The general solution of a two-dimensional linear dynamical system \begin{align*} \dot{x}_1 &= f(x_1, x_2) \\ \dot{x}_2 &= g(x_1, x_2) \end{align*} is given by \begin{align*} c_1e^{\lambda_1 t}\mathbf{v}_1 + c_2e^{\lambda_2 t}\mathbf{v}_2 \end{align*} where $\mathbf{v}_i$ are eigenvectors of the real valued coefficient matrix $A$ obtained from writing the system as \begin{equation*} \mathbf{\dot{x}} = A\mathbf{x}\end{equation*} and $\lambda_i$ are the eigenvalues. Suppose that $\lambda_i$ are complex valued. Given the initial condition $ \mathbf{x_0} = [x^1_0, x^2_0]$ where $x^i_0$ are real valued numbers, the coefficients $c_i$ of the general solution must be complex conjugates in order to satisfy the initial condition.

If we suppose that this dynamical system arose in the context of modeling physical phenomena, could there ever be a situation in which the initial condition $\mathbf{x_0}$ has complex valued elements resulting in real valued coefficients $c_i$?

  • 1
    $\begingroup$ If your system is linear you should simply write it as a linear system. Your formulation is confusing. And the solution for a real physical problem can only be real. Remember that you can always deduce from the complex fundamental system the real fundamental system, hence avoid complex basis solution. $\endgroup$ – MrYouMath Oct 10 '17 at 17:15
  • $\begingroup$ If your system has real-valued coefficients there is always a basis that consists of real-valued functions. Almost all textbooks cover how to go from easy-to-know complex-valued basis functions for complex eigenvalues to real-valued. Or they just write real-valued basis right from the start. $\endgroup$ – Evgeny Oct 10 '17 at 19:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.