I am trying to compute the maximum value of the solution of a system with two ODEs using Octave. I have firstly solved the system itself:
function xdot = f (x,t)
a1=0.00875;
a2=0.075;
b1=7.5;
b2=2.5;
d1=0.0001;
d2=0.0001;
g=4*10^(-8);
K1=5000;
K2=2500;
n=2;
m=2;
xdot = zeros(2,1);
xdot(1) = a1+b1*x(1)^n/(K1^n+x(1)^n)-g*x(1)*x(2)-d1*x(1);
xdot(2) = a2+b2*x(1)^m/(K2^m+x(1)^m)-d2*x(2);
endfunction
t = linspace(0, 5000, 200)';
x0 = [1000; 1000];
x = lsode ("f", x0, t);
set term dumb;
plot(t,x);
But now I do not know how to compute the maximum value of the two functions (numerical ones) obtained as solutions of the system. I have searched in the Internet but I haven't found what I want... I only have found the function fminbnd for the minimum of a function on an interval...
Is it possible to compute the maximum value of a numeric function with Octave?
Generally, if you know how to find the minimum of a function, you also know how to find its maximum: just look for the minimum of -f.
However, fminbnd is designed for functions that can be evaluated at any given point. What you have is just a vector of 200 points. In principle, you could use interpolation to get a function and then maximize that. But this is not really needed, because all the information you have is in that matrix x anyway, so it makes sense to just take the maximum value there. Like this:
[x1m, i1] = max(x(:,1));
[x2m, i2] = max(x(:,2));
disp(sprintf('Maximum of x1 is %f attained at t = %f', x1m, t(i1)));
disp(sprintf('Maximum of x1 is %f attained at t = %f', x2m, t(i2)));
Related
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:
I am attempting to use Octave to solve for a differential equation using Euler's method.
The Euler method was given to me (and is correct), which works for the given Initial Value Problem,
y*y'' + (y')^2 + 1 = 0; y(1) = 1;
That initial value problem is defined in the following Octave function:
function [YDOT] = f(t, Y)
YDOT(1) = Y(2);
YDOT(2) = -(1 + Y(2)^2)/Y(1);
The question I have is about this function definition. Why is YDOT(1) != 1? What is Y(2)?
I have not found any documentation on the definition of a function using function [YDOT] instead of simply function YDOT, and I would appreciate any clarification on what the Octave code is doing.
First things first: You have a (non linear) differential equation of order two which will require you to have two initial conditions. Thus the given information from above is not enough.
The following is defined for further explanations: A==B means A is identical to B; A=>B means B follows from A.
It seems you are mixing a few things. The guy who gave you the files rewrote the equation in the following way:
y*y'' + (y')^2 + 1 = 0; y(1) = 1; | (I) y := y1 & (II) y' := y2
(I) & (II)=>(III): y' = y2 = y1' | y2==Y(2) & y1'==YDOT(1)
Ocatve is "matrix/vector oriented" so we are writing everything in vectors or matrices. Rather writing y1=alpha and y2=beta we are writing y=[alpha; beta] where y(1)==y1=alpha and y(2)==y2=beta. You will soon realize the tremendous advantage of using especially this mathematical formalization for ALL of your problems.
(III) & f(t,Y)=>(IV): y2' == YDOT(2) = y'' = (-1 -(y')^2) / y
Now recall what is y' and y from the definitions in (I) and (II)!
y' = y2 == Y(2) & y = y1 == Y(1)
So we can rewrite equation (IV)
(IV): y2' == YDOT(2) = (-1 -(y')^2) / y == -(1 + Y(2)^2)/Y(1)
So from equation (III) and (IV) we can derive what you already know:
YDOT(1) = Y(2)
YDOT(2) = -(1 + Y(2)^2)/Y(1)
These equations are passed to the solver. Differential equations of all types are solved numerically by retrieving the "next" value in a near neighborhood to some "previously known" value. (The step size inside this neighborhood is one of the key questions when writing solvers!) So your solver uses your initial condition Y(1)==y(1)=1 to make the next step and calculate the "next" value. So right at the start YDOT(1)=Y(2)==y(2) but you didn't tell us this value! But from then on YDOT(1) is varied by the solver in dependency to the function shape to solve your problem and give you ONE unique y(t) solution.
It seems you are using Octave for the first time so let's make a last comment on function [YDOT] = f(t, Y). In general a function is defined in this way:
function[retVal1, retVal2, ...] = myArbitraryName(arg1, arg2, ...)
Where retVal is the return value or output and arg is the argument or input.
I'm trying to compute the Fourier coefficients for a waveform using MATLAB. The coefficients can be computed using the following formulas:
T is chosen to be 1 which gives omega = 2pi.
However I'm having issues performing the integrals. The functions are are triangle wave (Which can be generated using sawtooth(t,0.5) if I'm not mistaking) as well as a square wave.
I've tried with the following code (For the triangle wave):
function [ a0,am,bm ] = test( numTerms )
b_m = zeros(1,numTerms);
w=2*pi;
for i = 1:numTerms
f1 = #(t) sawtooth(t,0.5).*cos(i*w*t);
f2 = #(t) sawtooth(t,0.5).*sin(i*w*t);
am(i) = 2*quad(f1,0,1);
bm(i) = 2*quad(f2,0,1);
end
end
However it's not getting anywhere near the values I need. The b_m coefficients are given for a
triangle wave and are supposed to be 1/m^2 and -1/m^2 when m is odd alternating beginning with the positive term.
The major issue for me is that I don't quite understand how integrals work in MATLAB and I'm not sure whether or not the approach I've chosen works.
