Systems of coupled linear first-order ODEs

Differential equations: Systems of differential equations

Systems of coupled linear first-order ODEs

For a brief look at more than one differential equation with several unknown functions, we consider the following system of coupled homogeneous linear first-order differential equations:

$\eqs{\dfrac{\dd x}{\dd t} &= &a\cdot x+ b\cdot y\cr \dfrac{\dd y}{\dd t} &=& c\cdot x + d\cdot y}$ where $x$ and $y$ are unknown functions of $t$ , and $a$ , $b$ , $c$ , $d$ are constants. This system can be solved by use of the theory of homogeneous linear second-order ODEs with constant coefficients.

Conversion of two coupled homogeneous linear first-order differential equations to a homogeneous linear second-order equation

Let $a$ , $b$ , $c$ , $d$ , $p$ , $q$ be constants, and let $x$ , $y$ , $z$ be functions of $t$ .

If $\rv{x,y}$ is a solution of the system of two coupled first order differential equations homogeneous linear $\eqs{x' &= &a\cdot x+ b\cdot y\cr y' &=& c\cdot x + d\cdot y}$ then both $x$ and $y$ are solutions of the homogeneous linear second order differential equation $z''-(a+d)\cdot z'+(a\cdot d-b\cdot c)\cdot z =0$
If $x$ and $y$ are linearly independent solutions of the linear second-order differential equation $z''+p\cdot z'+q\cdot z =0$ then there are constants $a_1$ , $b_1$ , $c_1$ , $d_1$ such that $\eqs{x' &= &a_1\cdot x+ b_1\cdot y\cr y' &=& c_1\cdot x + d_1\cdot y}$

The first statement is easy to check by expressing the second derivatives of $x$ (and $y$ ) in terms of $x$ and $x'$ (respectively, $y$ and $y'$ ). We carry out the calculation only for $x$ :

$\begin{array}{rcl} x''&=& \dfrac{\dd }{\dd t} (x') \\ &&\phantom{x}\color{blue}{\text{definition second deivative}}\\ &=& \dfrac{\dd }{\dd t} \left(a\cdot x+b\cdot y\right) \\ &&\phantom{x}\color{blue}{\text{first equation used here}}\\ &=& a\cdot x'+b\cdot y' \\ &&\phantom{x}\color{blue}{\text{extended sum rule for differentiation}}\\ &=& a\cdot x'+b\cdot \left(c\cdot x+d\cdot y\right)\\ &&\phantom{x}\color{blue}{\text{second equation used here}}\\ &=& a\cdot x'+b\cdot c\cdot x+d\cdot ( b\cdot y)\\&&\phantom{x}\color{blue}{\text{terms rearranged}}\\ &=& a\cdot x'+b\cdot c\cdot x+d\cdot \left(x'-a\cdot x\right)\\ &&\phantom{x}\color{blue}{\text{first equation used here}}\\ &=& (a+d)\cdot x'+(b\cdot c -a\cdot d)\cdot x\\&&\phantom{x}\color{blue}{\text{terms rearranged}}\end{array}$

If we carry all terms to the left, we see that $x$ satisfies the second-order differential equation in $x$ of the first statement of the theorem.

The second statement follows from the Uniqueness theorem of specific solutions of linear second-order differential equations. After all, if $x$ and $y$ are linearly independent solutions of the second-order differential equation in $z$ , then, due to the theorem, any other solution is a linear combination of the two. Notice that $x'$ is also a solution of the second-order differential equation in $z$ . (This can be seen by differentiating all terms of the differential equation and replacing $z$ by $x$ .) If we apply the above theorem to $x'$ , we find that there are constants $a_1$ and $b_1$ such that $x'=a_1\cdot x+ b_1\cdot y$ . Similarly, we find that there are constants $c_1$ and $d_1$ such that $x'=c_1\cdot x+ d_1\cdot y$ .

The second statement shows that there is also a way back: solutions of a homogeneous linear second-degree equation with constant coefficients are also solutions of a system of linear first-order coupled ODEs.

The coupled system is autonomous, in the sense that the order and the degree of the differential equation is equal to $1$ (for both functions $x$ and $y$ ), and that the right-hand sides do not depend on the independent variable.

We know that the type of solution of the above second-order ODE depends on the discriminant of the corresponding characteristic polynomial

$\lambda^2 -(a+d)\lambda+(ad-bc)$ This discriminant $D$ can be rewritten in terms of the system of coupled first-order differential equations:

$D=(a+d)^2-4(a\cdot d-b\cdot c)=(a-d)^2+4b\cdot c$ We now distinguish again three cases (statements regarding $x(t)$ also hold for $y(t)$ ):

If $(a-d)^2+4b\cdot c\gt0$ then there are two real solutions of the characteristic equation: $\lambda_{1,2}=\frac{a+d\pm\sqrt{(a-d)^2+4b\cdot c}}{2}$ and the general solution for $x(t)$ is $x(t)=c_1\cdot \e^ {\lambda_1t}+c_2\cdot \e^ {\lambda_2t}$
If $(a-d)^2+4b\cdot c=0$ then there is a single real solution of the characteristic equation: $\lambda=\frac{a+d}{2}$ and the general solution for $x(t)$ is $x(t)=(c_1+c_2 t)\cdot \e^ {\lambda t}$
If $(a-d)^2+4b\cdot c\lt0$ then there are no real solutions and there are two complex solutions of the characteristic equation: $\lambda_{1,2}=\alpha\pm{\ii}\,\beta$ with $\alpha=\frac{a+d}{2}\quad\text{and}\quad\beta=\frac{\sqrt{-(a-d)^2-4b\cdot c}}{2}$ and the general solution for $x(t)$ is $x(t)=\e^ {\alpha t}\cdot\bigl(c_1\cdot \cos(\beta t)+c_2\cdot \sin(\beta t)\bigr)$

We show how the first statement of the theorem can be used to solve the coupled system.

