Asymptotes of an implicit function?

[lavoisier]

Hi everyone,
I've been working on a problem for some time and I seem to be able to solve only half of it.
I wonder if anyone can help me with it.

I want to calculate the right asymptote of a function F(t), i.e. a line p = m t + b such that F(t) approaches it for t$\rightarrow \infty$.

This would be pretty straightforward if I had an explicit definition p = F(t).
I don't.
What I have is a differential equation for another function G(t):

$- \frac{dG(t)}{dt} = \frac{a + Q G(t) + c - \sqrt{(a + Q G(t) + c)^{2} - 4 Q G(t) c}}{2 V}$

where F(t) = Ln(G(t))
and a, Q, c and V are positive real constants.
The d.e. doesn't seem to be (easily) solvable.
So I tried another approach: leaving the function implicit.

The part that I solved by this approach was finding m (the slope of the asymptote).

I considered that:

$- \frac{dF(t)}{dt} = - \frac{d Ln(G(t))}{dt} = - \frac{1}{G(t)}\frac{dG(t)}{dt}$

And note that from the theory that generated the above differential equation, I know that:

$lim_{t \rightarrow \infty} G(t) = 0$

So I thought, as m is also F'(t) for t$\rightarrow \infty$, then:

$m = lim_{t \rightarrow \infty} \frac{dF(t)}{dt} = lim_{t \rightarrow \infty} \frac{1}{G(t)}\frac{dG(t)}{dt} = - lim_{G(t) \rightarrow 0} \frac{a + Q G(t) + c - \sqrt{(a + Q G(t) + c)^{2} - 4 Q G(t) c}}{2 V G(t)}$

I don't know if this is mathematically 'legal', but I did it anyway. I submitted the expression to Maxima and this is what came out:

$m = - \frac{Q c}{V (a + c)}$

which seems correct when I try it out graphically.

Now, the part that I don't know how to do is to find the general expression for the intercept q.

What I know from the theory is that:

$q = lim_{t \rightarrow \infty} F(t) - m t$

but as I don't have F(t), I'm stuck.
I thought of substituting F(t) = Ln(G(t)), but that of course gives me -∞ +∞.
I wondered if I could use the Hopital rule, but I think that only applies to ratios.

So... any suggestions on how I could do this (or if it's at all possible in this case)?

Thanks!
L

 

[pasmith]

Hi everyone,
I've been working on a problem for some time and I seem to be able to solve only half of it.
I wonder if anyone can help me with it.

I want to calculate the right asymptote of a function F(t), i.e. a line p = m t + b such that F(t) approaches it for t$\rightarrow \infty$.

This would be pretty straightforward if I had an explicit definition p = F(t).
I don't.
What I have is a differential equation for another function G(t):

$- \frac{dG(t)}{dt} = \frac{a + Q G(t) + c - \sqrt{(a + Q G(t) + c)^{2} - 4 Q G(t) c}}{2 V}$

where F(t) = Ln(G(t))
and a, Q, c and V are positive real constants.
The d.e. doesn't seem to be (easily) solvable.
So I tried another approach: leaving the function implicit.

The part that I solved by this approach was finding m (the slope of the asymptote).

I considered that:

$- \frac{dF(t)}{dt} = - \frac{d Ln(G(t))}{dt} = - \frac{1}{G(t)}\frac{dG(t)}{dt}$

And note that from the theory that generated the above differential equation, I know that:

$lim_{t \rightarrow \infty} G(t) = 0$

So I thought, as m is also F'(t) for t$\rightarrow \infty$, then:

$m = lim_{t \rightarrow \infty} \frac{dF(t)}{dt} = lim_{t \rightarrow \infty} \frac{1}{G(t)}\frac{dG(t)}{dt} = - lim_{G(t) \rightarrow 0} \frac{a + Q G(t) + c - \sqrt{(a + Q G(t) + c)^{2} - 4 Q G(t) c}}{2 V G(t)}$

I don't know if this is mathematically 'legal', but I did it anyway. I submitted the expression to Maxima and this is what came out:

$m = - \frac{Q c}{V (a + c)}$

which seems correct when I try it out graphically.

