I'm trying to solve 4 nonlinear equations in matlab.
i wrote a code using fsolve but it failed to proceed, and wrote another one using solve but after 4 hours of running without a result I stop it myself.
I dont know what else i can use to solve it?
is it solvable at all?
this is the code I wrote using fsolve
x and y are vectors and n is the length of this vectors equal to 1200. nu is degree of freedom equal to 1196.
p(1), p(2)... and p(4) are the unknowns
x = data1(:,1);
y = data1(:,2);
n = length(x);
p0 = [0 0 0 0];
options = optimset('Display','iter');
[p,fval] = fsolve(#myfunc,p0,options);
myfunc:
function F = myfunc(p)
global x y n
r = y-p(1)*x.^3-p(2)*x.^2-p(3)*x-p(4);
F = [sum(x.^3)-(n-4-2)*sum(x.^3./r);
sum(x.^2)-(n-4-2)*sum(x.^2./r);
sum(x.^1)-(n-4-2)*sum(x.^1./r);
sum(x.^0)-(n-4-2)*sum(x.^0./r)];
Sounds like the problem is too stiff for fsolve, or you've got a poor starting guess. The code you have looks fine, however, if you may want to use a parametrized anonymous function:
x = data1(:,1);
y = data1(:,2);
n = length(x);
p0 = [0 0 0 0];
options = optimset('Display','iter');
[p,fval] = fsolve(#(p) myfunc(p,x,y,n),p0,options);
then myfunc:
function F = myfunc(p,x,y,n)
r = y-p(1)*x.^3-p(2)*x.^2-p(3)*x-p(4);
F = [sum(x.^3)-(n-4-2)*sum(x.^3./r);
sum(x.^2)-(n-4-2)*sum(x.^2./r);
sum(x.^1)-(n-4-2)*sum(x.^1./r);
sum(x.^0)-(n-4-2)*sum(x.^0./r)];
end
Related
I am using Matlab to try and solve a system of three first-order ODEs, but the error message I get is 'syms' requires Symbolic Math Toolbox.
Error in spiders (line 1)
syms f(t) s(t) v(t) r W c h q a k b H K e
On a previous occasion, I received an error saying that this system of ODEs cannot be solved explicitly (i.e. in closed form). I think that numerical integration is the only way. The r,W,c,h, etc are parameters. Could someone please tell me how I can simulate/solve and plot the ODEs below?
syms f(t) s(t) v(t) r W c h q a k b H K e
r = 1;
W = 0.5;
c = 0.4;
h = 0.9;
q = 9;
a = 5;
k = 0.8;
b = 6;
H = 3;
K = 1.3;
e = 2;
ode1 = diff(f) == r*f*(1 - f/W) - c*s*f - h*(1 - q)*f;
ode2 = diff(s) == s*(-a + k*b*v/(H + v) + k*c*f) - h*K*q*s;
ode3 = diff(v) == v*(e - b*s/(H + v)) - h*q*v;
odes=[ode1;ode2;ode3]
S = dsolve(odes)
plot(S);
Looks like you where previously using a different MATLAB license, which included the Symbolic Math Toolbox. I assume you no longer have it available and you are now looking for numeric alternatives.
You have to define functions for the equations you are trying to solve. Then you can call one of the ode solvers. An example from the documentation:
y0 = [1; 0; 0];
tspan = [0 4*logspace(-6,6)];
M = [1 0 0; 0 1 0; 0 0 0];
options = odeset('Mass',M,'RelTol',1e-4,'AbsTol',[1e-6 1e-10 1e-6]);
[t,y] = ode15s(#robertsdae,tspan,y0,options);
Where robertsdae is the equation to solve. The full example including further explanations is available here.
Given a set of datapoints I'm trying to approximate the coefficients a,b in the function U(x)=8-ax^b using Newtons method in MATLAB.
x = [150 200 300 500 1000 2000]';
y = [2 3 4 5 6 7]';
a=170; b=-0.7; iter = 0;
for iter=1:5
f=8-a*x.^(b) -y;
J = [-x.^b -a*b*x.^(b-1)]; %Jacobis matrix
h=J\f;
a=a-h(1); b=b-h(2);
disp(norm(f))
iter = iter+1;
end
The results are incorrect and I've not been sucessful of finding the misstep. All help is appreciated.
