Systems of Differential Equations
6.4: Linear Systems of Differential Equations
A. Introduction
After exploring first-order and second-order differential equations, we now turn our attention to systems of differential equations. These systems are instrumental in modeling scenarios with multiple interdependent processes, common in complex real-world situations.
For instance, in an ecosystem with interacting species like prey and predators, the rate of change in each species’ population depends not only on its size but also on the populations of other species. This interaction leads to a system of differential equations where each equation represents the growth rate of one species, encapsulating their interrelations. Similarly, in mixing problems with interconnected tanks, the concentration in one tank affects and is affected by concentrations in connected tanks. In mechanical systems, such as a mass-spring system with multiple masses and springs, each mass’s displacement is influenced by its neighbors, forming a system of interconnected differential equations.
B. Systems of Linear First-Order Differential Equations
In this section, we introduce the matrix method for solving systems of linear first-order differential equations. These systems are characterized by each equation being first-order and linear. Such systems can be written in the following form.
[asciimath]{(y'_1=a_11y_1+a_12y_2+...+a_(1n)y_n+f_1(t)),(y'_2=a_21y_1+a_22y_2+...+a_(2n)y_n+f_2(t) ),(vdots),(y'_n=a_(n1)y_1+a_(n2)y_2+...+a_(n\n)y_n+f_n(t) ) :}[/asciimath]
Matrix notation simplifies the characterization and solution of these systems, similar to how systems of algebraic equations are handled. A linear first-order system can be expressed in matrix form as
[asciimath][[y'_1],[y'_2], [vdots] ,[y'_n] ]=[/asciimath] [asciimath][[a_11, a_12 ,..., a_(1n)],[a_21,a_22, ..., a_(2n)], [vdots,vdots ,ddots, vdots] ,[a_(n1), a_(n2) , ..., a_(n\n)] ][/asciimath] [asciimath][[y_1],[y_2], [vdots] ,[y_n] ]+[/asciimath] [asciimath][[f_1],[f_2], [vdots] ,[f_n] ][/asciimath]
In vector notation, the system is written as
[asciimath]bb(y')= A(t)bb(y) + bb(f)(t)[/asciimath] (6.4.1)
Here matrix [asciimath]A(t)[/asciimath] is the coefficient matrix and [asciimath]bb(f)[/asciimath] is the forcing function vector. [asciimath]A(t)[/asciimath] and [asciimath]bb(f)[/asciimath] are continuous if their entries are continuous. If [asciimath]bb(f(t))=0[/asciimath] in Equation 6.4.1, the system is homogeneous; otherwise, it is nonhomogeneous.
An initial value problem involves finding a solution for
[asciimath]bb(y') = A(t)bb(y) + bb(f)(t),[/asciimath] [asciimath]bb(y)(t_0) = bb(k)[/asciimath] (6.4.2)
where [asciimath]veck[/asciimath] is a constant vector representing the initial condition.
[asciimath]veck=[[k_1],[k_2], [vdots] ,[k_n] ][/asciimath]
Write the given system of differential equations in matrix form.
[asciimath]{(y'_1=2y_1+y_2-3e^(2t)),(y'_2=y_1+y_2+e^(2t)) :}[/asciimath]
Show/Hide Solution
The system can be written in matrix form as
[asciimath][(y'_1),(y'_2)] =[(2,1),(1,1)][(y_1),(y_2)]+[(-3e^(2t)),(e^(2t))][/asciimath]
[asciimath]bb (y)'=[(2,1),(1,1)]bb"y"+[(-3),(1)]e^(2t)[/asciimath]
An initial value problem for the system can be written as
[asciimath]bb"y'"=[(2,1),(1,1)]bb"y"+[(-3),(1)]e^(2t),[/asciimath] [asciimath]"y"(t_0)=[(k_0),(k_1)][/asciimath]
Try an Example
Existence and Uniqueness Theorem. If the coefficient matrix [asciimath]A(t)[/asciimath] and the forcing function [asciimath]bb"f"(t)[/asciimath] are continuous on an open interval containing[asciimath]t_0[/asciimath], then there exists a unique solution to the following initial value problem on that interval.
