Systems of Differential Equations
6.8 Constant-Coefficient Homogeneous Systems: Repeated Eigenvalues
In this section, we explore solutions to the homogeneous system with constant coefficients when the eigenvalues of the coefficient matrix are repeated. Specifically, we encounter a unique challenge when an eigenvalue’s algebraic multiplicity (the number of times it appears as a root of the characteristic polynomial) exceeds its geometric multiplicity (the number of linearly independent eigenvectors associated with it). This discrepancy necessitates a specific approach to find all the linearly independent solutions necessary for a complete solution to the system. Our focus here is on the case where an eigenvalue has an algebraic multiplicity of two but a geometric multiplicity of only one. In such situations, the concept of generalized eigenvectors becomes crucial to developing a comprehensive solution.
Consider a homogeneous system denoted as
[asciimath]bb{y}' = Abb{y}[/asciimath](6.8.1)
where matrix [asciimath]A[/asciimath] has an eigenvalue [asciimath]lambda[/asciimath] that is repeated twice (i.e., it has an algebraic multiplicity of two).
Theorem. If an [asciimath]n xx n[/asciimath] matrix [asciimath]A[/asciimath] has an eigenvalue [asciimath]lambda[/asciimath] with an algebraic multiplicity of two, but only one linearly independent eigenvector associated with it (i.e., a geometric multiplicity of one), the system will have additional solutions derived from generalized eigenvectors.
Finding Generalized Eigenvectors
For an eigenvalue [asciimath]lambda[/asciimath] with only one independent standard eigenvector [asciimath]mathbf{u}[/asciimath], we need to find a generalized eigenvector [asciimath]mathbf{v}[/asciimath] by solving the equation:
[asciimath](A - lambda I)mathbf{v} = mathbf{u}.[/asciimath]
This generalized eigenvector [asciimath]mathbf{v}[/asciimath] is not a solution to [asciimath](A - lambda I)mathbf{u} = mathbf{0}[/asciimath] but does satisfy the above equation.
Constructing the Solution
The solution for the eigenvalue [asciimath]lambda[/asciimath] includes terms involving both the standard and generalized eigenvectors. The two solutions are linearly independent.
1. [asciimath]bb{y}_1= e^{lambda t}bb{u}[/asciimath] – associated with the standard eigenvector.
2. [asciimath]bb{y}_2= e^{lambda t}tbb{u} +e^{lambda t}bb{v}[/asciimath] – associated with the generalized eigenvector.
General Solution for the System
The general solution to the system 6.8.1 combines these solutions.
[asciimath]bb{y}(t)=c_1e^{lambda t}bb{u} + c_2e^{lambda t} (tbb{u} +bb{v} )[/asciimath](6.8.2)
where [asciimath]c_1[/asciimath] and [asciimath]c_2[/asciimath] are arbitrary constants.
Solve the system of differential equations with the given initial values.
[asciimath]bb"y'"=[(3,-1),(1,5)]bb"y",[/asciimath] [asciimath]bb"y"(0)=[(-3),(2)][/asciimath]
Show/Hide Solution
1. First we need to find the eigenvalues of the coefficient matrix [asciimath]A[/asciimath].
The characteristic polynomial of [asciimath]A[/asciimath] is given by
[asciimath]c_A(lambda)=det(A-lambda I)[/asciimath]
[asciimath]=|(3-lambda,-1),(1,5-lambda) |[/asciimath]
[asciimath]=(3-lambda)(5-lambda)+1[/asciimath]
[asciimath]=lambda^2-8lambda+16[/asciimath]
[asciimath]=(lambda-4)^2[/asciimath]
The characteristic polynomial [asciimath]c_A(lambda)[/asciimath] has a repeated root. Thus [asciimath]lambda_(1,2)=4[/asciimath] is the eigenvalue of [asciimath]A[/asciimath] with multiplicity of two.
2. To find the corresponding standard eigenvectors, we need to find the solution to the equation [asciimath](A-lambda I)bb"u"=bb"0"[/asciimath].
