Matlab integration error: Integrand output size does not match the input size

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JJ on 27 Sep 2014
Commented: JJ on 28 Sep 2014
I am attempting a 4th order nested integration in Matlab. The function is as follows:
second_integral = @(r_dash, phi_dash, r, phi) ( (exp(-1i * k * r_dash * cos(phi_dash) * sin(theta)) .* exp(-1i * k * sqrt(r.^2 + r_dash.^2 - 2*( r * r_dash .* cos (phi_dash - phi))))) / (2 .* pi .* sqrt(r.^2 + r_dash.^2 - 2.*( r .* r_dash .* cos (phi_dash - phi))) ) );
With the constants in this particular example being:
a_1 = 7e-5;
a_2 = 5e-5;
k = 1;
theta = 0; % can also be 0.5, 1, 1.5
The integration is being done as
integrated_second_integral =integral(@(r_dash)integral3(@(phi_dash,r,phi)second_integral(r_dash,phi_dash,r,phi),0,(2*pi),0,a_2,0,(2*pi)),0,a_1,'ArrayValued',true);
When the integration command is run I get the following errors:
Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 2.725290e-19.
> In @(r_dash,phi_dash,r,phi)((exp(-1i*k*r_dash*cos(phi_dash)*sin(theta)).*exp(-1i*k*sqrt(r.^2+r_dash.^2-2*(r*r_dash.*cos(phi_dash-phi)))))/(2.*pi.*sqrt(r.^2+r_dash.^2-2.*(r.*r_dash.*cos(phi_dash-phi)))))
In @(phi_dash,r,phi)second_integral(r_dash,phi_dash,r,phi)
In integral3>@(y,z)FUN(x(1)*ones(size(z)),y,z) at 139
In funfun/private/integral2Calc>integral2t/tensor at 229
In funfun/private/integral2Calc>integral2t at 56
In funfun/private/integral2Calc at 10
In integral3>innerintegral at 138
In funfun/private/integralCalc>iterateScalarValued at 314
In funfun/private/integralCalc>vadapt at 133
In funfun/private/integralCalc at 76
In integral3 at 122
In @(r_dash)integral3(@(phi_dash,r,phi)second_integral(r_dash,phi_dash,r,phi),0,(2*pi),0,a_2,0,(2*pi))
In funfun/private/integralCalc>iterateArrayValued at 157
In funfun/private/integralCalc>vadapt at 131
In funfun/private/integralCalc at 76
In integral at 89
In this_matlab_script at 61
Error using integral2Calc>integral2t/tensor (line 242)
Integrand output size does not match the input size.
Error in integral2Calc>integral2t (line 56)
[Qsub,esub] = tensor(thetaL,thetaR,phiB,phiT);
Error in integral2Calc (line 10)
[q,errbnd] = integral2t(fun,xmin,xmax,ymin,ymax,optionstruct);
Error in integral3/innerintegral (line 138)
Q1 = integral2Calc( ...
Error in integralCalc/iterateScalarValued (line 314)
fx = FUN(t);
Error in integralCalc/vadapt (line 133)
[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);
Error in integralCalc (line 76)
[q,errbnd] = vadapt(@AtoBInvTransform,interval);
Error in integral3 (line 122)
Q = integralCalc(@innerintegral,xmin,xmax,integralOptions);
Error in
Error in integralCalc/iterateArrayValued (line 157)
fxj = FUN(t(1)).*w(1);
Error in integralCalc/vadapt (line 131)
[q,errbnd] = iterateArrayValued(u,tinterval,pathlen);
Error in integralCalc (line 76)
[q,errbnd] = vadapt(@AtoBInvTransform,interval);
Error in integral (line 89)
Q = integralCalc(fun,a,b,opstruct);
Error in this_matlab_script (line 61)
The source of this error is a complete mystery to me. Could someone kindly shed some light on what may be happening here and what I can do to remedy it?

Answers (1)

Mike Hosea
Mike Hosea on 27 Sep 2014
When you see that error, it means that you used a / instead of ./ . The integrator passed in an array for "vectorized" element-wise evaluation, and instead of just processing it elementwise, your integrand ended trying to solve a linear system with it. This problem is a little bit on the edge of what the routines can solve. I had some problems using the default tolerances, so I had to loosen them a bit and then see how far I could tighten them back up.. After fixing the integrand to be vectorized in all variables, I got this
a_1 = 7e-5;
a_2 = 5e-5;
k = 1;
theta = 0;%can also be 0.5,1,1.5
second_integral=@(r_dash,phi_dash,r,phi)((exp(-1i*k*r_dash.*cos(phi_dash)*sin(theta)).*exp(-1i*k*sqrt(r.^2 + r_dash.^2 - 2*(r.*r_dash.*cos(phi_dash - phi)))))./(2.*pi.*sqrt(r.^2 + r_dash.^2 - 2.*(r.*r_dash.*cos(phi_dash - phi)))));
integrated_second_integral = integralN(second_integral,0,a_1,0,2*pi,0,a_2,0,2*pi,'AbsTol',1e-8,'RelTol',1e-3)
Although theoretically the integral of integral3 approach should work, it's somewhat more efficient to do integral2 of integral2. Setting that up is a little more work because integral2 doesn't have an 'ArrayValued' option, so I used the integralN routine I put on the file exchange which does exactly that.
>> integrated_second_integral = integralN(second_integral,0,a_1,0,2*pi,0,a_2,0,2*pi,'AbsTol',1e-8,'RelTol',1e-3)
integrated_second_integral =
8.479042788173304e-04 - 2.199114856608778e-08i_
  1 Comment
JJ on 28 Sep 2014
Mike, thank you for your answer - I have copied the code and module and it works as expected. I now just need to dig into integralN to understand what it's doing :p Thanks again!

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