Wrong plot upon right calculation

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Hexe
Hexe am 6 Dez. 2022
Kommentiert: Star Strider am 7 Dez. 2022
Hi,
I have the next problem. Here is a code for calculation of the function K(s) with integral fun1. When s=0 the K(s)=1 at all values of parameters n,t,r1. And when I calculate it by vpa for s=0 I obtain this K=1 at all parameters. Thus, the curve K(s) must always start from the point (0,1). But when I make a plot, the curve start from some different value. Please, help me: what do I do wrong? Calculation of this integral by syms package gives right plots, but it calculates too long and I want to do it by the code below. Thank you.
clear all, close all
D=1;
n=0.01;
r1=1;
t=1;
r=(r1^2+4*t/3);
%s=0;
s=0:0.1:10;
for i = 1:length(s)
k=s(i);
fun1 = @(x)((((1-(((1-x.^2).*k.^2)/(6*r))).*besseli(0,(((1-x.^2).*k.^2)/(6*r))/2))+(((((1-x.^2).*k.^2)/(6*r))).*besseli(1,(((1-x.^2).*k.^2)/(6*r))/2))).*exp(((k.^2)/(12*r)-2*n*t).*x.^2));
f1(i,:)=integral(fun1,-1,1);
K=(exp(-k.^2/(12*r))*D*sqrt(2*n*t)/(sqrt(pi)*erf(sqrt(2*n*t))))*f1;
end
plot(s,K,'b-');
%vpa(Fun,5) % calculation of K(s,t) at certain s. At s=0 must be next K(s,t)=1

Akzeptierte Antwort

Star Strider
Star Strider am 6 Dez. 2022
If it needs to begin at (0,1), the only changes required are that in the ‘K’ assignment, the current value of ‘f1’ needs to be referenced specifically and ‘K’ needs to be indexed:
K(i) = (exp(-k.^2/(12*r))*D*sqrt(2*n*t)/(sqrt(pi)*erf(sqrt(2*n*t))))*f1(i);
See if those changes produce the result you want —
D=1;
n=0.01;
r1=1;
t=1;
r=(r1^2+4*t/3);
%s=0;
s=0:0.1:10;
for i = 1:length(s)
k=s(i);
fun1 = @(x)((((1-(((1-x.^2).*k.^2)/(6*r))).*besseli(0,(((1-x.^2).*k.^2)/(6*r))/2))+(((((1-x.^2).*k.^2)/(6*r))).*besseli(1,(((1-x.^2).*k.^2)/(6*r))/2))).*exp(((k.^2)/(12*r)-2*n*t).*x.^2));
f1(i,:)=integral(fun1,-1,1);
K(i)=(exp(-k.^2/(12*r))*D*sqrt(2*n*t)/(sqrt(pi)*erf(sqrt(2*n*t))))*f1(i);
end
plot(s,K,'b-');
.
  2 Kommentare
Hexe
Hexe am 7 Dez. 2022
Dear Star Strider,
Thank you very much for you help. Now it works as it has to.
Sincerely
Olha.
Star Strider
Star Strider am 7 Dez. 2022
As always, my pleasure!

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