Numerically integrating a symbolic equation in a for loop?

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Harvey Rael
Harvey Rael 2018-6-25
评论： Torsten 2018-6-26
Hi! I'll try to explain qualitatively first what I am trying to do before then giving code. I have an equation that depends on a variable (theta) that changes (though how it changes doesn't matter). However, theta's end point depends on another variable N, and N is significant. I have a function that needs to be numerically integrated in terms of theta, to its end point which is pi/2+2piN. I want to change N with a for loop and store the outputs to be plotted.
Here's my code:
clc;clear;
% Define constants
AU = astroConstants(2);
mu = astroConstants(4);
% Define start and finish parameters for the exponential sinusoid.
r1 = AU; % Initial radius
psi = pi/2; % Final polar angle of Mars/finish transfer
phi = pi/2;
r2 = 1.5*AU; %Final radius
k2=0.2;
N = 0:1:5;
k1 = zeros(size(N));
y = zeros(size(N));
for i = 1:length(N)
k1(i) = sqrt( ( (log(r1/r2) + sin(k2*(psi + 2*pi.*N(i)))*tan(0)/k2) / (1-cos(k2*(psi+2*pi.*N(i)))) )^2 +
tan(0)^2/k2^2 );
k0 = r1./exp(k1(i)*sin(phi)); %The parameters of the exponential sinusoid
R = @(theta) k0*exp(k1(i)*sin(k2.*theta + phi)); %Equation of the exponential sinusoid
theta_dot = @(theta) sqrt((mu./(R(theta).^3))./((tan(0))^2 + k1(i)*k2^2*sin(k2*theta + phi) + 1));
z = @(theta)1./theta_dot(theta); %change of variable name for ease of use
y(i) = integral(z,0,(psi+2.*pi.*N(i))) - 94608000; %function to be plotted
end
The error message I keep getting is :
"Error in integral (line 88) Q = integralCalc(fun,a,b,opstruct);
Error in Untitled2 (line 23) y(i) = integral(theta_dot,0,(psi+2.*pi.*N(i))) - 94608000; %function to be plotted"
Anyone know where I am going wrong? When I have N as a single number, everything works fine.
EDIT : As @Torsten pointed out I forgot add in a couple of (i)'s in my code - so I have added them in. Error message persists

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采纳的回答

Harvey Rael
Harvey Rael 2018-6-25
Fixed it - simply needed to flick the 'ArrayValued' setting in integral to true.

更多回答（1 个）

Torsten
Torsten 2018-6-25
In the definition of "R" and "thetadot", you use "k1" instead of "k1(i)".
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Harvey Rael
Harvey Rael 2018-6-25
Post your code then? Is y a vector or a single point solution? Because I contiually get errors
Torsten
Torsten 2018-6-26
function main
AU = 2.0;
mu = 0.5;
% Define start and finish parameters for the exponential sinusoid.
r1 = AU; % Initial radius
psi = pi/2; % Final polar angle of Mars/finish transfer
phi = pi/2;
r2 = 1.5*AU; %Final radius
k2 = 0.2;
N = 0:1:5;
k1 = zeros(size(N));
y = zeros(size(N));
for i = 1:length(N)
k1(i) = sqrt( ( (log(r1/r2) + sin(k2*(psi + 2*pi.*N(i)))*tan(0)/k2) / (1-cos(k2*(psi+2*pi.*N(i)))) )^2 +...
tan(0)^2/k2^2 );
k0 = r1./exp(k1(i)*sin(phi)); %The parameters of the exponential sinusoid
R = @(theta) k0*exp(k1(i)*sin(k2.*theta + phi)); %Equation of the exponential sinusoid
theta_dot = @(theta) sqrt((mu./(R(theta).^3))./((tan(0))^2 + k1(i)*k2^2*sin(k2*theta + phi) + 1));
z = @(theta)1./theta_dot(theta); %change of variable name for ease of use
y(i) = integral(z,0,(psi+2.*pi.*N(i))) - 94608000; %function to be plotted
end
end
Best wishes
Torsten.

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