Abdoulaye

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3 years, 278 days

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These are replies submitted by Abdoulaye

@Preben Alsholm Thanks it works

@Preben Alsholm Thanks !

your reasonning is very interesting, but in my case I have to choose some random 0<a<1. So I am looking the sup{ x > 0 / Z(x) = 0 }.

According to your graph, the family of solutions is increasing in C=-2000..2000, so we can choose the first curve (C=-2000). Also, since I am looking the x's such that Z(x)=0, we can also focus on C=-2000..0. 

Let me check your solution again. 

Thank you!

@Thomas Dean Thank you. This is a collocation method with one variable. It does not work with my example.

I am trying to fix a uniform grid X = [x0,...,xM], beta=[beta0,...,betaN]  and write the solution in the form below but it is not simple

Any suggestion ?

@Kitonum I have checked your collocation solution with a polynomial approach and it is correct. Thank you. Can we extend your solution to solve a Volterra Integral equation depending on two variables  ? More precisely, I would like to find a numerical solution of :

 

 

 

Z(x, beta) = (5/8)*beta+5/8-(51/8)*beta*exp(-2*x)-(51/8)*exp(-x)+(1/2)*(Int(exp(-z)*Z(x-z, 2*beta*exp(-z)), z = 0 .. x)) , with x>0, beta>0.

   
  In this case we will probably have :

@Kitonum Thank you !
This is what I was looking for and compare with the exact solution. Can you please check my code. The plot does not match the exact solution.

collocation.mw

 

I did not know that we can find an exact solution

@vv Thank you!

@Carl Love Thank you!

@tomleslie  Thank you for your explanation and attached code!

@ilyasweb  I don't understand what you wrote ??

@Kitonum Thanks  Kitonum 13190  !

@radaar 

Thanks @radaar 

I will check also the newton raphson method by approximating integral by middlesum.

@vv 5987

Yes it is true. An interpolation is needed to show the convergence.

@ vv 5982

I am thinking about Riemann Sum for the Integral term :   int ( F( x-z, 2*y*exp(-z) ), z=0..+infinity )

 

Thank you  vv 5982 

Let me check your compuation and see how we can compute F3(x,y). This is the main difficulty of the problem, since we cannot find S2(y) analytically. The ''fsolve or solve function'' will not work for S2(y).   This implies that we will not be able to compute U2(x,y) analytically, so F3(x,y) will not be analytic. 

 

This is the situation : 

F1(x,y)  [ is analytic ]      and     S1 (y) [ is analytic ]       

U1(x,y) [ is analytic ]   implies   F2(x,y)  [ is analytic ]     but     S2 (y) [ is not analytic ]   

U2(x,y) [ is not analytic ]   implies   F3(x,y)  [ is not analytic ]     and    S3 (y) [ is not analytic ]     

                                     . . . . . . . .

Un(x,y) [ is not analytic ]   implies   Fn+1(x,y)  [ is not analytic ]     and    Sn+1 (y) [ is not analytic ]     

 

N.B.  The approach for solving this problem is a Spline Interpolation solution and show convergence. 

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