Wrong start of the curve in double integral

Question

Hexe 2023-1-29

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1902675-wrong-start-of-the-curve-in-double-integral

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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Answer 1

C B 2023-1-29

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1902675-wrong-start-of-the-curve-in-double-integral#answer_1158885

在 MATLAB Online 中打开

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-')