dt}=Ky$ is $y=y_0{e}^{Kt}$ .]

Answer

Thus if $X=\left[\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right]Ẋ=\left[\begin{array}{c}\hfill ẋ\hfill \\ \hfill ẏ\hfill \end{array}\right]$ $A=\left[\begin{array}{cc}\hfill -2& \hfill 0\\ \hfill 0& \hfill -5\end{array}\right]$

the two equations (1) can be written as

$\left[\begin{array}{c}\hfill ẋ\hfill \\ \hfill ẏ\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill -2& \hfill 0\\ \hfill 0& \hfill -5\end{array}\right]\left[\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right]$

i.e. $Ẋ=AX$ .

Task!

Write in matrix form the pair of coupled differential equations

$\left.\begin{array}{c}ẋ=4x+2y\hfill \\ ẏ=-x+y\hfill \end{array}\right\}$ $\left(2\right)$

Answer

The essential difference between the two pairs of differential equations just considered is that the pair (1) were really separate equations whereas pair (2) were coupled:

• The first equation of $\left(1\right)$ involving only the unknown $x$ , the second involving only $y$ . In matrix terms this corresponded to a diagonal matrix $A$ in the system $Ẋ=AX$ .
• The pair $\left(2\right)$ were coupled in that both equations involved both $x$ and $y$ . This corresponded to the non-diagonal matrix $A$ in the system $Ẋ=AX$ which you found in the last Task.

Clearly the second system here is more difficult to deal with than the first and this is where we can use our knowledge of diagonalization.

Consider a system of differential equations written in matrix form: $Ẋ=AX$ where

$X=\left[\begin{array}{c}\hfill x\left(t\right)\hfill \\ \hfill y\left(t\right)\hfill \end{array}\right]\text{ and}Ẋ=\left[\begin{array}{c}\hfill ẋ\left(t\right)\hfill \\ \hfill ẏ\left(t\right)\hfill \end{array}\right]$

We now introduce a new column vector of unknowns $Y=\left[\begin{array}{c}\hfill r\left(t\right)\hfill \\ \hfill s\left(t\right)\hfill \end{array}\right]$ through the relation

$X=PY$

where $P$ is the modal matrix of $A$ . Then, since $P$ is a matrix of constants:

$Ẋ=PẎ\text{ so}Ẋ=AX\text{ becomes}PẎ=A\left(PY\right)$

$\text{Then, multiplying by}{P}^{-1}\text{ on the left,}Ẏ=\left(P^{-1}AP\right)Y$

But, because of the properties of the modal matrix, we know that $P^{-1}AP$ is a diagonal matrix . Thus if ${\lambda }_1,{\lambda }_2$ are distinct eigenvalues of $A$ then:

$P^{-1}AP=\left[\begin{array}{cc}\hfill {\lambda }_1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {\lambda }_2\hfill \end{array}\right]$

Hence $Ẏ=\left(P^{-1}AP\right)Y$ becomes

$\left[\begin{array}{c}\hfill ṙ\hfill \\ \hfill ṡ\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill {\lambda }_1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {\lambda }_2\hfill \end{array}\right]\left[\begin{array}{c}\hfill r\hfill \\ \hfill s\hfill \end{array}\right].$

That is, when written out we have

These equations are decoupled . The first equation only involves the unknown function $r\left(t\right)$ and has solution $r\left(t\right)=C{e}^{{\lambda }_{1}t}$ . The second equation only involves the unknown function $s\left(t\right)$ and has solution $s\left(t\right)=K{e}^{{\lambda }_{2}t}$ . [ $C,K$ are arbitrary constants.]

Once $r,s$ are known the original unknowns $x,y$ can be found from the relation $X=PY$ .

Note that the theory outlined above is more widely applicable as specified in the next Key Point:

Example 10

Find the solution of the coupled differential equations

$ẋ=4x+2y$

$ẏ=-x+y\text{ with initial conditions}x\left(0\right)=1y\left(0\right)=0$

Here $ẋ\equiv \frac{dx}{dt}$ and $ẏ\equiv \frac{dy}{dt}$ .

Solution

Here $A=\left[\begin{array}{cc}\hfill 4\hfill & \hfill 2\hfill \\ \hfill -1\hfill & \hfill 1\hfill \end{array}\right].$ It is easily checked that $A$ has distinct eigenvalues ${\lambda }_1=3{\lambda }_2=2$ and corresponding eigenvectors $X_1=\left[\begin{array}{c}\hfill -2\hfill \\ \hfill 1\hfill \end{array}\right],X_2=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill -1\hfill \end{array}\right]$ .

Therefore, taking $P=\left[\begin{array}{cc}\hfill -2\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -1\hfill \end{array}\right]\text{ then}{P}^{-1}AP=\left[\begin{array}{cc}\hfill 3\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill \end{array}\right]$

and using Key Point 3, $r\left(t\right)=C{e}^{3t}s\left(t\right)=K{e}^{2t}.$

Therefore $x=-2C{e}^{3t}+K{e}^{2t}$ and $y=C{e}^{3t}-K{e}^{2t}.$

We can now impose the initial conditions $x\left(0\right)=1$ and $y\left(0\right)=0$ to give

Thus $C=K=-1$ and the solution to the original system of differential equations is

The approach we have demonstrated in Example 10 can be extended to

1. Systems of first order differential equations with $n$ unknowns (Key Point 3)
2. Systems of second order differential equations (described in the next subsection).

The only restriction, as we have said, is that the matrix $A$ in the system $Ẋ=AX$ has distinct eigenvalues.