The following is my code. I try to model PFR in Matlab using ode23s. It works well with one component irreversible reaction. But when extending more dependent variables, 'Matrix dimensions must agree' problem shows. Have no idea how to fix it. Is possible to use other software to solve similar problems?
Thank you.
function PFR_MA_length
clear all; clc; close all;
function dCdt = df(t,C)
dCdt = zeros(N,2);
dCddt = [0; -vo*diff(C(:,1))./diff(V)-(-kM*C(2:end,1).*C(2:end,2)-kS*C(2:end,1))];
dCmdt = [0; -vo*diff(C(:,2))./diff(V)-(-kM*C(2:end,1).*C(2:end,2))];
dCdt(:,1) = dCddt;
dCdt(:,2) = dCmdt;
end
kM = 1;
kS = 0.5; % assumptions of the rate constants
C0 = [2, 2]; % assumptions of the entering concentration
vo = 2; % volumetric flow rate
volume = 20; % total volume of reactor, spacetime = 10
N = 100; % number of points to discretize the reactor volume on
init = zeros(N,2); % Concentration in reactor at t = 0
init(1,:) = C0; % concentration at entrance
V = linspace(0,volume,N)'; % discretized volume elements, in column form
tspan = [0 20];
[t,C] = ode23s(#(t,C) df(t,C),tspan,init);
end
'''
You can put a break point on the line that computes dCddt and observe that the size of the matrices C and V are different.
>> size(C)
ans =
200 1
>> size(V)
ans =
100 1
The element-wise divide operation, ./, between these two variables would then result in the error that you mentioned.
Per ode23s's help, the output of the call to dCdt = df(t,C) needs to be a vector. However, you are returning a matrix of size 100x2. In the next call to the same function, ode32s converts it to a vector when computing the value of C, hence the size 200x1.
In the GNU octave interpretation of Matlab behavior, one has to explicitly make sure that the solver only sees flat one-dimensional state vectors. These have to be translated forward and back in the application of the model.
Explicitly reading the object A as flat array A(:) forgets the matrix dimension information, these can be added back with the reshape(A,m,n) command.
function dCdt = df(t,C)
C = reshape(C,N,2);
...
dCdt = dCdt(:);
end
...
[t,C] = ode45(#(t,C) df(t,C), tspan, init(:));
Related
I am struggling with solving a problem as efficient as possible.
I have two equations and want to solve them together. Furthermore, I want to solve them for four different cases. I can solve them independently altering the code for each case. I can also solve them by accessing a vector that contains the desired value for one of them (e.g., the first value of A) by using arrayfun. However I can't manage to do all of them together.
My code:
clc
clear all
close all
% scalar parameters
g = 9.807; % Gravity constant, m/s2
d = 1.225; % Density of air, kg/m3
x = 1000; % Height, m
% vector parameters
%t = 0:2:10; % time, seconds
A = [1.2; 1.7; 1.8; 0.3]; % Area, m^2
m = [82; 84; 90; 25]; % Mass, kg
Cd = [0.3; 1.14; 0.29; 0.045]; % Drag coefficient, -
syms v t
eqn1 = 2*m./(A.*Cd*d).* log(abs(cosh(t.*sqrt(A.*Cd*d*g./(2*m))))) == x;
eqn2 = (2*g*m./(d*A.*Cd)).*tanh(t.* sqrt((g*d*Cd.*A)./(2*m))) - v == 0;
eqns=[eqn1 eqn2];
variables = [v t];
result = arrayfun(#vpasolve, eqns, 'uniform', 0)
%disp('v='),disp(eval(v));
%disp('t='),disp(eval(t));
I get a result for t (which is weirdly negative and I don't know why), but I only get a {1×1 struct} for v, which I don't want. I know I can solve this also by using a for loop, but I wanted to make it more efficient.
I tried the code written above and solved it in various forms, however not as desired.
The way you have it implemented you're effectively trying to solve four equations with two unknowns, which isn't supported by vpasolve except in the case of polynomial systems.
