1
$\begingroup$

If this should be on SE Physics instead, let me know, and I'll post there instead.

I have a diff eq. of the form $\ddot x + \beta \dot x^2 = 0$

Maybe there is a simple analytic solution, but I haven't been able to find it (can't use the method for Bernoulli diff eqs because $P(x) = 0 $). If it is the case that there is no analytic solution, this makes this problem even more interesting.

$\beta > 0$ here is a factor of units $[m^{-1}]$, and it looks something like $\beta = c\frac{A}{V}$, where $c$ is a dimensionless physical constant, $A$ is area, $V$ is volume.

Now, there is a simple solution to $\ddot x + \dot x ^2 = 0$, namely $$x = ln(t + c_1)+c_2$$ (I did not solve this myself and I'm not sure how I would, but it's easy to check that it works. I got this from Wolfram Alpha, which did not give me an analytic solution for the original problem*).

Now, depending on the object's actual size and shape, it should always (or at least in most cases) be possible to come up with a unit of distance such that measuring in this system gives $\frac{A}{V} = c^{-1}$.

If we let $x(0) = 0,\ \dot x(0) = v_0$, we get: $$c_1 = v_0^{-1}$$ $$c_2= ln(v_0)$$

Now if we convert from, say meters, to our new units, write that in to our $v_0$ and then convert our general result back to meters, it seems we've solved this problem.

So I have 2 questions:

1) Is this approach valid? The results I get don't make physical sense, but I don't see any flaws in the argument (there may be problems with my physical assumptions).

2) If this is a valid approach, does it reduce to any other method?Is it somehow equivalent to other approaches? It feels like something essentially different in some way. If it does have unique power, where does that power come from? It feels sort of like black magic, getting something out of nothing.

One other thing: if I'm totally wrong, and there is an easy way to solve this, I'd appreciate any approach. Also, if that's the case, are there any situations where this trick allows us to solve things we couldn't otherwise?

$\endgroup$
  • $\begingroup$ $\dot{x}=y$ then $\ddot{x}=y'$. $\endgroup$ – Nosrati Jun 25 '18 at 8:30
  • $\begingroup$ This is a separable first-order equation in $\dot x$ $\endgroup$ – Dylan Jun 25 '18 at 13:22
1
$\begingroup$

I'm gonna make an attempt but it was long time ago I solved these on paper.

You can solve by rewriting $y''=-by'^2$ and then taking logarithms: $\log(y'')=\log(-b)+2\log(y')$ and then substitute $y'=\exp(g)$ giving:

$$y''=/\text{ chain rule }/=g'\exp(g)$$ inserting $$\log(g')+g=\log(-b)+2g$$ simplifying $$g'=-b\exp(g)$$ which has solution $$g=-\log(bt+c_1)$$

now what remains is to substitute back from $g$ and find $y'(t)$ and from there $y(t)$.

You can use wolfram alpha to verify the solution: our solution, Wolfram's solution

The only thing differing is the name of the constants

  1. Wolfram has $c_1,c_2$
  2. We have $c,k_1$
$\endgroup$
  • $\begingroup$ This log trick is great! Just curious, how did You solve g' = -b exp(g)? $\endgroup$ – David Lalo Jun 25 '18 at 11:38
  • 1
    $\begingroup$ You can use for example logarithm and then after a few steps of rewriting you will see the famous logarithmic derivative $\log(h)' = \frac{h'}{h}$ en.wikipedia.org/wiki/Logarithmic_derivative it is also very useful trick when you do substitution in other differential equations. $\endgroup$ – mathreadler Jun 25 '18 at 11:44
  • $\begingroup$ This solution is fine, but I don't think the substitution is very necessary. It's a separable equation! $\endgroup$ – Dylan Jun 25 '18 at 13:18
  • $\begingroup$ @Dylan Substitution is a valuable trick to learn, as is separability. I see no reason to always stick with the easiest way. Much can often be learned in not always going the easiest way. $\endgroup$ – mathreadler Jun 25 '18 at 13:31
  • $\begingroup$ I agree with, but I think one should learn the easy way first. In this case, OP didn't know about the separable trick. $\endgroup$ – Dylan Jun 25 '18 at 15:07
3
$\begingroup$

Using @user108128's hint $$x''+\beta (x')^2=0 \implies y'+\beta y^2=0\implies y=\frac{1}{\beta t+c_1}$$ Then $$x'=\frac{1}{\beta t+c_1}\implies x=\frac 1 \beta \log(\beta t+c_1)+c_2$$

So, using the conditions at $t=0$ $$0=\frac 1 \beta \log(c_1)+c_2$$ $$v_0=\frac 1 {c_1}$$ then $$c_1=\frac 1 {v_0}\qquad \text{and} \qquad c_2=\frac 1 \beta \log(v_0)$$

$\endgroup$
  • $\begingroup$ Thanks, but it's actually x'' + b(x')^2 that I'm looking for, not x'' + bx' $\endgroup$ – David Lalo Jun 25 '18 at 11:34
  • 1
    $\begingroup$ The answer is still correct, he just made a typo. $\endgroup$ – Dylan Jun 25 '18 at 13:13
  • $\begingroup$ @Claude hope you don't mind my edit $\endgroup$ – Dylan Jun 25 '18 at 13:22

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.