I am trying to model a system of three differential equations. This is a droplet model, parametrized vs the arc length, s.
The equations are:
dx/ds=cos(theta)
dz/ds=sin(theta)
(theta)/ds=2*b+c*z-sin(theta)/x
The initial conditions are that x,z, and theta are all 0 at s=0. To avoid the singularity on d(theta)/ds, I also have the condition that, at s=0, d(theta)/ds=b. I have already written this code:
[s,x]=ode23(#(s,x)drpivp(s,x,p),sspan,x0);
%where p contains two parameters and x0 contains initial angle theta, x, z values.
%droplet ODE function:
function drpivp = drpivp(s,x,p);
%x(1)=theta
%x(2)=x
%x(3)=z
%b is curvature at apex
%c is capillarity constant
b=p(1);
c=p(2);
drpivp=[2/p(1)+p(2)*x(3)-sin(x(1))/x(2); cos(x(1)); sin(x(1))];
Which yields a solution that spirals out. Instead of creating one droplet profile, it creates many. Of course, here I have not initialized the equation properly, because I am not certain how to use a different equation for theta at s=0.
So the question is: How do I include the initial condition that d(theta)/ds=b instead of it's usual at s=0? Is this possible using the built-in solvers on matlab?
Thanks.
There are several ways of doing this, the easiest is to simply add an if statement into your equation:
function drpivp = drpivp(s,x,p);
%x(1)=theta
%x(2)=x
%x(3)=z
%b is curvature at apex
%c is capillarity constant
b=p(1);
c=p(2);
if (s == 0)
drpivp=[b; cos(x(1)); sin(x(1))];
else
drpivp=[2/p(1)+p(2)*x(3)-sin(x(1))/x(2); cos(x(1)); sin(x(1))];
end
Related
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 have a task of writing an M script that would set up (and finally solve) a set of 19 differential equations and all of the equation coefficients. I am not sure what is the best way to input those equations.
Example of some equations:
Example of coefficients:
I don't think that this would be suited for Simulink. Other examples use a function in the form of #(t,x) where t is time and x the vector of all variables.
There are loads of "examples" online, but they don't seem to be appropriate for such a large set of large equations.
For example this exmple of solving 3 equatons
Even for 3 simple equations as they have solved, the functions are getting messy:
f = #(t,x) [-x(1)+3*x(3);-x(2)+2*x(3);x(1)^2-2*x(3)];
Using this notation and getting to the x(19) and cross-referencing all instances of x would be a mess.
I would like your help, and a simple example of how could I write these equations line by line, maybe using the symbolic toolbox, and then put them in an array that I can then forward to the solver.
As I said, I know there are examples online, but they tackle only the most basic system, and really they don't scale well if you want a clean and easily readable code.
I would like to have a similar to Wolfram alpha, where you type variable names as they are (instead od x(1), x(2), ... m x(19)) and if it's possible to get all solution vectors with their variable names.
You don't have to use an anonymous function handle as an ode function, you can create a separate function file (as shown in the odefun section of ode45.
For example, your odefun can look like:
function dy = myode(t,y)
% first unpack state variables
i_d2 = y(1);
i_q2 = y(2);
...
gamma2 = y(end-1);
omega2 = y(end);
% determine all constants
c34 = expression_for_c34;
...
c61 = expression_for_61;
% determine state derivative
i_d2_dot = expression;
...
omega2_dot = expression;
% pack state derivative based on order in state vector
dy(1) = i_d2_dot;
...
dy(end) = omega2_dot;
end
From this myode function you can also call other functions to e.g. determine the value for some coefficients based on the current state. Next, integrate the system using a suitable ode solver:
[t,y] = ode45(#myode,tspan,y0);
this a question that envolves both programming and mathematics. So, I'm trying to write a code that computes the general solution of a system of linear ODEs described by . The mathematical formula it's shown above:
where the greek symbol \PHI that appers in the equation is the expm(A*t)
clear all
A=[-2]; %system matrix
t0=1; %initial time of simulation
tf=2; %final time of simulation
syms t x_0
x0=x_0;
hom=expm(A*t); %hom means "homogeneous solution"
hom_initialcond=hom*x0;%this is the homogeneous solution multiplied by the initial conditon
invhom=inv(hom); %this is the inverse of the greek letter at which, multiplied by the input of the system, composes the integrand of the integral
g=5*cos(2*t); %system input
integrand=invhom*g; %computation of the integrand
integral=int(integrand,t0,t); %computation of the definite integral from t0 to t, as shown by the math formula
partsol=hom*integral; %this is the particular solution
gen_sol=partsol+hom_initialcond %this is the general solution
x_0=1; %this is the initial condition
t=linspace(t0,tf); %vector of time from t0 to tf
y=double(subs(gen_sol)); %here I am evaluating my symbolic expression
plot(t,y)
The problem is that my plot of the ODE's solution it's not looking well, as you can see:
The solution it's wrong because the curve shown in the graph doesnt start at the initial value equals 1. But the shape it's very similar from the plot gave by the MATLAB ODE solver:
However, if I set t0=0 then the plot gave by my code and by MATLAB solver it's exacly equal to each other. So, my code it's fine for t0=0 but with any other values my code goes wrong.
