Solutions of First Order Linear Systems

An n-dimensional first order linear system is a list of n first order, linear, differential equations,in n unknown functions of one variable. It may be written as the matrix/vector equation

x = Ax+ f

where x is an n-dimensional vector containing the unknown functions and A is the nn coefficientmatrix. For example, the 3-dimensional linear system

x1 = x1 + 5x2 + 3x3 + et

x2 = x2 + x3x3 = 2x2 2x3 t2

has

x =

x1x2x3

and A = 1 5 30 1 1

0 2 2

and f = et0

t2

Note that the book uses ys instead of xs in some sections, so you should be comfortable usingeither letter. We will almost always use t for the variable.

(1) Homogeneous Solution: Solve the homogeneous equation

x = Ax

to find n linearly independent vector solutions x1,x2, . . . ,xn, and hence the homogeneoussolution xh = c1x1+ c2x2+ + cnxn. You can check whether x1,x2, . . . ,xn are linearlyindependent by calculating the Wronksian

W (t0) = det(x1(t0) x2(t0) . . . xn(t0)

)at any point t0 in the domain of x1,x2, . . . ,xn, and showing that it is non-zero. Note thatthe Wronskian is formed by first placing the column vector solutions next to each otherto form a square matrix, substituting t0 and then taking the determinant of the matrix.

To find the n linearly independent vector solutions, you first need to find the eigenvalues of A, by solving the characteristic equation det(A I) = 0 for , and then find thecorresponding eigenvectors v, by solving the vector equation (AI)v = 0 for v. If thereare repeated eigenvalues you may need to also find generalized eigenvectors - this case willbe explained below.(a) Distinct Real Eigenvalues: If is a real eigenvalue which appears once as a root

of the characteristic equation, and v is the corresponding eigenvector then x = etvis a vector solution of the homogeneous equation.

(b) Distinct Complex Eigenvalues: If = + i is a complex eigenvalue then so isits complex conjugate = i. Each of these eigenvalues will provide one vectorsolution, but since the eigenvalues are closely related, it is probably easier to use = + i to find both solutions. Find the corresponding eigenvector v and split itusing vector addition as v = v1 + iv2, where v1 and v2 have real entries. Then thetwo vector solutions are

x1 = et(cos(t)v1 sin(t)v2)and

x2 = et(sin(t)v1 + cos(t)v2)

1

2

Example:

x =(

1 11 1

)x

The eigenvalues are 1 + i, and 1 i. Use the eigenvalue 1 + i to find the eigenvectorby solving (

i 11 i

)v = 0

Then

v =(

1i

)=(

10

)+ i

(01

)= v1 + iv2

and so the two corresponding vector solutions are

x1 = et(cos(t)(

10

) sin(t)

(01

))

and

x2 = et(sin(t)(

10

)+ cos(t)

(01

))

(c) Repeated Eigenvalues: If an eigenvalue is repeated we need to analyse the matrixA more carefully to find the corresponding vector solutions.

Definition 1. The Algebraic Multiplicity (AM) of an eigenvalue is the number oftimes it appears as a root of the characteristic equation det(A I) = 0.

For each eigenvalue , we need to find the same number of vector solutions as itsAlgebraic Multiplicity. If AM = 1, then we are in one of the situations describedabove so the more difficult (and interesting) case is when AM > 1.

Definition 2. The Geometric Multiplicity (GM) of an eigenvalue is the num-ber of linearly independent eigenvectors corresponding to (or the dimension of theeigenspace). An equivalent (and probably more helpful) definition is that the Geo-metric Multiplicity is the number of degrees of freedom in the eigenvector equation(A I)v = 0.

These two numbers always have the property that 1 GM AM . In the casewhere GM = AM , we have enough eigenvectors to produce solutions of the formx = etv corresponding to and we procede as above.Thus the most interesting case is when GM < AM . To see what happens in general,you are referred to the textbook, however to avoid being cast adrift in a sea ofconfusing abstraction and notation we shall only consider a special case here.

Definition 3. If is a an eigenvalue for A, then v is a corresponding generalizedeigenvector if (A I)dv = 0 for some positive integer d.

Note that an eigenvector is also a generalized eigenvector (take d = 1) but as dincreases there will be more solutions which are not eigenvectors.Consider a real eigenvalue for which AM = 2 and GM = 1, so that we are lookingfor two vector solutions corresponding to but we only have one linearly independenteigenvector v1, and hence only one vector solution x1 = etv1.Solve the equation (A I)2v = 0 which will have two degrees of freedom, so therewill be two linearly independent generalized eigenvectors. It is strongly recommended(though not necessary) that you choose the eigenvector v1 as one of the generalized

3

eigenvectors, since this gives us the solution x1 = etv1 which we already know. Ifv2 is the other generalized eigenvector then, the second solution corresponding to is

x2 = et(I + t(A I))v2.For more details on how this solution arises, see the Appendix on matrix exponentials.

(2) Particular Solution: To find the particular solution xp for an inhomogeneous first orderlinear system

x = Ax+ fwe use the Variation of Parameters method.

Definition 4. The Fundamental Matrix

Y =(x1 x2 . . . xn

)is the n n matrix formed by writing the n column vector solutions of the homogeneousequation next to each other

It is easy to check that Y = AY and that Y is invertible because its determinant is theWronskian and hence non-zero. It can then be shown that the particular vector solutionis

xp = Y

Y 1f dt

You should compute xp in the following order:

(a) Find Y 1 which is an n n matrix.(b) Find Y 1f which is a vector.

