Solving coupled non linear differential equation by Mat-lab or by calculations
equation 1: x'(t) = -a* x(t) /(x(t) + y(t))
equation 2: y'(t) = -b* y(t) /(x(t) + y(t))
I tried in mathematica but got a very comlicated solution.
Solve[{x'[t] == -a* x[t] /(x[t] + y[t]), y'[t] == -b* y[t] /(x[t] + y[t])}, {x, y}, t]
How can I plot it?
My initial conditions are
x(0) = xo
y(0) = yo
Also, a and b are constants.
I have to plot x and y wrt t after inserting values of a and b . ( a= 2 , b =5 say )
A lot of things to note in this situation:
You need to create a function that contains both a and b:
function dy =soProblem(t,y,a,b)
dy=[-a*y(1)/(y(1)+y(2)); -b*y(2)/(y(1)+y(2))];
end
Call the standard ode function:
a = 2;
b = 5; tend = 10; x0 = 1; y0 = 2;
[T,Y] = ode45(#(t,y)soProblem(t,y,a,b),[0 tend],[x0 y0]);
plot (T,Y)
Realize you may have a stiff equation on your hands.
Have fun identifying the ideal function call:
[T15,Y15] = ode15s(#(t,y)soProblem(t,y,a,b),[0 tend],[x0 y0]);
[T23t,Y23t] = ode23t(#(t,y)soProblem(t,y,a,b),[0 tend],[x0 y0]);
[T23tb,Y23tb] = ode23tb(#(t,y)soProblem(t,y,a,b),[0 tend],[x0 y0]);
%note ode23s doesn't converge (or at least takes forever)
plot (T,Y,T15,Y15,T23t,Y23t,T23tb,Y23tb)
Understand why mathematica becomes restless
In mathematica:
Try ndsolve
In matlab:
Create a function file yourfunction.m:
function [Y_prime]=yourfunction(t, Y)
Y_prime=[-2*Y(1)./(Y(1) + Y(2)) -5*Y(2)./(Y(1) + Y(2))];
end
and then
[T,Y] = ode45(yourfunction,[0 t_end],[x0 y0]);
plot(T,Y(:,1));
hold on
plot(T,Y(:,2));
Related
For example, FX = x ^ 2 + sin (x)
Just for curiosity, I don't want to use the CVX toolbox to do this.
You can check this within some interval [a,b] by checking if the second derivative is nonnegative. For this you have to define a vector of x-values, find the numerical second derivative and check whether it is not too negative:
a = 0;
b = 1;
margin = 1e-5;
point_count = 100;
f=#(x) x.^2 + sin(x);
x = linspace(a, b, point_count)
is_convex = all(diff(x, 2) > -margin);
Since this is a numerical test, you need to adjust the parameter to the properties of the function, that is if the function does wild things on a small scale we might not be able to pick it up. E.g. with the parameters above the test will falsely report the function f=#(x)sin(99.5*2*pi*x-3) as convex.
clear
syms x real
syms f(x) d(x) d1(x)
f = x^2 + sin(x)
d = diff(f,x,2)==0
d1 = diff(f,x,2)
expSolution = solve(d, x)
if size(expSolution,1) == 0
if eval(subs(d1,x,0))>0
disp("condition 1- the graph is concave upward");
else
disp("condition 2 - the graph is concave download");
end
else
disp("condition 3 -- not certain")
end
I'm having some issues getting my RK2 algorithm to work for a certain second-order linear differential equation. I have posted my current code (with the provided parameters) below. For some reason, the value of y1 deviates from the true value by a wider margin each iteration. Any input would be greatly appreciated. Thanks!
Code:
f = #(x,y1,y2) [y2; (1+y2)/x];
a = 1;
b = 2;
alpha = 0;
beta = 1;
n = 21;
h = (b-a)/(n-1);
yexact = #(x) 2*log(x)/log(2) - x +1;
ye = yexact((a:h:b)');
s = (beta - alpha)/(b - a);
y0 = [alpha;s];
[y1, y2] = RungeKuttaTwo2D(f, a, b, h, y0);
error = abs(ye - y1);
function [y1, y2] = RungeKuttaTwo2D(f, a, b, h, y0)
n = floor((b-a)/h);
y1 = zeros(n+1,1); y2 = y1;
y1(1) = y0(1); y2(1) = y0(2);
for i=1:n-1
ti = a+(i-1)*h;
fvalue1 = f(ti,y1(i),y2(i));
k1 = h*fvalue1;
fvalue2 = f(ti+h/2,y1(i)+k1(1)/2,y2(i)+k1(2)/2);
k2 = h*fvalue2;
y1(i+1) = y1(i) + k2(1);
y2(i+1) = y2(i) + k2(2);
end
end
Your exact solution is wrong. It is possible that your differential equation is missing a minus sign.
y2'=(1+y2)/x has as its solution y2(x)=C*x-1 and as y1'=y2 then y1(x)=0.5*C*x^2-x+D.
