Wrong start of the curve in double integral
6 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Hexe
am 29 Jan. 2023
Kommentiert: Hexe
am 29 Jan. 2023
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-')
0 Kommentare
Akzeptierte Antwort
Chetan Bhavsar
am 29 Jan. 2023
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-')
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Calculus finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!

