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I have to solve the following system:
X'(t) = -D(t)x(t)+μ(s(t), p(t))x(t);
S'(t) = D(t)(s(t)^in - s(t)) - Yxsμ(s(t), p(t))x(t)
p'(t) = -D(t)p(t)+(aμ(s(t), p(t))+b)x(t)
where
μ(s(t), p(t)) = Μmax ((1 - (p(t)/pm))s(t)) / (km+s(t)+(s(t) ^ 2)/ki)
where Yxs, a,b, Mmax, Pm, km, ki are constant variables, then I have to linearizate the system and find the balance Points of thiw system. any suggestion how to do it with Matlab or Mathematica??
Matlab can help with some steps, but there might be few where you do have to write down some equations yourself.
To start with a simple side note: Matlabs ODE45 function allows to simulate any function of the form dx/dt = f(x,u), regardless of how non-linear or time variant they might become.
to linearize such a system, you need to derive a jacobian matrix and substitute the linearization point in this matrix. This linearization point is any point where all state derivatives equal 0, it does not need to be a balance point. However, it is desired to have it a balance point as this means the linearization point is a stable equilibrium.
So in MATLAB:
create symbolic variables for all states and inputs (so x(t), s(t) and p(t))
create symbolic state equations dx/dt = f(x,u) and output equation y = g(x,u)
derive symbolic State space matrices A,B,C,D using the "jacobian" function
substitute linearization point in these symbolic state space matrices using "subs"
use eval(symbolic matrix) to retrieve a numeric matrix.
Depending on the non-linear complexity and the chosen linearization point, the linearized system might only within acceptable bounds from the actual system for a very tight region, so be aware of that.
I'm using octave 3.8.1 which works like matlab.
I have an array of thousands of values I've only included three groupings as an example below:
(amp1=0.2; freq1=3; phase1=1; is an example of one grouping)
t=0;
amp1=0.2; freq1=3; phase1=1; %1st grouping
amp2=1.4; freq2=2; phase2=1.7; %2nd grouping
amp3=0.8; freq3=5; phase3=1.5; %3rd grouping
The Octave / Matlab code below solves for Y so I can plug it back into the equation to check values along with calculating values not located in the array.
clear all
t=0;
Y=0;
a1=[.2,3,1;1.4,2,1.7;.8,5,1.5]
for kk=1:1:length(a1)
Y=Y+a1(kk,1)*cos ((a1(kk,2))*t+a1(kk,3))
kk
end
Y
PS: I'm not trying to solve for Y since it's already solved for I'm trying to solve for Phase
The formulas located below are used to calculate Phase but I'm not sure how to put it into a for loop that will work in an array of n groupings:
How would I write the equation / for loop for finding the phase if I want to find freq=2.5 and amp=.23 and the phase is unknown I've looked online and it may require writing non linear equations which I'm not sure how to convert what I'm trying to do into such an equation.
phase1_test=acos(Y/amp1-amp3*cos(2*freq3*pi*t+phase3)/amp1-amp2*cos(2*freq2*pi*t+phase2)/amp1)-2*freq1*pi*t
phase2_test=acos(Y/amp2-amp3*cos(2*freq3*pi*t+phase3)/amp2-amp1*cos(2*freq1*pi*t+phase1)/amp2)-2*freq2*pi*t
phase3_test=acos(Y/amp3-amp2*cos(2*freq2*pi*t+phase2)/amp3-amp1*cos(2*freq1*pi*t+phase1)/amp3)-2*freq2*pi*t
Image of formula below:
I would like to do a check / calculate phases if given a freq and amp values.
I know I have to do a for loop but how do I convert the phase equation into a for loop so it will work on n groupings in an array and calculate different values not found in the array?
Basically I would be given an array of n groupings and freq=2.5 and amp=.23 and use the formula to calculate phase. Note: freq will not always be in the array hence why I'm trying to calculate the phase using a formula.
Ok, I think I finally understand your question:
you are trying to find a set of phase1, phase2,..., phaseN, such that equations like the ones you describe are satisfied
You know how to find y, and you supply values for freq and amp.
In Matlab, such a problem would be solved using, for example fsolve, but let's look at your problem step by step.
For simplicity, let me re-write your equations for phase1, phase2, and phase3. For example, your first equation, the one for phase1, would read
amp1*cos(phase1 + 2 freq1 pi t) + amp2*cos(2 freq2 pi t + phase2) + amp3*cos(2 freq3 pi t + phase3) - y = 0
Note that ampX (X is a placeholder for 1, 2, 3) are given, pi is a constant, t is given via Y (I think), freqX are given.
