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
Related
I am trying to solve two equations with complex coefficients using ode45.
But iam getting an error message as "Inputs must be floats, namely single or
double."
X = sym(['[',sprintf('X(%d) ',1:2),']']);
Eqns=[-(X(1)*23788605396486326904946699391889*1i)/38685626227668133590597632 + (X(2)*23788605396486326904946699391889*1i)/38685626227668133590597632; (X(2)*23788605396486326904946699391889*1i)/38685626227668133590597632 + X(1)*(- 2500000 + (5223289665997855453060886952725538686654593059791*1i)/324518553658426726783156020576256)] ;
f=#(t,X)[Eqns];
[t,Xabc]=ode45(f,[0 300*10^-6],[0 1])
How can i fix this ? Can somebody can help me ?
Per the MathWorks Support Team, the "ODE solvers in MATLAB 5 (R12) and later releases properly handle complex valued systems." So the complex numbers are the not the issue.
The error "Inputs must be floats, namely single or double." stems from your definition of f using Symbolic Variables that are, unlike complex numbers, not floats. The easiest way to get around this is to not use the Symbolic Toolbox at all; just makes Eqns an anonymous function:
Eqns= #(t,X) [-(X(1)*23788605396486326904946699391889*1i)/38685626227668133590597632 + (X(2)*23788605396486326904946699391889*1i)/38685626227668133590597632; (X(2)*23788605396486326904946699391889*1i)/38685626227668133590597632 + X(1)*(- 2500000 + (5223289665997855453060886952725538686654593059791*1i)/324518553658426726783156020576256)] ;
[t,Xabc]=ode45(Eqns,[0 300*10^-6],[0 1]);
That being said, I'd like to point out that numerically time integrating this system over 300 microseconds (I assume without units given) will take a long time since your coefficient matrix has imaginary eigenvalues on the order of 10E+10. The extremely short wavelength of those oscillations will more than likely be resolved by Matlab's adaptive methods, and that will take a while to solve for a time span just a few orders greater than the wavelength.
I'd, therefore, suggest an analytical approach to this problem; unless it is a stepping stone another problem that is non-analytically solvable.
Systems of ordinary differential equations of the form
,
which is a linear, homogenous system with a constant coefficient matrix, has the general solution
,
where the m-subscripted exponential function is the matrix exponential.
Therefore, the analytical solution to the system can be calculated exactly assuming the matrix exponential can be calculated.
In Matlab, the matrix exponential is calculate via the expm function.
The following code computes the analytical solution and compares it to the numerical one for a short time span:
% Set-up
A = [-23788605396486326904946699391889i/38685626227668133590597632,23788605396486326904946699391889i/38685626227668133590597632;...
-2500000+5223289665997855453060886952725538686654593059791i/324518553658426726783156020576256,23788605396486326904946699391889i/38685626227668133590597632];
Eqns = #(t,X) A*X;
X0 = [0;1];
% Numerical
options = odeset('RelTol',1E-8,'AbsTol',1E-8);
[t,Xabc]=ode45(Eqns,[0 1E-9],X0,options);
% Analytical
Xana = cell2mat(arrayfun(#(tk) expm(A*tk)*X0,t,'UniformOutput',false)')';
k = 1;
% Plots
figure(1);
subplot(3,1,1)
plot(t,abs(Xana(:,k)),t,abs(Xabc(:,k)),'--');
title('Magnitude');
subplot(3,1,2)
plot(t,real(Xana(:,k)),t,real(Xabc(:,k)),'--');
title('Real');
ylabel('Values');
subplot(3,1,3)
plot(t,imag(Xana(:,k)),t,imag(Xabc(:,k)),'--');
title('Imaginary');
xlabel('Time');
The comparison plot is:
The output of ode45 matches the magnitude and real parts of the solution very well, but the imaginary portion is out-of-phase by exactly π.
However, since ode45's error estimator only looks at norms, the phase difference is not noticed which may lead to problems depending on the application.
It will be noted that while the matrix exponential solution is far more costly than ode45 for the same number of time vector elements, the analytical solution will produce the exact solution for any time vector of any density given to it. So for long time solutions, the matrix exponential can be viewed as an improvement in some sense.
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'm trying to teach myself how to use MATLAB for solving state-space systems, I have what seems to be a pretty straight-forward system but have been unable to find any decent straight-forward examples for a novice thus far.
I'd like a simple walk-through of how to translate the system into MATLAB, what variables to set, and how to solve for about 50(?) seconds (from t=0 to 50 or any value really).
I'd like to use ode45 since it's a 4th order method using a Runge-Kutta variant.
Here's the 2nd-order equation:
θ''+0.03|θ'|θ'+4pi^2*sinθ=0
The state-space:
x_1'=x_2
x_2'=-4pi^2*sin(x_1)-0.03|x_2|x_2
x_1 = θ, x_2 = θ'
θ(0)=pi/9 rad, θ'(0)=0, h(step)=1
You need a derivative function function, which, given the current state of the system and the current time, returns the derivative of all of the state variables. Generally this function is of the form
function xDash=derivative(t,x)
and xDash is a vector with the derivative of each element, and x is a vector of the state variables. If your variables are called x_1, x_2 etc. it's a good idea to put x_1 in x(1), etc. Then you need a formula for the derivative of each state variable in terms of the other state variables, for example you could have xDash_1=x_1-x_2 and you would code this as xDash(1)=x(1)-x(2). Hopefully that clears something up.
