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I would like to solve the following system

\begin{equation*} \begin{cases} \ddot{ u}_x(z) = \gamma_1 u_x(z) - i\gamma_2 \dot{u}_z(z) & \\ \ddot{u}_z(z) = \gamma_3 u_z(z) - i\gamma_4 \dot{u}_x(z) & \\ \end{cases} \end{equation*} I would like to make use of the following notation $\textbf{x}_1 = u_x(z) $ , $\textbf{x}_2=\dot{u}_x(z)$ and $\textbf{z}_1 = u_z(z) $ , $\textbf{z}_2 = \dot{u}_z(z)$.Hence I can rewrite the above system as a 4 dimensional system \begin{equation*} \begin{cases} \dot{\textbf{x}}_1=\textbf{x}_2 & \\ \dot{\textbf{x}}_2 = \gamma_1 \textbf{x}_1 - i\gamma_2 \textbf{z}_2 & \\ \dot{\textbf{z}}_1=\textbf{z}_2 & \\ \dot{\textbf{z}}_2 = \gamma_3 \textbf{z}_1 - i\gamma_4 \textbf{x}_2 & \\ \end{cases} \end{equation*} Or more clearly as \begin{eqnarray} \begin{bmatrix} \dot{\textbf{x}}_1 \\ \dot{\textbf{x}}_2 \\ \dot{\textbf{z}}_1 \\ \dot{\textbf{z}}_2 \end{bmatrix} & = & \begin{pmatrix}0 & 1 & 0 & 0 \\ \gamma_1 & 0 & 0 & -i\gamma_2 \\ 0 & 0 & 0 & 1 \\ 0 & -i\gamma_4 & \gamma_3 & 0\end{pmatrix} \begin{bmatrix} \textbf{x}_1 \\ \textbf{x}_2 \\ \textbf{z}_1 \\ \textbf{z}_2 \end{bmatrix} \end{eqnarray}

$\gamma_2$ and $\gamma_4$ are always positive and real. $\gamma_1$ and $\gamma_3$ may be real or complex,but for now I will take them as real.

In most textbooks people assume that the above matrix has only real entries. Hence,my first approach to the problem was to turn the above complex matrix into a real matrix as follows. Suppose one has a complex matrix of the form

\begin{equation*} Z=X+iY \in M_{n}(C) \end{equation*} with X,Y $\in$ $M_n$(R). We can make a correspondence with a real matrix in the following way

\begin{equation*} Z\rightarrow \begin{pmatrix}X & -Y \\ Y & X \end{pmatrix} \in M_{2n}(R) \end{equation*}

Hence the above ode linear system becomes

\begin{eqnarray} \begin{bmatrix} Re(\dot{\textbf{Z}}) \\ Im(\dot{\textbf{Z}}) \end{bmatrix} & = & \begin{pmatrix}X & -Y \\ Y & X \end{pmatrix} \begin{bmatrix} Re(\textbf{Z}) \\ Im(\textbf{Z}) \end{bmatrix} \end{eqnarray}

where

\begin{eqnarray} \textbf{Z}& = & \begin{bmatrix} \textbf{x}_1 \\ \textbf{x}_2 \\ \textbf{z}_1 \\ \textbf{z}_2 \end{bmatrix} \end{eqnarray}

$\textbf{QUESTION} $

Is this a sound way to solve the linear ode system with complex coefficients? Are there other ways to solve this type of system?

Can you please also provide some references to how this type of systems can be solved?

Thank you very much

  • 1
    Can't you use the usual ansatz of eigenvector-exponentials? Of course the system has complex eigenvalues and complex eigenvectors, but that does not invalid the method. – 2017-01-09
  • 0
    Excuse me, but I don't follow:). What ansatz? Can you be a little more specific please – 2017-01-09
  • 0
    Given a Matrix $A$ mit eigenvalues $\lambda_i$ to eigenvector $v_i$ the solution of the ODE $\dot{x} = Ax$ is (basically) given by $x(t) = \sum_i C_i e^{\lambda_i \cdot t}\cdot v_i$, wehere $C_i$ is the integration-Constant to be determined by initial-value/boundary/etc. This gets more complicated, when your Matrix is not diagonizable, but that's also to find in ODE textbooks – 2017-01-09
  • 0
    I don't think this is efficient as it doubles the number of terms. On the opposite, keep the complex coefficients and use the same formulas as with real coefficients. – 2017-01-09
  • 0
    Laray, I see what you mean. That was one my first thoughts,to apply the same methods from the real case but I wans't sure if they would lead to the correct results in the complex case. – 2017-01-09
  • 0
    In that case, I wuould just try it with an example and check, if the plots match with an Matalb ode45 solution. @YvesDaoust What do you mean by "doubling the terms"? I will still have just 4 complex terms. That could be seperated into 4 growing and 4 periodic ones. – 2017-01-09
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    @Laray: real + imaginary terms = 2 terms – 2017-01-09

0 Answers 0