Abdoulaye

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These are questions asked by Abdoulaye

 

Hello,

How to show the y-axis numbers in the second plot ?   y is in [0,3]. 

I try the following but it shows a negative axis. 

display(P, L, axis = [tickmarks = [5, subticks = 3], thickness = 2]);  this is not correct

The file is attached.

Thanks


 

Download Curves.mw

 

 

Hello,

I need help to approximate Fn(x,y) and Sn(y) for n>=3 to see the convergence. 

Here is the problem:

I have an integral equation Fn(x, y) defined by

F0(x,y) = 0,

F1(x,y) = (9/8)*(y+1) - (21/8)*( y*exp(-2*x) + exp(-x) )

Fn+1(x,y) = (9/8)*(y+1) - (21/8)*( y*exp(-2*x) + exp(-x) ) + (3/4)*int ( Fn( x-z, 2*y*exp(-z) ), z=0..Un(x) ), where the function Un(x) is defined below.

where I suppose that Fn(x,y) >=0 and y>=0 and x>=0. I am iterating the integral equation and find the zeros Sn(y) solutions of Fn(S(y) , y) = 0. The goal is to show that the sequence Fn(x,y) converges to some F(x,y) and Sn(y) converges to some S(y). We can also plot the curves Sn(y) to show the convergence. Notice that Sn(y) >= 0 is increasing in n and decreasing in y.

What I did :

If n = 0, I solve F1(x,y) = 0 and find analytically x = S1(y) = -ln((1/14)*(-7+sqrt(84*y^2+84*y+49))/y). Notice that F1(x,y) = 0 is quadratic in exp(-x) and then F1(x,y) = 0 is easy to solve. S1(y) is the positive function for y>=0.

If n = 1, we need to solve F2(x,y) = (9/8)*(y+1) -(21/8)*( y*exp(-2*x) + exp(-x) ) + (3/4)*int ( F1( x-z, 2*y*exp(-z) ), z, 0, U1(x) ). The function U1(x) is obtained as follows :

i) We define g( U ) = U + S1( 2*y*exp(-U) )

ii) We find the function U1(x,y) solution of g( U1(x,y) ) = x

 

A generalization of Un(x) is given by

i) Define g( U ) = U + Sn( 2*y*exp(-U) )

ii) Find the function Un(x,y) solution of g( Un(x,y) ) = x

The curve S2(y) :

I find the analytic function F2(x, y), but it is impossible to solve F2(x,y) = 0 and find x = S2(y) analytically. So I choose an interval [0, 10] for y and discretize that interval. Now at each point yi in [0, 10], the function F2(x, yi) depends only on x. I apply the bisection method (dichotomic method) and find the value xi such that F2(xi, yi) = 0. The data (yi, xi) gives the curve S2(y). I plot the data (yi, xi) and (y, S1(y)) in the same curve.

Question:

It is impossible to iterate the process below and find F3(x, y), F4(x,y), etc... because we cannot find U2(x), U3(x), etc... analytically. Any suggestions or helps to plot the curves (y, Sn(y)) for n = 3, 4, etc. to see the convergence ?

Thanks !

Hello,

How to write the Newton's method and Bisection method in Maple of a function depending on a parameter ? The goal is to solve the equation F(y, eta) = 0 and find the root y(eta) using the Newton's method or Bisection method. An example is attached. 

Thank you

 

Hi, I would like to plot multiple curves with arrows pointed at each curve. I don't know how to plot the arrows with values. A similar figure is attached. Thanks for helping . . .

Hello,

Questions

1)  Can we compute the error  ( x(t-2*Pi) - x(t) )  without plotting  abs(  x( t-2*Pi ) - x(t) ) ?

     i.e. using norm 1, 2 or infinity  ?

 

2)  what is op( [ a, b, c] , F ),  where F is a piecewise function ?


Download example.mw

 

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