0
$\begingroup$

$$ \begin{align} i x'(t)= c A(t)(e^{-i q(t)}) y(t) \end{align} $$ \begin{align} i y'(t)= c A(t)(e^{i q(t)}) x(t) \end{align} $$ when A(t) and q(t) are functions as follows:

\begin{align} A(t)=a cos(wt-r),\\ q(t)=b+dt+k sin(wt+r) \end{align}

where all of \begin{align} a,b,c,d,k,w,r \end{align} are constants .

and the boundary condition is as follows: $$ X(t=0)=1,\\ Y(t=0)=0 \\ $$ It might be possible to slightly simplify the solutions by considering special cases. For example assume that when one of the parameters of the theory, say $d$ has some special values say $d_0=0$ or $1$ or something. Then if in this case an analytical solution $x_0(t)$ and $y_0(t)$ can be found, in the next step you may assume that the solutions $x(t)=x_0(t)\cdot X(t)$ and $y(t)=y_0(t)\cdot Y(t)$. Hopefully the equations for $X(t)$ and $Y(t)$ can be much simpler??

Improve this question
$\endgroup$
1
  • $\begingroup$ I took the liberty of fixing your formatting $\endgroup$
    – Yemon Choi
    Commented Jan 30, 2013 at 20:38

1 Answer 1

Reset to default
1
$\begingroup$

I couldn't get a closed-form solution in general, but with $d=0$ and $r=0$ Maple finds the following solution, where $s = \sqrt{4 a^2 c^2 + k^2 w^2}$:

$$ \eqalign{ x(t) &= \frac{s+kw}{2s} e^{-i (kw - s) \sin(w t)/(2w)} + \frac{s-kw}{2s} e^{-i(kw+s) \sin(wt)/(2w)} \cr y(t) &=\frac{ac}{s} e^{ib} \left(e^{i(k w - s) \sin(w t)/(2w)} - e^{i(k w + s) \sin(w t)/(2w)}\right)} $$

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks a lot Robert . But one more question: can we have a solution in the form below for example for x(t) : # x(t)= Xn e^(-inwt) where # shows the sigma notation and Xn are some coefficients . I need a solution in this form to approximate the Xn for n=1,2,3,4,5. Am I right? $\endgroup$
    – hossein
    Commented Jan 31, 2013 at 11:21
  • $\begingroup$ You may note that (if $w$ is real) the solution above is periodic with period $2\pi/w$, thus can be expanded in a Fourier series. Your $X_n$ are coefficients of this series. $\endgroup$
    – Robert Israel
    Commented Feb 1, 2013 at 8:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .