4. Homogeneous solutions¶
Given
where \(\mathbf{x}\in\mathbb{R}^{n}\) and \(\mathbf{A}\in\mathbb{R}^{n\times n}\), we can easily show in the usual way that any linear combination of solutions is also a solution. Thus our major task is to find a basis for the homogeneous solutions.
Attention
You need to read and write carefully when in vector-land. In particular, note that \(\mathbf{x}_1\) refers to the first vector of a collection, while \(x_1\) means the first component of a vector \(\mathbf{x}\).
4.1. Fundamental matrix¶
Suppose \(\mathbf{x}_1,\ldots,\mathbf{x}_m\) are homogeneous solutions of the ODE, and we use a linear combination of them to satisfy an initial condition \(\mathbf{x}(t_0)=\mathbf{x}_0\):
Using the equivalence of linear combination with matrix-vector multiplication, we define the \(n\times m\) matrix
so that
This is a linear algebraic system for the coefficients \(c_i\). We can expect a unique solution if and only if \(m=n\) and \(\bfX(t_0)\) is invertible.
The \(n\times n\) matrix \(\bfX(t)\) is a fundamental matrix of the homogeneous system \(\mathbf{x}' = \mathbf{A}(t)\mathbf{x}\) if its columns satisfy
and \(\bfX(t)\) is invertible at all times in an open interval \(I\) where \(\bfA\) is continuous.
Note
A fundamental matrix of a particular system is not unique.
Because matrix-matrix multiplication can be interpreted columnwise, a statement equivalent to “\(\mathbf{x}_j'=\mathbf{A}\mathbf{x}_j\) for all \(j\)” is that \(\bfX'=\bfA\bfX\).
The fundamental matrix gives all we need for the general homogeneous solution.
If \(\bfX\) is a fundamental matrix for \(\bfx'=\bfA(t)\bfx\), then the general solution of this system is
for an arbitrary constant vector \(\mathbf{c}\).
4.2. Wronskian¶
Since invertibility is important to the general solution, we should not be surprised to see a determinant popping up.
The Wronskian of a collection of homogeneous solutions \(\mathbf{x}_1,\ldots,\mathbf{x}_n\) is
Thus, a solution set with a nonzero Wronskian can be used as the columns of a fundamental matrix.
Example
Suppose
where \(\mathbf{v}_1\) and \(\mathbf{v}_2\) are columns of an invertible \(2\times 2\) constant matrix. Show that the Wronskian of these solutions is nonzero at all times.
Solution
Say \(\mathbf{v}_1=[a,\,c]\) and \(\mathbf{v}_2=[b,\,d]\). Then
The exponential function is never zero, and the problem statement guarantees that the last determinant is nonzero. Hence \(W\) is nonzero as well.
There is a remarkable result known as Abel’s formula that simplifies the issue of a nonzero Wronskian. We state a limited form of it here.
Suppose that \(\mathbf{A}(t)\) is continuous on an open interval \(I\), and let \(\mathbf{x}_1,\ldots,\mathbf{x}_n\) be solutions of \(\mathbf{x}'=\mathbf{A}\mathbf{x}\) whose Wronskian is \(W(t)\). Then if \(W\) is nonzero at any time in \(I\), it is nonzero at all times in \(I\).
Hence it suffices to check the Wronskian at any one time, since the conclusion about whether it is zero will then apply to all times.