Equation simplifies if a certain parameter is small

[Derivator]

Hi folks,

I have a differential equation which looks like

$$sin(f) + \frac{1}{Q}\frac{d(f)}{dt} + \frac{d^2(f)}{d t^2} = g(t)$$

Now for Q << 1 this should, according to our lecture, simplify to

$$sin(f) + \frac{1}{Q}\frac{d(f)}{dt}}{d t^2} = g(t)$$

Why that?

I mean, obviously for Q<<1, 1/Q >> 1. But why do we negelct only $$\frac{d^2(f)}{d t^2}$$?derivator

 

[CompuChip]

Unfortunately I cannot see the attachment, I suspect that you forgot to delete a dt².
Probably there is some other hidden assumption somewhere, like d²f/dt² varying slowly, or df/dt and d²f/dt² being of similar magnitude, such that the second derivative can be neglected.

I am also a bit puzzled why the sine term should not be neglected as well. Possibly because the equation can be solved exactly with it, perhaps because df/dt is also quite large and the quickly oscillating sine gives a non-trivial contribution?

 

[Derivator]

I deleted the attachement, since I tried to post a more abstract case, which isn't related to any physical problem.

The original problem can be found in the following screenshot:

(please not: not eqn 2.6 should reduce to 2.8. in fact, 2.7 should reduce to 2.8 )

which is from page 13 of this pdf-file: http://www.weizmann.ac.il/condmat/superc/theses_files/8.pdf

 

[CompuChip]

If I throw away the second derivative and solve for the first derivative, I don't get equation (2.8).
So it seems like something more is going on here.
Unfortunately I'll have to go out for an hour or so, but I will take a better look when I get back and see if I can understand what's going on here on a purely mathematical basis.

 

[CompuChip]

I now got equation (2.8), initially I missed that they switched back to a t-derivative rather than $\tau$-derivative.
Indeed the only approximation they make is ignoring the $$\frac{d^2\gamma}{d\tau^2}$$

Let's follow through with the second derivative intact: we start from
$$\frac{d^2\gamma}{d\tau^2} + \frac{1}{Q} \frac{d\gamma}{d\tau} + \sin\gamma = \frac{I}{I_0}$$
Since $\tau = Q \frac{t}{RC}$,
$$\frac{d\gamma}{d\tau} = \frac{d\gamma}{dt} \frac{dt}{d\tau} = \frac{RC}{Q} \frac{d\gamma}{dt}$$
and
$$\frac{d^2\gamma}{d\tau} = \left(\frac{RC}{Q}\right)^2 \frac{d^2\gamma}{dt^2}.$$

The equation thus becomes
$$\left( \frac{RC}{Q} \right)^2 \frac{d^2\gamma}{dt^2} + \frac{RC}{Q^2} \frac{d\gamma}{dt} = \frac{I}{I_0} - \sin\gamma.$$
Isolating the first derivative,
$$\frac{d\gamma}{dt} = \frac{Q^2}{R C} \left( \frac{I}{I_0} - \sin\gamma \right) - Q \frac{d^2\gamma}{dt^2}.$$

So the question is basically: why can we neglect Q with respect to Q²/RC?
In other words, why is
$$\sqrt{\frac{2eI_0}{\hbar C}} R C \ll \frac{2eI_0}{\hbar C} (RC)^2 / RC$$
That is,
$$1 \ll \sqrt{\frac{2eI_0}{\hbar C}}$$
So apparently, there is the silent assumption that the RC time is very short here, i.e. RC << Q << 1.

How sensible this is I cannot tell you, I can do the math but I don't know much about Josephson junctions.