The jacobi matrix is wrong. Using Newton's method, you're trying to find the values of a and bthat would solve the equations 8-ax^b - y = 0. So, your Jacobi should be the derivatives of f with respect to a and b. That is J = [df/da df/db], resulting in:
J = [-x.^b -a.*x.^b.*log(x)]
and you will get the following curve for the 5 iterations:
Note that you can easily linearize your model with
Log(8 - U(x)) = Log(a) + b Log(x)
I'm trying to solve a system of ordinary differential equations in MATLAB.
I have a simple equation:
dy = -k/M *x - c/M *y+ F/M.
This is defined in my ode function test2.m, dependant on the values X and t. I want to trig 'F' with a signal, generated by my custom function squaresignal.m. The output hereof, is the variable u, spanding from 0 to 1, as it is a smooth heaviside function. - Think square wave. The inputs in squaresignal.m, is t and f.
u=squaresignal(t,f)
These values are to be used inside my function test2, in order to enable or disable variable 'F' with the value u==1 (enable). Disable for all other values.
My ode function test2.m reads:
function dX = test2(t ,X, u)
x = X (1) ;
y = X (2) ;
M = 10;
k = 50;
c = 10;
F = 300;
if u == 1
F = F;
else
F = 0,
end
dx = y ;
dy = -k/M *x - c/M *y+ F/M ;
dX = [ dx dy ]';
end
And my runscript reads:
clc
clear all
tstart = 0;
tend = 10;
tsteps = 0.01;
tspan = [0 10];
t = [tstart:tsteps:tend];
f = 2;
u = squaresignal(t,f)
for ii = 1:length(u)
options=odeset('maxstep',tsteps,'outputfcn',#odeplot);
[t,X]=ode15s(#(t,X)test2(t,X,u(ii)),[tstart tend],[0 0],u);
end
figure (1);
plot(t,X(:,1))
figure (2);
plot(t,X(:,2));
However, the for-loop does not seem to do it's magic. I still only get F=0, instead of F=F, at times when u==1. And i know, that u is equal to one at some times, because the output of squaresignal.m is visible to me.
So the real question is this. How do i properly pass my variable u, to my function test2.m, and use it there to trig F? Is it possible that the squaresignal.m should be inside the odefunction test2.m instead?
Here's an example where I pass a variable coeff to the differential equation:
function [T,Q] = main()
t_span = [0 10];
q0 = [0.1; 0.2]; % initial state
ode_options = odeset(); % currently no options... You could add some here
coeff = 0.3; % The parameter we wish to pass to the differential eq.
[T,Q] = ode15s(#(t,q)diffeq(t,q,coeff),t_span,q0, ode_options);
end
function dq = diffeq(t,q,coeff)
% Preallocate vector dq
dq = zeros(length(q),1);
% Update dq:
dq(1) = q(2);
dq(2) = -coeff*sin(q(1));
end
EDIT:
Could this be the problem?
tstart = 0;
tend = 10;
tsteps = 0.01;
tspan = [0 10];
t = [tstart:tsteps:tend];
f = 2;
u = squaresignal(t,f)
Here you create a time vector t which has nothing to do with the time vector returned by the ODE solver! This means that at first we have t[2]=0.01 but once you ran your ODE solver, t[2] can be anything. So yes, if you want to load an external signal source depending on time, then you need to call your squaresignal.m from within the differential equation and pass the solver's current time t! Your code should look like this (note that I'm passing f now as an additional argument to the diffeq):
function dX = test2(t ,X, f)
x = X (1) ;
y = X (2) ;
M = 10;
k = 50;
c = 10;
F = 300;
u = squaresignal(t,f)
if u == 1
F = F;
else
F = 0,
end
dx = y ;
dy = -k/M *x - c/M *y+ F/M ;
dX = [ dx dy ]';
end
Note however that matlab's ODE solvers do not like at all what you're doing here. You are drastically (i.e. non-smoothly) changing the dynamics of your system. What you should do is to use one of the following:
a) events if you want to trigger some behaviour (like termination) depending on the integrated variable x or
b) If you want to trigger the behaviour based on the time t, you should segment your integration into different parts where the differential equation does not vary during one segment. You can then resume your integration by using the current state and time as x0 and t0 for the next run of ode15s. Of course this only works of you're external signal source u is something simple like a step funcion or square wave. In case of the square wave you would only integrate for a timespan during which the wave does not jump. And then exactly at the time of the jump you start another integration with altered differential equations.
In Matlab I want to create the partial derivative of a cost function called J(theta_0, theta_1) (in order to do the calculations necessary to do gradient descent).