Solution of the coupled system of first-order equations by using the second-order equation

The general solution of the coupled system of differential equations

$\eqs{x' &= &a\cdot x+ b\cdot y\cr y' &=& c\cdot x + d\cdot y}$

can be found as follows:

Solve the linear second-order equation $z''-(a+d)\cdot z'+(a\cdot d-b\cdot c)\cdot z =0$ ; this provides a pair of linearly independent solutions $u$ and $v$ of that equation.
Consequently, solutions $x$ (and $y$ ) of the system have the form $x=A\cdot u+B\cdot v$ (and $y=C\cdot u+D\cdot v$ , respectively), for yet to be determined constants $A$ , $B$ , $C$ , $D$ . Substitute these expressions for $x$ and $y$ in the system.
The result of the previous step is a pair of ordinary linear equations in the unknowns $A$ , $B$ , $C$ , and $D$ . Use these equations to express two of them in the other two.
Substitute the expressions for two of the four constants found in the previous step in the equations of step 2 in order to find the general form for $x$ and for $y$ .

Solve the following system of coupled homogeneous linear first-order differential equations:

$\eqs{x' &= &-5 x+ 4 y\cr y' &=& x -2 y}$
Enter your answer in the form $x=f(t)\land y=g(t)$ , where $f(t)$ and $g(t)$ are function rules that contain $C$ and $D$ as integration constants.

$x=D\cdot \euler^ {- t }-4\cdot C\cdot \euler^ {- 6\cdot t } \land y=D\cdot \euler^ {- t }+C\cdot \euler^ {- 6\cdot t } \\$

According to the Conversion of two coupled homogeneous linear first-order differential equations to a homogeneous linear second-order equation, the coordinates $\rv{x,y}$ of a solution of this system are also solutions of the linear second-order ODE $z''-(-5-2) z'+ (-5\cdot-2-4\cdot1) z = 0$ , which can be simplified to

$z''+ 7 z'+ 6 z = 0$
The characteristic equation $\lambda^2+7\lambda + 6 = 0$ has solution $\lambda=-6\lor\lambda=-1$ , so the general solution of this second-order equation is

$z=A\cdot \euler^ {- 6\cdot t }+B\cdot \euler^ {- t }$
Because $x$ and $y$ are also solutions, we can write

$\eqs{x &= &B\cdot \euler^ {- t }+A\cdot \euler^ {- 6\cdot t } \cr y &=&D\cdot \euler^ {- t }+C\cdot \euler^ {- 6\cdot t }\cr}$
for numbers $A$ , $B$ , $C$ , $D$ , that are yet to be determined.
If we fill these expressions for $x$ and $y$ into the coupled system, then we find a system of linear equations from which the following constants to be determined can be found:

$\begin{array}{rcl} \dfrac{\dd}{\dd t}\left(B\cdot \euler^ {- t }+A\cdot \euler^ {- 6\cdot t }\right) &=& -5 \left(B\cdot \euler^ {- t }+A\cdot \euler^ {- 6\cdot t }\right) + 4 \left( D\cdot \euler^ {- t }+C\cdot \euler^ {- 6\cdot t }\right)\\ \dfrac{\dd}{\dd t}\left(D\cdot \euler^ {- t }+C\cdot \euler^ {- 6\cdot t }\right) &=& \left(B\cdot \euler^ {- t }+A\cdot \euler^ {- 6\cdot t }\right) -2 \left( D\cdot \euler^ {- t }+C\cdot \euler^ {- 6\cdot t } \right)\\ &&\phantom{xx}\color{blue}{\text{the solutions found}}\\ -B\cdot \euler^ {- t }-6\cdot A\cdot \euler^ {- 6\cdot t }&=&4\cdot D\cdot \euler^ {- t }-5\cdot B\cdot \euler^ {- t }+4\cdot C\cdot \euler^ {- 6\cdot t }-5\cdot A\cdot \euler^ {- 6\cdot t } \\ -D\cdot \euler^ {- t }-6\cdot C\cdot \euler^ {- 6\cdot t }&=&-2\cdot D\cdot \euler^ {- t }+B\cdot \euler^ {- t }-2\cdot C\cdot \euler^ {- 6\cdot t }+A\cdot \euler^ {- 6\cdot t } \\ &&\phantom{xx}\color{blue}{\text{derivative calculated and brackets cleared}}\\ -6\cdot A&=&4\cdot C-5\cdot A \\ -6\cdot C&=&A-2\cdot C \\ -B&=&4\cdot D-5\cdot B \\ -D&=&B-2\cdot D \\ &&\phantom{xx}\color{blue}{\text{the coefficients of }\euler^ {- 6\cdot t }\text{ and }\euler^ {- t }\text{ compared}}\\ A&=&\displaystyle -4\cdot C \\ B&=&\displaystyle D \\ &&\phantom{xx}\color{blue}{\text{linear equations in }A,B,C,D\text{ solved}}\\ x&=&\displaystyle D\cdot \euler^ {- t }-4\cdot C\cdot \euler^ {- 6\cdot t } \\ y&=&\displaystyle D\cdot \euler^ {- t }+C\cdot \euler^ {- 6\cdot t } \\ &&\phantom{xx}\color{blue}{\text{solution substituted into function rules for }x\text{ and }y}\\ \end{array}$

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