For your example, the derivative function will look like
function xDash=derivative(t,x)
xDash=zeros(2,1);
xDash(1)=x(2);
xDash(2)=-4*pi^2*sin(x(1))-0.03*abs(x(2))*x(2);
end
and you would solve the system using
[T,X]=ode45(#derivative,0:50,[pi/9 0]);
This gives output at t=0,1,2,...,50.
I'm not too familiar with MATLAB or computational mathematics so I was wondering how I might solve an equation involving the sum of squares, where each term involves two vectors- one known and one unknown. This formula is supposed to represent the error and I need to minimize the error. I think I'm supposed to use least squares but I don't know too much about it and I'm wondering what function is best for doing that and what arguments would represent my equation. My teacher also mentioned something about taking derivatives and he formed a matrix using derivatives which confused me even more- am I required to take derivatives?
The problem that you must be trying to solve is
Min u'u = min \sum_i u_i^2, u=y-Xbeta, where u is the error, y is the vector of dependent variables you are trying to explain, X is a matrix of independent variables and beta is the vector you want to estimate.
Since sum u_i^2 is diferentiable (and convex), you can evaluate the minimal of this expression calculating its derivative and making it equal to zero.
If you do that, you find that beta=inv(X'X)X'y. This maybe calculated using the matlab function regress http://www.mathworks.com/help/stats/regress.html or writing this formula in Matlab. However, you should be careful how to evaluate the inverse (X'X) see Most efficient matrix inversion in MATLAB
This question has already confused me several days. While I referred to senior students, they also cannot give a reply.
We have ten ODEs, into which each a noise term should be added. The noise is defined as follows. since I always find that I cannot upload a picture, the formula below maybe not very clear. In order to understand, you can either read my explanation or go the this address: Plos one. You could find the description of the equations directly above the Support Information in this address
The white noise term epislon_i(t) is assumed with Gaussian distribution. epislon_i(t) means that for equation i, and at t timepoint, the value of the noise.
the auto-correlation of noise are given:
(EQ.1)
where delta(t) is the Dirac delta function and the diffusion matrix D is defined by
(EQ.2)
Our problem focuses on how to explain the Dirac delta function in the diffusion matrix. Since the property of Dirac delta function is delta(0) = Inf and delta(t) = 0 if t neq 0, we don't know how to calculate the epislonif we try to sqrt of 2D(x, t)delta(t-t'). So we simply assume that delta(0) = 1 and delta(t) = 0 if t neq 0; But we don't know whether or not this is right. Could you please tell me how to use Delta function of diffusion equation in MATLAB?
This question associates with the stochastic process in MATLAB. So we review different stochastic process to inspire our ideas. In MATLAB, the Wienner process is often defined as a = sqrt(dt) * rand(1, N). N is the number of steps, dt is the length of the steps. Correspondingly, the Brownian motion can be defined as: b = cumsum(a); All of these associate with stochastic process. However, they doesn't related to the white noise process which has a constraints on the matrix of auto-correlation, noted by D.
Then we consider that, we may simply use randn(1, 10) to generate a vector representing the noise. However, since the definition of the noise must satisfy the equation (2), this cannot enable noise term in different equation have the predefined partial correlation (D_ij). Then we try to use mvnrnd to generate a multiple variable normal distribution at each time step. Unfortunately, the function mvnrnd in MATLAB return a matrix. But we need to return a vector of length 10.
We are rather confused, so could you please give me just a light? Thanks so much!
NOTE: I see two hazy questions in here: 1) how to deal with a stochastic term in a DE and 2) how to deal with a delta function in a DE. Both of these are math related questions and http://www.math.stackexchange.com will be a better place for this. If you had a question pertaining to MATLAB, I haven't been able to pin it down, and you should perhaps add code examples to better illustrate your point. That said, I'll answer the two questions briefly, just to put you on the right track.
What you have here are not ODEs, but Stochastic differential equations (SDE). I'm not sure how you're using MATLAB to work with this, but routines like ode45 or ode23 will not be of any help. For SDEs, your usual mathematical tools of separation of variables/method of characteristics etc don't work and you'll need to use Itô calculus and Itô integrals to work with them. The solutions, as you might have guessed, will be stochastic. To learn more about SDEs and working with them, you can consider Stochastic Differential Equations: An Introduction with Applications by Bernt Øksendal and for numerical solutions, Numerical Solution of Stochastic Differential Equations by Peter E. Kloeden and Eckhard Platen.
Coming to the delta function part, you can easily deal with it by taking the Fourier transform of the ODE. Recall that the Fourier transform of a delta function is 1. This greatly simplifies the DE and you can take an inverse transform in the very end to return to the original domain.