MATLAB: Index exceeds matrix dimensions.

matrix

Hello dear Matlab Community,

I would like to request your help again (it's really difficult to resist, seeing how helpful you were in the last days…and how long it takes me to not figure out the solution of these problems!)

I know that this post looks quite alike one of yesterday but I unfortunately got stuck again: now my initial condition y0 for solving a system of two ODE is an array with two values.

But when running the code below I get: Index exceeds matrix dimensions. Error in NewTry>myODE (line 30) Tp=y(2);

I suspect that this might be related to a syntax issue and the way the initial condition is written as "y0" in my ODE solving equation because if I replace y0 with it s actual initla values (as follows) matlab solves the problem without complaining.

[L,y]=ode45(@(L,y) myODE(L,y,Aap,Acp),Pool(k:k+1),[30,5.2]);

So I don't understand why Tp can't be associated to the second element of the array of y when y0 has clearly been declared with two elements…

Merci!!

a =1;
b =2;
Pool = [0,a,b];
y0=[30,5.2];
LAll=[];
yAll=[];
for k=1:numel(Pool)-1
    switch k
        case rem(k,2)==1
            Aap=a; Acp=Aap;
        otherwise
            Aap=2*b; Acp=0; 
    end
 
[L,y]=ode45(@(L,y) myODE(L,y,Aap,Acp),Pool(k:k+1),y0);
LAll=cat(1,LAll,L);
yAll=cat(1,yAll,y);
y0=y(end,1);
end
plot(LAll,yAll(:,:));
function dy = myODE(L,y,Aap,Acp)
global u
u=y(1);
global Tp
Tp=y(2);
dudL = myODE2(L,u,Tp,Aap,Acp);  
dTpdL = myODE1(L,u,Tp,Aap,Acp);
dy =[dudL;dTpdL];
end
function dudL = myODE2(~,u,Tp,Aap,~)            
Psatp = 0.13*exp(18-3800/(Tp+227.03));
Phi= 1-exp(-(47*u^2+0.1*Tp*u^1.06));
Ppw= Psatp*Phi;
dudL=-(Aap)*(Ppw/(Tp+273.15)-0.19);
end
function dTpdL=myODE1(L,u,Tp,Aap,Acp)
dTpdL=(140-Tp)*Acp+(100-Tp)*Aap-Aap/(Tp+273);                                                                                           
end

Best Answer

y0=y(end,:);

instead of

y0=y(end,1);

Related Solutions

MATLAB: Solving a system of ODE over different intervals (with different conditions on parameters)

Avoid using globals to provide parameters. See <Answers: Anonymous for params>

You can integerate over different time intervals by using a loop:

tPool = [0,1,2,4,8];
y0    = 0;  % Your initial value

tAll  = [];
yAll  = [];
for k = 1:numel(tPool) - 1
    % Your parameter depending on the time step:

    switch k
        case 1
          P = 17;
        case 2
          P = 23;
        case 3
          P = 10992;
        otherwise
          P = 0;
    end
    
    % Call integrator, provide current parameter by anonymous function:

    [t,y] = ode45((t,y) @fcn(t,y,P), [tPool(k:k+1)], y0)
    
    tAll  = cat(1, tAll, t);  % Append new solution to total solution

    yAll  = cat(1, yAll, y);
    
    y0    = y(end, 1);  % Update the initial value

end

If the zones are not predefined by time steps, but by positions, use a while loop until the final time is reached and an event function set by odeset:

t0    = 0;
tEnd  = 8;
y0    = 0;  % Your initial value
tAll  = [];
yAll  = [];
while t0 < tEnd
    % Your parameter depending on the time step:
    if y(1) < 10
      P = 17;
      eventP = 20
    elseif y(1) < 20
      P = 23;
      eventP = 30;
    else   
      P = 10992;
      eventP = 100;
    end
    
    % Create new event function, which stops the integration at the
    % next step:
    opt = odeset('Events', @(t,y) @myEventsFcn(t,y,eventP));
    
    % Call integrator, provide current parameter by anonymous function:
    % tEnd is not reached, because the integration is stopped by the
    % event function except for the last step:
    [t,y] = ode45((t,y) @fcn(t,y,P), [t0, tEnd], y0, opt)
    
    tAll  = cat(1, tAll, t);  % Append new solution to total solution
    yAll  = cat(1, yAll, y);
    
    t0    = t(end);
    y0    = y(end, 1);  % Update the initial value
end
function [position,isterminal,direction] = yourEventFcn(t,y,eventP)
  position   = y(1) - eventP; % The value that we want to be zero
  isterminal = 1;   % Halt integration 
  direction  = 0;   % The zero can be approached from either direction
end

Both examples are just a demonstration and they do not run!!!

MATLAB: Index exceeds matrix dimensions.

See my answer to your former question: https://www.mathworks.com/matlabcentral/answers/439728-solving-a-system-of-ode-over-different-intervals-with-different-conditions-on-parameters#answer_356399 :

tPool = [0,1,2,4,8];
...
for k = 1:numel(tPool) - 1
    %                  ^^^

Concerning: "If I add the parameters Aap,Acp as input parameters of my anonymous function then I get: Not enough input arguments. Error in verysimple3>@(l,Tp,Aap,Acp)myODE(l,Tp,Aap,Acp)"

The ODE integrator calls the function to be integrated with 2 input arguments. The anonymous function is created to add further inputs. Therefore this works:

[l, Tp]=ode45(@(l,Tp) myODE(l,Tp,Aap,Acp),Pool(k:k+1),y0);

because here ODE45 provides the 2 inputs (l, Tp), the anonymous function appends 2 further inputs. With:

[l, Tp]=ode45(@(l,Tp,Aap,Acp) myODE(l,Tp,Aap,Acp), Pool(k:k+1), y0);

ODE45 is instructed to provide 4 inputs, but it can offer 2 only.

By the way, using "l" (lowercase L) as variable causes troubles freuqently, when it is confused with a "1" (one) or "I" (uppercase i).

Best Answer

Related Solutions

MATLAB: Solving a system of ODE over different intervals (with different conditions on parameters)

MATLAB: Index exceeds matrix dimensions.

Related Question