Solving a differential system of equations in matrix form

Question

I have a basic question about eigenvector.

If I have the following system:

$ \begin{pmatrix} \dot{\eta}_1 \\ \dot{\eta}_2 \\ \end{pmatrix} = {\cal{B}} \begin{pmatrix} \eta_1 \\ \eta_2 \\ \end{pmatrix} $

Where ${\cal{B}}$ is a matrix with constant parameters. For finding the solution to this system I do $\eta = P f$ where P is the eigenvector of ${\cal{B}}$, in this case, I can write:

$ \begin{pmatrix} \dot{f}_1 \\ \dot{f}_2 \\ \end{pmatrix} = {\cal{D}} \begin{pmatrix} f_1 \\ f_2 \\ \end{pmatrix} $

Where ${\cal{D}}$ is a diagonal matrix. I have two eigenvalue ($\lambda_1 $ and $\lambda_2$) and two corresponding eigenvectors ($v_1$ and $v_2$), the vector $v_1$ have components $v_{11}$ and $v_{12}$. The solution of this equation is:

$f_1 = c_0 e^{\lambda_1 t}$ and $f_1 = c_0 e^{\lambda_2 t}$,

If for example I select the first eigenvalue $\lambda_1$, then the $\eta_1$ and $\eta_2$ should be written like this

$$\eta_1 = c_0 v_{11} e^{\lambda_1 t} $$

$$\eta_1 = c_0 v_{12} e^{\lambda_1 t} $$

Is this correct?

And the general solution should be

$$\eta_1 = c_0 v_{11} e^{\lambda_1 t} +c_{01} v_{21} e^{\lambda_2 t} $$

$$\eta_1 = c_0 v_{12} e^{\lambda_1 t} +c_{02} v_{22} e^{\lambda_2 t}$$

Am I right? I have severe doubts about this.

Thanks

Answer 1

Consider the system $\dot{x} = Ax$, where $A\in\mathbb{R}^{n\times n}$. If you assume $A$ to be diagonalizable (i.e. the geometric multiplicity equals the algebraic multiplicity for every eigenvalue of $A$), then you can find an invertible transformation $T\in\mathbb{R}^{n\times n}$ in terms of eigenvectors of $A$ such that $A = T \Lambda T^{-1}$, where $\Lambda = \text{diag}(\lambda_1,\lambda_2,\dots,\lambda_n)$. Setting $z = T^{-1}x$, the differential equation becomes \begin{equation*} \dot{z} = \Lambda z. \end{equation*} Now, as you found, the decoupled system in the $z$-coordinates makes solving the differential equation very simple, namely: \begin{equation*} z_i(t) = e^{\lambda_i t}z_i(0) ~ \text{for all $i\in\{1,2,\dots,n\}$}, \end{equation*} or, equivalently, \begin{equation*} z(t) = \text{diag}(e^{\lambda_1 t},e^{\lambda_2 t},\dots,e^{\lambda_n t}) z(0). \end{equation*} To find the solution in the original $x$-coordinates, simply reverse the transformation: \begin{equation*} x(t) = Tz(t) = T\text{diag}(e^{\lambda_1 t},e^{\lambda_2 t},\dots,e^{\lambda_n t}) T^{-1}x(0). \end{equation*}

Edit: A great resource for learning more about systems of linear ordinary differential equations (linear systems theory) is Stanford's EE263 course.