Warning: Non-finite result. The integration was unsuccessful. Singularity likely.

Hi i run a code thta include two doyble integration i recieve e message Warning: Non-finite result. The integration was unsuccessful.

but the final resulta are finite what happen ? the results are reliably ?

the main code is

currentMoM()

f = 300000000

N = 40

ra = 1

k0 = 6.2832

Z0 = 376.9911

lambda = 1

ans = NaN +    NaNi

Warning: Non-finite result. The integration was unsuccessful. Singularity likely.

ans = NaN +    NaNi

Unrecognized function or variable 'Efieldin'.

Error in solution>@(x)(4./(Z0.*k0)).*triangle_basisn(x,index_i).*Efieldin(x) (line 37)
func=@(x)(4./(Z0.*k0)).*triangle_basisn(x,index_i).*Efieldin(x);

Error in integralCalc>iterateScalarValued (line 334)
fx = FUN(t);

Error in integralCalc>vadapt (line 148)
[q,errbnd] = iterateScalarValued(u,tinterval,pathlen, ...

Error in integralCalc (line 77)
[q,errbnd] = vadapt(vfunAB,interval, ...

Error in integral (line 87)
Q = integralCalc(fun,a,b,opstruct);

Error in solution>currentMoM (line 38)
gm(index_i) =integral(func,Phi0(index_i),Phi0(index_i)+2*pi/N);

function  [Is]=currentMoM()
%UNTITLED2 Summary of this function goes here
%   Detailed explanation goes here
[f,N,ra,k0,Z0,lambda] = parameter()
gamma_const=1.781;
%Phi0=zeros(N);
e=exp(1);
dftm=2.*pi./N;
for jj = 1:N
 Phi0(jj)=(jj-1).*dftm;
  
end
    
   % delta_c(i) = sqrt((pos(i,1) - pos(i+1,1))^2 + (pos(i,2) - pos(i+1,2))^2);
lmn = zeros(N);
%zmn = zeros(N);
gm = zeros(1,N);
zmn = zeros(N);
%vim = zeros(1,N);
%vsn = zeros(1,N);
coeif=(Z0.*k0./4).*ra.*dftm;
coeifn=(Z0./2).*sin(k0.*ra.*dftm./2);
for index_i = 1:N
    for index_j = 1:N
        if   index_i == index_j
            funa = @(x,y)triangle_basisn(x,index_i).*triangle_basisn(y,index_j).*ra.*(1-j.*(2/pi).*log((gamma_const./2).*k0.*ra.*sqrt(2-2.*cos(x-y)))) ;
            funa(Phi0(index_i),Phi0(index_j))
            lmn(index_i,index_j) =integral2(funa,Phi0(index_i),Phi0(index_i)+2*pi/N,Phi0(index_j),Phi0(index_j)+2*pi/N);
        lmn(index_i,index_j)
        else
            funb = @(x,y)triangle_basisn(x,index_i).*triangle_basisn(y,index_j).*ra.*besselh(0,2,k0.*ra.*sqrt(2-2.*cos(x-y))); 
         
            lmn(index_i,index_j) =integral2(funb,Phi0(index_i),Phi0(index_i)+2*pi/N,Phi0(index_j),Phi0(index_j)+2*pi/N);
         
           
        func=@(x)(4./(Z0.*k0)).*triangle_basisn(x,index_i).*Efieldin(x);
        gm(index_i) =integral(func,Phi0(index_i),Phi0(index_i)+2*pi/N);
        zmn(index_i,index_j) = lmn(index_i,index_j) ;
        end
      
    %vim(index_i) = delta_c(index_i) * exp(j*k0*(xm(index_i)*cos(phi_i)+ym(index_i)*sin(phi_i)));
    %vsn(index_i) = delta_c(index_i) * exp(j*k0*(xm(index_i)*cos(phi_s)+ym(index_i)*sin(phi_s)));
    end
end
W = linsolve(zmn,gm');
for ii=1:N 
Is(ii)=W(ii); 
end 
y= Is;
end

function [f,N,ra,k0,Z0,lambda] = parameter()
%UNTITLED Summary of this function goes here
c0=3e8;
Z0=120.*pi;
ra=1;
N=40;
f=300e6;
lambda=c0./f;
k0=2*pi./lambda;
end

function   z=triangle_basisn(phi,kk)
[~,N,ra,k0,Z0,lambda] = parameter();
%Phin=zeros(N)
dftm=2.*pi./N;
%for jj = 1:N+1
%Phi0(jj)=(jj-1).*dftm; 
%end
%Phin=Phi0
if ( phi >=(kk-1).*dftm ) & ( phi <=kk.*dftm);
z=(phi-(kk-1).*dftm)./dftm;
 elseif (phi >= kk.*dftm) &  (phi <= (kk+1).*dftm);
 z=((kk+1).* dftm -phi)./dftm;
else
    z=0 ;
end
end

thank you

Goerge

[Torsten]

Did you make any changes to your old code ? The same problem remained (see above).

Your function becomes NaN for

funa(Phi0(index_i),Phi0(index_j))

[george veropoulos]

hi no the i receive a finitr results

i musts changre the limit ?