It looks like you're trying to solve the same system for four different combinations of values for the parameters A, m, and Cd. The easiest way to do that is to just break this out as a for loop (there is simply no benefit to using arrayfun here – speed or otherwise). You'll need to create some temporary symbolic variable for the parameter (there are a few ways this could be done). Here is example code to do just that:
% scalar parameters
g = 9.807; % Gravity constant, m/s2
d = 1.225; % Density of air, kg/m3
x = 1000; % Height, m
% vector parameters
%t = 0:2:10; % time, seconds
A = [1.2; 1.7; 1.8; 0.3]; % Area, m^2
m = [82; 84; 90; 25]; % Mass, kg
Cd = [0.3; 1.14; 0.29; 0.045]; % Drag coefficient, -
syms v t real
syms A_sym m_sym Cd_sym real % Temporary symbolic variables
eqn1 = 2*m_sym./(A_sym.*Cd_sym*d).* log(abs(cosh(t.*sqrt(A_sym.*Cd_sym*d*g./(2*m_sym))))) == x;
eqn2 = (2*g*m_sym./(d*A_sym.*Cd_sym)).*tanh(t.* sqrt((g*d*Cd_sym.*A_sym)./(2*m_sym))) - v == 0;
eqns=[eqn1 eqn2];
variables = [v t];
n = length(A); % Number of prameter sets
result = cell(n,1); % Preallocate resultant cell array
for i = 1:n
% Substitute in numeric values for i-th parameter set
eqns_sub = subs(eqns,{A_sym,m_sym,Cd_sym},{A(i),m(i),Cd(i)});
% Solve and store in cell array
result{i} = vpasolve(eqns_sub,variables);
end
The output, result, is a 4-by-1 cell array of structs with your two variables, v and t, as VPA-valued fields for each.
You could replace the last few lines with the following if what you want is a numeric array as the output (this assumes that only one solution is found for each parameter set):
n = length(A); % Number of prameter sets
result = zeros(n,length(variables)); % Preallocate resultant array
for i = 1:n
% Substitute in numeric values for i-th parameter set
eqns_sub = subs(eqns,{A_sym,m_sym,Cd_sym},{A(i),m(i),Cd(i)});
% Solve, convert to double precision, and store in array
out = vpasolve(eqns_sub,variables);
result(i,:) = double([out.v out.t]);
end
I have written a script to compute and solve a simple inverted pendalum system.Now suppose that I want to solve the nonlinear dynamic equation of the system with ODE45 function with different values of initial conditions.How could I use a for loop to solve for state vector of X for different values of initial conditions?I wrote a for loop to do that but I could not get the answer I wanted.Help me please.Here are my function and mfile as follows:
function xDot = of(x,g,L,u)
xDot = zeros(2,1);
xDot(1) = x(2);
xDot(2) = ((g./L)*sin(x(1)))+u;
end
And this is my main code:
clc;
clear;close all;
%% Solve The Nonlinear Equation
L = 1;
g = 9.81;
h = 0.25;
t = [0:h:5];
A = [0 1;(g/L) 0];
B =[0 1]';
Ics = [pi,0;pi/2 0;pi/5 0;0.001 0;pi 0.5;pi/2 0.5;pi/5 0.5;0.001 0.5];
[Poles,~] = eig(A); %Poles Of Closed LOop System
R = 0.01;
Q = eye(2);
K = lqr(A,B,Q,R);
u = #(x)-K*(x);
for i=1:size(Ics,1)
[~,X] = ode45(#(t,x)of(x,g,L,u(x)),t,Ics(i,:));
end
Also note that I want the first column of X vector which is the angular displacements of the pendulum in each iteration because the second column of X vector in ODE45 is always the Derivative of the main state vector.
You can store all the integration outputs for the different initial conditions in a 3D array.
The number of rows of Xout will equal the number of time steps at which you want to evaluate your solution, so numel(t). The number of columns is the number of states, and then the third dimension will be the number of initial conditions you want to test.
Xout = zeros(numel(t), size(Ics, 2), size(Ics, 1)); % initialize the 3D array
for k = 1:size(Ics, 1)
[~, Xout(:, :, k)] = ode45(#(t, x)of(x, g, L, u(x)), t, Ics(k, :));
end
I'm new to this concept of parallel pooling on MATLAB (I'm using the version 2019 a) and coding. This code that I'm going to share with you was available on the net, with some few modifications that I've made it for my requirements.
