I have a ode system that I solve with Matlab. I want to find the steady state of the system and to do so, I use the event function as described here.
But some times, the solver doesn't stop even if the criterium is achieved.
For example, if you solve the following system with x0 = 10 the solver will stop before 2000 but with x0 = 0.0001 it won't.
The event function (eventfun_t.m)
function [x,isterm,dir] = eventfun_t(t,y)
dy = test_systeme(t,y);
x = norm(dy) - 1e-3;
isterm = 1;
dir = 0; %or -1, doesn't matter
end
The system (test_systeme.m)
function dx = test_systeme(t,x)
v = x./(x+1);
dx = -v;
end
Solve the system
x0 = 10;
eventfonction = #(t,y) eventfun_t(t,y);
optionsode=odeset('Events',eventfonction);
[t x]=ode15s(#(t,x) test_systeme(t,x),[0 2000],x0,optionsode);
I suppose it's because with x0 = 0.0001 norm(dy) is already lower than 1e-3 but in that case, how can I stop the solver without checking by myself ?
The events function checks for sign changes in value. So if value(t=0)<0 and value(0 < t < t_end)<0, then it will never trigger.
A way I've gotten around this is to use conditional statements:
value = 1;
if norm(dy)<1e-3&&t~=0
value = -1;
end
The t~=0 statement allows value to change sign after the first step if it's already less than the theshold.
Related
The function below, bisection, is supposed to find a root given three inputs: a function f, and an interval defined using the two parameters a and b. The intention is that the value of a and b are changed within the function to approach a common point, as long as their signs are different.
When I call my function like this:
bisection( #(x)x-1 ,-2,3)
no output is returned. What am I doing wrong?
function X = bisection(f,a,b)
if ge((f(a)*f(b)),0)
disp('Wrong')
return;
end
X = (a+b)/2;
while abs(X)>0.01
if f(X)*f(a)>0
X=a;
else
X=b;
end
end
Enter the Infinite!
Well done! You've written your first (and not last, believe me) infinite loop. The problem is threefold. Firstly your stop condition was abs(X) and should have been abs(f(X)) - you don't care for X to be zero, you want f(X) to be zeros. Secondly you don't update your your X correctly so your break condition is never hit (unless you are lucky to give this function symmetrical a,b bounds around the zero of the function). You could see this easily by adding a line like disp(f(X)); pause(0.5); somewhere in the while-loop.
In general try to avoid infinite loops with some explicit stop condition. In my code below I've put in the interaction limit past which the algorithm will just stop (it would be more elegant to catch that condition and warn the user of hitting the iteration limit...).
function x0 = bisection(f,a,b)
assert(f(a)*f(b)<0,'Invalid f(x) range. f(a)*f(b) >= 0');
tol = 0.00001; % Tolerance
iter_limit = 10000; % Limit of number of iterations
iter = 0;
x0 = (a+b)/2; % Midpoint
while abs(f(x0)) > tol && iter < iter_limit
if f(x0)*f(a) > 0
a = x0; % Zero to the right of the midpoint
else
b = x0; % Zero to the left
end
x0 = (a+b)/2; % Recalculate midpoint
iter = iter + 1;
end
end
This should work no problem with
f = #(x)x-1;
bisection(f,-2,3);
I get something like 0.999992370... which is within specified tolerance from the actual answer (1).
The issue I have is having to compute the derivative (in real time) of the solution produced by ode45 within the events function.
Some pseudo-code to explain what I'm mean is,
function dx = myfunc(t,x,xevent,ie)
persistent xs
persistent dx2
% xevent is the solution at the event
if isempty(dx2) == 1
dx2 = 0;
end
k = sign(dx2*x(1));
if ie(end) == 1
xs = xevent
elseif ie(end) == 2
xs = xevent
end
dx(1) = x(2);
dx(2) = function of k,x(1),x(2), and xs;
dx2 = dx(2);
end
function [value,isterminal,direction] = myeventfcn(t,x)
% dx2 = some function of x
if dx2*x(1)>=0
position = [dx2*x(1); dx2*x(1)];
isterminal = [1; 1];
direction = [1 ; -1 ];
elseif dx2*x(1)<0
position = [dx2*x(1); dx2*x(1)];
isterminal = [1; 1];
direction = [-1 ; 1 ];
end
I know that if I didn't need to use the solution at the event within myfunc I could just compute dx=myfunc(t,x) within my event function and use dx(2), yet since xevent is used within myfunc I can't input xevent.
