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4.4: Constant Coefficient Homogeneous Systems I


Constant Coefficient Homogenous Systems I

We'll now begin our study of the homogeneous system

egin{equation}label{eq:4.4.1}
{f y}'=A{f y},
end{equation}

where (A) is an (n imes n) constant matrix. Since (A) is continuous on ((-infty,infty)), Theorem ((4.2.1)) implies that all solutions of eqref{eq:4.4.1} are defined on ((-infty,infty)). Therefore, when we speak of solutions of ({f y}'=A{f y}), we'll mean solutions on ((-infty,infty)).

In this section we assume that all the eigenvalues of (A) are real and that (A) has a set of (n) linearly independent eigenvectors. In the next two sections we consider the cases where some of the eigenvalues of (A) are complex, or where (A) does not have (n) linearly independent eigenvectors.

In Example ((4.3.2)) we showed that the vector functions

egin{eqnarray*}
{f y}_1 = left[ egin{array} -e^{2t} 2e^{2t} end{array} ight] quad mbox{and} quad {f y}_2 = left[ egin{array} -e^{-t} e^{-t} end{array} ight]
end{eqnarray*}

form a fundamental set of solutions of the system

egin{equation}label{eq:4.4.2}
{f y}' = left[ egin{array} {-4} & {-3} 6 & 5 end{array} ight] {f y},
end{equation}

but we did not show how we obtained ({f y}_1) and ({f y}_2) in the first place. To see how these solutions can be obtained we write eqref{eq:4.4.2} as

egin{equation}label{eq:4.4.3}
egin{array}{ccc}
y_1'&=&-4y_1-3y_2y_2'&=&phantom{-}6y_1+5y_2end{array}
end{equation}

and look for solutions of the form

egin{equation}label{eq:4.4.4}
y_1=x_1e^{lambda t} quad mbox{and} quad y_2=x_2e^{lambda t},
end{equation}

where (x_1), (x_2), and (lambda) are constants to be determined. Differentiating eqref{eq:4.4.4} yields

egin{eqnarray*}
y_1' = lambda x_1e^{lambda t} quad mbox{and} quad y_2'=lambda x_2e^{lambda t}.
end{eqnarray*}

Substituting this and eqref{eq:4.4.4} into eqref{eq:4.4.3} and canceling the common factor (e^{lambda t}) yields

egin{eqnarray*}
egin{array}{ccc}-4x_1-3x_2&=&lambda x_1
6 x_1+5x_2&=&lambda x_2.end{array}
end{eqnarray*}

For a given (lambda), this is a homogeneous algebraic system, since it can be rewritten as

egin{equation}label{eq:4.4.5}
egin{array}{rcl} (-4-lambda) x_1-3 x_2&=&0
6 x_1+(5-lambda) x_2&=&0.end{array}
end{equation}

The trivial solution (x_1=x_2=0) of this system isn't useful, since it corresponds to the trivial solution (y_1equiv y_2equiv0) of eqref{eq:4.4.3}, which can't be part of a fundamental set of solutions of eqref{eq:4.4.2}. Therefore we consider only those values of (lambda) for which eqref{eq:4.4.5} has nontrivial solutions. These are the values of (lambda) for which the determinant of eqref{eq:4.4.5} is zero; that is,

egin{eqnarray*}
left|egin{array}{cc}-4-lambda&-36&5-lambdaend{array} ight|&=&
(-4-lambda)(5-lambda)+18&=&lambda^2-lambda-2
&=&(lambda-2)(lambda+1)=0,
end{eqnarray*}

which has the solutions (lambda_1=2) and (lambda_2=-1).