[Torsten]

Why is

lmn(1,1) = NaN + i*NaN

a finite result (see above) ?

[george veropoulos]

the code give a finete results for the is

[Torsten]

In the computation of zmn, results from the case index_i = index_j (thus the case where NaN results are computed) are not used. Thus - although you get the warning: Non-finite result. The integration was unsuccessful, your results for Is might be reliable. Which raises the question: why do you compute lmn for index_i = index_j as below

        if   index_i == index_j
            funa = @(x,y)triangle_basisn(x,index_i).*triangle_basisn(y,index_j).*ra.*(1-j.*(2/pi).*log((gamma_const./2).*k0.*ra.*sqrt(2-2.*cos(x-y)))) ;
            lmn(index_i,index_j) =integral2(funa,Phi0(index_i),Phi0(index_i)+2*pi/N,Phi0(index_j),Phi0(index_j)+2*pi/N);            

at all if the result is nowhere used ?

[george veropoulos]

hi

the element of z or l matrix with index_i == index_j are the diagonal element

in this cace i use a small approximation (1-j.*(2/pi).*log(A.*sqrt(2-2.*cos(x-y))))

of Hankel function besselh(0,2,kR)

[Torsten]

But you never write the results of the integration for index_i = index_j into the zmn matrix which is used to compute Is. So zmn(i,i) (the diagonal) remains at its initial values given to them by preallocation:

zmn = zeros(N);

[george veropoulos]

the correcr place is before ending second for loop ?

[george veropoulos]

You are write.. i use an approximation diagonal teem log(A)+log(x-y+1e-6) i add a small number

idont receive any message from integral but receive a warning from matrix "Matrix is singular to working precision"

currentMoM()

f = 300000000

N = 40

ra = 1

k0 = 6.2832

Z0 = 376.9911

lambda = 1

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.

ans = 
  Columns 1 through 9

   0.0031 + 0.0015i   0.0023 + 0.0014i   0.0011 + 0.0010i   0.0001 + 0.0003i  -0.0004 + 0.0000i  -0.0011 + 0.0005i  -0.0025 + 0.0007i  -0.0035 - 0.0007i  -0.0019 - 0.0026i

  Columns 10 through 18

   0.0019 - 0.0023i   0.0042 + 0.0012i   0.0024 + 0.0046i  -0.0016 + 0.0041i  -0.0036 + 0.0000i  -0.0022 - 0.0044i   0.0012 - 0.0064i   0.0042 - 0.0059i   0.0057 - 0.0042i

  Columns 19 through 27

   0.0061 - 0.0027i   0.0061 - 0.0019i   0.0061 - 0.0020i   0.0061 - 0.0031i   0.0054 - 0.0048i   0.0033 - 0.0063i   0.0000 - 0.0060i  -0.0030 - 0.0031i  -0.0033 + 0.0016i

  Columns 28 through 36

  -0.0003 + 0.0048i   0.0034 + 0.0038i   0.0038 - 0.0002i   0.0006 - 0.0028i  -0.0027 - 0.0021i  -0.0034 - 0.0000i  -0.0020 + 0.0007i  -0.0008 + 0.0003i  -0.0003 + 0.0001i

  Columns 37 through 40

   0.0004 + 0.0006i   0.0015 + 0.0012i   0.0026 + 0.0015i   0.0032 + 0.0015i

function   [Is]=currentMoM()
%UNTITLED2 Summary of this function goes here
%   Detailed explanation goes here
[f,N,ra,k0,Z0,lambda] = parameter()
gamma_const=1.781;
Phi0=zeros(N);
e=exp(1);
dftm=2.*pi./N;
for jj = 1:N
    Phi0(jj)=(jj-1).*dftm;
    
end

lmn = zeros(N,N);
gm = zeros(1,N);
zmn = zeros(N,N);
%vim = zeros(1,N);
%vsn = zeros(1,N);
coeif=(Z0.*k0./4).*ra.*dftm;
coeifn=(Z0./2).*sin(k0.*ra.*dftm./2);
for index_i = 1:N
    for index_j = 1:N
        if   index_i == index_j
            A=(gamma_const./2).*k0.*ra ;
            %funa = @(x,y)triangle_basisn(x,index_i).*triangle_basisn(y,index_j).*ra.*(1-j.*(2/pi).*ra.*(1-j.*(2/pi).*ra.*log(A.*sqrt(2-2.*cos(x-y))))) ;
            funa = @(x,y)triangle_basisn(x,index_i).*triangle_basisn(y,index_j).*ra.*(1-j.*(2/pi).*(log(A)+log(x-y+1e-8)));
            lmn(index_i,index_j) =integral2(funa,Phi0(index_i),Phi0(index_i)+2*pi/N,Phi0(index_j),Phi0(index_j)+2*pi/N);
            zmn(index_i,index_j) = lmn(index_i,index_j) ;
        else
            funb = @(x,y)triangle_basisn(x,index_i).*triangle_basisn(y,index_j).*ra.*besselh(0,2,k0.*ra.*sqrt(2-2.*cos(x-y))); 
            
