I wrote the Lotka Voterra model (prey-predator) as a Scilab function and solved it with ODE. My problem is that I want to see the flows evolution. I found a solution by including flows in the resolution (in the script below only for one) but it’s really “heavy”. Anyone has a better solution?
//parameters
x=[0.04,0.005,0.3,0.2]
n = x(1);//birth rate of prey
c = x(2);//capture rate
e = x(3);//energy from capture
m = x(4);//death rate or predator
Tmax=100; // maximum time of simulation
dt=0.01; // step of time
t=0:dt:Tmax; //simulation time definition
Ci = [20,30,0]'; //initial conditions (including for 1 flow)
//Lotka Volterra model
function [dy]=LV(t, y, n, c, e, m)
GrowthP = n * y(1)
IngestC = c * y(1) * y(2)
MortC = m * y(2)
dy(1) = GrowthP - IngestC
dy(2) = IngestC * e - MortC
dy(3) = IngestC //here one flow in ode
endfunction
//resolution
sol=ode(Ci,0,t,LV)
//dataframe creation to stock data
soldef=zeros(3,10001);
//for the line 3 and form 2 to the last data it take the value,
//remove the one of the previous time step and store it in soldef
for i = 2:length(sol(3,:))
soldef(3,i)=sol(3,i)-sol(3,i-1);
end
//complete the dataframe with the stock
soldef(1:2,:)=sol(1:2,:)
Thanks for the interest and the time you give to my problem (and sorry for my English)
If you want animated display of your results, you can use the code below to plot more and more data on a graph:
[row,col]=size(soldef);
scf(0);
clf(0);
plot2d(soldef(:,1:2)',rect=[0,0,col,max(soldef)]); //plot first 2 points
en=gce(); //handle of the graphic entities
for i=3:col //plot more and more points
en.children(1).data=[1:i;soldef(3,1:i)]'; //update data matrices
en.children(2).data=[1:i;soldef(2,1:i)]';
en.children(3).data=[1:i;soldef(1,1:i)]';
//you can save the frames here...
xpause(1e4); //wait some time if amination is too fast
end
If you save all the graphs (with e.g.xs2gif()), with an external application (google for "gif maker") you can even make a .gif animation of your results to embedd in a webpage or presentation.
Related
In Julia 1.0.5, I have a function f(x::Vector{<:Real}), defined as
f(x::Vector{<:Real}) = (x[1] - 2)^2 + ( x[2] - 1 )^2
The signature is like this, because I would like to use it with the ForwardDiff package, and it works with it just fine. I give the function to ForwardDiff.gradient, and everything works like a charm.
However, I would also like to do some visualizations with PyPlot, using this same function f. Namely, I would like to draw its contour with contourf. For this purpose, I have constructed two vectors X::Vector{<:Real} and Y::Vector{<:Real}, and would like to call the same function f with them to produce the contour.
However, making the call f.([X, Y]) is not broadcasting the vectors as I would like, as I get the error
LoadError: MethodError: no method matching (::getfield(Main, Symbol("#f#1044")))(::Int64)
Closest candidates are:
f(!Matched::Array{#s25,1} where #s25<:Real)
This of course prevents me from using the contourf function, as it needs the values of f on a 2D-grid.
Do I need to define an entirely different f(x::Vector{<:Real}, y::Vector{<:Real}) to be able to plot the contour as I would like, or is there an alternative where I can avoid this?
this problem can be resolved by the power of multiple dispatch:
f(x::Vector{<:Real}) = (x[1] - 2)^2 + ( x[2] - 1 )^2
f(x::Real,y::Real) = f([x,y])
nx = 10
ny = 20
X = rand(nx) #mesh of x points
Y = rand(ny) #mesh of y points
Z = f.(transpose(X),Y) #nx x ny matrix
for the two argument gradient:
two_point_gradient(f,x,y) = ForwardDiff.gradient(f,[x,y])
G = two_point_gradient.(f,transpose(X),Y) #returns a vector of gradients, where G[i..] = gradient(f,X[i..],Y[i...])