Taking (lambda=2) in eqref{eq:4.4.5} yields

egin{eqnarray*}
-6 x_1-3 x_2&=&0
6 x_1+3 x_2&=&0,
end{eqnarray*}

which implies that (x_1=-x_2/2), where (x_2) can be chosen arbitrarily. Choosing (x_2=2) yields the solution (y_1=-e^{2t}),
(y_2=2e^{2t}) of eqref{eq:4.4.3}. We can write this solution in vector form as

egin{equation}label{eq:4.4.6}
{f y}_1= left[ egin{array} -1 {phantom{-}2} e^{2t} end{array} ight].
end{equation}

Taking (lambda=-1) in eqref{eq:4.4.5} yields the system

egin{eqnarray*}
-3 x_1-3 x_2&=&0
phantom{-}6 x_1+6 x_2&=&0,
end{eqnarray*}

so (x_1=-x_2). Taking |(x_2=1) here yields the solution (y_1=-e^{-t}), (y_2=e^{-t}) of eqref{eq:4.4.3}. We can write this solution in vector form as

egin{equation}label{eq:4.4.7}
{f y}_2= left[ egin{array} -1 {phantom{-}1}e^{-t} end{array} ight].
end{equation}

In eqref{eq:4.4.6} and eqref{eq:4.4.7} the constant coefficients in the arguments of the exponential functions are the eigenvalues of the coefficient matrix in eqref{eq:4.4.2}, and the vector coefficients of the exponential functions are associated eigenvectors. This illustrates the next theorem.

Theorem (PageIndex{1})

Suppose the (n imes n) constant matrix (A) has (n) real eigenvalues (lambda_1,lambda_2,ldots,lambda_n) (which need not be distinct) with associated linearly independent eigenvectors ({f x}_1,{f x}_2,ldots,{f x}_n). Then the functions

egin{eqnarray*}
{f y}_1 = {f x}_1 e^{lambda_1 t}, , {f y}_2 = {f x}_2 e^{lambda_2 t}, , dots, , {f y}_n = {f x}_n e^{lambda_n t}
end{eqnarray*}

form a fundamental set of solutions of ({f y}'=A{f y};) that is, the general solution of this system is

egin{eqnarray*}
{f y} = c_1 {f x}_1 e^{lambda_1 t} + c_2 {f x}_2 e^{lambda_2 t} + cdots + c_n {f x}_n e^{lambda_n t}.
end{eqnarray*}

Proof

Differentiating ({f y}_i={f x}_ie^{lambda_it}) and recalling that (A{f x}_i=lambda_i{f x}_i) yields

egin{eqnarray*}
{f y}_i' = lambda_i {f x}_i e^{lambda_i t} = A {f x}_i e^{lambda_i t} = A {f y}_i.
end{eqnarray*}

This shows that ({f y}_i) is a solution of ({f y}'=A{f y}).

The Wronskian of ({{f y}_1,{f y}_2,ldots,{f y}_n}) is

egin{eqnarray*}
left| egin{array} x_{11} e^{lambda_1 t} & x_{12} e^{lambda_2 t} & cdots & x_{1n} e^{lambda_n t} x_{21} e^{lambda_1 t} & x_{22} e^{lambda_2 t} & cdots & x_{2n} e^{lambda_n t} vdots & vdots & ddots & vdots x_{n1} e^{lambda_1 t} & x_{n2} e^{lambda_2 t} & cdots & x_{nn} e^{lambda x_n} end{array} ight| = e^{lambda_1 t} e^{lambda_2 t} cdots e^{lambda_n t} left| egin{array} x_{11} & x_{12} & cdots & x_{1n} x_{21} & x_{22} & cdots & x_{2n} vdots & vdots & ddots & vdots x_{n1} & x_{n2} & cdots & x_{nn} end{array} ight|.
end{eqnarray*}

Since the columns of the determinant on the right are ({f x}_1), ({f x}_2), (dots), ({f x}_n), which are assumed to be linearly independent, the determinant is nonzero. Therefore Theorem ((4.3.3)) implies that ({{f y}_1,{f y}_2,ldots,{f y}_n}) is a fundamental set of solutions of ({f y}'=A{f y}).

