Wrong start of the curve in double integral

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Hexe
Hexe 2023-1-29
评论： Hexe 2023-1-29
Hi! I solve the double integral and it shows a right behavior of the curve, but it starts from different points at different parameters. But it should always start from the point (0,1). What is wrong?
n = 0.1 ;
t = 1;
r = 1;
s = 0:0.01:1;
b=sqrt(2*t)/r;
fun = @(x,z,k) exp(-2.*n.*t.*x.^2).*exp(-z.^2).*(erf(((z+x.*k./r)./(2.*b)))+erf(((z-x.*k./r)./(2.*b)))-z./(sqrt(pi).*b).*(exp(-((z+x.*k./r)./(2.*b)).^2)+exp(-((z-x.*k./r)./(2.*b)).^2)));
f3 = arrayfun(@(k)integral2(@(x,z)fun(x,z,k),0,Inf,0,1),s);
Cor = ((sqrt(2*n*t))/(erf(sqrt(2*n*t))*(atan(1/(2*b))-(b/(2*(b^2+0.25))))))*f3;
plot(s,Cor,'b-')

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采纳的回答

C B
C B 2023-1-29
Ran in:
n = 0.1 ;
t = 1;
r = 1;
s = 0:0.01:1;
b=sqrt(2*t)/r;
fun = @(x,z,k) exp(-2.*n.*t.*x.^2).*exp(-z.^2).*(erf(((z+x.*k./r)./(2.*b)))+erf(((z-x.*k./r)./(2.*b)))-z./(sqrt(pi).*b).*(exp(-((z+x.*k./r)./(2.*b)).^2)+exp(-((z-x.*k./r)./(2.*b)).^2)));
f3 = arrayfun(@(k)integral2(@(x,z)fun(x,z,k),0,Inf,0,1),s);
Cor = ((sqrt(2*n*t))/(erf(sqrt(2*n*t))*(atan(1/(2*b))-(b/(2*(b^2+0.25))))))*f3;
plot(s,Cor,'b-')
n = 0.1 ;
t = 1;
r = 1;
s = 0:0.01:1;
b=sqrt(2*t)/r;
fun = @(x,z,k) exp(-2.*n.*t.*x.^2).*exp(-z.^2).*(erf(((z+x.*k./r)./(2.*b)))+erf(((z-x.*k./r)./(2.*b)))-z./(sqrt(pi).*b).*(exp(-((z+x.*k./r)./(2.*b)).^2)+exp(-((z-x.*k./r)./(2.*b)).^2)));
f3 = arrayfun(@(k)integral2(@(x,z)fun(x,z,k),0,1,0,1),s);
Cor = ((sqrt(2*n*t))/(erf(sqrt(2*n*t))*(atan(1/(2*b))-(b/(2*(b^2+0.25))))))*f3;
Cor = Cor + (1 - Cor(1));
plot(s,Cor,'b-')
  1 个评论
Hexe
Hexe 2023-1-29
Dear Chetan Bhavsar!
Thank you very much. Now it works as it should.
Sincerely
Olha.

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