Problem Statement: I'm having a non-linear system (Rossler equation) & I have to plot its Bifurcation diagram, I tried to do it normally using for loop but its computation time was too much and my computer got hanged several times, so I got an advice to parallel pool my code in order to come out of this problem. I tried to learn how to parallel pool using MATLAB on the net but still I'm not able to resolve my Issues as there are still some problems since there are 2 parfor loops in my code I'm having problems with Indexing and in assignment of the global parameter (Please note: This code is written for normal execution without using parallel pooling).
I'm attaching my code below here, please excuse if I've mentioned a lot many lines of codes.
clc;
a = 0.2; b = 0.2; global c;
crange = 1:0.05:90; % Range for parameter c
k = 0; tspan = 0:0.1:500; % Time interval for solving Rossler system
xmax = []; % A matrix for storing the sorted value of x1
for c = crange
f = #(t,x) [-x(2)-x(3); x(1)+a*x(2); b+x(3)*(x(1)-c)];
x0 = [1 1 0]; % initial condition for Rossler system
k = k + 1;
[t,x] = ode45(f,tspan,x0); % call ode() to solve Rossler system
count = find(t>100); % find all the t values which is >10
x = x(count,:);
j = 1;
n = length(x(:,1)); % find the length of vector x1(x in our problem)
for i=2 : n-1
% check for the min value in 1st column of sol matrix
if (x(i-1,1)+eps) < x(i,1) && x(i,1) > (x(i+1,1)+eps)
xmax(k,j)=x(i,1); % Sorting the values of x1 in increasing order
j=j+1;
end
end
% generating bifurcation map by plotting j-1 element of kth row each time
if j>1
plot(c,xmax(k,1:j-1),'k.','MarkerSize',1);
end
hold on;
index(k)=j-1;
end
xlabel('Bifuracation parameter c');
ylabel('x max');
title('Bifurcation diagram for c');
This can be made compatible with parfor by taking a few relatively simple steps. Firstly, parfor workers cannot produce on-screen graphics, so we need to change things to emit a result. In your case, this is not totally trivial since your primary result xmax is being assigned-to in a not-completely-uniform manner - you're assigning different numbers of elements on different loop iterations. Not only that, it appears not to be possible to predict up-front how many columns xmax needs.
Secondly, you need to make some minor changes to the loop iteration to be compatible with parfor, which requires consecutive integer loop iterates.
So, the major change is to have the loop write individual rows of results to a cell array I've called xmax_cell. Outside the parfor loop, it's trivial to convert this back to matrix form.
Putting all this together, we end up with this, which works correctly in R2019b as far as I can tell:
clc;
a = 0.2; b = 0.2;
crange = 1:0.05:90; % Range for parameter c
tspan = 0:0.1:500; % Time interval for solving Rossler system
% PARFOR loop outputs: a cell array of result rows ...
xmax_cell = cell(numel(crange), 1);
% ... and a track of the largest result row
maxNumCols = 0;
parfor k = 1:numel(crange)
c = crange(k);
f = #(t,x) [-x(2)-x(3); x(1)+a*x(2); b+x(3)*(x(1)-c)];
x0 = [1 1 0]; % initial condition for Rossler system
[t,x] = ode45(f,tspan,x0); % call ode() to solve Rossler system
count = find(t>100); % find all the t values which is >10
x = x(count,:);
j = 1;
n = length(x(:,1)); % find the length of vector x1(x in our problem)
this_xmax = [];
for i=2 : n-1
% check for the min value in 1st column of sol matrix
if (x(i-1,1)+eps) < x(i,1) && x(i,1) > (x(i+1,1)+eps)
this_xmax(j) = x(i,1);
j=j+1;
end
end
% Keep track of what's the maximum number of columns
maxNumCols = max(maxNumCols, numel(this_xmax));
% Store this row into the output cell array.
xmax_cell{k} = this_xmax;
end
% Fix up xmax - push each row into the resulting matrix.
xmax = NaN(numel(crange), maxNumCols);
for idx = 1:numel(crange)
this_max = xmax_cell{idx};
xmax(idx, 1:numel(this_max)) = this_max;
end
% Plot
plot(crange, xmax', 'k.', 'MarkerSize', 1)
xlabel('Bifuracation parameter c');
ylabel('x max');
title('Bifurcation diagram for c');
The equations can be found here. As you can see it is set of 8 scalar equations closed to 3 matrix ones. In order to let Matlab know that equations are matrix - wise, I declare variable time dependent vector functions as:
syms t p1(t) p2(t) p3(t)
p(t) = symfun([p1(t);p2(t);p3(t)], t);
p = formula(p(t)); % allows indexing for vector p
% same goes for w(t) and m(t)...
Known matrices are declared as follows:
A = sym('A%d%d',[3 3]);
Fq = sym('Fq%d%d',[2 3]);
Im = diag(sym('Im%d%d',[1 3]));
The system is now ready to be modeled according to guide:
eqs = [diff(p) == A*w + Fq'*m,...
diff(w) == -Im*p,...
Fq*w == 0];
vars = [p; w; m];
At this point, when I try to reduce index (since it equals 2), I receive following error:
[DAEs,DAEvars] = reduceDAEIndex(eqs,vars);
Error using sym/reduceDAEIndex (line 95)
Expecting as many equations as variables.
The error would not arise if we had declared all variables as scalars:
syms A Im Fq real p(t) w(t) m(t)
Quoting symfun documentation (tips section):
Symbolic functions are always scalars, therefore, you cannot index into a function.
However it is hard for me to believe that it's not possible to solve these equations matrix - wise. Obviously, one can expand it to 8 scalar equations, but the multi body system concerned here is very simple and the aim is to be able to solve complex ones - hence the question: is it possible to solve matrix DAE in Matlab, and if so - what has to be fixed in order for this to work?
Ps. I have another issue with Matlab DAE solver: input variables (known coefficient functions) for my model are time variant. As far as example is concerned, they are constant in all domain, however for my problem they change in time. This problem has been brought out here. I would be grateful if you referred to it, should you have any solution.
Finally, I managed to find correct syntax for this problem. I made a mistake of treating matrix variables (such as A, Fq) as a single entity. Below I present code that utilizes matrix approach and solves this particular DAE:
% Define symbolic variables.
q = sym('q%d',[3 1]); % state variables
a = sym('a'); k = sym('k'); % constant parameters
t = sym('t','real'); % independent variable
% Define system variables and group them in vectors:
p1(t) = sym('p1(t)'); p2(t) = sym('p2(t)'); p3(t) = sym('p3(t)');
w1(t) = sym('w1(t)'); w2(t) = sym('w2(t)'); w3(t) = sym('w3(t)');
m1(t) = sym('m1(t)'); m2(t) = sym('m2(t)');
pvect = [p1(t); p2(t); p3(t)];
wvect = [w1(t); w2(t); w3(t)];
mvect = [m1(t); m2(t)];
% Define matrices:
mass = diag(sym('ms%d',[1 3]));
Fq = [0 -1 a;
0 0 1];
A = [1 0 0;
0 1 a;
0 a -q(1)*a] * k;
% Define sets of equations and aggregate them into one set:
set1 = diff(pvect,t) == A*wvect + Fq'*mvect;
set2 = mass*diff(wvect,t) == -pvect;
set3 = Fq*wvect == 0;
eqs = [set1; set2; set3];
% Close all system variables in one vector:
vars = [pvect; wvect; mvect];
% Reduce index of the system and remove redundnat equations:
[DAEs,DAEvars] = reduceDAEIndex(eqs,vars);
[DAEs,DAEvars] = reduceRedundancies(DAEs,DAEvars);
[M,F] = massMatrixForm(DAEs,DAEvars);
We receive very simple 2x2 ODE for two variables p1(t) and w1(t). Keep in mind that after reducing redundancies we got rid of all elements from state vector q. This means that all left variables (k and mass(1,1)) are not time dependent. If there had been time dependency of some variables within the system, the case would have been much harder to solve.
