Characteristic time?

Solution 1:

The characteristic time is usually defined to be the time in which a quantity, in this case the deviation from $(0,b,0)$, decreases by $1/\mathrm e$. In the present case the functions don't decay exactly exponentially, but it still makes sense to speak of the characteristic time with respect to the leading terms for $t\to\infty$. These are

$$\operatorname{sech} t=\frac2{\mathrm e^t-\mathrm e^{-t}}\approx2\mathrm e^{-t}$$

and

$$\operatorname{tanh} t-1=\frac{\mathrm e^t+\mathrm e^{-t}}{\mathrm e^t-\mathrm e^{-t}}-1=\frac{2\mathrm e^{-t}}{\mathrm e^t-\mathrm e^{-t}}\approx2\mathrm e^{-2t}\;.$$

Thus the $y$ component decays twice as quickly as the other two, and the characteristic time of the approach could be said to be that of $\mathrm e^{-t}$, which is $1$.

(This is basically the same answer as Henning's, just with less rigour and more numbers.)

Solution 2:

What is meant in this particular context is almost certainly that for large $t$, we can approximate $$\begin{pmatrix} a \operatorname{sech} t \\ b\tanh t \\ c \operatorname{sech} t \end{pmatrix} \sim \begin{pmatrix}0 \\ b \\ 0\end{pmatrix} + e^{-t/k} \begin{pmatrix}p \\ q \\ r\end{pmatrix} + o(e^{-t/k})$$ for appropriate constants $k$, $p$, $q$, $r$. The characteristic time of the approach to $(0,b,0)$ is then $k$.

(And your $b$ had better be $1$ for the question to make sense).

Hmm ... alternatively, "align with $(0,1,0)$" could mean the process of the direction of the vector approaching the direction of the positive $y$ axis. In that case the relevant approximation would be something like $$ \frac{ \sqrt{a^2+c^2} \operatorname{sech} t }{ b \tanh t } \sim e^{-t/k} s + o(e^{-t/k}) $$ for constants $k$ and $s$. Again $k$ is the characteristic time.

Here the left-hand side of this represents the angle between the vector and $(0,1,0)$ as seen from the origin, rather than the distance between your vector and $(0,1,0)$. Strictly speaking the left-hand side should arguably be $\tan^{-1}\left(\frac{\sqrt{a^2+c^2}\operatorname{sech}t}{b\tanh t}\right)$, but since $\tan^{-1}(x)\sim x$ for small $x$ anyway, it doesn't matter for the result.

(So much for "almost certainly").