Abdoulaye

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These are replies submitted by Abdoulaye

@vv 5982. Thanks !

Your idea is interesting but I am not familiar with a Spline Interpolation. Here Fn at the data points (xi, yi) is of size M*N but can we get analytical expressions for all  Fn(x,y)  and Un(x,y) ?  Can you please Show F2(x,y), U1(x,y) and S2(y) with for example  the following  data ? 

 

 

@Rouben Rostamian  OK. Thank you !

@Rouben Rostamian  Do you have an idea to how to iterate numerically this process ? 

@Rouben Rostamian  Thank you. Yes, U1 depends on (x,y). I have attached the file
 

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researt:

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F1 := proc (x, y) options operator, arrow; (9/8)*y+9/8-(21/8)*y*exp(-2*x)-(21/8)*exp(-x) end proc:

(9/8)*y+9/8-(21/8)*y*exp(-2*x)-(21/8)*exp(-x)

(1)

S1 := proc (y) options operator, arrow; solve(F1(x, y), x)[1] end proc:

-ln((1/14)*(-7+(84*y^2+84*y+49)^(1/2))/y)

(2)

``

g := proc (U) options operator, arrow; U+S1(2*y*exp(-U)) end proc;

proc (U) options operator, arrow; U+S1(2*y*exp(-U)) end proc

(3)

U1 := proc (x) options operator, arrow; solve(g(U) = x, U)[1] end proc:

ln((1/14)*(3*exp(x)+(9*(exp(x))^2+168*y*exp(x)+336*y^2)^(1/2))*exp(x)/(2*y+exp(x)))

(4)

``

F2 := proc (x, y) options operator, arrow; (9/8)*y+9/8-(21/8)*y*exp(-2*x)-(21/8)*exp(-x)+(3/4)*(int(F1(x-z, 2*y*exp(-z)), z = 0 .. U1(x))) end proc;

proc (x, y) options operator, arrow; (9/8)*y+9/8-(21/8)*y*exp(-2*x)-(21/8)*exp(-x)+(3/4)*(int(F1(x-z, 2*y*exp(-z)), z = 0 .. U1(x))) end proc

(5)

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F2(x, y);

(9/8)*y+9/8-(21/8)*y*exp(-2*x)-(21/8)*exp(-x)+(9/32)*(3*(9*exp(2*x)+168*y*exp(x)+336*y^2)^(1/2)*exp(2*x)*ln((3*exp(x)+(9*exp(2*x)+168*y*exp(x)+336*y^2)^(1/2))*exp(x)/(2*y+exp(x)))-3*(9*exp(2*x)+168*y*exp(x)+336*y^2)^(1/2)*ln(2)*exp(2*x)-3*(9*exp(2*x)+168*y*exp(x)+336*y^2)^(1/2)*ln(7)*exp(2*x)+6*(9*exp(2*x)+168*y*exp(x)+336*y^2)^(1/2)*exp(2*x)*y+9*exp(3*x)*ln((3*exp(x)+(9*exp(2*x)+168*y*exp(x)+336*y^2)^(1/2))*exp(x)/(2*y+exp(x)))-9*ln(2)*exp(3*x)-9*ln(7)*exp(3*x)+18*exp(3*x)*y-3*(9*exp(2*x)+168*y*exp(x)+336*y^2)^(1/2)*exp(2*x)-9*exp(3*x)-168*y*exp(2*x)-336*y^2*exp(x)+7*(9*exp(2*x)+168*y*exp(x)+336*y^2)^(1/2)*exp(x)+14*(9*exp(2*x)+168*y*exp(x)+336*y^2)^(1/2)*y+21*exp(2*x)+42*y*exp(x))*exp(-2*x)/(3*exp(x)+(9*exp(2*x)+168*y*exp(x)+336*y^2)^(1/2))

(6)

``


 

Download question.mw

@Rouben Rostamian  

It is a typo. I have corrected it 

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

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

 

Thanks !

 

@Carl Love Thank you Carl,

 

Can you please check the infnorm from the worksheet, I got   0.3096747114  ? See attached the file norm.mw

 norm.mw

@Carl Love  That's true ! Thanks !

Questions

1)  Can we compute the error  ( x(t-2IPi) - 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 ?

@Carl Love OK I see... 

1) Do I need to add  1e-7  if I  alternate <= with < ?

@Carl Love  Yes it is true but I don't understand your interval of time for the graph ?  It is related to the derivative ?

@Carl Love  Thank you Carl !   I want to show that the function is close enough to being 2*Pi periodic and then find the error :

abs ( X(t-2*Pi) - X(t) ) : 

For example, if I take the last curve, I will obtain:

 

Why  X(t-2*Pi) - X(t)    is not small ?  why it depends to t  ?

@Carl Love  Thank you Carl, here is the function but I don't know how to prove that is

2π-periodic ? 

@Carl Love  Thanks it works !!!

@vv  Thanks ,  it works !

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