Talk-Polywell.org

Hello All,

In Joe khackan’s latest paper – he models the current drawn by a Langmuir probe.

Part of doing this means dealing with some nasty math problems.

One expression from the paper is a nasty double integration.

I am struggling it to solve it. If anyone out there is sharp with Mathematica or MATLAB. I would appreciate an analytical solution (algebra) or some code, or excel file which deals with this. I have made some attempts with software to get at it, and the problem comes when I try to integrate to infinity, or just a really high number... I have not looked at limits yet.

Ideally, the code would take an applied voltage, and spit back an expected current. The current would be the solution to this double integration.

I do not think there is an analytical solution. The paper did reference another math paper dealing with this equation, but it converged too slowly. I got that paper and took this to an expert and they are still working on it.

Here it is:



I need to take this and make it algebra. Fortunately I have estimates of what the solution should be. Roughly speaking:

Applied: Solution:
-11 --- 0.1
-9 --- 0.1
-7 --- 0.1
-5 --- 0.3
-3 --- 0.3
-1 --- 0.4
1 --- 0.6
3 --- 0.8
5 --- 1
7 --- 1.7
9 --- 2.3
11 --- 3.1
13 --- 4.2
15 --- 5.2
17 --- 6.5
19 --- 7.6
21 --- 8.8
23 --- 10.3
25 --- 11.5
27 --- 12.9
29 --- 14.3

Can anyone help?

taking a stab n the dark here - and my calculus is very rusty, but looks like the solution might involve changing the bounds of integration
http://math.ucsd.edu/~wgarner/math20b/change_limits.htm


move the terms into the exponent to individual ones? (e^(a+b) to e^a*e^b))?
then move any constant multilipliers, like C, e^(D*E), etc. out to the left of the integral
try changing the bounds of integration, maybe get that cos(theta) out of the exponent? (that reminds me of e^(i(theta))=cos(theta)+i*sin(theta))
that square root term ... pythagorean, or the result of some inner product of some sort?
* you got both R and dR in the integral. perhaps thats a hint.
* you got an e^(in the integral). that looks like the main thing - the base of the integral, so to speak; looks like its an integral of an exponential function. perhaps a rule related to that, also considering the bullet just above.
** that is, maybe you got something like d(x) = f(d^2(x)).
* i'm looking for an integration by parts solution and i don't see it.
**in any case the radical is really messing me up. maybe chaging the bounds of integration, somehow) ( http://math.ucsd.edu/~wgarner/math20b/change_limits.htm )

maybe reverse the order of integration - do theta first, and check out the last two on http://en.wikipedia.org/wiki/List_of_in ... _functions

... and get back a modified bessel function. great.
yeah, no, that one beyond me. it would seem matlab would be the way to go. the modern version of "just use a calculator".
there's a free alternative to matlab, "octave": http://www.gnu.org/software/octave/

You might try not integrating to infinity, but only up to the Debye length. That is sufficiently messy looking that I would do it numerically. But I'm pretty lazy when it comes to integrating. lol.

[edit 1] I just noticed the angular dependence. What is the angle wrt? For that matter, is R a distance or just some variable he introduced?

[edit 2] Also, I can one constant that can be eliminated: E. Why not just absorb that into the C constant?

Have you tried Wolfram Alpha? If you sign up for the pro trial, you get additional processing time, which may be able to solve it for you.

http://www.wolframalpha.com/input/?i=integral

I took your equation and turned it into the following, replacing E with G because it thinks any standalone e or E means Euler's number, and theta with T, because I wasn't sure how well it handles Greek:
Just plug that into the "function to integrate" to solve the first integral. Then, assuming it actually gets a result for you, plug the result back into the solver, replacing x with y, and T with x, because the default expects x to be the value of integration. (You can also click the little "variable" text and change it.)

x*(x^2+B*(125-V))^0.5*C*e^(D*x^2)*e^(D*G)*e^(D*F*x*cos(T)))

H=C*e^(D*G), I = D*F, J = B*(125-V)

H*x*(x^2+J)^0.5*e^(D*x^2)*e^(I*x*cos(T)))

cleaner.

what happens when you change the limit of integration - R to what the bound would be at a hypothetical voltage (say, Vh)?

R=(125-Vh)^(1/2)*B(1/2)
R^2/B=125-Vh
125-R^2/B=Vh

and then you'd integrate with respect to dVh*d(theta) over 0<Vh<Va or Va<Vh<infinity, depending on whether Va<Vh

also curious if you change the limits from polar to cartesian. i.e. change dRdT to dXdY, thus turning the Rcos(T) term into simply "X".

Edit: dVh/dR = -2R/B = -2(125-Vh)^(1/2)*B^(-1/2)

So to convert the integral to with respect to dVh, one has to "absorb" the right side of the equation, which is the same thing as dividing by the right side times a dR, and multiplying by Vh ( thus multiplying by x/x = 1)....

And then we substitute the R's with what they are in Terms of Vh, and see if we can simplify
Just thinking out loud here.

When I try to visualize it, R is a radius. It's 3-d, though. The detector is directional in nature. Theta is the angle from the line that represents its direction.

There's two functions here, the e^ part is a "transmission" function, so to speak - the part in the ^() is about how much information is lost as the distance it has to travel increases. And you'll notice one term is r^2 - a function if the surface area. And d is negative, so it's a decreasing one. I'm curious what e^(-d) is. And then there's a term that shows better ( or worse) transmission along the direction of the probe.

The other function - the one not in the e - is related to the data actually being transmitted - what it's value is at that point in space - it's the "original" that gets copied at exponentially decaying strength with each copy. And you'll notice it's close to r^2 - the surface area, and get asymptotically closer as r increases.

I want to change the limits of integration show that the part in the e^() is constant, and thus I can drop it. But that means moving the dx's up there, which is well, unconventional. And it involves the chain rule somehow.