I was wondering if it's possible to combine for-loops/while-loops and ODE45?, please look at the example below:
I have a function (ode) which I want to solve at different ic (initial conditions):
ser = #(x) x.^(-0.3) - x.^(1.8); % the function
tspan = 1:0.02:2;
x0 = 0.5;
% x0 = 0.8;
% x0 = 1.2;
% x0 = 1.8;
% x0 = 2;
% x0 = 2.5;
[~, x_t] = ode45(#(t,x) ser(x), tspan, x0);
plot(tspan,x_t,'r-')
And plot the solution curves at the end.
Is there a way to pass the other icwithout doing it manually, like a loop? or any other way to optimize this step? If I could receive some help with this, it will be very appreciated because I have to compute many ODEs (more complex ones) at 15 to 25 different ic.
Thanks in advance!
PS. If the code has to change (e.g. different names for the x0s or solutions x_t) it will be OK!
If you have a predefined no. of ic you can use this for loop:
for i=1:NoOfic
test_mat(i,:)=test_ode_45(x(i),tspan);
end
You should predefine your matrix (test_mat) in a sufficient dimension e.g.
test_mat = zeros(NoOfic, SizeOftspan)
Obviously I have defined the variables NoOficand SizeOftspanto the specific numbers
My whole test-code is:
tspan = 1:0.02:2;
x= [0.4 0.8 1.2 1.5 2.6];
sizeOftspan = size(tspan);
sizeOfFamily= size(x);
test_mat = zeros(sizeOfFamily(2),sizeOftspan(2));
test_tes= test_ode_45(x(1),tspan);
for i=1:sizeOfFamily(2)
test_mat(i,:)=test_ode_45(x(i),tspan);
end
plot(tspan, test_mat)
Where my test_ode_45 function is a simple x*sin(t) function
Related
I'm solving a set of nonlinear simultaneous equations using Matlab's fsolve to find unknown parameters x1 and x2. The simultaneous equations have two independent parameters a and b, as defined in the root2d function:
function F = root2d(x,a,b)
F(1) = exp(-exp(-(x(1)+x(2)))) - x(2)*(a+x(1)^2);
F(2) = x(1)*cos(x(2)) + x(2)*sin(x(1)) - b;
end
I use the following code to solve the simultaneous equations, and plot the results as a 2d figure using pcolor.
alist = linspace(0.6,1.2,10);
blist = linspace(0.4,0.8,5);
% results
x1list = zeros(length(blist),length(alist));
x2list = zeros(length(blist),length(alist));
% solver options
options = optimoptions('fsolve','Display','None');
for ii = 1:length(blist)
b = blist(ii);
for jj = 1:length(alist)
a = alist(jj);
x0 = [0 0]; % init guess
[xopt,yopt,exitflag] = fsolve(#(x0)root2d(x0,a,b),x0,options);
% optimised values
x1list(ii,jj) = xopt(1);
x2list(ii,jj) = xopt(2);
success(ii,jj) = exitflag; % did solver succeed?
end
end
% plotting
figure
s = pcolor(alist(success>0),blist(success>0),x1list(success>0));
xlabel('a')
ylabel('b')
title('my data x_1')
figure
s = pcolor(alist(success>0),blist(success>0),x2list(success>0));
xlabel('a')
ylabel('b')
title('my data x_2')
However I only want to plot the x1 and x2 where the solver has successfully converged to a solution. This is where the success matrix element (or exitflag) has a value greater than 0. Usually you just write x1list(success>0) when using the plot function and Matlab omits any solutions where (success<=0), but pcolor doesn't have that functionality.
Is there a way around this? For example, displaying all (success<=0) solutions as a black area.
Yes there is!
The easiest way is to make them NaN, so they are simply not drawn.
just do
x1list(~success)=NaN;
pcolor(alist,blist,x1list)
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 am trying to solve the following least squares problem:
b(alpha)=A(alpha,beta)x(beta)
I am trying to use an alternative approach, which is to assume the functional form of x(beta) through the use of tunable parameters, say x(beta, a, c). How can I solve this problem in MATLAB for a least squares solution for those parameters?
I second the comments -this would be much easier if you gave a slightly more verbose description of your problem and most importantly add a minimal working example.
