Skip to main content
1-Visitor
September 26, 2018
Solved

Using rkfixed in mathcad prime 2.0

  • September 26, 2018
  • 3 replies
  • 9748 views

Greetings!

I want to solve this problem using rkfixed function in mathcad prime 2.0. Help would be highly appreciated!

I am having trouble understanding the derivative matrix of this function

Best answer by LucMeekes

OK I found them (they are there today, they weren't yesterday. I wonder where you grabbed the second from...because both equations define v(z)*x'(z). And I wonder why you use d v(z)/dz on one occasion, and all other derivatives use prime notation:  v'(z) and x'(z)...?

Anyway, here goes:

LM_20180927_rkfixed1.png

LM_20180927_rkfixed2.png

LM_20180927_rkfixed3.png

With such results choosing 47, 470 or 4700 steps makes no difference in the graphs.

Possibly not the solution you expected.  I suggest you check your differential equations once more.

 

Success!
Luc

3 replies

23-Emerald IV
September 26, 2018

I think first you have to clear up what the function x'(z) is. It is not defined. Do you need to solve two functions ( v and x) simultaneously (which isn't impossible, but requires a set of functions to rkfixed.)?

I suggest you look up the example  for "Solving a First-Order ODE Initial Value Problem" in the help. That should explain the usage of rkfixed.

 

Success!
Luc

Asad1-VisitorAuthor
1-Visitor
September 27, 2018

Right! I have two equations which contain v(z), v'(z), x(z) and x'(z)

the rest constants are known

initial conditions are known: v(0)=1; and x(0)=0.013

I need to solve it from ini=0 to end=0.47

I need assistance in forming the derivative matrix. It would be highly appreciated!

 

Further in the first equation we can assume the value for variable x'(z) as a constant as 0.5 and put the values of v(z) and v'(z) in the second equation to obtain x(z) and x'(z) then iterate the procedure.

I need to use rkfixed to do this.

 

Thanks!

23-Emerald IV
September 27, 2018
If x'(z)=0.5 then x(z)=0.5z+c
With x(0)=0.013 that makes c=0.013.
So x(z)=0.5z+0.013.
Who needs rkfixed?

Luc.
24-Ruby IV
September 27, 2018

Why do you not use units?

Units.png

Asad1-VisitorAuthor
1-Visitor
September 27, 2018

because most of the times they cause a hindrance

Asad1-VisitorAuthor
1-Visitor
September 27, 2018

I tried to solve the first equation but it gives an error saying about derivative function D that this function cannot be used here.

 

File is attached.

24-Ruby IV
September 27, 2018

One more try - with Odesolve - why not!

Odesolve.png