If the sign in the y2 equation were flipped, y2'=-(1+y2)/x, one would get y2(x)=C/x-1 with integral y1(x)=C*log(x)-x+D, which contains the given exact solution.
0=y1(1) = -1+D ==> D=1
1=y1(2) = C*log(2)-1 == C=1/log(2)
Additionally, the arrays in the integration loop have length n+1, so that the loop has to be from i=1 to n. Else the last element remains zero, which gives wrong residuals for the second boundary condition.
Correcting that and enlarging the computation to one secant step finds the correct solution for the discretization, as the ODE is linear. The error to the exact solution is bounded by 0.000285, which is reasonable for a second order method with step size 0.05.
Introduction
NOTE IN CODE AND DISUSSION:
A single d is first derivative A double d is second derivative
I am using Matlab to simulate some dynamic systems through numerically solving the governing LaGrange Equations. Basically a set of Second Order Ordinary Differential Equations. I am using ODE45. I found a great tutorial from Mathworks (link for tutorial below) on how to solve a basic set of second order ordinary differential equations.
https://www.mathworks.com/academia/student_center/tutorials/source/computational-math/solving-ordinary-diff-equations/player.html
Based on the tutorial I simulated the motion for an elastic spring pendulum by obtaining two second order ordinary differential equations (one for angle theta and the other for spring elongation) shown below:
theta double prime equation:
M*thetadd*(L + del)^2 + M*g*sin(theta)*(L + del) + M*deld*thetad*(2*L + 2*del) = 0
del (spring elongation) double prime equation:
K*del + M*deldd - (M*thetad^2*(2*L + 2*del))/2 - M*g*cos(theta) = 0
Both equations above have form ydd = f(x, xd, y, yd)
I solved the set of equations by a common reduction of order method; setting column vector z to [theta, thetad, del, deld] and therefore zd = [thetad, thetadd, deld, deldd]. Next I used two matlab files; a simulation file and a function handle file for ode45. See code below of simulation file and function handle file:
Simulation File
%ElasticPdlmSymMainSim
clc
clear all;
%Define parameters
global M K L g;
M = 1;
K = 25.6;
L = 1;
g = 9.8;
% define initial values for theta, thetad, del, deld
theta_0 = 0;
thetad_0 = .5;
del_0 = 1;
deld_0 = 0;
initialValues = [theta_0, thetad_0, del_0, deld_0];
% Set a timespan
t_initial = 0;
t_final = 36;
dt = .01;
N = (t_final - t_initial)/dt;
timeSpan = linspace(t_final, t_initial, N);
% Run ode45 to get z (theta, thetad, del, deld)
[t, z] = ode45(#OdeFunHndlSpngPdlmSym, timeSpan, initialValues);
Here is the function handle file:
function dz = OdeFunHndlSpngPdlmSym(~, z)
% Define Global Parameters
global M K L g
% Take output from SymDevFElSpringPdlm.m file for fy1 and fy2 and
% substitute into z2 and z4 respectively
% z1 and z3 are simply z2 and z4
% fy1=thetadd=z(2)= -(M*g*sin(z1)*(L + z3) + M*z2*z4*(2*L + 2*z3))/(M*(L + z3)^2)
% fy2=deldd=z(4)=((M*(2*L + 2*z3)*z2^2)/2 - K*z3 + M*g*cos(z1))/M
% return column vector [thetad; thetadd; deld; deldd]
dz = [z(2);
-(M*g*sin(z(1))*(L + z(3)) + M*z(2)*z(4)*(2*L + 2*z(3)))/(M*(L + z(3))^2);
z(4);
((M*(2*L + 2*z(3))*z(2)^2)/2 - K*z(3) + M*g*cos(z(1)))/M];
Question
However, I am coming across systems of equations where the variables can not be solved for explicitly as is the case with spring pendulum example. For one case I have the following set of ordinary differential equations:
y double prime equation
ydd - .5*L*(xdd*sin(x) + xd^2*cos(x) + (k/m)*y - g = 0
x double prime equation
.33*L^2*xdd - .5*L*ydd*sin(x) - .33*L^2*C*cos(x) + .5*g*L*sin(x) = 0
L, g, m, k, and C are given parameters.
