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4.3: Basic Theory of Homogeneous Linear System

This page is a draft and is under active development. 

( \newcommand{\kernel}{\mathrm{null}\,}\)

In this section we consider homogeneous linear systems y=A(t)y, where A=A(t) is a continuous n×n matrix function on an interval (a,b). The theory of linear homogeneous systems has much in common with the theory of linear homogeneous scalar equations, which we considered in Sections 2.1 and 3.1.

Whenever we refer to solutions of y=A(t)y we'll mean solutions on (a,b). Since y0 is obviously a solution of y=A(t)y, we call it the trivial solution. Any other solution is nontrivial.

If y1, y2, , yn are vector functions defined on an interval (a,b) and c1, c2, , cn are constants, then

y=c1y1+c2y2++cnyn

is a linear combination of  y1, y2, , yn. It's easy show that if y1, y2, , yn are solutions of y=A(t)y on (a,b), then so is any linear combination of y1, y2, , yn (Exercise (4.3E.1)). We say that {y1,y2,,yn} is a fundamental set of solutions of y=A(t)y on (a,b) on if every solution of y=A(t)y on (a,b) can be written as a linear combination of y1, y2, , yn, as in (???). In this case we say that (???) is the general solution of y=A(t)y on (a,b).

It can be shown that if A is continuous on (a,b) then y=A(t)y has infinitely many fundamental sets of solutions on (a,b) (Exercises (4.3E.15) and (4.3E.16)). The next definition will help to characterize fundamental sets of solutions of y=A(t)y.

We say that a set {y1,y2,,yn} of n-vector functions is linearly independent on (a,b) if the only constants c1, c2, , cn such that

c1y1(t)+c2y2(t)++cnyn(t)=0,a<t<b,

are c1=c2==cn=0. If (???) holds for some set of constants c1, c2, , cn that are not all zero, then {y1,y2,,yn} is linearly dependent on (a,b).

The next theorem is analogous to Theorems (2.1.3) and (3.1.2).

Theorem 4.3.1

Suppose the n×n matrix A=A(t) is continuous on (a,b). Then a set {y1,y2,,yn} of n solutions of y=A(t)y on (a,b) is a fundamental set if and only if it's linearly independent on (a,b).

Proof

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Example 4.3.1

Show that the vector functions

y1=[et0et],y2=[0e3t1],andy3=[e2te3t0]

are linearly independent on every interval (a,b).

Answer

Suppose

c1[et0et]+c2[0e3t1]+c3[e2te3t0]=[000],a<t<b.

We must show that c1=c2=c3=0. Rewriting this equation in matrix form yields

[et0e2t0e3te3tet10][c1c2c3]=[000],a<t<b.

Expanding the determinant of this system in cofactors of the entries of the first row yields

|et0e2t0e3te3tet10|=et|e3te3t10|0|0e3tet0|+e2t|0e3tet1|=et(e3t)+e2t(e2t)=2e4t.

Since this determinant is never zero, c1=c2=c3=0.

We can use the method in Example (4.3.1) to test n solutions {y1,y2,,yn} of any n×n system y=A(t)y for linear independence on an interval (a,b) on which A is continuous. To explain this (and for other purposes later), it's useful to write a linear combination of y1, y2, , yn in a different way. We first write the vector functions in terms of their components as

y1=[y11y21yn1],y2=[y12y22yn2],,yn=[y1ny2nynn].

If

y=c1y1+c2y2++cnyn

then

y=c1[y11y21yn1]+c2[y12y22yn2]++cn[y1ny2nynn]=[y11y12y1ny21y22y2ny1ny2nynn].

This shows that

c1y1+c2y2++cnyn=Yc,

where

c=cn

and

Y=[y1y2yn]=[y11y12y1ny21y22y2nyn1yn2ynn];

that is, the columns of Y are the vector functions y1,y2,,yn.

For reference below, note that

Y=[y1y2yn]=[Ay1Ay2Ayn]=A[y1y2yn]=AY;

that is, Y satisfies the matrix differential equation

Y=AY.