            lmn(index_i,index_j) =integral2(funb,Phi0(index_i),Phi0(index_i)+2*pi/N,Phi0(index_j),Phi0(index_j)+2*pi/N);
            
            zmn(index_i,index_j) = lmn(index_i,index_j) ;
            func=@(x)triangle_basisn(x,index_i).*(4./(Z0.*k0)).*Efieldin(x);
            gm(index_i) =integral(func,Phi0(index_i) ,Phi0(index_i)+2*pi/N);
            
        end
        
        
    end
end
    W = linsolve(zmn,gm');
    for ii=1:N 
        Is(ii)=W(ii); 
    end 
    y= Is;
end
function [f,N,ra,k0,Z0,lambda] = parameter()
%UNTITLED Summary of this function goes here
c0=3e8;
Z0=120.*pi;
ra=1;
N=40;
f=300e6;
lambda=c0./f;
k0=2*pi./lambda;
end
function   z=triangle_basisn(phi,kk)
[~,N,ra,k0,Z0,lambda] = parameter();
%Phin=zeros(N)
dftm=2.*pi./N;
%for jj = 1:N+1
%Phi0(jj)=(jj-1).*dftm; 
%end
%Phin=Phi0
if ( phi >=(kk-1).*dftm ) & ( phi <=kk.*dftm);
z=(phi-(kk-1).*dftm)./dftm;
 elseif (phi >= kk.*dftm) &  (phi <= (kk+1).*dftm);
 z=((kk+1).* dftm -phi)./dftm;
else
    z=0 ;
end
end
function   z=Efieldin(phi)
%amplitude 
[f,N,ra,k0,Z0] = parameter();
E0=1;
%z=E0.*exp(j.*k0.*ra.*cos(phi)+j.*k0.*ra.*sin(phi));
z=E0.*exp(j.*k0.*ra.*cos(phi));
end

[Torsten]

idont receive any message from integral but receive a warning from matrix "Matrix is singular to working precision"

In R2024b, I get many warning that some of the integrals could not be computed reliably, but no warning about a matrix that is singular to working precision (see above). Do we use the same code ? What MATLAB release are you working with ?

[george veropoulos]

i receive the same message ... but also this the message with matrix i use 2024a

[Torsten]

It seems you try to integrate a discontinuous function. This will almost always fail.

if ( phi >=(kk-1).*dftm ) & ( phi <=kk.*dftm);
z=(phi-(kk-1).*dftm)./dftm;
 elseif (phi >= kk.*dftm) &  (phi <= (kk+1).*dftm);
 z=((kk+1).* dftm -phi)./dftm;
else
    z=0 ;
end

[george veropoulos]

this is a screen shot

[george veropoulos]

if i remove the function triangle_basisn i receive the same message !!

[Torsten]

I understand that your computation doesn't succeed. And I gave you the most probable reason: the function you try to integrate is discontinuous. There can be other reasons: there is an error in your code, the parameters are too extreme and so on.

If I were you, I'd first make a surface plot of the real and imaginary parts of the functions you try to integrate and this way find out where "integral2" might encounter problems.

[george veropoulos]

what parameter are too extreme ?

[Torsten]

I didn't check whether you use f = 300e6 somewhere - at least it's an output from your parameter function.

But it was just an example to show that the reasons for failure can be multifold.

[george veropoulos]

i set triangle_basisn(phi,kk) equal to 1 . Ι receive problem from the integral in funa with the log singularity!

[george veropoulos]

Warning: Reached the maximum number of function evaluations (10000). The

result fails the global error test.

> In integral2Calc>integral2t (line 139)

In integral2Calc (line 9)

In integral2 (line 105)

In currentMoM (line 31)

this is the line 31

funa = @(x,y)triangle_basisn(x,index_i).*triangle_basisn(y,index_j).*ra.*(1-j.*(2/pi).*(log(A)+log(x-y+1e-9)));

[Torsten]

And this surprises you ? I don't know if this happens in your box of integration, but if x = y, you get log(0) = -Inf.

So again: make a surface plot of the real and imaginary parts of the functions you try to integrate.

[george veropoulos]

no. ! how i can remove this singularity is the problem

[george veropoulos]

It is possible to make. A monte Carlo. Integration?

[Torsten]

Of course you can do a Monte-Carlo integration. But if you change the integration method, the mathematical problem about the integrand won't vanish. In the Monte-Carlo integration, the problem will show up that you don't get a converged solution for the integral for an increasing number of random test points.

[george veropoulos]

[Walter Roberson]

And you expect to be able to integrate that ????

[george veropoulos]

with triangle_basisn

[george veropoulos]

this integral come out from the method of moment in EM...

[Torsten]

@george veropoulos

The white diagonal seems to indicate that you get NaN values for x = y.

And the varying colors between green and blue show discontinuous behaviour ?

I think you have no chance to integrate this function with a numerical method.

EM is ElectroMagnetism ?

[george veropoulos]

i dont know if is possible to find analyticlally this integral

[Torsten]

Do you have a literature link where this integral is written out in a mathematical way ?

[george veropoulos]

from the book of Gibson

[george veropoulos]

it work!!!!