initial condition vector is 4 but returns a vector of 1

Error using odearguments (line 95)

CT_ACETONE_EXAM2 returns a vector of length 1, but the length of initial conditions vector is 4. The vector returned by CT_ACETONE_EXAM2 and the initial conditions vector must have the same number of elements.

Error in ode15s (line 150)

odearguments(FcnHandlesUsed, solver_name, ode, tspan, y0, options, varargin);

Error in CT_Acetone_exam2_DAE (line 21)

[t,y] = ode15s(@CT_Acetone_exam2, tspan, y0, Options);

function UV = CT_Acetone_exam2 (t,y)

% Beer Lambert Data
phi = 155;                       %Quantum Yield (mol/Ein)
Io = -1.225806452E7;    %Intensity = k3/-phi **zero order kinetics**
f = 1;                              %Acetone is the only Chromophore
ep = 8.0;                        %Molar absorption coefficient (M-1cm-1)
L = 2.54;                        %Path length (cm)
PROP = 2;                    %Concentration of Propanol - 2 mol/L
% Kinetic constant
k2 = 2.5E10;                 % M-1 sec-1
k3 = 1.9E9;                   % M-1 sec-1

UV =[-phi.*Io.*f.*(1-exp(-2.303.*ep.*L.*y(1)))+k3.*y(3).*y(4)+(phi.*Io.*f.*(1-exp(-2.303.*ep.*L.*y(1)))-k2.*y(2).*PROP)+(2*k2.*y(2).*PROP-k3.*y(3).*y(4))-(k3.*y(4).*y(3))];


tic; %Clock starts
clc
clf
clear all

AC = 69E-3; %Initial conc in moles/L
CT = 3E-3; %Target Initial concentration
tspan = [0 60];
y0 = [AC 0 0 CT];

% Option: Tolerances (AC, AC*, RAD, CT, Abs)
  M = diag([1 1 1 1 1]);
  Options = odeset('Mass', M,'Reltol', 1E-8, 'AbsTol', [1E-6 1E-20 1E-16 1E-9]);

% ODE's Stiff solver
  [t,y] = ode15s(@CT_Acetone_exam2, tspan, y0, Options);

% Plotting results
  hold all

figure(1)
  plot(t,1E3*y(:,3),'-','LineWidth',3,'Color',[0 1 0]);
  xlabel('time (sec)');ylabel('AC* (mM)');
  title('RAD vs Time');

figure(2)
  plot(t,1E3*y(:,4),'-','LineWidth',3,'Color',[1 0 1]);
  xlabel('time (sec)');ylabel('CT (mM)');
  title('CT vs Time');
  toc; %Clock Stops

Answer 1

[Walter Roberson]

The message is correct.

Your odefun uses four states, y(1), y(2), y(3), y(4).

Your odefun must return a column vector

[dy(1)/dt; dy(2)/dt; dy(3)/dt; dy(4)/dt]

[Ted Kim]

Thank you Walter,

I'm not sure where I need to edit the code to return the results as a column vecotr. I am pretty new to MATLAB and coding

[Walter Roberson]

UV =[-phi.*Io.*f.*(1-exp(-2.303.*ep.*L.*y(1)))+k3.*y(3).*y(4)+(phi.*Io.*f.*(1-exp(-2.303.*ep.*L.*y(1)))-k2.*y(2).*PROP)+(2*k2.*y(2).*PROP-k3.*y(3).*y(4))-(k3.*y(4).*y(3))];

You used [] as if you think you are creating a vector, but you are calculating only one expression.

If you happen to have a system of the form

y' = function of y, y', y'', y'''

without any explicit expression for y'' or y'''

then you should return

[expression;
 y(2);
 y(3);
 y(4)]

I think.

[Ted Kim]

function UV = CT_Acetone_exam2 (t,y)

% Beer Lambert Data

phi = 155; %Quantum Yield (mol/Ein)

Io = 1.2258064E07; %Intensity = k/phi

f = 1; %Acetone is the only Chromophore

ep = 8.0; %Molar absorption coefficient (M-1cm-1)

L = 2.54; %Path length (cm)

PROP = 2; %Concentration of Propanol - 2 mol/L

% Kinetic constant

k2 = 2.5E10; % M-1 sec-1

k3 = 1.9E9; % M-1 sec-1

y(1) = -phi.*Io.*f.*(1-exp(-2.303.*ep.*L.*y(1)));

y(2) = phi.*Io.*f.*(1-exp(-2.303.*ep.*L.*y(1)))-k2.*y(2).*(PROP);

y(3) = 2*k2.*y(2).*(PROP)-k3.*y(3).*y(4);

y(4) = -k3.*y(4).*y(3);

UV = [-phi.*Io.*f.*(1-exp(-2.303.*ep.*L.*y(1)))+k3.*y(3).*y(4)+phi.*Io.*f.*(1-exp(-2.303.*ep.*L.*y(1)))-k2.*y(2).*(PROP)+2*k2.*y(2).*(PROP)-k3.*y(3).*y(4)-k3.*y(4).*y(3)];

So I defined the y expressions for the function, but I'm still getting the same error.

[Ted Kim]

Each of the y expression need to yield a vector, based on the variable t from the function function_name(t,y), but I'm not sure what to change

[Walter Roberson]

y(1) = -phi.*Io.*f.*(1-exp(-2.303.*ep.*L.*y(1)));
y(2) = phi.*Io.*f.*(1-exp(-2.303.*ep.*L.*y(1)))-k2.*y(2).*(PROP);
y(3) = 2*k2.*y(2).*(PROP)-k3.*y(3).*y(4);
y(4) = -k3.*y(4).*y(3);

At this point y exists and is the boundary conditions input. You then alter every component of the vector, in some cases using the old (input) values but in other cases using the values you just computed. This is confusing for the readers, who would have good reason to doubt that the code has been implemented properly. If you need a vector of coefficients calculated from the inputs, then use a different variable name to store the values, so as to clearly distinguish between old and new values.

[Walter Roberson]

Your output is not y. Your output is UV

UV = [-phi.*Io.*f.*(1-exp(-2.303.*ep.*L.*y(1)))+k3.*y(3).*y(4)+phi.*Io.*f.*(1-exp(-2.303.*ep.*L.*y(1)))-k2.*y(2).*(PROP)+2*k2.*y(2).*(PROP)-k3.*y(3).*y(4)-k3.*y(4).*y(3)];

and although you wrote that with [] as if you are producing a vector output, you are only calculating a scalar there.

Your CT_ACETONE_exam2 must return a column vector [dy(1)/dt; dy(2)/dt; dy(3)/dt; dy(4)/dt]

No matter how complicated the calculation is, this is required.

You look to be doing some kind complicated chemical reactions. If the four boundary conditions are concentrations of four different species, then the four outputs would be the derivatives of how those species each change.