Note that x'' term appears in y'' equation and y'' term appears in x'' equation so I am not able to use reduction of order method. Can I use Matlab ODE45 to solve the set of ordinary differential equations in the second example in a manner similar to first example?
Thanks!
This problem can be solved by working out some of the math by hand. The equations are linear in xdd and ydd so it should be straightforward to solve.
ydd - .5*L*(xdd*sin(x) + xd^2*cos(x)) + (k/m)*y - g = 0
.33*L^2*xdd - .5*L*ydd*sin(x) - .33*L^2*C*cos(x) + .5*g*L*sin(x) = 0
can be rewritten as
-.5*L*sin(x)*xdd + ydd = -.5*L*xd^2*cos(x) - (k/m)*y + g
.33*L^2*xdd - .5*L*sin(x)*ydd = .33*L^2*C*cos(x) - .5*g*L*sin(x)
which is the form A*x=b.
For more complex systems, you can look into the fsolve function.
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.
I am trying to solve a differential equation with the ode solver ode45 with MATLAB. I have tried using it with other simpler functions and let it plot the function. They all look correct, but when I plug in the function that I need to solve, it fails. The plot starts off at y(0) = 1 but starts decreasing at some point when it should have been an increasing function all the way up to its critical point.
function [xpts,soln] = diffsolver(p1x,p2x,p3x,p1rr,y0)
syms x y
yp = matlabFunction((p3x/p1x) - (p2x/p1x) * y);
[xpts,soln] = ode45(yp,[0 p1rr],y0);
p1x, p2x, and p3x are polynomials and they are passed into this diffsolver function as parameters.
p1rr here is the critical point. The function should diverge after the critical point, so i want to integrate it up to that point.
EDIT: Here is the code that I have before using diffsolver, the above function. I do pade approximation to find the polynomials p1, p2, and p3. Then i find the critical point, which is the root of p1 that is closest to the target (target is specified by user).
I check if the critical point is empty (sometimes there might not be a critical point in some functions). If its not empty, then it uses the above function to solve the differential equation. Then it plots the x- and y- points returned from the above function basically.
function error = padeapprox(m,n,j)
global f df p1 p2 p3 N target
error = 0;
size = m + n + j + 2;
A = zeros(size,size);
for i = 1:m
A((i + 1):size,i) = df(1:(size - i));
end
for i = (m + 1):(m + n + 1)
A((i - m):size,i) = f(1:(size + 1 - i + m));
end
for i = (m + n + 2):size
A(i - (m + n + 1),i) = -1;
end
if det(A) == 0
error = 1;
fprintf('Warning: Matrix is singular.\n');
end
V = -A\df(1:size);
p1 = [1];
for i = 1:m
p1 = [p1; V(i)];
end
p2 = [];
for i = (m + 1):(m + n + 1)
p2 = [p2; V(i)];
end
p3 = [];
for i = (m + n + 2):size
p3 = [p3; V(i)];
end
fx = poly2sym(f(end:-1:1));
dfx = poly2sym(df(end:-1:1));
p1x = poly2sym(p1(end:-1:1));
p2x = poly2sym(p2(end:-1:1));
p3x = poly2sym(p3(end:-1:1));
p3fullx = p1x * dfx + p2x * fx;
p3full = sym2poly(p3fullx); p3full = p3full(end:-1:1);
p1r = roots(p1(end:-1:1));
p1rr = findroots(p1r,target); % findroots eliminates unreal roots and chooses the one closest to the target
if ~isempty(p1rr)
[xpts,soln] = diffsolver(p1x,p2x,p3fullx,p1rr,f(1));
if rcond(A) >= 1e-10
plot(xpts,soln); axis([0 p1rr 0 5]); hold all
end
end
I saw some examples using another function to generate the differential equation but i've tried using the matlabFunction() method with other simpler functions and it seems like it works. Its just that when I try to solve this function, it fails. The solved values start becoming negative when they should all be positive.
I also tried using another solver, dsolve(). But it gives me an implicit solution all the time...
Does anyone have an idea why this is happening? Any advice is appreciated. Thank you!
Since your code seems to work for simpler functions, you could try to increase the accuracy options of the ode45 solver.
This can be achieved by using odeset:
options = odeset('RelTol',1e-10,'AbsTol',1e-10);
[T,Y] = ode45(#function,[tspan],[y0],options);