Let $$f(G) = -\frac{a + Q G(t) + c - \sqrt{(a + Q G(t) + c)^{2} - 4 Q G(t) c}}{2 V}$$ so that $$\frac{dG}{dt} = f(G).$$ What you've done is to compute $$f'(0) = \lim_{G \to 0} \frac{f(G) - f(0)}{G} = \lim_{G \to 0} \frac{f(G)}{G} = -\frac{Qc}{V(a + c)}$$ since $f(0) = 0$.

The more rigorous method of obtaining $f'(0)$ as the slope of the asymptote of $F$ is to observe that since, for non-zero values of the parameters, $f(G) < 0$ for $G > 0$ and $f(0) = 0$, we must have $G(t) \to 0$ if $G(0) > 0$. Expanding $f$ in a taylor series about $G = 0$ gives $$\frac{dG}{dt} = f'(0)G + \frac{f''(0)}2 G^2 + \dots.$$ Hence to leading order $$\frac{dF}{dt} = \frac{1}{G}\frac{dG}{dt} \sim f'(0).$$

Now, the part that I don't know how to do is to find the general expression for the intercept q.

What I know from the theory is that:

$q = lim_{t \rightarrow \infty} F(t) - m t$

but as I don't have F(t), I'm stuck.

You're stuck; the intercept will in general depend on $F(0)$.

 

[lavoisier]

Thanks pasmith.
Could you please just clarify, from your analysis, if you're saying that q exists but can't be calculated from the information I have, or that it doesn't exist as a single analytic expression?
FYI, I found that in a similar but simpler case:
$q = ln(G(0))+G(0) \frac{Q}{a}$
Would dimensional analysis help perhaps?
Thanks
L

 

[lavoisier]

Just wanted to report my progress with this problem.

I found that it's actually possible to solve the d.e. in F(t):

$- \frac{dF(t)}{dt} = \frac{a + Q e^{F(t)} + c - \sqrt{(a + Q e^{F(t)} + c)^{2} - 4 Q e^{F(t)} c}}{2 V e^{F(t)}}$

What I get is an implicit expression:

$t + constant = T(F)$

I plugged in the initial conditions to find;

$t = T(F) - T(F_{0})$

T(F) is a rather complicated function, where F, exp(F), a square root of a function of exp(F) and two asinh of a function of exp(F) are summed. I tried to simplify it (e.g. by expressing asinh as the logarithm), but with no success. I only got more complicated stuff.
Bottom line is, I don't think the solution of the d.e. can be inverted to give F(t).

However, I thought that if I could find the left asymptote of T(F) (because F tends to -∞ when G tends to 0⁺), the asymptote of its inverse would just be found by swapping t and F.

I 'did the maths', and indeed, I could find the slope and intercept for T(F).
The slope was the same that I had derived before. The intercept was a rather complicated function, similar to T(F₀).

I checked it numerically, and it worked!

As I said before, I don't know how 'legal' any of these operations are in terms of pure mathematics, but hey, I'm taking a pragmatic approach - if it does the job...

The next thing I need to do is to find this integral:

$\int^{+∞}_{0} G(t) dt$

Not sure how I might do that, when I haven't got F(t). I don't suppose one can integrate an implicit expression and then invert it in some way.
Taylor wouldn't work on integrals, would it?

Thanks
L

 

[CellCoree]

una

Hello Luna,

Thank you for sharing your problem with us. Asymptotes of implicit functions can be tricky to find, but there are some methods you can use to approach this problem.

First, I would recommend trying to find an explicit form for the function G(t) by solving the differential equation. This may involve using numerical methods or approximations, but having an explicit form for G(t) will make it easier to find the asymptote of F(t).

If that is not possible, another approach is to use the concept of a horizontal asymptote. A horizontal asymptote is a line that the function approaches as t increases or decreases without bound. In this case, you can use the fact that F(t) = Ln(G(t)) and the limit of G(t) as t approaches infinity is 0. This means that the horizontal asymptote of F(t) will be at y = -∞, since Ln(0) = -∞. Similarly, the horizontal asymptote for F(t) as t approaches negative infinity will be at y = ∞.

Another method you can try is to use the Taylor series expansion of F(t) around the point t = ∞. This will give you an expression for F(t) in terms of G(t) and its derivatives. From there, you can use the known value of m (the slope of the asymptote) to solve for the intercept q.

I hope these suggestions help you in finding the asymptotes of your implicit function. Good luck with your problem!