Find the root of integral equation

Click For Summary
SUMMARY

The forum discussion revolves around solving the integral equation ∫ dR / (1 + c*r)^(a/r) = d, with limits from 0 to Rmax, where Rmax^2 = [u]^2 - α^2. The user seeks to implement a C function to find the value of α given d, acknowledging the complexity due to α's presence in both the integral and its limits. The recommended approach involves using iterative methods, specifically the Newton-Raphson method, to estimate α by differentiating with respect to α and adjusting based on the resulting error.

PREREQUISITES
  • Understanding of integral calculus and limits of integration
  • Familiarity with numerical methods, particularly iterative methods
  • Knowledge of the Newton-Raphson method for root finding
  • Basic programming skills in C for implementing the solution
NEXT STEPS
  • Research the Newton-Raphson method for root finding in numerical analysis
  • Study differentiation techniques for integrals involving variable limits
  • Explore iterative methods for solving equations with unknowns in integrals
  • Learn about the fundamental theorem of calculus and its applications in numerical methods
USEFUL FOR

Mathematicians, physicists, and software developers involved in numerical analysis, particularly those working with integral equations and iterative root-finding methods.

TheSource007
Messages
14
Reaction score
0

Homework Statement


Hi everyone.
I have encountered a weird equation while doing some research and I have no idea how to solve it.
The equation goes like this

∫ dR / (1+ c*r) ^ (a/r) = d, limits of integration are from 0 to Rmax,
where Rmax ^2 = ^2 - α^2, where u is a constant value of r.
and r^2 = R^2 + α^2,
c is a constant and d is my independent variable.

Homework Equations



The Attempt at a Solution


I want to do a C function that takes one value of d and returns a value of α. I thought of doing
∫ (1 / (1+ c*r) ^ (a/r) ) - d = 0 and try to find the root, or do minimization. The problem is that α also appears on the limits of integration, and even if it wasn't there, I still don't know how to do either.
Also, for any value of d, there is a value of α, but r changes from r = α to r= u.

I would appreciate any help. Thanks
 
Physics news on Phys.org
Even though alpha appears both inside the integral and in the limit, you can differentiate wrt alpha. That allows you to use standard iterative methods: given a trial alpha, find the resulting d, calculate the error and use the derivative to estimate an adjustment to alpha.
... and it will have the standard pitfalls.
 
Could you refer me to sources where they explain more on the iterative methods that can do this sort of problem? I have no experience in finding a numerical value for an unknown that appears inside an integral (and inside the limits for that matter)
 
TheSource007 said:
Could you refer me to sources where they explain more on the iterative methods that can do this sort of problem? I have no experience in finding a numerical value for an unknown that appears inside an integral (and inside the limits for that matter)

A basic method when you know the derivative function is Newton-Raphson. See e.g. http://en.wikipedia.org/wiki/Newton's_method.
You understand how to expand ##\frac d{dx}\int^{f(x)}g(x, z).dz##, right?
 
haruspex said:
You understand how to expand ##\frac d{dx}\int^{f(x)}g(x, z).dz##, right?
Is that just the fundamental theorem of calculus?

And also, is there any method to find the root that does not involve a first trial of alpha?
Thanks
 
TheSource007 said:
Is that just the fundamental theorem of calculus?
It's a little more complicated because x is in two places.
And also, is there any method to find the root that does not involve a first trial of alpha?
Thanks
No.
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
Replies
1
Views
2K
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K