I need to solve this differential equation using Runge-Kytta 4(5) on Scilab:
The initial conditions are above. The interval and the h-step are:
I don't need to implement Runge-Kutta. I just need to solve this and plot the result on the plane:
I tried to follow these instructions on the official "Scilab Help":
https://x-engineer.org/graduate-engineering/programming-languages/scilab/solve-second-order-ordinary-differential-equation-ode-scilab/
The suggested code is:
// Import the diagram and set the ending time
loadScicos();
loadXcosLibs();
importXcosDiagram("SCI/modules/xcos/examples/solvers/ODE_Example.zcos");
scs_m.props.tf = 5000;
// Select the solver Runge-Kutta and set the precision
scs_m.props.tol(6) = 6;
scs_m.props.tol(7) = 10^-2;
// Start the timer, launch the simulation and display time
tic();
try xcos_simulate(scs_m, 4); catch disp(lasterror()); end
t = toc();
disp(t, "Time for Runge-Kutta:");
However, it is not clear for me how I can change this for the specific differential equation that I showed above. I have a very basic knowledge of Scilab.
The final plot should be something like the picture bellow, an ellipse:
Just to provide some mathematical context, this is the differential equation that describes the pendulum problem.
Could someone help me, please?
=========
UPDATE
Based on #luizpauloml comments, I am updating this post.
I need to convert the second-order ODE into a system of first-order ODEs and then I need to write a function to represent such system.
So, I know how to do this on pen and paper. Hence, using z as a variable:
OK, but how do I write a normal script?
The Xcos is quite disposable. I only kept it because I was trying to mimic the example on the official Scilab page.
To solve this, you need to use ode(), which can employ many methods, Runge-Kutta included. First, you need to define a function to represent the system of ODEs, and Step 1 in the link you provided shows you what to do:
function z = f(t,y)
//f(t,z) represents the sysmte of ODEs:
// -the first argument should always be the independe variable
// -the second argument should always be the dependent variables
// -it may have more than two arguments
// -y is a vector 2x1: y(1) = theta, y(2) = theta'
// -z is a vector 2x1: z(1) = z , z(2) = z'
z(1) = y(2) //first equation: z = theta'
z(2) = 10*sin(y(1)) //second equation: z' = 10*sin(theta)
endfunction
Notice that even if t (the independent variable) does not explicitly appear in your system of ODEs, it still needs to be an argument of f(). Now you just use ode(), setting the flag 'rk' or 'rkf' to use either one of the available Runge-Kutta methods:
ts = linspace(0,3,200);
theta0 = %pi/4;
dtheta0 = 0;
y0 = [theta0; dtheta0];
t0 = 0;
thetas = ode('rk',y0, t0, ts, f); //the output have the same order
//as the argument `y` of f()
scf(1); clf();
plot2d(thetas(2,:),thetas(1,:),-5);
xtitle('Phase portrait', 'theta''(t)','theta(t)');
xgrid();
The output:
Apologies for the rather long name, but I wanted to be specific. I am rather new to R and coding so please go easy on me.
I have a function as follows:
myfun = function(x, y, g) {return(1 / (1 + exp(y*g%*%x)))}
where x is any data frame with n rows and d columns, y is a scalar and integer, and g is a vector of length d (i.e. same as x). I want to run this function for each row of x without using loops.
I have tried various function in the apply family similar to the code below:
apply(x = a, 1, myfun(y = 1, g = b)
where a is a 3x3 data frame and b is a vector 3 elements long. The above code gives an error that I am missing an argument from myfun, but I am obviously clueless on what to try.
Thanks for any help in advance!
Edit: My actual data frame is huge, sparse, and not very straight forward (I think), so I will include an example data frame and other variables:
a = data.frame(c1 = seq(1,3,1), c2 = seq(4,6,1), c3 = seq(7,9,1))
b = c(1,2,3)
c = 1
Also, I think I may have not clearly stated an important piece of information. I want to actually do a summation of myfun over all the rows and values of b, so I actually want the following:
answer = myfun(a[1,], c, b[1]) + myfun(a[2,], c, b[2]) + myfun(a[3,], c, b[3])
In other words, a[1,] should be applied to myfun with b[1] as they are grouped together. I also made an edit to the function above because I forgot to include return(). Hopefully, this makes things more clear. Apologies for the confusion!
I am doing a simple task on ocatve.
I have to administer drug dose at day 1,7,21 and 28..
i wrote a function like that:
function xdot = f (x,t)
a=[1;7;21;28]
drug=0; ### initially drug is zero
if (t==a)
drug=drug+57.947;
else
drug=drug+0;
endif
xdot=(-0.4077)*(x)+ drug; passing the value of drug to differential equation
endfunction
In the main file i called this function in lsode:
t=linspace(0,30,30);
x0=0;
y=lsode(#ex,x0,t); ### ex is the file name where function is written
plot(t,y,'o')
This program doesn't work.. it displays all the time zero value for drug. Can anybody help me that how to administer dug dose with certain time step by manipulating linspace function.
It looks like you have a simple clearance model, and at each time in a, you want the dose to be delivered instantaneously. That is, at each time in a, the amount of drug in the subject increases by 57.947.
If that is the model that you have in mind, implementing it in the formula for xdot will not work well. You would actually need to implement it as a "delta function", and lsode won't work with that.
Instead, I recommend solving the problem in stages, corresponding to the time intervals [0, 1], [1, 7], [7, 21], and [21, 28]. During one stage, the differental equation is simply xdot = -0.4077*x. In the first stage, the initial condition is 0. In the next stage, the initial condition is the final value of the previous stage plus the dosage amount 57.947.
Here's a script:
dose = 57.947;
a = [1 7 21 28 30];
x0 = 0.0;
t0 = 0;
t = [];
y = [];
for t1 = a
tinterval = linspace(t0, t1, 2*(t1 - t0) + 1);
yinterval = lsode(#(x, t) -0.4077*x, x0, tinterval);
t = [t tinterval];
y = [y yinterval'];
t0 = t1;
x0 = yinterval(end) + dose;
end
plot(t, y, "linewidth", 2);
The script creates this plot:
Note that the differential equation xdot = -k*x has the solution x0*exp(-k*(t-t0)), so the call to lsode could be replaced with
yinterval = x0*exp(-0.4077*(tinterval - t0));
If you do this, also remove the transpose from yinterval two lines below:
y = [y yinterval];
If you want to keep the implementation of the administration of the drug dose within the formula for xdot, you'll need to distribute it over a small time interval. It could be implemented as a short rectangular pulse of width w and height 57.974/w. You'll also need to ensure that lsode takes sufficiently small internal time steps (less than w) so that it "sees" the drug dose.
You probably want replace t==a by ismember (t, a). Also, you should erase the else clause as it has no effect on the answer.
UPDATE
Consider rewriting your function as:
function xdot = f (x,t)
xdot = -0.4077*x + 57.947*ismember (t, [1 7 21 28])
endfunction
I'm trying to solve Kepler's Equation as a step towards finding the true anomaly of an orbiting body given time. It turns out though, that Kepler's equation is difficult to solve, and the wikipedia page describes the process using calculus. Well, I don't know calculus, but I understand that solving the equation involves an infinite number of sets which produce closer and closer approximations to the correct answer.
I can't see from looking at the math how to do this computationally, so I was hoping someone with a better maths background could help me out. How can I solve this beast computationally?
FWIW, I'm using F# -- and I can calculate the other elements necessary for this equation, it's just this part I'm having trouble with.
I'm also open to methods which approximate the true anomaly given time, periapsis distance, and eccentricity
This paper:
A Practical Method for Solving the Kepler Equation http://murison.alpheratz.net/dynamics/twobody/KeplerIterations_summary.pdf
shows how to solve Kepler's equation using an iterative computing method. It should be fairly straightforward to translate it to the language of your choice.
You might also find this interesting. It's an ocaml program, part of which claims to contain a Kepler Equation solver. Since F# is in the ML family of languages (as is ocaml), this might provide a good starting point.
wanted to drop a reply in here in case this page gets found by anyone else looking for similar materials.
The following was written as an "expression" in Adobe's After Effects software, so it's javascriptish, although I have a Python version for a different app (cinema 4d). The idea is the same: execute Newton's method iteratively until some arbitrary precision is reached.
Please note that I'm not posting this code as exemplary or meaningfully efficient in any way, just posting code we produced on a deadline to accomplish a specific task (namely, move a planet around a focus according to Kepler's laws, and do so accurately). We don't write code for a living, and so we're also not posting this for critique. Quick & dirty is what meets deadlines.
In After Effects, any "expression" code is executed once - for every single frame in the animation. This restricts what one can do when implementing many algorithms, due to the inability to address global data easily (other algorithms for Keplerian motion use interatively updated velocity vectors, an approach we couldn't use). The result the code leaves behind is the [x,y] position of the object at that instant in time (internally, this is the frame number), and the code is intended to be attached to the position element of an object layer on the timeline.
This code evolved out of material found at http://www.jgiesen.de/kepler/kepler.html, and is offered here for the next guy.
pi = Math.PI;
function EccAnom(ec,am,dp,_maxiter) {
// ec=eccentricity, am=mean anomaly,
// dp=number of decimal places
pi=Math.PI;
i=0;
delta=Math.pow(10,-dp);
var E, F;
// some attempt to optimize prediction
if (ec<0.8) {
E=am;
} else {
E= am + Math.sin(am);
}
F = E - ec*Math.sin(E) - am;
while ((Math.abs(F)>delta) && (i<_maxiter)) {
E = E - F/(1.0-(ec* Math.cos(E) ));
F = E - ec * Math.sin(E) - am;
i = i + 1;
}
return Math.round(E*Math.pow(10,dp))/Math.pow(10,dp);
}
function TrueAnom(ec,E,dp) {
S=Math.sin(E);
C=Math.cos(E);
fak=Math.sqrt(1.0-ec^2);
phi = 2.0 * Math.atan(Math.sqrt((1.0+ec)/(1.0-ec))*Math.tan(E/2.0));
return Math.round(phi*Math.pow(10,dp))/Math.pow(10,dp);
}
function MeanAnom(time,_period) {
curr_frame = timeToFrames(time);
if (curr_frame <= _period) {
frames_done = curr_frame;
if (frames_done < 1) frames_done = 1;
} else {
frames_done = curr_frame % _period;
}
_fractime = (frames_done * 1.0 ) / _period;
mean_temp = (2.0*Math.PI) * (-1.0 * _fractime);
return mean_temp;
}
//==============================
// a=semimajor axis, ec=eccentricity, E=eccentric anomaly
// delta = delta digits to exit, period = per., in frames
//----------------------------------------------------------
_eccen = 0.9;
_delta = 14;
_maxiter = 1000;
_period = 300;
_semi_a = 70.0;
_semi_b = _semi_a * Math.sqrt(1.0-_eccen^2);
_meananom = MeanAnom(time,_period);
_eccentricanomaly = EccAnom(_eccen,_meananom,_delta,_maxiter);
_trueanomaly = TrueAnom(_eccen,_eccentricanomaly,_delta);
r = _semi_a * (1.0 - _eccen^2) / (1.0 + (_eccen*Math.cos(_trueanomaly)));
x = r * Math.cos(_trueanomaly);
y = r * Math.sin(_trueanomaly);
_foc=_semi_a*_eccen;
[1460+x+_foc,540+y];
You could check this out, implemented in C# by Carl Johansen
Represents a body in elliptical orbit about a massive central body
Here is a comment from the code
True Anomaly in this context is the
angle between the body and the sun.
For elliptical orbits, it's a bit
tricky. The percentage of the period
completed is still a key input, but we
also need to apply Kepler's
equation (based on the eccentricity)
to ensure that we sweep out equal
areas in equal times. This
equation is transcendental (ie can't
be solved algebraically) so we
either have to use an approximating
equation or solve by a numeric method.
My implementation uses
Newton-Raphson iteration to get an
excellent approximate answer (usually
in 2 or 3 iterations).