The determinant of Y,

W=|y11y12y1ny21y22y2nyn1yn2ynn|

is called the Wronskian of {y1,y2,,yn}. It can be shown (Exercises (4.3E.2) and (4.3E.3)) that this definition is analogous to definitions of the Wronskian of scalar functions given in Sections 2.1 and 3.1. The next theorem is analogous to Theorems (2.1.4) and (3.1.3). The proof is sketched in Exercise (4.3E.4) for n=2 and in Exercise (4.3E.5) for general n.

Theorem - ABEL'S FORMULA 4.3.2

Suppose the n×n matrix A=A(t) is continuous on (a,b), let y1, y2, , yn be solutions of y=A(t)y on (a,b), and let t0 be in (a,b). Then the Wronskian of {y1,y2,,yn} is given by

W(t)=W(t0)exp(tt0[a11(s)+a22(s)++ann(s)]ds),a<t<b.

Therefore, either W has no zeros in (a,b) or W0 on (a,b).

Proof

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The sum of the diagonal entries of a square matrix A is called the trace of A, denoted by tr(A). Thus, for an n×n matrix A,

tr(A)=a11+a22++ann,

and (???) can be written as

W(t)=W(t0)exp(tt0tr(A)(s))ds),a<t<b.

The next theorem is analogous to Theorems (2.1.6) and (3.1.4).

Theorem 4.3.3

Suppose the n×n matrix A=A(t) is continuous on (a,b) and let y1, y2, , yn be solutions of y=A(t)y on (a,b). Then the following statements are equivalent; that is, they are either all true or all false:

(a) The general solution of y=A(t)y on (a,b) is y=c1y1+c2y2++cnyn, where c1, c2, , cn are arbitrary constants.

(b) {y1,y2,,yn} is a fundamental set of solutions of y=A(t)y on (a,b).

(c) {y1,y2,,yn} is linearly independent on (a,b).

(d) The Wronskian of {y1,y2,,yn} is nonzero at some point in (a,b).

(e) The Wronskian of {y1,y2,,yn} is nonzero at all points in (a,b).

Proof

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We say that Y in (???) is a fundamental matrix for y=A(t)y if any (and therefore all) of the statements (a)(e) of Theorem (4.3.2) are true for the columns of Y. In this case, (???) implies that the general solution of y=A(t)y can be written as y=Yc, where c is an arbitrary constant n vector.

Example 4.3.2

The vector functions

y1=[e2t2e2t]andy2=[etet]

are solutions of the constant coefficient system

y=[4365]y

on (,). (Verify.)

(a) Compute the Wronskian of {y1,y2} directly from the definition (???)

(b) Verify Abel's formula (???) for the Wronskian of {y1,y2}.

(c) Find the general solution of (???).

(d) Solve the initial value problem

y=[4365]y,y(0)=[45].

Answer

(a) From (???)

W(t)=|e2tet2e2tet|=e2tet[1121]=et.

(b) Here

A=[4365]

so tr(A)=4+5=1. If t0 is an arbitrary real number then (???) implies that

W(t)=W(t0)exp(tt01ds)=|e2t0et02e2t0et0|e(tt0)=et0ett0=et,

which is consistent with (4.3.19).

(c) Since W(t)0, Theorem (4.3.3) implies that {y1,y2} is a fundamental set of solutions of (???) and

Y=[e2tet2e2tet]

is a fundamental matrix for (???). Therefore the general solution of (???) is

y=c1y1+c2y2=c1[e2t2e2t]+c2[etet]=[e2tet2e2tet][c1c2].

(d) Setting t=0 in (4.3.20) and imposing the initial condition in (???) yields

C1[12]+c2[11]=[45].

Thus,

c1c2=42c1+c2=5.

The solution of this system is c1=1, c2=3. Substituting these values into (4.3.20) yields

y=[e2t2e2t]3[etet]=[e2t+3et2e2t3et]

as the solution of (???).


This page titled 4.3: Basic Theory of Homogeneous Linear System is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by William F. Trench.

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