Compute the fundamental matrix of a given linear system of first-order differential equations

Let Y and A be matrices as defined in this previous exercise, with \frac{d}{dt} Y = AY. Compute the fundamental matrix of this differential equation where A is one of the following matrices:

  1. A_1 = \begin{bmatrix} 2 & -2 & 14 \\ 0 & 3 & -7 \\ 0 & 0 & 2 \end{bmatrix}
  2. A_2 = \begin{bmatrix} 9 & 4 & 5 \\ -4 & 0 & -3 \\ -6 & -4 & -2 \end{bmatrix}
  3. A_3 = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix}

Let A be a matrix, and let P^{-1}AP = B be in Jordan canonical form. Using this previous exercise, we have that \mathsf{exp}(At) = P\mathsf{exp}(Bt)P^{-1} is a fundamental matrix, and by this previous exercise, P\mathsf{exp}(Bt) is a fundamental matrix of our differential equation. We can use this previous result (and part 2 of this previous exercise) to compute \mathsf{exp}(Bt).

We computed the Jordan canonical forms (and conjugators) for these three matrices in this previous exercise.

For the matrix A_1, we have P_1 = \begin{bmatrix} -7 & -1 & -2 \\ 7 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix} and P_1^{-1}A_1P_1 = B_1 = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}. Now \mathsf{exp}(Bt) = \mathsf{exp}([2]t \oplus [2]t \oplus [3]t) = \begin{bmatrix} e^{2t} & 0 & 0 \\ 0 & e^{2t} & 0 \\ 0 & 0 & e^{3t} \end{bmatrix}, and P_1 \mathsf{exp}(B_1t) = \begin{bmatrix} -7e^{2t} & -e^{2t} & -2e^{3t} \\ 7e^{2t} & 0 & e^{3t} \\ e^{2t} & 0 & 0 \end{bmatrix}. We can easily verify that this is indeed a fundamental matrix.

For the matrix A_2, as computed in this previous exercise, if we let P_2 = \begin{bmatrix} -2 & 5 & 3 \\ 1 & -3 & -2 \\ 2 & -5 & -2 \end{bmatrix} then P_2^{-1}A_2P_2 = B_2 = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} is in Jordan canonical form. Now \mathsf{exp}(B_2t) = \mathsf{exp}(\begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix} t) \oplus \mathsf{exp}([3]t) = \begin{bmatrix} e^{2t} & te^{2t} & 0 \\ 0 & e^{2t} & 0 \\ 0 & 0 & e^{3t} \end{bmatrix}. So P_2\mathsf{exp}(B_2t) = \begin{bmatrix} -2e^{2t} & -2te^{2t} + 5e^{2t} & 3e^{3t} \\ e^{2t} & te^{2t} - 3e^{2t} & -2e^{3t} \\ 2e^{2t} & 2te^{2t} - 5e^{2t} & -2e^{3t} \end{bmatrix} is a fundamental matrix.

Finally, consider A_3. Letting P_3 = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 \end{bmatrix} (Seethis previous exercise), we have P_3^{-1}A_3P_3 = B_3 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \oplus \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} in Jordan canonical form. Then \mathsf{exp}(B_3t) = \begin{bmatrix} e^t & te^t \\ 0 & e^t \end{bmatrix} \oplus \begin{bmatrix} e^t & te^t \\ 0 & e^t \end{bmatrix}, and hence P_3\mathsf{exp}(B_3t) = \begin{bmatrix} 0 & 0 & e^t & te^t \\ 0 & 0 & 0 & e^t \\ e^t & te^t & 0 & 0 \\ 0 & e^t & e^t & te^t + e^t \end{bmatrix} is a fundamental matrix.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: