MATLAB: Storing values from loops

loops

how do i store the values of MJ_V12, they are being overwritten in each loop. Should end up with a 41 x 100 matrix. i've tried

MJ_V12=zeros(41,100);

but didn't work for me.

for a=[1:100];
%coding here left out
FMJ_cars_wait_2 = @(T12,MJ_V12) (lambdaMJ_sec(a) - MJ_Flow);%
[T12,MJ_V12] = ode45(FMJ_cars_wait_2,[MJ_Red,Total_cycle],R_G_MJ);
end

Best Answer

Use a temporary variable:

for a=[1:100];
  %coding here left out
  FMJ_cars_wait_2 = @(T12,MJ_V12) (lambdaMJ_sec(a) - MJ_Flow);%
  [T12,temp_var] = ode45(FMJ_cars_wait_2,[MJ_Red,Total_cycle],R_G_MJ);
   MJ_V12(:,a) = temp_var;
end

Related Solutions

MATLAB: Unable to find explicit solution to differential eq

Hi Ditlev,

I believe this does have an analytic solution that is not overly complicated, of the form

x = (a+be^(-ct)) / (f+ge^(-ct))

First of all the scaling of this problem is not too good so I went to the variables (I used y in place of your x2)

u = km*t         x = (kp/km) x_old        y = (kp/km) y_old

in which case the equations are

x' = 2y - 2x^2
y' = -y + x^2

where differentiation is d/du. The starting value for the new x is

(kp/km)*1e-9 = 740.7

The most complicated the algebra gets is the roots of a quadratic eqn. The code below plots both a numerical and analytic solution.

Eventually the solution gets to the fixed point

x = r1 = 18.9966      y = -r1/2 + C = 360.8721

at which point x^2 = y and both derivatives are zero.

kp = 2*10^7;
km = 0.27*10^-4;
x0 = (kp/km)*1e-9   % starting point for x
y0 = 0;
% numerical version
[unum xynum] = ode45(@(u,xy) fun(u,xy), [0 .01], [x0 y0]);         % xy = [x; y]
%analytic version
C = x0/2;
r = roots([1,1/2,-C]);
r1 = max(r);
r2 = min(r);
alpha = 2*(r1-r2);
D = (x0-r1)/(x0-r2);
u = 0:1e-4:.01;
x = (r1-D*r2*exp(-alpha*u))./(1-D*exp(-alpha*u));   % solution
y = -x/2 + C;
figure(1)
plot(unum,xynum,u,x,'o',u,y,'o')
grid on
function dxy = fun(u,xy)
  dxy = zeros(2,1);
  dxy(1) = 2*xy(2)-2*xy(1)^2;
  dxy(2) = -xy(2) + xy(1)^2;
end

MATLAB: ??? Subscripted assignment dimension mismatch

First:

 doc interp1

and once you understand roughly what it does and its input parameters, I guess that you could modify your code a little bit as follows (untested):

lambdaMJ = 7.2;                               
MJ_Flow = (30/60)*2;
A = 100;                                  
B = 1;                                  
C = 70;                                      
D = 4;                                    
E = C + D;                       
F = A - C - D;     
MJ_start=0;
% NEW
n            = 100 ;
step         = 0.5 ;
tGrid        = (0:step:F).' ;
lambdaMJ_sec = zeros(1, n) ;
Major_end    = zeros(1, n) ;
Major1       = zeros(length(tGrid), n) ;
for a = 1:n                               % Updated: upper bound.
Num_cylces = 1;
XMJ=(poissrnd(lambdaMJ, A, Num_cylces));
lambdaMJ_sec(a) = mean(XMJ)/10;
% 
FMJ_cars_wait = @(T11,MJ_V11) (lambdaMJ_sec(a));
[T11,MJ_V11] = ode45(FMJ_cars_wait,[0,F],MJ_start);
R_G_MJ = MJ_V11(end:end);
%

FMJ_cars_wait_2 = @(T12,MJ_V12) (lambdaMJ_sec(a) - MJ_Flow);
[T12,MJ_V12] = ode45(FMJ_cars_wait_2,[F,A],R_G_MJ);
%
Major_end(a)=MJ_V12(end); 
MJ_start=MJ_V12(end);          
if MJ_start < 0            
    MJ_start=0;
end                                              % Updated: no else clause.
%   
Major1(:,a) = interp1(T11, MJ_V11, tGrid);       % Updated: interpolate.
end

Best Answer

Related Solutions

MATLAB: Unable to find explicit solution to differential eq

MATLAB: ??? Subscripted assignment dimension mismatch

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