Optimizing Decompression Stop Depth with Partial Derivatives in PDE Simulation

  • Thread starter Thread starter rmurray
  • Start date Start date
  • Tags Tags
    Pde
rmurray
Messages
2
Reaction score
0
I haven't worked with partial derivatives since high school 25 years ago so I'm quite a bit rusty and need a little guidance. I'm reading through a paper and would like to write a program to simulate it. The equations are:

Eq 1. Eq 13. M=f*Po+f*(Pb-Po)*(1-ln(e)^(-0.693*time/Halftime)
Eq 10. a=3.25*Halftime^-0.25
Eq 11. Mo=152.7*Halftime^-0.25
Eq 12. D=(M-Mo)/a
Eq 13. D=(f*Po+f*(Pb-Po)*(1-ln(e)^(-0.693*time/Halftime)-Mo)/a

Where:
f= gas fraction
Po= initial pressure
Pb= bottom pressure

I'm having problems setting up equation 13 where:
"The partial derivative of Eq 13 with respect to Halftime is the rate of change of hte decompression stop depth with respect to the tissue halftime for any bottom time(time), depth(Pb), and M value. When this partial derivative equals zero, the stop depth is minimumized and we get the Halftime of hte deepest stop depth. When we substitute the critical halftime value in Eq 13 we get the deepest stop depth."

So Halftime can be expressed as a function of time, but I can not figure out where to get to this point. I can post a link to the paper if anyone needs further clarification.

Any help would be greatly appreciated!
Thank you,
Rob
 
Last edited by a moderator:
Physics news on Phys.org
So after going through and trying a few things, this is the best I can come up with. I'm pretty sure I've gone in the wrong direction, but it was an attempt to solve this:

(10785*x*12833^(693*y/1000*x)+(-29896*y-10785*x)*4721^(693*y/1000*x))/(140205*x^(7/4)*12833^(693*y/1000*x)

Does not seem any easier to solve, so I'm rethinking the process. Solving for x is actually solving for the half-life with y being a function of time.
 
This appears to be a problem in differentiating. There is no "differential equation" here.
 

Similar threads

Replies
38
Views
46K
Back
Top