Examples

Verifying a fundamental matrix

Verifying a fundamental matrix

Problem: Consider the ODE system:

$${\mathbf{y}}^{\prime }(t)=\left(\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right)\mathbf{y}(t).$$

Verify that the following is a fundamental matrix for this ODE:

$$\mathbf{Y}=\left(\begin{array}{ccc}{e}^{2t}& -e^{-t}& 0\\ e^{2t}& 0& e^{-t}\\ e^{2t}& e^{-t}& -e^{-t}\end{array}\right).$$

Solution: First we must check that each column of $\mathbf{Y}$ is a solution vector for the ODE:

$$\begin{array}{rl}{\mathbf{y}}_1^{\prime }={\left(\begin{array}{c}{e}^{2t}\\ e^{2t}\\ e^{2t}\end{array}\right)}^{\prime }& =\left(\begin{array}{c}2e^{2t}\\ 2e^{2t}\\ 2e^{2t}\end{array}\right)=2\left(\begin{array}{c}{e}^{2t}\\ e^{2t}\\ e^{2t}\end{array}\right),\\ & \\ \mathbf{A}{\mathbf{y}}_1=\left(\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right)\left(\begin{array}{c}{e}^{2t}\\ e^{2t}\\ e^{2t}\end{array}\right)& =\left(\begin{array}{c}0+e^{2t}+e^{2t}\\ e^{2t}+0+e^{2t}\\ e^{2t}+e^{2t}+0\end{array}\right)=2\left(\begin{array}{c}{e}^{2t}\\ e^{2t}\\ e^{2t}\end{array}\right).\end{array}$$

Similarly for the second column vector:

$$\begin{array}{rl}{\left(\begin{array}{c}-e^{-t}\\ 0\\ e^{-t}\end{array}\right)}^{\prime }& =\left(\begin{array}{c}{e}^{-t}\\ 0\\ -e^{-t}\end{array}\right)\\ & \\ \left(\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right)\left(\begin{array}{c}-e^{-t}\\ 0\\ e^{-t}\end{array}\right)& =\left(\begin{array}{c}0+0+e^{-t}\\ -e^{-t}+0+e^{-t}\\ -e^{-t}+0+0\end{array}\right)=\left(\begin{array}{c}{e}^{-t}\\ 0\\ -e^{-t}\end{array}\right).\end{array}$$

And similarly for the third column vector.

Secondly, we must verify that these solutions are independent. For that, we compute the determinant of the Wronskian matrix using the formula for $3\times 3$ matrices:

$$\begin{array}{rl}W(t)=\det \mathbf{Y}& =\left|\begin{array}{ccc}{e}^{2t}& -e^{-t}& 0\\ e^{2t}& 0& e^{-t}\\ e^{2t}& e^{-t}& -e^{-t}\end{array}\right|\\ & \\ & =e^{2t}\left|\begin{array}{cc}0& e^{-t}\\ e^{-t}& -e^{-t}\end{array}\right|--e^{-t}\left|\begin{array}{cc}{e}^{2t}& e^{-t}\\ e^{2t}& -e^{-t}\end{array}\right|+0\left|\begin{array}{cc}{e}^{2t}& 0\\ e^{2t}& e^{-t}\end{array}\right|\\ & \\ & =e^{2t}(-e^{-2t})+e^{-t}(-2e^t)+0=-1-2=-3.\end{array}$$

Therefore $W(t)=-3$ and is never $0$, so the columns of $\mathbf{Y}$ are independent. Combining this fact with the first fact, we know that $\mathbf{Y}$ is a fundamental matrix.