For [asciimath]lambda_(1,2)=4[/asciimath], we have
[asciimath](A-lambda_(1,2) I)bb"u"=bb"0"[/asciimath]
[asciimath]=[(3-4,-1),(1,5-4)][/asciimath] [asciimath][(u_1),(u_2)]=[(0),(0)][/asciimath]
[asciimath][(-1,-1),(1,1) ][/asciimath] [asciimath][(u_1),(u_2)]=[(0),(0)][/asciimath]
To solve the system, we form the augmented matrix and bring it to RREF using row operations.
[asciimath][(-1,-1,|,0),(1,1,|,0) ]~[/asciimath][asciimath][(1,1,|,0),(0,0,|,0) ][/asciimath]
Therefore, the eigenvectors corresponding to [asciimath]lambda_(1,2)=4[/asciimath] are [asciimath]bb"u"_1=t[(-1),(1)][/asciimath]. Taking [asciimath]t=1[/asciimath], a basic eigenvector corresponding to [asciimath]lambda_(1,2)=4[/asciimath] is [asciimath]bb"u"_1[/asciimath] [asciimath]=[(-1),(1)][/asciimath].
3. We need to find a generalized eigenvector [asciimath]v[/asciimath] such that
[asciimath](A - lambda_(1,2) I)bb{v} =bb{u}_1[/asciimath]
[asciimath][(-1,-1),(1,1) ][/asciimath][asciimath][(v_1),(v_2)]=[(-1),(1)][/asciimath]
To solve the system, we form the augmented matrix and bring it to RREF using row operations.
[asciimath][(-1,-1,|,-1),(1,1,|,1) ]~[/asciimath][asciimath][(1,1,|,1),(0,0,|,0) ][/asciimath]
The solution to this is [asciimath]bb"v"=[(1-t),(t)][/asciimath]. Taking [asciimath]t=1[/asciimath], a generalized eigenvector is [asciimath]bb"v"[/asciimath] [asciimath]=[(0),(1)][/asciimath].
4. A general solution to the system is given by Equation 6.8.2.
[asciimath]bb{y}(t)=c_1e^{4 t}[(-1),(1)]+ c_2e^(4t)(t[(-1),(1)] +[(0),(1)])[/asciimath]
5. We apply the initial conditions to find constants [asciimath]c_1[/asciimath] and [asciimath]c_2[/asciimath].
[asciimath]bb"y"(0)=[(-3),(2)][/asciimath]
[asciimath]c_1[(-1),(1)]+ c_2[(0),(1)]=[/asciimath] [asciimath][(-3),(2)][/asciimath]
This gives a system of two equations and two unknowns.
[asciimath]{(-c_1=-3),(c_1+c_2=2 ):}[/asciimath]
Solving the system gives
[asciimath]c_1=3,[/asciimath] [asciimath]c_2=-1[/asciimath]
Therefore, the solution to the initial value problem is
[asciimath]bb{y}(t)=3e^{4 t}[(-1),(1)]-e^(4t)(t[(-1),(1)] +[(0),(1)])[/asciimath]
Try an Example
Solve the system of differential equations with the given initial values.
[asciimath]{(y_1' = 3 y_1 +y_2),(y_2' = 3 y_2 ),(y_3' = 5y_1- y_2 -2y_3) :},[/asciimath] [asciimath]y_1(0) = -2, \ y_2(0) = 5, \ y_3(0)=-5[/asciimath]
Show/Hide Solution
1. We first express the IVP in the matrix notation.
[asciimath]bb"y"'=Abb(y),[/asciimath] [asciimath]bb(y)(0)=[(-2),(5),(-5)][/asciimath]
where [asciimath]A=[(3,1,0),(0,3,0),(5,-1,-2) ][/asciimath].
2. We find the eigenvalues of the coefficient matrix [asciimath]A[/asciimath].
The characteristic polynomial of [asciimath]A[/asciimath] is given by
[asciimath]c_A(lambda)=det(A-lambda I)[/asciimath]
[asciimath]=|(3-lambda,1,0),(0,3-lambda,0),(5,-1,-2-lambda) |[/asciimath]
[asciimath]=(-2-lambda)|(3-lambda,1),(0,3-lambda)|[/asciimath]
[asciimath]=(-2-lambda)(3-lambda)^2[/asciimath]
The eigenvalues are [asciimath]lambda_1=-2[/asciimath] with a multiplicity of one and [asciimath]lambda_(2,3)=3[/asciimath] with a multiplicity of two.
3. To find the corresponding standard eigenvectors, we solve [asciimath](A-lambda I)bb"u"=bb"0"[/asciimath].
For [asciimath]lambda_1=-2[/asciimath], we have
[asciimath](A-lambda_(1) I)bb"u"=bb"0"[/asciimath]
[asciimath]=[(5,1,0),(0,5,0),(5,-1,0) ][/asciimath][asciimath][(u_1),(u_2),(u_3)]=[(0),(0),(0)][/asciimath]
To solve the system, we form the augmented matrix and bring it to RREF using row operations.
[asciimath][(5,1,0),(0,5,0),(5,-1,0) ]~[/asciimath] [asciimath][(1,0,0),(0,1,0),(0,0,0) ][/asciimath]
Therefore, the eigenvectors corresponding to [asciimath]lambda_(1)=-2[/asciimath] are [asciimath]bb"u"_1=t[(0),(0),(1)][/asciimath]. Taking [asciimath]t=1[/asciimath], a basic eigenvector corresponding to [asciimath]lambda_(1)=-2[/asciimath] is [asciimath]bb"u"_1[/asciimath] [asciimath]=[(0),(0),(1)][/asciimath].
For [asciimath]lambda_(2,3)=3[/asciimath], we have
[asciimath](A-lambda_(2,3) I)bb"u"=bb"0"[/asciimath]
[asciimath]=[(0,1,0),(0,0,0),(5,-1,-5) ][/asciimath] [asciimath][(u_1),(u_2),(u_3)]=[(0),(0),(0)][/asciimath]
To solve the system, we form the augmented matrix and bring it to RREF using row operations.
[asciimath][(0,1,0,|,0),(0,0,0,|,0) ,(5,-1,-5,|,0) ]~[/asciimath] [asciimath][(1,0,-1,|,0),(0,1,0,|,0) ,(0,0,0,|,0) ][/asciimath]
Therefore, the eigenvectors corresponding to [asciimath]lambda_(2,3)=3[/asciimath] are [asciimath]bb"u"_2=t[(1),(0),(1)][/asciimath]. Taking [asciimath]t=1[/asciimath], a basic eigenvector corresponding to [asciimath]lambda_(2,3)=3[/asciimath] is [asciimath]bb"u"_2[/asciimath] [asciimath]=[(1),(0),(1)][/asciimath].
For [asciimath]lambda_(2,3)=3,[/asciimath] the geometric multiplicity is one and thus less than the algebraic multiplicity (which is 2). This means that the dimension of the eigenspace associated with [asciimath]lambda_(2,3)[/asciimath] is one (all eigenvectors are spanned by the only vector [asciimath]bb(u)_2[/asciimath]).
4. Therefore, we need to find a generalized vector [asciimath]bb(v)[/asciimath] such that
[asciimath](A - lambda_(2,3) I)bb{v} =bb{u}_2[/asciimath]
[asciimath][(0,1,0),(0,0,0),(5,-1,-5) ] [(v_1),(v_2),(v_3)]=[(1),(0),(1)][/asciimath]
To solve the system, we form the augmented matrix and bring it to RREF using row operations.