(c) FindY 1f dt which is a vector. You compute the integral of a vector by finding an

anti-derivative of each entry, and this is a case where you can choose the constantsof integration to be 0 because we are only looking for one particular solution.

(d) Find xp = YY 1f dt which is a vector.

(3) General Solution: The general solution is the vector

x = xh + xp = c1x1 + c2x2 + + cnxn + xp

(4) Initial Value Problems: An IVP will have the form

x = Ax+ f , x(t0) = x0In this case you can solve for the constants c1, c2, . . . , cn. Substitute the initial conditioninto the general solution, and this will give you n equations in the n unknown constants.

4

Examples: See if you can verify the solutions to the following problems.

(1) In this example we have a repeated eigenvalue but there are sufficient eigenvectors

x =

2 2 42 1 24 2 6

xThe characteristic equation for the matrix is

(+ 1)(+ 2)2 = 0so the eigenvalues are 1 and 2. First consider the eigenvalue = 2, which hasAlgebraic Multiplicity 2, and for which the eigenvector equation is 4 2 42 1 2

4 2 4

v = 0.This equation has two degrees of freedom so the eigenvalue = 2 has Geometric Multi-plicity 2 and thus it has two eigenvectors. For example we could choose

v1 =

120

, v2 = 10

1

The eigenvalue = 1 has Algebraic Multiplicity 1 and a corresponding eigenvector

v3 =

212

.The general solution of the system is

x = c1e2t

120

+ c2e2t 10

1

+ c3et 21

2

(2) In this example there is a repeated eigenvalue and we need to find a generalized eigenvector.

x =

1 2 10 1 01 3 3

xThe characteristic equation for the matrix is (as in the previous example)

(+ 1)(+ 2)2 = 0and so the eigenvalues are 1 and 2. First consider the eigenvalue = 2, which hasAlgebraic Multiplicity 2, and for which the eigenvector equation is 1 2 10 1 0

1 3 1

v = 0which only has one degree of freedom. We can choose the eigenvector

v1 =

101

5

but now we need to also find a generalized eigenvector. Compute

(A I)2 =

1 2 10 1 01 3 1

2 = 0 1 00 1 0

0 2 0

and then solve the equation 0 1 00 1 0

0 2 0

v = 0which has two degrees of freedom. As always, we should choose the eigenvector v1 asone solution. The other solution v2 is a generalized eigenvector and must be linearlyindependent with v1, so for example we can choose

v2 =

100

The two vector solutions of the system are then

x1 = e2t

101

and

x2 = e2t(I + t

1 2 10 1 01 3 1

) 10

0

= e2t 1 + t0

t

The eigenvalue = 1 has Algebraic Multiplicity 1 and a corresponding eigenvector

v3 =

112

and so the general solution of the system is

x = c1e2t

101

+ c2e2t 1 + t0

t

+ c3et 11

2

(3) Solve the initial value problem for following system of equations

x = x+ y + t, y = x y t, x(0) = 1, y(0) = 1to find formulas for x(t) and y(t).First put the system of equations into matrix/vector form x = Ax+ f where

A =(1 11 1

)and f =

(t

t

)and then find the eigenvalues of A which are 0 and 2. The corresponding eigenvectorsare

v1 =(

11

)and v2 =

(11

)so a Fundamental Matrix for the homogeneous system is

Y =(

1 e2t1 e2t

).

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Now we compute the particular solution xp

(a) Y 1 = 12

(1 1

e2t e2t)

(b) Y 1f =(

0e2tt

)

(c)Y 1f dt =

(0

12e2tt+ 14e

2t

)

(d) xp = YY 1f dt =

(t2

14

t2 +14

)The general solution is

x = xh + xp = c1

(11

)+ c2e2t

(11

)+(

t2

14

t2 +14

)Finally, remember that x =

(xy

)and substitute the initial values to find c1 = 1 and

c2 = 1/4. The solutions are

x =34+

14e2t +

t

2

y =54 1

4e2t t

2

Appendix: Matrix Exponentials

Definition 5. Let A be a square matrix and let t be a variable then

etA = I + tA+t2

2!A2 +

t3

3!A3 +

This is the definition but we NEVER (well almost never) use it to actually find etA becausecomputing an infinite series of matrices is too hard. Usually we want to compute etAv since thisis a solution of x = Ax for any vector v.

Let be an eigenvalue of A, and note that tA = t(I +A I). Then, because tI is a diagonalmatrix, we can use the usual law of exponents and so

etA = etIet(AI)

= et(I + t(A I) + t2

2!(A I)2 + t

3

3!(A I)3 + )

Now if v1 is an eigenvector, then (A I)v1 = 0, soetAv1 = etIv1 = etv1

and if v2 is a generalized eigenvector, then (A I)dv2 = 0, so

etAv2 = et(I + t(A I) +t2

2!(A I)2 + t

3

3!(A I)3 + + t

d1

(d 1)!(A I)d1)v2

Finally, if you ever need to compute a matrix exponential, find a Fundamental Matrix Y for theequation x = Ax and then use the formula

etA = Y Y (0)1