Undefined function 'matlabFunction' for input arguments of type 'function_handle'.

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Yang Liu
Yang Liu 2018-7-7
评论： Walter Roberson 2018-7-7
Hello everyone, I have problem about integrate function and convert the sym to number...the following is my code
it always shows
"Undefined function ' *matlabFunction*' for input arguments of type 'function_handle'."..
I am not sure whether I could use 'matlabFunction' or not.. thank you for your help:)
ka = 1.38e-23; % Boltzmann constant, unit m^2*kg*s^(-2)*K^(-1)
T = 273+25; % Room temperature, where experiments were conducted, unit K
u_0 = pi*4e-7; % Permeability of free space, unit Henry/m
d_pb = 11e-9; % mean magnetic nanoparticle diameter in microbeads
d_pm=9.9e-9;% median magnetic nanoparticle diameter in microbeads (must use nm unit)
M_db = 370e+3; % Magnetic nanoparticle bulk magnetization
H=1.45/u_0; %T magnetic flux density
HG =20 ; %T/m field gradient
alpha=u_0*pi*d_pb^3*M_db*H/6/ka/T;
Lang=coth(alpha)-1/alpha;
% Fm=u_0*pi*d_pb^3/6*M_db*Lang*HG;
syms D %polydisperse
lamda_P=6*ka*T/pi./D.^3./M_db/Lang/HG;
Fm_P=u_0*pi.*D.^3./6*M_db*Lang*HG;
u=1/3/pi/eta_f./D; % mobility given by the sokes formula
D_0=u*ka*T;
A=u.*Fm_P;
s =0.6; % poly dispersity index
beta = (log(1+s^2))^0.5;
w= 1200e-6; % channel width
l=lamda_P;
n0 =3e16;
Z=linspace(0,1200,5);
T=linspace(1e5,1e5,1);
syms k
for i = 1:length(T)
t=T(i);
for j = 1:length(Z)
z=Z(j)*1e-6;
KV=(k^2*(1-(-1)^k*exp(w/2./l))/(pi^2*k^2+ w^2/4./l.^2)^2*(cos(pi*k*z/w)-w/2/pi/k./l.*sin(pi*k*z/w))*exp(-1*(pi^2*k^2+w^2/4./l.^2)*D_0*t/w^2));
fun1_1 = @(D)(n0*w*exp(-z./l)./l./(1-exp(-w./l))-2*pi^2*n0*w./l.*exp(-z/2/l).*double(symsum(KV,k,1,1e3))).*exp(-(log(D./d_pm)).^2./2./beta^2)./beta./D./(2*pi)^0.5; % number concentration distribution
fun1fcn=matlabFunction(fun1_1);
N_zt(i,j)=integral(fun1fcn,0,Inf)
end
end
plot(Z,N_zt)

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回答（1 个）

Walter Roberson
Walter Roberson 2018-7-7
Remove the @(D) from the definition of fun1_1
  2 个评论
Yang Liu
Yang Liu 2018-7-7
编辑：Walter Roberson 2018-7-7
thank you for your response. matlab could process the code now. but the result is NaN.
my revised code is:
fun1_1 =(n0*w*exp(-z./l)./l./(1-exp(-w./l))-2*pi^2*n0*w./l.*exp(-z/2./l).*symsum(KV,k,1,1e3)).*exp(-(log(D./d_pm)).^2./2./beta/beta)./beta./D./(2*pi)^0.5
fun1fcn=matlabFunction(fun1_1);
N_zt(i,j)=vpa(integral(fun1fcn,0,Inf))
I am totally confused now...sign..
Walter Roberson
Walter Roberson 2018-7-7
You should not go back and forth between symbolic and numeric. If you want to end up with symbolic you should use
fun1_1 = ... etc
N_zt(i, j) = vpaintegral(fun1_1, D, 0, Inf);
However this will still have problems.
Your fun1_1 multiplies and divides by D in a number of places, so at your starting point of 0, any numeric evaluation is going to end up with places in the code with division by 0, or with 0 / 0 . The meaning of the integral at 0 therefore has to involve careful evaluation of limits. Unfortunately calculating the limit at 0 turns out to be fairly time consuming; I am not sure how many hours it will take to compute it.

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