The general solution in terms of fundamental matrices is
or more often seen as
But since the initial time is often taken to be zero, the inverse of the fundamental matrix is often omitted since it is the identity for linear, constant coefficient problems at zero (i.e., expm(zeros(n)) == eye(n)) and the c vector is equivalent to the initial condition vector.
Swapping some of the lines around near your symbolic declaration to this
syms t x_0 c_0
hom = expm(A*t) ;
invhom = inv(hom) ;
invhom_0 = subs(invhom,t,sym(t0)) ;
c_0 = invhom_0 * x_0 ;
hom_initialcond = hom * c_0 ;
should provide the correct solution for non-zero initial time.
I have the following differential equation which I'm not able to solve.
We know the following about the equation:
D(r) is a third grade polynom
D'(1)=D'(2)=0
D(2)=2D(1)
u(1)=450
u'(2)=-K * (u(2)-Te)
Where K and Te are constants.
I want to approximate the problem using a matrix and I managed to solve
the similiar equation: with the same limit conditions for u(1) and u'(2).
On this equation I approximated u' and u'' with central differences and used a finite difference method between r=1 to r=2. I then placed the results in a matrix A in matlab and the limit conditions in the vector Y in matlab and ran u=A\Y to get how the u value changes. Heres my matlab code for the equation I managed to solve:
clear
a=1;
b=2;
N=100;
h = (b-a)/N;
K=3.20;
Ti=450;
Te=20;
A = zeros(N+2);
A(1,1)=1;
A(end,end)=1/(2*h*K);
A(end,end-1)=1;
A(end,end-2)=-1/(2*h*K);
r=a+h:h:b;
%y(i)
for i=1:1:length(r)
yi(i)=-r(i)*(2/(h^2));
end
A(2:end-1,2:end-1)=A(2:end-1,2:end-1)+diag(yi);
%y(i-1)
for i=1:1:length(r)-1
ymin(i)=r(i+1)*(1/(h^2))-1/(2*h);
end
A(3:end-1,2:end-2) = A(3:end-1,2:end-2)+diag(ymin);
%y(i+1)
for i=1:1:length(r)
ymax(i)=r(i)*(1/(h^2))+1/(2*h);
end
A(2:end-1,3:end)=A(2:end-1,3:end)+diag(ymax);
Y=zeros(N+2,1);
Y(1) =Ti;
Y(2)=-(Ti*(r(1)/(h^2)-(1/(2*h))));
Y(end) = Te;
r=[1,r];
u=A\Y;
plot(r,u(1:end-1));
My question is, how do I solve the first differential equation?
As TroyHaskin pointed out in comments, one can determine D up to a constant factor, and that constant factor cancels out in D'/D anyway. Put another way: we can assume that D(1)=1 (a convenient number), since D can be multiplied by any constant. Now it's easy to find the coefficients (done with Wolfram Alpha), and the polynomial turns out to be
D(r) = -2r^3+9r^2-12r+6
with derivative D'(r) = -6r^2+18r-12. (There is also a smarter way to find the polynomial by starting with D', which is quadratic with known roots.)
I would probably use this information right away, computing the coefficient k of the first derivative:
r = a+h:h:b;
k = 1+r.*(-6*r.^2+18*r-12)./(-2*r.^3+9*r.^2-12*r+6);
It seems that k is always positive on the interval [1,2], so if you want to minimize the changes to existing code, just replace r(i) by r(i)/k(i) in it.
By the way, instead of loops like
for i=1:1:length(r)
yi(i)=-r(i)*(2/(h^2));
end
one usually does simply
yi=-r*(2/(h^2));
This vectorization makes the code more compact and can benefit the performance too (not so much in your example, where solving the linear system is the bottleneck). Another benefit is that yi is properly initialized, while with your loop construction, if yi happened to already exist and have length greater than length(r), the resulting array would have extraneous entries. (This is a potential source of hard-to-track bugs.)
I have to analyze 802.11 saturation throughput using matlab, and here is my problem. I'm trying to solve parametric equations below (parameters are m,W,a) using solve function and i get
Warning: Explicit solution could not be found
How could I solve above equations using matlab?
I guess you were trying to find an analytical solution for tau and p using symbolic math. Unless you're really lucky with your parameters (e.g. m=1), there won't be an analytical solution.
If you're interested in numerical values for tau and p, I suggest you manually substitue p in the first equation, and then solve an equation of the form tau-bigFraction=0 using, e.g. fzero.
Here's how you'd use fzero to solve a simple equation kx=exp(-x), with k being a parameter.
k = 5; %# set a value for k
x = fzero(#(x)k*x-exp(-x),0); %# initial guess: x=0