% Replace symbolic variables with numeric ones:
M = odeFunction(M, DAEvars,mass(1,1));
F = odeFunction(F, DAEvars, k);
k = 2000; numericMass = 4;
F = #(t, Y) F(t, Y, k);
M = #(t, Y) M(t, Y, numericMass);
% set the solver:
opt = odeset('Mass', M); % Mass matrix of the system
TIME = [1; 0]; % Time boundaries of the simulation (backwards in time)
y0 = [1 0]'; % Initial conditions for left variables p1(t) and w1(t)
% Call the solver
[T, solution] = ode15s(F, TIME, y0, opt);
% Plot results
plot(T,solution(:,1),T,solution(:,2))
I have a function f(t) and want to get all the points where it intersects y=-1 and y=1 in the range 0 to 6*pi.
The only way I cold do it is ploting them and trying to locate the x-axis pt where f(t) meets the y=1 graph. But this doesn't give me the exact point. Instead gives me a near by value.
clear;
clc;
f=#(t) (9*(sin(t))/t) + cos(t);
fplot(f,[0 6*pi]);
hold on; plot(0:0.01:6*pi,1,'r-');
plot(0:0.01:6*pi,-1,'r-');
x=0:0.2:6*pi; h=cos(x); plot(x,h,':')
You are essentially trying to solve a system of two equations, at least in general. For the simple case where one of the equations is a constant, thus y = 1, we can solve it using fzero. Of course, it is always a good idea to use graphical means to find a good starting point.
f=#(t) (9*(sin(t))./t) + cos(t);
y0 = 1;
The idea is if you want to find where the two curves intersect, is to subtract them, then look for a root of the resulting difference.
(By the way, note that I used ./ for the divide, so that MATLAB won't have problem for vector or array input in f. This is a good habit to develop.)
Note that f(t) is not strictly defined in MATLAB at zero, since it results in 0/0. (A limit exists of course for the function, and can be evaluated using my limest tool.)
limest(f,0)
ans =
10
Since I know the solution is not at 0, I'll just use the fzero bounds from looking there for a root.
format long g
fzero(#(t) f(t) - y0,[eps,6*pi])
ans =
2.58268206208857
But is this the only root? What if we have two or more solutions? Finding all the roots of a completely general function can be a nasty problem, as some roots may be infinitely close together, or there may be infinitely many roots.
One idea is to use a tool that knows how to look for multiple solutions to a problem. Again, found on the file exchange, we can use research.
y0 = 1;
rmsearch(#(t) f(t) - y0,'fzero',1,eps,6*pi)
ans =
2.58268206208857
6.28318530717959
7.97464518075547
12.5663706143592
13.7270312712311
y0 = -1;
rmsearch(#(t) f(t) - y0,'fzero',1,eps,6*pi)
ans =
3.14159265358979
5.23030501095915
9.42477796076938
10.8130654321854
15.707963267949
16.6967239156574
Try this:
y = fplot(f,[0 6*pi]);
now you can analyse y for the value you are looking for.
[x,y] = fplot(f,[0 6*pi]);
[~,i] = min(abs(y-1));
point = x(i);
this will find one, nearest crossing point. Otherwhise you going through the vector with for
Here is the variant with for I often use:
clear;
clc;
f=#(t) (9*(sin(t))/t) + cos(t);
fplot(f,[0 6*pi]);
[fx,fy] = fplot(f,[0 6*pi]);
hold on; plot(0:0.01:6*pi,1,'r-');
plot(0:0.01:6*pi,-1,'r-');
x=0:0.2:6*pi; h=cos(x); plot(x,h,':')
k = 1; % rising
kt = 1; % rising
pn = 0; % number of crossings
fy = abs(fy-1);
for n = 2:length(fx)
if fy(n-1)>fy(n)
k = 0; % falling
else
k = 1; % rising
end
if k==1 && kt ==0 % change from falling to rising
pn = pn +1;
p(pn) = fx(n);
end
kt = k;
end
You can make this faster, if you make an mex-file of this...