Vector Representations of Solutions
Consider the linear system of differential equations
This system may be rewritten using matrix-notation. Indeed, set
,
then the above system is equivalent to the matricial equation
.
Using the matrix product, we get
.
The matrix
is called the coefficient matrix of the system. Note that the coefficients of the matrix A can be constant or not. The
vector function
is called the nonhomogeneous term.
Remark: One may think that the equation above is only valid for linear systems of two equations. However, that is
not the case. For example, consider the linear system
Then, in matricial notation, the system is equivalent to
,
where
.
Equilibrium Points of Homogeneous Linear Systems
Consider the homogeneous linear system
The equilibrium points are given by the equations
Clearly, x=0 and y=0 give a trivial solution. Hence, the function
gives a constant solution to the
linear system. We call it the trivial solution. In general, the equilibrium points are the intersection between two
lines. Since the two lines intersect, they are the same (if parallel) or the intersection is reduced to one point. So,
the set of equilibrium points is the entire line ax+by=0, or the trivial point (0,0). This conclusion is related to the
determinant of the matrix coefficient. Indeed, if
is not equal to 0 (zero), then we have one equilibrium point (the trivial one).
The Linearity Principle
This is may be the most important property for linear systems. Consider the homogeneous linear system
,
then
1.
if Y(t) is a solution and k is a constant, then k Y(t) is also a solution;
2.
if
and
are two solutions, then
This clearly implies that if
and
is also a solution.
are two solutions and
and
are two arbitrary constants, then
is also a solution. This conclusion is also known as the Principle of Superposition.
Clearly, from the Principle of Superposition, we may generate plenty of solutions once two solutions are known.
The natural question to ask therefore, is whether we have obtained all the solutions. In order to better appreciate
this problem let's consider the following example.
Example: Consider the linear system
Show that any solution Y to this system is given as
,
where
,
and
and
are two constants.
Answer: It is easy to check that indeed
and
are solutions to the given system. Let Y be any solution. Set
.
By the uniqueness and existence theorem, Y is the only solution to the IVP
.
Let us find
and
such that
. If this is the case, we should have
, which gives
,
which implies
Clearly, this gives
.
Consider the function
.
The linearity principle implies that
is a solution. And, since
,
the uniqueness and existence theorem implies that in fact
gives the desired conclusion.
Remark: When you look at the above example you will notice that what made the conclusion work is that we were
able to solve the algebraic system
and this was possible because the two vectors
are linearly independent. In fact, the above conclusion is always valid whenever we have a linear independence
around.
Theorem: The General Solution
Suppose
and
are two solutions to the linear system
.
Assume that the vectors
and
are linearly independent. Then, the solution to the IVP
,
is given by
,
for some constants
and
. In this case, the two-parameter family
,
where
and
and
are arbitrary constants, is called the general solution of the system. Then, the two solutions
are said to be linearly independent.
Example: Consider the undamped harmonic oscillator
.
Show that any solution x is given by
.
Answer: Consider the associated linear system
Set
. Note that the second component is just the derivative of the first one. Consider the two vector
functions
It is easy to check that these two vector functions are in fact solutions to the given system. Also, you may check
that the two vectors
are linearly independent. Therefore, any solution Y of the system is given by
,
where
and
are two constants. Using the first component of Y, we see that any solution x(t) of the equation is
given by
,
where
and
are two arbitrary constants.
Clearly, the main problem now is how to find the two linearly independent solutions. This problem will be
discussed in the next section and will use the eigenvalue and eigenvector technique.