Minimums step size reached, while integrating.

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Jay
Jay 2017-9-21
回答： Walter Roberson 2017-9-21
Hello everyone, so I ran this piece of code and I got multiple warnings on reaching the minimum step size. I get a very dubious output. So i tried changing the sample size, but irrespective of me increasing or decreasing, I get the same warnings and get the same unsure output.
Here is the code I'm running -
f=15e8:(20e9-10e8)/100:16e9;
s=length(f);
epso = 8.8 *10.^-12;
epsr=5.7;
ro=3.63*(10.^-3);
epsl = 2.5;
ri=1.13*(10.^-3);
copen = (epso/sqrt(epsr)*log(ro/ri));
cload= (epsl/sqrt(epsr)*log(ro/ri));
t = zeros(1,s);
for i = 1:length(f)
fun = @(x)((((((besselj(0,(2*pi*f(i))/(3e8))*x*ro)-besselj(0,(2*pi*f(i))/(3e8))*x*ri)).^2)./x).*freload(x,f(i)));
y(i) = copen*abs(integral(fun,0,Inf,'arrayvalued',true));
t(i)=(1-y(i))/(1+y(i));
end
plot(f,abs(t));
The function called is as follows -
function f = freload(x,q)
epsl=3.4-0.004j;
f=(1./sqrt(epsl-(x.^2))).*(1+exp(-2*1i*2*pi*q/(3e8).*sqrt(epsl-(x.^2))))./((1-exp(-2*1i*2*pi*q/(3e8).*sqrt(epsl-(x.^2)))));
end
The full warning I get is as follows -
Warning: Minimum step size reached near x = 1.75484e+22. There may be a singularity, or the tolerances
may be too tight for this problem.
> In integralCalc/checkSpacing (line 456)
In integralCalc/iterateScalarValued (line 319)
In integralCalc/vadapt (line 132)
In integralCalc (line 83)
In integral (line 88)
In trial (line 14)
And the number of times I get this depends on the number of samples taken. Does anyone know why this problem occurs, or how I could go about overcoming this ? Any help or suggestion would be greatly appreciated.
Thanks.
  1 个评论
Jay
Jay 2017-9-21
I tried executing the integral outside the for loop only for one value of f, and while I get an output is returned I still get the same warning, that a minimum step size is reached near a certain value for x. Moreover this value for x does not change for different values of f. So, could this be a problem while integrating the bessel function ? It reaches the minimum step size at 1.75e+22. Could it be a problem of using large numbers ?

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回答（1 个）

Walter Roberson
Walter Roberson 2017-9-21
On your first iteration, i = 1, if you switch to the symbolic toolbox
syms x
g = fun(x)
then limit(imag(g), x, inf) is non-zero . Therefore if the imaginary component is convergent it is infinite anyhow, so the solution for the integral is going to have to be either non-convergent or a complex number involving infinity.

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