Example (PageIndex{1})

(a) Find the general solution of

egin{equation}label{eq:4.4.8}
{f y}'= left[ egin{array} 2 & 4 4 & 2 end{array} ight] {f y}.
end{equation}

(b) Solve the initial value problem

egin{equation}label{eq:4.4.9}
{f y}'= left[ egin{array} 2 & 4 4 & 2 end{array} ight] {f y}, quad {f y}(0)= left[ egin{array} 5 -1 end{array} ight].
end{equation}

Answer

(a) The characteristic polynomial of the coefficient matrix (A) in eqref{eq:4.4.8} is

egin{eqnarray*}
left|egin{array}{cc} 2-lambda&44&2-lambdaend{array} ight|
&=& (lambda-2)^2-16
&=& (lambda-2-4)(lambda-2+4)
&=& (lambda-6)(lambda+2).
end{eqnarray*}

Hence, (lambda_1=6) and (lambda_2 =-2) are eigenvalues of (A). To obtain the eigenvectors, we must solve the system

egin{equation}label{eq:4.4.10}
left[egin{array}{cc} 2-lambda&44&2-lambdaend{array} ight]
left[egin{array}{c} x_1x_2end{array} ight]=
left[egin{array}{c} 0end{array} ight]
end{equation}

with (lambda=6) and (lambda=-2). Setting (lambda=6) in eqref{eq:4.4.10} yields

egin{eqnarray*}
left[ egin{array} -4 & 4 4 & -4 end{array} ight] left[ egin{array} x_1 x_2 end{array} ight] = left[ egin{array} 0 0 end{array} ight],
end{eqnarray*}

which implies that (x_1=x_2). Taking (x_2=1) yields the eigenvector

egin{eqnarray*}
{f x}_1 = left[ egin{array} 1 1 end{array} ight],
end{eqnarray*}

so

egin{eqnarray*}
{f y}_1 = left[ egin{array} 1 1 end{array} ight] e^{6t}
end{eqnarray*}

is a solution of eqref{eq:4.4.8}. Setting (lambda=-2) in eqref{eq:4.4.10} yields

egin{eqnarray*}
left[ egin{array} 4 & 4 4 & 4 end{array} ight] left[ egin{array} x_1 x_2 end{array} ight] = left[ egin{array} 0 0 end{array} ight],
end{eqnarray*}

which implies that (x_1=-x_2). Taking (x_2=1) yields the eigenvector

egin{eqnarray*}
{f x}_2 = left[ egin{array} -1 1 end{array} ight],
end{eqnarray*}

so

egin{eqnarray*}
{f y}_2 = left[ egin{array} -1 1 end{array} ight] e^{-2t}
end{eqnarray*}

is a solution of eqref{eq:4.4.8}. From Theorem ((4.4.1)), the general solution of eqref{eq:4.4.8} is

egin{equation}label{eq:4.4.11}
{f y}=c_1{f y}_1+c_2{f y}_2=c_1left[egin{array}{r}11end{array} ight]e^{6t}+c_2left[egin{array}{r}-11
end{array} ight]e^{-2t}.
end{equation}

(b) To satisfy the initial condition in eqref{eq:4.4.9}, we must choose (c_1) and (c_2) in eqref{eq:4.4.11} so that

egin{eqnarray*}
c_1 left[ egin{array} 1 1 end{array} ight] + c_2 left[ egin{array} -1 1 end{array} ight] = left[ egin{array} 5 -1 end{array} ight].
end{eqnarray*}

This is equivalent to the system

egin{eqnarray*}
c_1-c_2&=&phantom{-}5
c_1+c_2&=&-1,
end{eqnarray*}

so (c_1=2, c_2=-3). Therefore the solution of eqref{eq:4.4.9} is

egin{eqnarray*}
{f y} = 2 left[ egin{array} 1 1 end{array} ight] e^{6t} -3 left[ egin{array} -1 1 end{array} ight] e^{-2t},
end{eqnarray*}

or, in terms of components,

egin{eqnarray*}
y_1 = 2e^{6t} + 3e^{-2t}, quad y_2 = 2e^{6t} - 3e^{-2t}.
end{eqnarray*}

