Abel's Fundamental Matrix Formula

Recall from the Fundamental Matrices to a Linear Homogeneous System of First Order ODEs that if $\{ \phi^{[1]}, \phi^{[2]}, ..., \phi^{[n]} \}$ is a fundamental set of solutions to a linear homogeneous system of $n$ first order ODEs $\mathbf{x}' = A(t) \mathbf{x}$ then the corresponding fundamental matrix is defined as the $n \times n$ matrix:

(1)

\begin{align} \quad \Phi = \begin{bmatrix} \phi^{[1]} & \phi^{[2]} & \cdots & \phi^{[n]} \end{bmatrix} = \begin{bmatrix} \phi_1^{[1]} & \phi_1^{[2]} & \cdots & \phi_1^{[n]} \\ \phi_2^{[1]} & \phi_2^{[2]} & \cdots & \phi_2^{[n]} \\ \vdots & \vdots & \ddots & \vdots \\ \phi_n^{[1]} & \phi_n^{[2]} & \cdots & \phi_n^{[n]} \\ \end{bmatrix} \end{align}

We will now prove an important result known as Abel's Fundamental Matrix formula.

Theorem 1 (Abel's Fundamental Matrix Formula): If $\Phi$ is a fundamental matrix to the linear homogeneous system $\mathbf{x}' = A(t)\mathbf{x}$ then $\Phi$ is a solution to the matrix equation $X' = A(t)X$.

We define $X' = [x_{i,j}']$ where $X = [x_{i,j}]$.

• Proof: Let $\Phi$ be a fundamental matrix to the linear homogeneous system $\mathbf{x}' = A(t)\mathbf{x}$. Then each of the columns of $\Phi$ is a solution to $\mathbf{x}' = A(t)\mathbf{x}$ on some predescribed interval $J = (a, b)$. So:

(2)

\begin{align} \quad \Phi' &= \begin{bmatrix} \phi^{[1]'} & \phi^{[2]'} & \cdots & \phi^{[n]'} \end{bmatrix} \\ &= \begin{bmatrix} A(t) \phi^{[1]} & A(t) \phi^{[2]} & \cdots & A(t)\phi^{[n]} \end{bmatrix} \\ &= A(t) \begin{bmatrix} \phi^{[1]}& \phi^{[2]} & \cdots & \phi^{[n]} \end{bmatrix} \\ &= A(t) \Phi \end{align}

• Hence $\Phi$ is a solution to the matrix equation $X' = A(t)X$. $\blacksquare$

We have already looked at the following linear homogeneous system of $2$ first order ODEs:

(3)

\begin{align} \quad x_1' &= x_1 \\ \quad x_2' &= 2x_2 \end{align}

We have that $A(t) = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$.

We have found a fundamental matrix for the system:

(4)

\begin{align} \quad \Phi = \begin{bmatrix} e^t & 0 \\ 0 & e^{2t} \end{bmatrix} \end{align}

We will show that $\Phi$ is the solution to the matrix equation $X' = A(t) X$, i.e., $\Phi' = A(t)\Phi$. We first compute the lefthand side of this equation:

(5)

\begin{align} \quad \Phi' = \begin{bmatrix} e^t & 0 \\ 0 & 2e^{2t} \end{bmatrix} \end{align}

And now the righthand side of this equation:

(6)

\begin{align} \quad A(t) \Phi &= \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} e^t & 0 \\ 0 & e^{2t} \end{bmatrix} \\ &= \begin{bmatrix} e^t & 0 \\ 0 & 2e^{2t} \end{bmatrix} \end{align}

We see that indeed $\Phi' = A(t)\Phi$, i.e., $\Phi$ is a solution to the matrix equation $X' = A(t)X$.