Hence, you are, in fact, dealing with a non-linear vector equation of the form
F(phase) = 0
where F is a multi-dimensional (vector) function taking a multi-dimensional (vector) input variable phase (comprised of phase1, phase2,..., phaseN). And you are looking for the set of phaseX, where all of the components of your vector function F are zero. N.B. F is a shorthand for your functions. Therefore, the first component of F, called f1, for example, is
f1 = amp1*cos(phase1+...) + amp2*cos(phase2+...) + amp3*cos(phase3+...) - y = 0.
Hence, f1 is a one-dimensional function of phase1, phase2, and phase3.
The technical term for what you are trying to do is find a zero of a non-linear vector function, or find a solution of a non-linear vector function. In Matlab, there are different approaches.
For a one-dimensional function, you can use fzero, which is explained at http://www.mathworks.com/help/matlab/ref/fzero.html?refresh=true
For a multi-dimensional (vector) function as yours, I would look into using fsolve, which is part of Matlab's optimization toolbox (which means I don't know how to do this in Octave). The function fsolve is explained at http://www.mathworks.com/help/optim/ug/fsolve.html
If you know an approximate solution for your phases, you may also look into iterative, local methods.
In particular, I would recommend you look into the Newton's Method, which allows you to find a solution to your system of equations F. Wikipedia has a good explanation of Newton's Method at https://en.wikipedia.org/wiki/Newton%27s_method . Newton iterations are very simple to implement and you should find a lot of resources online. You will have to compute the derivative of your function F with respect to each of your variables phaseX, which is very simple to compute since you're only dealing with cos() functions. For starters, have a look at the one-dimensional Newton iteration method in Matlab at http://www.math.colostate.edu/~gerhard/classes/331/lab/newton.html .
Finally, if you want to dig deeper, I found a textbook on this topic from the society for industrial and applied math: https://www.siam.org/books/textbooks/fr16_book.pdf .
As you can see, this is a very large field; Newton's method should be able to help you out, though.
Good luck!
I have a set of 4 PDEs:
du/dt + A(u) * du/dx = Q(u)
where,u is a matrix and contains:
u=[u1;u2;u3;u4]
and A is a 4*4 matrix. Q is 4*1. A and Q are function of u=[u1;u2;u3;u4].
But my questions are:
How can I solve above equation in MATLAB?
If I solved it by PDE functions of Matlab,can I convert it to a
simple function that is not used from ready functions of Matlab?
Is there any way that I calculate A and Q explicitly. I mean that in
every time step, I calculate A and Q from data of previous time step
and put new value in the equation that causes faster run of program?
PDEs require finite differences, finite elements, boundary elements, etc. You can also turn them into ODEs using transforms like Laplace, Fourier, etc. Solve those using ODE functions and then transform back. Neither one is trivial.
Your equation is a non-linear transient diffusion equation. It's a parabolic PDE.
The equation you posted has the additional difficulty of being non-linear, because both the A matrix and Q vector are functions of the independent variable q. You'll have to start by linearizing your equations. Solve for increments in u rather than u itself.
Once you've done that, discretize the du/dx term using finite differences, finite elements, or boundary elements. You should start with a weighted residual integral formulation.
You're almost done: Next to integrate w.r.t. time using the method of your choice.
It's not trivial.
Google found this: maybe it will help you.
http://www.mathworks.com/matlabcentral/fileexchange/3710-nonlinear-diffusion-toolbox
I am doing a project where i find an approximation of the Sine function, using the Least Squares method. Also i can use 12 values of my own choice.Since i couldn't figure out how to solve it i thought of using Taylor's series for Sine and then solving it as a polynomial of order 5. Here is my code :
%% Find the sine of the 12 known values
x=[0,pi/8,pi/4,7*pi/2,3*pi/4,pi,4*pi/11,3*pi/2,2*pi,5*pi/4,3*pi/8,12*pi/20];
y=zeros(12,1);
for i=1:12
y=sin(x);
end
n=12;
j=5;
%% Find the sums to populate the matrix A and matrix B
s1=sum(x);s2=sum(x.^2);
s3=sum(x.^3);s4=sum(x.^4);
s5=sum(x.^5);s6=sum(x.^6);
s7=sum(x.^7);s8=sum(x.^8);
s9=sum(x.^9);s10=sum(x.^10);
sy=sum(y);
sxy=sum(x.*y);
sxy2=sum( (x.^2).*y);
sxy3=sum( (x.^3).*y);
sxy4=sum( (x.^4).*y);
sxy5=sum( (x.^5).*y);
A=[n,s1,s2,s3,s4,s5;s1,s2,s3,s4,s5,s6;s2,s3,s4,s5,s6,s7;
s3,s4,s5,s6,s7,s8;s4,s5,s6,s7,s8,s9;s5,s6,s7,s8,s9,s10];
B=[sy;sxy;sxy2;sxy3;sxy4;sxy5];
Then at matlab i get this result
>> a=A^-1*B
a =
-0.0248
1.2203
-0.2351
-0.1408
0.0364
-0.0021
However when i try to replace the values of a in the taylor series and solve f.e t=pi/2 i get wrong results
>> t=pi/2;
fun=t-t^3*a(4)+a(6)*t^5
fun =
2.0967
I am doing something wrong when i replace the values of a matrix in the Taylor series or is my initial thought flawed ?