Example (PageIndex{2})

(a) Find the general solution of

egin{equation}label{eq:4.4.12}
{f y}'=left[egin{array}{rrr}3&-1&-1-2& 3& 24&-1&-2end{array} ight]{f y}.
end{equation}

(b) Solve the initial value problem

egin{equation}label{eq:4.4.13}
{f y}'=left[egin{array}{rrr}3&-1&-1-2&3&
24&-1&-2end{array}
ight]{f y},quad{f y}(0)=left[egin{array}{r}2
-18end{array} ight].
end{equation}

Answer

(a) The characteristic polynomial of the coefficient matrix (A) in eqref{eq:4.4.12} is

egin{eqnarray*}
left| egin{array} 3-lambda & -1 & -1 -2 & 3-lambda & 2 4 & -1 & -2-lambda end{array} ight| = -(lambda-2)(lambda-3)(lambda+1).
end{eqnarray*}

Hence, the eigenvalues of (A) are (lambda_1=2), (lambda_2=3), and (lambda_3=-1). To find the eigenvectors, we must solve the system

egin{equation}label{eq:4.4.14}
left[egin{array}{ccc}3-lambda&-1&-1-2&3-lambda& 24&-1& -2-lambdaend{array} ight]left[egin{array}{c} x_1x_2x_3 end{array} ight]=left[egin{array}{r}0end{array} ight]
end{equation}

with (lambda=2), (3), (-1). With (lambda=2), the augmented matrix of eqref{eq:4.4.14} is

egin{eqnarray*}
left[ egin{array} 1 & -1 & -1 & vdots & 0 -2 & 1 & 2 & vdots & 0 4 & -1 & -4 & vdots & 0 end{array} ight],
end{eqnarray*}

which is row equivalent to

egin{eqnarray*}
left[ egin{array} 1 & 0 & -1 & vdots & 0 0 & 1 & 0 & vdots & 0 0 & 0 & 0 & vdots & 0 end{array} ight].
end{eqnarray*}

Hence, (x_1=x_3) and (x_2=0). Taking (x_3=1) yields

egin{eqnarray*}
{f y}_1 = left[ egin{array} 1 0 1 end{array} ight] e^{2t}
end{eqnarray*}

as a solution of eqref{eq:4.4.12}. With (lambda=3), the augmented matrix of eqref{eq:4.4.14} is

egin{eqnarray*}
left[ egin{array} 0 & -1 & -1 & vdots & 0 -2 & 0 & 2 & vdots & 0 4 & -1 & -5 & vdots & 0 end{array} ight],
end{eqnarray*}

which is row equivalent to

egin{eqnarray*}
left[ egin{array} 1 & 0 & -1 & vdots & 0 0 & 1 & 1 & vdots & 0 0 & 0 & 0 & vdots & 0 end{array} ight].
end{eqnarray*}

Hence, (x_1=x_3) and (x_2=-x_3). Taking (x_3=1) yields

egin{eqnarray*}
{f y}_2 = left[ egin{array} 1 -1 1 end{array} ight] e^{3t}
end{eqnarray*}

as a solution of eqref{eq:4.4.12}. With (lambda=-1), the augmented matrix of eqref{eq:4.4.14} is

egin{eqnarray*}
left[ egin{array} 4 & -1 & -1 & vdots & 0 -2 & 4 & 2 & vdots & 0 4 & -1 & -1 & vdots & 0 end{array} ight],
end{eqnarray*}

which is row equivalent to

egin{eqnarray*}
left[ egin{array} 1 & 0 & -{1 over 7} & vdots & 0 0 & 1 & {3 over 7} & vdots & 0 0 & 0 & 0 & vdots & 0 end{array} ight].
end{eqnarray*}

