I am getting error:Warning: Failure at t=0.000000e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.905050e-323)

This is my code

clc
clear
global datei Temp c0
datei='input_data.m';
run(datei);
zspan=[0 V/A_hc];                                                     
%intergration of the system 
options=odeset('RelTol',1e-12,'AbsTol',1e-12);%,'NonNegative',2:7);         
for n=1:size(c,1)
    switch dataset
        case 1
            name=[experiment ' ' num2str(y0_nh3_scr(n,2)) ' ppm NH3 ' num2str(y0_nh3_scr(n,1)) ' vol NO ' num2str(y0_nh3_scr(n,5)) '% O2' ];
        case 2
            name=[experiment ' ' num2str(y0_nh3_ox(n,2)) ' ppm NH3 ' num2str(y0_nh3_ox(n,5)) ' vol O2'];
        case 3
            name=[experiment ' ' num2str(y0_hcho_ox(n,3)) ' ppm HCHO ' num2str(y0_hcho_ox(n,5)) ' vol O2'];
    end
    results=zeros(length(T),size(c,2)+2);
    for m=1:length(T)
        c0=c(n,1:4).*T_N./T(m);
        c_o2=c0_o2(n).*T_N./T(m);
        p_o2=p0_o2(n);
        y0=[c0 0];                                                      
        Temp=T(m);
        tic
        disp(['Calculating Step ' num2str(m) ' of ' num2str(length(T))])
        [t,y]=ode15s(@massbalance,zspan,y0,options);                   
        toc
        results(m,:)=[T(m)-273.15 y(end,:)];   
    end
end

Calculating Step 1 of 10

Warning: Failure at t=0.000000e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.905050e-323) at time t.

Elapsed time is 8.653056 seconds.

Calculating Step 2 of 10

Warning: Failure at t=0.000000e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.905050e-323) at time t.

Elapsed time is 8.252577 seconds.

Calculating Step 3 of 10

Warning: Failure at t=0.000000e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.905050e-323) at time t.

Elapsed time is 8.240095 seconds.

Calculating Step 4 of 10

Warning: Failure at t=0.000000e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.905050e-323) at time t.

Elapsed time is 8.303395 seconds.

Calculating Step 5 of 10

Warning: Failure at t=0.000000e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.905050e-323) at time t.

Elapsed time is 8.329335 seconds.

Calculating Step 6 of 10

Warning: Failure at t=0.000000e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.905050e-323) at time t.

Elapsed time is 8.261818 seconds.

Calculating Step 7 of 10

Warning: Failure at t=0.000000e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.905050e-323) at time t.

Elapsed time is 8.208911 seconds.

Calculating Step 8 of 10

Warning: Failure at t=0.000000e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.905050e-323) at time t.

Elapsed time is 8.235031 seconds.

Calculating Step 9 of 10

Warning: Failure at t=0.000000e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.905050e-323) at time t.

Elapsed time is 8.199416 seconds.

Calculating Step 10 of 10

Warning: Failure at t=0.000000e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.905050e-323) at time t.

Elapsed time is 8.179562 seconds.

results(:,2:5)=results(:,2:5).*Vm.*1e6.*(results(:,1)+273.15)./T_N;
switch dataset
    case 1
        results(:,2:3)=(1-results(:,2:3)./y0_nh3_scr(n,1:2)).*100;
    case 2
        results(:,3)=(1-results(:,3)./y0_nh3_ox(n,2)).*100;
    case 3
        results(:,4)=(1-results(:,4)./y0_hcho_ox(n,3)).*100;
end
name=[name ' ' char(num2str(fix(clock)))];
save(name,'results','-double','-ascii')
plot(results(:,1),results(:,2),'b',results(:,1),results(:,3),'r',results(:,1),results(:,6),'g')
xlabel('T / ^oC')
ylabel('y / ppm')
legend('NO','NH3','N2O')

function dy_dz = massbalance(takt,y)
global datei Temp c0
run(datei)
F=F*Temp/T_N;                                                               
u=F/A_hc;                                                                    
if c0(1)==0
    kin=kin_ox;
else
    kin=kin_scr;
end
r1=kin(1,1).*exp(-kin(1,2)./R.*(1/Temp-1/kin(1,3))).*y(2).*(1-y(5));                       
r2=kin(2,1).*exp(-(kin(2,2)-y(5).*kin(2,3))./(R.*Temp)).*y(5);              
r3=kin(3,1).*exp(-kin(3,2)./(R.*Temp)).*y(1).*y(5);                     
r4=kin(4,1).*exp(-kin(4,2)./(R.*Temp)).*y(5);                                   
r5=kin(5,1).*exp(-kin(5,2)./(R.*Temp)).*y(4);                                
r6=kin(6,1).*exp(-kin(6,2.)/(R.*Temp)).*y(3);
%% mass balances
dy_dz=[1/u*(-r3)                                               
    1/u*(-r1+r2)                                                
    1/u*(-r6)                                                  
    1/u*(-r5)                                                  
    (r1-r2-r3-r4)/(u*gam)]; 
end

I am not able to figure out where I am making a mistake. I appricate all the help in advance.

Answer 1

[Torsten]

If you remove the semicolon behind

%% mass balances
dy_dz=[1/u*(-r3)                                               
    1/u*(-r1+r2)                                                
    1/u*(-r6)                                                  
    1/u*(-r5)                                                  
    (r1-r2-r3-r4)/(u*gam)];

and execute the code, you will see from what MATLAB writes to screen that the vector dy_dz contains NaN values right from the beginning of the integration.

And remember the time MATLAB needs to run "input_data.m" each time the function "massbalance" is called. Run "input_data.m" once as you already do in your script and pass the necessary parameters and arrays to "massbalance" via your call to ode15s:

[t,y] = ode15s(@(t,y)massbalance(t,y,parameter1,parameter2,...,array1,array2,...),zspan,y0,options);
...

function dy = massbalance(t,y,parameter1,parameter2,...,array1,array2,...)
    ...
end

Answer 2

[Joshua Levin Kurniawan]

Error "Unable to meet integration tolerances without reducing the step size below the smallest value allowed " usually appears when there is a sudden discontinuities/sharp profile in the reference signal. We cannot avoid this error unless you set the relative and absolute tolerance to a bigger value.