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Problem of integration, the code run but the solution is not satisfying

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Tommaso Attili
Tommaso Attili 2019 年 2 月 18 日
閉鎖済み: MATLAB Answer Bot 2021 年 8 月 20 日
Hi all! I coded the Eq.C.4 to find η(t) and the code works, but, as you can see from the fig. attached, my solution does not match with the one proposed by a scientific paper for the same equation and input data. Any suggestions?
Screenshot (13).png
Roi=916;
Row=1024;
R=Row/Roi;
r=50;
d=50;
%column geometry
H0=50;
l=5;
b=5;
a=(2*l*H0/pi)^0.5;
Area=pi*a^2;
Vol=Area*H0;
%Dinamic of the block
Omega=1;
va=(3*(pi)^(3/2))/(16*sqrt(2))*(Omega*sqrt(l*H0))/(1+pi/4*R*(l/b));
%Surfaceelevation
E=zeros(1,301);
t=0:0.1:30;
for i=1:301
O=@(x,y)(besselj(0,x.*y.*a/d)).*((1-y.^2).^0.5).*y.*besselj(0,x*r/d).*((x.*tanh(x)).^0.5).*sin((((9.81/d*t(i).^2).*x).*tanh(x)).^0.5).*x;
E(i)=integral2(O,0,2000,0,1,'Method','iterated','AbsTol',1e-10,'RelTol',1e-10)
end
E1=-2*(a.^3)/(d^3.5)/(9.81^0.5)*va*d*E;
  6 件のコメント
Are Mjaavatten
Are Mjaavatten 2019 年 2 月 20 日
I am not able to reproduce the figure from the paper either. Your curve at least resembles the original. Mine is totally different. Here is my attempt:
h = 50;
b = 25.2;
Va = 6.21;
g = 9.81;
r = 50;
a = b/2/h;
int1 = @(x) besselj(0,r/h*x).*(sin(a*x)-a*x.*cos(a*x))/a^3./x.^2;
int2 = @(x,t) sqrt(x.*tanh(x)).*sin(g*t^2/h*x.*tanh(x));
integrand = @(x,t) int1(x).*int2(x,t);
N = 200;
t = linspace(0,5,N);
I = zeros(1,N);
for i = 1:N
fun = @(x) integrand(x,t(i));
I(i) = integral(fun,0,200);
end
eta = -2*a^3*Va/sqrt(g*h)*I;
figure; plot(t,eta)
Increasing the upper integration limit dampens out the wiggles, but the general shape is unchanged. There seems to be two possible explanations:
1) I have made some mistake (misunderstanding or bug).
2) The figure in the original paper is based on different parameters and/or equations.
I leave it to you to try to figure this out.
Tommaso Attili
Tommaso Attili 2019 年 2 月 20 日
Thanks! Yes I plotted your solution. At this point I really don't have idea. Thanks for that.

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