Hence, (x_1=x_3/7) and (x_2=-3x_3/7). Taking (x_3=7) yields

egin{eqnarray*}
{f y}_3 = left[ egin{array} 1 -3 7 end{array} ight] e^{-t}
end{eqnarray*}

as a solution of eqref{eq:4.4.12}. By Theorem ((4.4.1)), the general solution of eqref{eq:4.4.12} is

egin{eqnarray*}
{f y} = c_1 left[ egin{array} 1 0 1 end{array} ight] e^{2t} + c_2 left[ egin{array} 1 -1 1 end{array} ight] e^{3t} + c_3 left[ egin{array} 1 -3 7 end{array} ight] e^{-t},
end{eqnarray*}

which can also be written as

egin{equation}label{eq:4.4.15}
{f y}=left[egin{array}{crc}e^{2t}&e^{3t}&e^{-t}
&-e^{3t}&
-3e^{-t}e^{2t}&e^{3t}&phantom{-}7e^{-t}end{array}
ight]left[egin{array}{c} c_1c_2c_3end{array} ight].
end{equation}

(b) To satisfy the initial condition in eqref{eq:4.4.13} we must choose (c_1), (c_2), (c_3) in eqref{eq:4.4.15} so that

egin{eqnarray*}
left[ egin{array} 1 & 1 & 1 0 & -1 & -3 1 & 1 & 7 end{array} ight] left[ egin{array} c_1 c_2 c_3 end{array} ight] = left[ egin{array} 2 -1 8 end{array} ight].
end{eqnarray*}

Solving this system yields (c_1=3), (c_2=-2), (c_3=1). Hence, the solution of eqref{eq:4.4.13} is

egin{eqnarray*}
{f y}&=&left[egin{array}{ccc}e^{2t}&e^{3t}&
e^{-t}&-e^{3t}
&-3e^{-t}e^{2t}&e^{3t}&7e^{-t}end{array}
ight]
left[egin{array}{r}3-21end{array} ight]
&=&3left[egin{array}{r}11end{array} ight]e^{2t}-2
left[egin{array}{r}1-11end{array} ight]
e^{3t}+left[egin{array}{r}1-37end{array}
ight]e^{-t}.
end{eqnarray*}

Example (PageIndex{3})

Find the general solution of

egin{equation}label{eq:4.4.16}
{f y}'=left[egin{array}{rrr}-3&2&2
2&-3&2\2&2&-3
end{array} ight]{f y}.
end{equation}

Answer

The characteristic polynomial of the coefficient matrix (A) in eqref{eq:4.4.16} is

egin{eqnarray*}
left[ egin{array} -3-lambda & 2 & 2 2 & -3-lambda & 2 2 & 2 & -3-lambda end{array} ight] = -(lambda-1)(lambda+5)^2.
end{eqnarray*}

Hence, (lambda_1=1) is an eigenvalue of multiplicity (1), while (lambda_2=-5) is an eigenvalue of multiplicity (2). Eigenvectors associated with (lambda_1=1) are solutions of the system with augmented matrix

egin{eqnarray*}
left[ egin{array} -4 & 2 & 2 & vdots & 0 2 & -4 & 2 vdots & 0 2 & 2 & -4 & vdots & 0 end{array} ight],
end{eqnarray*}

which is row equivalent to

egin{eqnarray*}
left[ egin{array} 1 & 0 & -1 & vdots & 0 0 & 1 & -1 & vdots & 0 0 & 0 & 0 & vdots & 0 end{array} ight].
end{eqnarray*}

Hence, (x_1=x_2=x_3), and we choose (x_3=1) to obtain the solution

egin{equation}label{eq:4.4.17}
{f y}_1=left[egin{array}{r}111end{array} ight]e^t
end{equation}

of eqref{eq:4.4.16}. Eigenvectors associated with (lambda_2=-5) are solutions of the system with augmented matrix

egin{eqnarray*}
left[ egin{array} 2 & 2 & 2 & vdots & 0 2 & 2 & 2 & vdots & 0 2 & 2 & 2 & vdots & 0 end{array} ight].
end{eqnarray*}

