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Note: less than 1 lecture, second part of §5.1 in [EP], §7.4 in [BD]

First let us talk about matrix or vector valued functions. Such a function is just a matrix whose entries depend on some variable. If is the independent variable, we write a vector valued function as

Similarly a matrix valued function is

We can talk about the derivative or . This is just the matrix valued function whose entry is .

Rules of differentiation of matrix valued functions are similar to rules for normal functions. Let and be matrix valued functions. Let a scalar and let be a constant matrix. Then

Note the order of the multiplication in the last two expressions.A first order linear system of ODEs is a system that can be written as the vector equation

where is a matrix valued function, and and are vector valued functions. We will often suppress the dependence on and only write . A solution of the system is a vector valued function satisfying the vector equation.

For example, the equations

can be written asWe will mostly concentrate on equations that are not just linear, but are in fact constant coefficient equations. That is, the matrix will be constant; it will not depend on .

When (the zero vector), then we say the system is homogeneous. For homogeneous linear systems we have the principle of superposition, just like for single homogeneous equations.

Theorem 3.3.1 (Superposition). Let be a linear homogeneous system of ODEs. Suppose that are solutions of the equation, then

(3.1) |

is also a solution. Furthermore, if this is a system of equations ( is ), and are linearly independent, then every solution can be written as (3.1).

Linear independence for vector valued functions is the same idea as for normal functions. The vector valued functions are linearly independent when

has only the solution , where the equation must hold for all .

Example 3.3.1: , , are linearly depdendent because , and this holds for all . So , , and above will work.

On the other hand if we change the example just slightly , , , then the functions are linearly independent. First write and note that it has to hold for all . We get that

In other words and . If we set , then the second equation becomes . But then the first equation becomes for all and so . Thus the second equation is just , which means . So is the only solution and , , and are linearly independent.

The linear combination could always be written as

where is the matrix with columns , and is the column vector with entries . The matrix valued function is called the fundamental matrix, or the fundamental matrix solution.

To solve nonhomogeneous first order linear systems, we use the same technique as we applied to solve single linear nonhomogeneous equations.

Theorem 3.3.2. Let be a linear system of ODEs. Suppose is one particular solution. Then every solution can be written as

where is a solution to the associated homogeneous equation ().

So the procedure for systems is the same as for single equations. We find a particular solution to the nonhomogeneous equation, then we find the general solution to the associated homogeneous equation, and finally we add the two together.

Alright, suppose you have found the general solution of . Next suppose you are given an initial condition of the form for some constant vector . Let be the fundamental matrix solution of the associated homogeneous equation (i.e. columns of are solutions). The general solution can be written as

We are seeking a vector such that

In other words, we are solving for the nonhomogeneous system of linear equations

Example 3.3.2: In § 3.1 we solved the system

with initial conditions , . Let us consider this problem in the language of this section.The system is homogeneous, so . We write the system and the initial conditions as

We found the general solution was and . Letting and , we obtain the solution . Letting and , we obtain . These two solutions are linearly independent, as can be seen by setting , and noting that the resulting constant vectors are linearly independent. In matrix notation, the fundamental matrix solution is, therefore,

To solve the initial value problem we solve for the equation

or in other words,

A single elementary row operation shows . Our solution is

This new solution agrees with our previous solution from § 3.1.

Exercise 3.3.2: a) Verify that the system has the two solutions and . b) Write down the general solution. c) Write down the general solution in the form , (i.e. write down a formula for each element of the solution).