Note: i can't use any built-in function
If you need a least-squares approximation, simply decide on a fixed interval that you want to approximate on and generate some x abscissae on that interval (possibly equally spaced abscissae using linspace - or non-uniformly spaced as you have in your example). Then evaluate your sine function at each point such that you have
y = sin(x)
Then simply use the polyfit function (documented here) to obtain least squares parameters
b = polyfit(x,y,n)
where n is the degree of the polynomial you want to approximate. You can then use polyval (documented here) to obtain the values of your approximation at other values of x.
EDIT: As you can't use polyfit you can generate the Vandermonde matrix for the least-squares approximation directly (the below assumes x is a row vector).
A = ones(length(x),1);
x = x';
for i=1:n
A = [A x.^i];
end
then simply obtain the least squares parameters using
b = A\y;
You can clearly optimise the clumsy Vandermonde generation loop above I have just written to illustrate the concept. For better numerical stability you would also be better to use a nice orthogonal polynomial system like Chebyshev polynomials of the first kind. If you are not even allowed to use the matrix divide \ function then you will need to code up your own implementation of a QR factorisation and solve the system that way (or some other numerically stable method).
Starting from the plot of one curve, it is possible to obtain the parametric equation of that curve?
In particular, say x={1 2 3 4 5 6....} the x axis, and y = {a b c d e f....} the corresponding y axis. I have the plot(x,y).
Now, how i can obtain the equation that describe the plotted curve? it is possible to display the parametric equation starting from the spline interpolation?
Thank you
If you want to display a polynomial fit function alongside your graph, the following example should help:
x=-3:.1:3;
y=4*x.^3-5*x.^2-7.*x+2+10*rand(1,61);
p=polyfit(x,y,3); %# third order polynomial fit, p=[a,b,c,d] of ax^3+bx^2+cx+d
yfit=polyval(p,x); %# evaluate the curve fit over x
plot(x,y,'.')
hold on
plot(x,yfit,'-g')
equation=sprintf('y=%2.2gx^3+%2.2gx^2+%2.2gx+%2.2g',p); %# format string for equation
equation=strrep(equation,'+-','-'); %# replace any redundant signs
text(-1,-80,equation) %# place equation string on graph
legend('Data','Fit','Location','northwest')
Last year, I wrote up a set of three blogs for Loren, on the topic of modeling/interpolationg a curve. They may cover some of your questions, although I never did find the time to add another 3 blogs to finish the topic to my satisfaction. Perhaps one day I will get that done.
The problem is to recognize there are infinitely many curves that will interpolate a set of data points. A spline is a nice choice, because it can be made well behaved. However, that spline has no simple "equation" to write down. Instead, it has many polynomial segments, pieced together to be well behaved.
You're asking for the function/mapping between two data sets. Knowing the physics involved, the function can be derived by modeling the system. Write down the differential equations and solve it.
Left alone with just two data series, an input and an output with a 'black box' in between you may approximate the series with an arbitrary function. You may start with a polynomial function
y = a*x^2 + b*x + c
Given your input vector x and your output vector y, parameters a,b,c must be determined applying a fitness function.
There is an example of Polynomial Curve Fitting in the MathWorks documentation.
Curve Fitting Tool provides a flexible graphical user interfacewhere you can interactively fit curves and surfaces to data and viewplots. You can:
Create, plot, and compare multiple fits
Use linear or nonlinear regression, interpolation,local smoothing regression, or custom equations
View goodness-of-fit statistics, display confidenceintervals and residuals, remove outliers and assess fits with validationdata
Automatically generate code for fitting and plottingsurfaces, or export fits to workspace for further analysis