Hence, the components of these eigenvectors need only satisfy the single condition

egin{eqnarray*}
x_1 + x_2 + x_3 = 0.
end{eqnarray*}

Since there's only one equation here, we can choose (x_2) and (x_3) arbitrarily. We obtain one eigenvector by choosing (x_2=0) and (x_3=1), and another by choosing (x_2=1) and (x_3=0). In both cases (x_1=-1). Therefore

egin{eqnarray*}
left[ egin{array} -1 0 1 end{array} ight] quad mbox{and} quad left[ egin{array} -1 1 0 end{array} ight]
end{eqnarray*}

are linearly independent eigenvectors associated with (lambda_2= -5), and the corresponding solutions of eqref{eq:4.4.16} are

egin{eqnarray*}
{f y}_2 = left[ egin{array} -1 0 1 end{array} ight] e^{-5t} quad mbox{and} quad {f y}_3 = left[ egin{array} -1 1 0 end{array} ight] e^{-5t}.
end{eqnarray*}

Because of this and eqref{eq:4.4.17}, Theorem ((4.4.1)) implies that the general solution of eqref{eq:4.4.16} is

egin{eqnarray*}
{f y} = c_1 left[ egin{array} 1 1 1 end{array} ight] e^t + c_2 left[ egin{array} -1 0 1 end{array} ight] e^{-5t} + c_3 left[ egin{array} -1 1 0 end{array} ight] e^{-5t}.
end{eqnarray*}

Geometric Properties of Solutions when (n=2)

We'll now consider the geometric properties of solutions of a (2 imes 2) constant coefficient system

egin{equation} label{eq:4.4.18}
left[ egin{array} {y_1'} {y_2'} end{array} ight] = left[egin{array}{cc}a_{11}&a_{12}a_{21}&a_{22}
end{array} ight] left[ egin{array} {y_1} {y_2} end{array} ight].
end{equation}

It is convenient to think of a ``(y_1)-(y_2) plane," where a point is identified by rectangular coordinates ((y_1,y_2)). If ({f y}=displaystyle{ left[ egin{array} {y_1} {y_2} end{array} ight]}) is a non-constant solution of eqref{eq:4.4.18}, then the point ((y_1(t),y_2(t))) moves along a curve (C) in the (y_1)-(y_2) plane as (t) varies from (-infty) to (infty). We call (C) the ( extcolor{blue}{mbox{trajectory}} ) of ({f y}). (We also say that (C) is a trajectory of the system eqref{eq:4.4.18}.) It's important to note that (C) is the trajectory of infinitely many solutions of eqref{eq:4.4.18}, since if ( au) is any real number, then ({f y}(t- au)) is a solution of eqref{eq:4.4.18} (Exercise ((4.4E.28)) part (b)), and ((y_1(t- au),y_2(t- au))) also moves along (C) as (t) varies from (-infty) to (infty). Moreover, Exercise ((4.4E.28)) part (c) implies that distinct trajectories of eqref{eq:4.4.18} can't intersect, and that two solutions ({f y}_1) and ({f y}_2) of eqref{eq:4.4.18} have the same trajectory if and only if ({f y}_2(t)={f y}_1(t- au)) for some ( au).

From Exercise ((4.4E.28)) part (a), a trajectory of a nontrivial solution of eqref{eq:4.4.18} can't contain ((0,0)), which we define to be the trajectory of the trivial solution ({f y}equiv0). More generally, if ({f y}=displaystyle{ left[ egin{array} {k_1} {k_2} end{array} ight]} e{f 0}) is a constant solution of eqref{eq:4.4.18} (which could occur if zero is an eigenvalue of the matrix of eqref{eq:4.4.18}), we define the trajectory of ({f y}) to be the single point ((k_1,k_2)).