As far as I understand though, you want to solve a linear system of equations with some additional assumptions about the fitted parameters. This can be done by expressing them as an optimisation problem.
Here for example I've fitted a quadratic where the coefficients of x^0 and x^1 are both dependant on some other arbitrary parameter a (for this example a = 6 - that's what we're trying to recover from the data).
There are 2 different approaches plotted here - unconstrained and constrained optimisation. You can see that all of them approximate our data well, but only the constrained optimisation recovers a value of a close to 6 (5.728). Anyway, have a look at the code and I hope this helps with your problem somewhat. If you can, try to use the reduced number of parameters approach. It is always better to reduce your fitting problems to lower dimensional spaces if possible - much less risk of local minima and much faster solutions.
Here is the code:
close all; clear; clc;
%% Generate test data
x = 1:100;
rng(0); % Seed rng
% Polynomial where we know something about the parameters - we know that if
% the coefficient of x^0 is 'a'm then the coefficient of x^1 is (1-a).
a = 6;
y = a + (1-a).*x + 0.1*x.^2;
y = y + 30*randn(size(x)); % Add some noise
%% Fit with mrdivide and Vandermonde matrix
A = vander(x); A = A(:,end-2:end)';
b = y;
k1 = b/A;
%% Fit with an unconstrained optimiser
f = #(k) optimfun1(x,y,k);
k0 = [1 1 1]; % Starting point
k2 = fminsearch(f,k0);
%% Fit with a constrained optimiser
f = #(k) optimfun1(x,y,k);
k0 = [1 1 1];
Aeq = [0 1 1]; beq = 1; % Constrain k2 = 1 - k3 which is equivalent to k2+k3 = 1
k3 = fmincon(f,k0,[],[],Aeq,beq);
%% Fit with a reduced number of parameters
f = #(k) optimfun2(x,y,k);
k0 = [1 1];
k4 = fminsearch(f,k0);
k4 = [k4 1-k4(2)]; % Infer the last coeff.
%% Plot
plot(x,y,'ko');
hold on
plot(x,polyval(k1,x));
plot(x,polyval(k2,x));
plot(x,polyval(k3,x));
plot(x,polyval(k4,x));
legend('k^{dat} = [6.000 -5.000 0.100];',...
sprintf('k^{unc}_1 = [%.3f %.3f %.3f]',flipud(k1(:))),...
sprintf('k^{unc}_2 = [%.3f %.3f %.3f]',flipud(k2(:))),...
sprintf('k^{cns}_1 = [%.3f %.3f %.3f]',flipud(k3(:))),...
sprintf('k^{cns}_2 = [%.3f %.3f %.3f]',flipud(k4(:))),...
'location','northwest');
grid on;
xlabel('x');
ylabel('f(x)');
title(sprintf('f(x) = a + (1-a)x + 0.1x^2; a = %d',a));
function diff = optimfun1(x,y,k)
yfit = polyval(k,x);
dy = yfit-y;
diff = sum(dy.^2); % Sum of squared residuals
end
function diff = optimfun2(x,y,k)
k = [k 1-k(2)]; % Infer the last coeff.
yfit = polyval(k,x);
dy = yfit-y;
diff = sum(dy.^2);
end
Without knowing exactly how does the parameter works, it is difficult to figure out what to do. For example if the parameter is
x(beta, a, c) = a * x(beta) + c
Then your equation becomes
b(alpha)= A(alpha,beta) * (a * x(beta) + c)
b(alpha) - c*A(alpha,beta) = A(alpha,beta) * a * x(beta)
which then perhaps you can solve in the standard way (I'm treating b and A as numbers and x as the only variable here disregarding the alpha and beta). For more non-linear relation, it gets complex.
I have here an ODE to solve that has a test parameter a.
I would like a to be a function of temperature T (in this code it is solved as X(2) since it varies with time).
I would like to base it off experimental values of a and T.
Hence, I am thinking of doing it along the lines of interp1.
Here is my test dummy ODE.
function xprime = RabbitTemp(t,X)
% Model of Rabbit Population
% where,
% Xo = Initial Population of Rabbits
% X(1) = Population density of Rabbit
% X(2) = Temperature T (that varies with time)
% a = test parameter
% Interpolate here
Texptdata = [1 2 3 4];
aexptdata =[10 14 19 30];
a= interp1(Texptdata,aexptdata,X(2),'spline');
% ODE
dx = [0 0];
dx(1) = (X(1))*(1 - X(1)*a - 3*(X(2))));
dx(2) = sin(t);
xprime = [dx(1) dx(2)]';
end
I've tried running the code with my solver and it does compile fine.