I know there is a way of inputting constant parameters within the event function, but since it's the solution at the event location, which also changes, I'm not sure how to go about doing that.
My work around to this is to approximate dx(2) using the solution x. What I wanted to know is if it will be satisfactory to just use a finite difference approximation here, using a small fixed step size relative to the step size od45 takes, which is a variable step size.
As a note, the reason I have myeventfcn split by the if statement is to know what the direction the event is crossed, since it will update within myfunc.
Also, I need to use dx(2) of the previous successful time step and so that's why I have dx2 defined as a persistent variable. I believe having k=sign(dx(2)*x(1)) is okay to do, since my event depends on dx(2)*x(1), and so I won't be introducing any new discontinuities with the sign function.
Thanks for any help!
I have a Mechanical system with following equation:
xdot = Ax+ Bu
I want to solve this equation in a loop because in every step I need to update u but solvers like ode45 or lsim solving the differential equation for a time interval.
for i = 1:10001
if x(i,:)>= Sin1 & x(i,:)<=Sout2
U(i,:) = Ueq - (K*(S/Alpha))
else
U(i,:) = Ueq - (K*S)
end
% [y(i,:),t,x(i+1,:)]=lsim(sys,U(i,:),(time=i/1000),x(i,:));
or %[t,x] = ode45(#(t,x)furuta(t,x,A,B,U),(time=i/1000),x)
end
Do I have another ways to solve this equation in a loop for a single time(Not single time step).
There are a number of methods for updating and storing data across function calls.
For the ODE suite, I've come to like what is called "closures" for doing that.
A closure is basically a nested function accessing or modifying a variable from its parent function.
The code below makes use of this feature by wrapping the right-hand side function passed to ode45 and the 'OutputFcn' in a parent function called odeClosure().
You'll notice that I am using logical-indexing instead of an if-statement.
Vectors in if-statements will only be true if all elements are true and vice-versa for false.
Therefore, I create a logical array and use it to make the denominator either 1 or Alpha depending on the signal value for each row of x/U.
The 'OutputFcn' storeU() is called after a successful time step by ode45.
The function grows the U storage array and updates it appropriately.
The array U will have the same number of columns as the number of solution points requested by tspan (12 in this made-up example).
If a successful full step leaps over any requested points, the function is called with intermediate all requested times and their associated solution values (so x may be rectangular and not just a vector); this is why I used bsxfun in storeU and not in rhs.
Example function:
function [sol,U] = odeClosure()
% Initilize
% N = 10 ;
A = [ 0,0,1.0000,0; 0,0,0,1.0000;0,1.3975,-3.7330,-0.0010;0,21.0605,-6.4748,-0.0149];
B = [0;0;0.6199;1.0752 ] ;
x0 = [11;11;0;0];
K = 100;
S = [-0.2930;4.5262;-0.5085;1.2232];
Alpha = 0.2 ;
Ueq = [0;-25.0509;6.3149;-4.5085];
U = Ueq;
Sin1 = [-0.0172;-4.0974;-0.0517;-0.2993];
Sout2 = [0.0172 ; 4.0974; 0.0517; 0.2993];
% Solve
options = odeset('OutputFcn', #(t,x,flag) storeU(t,x,flag));
sol = ode45(#(t,x) rhs(t,x),[0,0.01:0.01:0.10,5],x0,options);
function xdot = rhs(~,x)
between = (x >= Sin1) & (x <= Sout2);
uwork = Ueq - K*S./(1 + (Alpha-1).*between);
xdot = A*x + B.*uwork;
end
function status = storeU(t,x,flag)
if isempty(flag)
% grow array
nAdd = length(t) ;
iCol = size(U,2) + (1:nAdd);
U(:,iCol) = 0 ;
% update U
between = bsxfun(#ge,x,Sin1) & bsxfun(#le,x,Sout2);
U(:,iCol) = Ueq(:,ones(1,nAdd)) - K*S./(1 + (Alpha-1).*between);
end
status = 0;
end
end
I am solving a set of ODEs (dy/dt) at t=0, all initial conditions t=0 y_0=(0,0,0). Can I add some number to the y values at different times (e.g., at t=10, y1 should be added to that number; at t=20, y2 should be added to that number, etc.) and solve the equations?
Inserting large discontinuities in your ODE in the way you suggest (and the way illustrated by #macduff) can lead to less precision and longer computation times (especially with ode45 - ode15s might be a better option or at least make sure that your absolute and relative tolerances are suitable). You've effectively produced a very stiff system. If you want to add some number to the ODEs starting at a specific time keep in mind that the solver only evaluates these equations at specific points in time of its own choosing. (Don't be mislead by the fact that you can obtain fixed step-size outputs by specifying tspan as more than two elements – all of Matlab's solvers are variable step-size solvers and choose their true steps based on error criteria.)
A better option is to integrate the system piecewise and append the resultant outputs from each run together:
% t = 0 to t = 10, pass parameter a = 0 to add to ODEs
a = 0;
tspan = [0 10];
[T,Y] = ode45(#(t,y)myfun(t,y,a),tspan,y_0);
% t = 10 to t = 20, pass parameter a = 10 to add to ODEs
a = 10;
[t,y] = ode45(#(t,y)myfun(t,y,a),tspan+T(end),Y(end,:));
T = [T;t(2:end)];
Y = [Y;y(2:end,:)];
% t = 20 to t = 30, pass parameter a = 20 to add to ODEs
a = 20;
[t,y] = ode45(#(t,y)myfun(t,y,a),tspan+T(end),Y(end,:));
T = [T;t(2:end)];
Y = [Y;y(2:end,:)];
The Matlab editor may complain about the array T and Y not being preallocated and/or growing, but it's fine in this case as they're growing in large chunks only a few times. Alternatively, if you want fixed output step-sizes, you can do this:
dt = 0.01;
T = 0:dt:30;
Y = zeros(length(T),length(y_0));
% t = 0 to t = 10, pass parameter a = 0 to add to ODEs
a = 0;
[~,Y(1:10/dt+1,:)] = ode45(#(t,y)myfun(t,y,a),T(1:10/dt+1),y_0);
% t = 10 to t = 20, pass parameter a = 10 to add to ODEs
a = 10;
[~,Y(10/dt+1:20/dt+1,:)] = ode45(#(t,y)myfun(t,y,a),T(10/dt+1:20/dt+1),Y(10/dt+1,:));
% t = 20 to t = 30, pass parameter a = 20 to add to ODEs
a = 20;
[~,Y(20/dt+1:end,:)] = ode45(#(t,y)myfun(t,y,a),T(20/dt+1:end),Y(20/dt+1,:));
One could easily convert both of the above blocks of code to more compact for loops if desired.
In both cases your ODE function myfun incorporates the parameter a this way:
function ydot = myfun(t,y,a)
y(1) = ... % add a however you like
...
Ok, like Simon McKenzie says we really need more info on your urgent issue, but I think I can help. From what you've given us, I'll assume you have a function myfun that you pass to something like ode45
y_0 = [0,0,0];
% here Tfinal is the time in seconds that you want to simulate to
% and specify the tspan so that you will get a solution at each whole
% number of seconds, I believe
[t,y] = ode45(#myfun,[0:0.1:Tfinal],y_0);
Somewhere you have defined your function, here I'll call it myfun
function dy = myfun(t,y)
% here let's check to see if it's time to add your offsets
% still, you may want to put a little fudge factor for the time, but
% you'll have to experiment, I'll set it up though
EPS = 1e-12;
if( t < 10 + EPS || t > 10 - EPS )
y(1) = y(1) + 10;
.
.
.
.
% this only makes sense if you then use y(1) in the compuatation
Otherwise, just add the offset to the returned solution vector, i.e.
idx10 = find( t == 10 ); % should be there since it's in the tspan
y(idx10:end) = y(idx10:end) + 10; % I guess you add it from that point on?
I'm running a set of ODEs with ode45 in MATLAB and I need to save one of the variables (that's not the derivative) for later use. I'm using the function 'assignin' to assign a temporary variable in the base workspace and updating it at each step. This seems to work, however, the size of the array does not match the size of the solution vector acquired from ode45. For example, I have the following nested function:
function [Z,Y] = droplet_momentum(theta,K,G,P,zspan,Y0)
options = odeset('RelTol',1e-7,'AbsTol',1e-7);
[Z,Y] = ode45(#momentum,zspan,Y0,options);
function DY = momentum(z,y)
DY = zeros(4,1);
%Entrained Total Velocity
Ve = sin(theta)*(y(4));
%Total Relative Velocity
Urs = sqrt((y(1) - y(4))^2 + (y(2) - Ve*cos(theta))^2 + (y(3))^2);
%Coefficients
PSI = K*Urs/y(1);
PHI = P*Urs/y(1);
%Liquid Axial Velocity
DY(1) = PSI*sign(y(1) - y(4))*(1 + (1/6)*(abs(y(1) - y(4))*G)^(2/3));
%Liquid Radial Velocity
DY(2) = PSI*sign(y(2) - Ve*cos(theta))*(1 + (1/6)*(abs(y(2) - ...
Ve*cos(theta))*G)^(2/3));
%Liquid Tangential Velocity
DY(3) = PSI*sign(y(3))*(1 + (1/6)*(abs(y(3))*G)^(2/3));
%Gaseous Axial Velocity
DY(4) = (1/z/y(4))*((PHI/z)*sign(y(1) - y(4))*(1 + ...
(1/6)*(abs(y(1) - y(4))*G)^(2/3)) + Ve*Ve - y(4)*y(4));
assignin('base','Ve_step',Ve);
evalin('base','Ve_out(end+1) = Ve_step');
end
end
In the above code, theta (radians), K (negative value), P, & G are constants and for the sake of this example can be taken as any value. Zspan is just the integration time step for the ODE solver and Y0 is the initial conditions vector (4x1). Again, for the sake of this example these can take any reasonable value. Now in the main file, the function is called with the following:
Ve_out = 0;
[Z,Y] = droplet_momentum(theta,K,G,P,zspan,Y0);
Ve_out = Ve_out(2:end);
This method works without complaint from MATLAB, but the problem is that the size of Ve_out is not the same as the size of Z or Y. The reason for this is because MATLAB calls the ODE function multiple times for its algorithm, so the solution is going to be slightly smaller than Ve_out. As am304 suggested, I could just simply calculated DY by giving the ode function a Z and Y vector such as DY = momentum(Z,Y), however, I need to get this working with 'assignin' (or similar method) because another version of this problem has an implicit dependence between DY and Ve and it would be too computationally expensive to calculate DY at every iteration (I will be running this problem for many iterations).
Ok, so let's start off with a quick example of an SSCCE:
function [Z,Y] = khan
options = odeset('RelTol',1e-7,'AbsTol',1e-7);
[Z,Y] = ode45(#momentum,[0 12],[0 0],options);
end
function Dy = momentum(z,y)
Dy = [0 0]';
Dy(1) = 3*y(1) + 2* y(2) - 2;
Dy(2) = y(1) - y(2);
Ve = Dy(1)+ y(2);
assignin('base','Ve_step',Ve);
evalin('base','Ve_out(end+1) = Ve_step;');
assignin('base','T_step',z);
evalin('base','T_out(end+1) = T_step;');
end
By running [Z,Y] = khan as the command line, I get a complete functional code that demonstrates your problem, without all the headaches associated. My patience for this has been exhausted: live and learn.
This seems to work, however, the size of the array does not match the
size of the solution vector acquired from ode45
Note that I added two lines to your code which extracts time variable. From the command prompt, one simply has to run the following to understand what's going on:
Ve_out = [];
T_out = [];
[Z,Y] = khan;
size (Z)
size (T_out)
size (Ve_out)
plot (diff(T_out))
ans =
109 1
ans =
1 163
ans =
1 163
Basically ode45 is an iterative algorithm, which means it will regularly course correct (that's why you regularly see diff(T) = 0). You can't force the algorithm to do what you want, you have to live with it.
So your options are
1. Use a fixed step algorithm
2. Have a function call that reproduces what you want after the ode45 algorithm has done its dirty work. (am304's solution)
3. Collects the data with the time function, then have an algorithm parse through everything to removes the extra data.
Can you not do something like that? Obviously check the sizes of the matrices/vectors are correct and amend the code accordingly.
[Z,Y] = droplet_momentum2(theta,K,G,P,zspan,Y0);
DY = momentum(Z,Y);
Ve = sin(theta)*(0.5*z*DY(4) + y(4));
i.e. once the ODE is solved, computed the derivative DY as a function of Z and Y (which have just been solved by the ODE) and finally Ve.