The function J(theta_0, theta_1) is defined as:
Lets say h_theta(x) = theta_1 + theta_2*x. Also: alpha is fixed, the starting values of theta_1 and theta_2 are given. Let's say in this example: alpha = 0.1 theta_1 = 0, theta_2 = 1. Also I have all the values for x and y in two different vectors.
VectorOfX =
5
5
6
VectorOfX =
6
6
10
Steps I took to try to solve this in Matlab: I have no clue how to solve this problem in matlab. So I started off with trying to define a function in Matlab and tried this:
theta_1 = 0
theta_2 = 1
syms x;
h_theta(x) = theta_1 + t2*x;
This worked, but is not what I really wanted. I wanted to get x^(i), which is in a vector. The next thing I tried was:
theta_1 = 0
theta_2 = 1
syms x;
h_theta(x) = theta_1 + t2*vectorOfX(1);
This gives the following error:
Error using sym/subsindex (line 672)
Invalid indexing or function definition. When defining a
function, ensure that the body of the function is a SYM
object. When indexing, the input must be numeric, logical or
':'.
Error in prog1>gradientDescent (line 46)
h_theta(x) = theta_1 + theta_2*vectorOfX(x);
I looked up this error and don't know how to solve it for this particular example. I have the feeling that I make matlab work against me instead of using it in my favor.
When I have to perform symbolic computations I prefer to use Mathematica. In that environment this is the code to get the partial derivatives you are looking for.
J[th1_, th2_, m_] := Sum[(th1 + th2*Subscript[x, i] - Subscript[y, i])^2, {i, 1, m}]/(2*m)
D[J[th1, th2, m], th1]
D[J[th1, th2, m], th2]
and gives
Coming back to MATLAB we can solve this problem with the following code
%// Constants.
alpha = 0.1;
theta_1 = 0;
theta_2 = 1;
X = [5 ; 5 ; 6];
Y = [6 ; 6 ; 10];
%// Number of points.
m = length(X);
%// Partial derivatives.
Dtheta1 = #(theta_1, theta_2) sum(2*(theta_1+theta_2*X-Y))/2/m;
Dtheta2 = #(theta_1, theta_2) sum(2*X.*(theta_1+theta_2*X-Y))/2/m;
%// Loop initialization.
toll = 1e-5;
maxIter = 100;
it = 0;
err = 1;
theta_1_Last = theta_1;
theta_2_Last = theta_2;
%// Iterations.
while err>toll && it<maxIter
theta_1 = theta_1 - alpha*Dtheta1(theta_1, theta_2);
theta_2 = theta_2 - alpha*Dtheta2(theta_1, theta_2);
it = it + 1;
err = norm([theta_1-theta_1_Last ; theta_2-theta_2_Last]);
theta_1_Last = theta_1;
theta_2_Last = theta_2;
end
Unfortunately for this case the iterations does not converge.
MATLAB is not very flexible for symbolic computations, however a way to get those partial derivatives is the following
m = 10;
syms th1 th2
x = sym('x', [m 1]);
y = sym('y', [m 1]);
J = #(th1, th2) sum((th1+th2.*x-y).^2)/2/m;
diff(J, th1)
diff(J, th2)
I am using the following code to produce a numerical solution to a system of ODE's with 6 boundary conditions.
I am using the initial conditions to obtain a solution but I must vary three other conditions in order to find the true solution. The function I have is as follows:
function diff = prob5diff(M,Fx,Fy)
u0 = [pi/2 0 0]';
sSpan = [0 13];
p = #(t,u) prob5(t,u,M,Fx,Fy);
options = odeset('reltol',1e-6,'abstol',1e-6);
[s,u] = ode45(p,sSpan,u0,options);
L = length(s);
x = u(:,2); y = u(:,3); theta = u(:,1);
diff(1) = x(L) - 5;
diff(2) = y(L);
diff(3) = theta(L) + pi/2;
end
Ultimately, different values of M,Fx, and Fy will produce different solutions and I would like a solution such that the values in diff are as close to zero as possible so I want fsolve to iterate through different values of M,Fx, and Fy
I am receiving the following error: when I call it in this way:
opt = optimset('Display','iter','TolFun',1e-6);
guess = [1;1;1];
soln = fsolve(#prob5diff,guess,opt);
Error in line:
soln = fsolve(#prob5diff,guess,opt);
Caused by:
Failure in initial user-supplied objective function evaluation. FSOLVE cannot continue.
Thanks!
One problem is that you have to call fsolve on prob5diff which takes a single vector input since your guess is a single vector:
prob5diff(x)
M = x(1);
Fx = x(2);
Fy = x(3);