Asymptotes for numerical solution of ordinary differential equation

[Davide82]

Hi!

I have a differential equation coming from Boltzmann transport equation which is a bit complicated and should be solved numerically instead of analytically.
I managed to get a plot using Runge-Kutta with software Derive.
The equation is in the second attachment.

In the papers of my teacher (where it states that solution has to be found by numerical integration) there is a plot where it shows a clear asymptote at y = 0.151 (attached).

How could I find this same result? Is there a way to find the asymptote? I can not let the computer calculate the function for x = infinite!

 

[Mark44]

Does the graph show the solution function? It's hard to tell because the vertical axis is labelled X_n and the horizontal axis is labelled x = \Delta m/T.

For other readers, the other thumbnail shows the differential equation to be
\frac{dy}{dx}= \frac{.255}{x^5}(12x + 6x^2 + x^3)[e^{-x} - y(1 + e^{-x})]

 

[Davide82]

First of all I apologize for not having written my equation with in-line TeX; lately I was writing in the computer software section of this forum where TeX is forbidden and so I forgot to use it here.

Yes the graph shows the "y" function.

I was wondering how I can say that my solution isn't going any lower than 0.151. Maybe there is a way to find the asymptote without having the analytical solution?

 

[hunt_mat]

One thing that you might try and do is look for terms in your equation which will be very small for large x, the exponential terms will be very small for large x and I might look at the equation:

<br /> \frac{dy}{dx}=-\frac{0.255}{x^{5}}(12x+6x^{2}+x^{3})y<br />

This can be solved more easily than the original equation.

 

[HallsofIvy]

But a horizontal asyptote of .151 is impossible. At a horizontal asymptote, the derivative must go to 0. As x goes to infinity, as mat_hunt said, the exponentials go to 0 so the derivative approaches
\frac{dy}{dx}= -\frac{0.255}{x^5}(12x+ 6x^2+ x^3)y
which does not go to 0 unless y is going to 0.

 

[Davide82]

Ok, I just tried to resolve that new equation; indeed Derive can solve analytically that equation and came with a solution but... this solution gives y = 10^{-478} when x = 10...

If there is no asymptote, you mean that the plot I posted is misleading and the asymptote is only an illusion but the function is infact decreasing forever?

 

[hunt_mat]

What you have seemed to have shown is that the asymptote is 0 rather than the one on the graph (what is you initial condition?).

HallofIvy was straight on the money, when he said that the horizontal asymptote is defined by dy/dx=0, you can plot the curve for y as a function of x as it reaches this asymptote an it appears to be:

<br /> y=\frac{e^{-x}}{1+e^{-x}}<br />

 

[Davide82]

Thank you to all. So the asymptote shown in the first image maybe was due to physical reasons and not to mathematical ones.