Error using mupadengin​e/feval_in​ternal First argument must be a condition.

Hi , I keep getting this error message "First argument must be condition" evidently regarding the set up in line 14 in the given code:

syms r x k z l m ph
% l=zeta1, m=zeta2
[ph,r] = meshgrid((0:5:360)*pi/180,0:.5:10);
[X,Y] = pol2cart(ph,r);
Z = X+1i*Y;
J = besselj(k,l.*r);
J2 = besselj(k,m.*r);
Y = bessely(k,l.*r);
Y2 = bessely(k,m.*r);
H = besselh(k,r);
F1=symsum(J.*exp(1i*k*ph),k,-5,5);
F2=symsum((J2+Y2).*exp(1i.*k.*ph),k,-5,5);
F3=symsum(H.*exp(1i.*k.*ph),k,-5,5);
pwu = piecewise(0<r & r<0.5, F1, 0.5<r & r<1, F2, r>1, F3);
U= subs(pwu, {l, m, r, ph}, {1.5, 3, 0<r & r<4, 0<ph & ph<2*pi  }); 
surf(X,Y,double(real(U)))
hold on
surf(X,Y,zeros(size(X)))
hold off


Error using mupadengine/feval_internal
First argument must be a condition.

Error in sym/piecewise (line 49)
pw = feval_internal(symengine, 'piecewise', lists{:}, 'ExclusiveConditions');

Error in piecewi_u (line 14)
pwu = piecewise(0<r & r<0.5, F1, 0.5<r & r<1, F2, r>1, F3);
 

I am using r as a variable with phi (ph), and even if I change this variable, I get the same error again.

Can anyone suggest a fix of this code?

Thanks!

Answer 1

[Walter Roberson]

syms k l m
% l=zeta1, m=zeta2
[ph, r] = meshgrid((0:5:360)*pi/180,[0:.05:0.5, 1:.5:10]);
[x, y] = pol2cart(ph,r);
Z = x + 1i*y;
J = besselj(k,l.*r);
J2 = besselj(k,m.*r);
Y = bessely(k,l.*r);
Y2 = bessely(k,m.*r);
H = besselh(k,r);

F1 = symsum((J).*exp(1i*k*ph),k,-5,5);
%F1 = symsum((J+Y).*exp(1i*k*ph),k,-5,5);

F2 = symsum((J2+Y2).*exp(1i.*k.*ph),k,-5,5);

F3 = symsum(H.*exp(1i.*k.*ph),k,-5,5);

pwu = nan(size(F1), 'like', F1);
mask = 0 <= r & r < 0.5;
pwu(mask) = F1(mask);
mask = 0.5 <= r & r < 1;
pwu(mask) = F2(mask);
mask = r >= 1;
pwu(mask) = F3(mask);

U = subs(pwu, {l, m}, {1.5, 3});

surf(x, y, real(double(U)))
hold on
surf(x, y, zeros(size(x)))
hold off

You created Y but you did not use it. By parallelism with F2 it looks like you would want (J+Y) for F1 instead of just J. I show that version in a comment. The difference is fairly noticable, because at r = 0, bessly(k,0) is infinite, and you get infinities and NaN showing up.

Your boundary conditions were broken: you excluded the cases where r was exactly the boundaries, but those were important cases.

[Sergio Manzetti]

Thank you very much. I will read this thoroughly beore asking eventual questions.