However I am just wondering if the ODE solver is calling the correct interpolated a corresponding to the correct corresponding value of X(2). I cannot seem to figure out how to verify this.
Can anyone help me verify this or point me to how I can go about doing it? Much appreciated!
How can I use matlab to solve the following Ordinary Differential Equations?
x''/y = y''/x = -( x''y + 2x'y' + xy'')
with two known points, such as t=0: x(0)= x0, y(0) = y0; t=1: x(1) = x1, y(1) = y1 ?
It doesn't need to be a complete formula if it is difficult. A numerical solution is ok, which means, given a specific t, I can get the value of x(t) and y(t).
If matlab is hard to do this, mathematica is also OK. But as I am not familiar with mathematica, so I would prefer matlab if possible.
Looking forward to help, thanks!
I asked the same question on stackexchange, but haven't get good answer yet.
https://math.stackexchange.com/questions/812985/matlab-or-mathematica-solve-ordinary-differential-equations
Hope I can get problem solved here!
What I have tried is:
---------MATLAB
syms t
>> [x, y] = dsolve('(D2x)/y = -(y*D2x + 2Dx*Dy + x*D2y)', '(D2y)/x = -(y*D2x + 2Dx*Dy + x*D2y)','t')
Error using sym>convertExpression (line 2246)
Conversion to 'sym' returned the MuPAD error: Error: Unexpected 'identifier'.
[line 1, col 31]
Error in sym>convertChar (line 2157)
s = convertExpression(x);
Error in sym>convertCharWithOption (line 2140)
s = convertChar(x);
Error in sym>tomupad (line 1871)
S = convertCharWithOption(x,a);
Error in sym (line 104)
S.s = tomupad(x,'');
Error in dsolve>mupadDsolve (line 324)
sys = [sys_sym sym(sys_str)];
Error in dsolve (line 186)
sol = mupadDsolve(args, options);
--------MATLAB
Also, I tried to add conditions, such as x(0) = 2, y(0)=8, x(1) = 7, y(1) = 18, and the errors are still similar. So what I think is that this cannot be solve by dsolve function.
So, again, the key problem is, given two known points, such as when t=0: x(0)= x0, y(0) = y0; t=1: x(1) = x1, y(1) = y1 , how I get the value of x(t) and y(t)?
Update:
I tried ode45 functions. First, in order to turn the 2-order equations into 1-order, I set x1 = x, x2=y, x3=x', x4=y'. After some calculation, the equation becomes:
x(1)' = x(3) (1)
x(2)' = x(4) (2)
x(3)' = x(2)/x(1)*(-2*x(1)*x(3)*x(4)/(1+x(1)^2+x(2)^2)) (3)
x(4)' = -2*x(1)*x(3)*x(4)/(1+x(1)^2+x(2)^2) (4)
So the matlab code I wrote is:
myOdes.m
function xdot = myOdes(t,x)
xdot = [x(3); x(4); x(2)/x(1)*(-2*x(1)*x(3)*x(4)/(1+x(1)^2+x(2)^2)); -2*x(1)*x(3)*x(4)/(1+x(1)^2+x(2)^2)]
end
main.m
t0 = 0;
tf = 1;
x0 = [2 3 5 7]';
[t,x] = ode45('myOdes',[t0,tf],x0);
plot(t,x)
It can work. However, actually this is not right. Because, what I know is that when t=0, the value of x and y, which is x(1) and x(2); and when t=1, the value of x and y. But the ode functions need the initial value: x0, I just wrote the condition x0 = [2 3 5 7]' randomly to help this code work. So how to solve this problem?