To be specific, this is the question: What do the trajectories look like, and how are they traversed? In this section we'll answer this question, assuming that the matrix

egin{eqnarray*}
A = left[ egin{array} a_{11} & a_{12} a_{21} & a_{22} end{array} ight]
end{eqnarray*}

of eqref{eq:4.4.18} has real eigenvalues (lambda_1) and (lambda_2) with associated linearly independent eigenvectors ({f x}_1) and ({f x}_2). Then the general solution of eqref{eq:4.4.18} is

egin{equation} label{eq:4.4.19}
{f y}= c_1{f x}_1
e^{lambda_1 t}+c_2{f x}_2e^{lambda_2 t}.
end{equation}

We'll consider other situations in the next two sections.

We leave it to you (Exercise ((4.4E.35))) to classify the trajectories of eqref{eq:4.4.18} if zero is an eigenvalue of (A). We'll confine our attention here to the case where both eigenvalues are nonzero. In this case the simplest situation is where (lambda_1=lambda_2 e0), so eqref{eq:4.4.19} becomes

egin{eqnarray*}
{f y} = (c_1 {f x}_1 + c_2 {f x}_2) e^{lambda_1 t}.
end{eqnarray*}

Since ({f x}_1) and ({f x}_2) are linearly independent, an arbitrary vector ({f x}) can be written as ({f x}=c_1{f x}_1+c_2{f x}_2). Therefore the general solution of eqref{eq:4.4.18} can be written as ({f y}={f x}e^{lambda_1 t}) where ({f x}) is an arbitrary (2)-vector, and the trajectories of nontrivial solutions of eqref{eq:4.4.18} are half-lines through (but not including) the origin. The direction of motion is away from the origin if (lambda_1>0) (Figure (4.4.1)), toward it if (lambda_1<0) (Figure (4.4.2)). (In these and the next figures an arrow through a point indicates the direction of motion along the trajectory through the point.)

Figure (4.4.1)

Trajectories of a (2 imes 2) system with a repeated positive

Figure (4.4.2)

Trajectories of a (2 imes 2) system with a repeated negative

Now suppose (lambda_2>lambda_1), and let (L_1) and (L_2) denote lines through the origin parallel to ({f x}_1) and ({f x}_2), respectively. By a half-line of (L_1) (or (L_2)), we mean either of the rays obtained by removing the origin from (L_1) (or (L_2)).

Letting (c_2=0) in eqref{eq:4.4.19} yields ({f y}=c_1{f x}_1e^{lambda_1 t}). If (c_1 e0), the trajectory defined by this solution is a half-line of (L_1). The direction of motion is away from the origin if (lambda_1>0), toward the origin if (lambda_1<0). Similarly, the trajectory of ({f y}=c_2{f x}_2e^{lambda_2 t}) with (c_2 e0) is a half-line of (L_2).

Henceforth, we assume that (c_1) and (c_2) in eqref{eq:4.4.19} are both nonzero. In this case, the trajectory of eqref{eq:4.4.19} can't intersect (L_1) or (L_2), since every point on these lines is on the trajectory of a solution for which either (c_1=0) or (c_2=0). (Remember: distinct trajectories can't intersect!). Therefore the trajectory of eqref{eq:4.4.19} must lie entirely in one of the four open sectors bounded by (L_1) and (L_2), but do not any point on (L_1) or (L_2). Since the initial point ((y_1(0),y_2(0))) defined by

egin{eqnarray*}
{f y}(0) = c_1 {f x}_1 + c_2 {f x}_2
end{eqnarray*}

is on the trajectory, we can determine which sector contains the trajectory from the signs of (c_1) and (c_2), as shown in Figure (4.4.3).

