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Solve for $\textbf{x}(t)$ from the system of differential equations $\textbf{x}' (t) = A \textbf{x}(t)$, where

$$ A =\begin{bmatrix} 0 & -1 & 0\\1 & 0 & 0\\ 0 & 0 & 1\\\end{bmatrix}$$

I did the following:

I found the characteristic polynomial, $|A - \lambda I| = (1-\lambda)(\lambda^2 +1) = 0$, thus $\lambda_1 = 1, \lambda_2 = i, \lambda_3 = -i$.

Now, for $\lambda_1$ the corresponding eigenvector is $\textbf{v}_1 = \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}$

For $\lambda_2$ the corresponding eigenvector is $\textbf{v}_2 = \begin{bmatrix} 1\\ -i\\ 0 \end{bmatrix}$

For $\lambda_3$ the corresponding eigenvector is $\textbf{v}_3 = \begin{bmatrix} 1\\ i\\ 0 \end{bmatrix}$

Thus $V = \begin{bmatrix} 0 & 1 & 1\\ 0 &-i & i\\ 1 &0 &0 \end{bmatrix}$

Next, I want to use this: $c_1 e^{\lambda_1 t}\textbf{v}_1 + c_2 e^{\lambda_2 t}\textbf{v}_2 + c_3 e^{\lambda_3 t}\textbf{v}_3$.

\begin{align}

\textbf{x}(t) &= c_1 e^{\lambda_1 t}\textbf{v}_1 + c_2 e^{\lambda_2 t}\textbf{v}_2 + c_3 e^{\lambda_3 t}\textbf{v}_3\

\textbf{x}(t) &= c_1 e^{t}$\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}$ + c_2 e^{i t}$\begin{bmatrix} 1\\ -i\\ 0 \end{bmatrix}$ + c_3 e^{-i t}$\begin{bmatrix} 1\\ i\\ 0 \end{bmatrix}$

\end{align}

GaMbiTaaaa
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3 Answers3

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The solution is $x(t)=\exp(tA) x(0)=V\exp(tD)V^{-1} x(0)$ where $$\exp(tD)=\begin{pmatrix}e^t&0&0\\0&e^{it}&0\\0&0&e^{-it}\end{pmatrix}$$

amine
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You have found that $$ A = V\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & -i\end{array}\right]V^{-1}. $$ Because $V$ is a constant matrix, the equation can be written as $$ \frac{d}{dt}(V^{-1}x(t))= \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & -i\end{array}\right](V^{-1}x(t)). $$ Therefore, $$ V^{-1}x(t) = \left[\begin{array}{c}ae^{t} \\ be^{it} \\ ce^{-it}\end{array}\right] $$ where $a$, $b$, $c$ are arbitrary constants. Solving for $x(t)$ gives $$ x(t) = V\left[\begin{array}{c}ae^{t} \\ be^{it} \\ ce^{-it}\end{array}\right] = ae^{t}{\bf v_{1}}+be^{it}{\bf v_2}+ce^{-it}{\bf v_3}. $$

Disintegrating By Parts
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Answer. $$ {\boldsymbol{x}}(t)=\mathrm{e}^{tA}{\boldsymbol{x}}(0), $$ where $$ \mathrm{e}^{tA}=\left(\begin{matrix} \cos t&-\sin t& 0\\ \sin t & \,\,\,\,\cos t & 0 \\ 0 & 0 &\mathrm{e}^t \end{matrix}\right). $$

Professional advice. Do not diagonalise this particular matrix $A$. It is much easier, for this one, to simply write down what its exponential is.

Yiorgos S. Smyrlis
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