Asymptotes of an implicit function?

  • Thread starter lavoisier
  • Start date
  • #1
177
24
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[itex]\rightarrow \infty[/itex].

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):

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

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:

[itex]- \frac{dF(t)}{dt} = - \frac{d Ln(G(t))}{dt} = - \frac{1}{G(t)}\frac{dG(t)}{dt} [/itex]

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

[itex]lim_{t \rightarrow \infty} G(t) = 0[/itex]

So I thought, as m is also F'(t) for t[itex]\rightarrow \infty[/itex], then:

[itex]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)}[/itex]

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:

[itex]m = - \frac{Q c}{V (a + c)}[/itex]

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:

[itex]q = lim_{t \rightarrow \infty} F(t) - m t[/itex]

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
 

Answers and Replies

  • #2
pasmith
Homework Helper
1,827
488
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[itex]\rightarrow \infty[/itex].

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):

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

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:

[itex]- \frac{dF(t)}{dt} = - \frac{d Ln(G(t))}{dt} = - \frac{1}{G(t)}\frac{dG(t)}{dt} [/itex]

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

[itex]lim_{t \rightarrow \infty} G(t) = 0[/itex]

So I thought, as m is also F'(t) for t[itex]\rightarrow \infty[/itex], then:

[itex]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)}[/itex]

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:

[itex]m = - \frac{Q c}{V (a + c)}[/itex]

which seems correct when I try it out graphically.
Let [tex]
f(G) = -\frac{a + Q G(t) + c - \sqrt{(a + Q G(t) + c)^{2} - 4 Q G(t) c}}{2 V}
[/tex] so that [tex]
\frac{dG}{dt} = f(G).
[/tex] What you've done is to compute [tex]
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)}
[/tex] since [itex]f(0) = 0[/itex].

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

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:

[itex]q = lim_{t \rightarrow \infty} F(t) - m t[/itex]

but as I don't have F(t), I'm stuck.
You're stuck; the intercept will in general depend on [itex]F(0)[/itex].
 
  • #3
177
24
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:
[itex]q = ln(G(0))+G(0) \frac{Q}{a}[/itex]
Would dimensional analysis help perhaps?
Thanks
L
 
  • #4
177
24
Just wanted to report my progress with this problem.

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

[itex]- \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)}}[/itex]

What I get is an implicit expression:

[itex]t + constant = T(F) [/itex]

I plugged in the initial conditions to find;

[itex]t = T(F) - T(F_{0}) [/itex]

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:

[itex]\int^{+∞}_{0} G(t) dt[/itex]

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
 

Related Threads on Asymptotes of an implicit function?

  • Last Post
Replies
6
Views
5K
  • Last Post
Replies
4
Views
602
  • Last Post
Replies
12
Views
4K
  • Last Post
Replies
14
Views
4K
  • Last Post
Replies
2
Views
24K
  • Last Post
Replies
7
Views
2K
Replies
4
Views
2K
Replies
5
Views
2K
  • Last Post
Replies
1
Views
2K
Replies
3
Views
4K
Top