The direction of ({f y}(t)) in eqref{eq:4.4.19} is the same as that of

egin{equation} label{eq:4.4.20}
e^{-lambda_2 t}{f y}(t)=
c_1{f x}_1e^{-(lambda_2-lambda_1)t}+c_2{f x}_2
end{equation}

and of

egin{equation} label{eq:4.4.21}
e^{-lambda_1 t}{f y}(t)=c_1{f
x}_1+c_2{f x}_2e^{(lambda_2-lambda_1)t}.
end{equation}

Since the right side of eqref{eq:4.4.20} approaches (c_2{f x}_2) as (t oinfty), the trajectory is asymptotically parallel to (L_2) as (t oinfty). Since the right side of eqref{eq:4.4.21} approaches (c_1{f x}_1) as (t o-infty), the trajectory is asymptotically parallel to (L_1) as (t o-infty).

The shape and direction of traversal of the trajectory of eqref{eq:4.4.19} depend upon whether (lambda_1) and (lambda_2) are both positive, both negative, or of opposite signs. We'll now analyze these three cases.

Henceforth (|{f u}|) denote the length of the vector ({f u}).

Figure (4.4.3)

Four open sectors bounded by (L_1) and (L_2)

Figure (4.4.4)

Two positive eigenvalues; motion away from origin

Case 1: (lambda_2>lambda_1>0)

Figure (4.4.4) shows some typical trajectories. In this case, (lim_{t o-infty}|{f y}(t)|=0), so the trajectory is not only asymptotically parallel to (L_1) as (t o-infty), but is actually asymptotically tangent to (L_1) at the origin. On the other hand, (lim_{t oinfty}|{f y}(t)|=infty) and

egin{eqnarray*}
lim_{t oinfty} left|{f y}(t)-c_2{f x}_2 e^{lambda_2 t} ight| = lim_{t oinfty}|c_1{f x_1} e^{lambda_1 t} | = infty,
end{eqnarray*}

so, although the trajectory is asymptotically parallel to (L_2) as (t oinfty), it's not asymptotically tangent to (L_2). The direction of motion along each trajectory is away from the origin.

Case 2: (0>lambda_2>lambda_1)

Figure (4.4.5) shows some typical trajectories. In this case, (lim_{t oinfty}|{f y}(t)|=0), so the trajectory is asymptotically tangent to (L_2) at the origin as (t oinfty). On the other hand, (lim_{t o-infty}|{f y}(t)|=infty) and

egin{eqnarray*}
lim_{t o-infty} left|{f y}(t)-c_1{f x}_1 e^{lambda_1 t} ight| = lim_{t o-infty} | c_2{f x}_2 e^{lambda_2 t} | = infty,
end{eqnarray*}

so, although the trajectory is asymptotically parallel to (L_1) as (t o-infty), it's not asymptotically tangent to it. The direction of motion along each trajectory is toward the origin.

Figure (4.4.5)

Two negative eigenvalues; motion toward the origin

Figure (4.4.6)

Eigenvalues of different signs

Case 3: (lambda_2>0>lambda_1)

Figure (4.4.6) shows some typical trajectories. In this case,

egin{eqnarray*}
lim_{t oinfty}|{f y}(t)| = infty quad mbox{and} quad lim{|t oinfty}left| {f y}(t)-c_2{f x}_2 e^{lambda_2 t} ight| = lim_{t oinfty}| c_1{f x}_1 e^{lambda_1 t}| = 0,
end{eqnarray*}

so the trajectory is asymptotically tangent to (L_2) as (t oinfty). Similarly,

egin{eqnarray*}
lim_{t oinfty}|{f y}(t)| = infty quad mbox{and} quad lim_{t oinfty}left|{f y}(t)-c_1{f x}_1 e^{lambda_1 t} ight| = lim_{t oinfty}|c_2{f x}_2 e^{lambda_2 t} | = 0,
end{eqnarray*}

so the trajectory is asymptotically tangent to (L_1) as (t o-infty). The direction of motion is toward the origin on (L_1) and away from the origin on (L_2). The direction of motion along any other trajectory is away from (L_1), toward (L_2).