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  Mathieu-like equation analysis

+ 5 like - 0 dislike
1119 views

Recently I've got Mathieu-like equation (it describes dynamics of vector-potential in presense of pseudo-scalar field):

$$
\tag 1 \frac{d^{2}A_{k}(\tau )}{d \tau} + \left(B_{k}(\tau ) - 2q_{k}(\tau )cos\left( 2\tau \right) \right)A_{k}(\tau ) = 0, \quad B_{k} = c + \frac{k^{2}}{8 \tau}, \quad q = \frac{Dk}{\tau^{\frac{5}{4}}},
$$

This is Mathieu-like equation, but with time-dependent coefficients. I need to analyze it on instabilities (exponentially growing solutions), but I don't know how to do it. Can you give me advise or literature sources, where I can reed about analysis of Eqs. like $(1)$?

asked Sep 5, 2015 in Mathematics by NAME_XXX (1,060 points) [ revision history ]
recategorized Sep 9, 2015 by Dilaton

1 Answer

+ 5 like - 0 dislike

One possible approximation is as follows.  When  $\tau$ tends to infinity, the original equation is reduced to

$${\frac {d^{2}}{d{\tau}^{2}}}A \left( \tau \right) + \left( c+{
\frac {{k}^{2}}{8\tau}} \right) A \left( \tau \right) =0$$

and the corresponding analytical solution is

$$A \left( \tau \right) =C_{1}
{{\rm \bf M}\left({\frac {-i{k}^{2}}{16\sqrt {c}}},\frac{1}{2},\,2\,i\sqrt {c}\tau\right)}
+C_{2}
{{\rm \bf W}\left({\frac {-i{k}^{2}}{16\sqrt {c}}},\frac{1}{2},2\,i\sqrt {c}\tau\right)}$$

where $M$ is the Whittaker $M$ function  and $W$ is the Whittaker $W$ function.

Making the change $c = -a^2$  and $a = -b$ we obtain

$$A \left( \tau \right) =C_{1}
{{\rm \bf M}\left({\frac {{k}^{2}}{16b}},\frac{1}{2} ,\,2\,b\tau\right)}+
C_{2}
{{\rm \bf W}\left({\frac {{k}^{2}}{16b}},\frac{1}{2},    \,2\,b\tau\right)}$$

Then for $k>0$ and $b>0$, the solution

$$A \left( \tau \right) =C_{1}
{{\rm \bf M}\left({\frac {{k}^{2}}{16b}},\frac{1}{2} ,\,2\,b\tau\right)}$$

gives a exponentially growing solution; and

$$A \left( \tau \right) =C_{2}
{{\rm \bf W}\left({\frac {{k}^{2}}{16b}},\frac{1}{2},    \,2\,b\tau\right)}$$

gives a exponential decaying solution.

answered Sep 6, 2015 by juancho (1,130 points) [ no revision ]

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