Edit:
To clairify, this is the form that I'm looking to write the function on when the coefficients have been determined:
Here's an attempt using fft:
function [ a0,am,bm ] = test( numTerms )
T=2*pi;
w=1;
t = [0:0.1:2];
f = fft(sawtooth(t,0.5));
am = real(f);
bm = imag(f);
func = num2str(f(1));
for i = 1:numTerms
func = strcat(func,'+',num2str(am(i)),'*cos(',num2str(i*w),'*t)','+',num2str(bm(i)),'*sin(',num2str(i*w),'*t)');
end
y = inline(func);
plot(t,y(t));
end
Looks to me that your problem is what sawtooth returns the mathworks documentation says that:
sawtooth(t,width) generates a modified triangle wave where width, a scalar parameter between 0 and 1, determines the point between 0 and 2π at which the maximum occurs. The function increases from -1 to 1 on the interval 0 to 2πwidth, then decreases linearly from 1 to -1 on the interval 2πwidth to 2π. Thus a parameter of 0.5 specifies a standard triangle wave, symmetric about time instant π with peak-to-peak amplitude of 1. sawtooth(t,1) is equivalent to sawtooth(t).
So I'm guessing that's part of your problem.
After you responded I looked into it some more. Looks to me like it's the quad function; not very accurate! I recast the problem like this:
function [ a0,am,bm ] = sotest( t, numTerms )
bm = zeros(1,numTerms);
am = zeros(1,numTerms);
% 2L = 1
L = 0.5;
for ii = 1:numTerms
am(ii) = (1/L)*quadl(#(x) aCos(x,ii,L),0,2*L);
bm(ii) = (1/L)*quadl(#(x) aSin(x,ii,L),0,2*L);
end
ii = 0;
a0 = (1/L)*trapz( t, t.*cos((ii*pi*t)/L) );
% now let's test it
y = ones(size(t))*(a0/2);
for ii=1:numTerms
y = y + am(ii)*cos(ii*2*pi*t);
y = y + bm(ii)*sin(ii*2*pi*t);
end
figure; plot( t, y);
end
function a = aCos(t,n,L)
a = t.*cos((n*pi*t)/L);
end
function b = aSin(t,n,L)
b = t.*sin((n*pi*t)/L);
end
And then I called it like:
[ a0,am,bm ] = sotest( t, 100 );
and I got:
Sweetness!!!
All I really changed was from quad to quadl. I figured that out by using trapz which worked great until the time vector I was using didn't have enough resolution, which led me to believe it was a numerical issue rather than something fundamental. Hope this helps!
To troubleshoot your code I would plot the functions you are using and investigate, how the quad function samples them. You might be undersampling them, so make sure your minimum step size is smaller than the period of the function by at least factor 10.
I would suggest using the FFTs that are built-in to Matlab. Not only is the FFT the most efficient method to compute a spectrum (it is n*log(n) dependent on the length n of the array, whereas the integral in n^2 dependent), it will also give you automatically the frequency points that are supported by your (equally spaced) time data. If you compute the integral yourself (might be needed if datapoints are not equally spaced), you might calculate frequency data that are not resolved (closer spacing than 1/over the spacing in time, i.e. beyond the 'Fourier limit').
I am a first time MATLAB user. I have a 2d array of t vs f in MATLAB. This 2d array correponds to a function, say f(t). I am using ode45 to solve a set of differential equations and f(t) is one of the coefficients i.e. I have a set of equations of the form x'(t)=f(t)x(t) or variations thereof. How do I proceed?
I need a way to convert my array of t vs f into a function that can be used in the ode45 method. Presumably, the conversion will do some linear interpolation and give me the equation of the best fit curve.
Or if there is another approach, I'm all ears!
Thank you!
This is a simple example with passing function as a parameter to right hand side of the ODE. Create files in the same directory and run main
main.m
t = 0:0.2:10; f = sin(t);
fun = #(xc) interp1(t, f, xc);
x0=0.5
% solve diff(x, t)=sin(t)
% pass function as parameter
[tsol, xsol] = ode45(#(t, x) diff_rhs(t, x, fun),[0.0 8.0], 0.5);
% we solve equation dx/dt = sin(x), x(0)=x0
% exact solution is x(t) = - cos(t) + x(0) + 1
plot(tsol, xsol, 'x');
hold on
plot(tsol, -cos(tsol) + x0 + 1, '-');
legend('numerical', 'exact');
diff_rhs.m
function dx = diff_rhs(t, x, fun)
dx = fun(t);
end
References in the documentation: interp1 and anonymous Functions.
How to find integrals from limits -infinity to +infinity in SCILAB ? ( Expression to be integrated are not directly integratable )
Change the variable of integration from x=(-inf,inf) to z=atan(x)
x=tan(z),
dx/dz = 1/(cos(z))^2
In the new variable z, the integration limits are from -%pi/2+eps to +%pi/2-eps, where eps is a very small positive number (else you will not be able to divide by the cos(z)) and
integral f(x) dx =
= integral f(x(z)) d(x(z))
= integral f(z) dx/dz dz
For example,
function y=Gaussian(x); y=exp(-x^2/2)/sqrt(2*%pi); endfunction;
intg(-10,10,Gaussian)
The same integration result is achieved with
function y=Gmodified(z); x=tan(z); y=Gaussian(x)/(cos(z))^2; endfunction;
intg(atan(-10),atan(10),Gmodified)
Interestingly, Scilab will take the above integral even for
intg(-%pi/2,%pi/2,Gmodified)
but this is only because Scilab evaluates 1/cos(%pi/2) as 1.633D+16 rather than infinity.