[asciimath][(0,1,0,|,1),(0,0,0,|,0),(5,-1,-5,|,1) ]~[/asciimath] [asciimath][(1,0,-1,|,2/5),(0,1,0,|,1),(0,0,0,|,0) ][/asciimath]
The solution to this is [asciimath]bb"v"=[(2/5+t),(1),(t)][/asciimath]. Taking [asciimath]t=0[/asciimath], a generalized eigenvector is [asciimath]bb"v"[/asciimath] [asciimath]=[(2/5),(1),(0)][/asciimath].
5. Three linearly independent solutions of the system are
-For [asciimath]lambda_1=-2[/asciimath] and standard eigenvector [asciimath]bb(u)_1[/asciimath]:
-For [asciimath]lambda_(2,3)=3[/asciimath] and the standard eigenvector [asciimath]bb(u)_2[/asciimath]:
-For [asciimath]lambda_(2,3)=3[/asciimath] and a generalized eigenvector [asciimath]bb(v)[/asciimath]:
[asciimath]bb(y)_3=e^(3t)(t[(1),(0),(1)]+[(2//5),(1),(0)])[/asciimath]
6. Therefore, a general solution to the system is given by the linear combination of the above solutions:
[asciimath]bb(y)(t)=c_1e^(-2t)[(0),(0),(1)] +[/asciimath] [asciimath]c_2e^(3t)[(1),(0),(1)]+[/asciimath] [asciimath]c_3e^(3t)(t[(1),(0),(1)]+[(2//5),(1),(0)])[/asciimath]
7. We apply the initial conditions to find the constants.
[asciimath]bb(y)(0)=[(-2),(5),(-5)][/asciimath]
[asciimath]c_1[(0),(0),(1)] +[/asciimath] [asciimath]c_2[(1),(0),(1)]+[/asciimath] [asciimath]c_3[(2//5),(1),(0)]=[/asciimath] [asciimath][(-2),(5),(-5)][/asciimath]
This gives a system of three equations and three unknowns.
[asciimath]{(c_2+2/5c_3=-2),(c_3=5 ),(c_1+c_2=-5) :}[/asciimath]
Solving the system gives
[asciimath]c_1=-1,[/asciimath] [asciimath]c_2=-4,[/asciimath] [asciimath]c_3=5[/asciimath]
Therefore, the solution to the initial value problem is
[asciimath]bb(y)(t)=-e^(-2t)[(0),(0),(1)]-[/asciimath] [asciimath]4e^(3t)[(1),(0),(1)]+[/asciimath] [asciimath]5e^(3t)(t[(1),(0),(1)]+[(2//5),(1),(0)])[/asciimath]
Try an Example
Section 6.8 Exercises
- Solve the system of differential equations.
[asciimath]bb"y"'=[(-3,1),(-1,-5)] bb"y"[/asciimath], [asciimath]bb"y"(t)=[(9 ),(-13)][/asciimath]
Show/Hide Answer
[asciimath]y_1(t)=13 e^(-4t)-4(1+t)e^(-4t)[/asciimath]
[asciimath]y_2(t)=-13 e^(-4t)+4te^(-4t)[/asciimath]
- Solve the system of differential equations.
[asciimath]bb"y"'=[(-6,2),(-2,-10)] bb"y"[/asciimath], [asciimath]bb"y"(t)=[(-14 ),(-26)][/asciimath]
Show/Hide Answer
[asciimath]y_1(t)= 26 e^(-8t)-40(1+2t)e^(-8t)[/asciimath]
[asciimath]y_2(t)= -26 e^(-8t)+80te^(-8t)[/asciimath]
- Solve the system of differential equations.
[asciimath]bb"y"'=[(2,1,0),(0,2,0),(4,-3,-1)] bb"y"[/asciimath], [asciimath]bb"y"(t)=[(-9 ),(-18),(27)][/asciimath]
Show/Hide Answer
[asciimath]bb(y)(t)=13e^(-t)[(0),(0),(1)]+[/asciimath] [asciimath]7/2e^(2t)[(3),(0),(4)]-[/asciimath] [asciimath]6e^(2t)(t[(3),(0),(4)]+[(13//4),(3),(0)])[/asciimath]