UPDATE:
I tried to use the function bvp4c after I realized that it is a boundary value problem and the following is my code (Suppose the two boundry value conditions are: when t=0: x=1, y=3; when t=1, x=6, y=9. x is x(1), y is x(2) ):
1. bc.m
function res = bc(ya,yb)
res = [ ya(1)-1; ya(2)-3; yb(1) - 6; yb(2)-9];
end
2. ode.m
function dydx = ode(t,x)
dydx = [x(3); x(4); x(2)/x(1)*(-2*x(1)*x(3)*x(4)/(1+x(1)^2+x(2)^2)); -2*x(1)*x(3)*x(4)/(1+x(1)^2+x(2)^2)];
end
3. mainBVP.m
solinit = bvpinit(linspace(0,6,10),[1 0 -1 0]);
sol = bvp4c(#ode,#bc,solinit);
t = linspace(0,6);
x = deval(sol,t);
plot(t,x(1,:));
hold on
plot(t,x(2,:));
plot(t,x(3,:));
plot(t,x(4,:));
x(1,:)
x(2,:)
It can work, but I don't know whether it is right. I will check it again to make sure it is the right code.
As mentioned, this isn't a math site, so try to give code or something showing some effort.
However, the first step you need to do is turn the DE into normal form (i.e., no 2nd derivatives). You do this by making a separate variable equal to the derivative. Then, you use
syms x y % or any variable instead of x or y
to define variables as symbolic. Use matlabfunction to create a symbolic function based on these variables. Finally, you can use the ode45 function to solve the symbolic function while passing variable values. I recommend you look up the full documentation in matlab in order to understand it better, but here is a very basic syntax:
MyFun= matlabFunction(eq,'vars',{x,y});
[xout,yout]=ode45(#(x,Y) MyFun(variables),[variable values],Options);
Hopefully this puts you in the right direction, so try messing around with it and provide code if you need more help.
EDIT:
This is how I would solve the problem. Note: I don't really like the matlabFunction creator but this is simply a personal preference for various reasons I won't go into.
% Seperate function of the first order state equations
function dz = firstOrderEqns(t,z)
dz(4,1) = 0;
dz(1) = -2.*z(3).*z(1).*z(4)./(1 + z(4).^2 + z(2).^2);
dz(2) = z(1);
dz(3) = -2.*z(2).*z(3).*z(1)./(1 + z(4).^2 + z(2).^2);
dz(4) = z(3);
end
% runfirstOrderEqns
%% Initial conditions i.e. # t=0
z1 = 5; % dy/dt = 5 (you didn't specify these initial conditions,
% these will depend on the system which you didn't really specify
z2 = 0; % y = 0
z3 = 5; % dx/dt = 5 (The same as for z1)
z4 = 0; % x = 0
IC = [z1, z2, z3, z4];
%% Run simulation
% Time vector: i.e closed interval [0,20]
t = [0,20]; % This is where you have to know about your system
% i.e what is it's time domain.
% Note: when a system has unstable poles at
% certain places the solver can crash you need
% to understand these.
% using default settings (See documentation ode45 for 'options')
[T,Y] = ode45(#firstOrderEqns,t,IC);
%% Plot function
plot(T,Y(:,1),'-',T,Y(:,2),'-.',T,Y(:,3),':',T,Y(:,4),'.');
legend('dy/dt','y','dx/dt','x')
As in my comments I have made a lot of assumtions that you need to fix for example, you didn't specify what the initial conditions for the first derivatives of the states are i.e. (z1, z3) which is important for the response of the system. Also you didn't specify the time interval your interested for the simulation etc.
Note: The second m file can be used with any state function in the correct format
The following is the answer we finally get #Chriso: use matlab bvp4c function to solve this boundary value problem (Suppose the two boundry value conditions are: when t=0: x=1, y=3; when t=1, x=6, y=9. x is x(1), y is x(2) ):
1. bc.m
function res = bc(ya,yb)
res = [ ya(1)-1; ya(2)-3; yb(1) - 6; yb(2)-9];
end
2. ode.m
function dydx = ode(t,x)
dydx = [x(3); x(4); x(2)/x(1)*(-2*x(1)*x(3)*x(4)/(1+x(1)^2+x(2)^2)); -2*x(1)*x(3)*x(4)/(1+x(1)^2+x(2)^2)];
end
3. mainBVP.m
solinit = bvpinit(linspace(0,6,10),[1 0 -1 0]);
sol = bvp4c(#ode,#bc,solinit);
t = linspace(0,6);
x = deval(sol,t);
plot(t,x(1,:));
hold on
plot(t,x(2,:));
plot(t,x(3,:));
plot(t,x(4,:));
x(1,:)
x(2,:)