[asciimath]bb"y"'=A(t)bb"y"+bb"f"(t),[/asciimath] [asciimath]bb"y"(t_0)=bb"k"[/asciimath]
a) Verify that
[asciimath]bb"y"=c_1[(-5),(3)]e^(2t)+c_2[(2),(-1)]e^(t)[/asciimath]
is a solution to the following system for any values of [asciimath]c_1[/asciimath] and [asciimath]c_2[/asciimath].
[asciimath]bb"y'"=[(-4,-10),(3,7)]bb"y"[/asciimath]
b) Find the solution to the initial condition
[asciimath]bb"y'"=[(-4,-10),(3,7)]bb"y",[/asciimath] [asciimath]bb"y"(0)=[(1),(-2)][/asciimath]
Show/Hide Solution
a) If [asciimath]bb(y)[/asciimath] is a solution to the system, then [asciimath]Abb(y)=bb(y)'[/asciimath].
[asciimath]LHS:[/asciimath]
[asciimath]bb"Ay"=[/asciimath] [asciimath]c_1[(-4,-10),(3,7)][(-5),(3)]e^(2t)[/asciimath] [asciimath]+c_2[(-4,-10),(3,7)] [(2),(-1)]e^(t)[/asciimath]
[asciimath]=c_1[(-10),(6)]e^(2t)[/asciimath] [asciimath]+c_2[(2),(-1)]e^(t)[/asciimath]
[asciimath]RHS:[/asciimath]
[asciimath]bb(y)'=[/asciimath] [asciimath]c_1[(-5),(3)]d/dt( e^(2t))+c_2[(2),(-1)]d/dt( e^(t) )[/asciimath]
[asciimath]=2c_1[(-5),(3)]e^(2t)+c_2[(2),(-1)]e^(t)[/asciimath]
[asciimath]=c_1[(-10),(6)]e^(2t)+c_2[(2),(-1)]e^(t)[/asciimath]
[asciimath]LHS = RHS[/asciimath]
b) Since the coefficient matrix is continuous for all real numbers [asciimath]RR[/asciimath], the Existence Theorem guarantees that the given initial value problem has a unique solution on [asciimath]RR[/asciimath]. To find constants [asciimath]c_1[/asciimath] and [asciimath]c_2[/asciimath], we apply the initial condition:
[asciimath]bb"y"(0)=[(1),(-2)][/asciimath]
[asciimath]c_1[(-5),(3)]e^(0)+c_2[(2),(-1)]e^(0)=[/asciimath] [asciimath][(1),(-2)][/asciimath]
[asciimath][(-5c_1+2c_2),(3c_1-c_2)]=[/asciimath] [asciimath][(1),(-2)][/asciimath]
This yields a system of two equations in two variables [asciimath]c_1[/asciimath] and [asciimath]c_2[/asciimath].
[asciimath]{(-5c_1+2c_2=1),(3c_1-c_2=-2 ):}[/asciimath]
Solving the system yields
[asciimath]c_1=-3,[/asciimath] [asciimath]c_2=-7[/asciimath]
Therefore the solution to the initial value problem is
[asciimath]bb(y)=-3[(-5),(3)]e^(2t)-7[(2),(-1)]e^(t)[/asciimath]
Try an Example
C. n-th Order Differential Equation as a System of n First-order Equations
Higher-degree differential equations can be transformed into systems of first-order differential equations. This conversion allows complex, higher-order problems to be analyzed using techniques and tools developed for first-order systems. This approach is widely used in numerical methods and theoretical analysis in various scientific and engineering applications. Here’s a step-by-step guide to this process.
How to Convert Single [asciimath]n[/asciimath]-th Order Differential Equations into a System of [asciimath]n[/asciimath] First-Order Equations
Consider a linear [asciimath]n[/asciimath]-th order differential equation:
[asciimath]a_n(t)y^{(n)} + a_{n-1}(t)y^{(n-1)} + cdots + a_1(t)y' + a_0(t)y = g(t)[/asciimath]
1. Introduce New Variables: Introduce [asciimath]n[/asciimath] new variables corresponding to the function [asciimath]y[/asciimath] and its derivatives up to order [asciimath]n-1[/asciimath]. Let
[asciimath]x_1 = y[/asciimath]
[asciimath]x_2 = y'[/asciimath]
[asciimath]x_3 = y''[/asciimath]
[asciimath]vdots[/asciimath]
[asciimath]x_n = y^{(n-1)}[/asciimath]
2. Express the Derivatives: Express the derivatives of these new variables in terms of the original differential equation.
