MATLAB: How do we plot differential Equation

differential equationskuramoto model

With theta1(0)=0.1, theta2(0)=0.3, theta3(0)=0.2 , I have to plot the solution in the range of [0.1] all in one graph.

I have no idea how to do this:

diff(theta1)=1+sin(theta2-theta1);

diff(theta2)=1+sin(theta1-theta2)+sin(theta3-theta2);

diff(theta3)=1+sin(theta2-theta3);

ezplot(

Please help

Best Answer

Use ode45() for estimating a numerical solution

IC = [0.1, 0.3, 0.2];
tspan = [0 1];
[t, thetas] = ode45(@odeFun, tspan, IC);
plot(t, thetas);
legend({'\theta_1', '\theta_2', '\theta_3'}, 'FontSize', 16, 'Location', 'best');
function dthetas = odeFun(t, thetas)
    theta1 = thetas(1);
    theta2 = thetas(2);
    theta3 = thetas(3);
    
    dtheta1 = 1+sin(theta2-theta1);
    dtheta2 = 1+sin(theta1-theta2)+sin(theta3-theta2);
    dtheta3 = 1+sin(theta2-theta3);
    
    dthetas = [dtheta1; dtheta2; dtheta3];
end

Related Solutions

MATLAB: Replace x with x(t) in a symbolic matrix

syms theta1 theta2 theta3
T = [  cos(theta1 + theta2 + theta3), sin(theta1 + theta2 + theta3), 0,   8*cos(theta1 + theta2 + theta3) + 5*cos(theta1 + theta2) + 10*cos(theta1)
    -sin(theta1 + theta2 + theta3), cos(theta1 + theta2 + theta3), 0, - 8*sin(theta1 + theta2 + theta3) - 5*sin(theta1 + theta2) - 10*sin(theta1)
    0,                             0, 1,                                                                           0
    0,                             0, 0,                                                                           1];
syms theta4(t) theta5(t) theta6(t)
temp = subs(T, {theta1, theta2, theta3}, {theta4, theta5, theta6});
syms theta1(t) theta2(t) theta3(t)
AT1 = subs(temp, {theta4, theta5, theta6}, {theta1, theta2, theta3});

"It is also fine if there is some way I can call elements of the matrix below,"

T10(t) =

[ cos(theta1(t)), sin(theta1(t)), 0, 10*cos(theta1(t))]

That is not a matrix: it is a symbolic function that returns a matrix. You cannot access individual elements of the function through normal indexing. You can, however, use formula(T10) which will return an array result that is not a function, and then you can index the array result.

MATLAB: Solving system of second order differential equations with ode45

I prefer to let the Symbolic MAth Toolbox do the ‘heavy lifting’:

syms theta1(t) theta2(t) t Y 
% theta1'' = -29.4sin(theta1)- 9.8sin(theta1 -2*theta2) - 2*sin(theta1-theta2)*(theta1'^(2)*cos(theta1-theta2)]/(3-cos(2*theta1-2*theta2))
% theta2'' = 2*sin(theta1-theta2)[(2*theta1')+19.6cos(theta1)+theta2'*cos(theta1-theta2)]]/(3-cos(2*theta1-2*theta1))
Dt11 = diff(theta1);
Dt12 = diff(Dt11);
Dt21 = diff(theta2);
Dt22 = diff(Dt21);
Eqn1 = Dt12 == (-29.4*sin(theta1)- 9.8*sin(theta1 -2*theta2) - 2*sin(theta1-theta2)*(Dt11^(2)*cos(theta1-theta2)))/(3-cos(2*theta1-2*theta2));
Eqn2 = Dt22 == (2*sin(theta1-theta2)*((2*theta1')+19.6*cos(theta1)+Dt21*cos(theta1-theta2)))/(3-cos(2*theta1-2*theta1));
Eqns = [Eqn1; Eqn2];
[VF,Sbs] = odeToVectorField(Eqns);
Sbs
odesfcn = matlabFunction(VF, 'Vars',{t, Y})

producing:

Sbs =
 
  theta2
 Dtheta2
  theta1
 Dtheta1
 
odesfcn =
  function_handle with value:
    @(t,Y)[Y(2);-sin(Y(1)-Y(3)).*(conj(Y(3)).*2.0+cos(Y(3)).*(9.8e+1./5.0)+cos(Y(1)-Y(3)).*Y(2));Y(4);-(sin(Y(1).*2.0-Y(3)).*(4.9e+1./5.0)-sin(Y(3)).*(1.47e+2./5.0)+cos(Y(1)-Y(3)).*sin(Y(1)-Y(3)).*Y(4).^2.*2.0)./(cos(Y(1).*2.0-Y(3).*2.0)-3.0)]

or:

    odesfcn = @(t,Y)[Y(2);-sin(Y(1)-Y(3)).*(conj(Y(3)).*2.0+cos(Y(3)).*(9.8e+1./5.0)+cos(Y(1)-Y(3)).*Y(2));Y(4);-(sin(Y(1).*2.0-Y(3)).*(4.9e+1./5.0)-sin(Y(3)).*(1.47e+2./5.0)+cos(Y(1)-Y(3)).*sin(Y(1)-Y(3)).*Y(4).^2.*2.0)./(cos(Y(1).*2.0-Y(3).*2.0)-3.0)];

Use the ‘Sbs’ vector in your legend call.

NOTE — I did not test this with an actual ode45 call. You may need to experiment with it to get it to work correctly.

Best Answer

Related Solutions

MATLAB: Replace x with x(t) in a symbolic matrix

MATLAB: Solving system of second order differential equations with ode45

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