[asciimath]x_1' = y' = x_2[/asciimath]
[asciimath]x_2' = y'' = x_3[/asciimath]
[asciimath]vdots[/asciimath]
[asciimath]x_{n-1}' = y^{(n-1)} = x_n[/asciimath]
[asciimath]x_n' = y^{(n)} = g(t) - a_{n-1}(t)y^{(n-1)} - cdots - a_1(t)y' - a_0(t)y[/asciimath]
Observe that the last equation is the original equation that is rearranged for the highest derivative of [asciimath]y[/asciimath]. In the last equation, substitute the new variables for [asciimath]y[/asciimath] and its derivatives:
[asciimath]x_n' = g(t) - a_{n-1}(t)x_n - cdots - a_1(t)x_2 - a_0(t)x_1[/asciimath]
3. Write the System of First-Order Equations: You now have a system of [asciimath]n[/asciimath] first-order linear differential equations:
[asciimath]x_1' = x_2[/asciimath]
[asciimath]x_2' = x_3[/asciimath]
[asciimath]vdots[/asciimath]
[asciimath]x_{n-1}' = x_n[/asciimath]
[asciimath]x_n' = g(t) - a_{n-1}(t)x_n - cdots - a_1(t)x_2 - a_0(t)x_1[/asciimath]
Write the given 2nd-order differential equation as a system of first-order linear differential equations.
[asciimath]3y''+2y'-6y=2sin(t),[/asciimath] [asciimath]y(0)=1,\ y'(0)=-1[/asciimath]
Show/Hide Solution
1. Introduce a new variable [asciimath]x_i[/asciimath] :
[asciimath]x_1=y[/asciimath]
[asciimath]x_2=y'[/asciimath]
2. Express the derivatives by differentiating the above equations and rearrange the original differential equation to isolate [asciimath]y''[/asciimath]:
[asciimath]x_1'=y'=x_2[/asciimath]
[asciimath]x_2'=y''[/asciimath] [asciimath]=2/3sin(t)-2/3y'+2y[/asciimath] [asciimath]->[/asciimath] [asciimath]x_2'[/asciimath] [asciimath]=2/3sin(t)-2/3x_2+2x_1[/asciimath]
We also express the initial conditions in terms of the new variables:
[asciimath]x_1(0)=y(0)=1[/asciimath]
[asciimath]x_2(0)=y'(0)=-1[/asciimath]
3. The system of first-order equations is then
[asciimath]x_1'=x_2[/asciimath]
[asciimath]x_2'[/asciimath] [asciimath]=2x_1-2/3x_2+2/3sin(t)[/asciimath]
[asciimath]x_1(0)=1, \ x_2(0)=-1[/asciimath]
Try an Example
Section 6.4 Exercises
- Write the system given system of differential equations in matrix form.
[asciimath]{(y_1' = 2 y_1 -2 y_2 -2 t^4),(y_2' = 6 y_1 + 3 y_2 + 5 t^4):}[/asciimath]
Show/Hide Answer
[asciimath][(y_1'),(y_2')]=[(2,-2),(6,3)][(y_1),(y_2)]+[(-2),(5)]t^4[/asciimath]
- Convert the given differential equation into a system of first-order equations by letting [asciimath]x=u, \ y = u'[/asciimath].
[asciimath]u''+5u'+2u = 2 e^(3t)[/asciimath]
Show/Hide Answer
[asciimath]x'=y[/asciimath]
[asciimath]y'=-5y-2x+2e^(3t)[/asciimath]
- Rewrite the system of linear equations
[asciimath][(x'),(y')] = [(3,-5),(2,4)] \ [(x),(y)][/asciimath]
As a single second-order differential equation for [asciimath]x[/asciimath].
Show/Hide Answer
[asciimath]x''-7x'+22x=0[/asciimath]
