MATLAB Answers


solve and plot a system of nonlinear 2nd order differential equations

Asked by Zhen Zhen on 25 Mar 2019
Latest activity Commented on by Zhen Zhen on 28 Mar 2019
hi there,
I'm trying to plot a graph of against with the following equations of motion:
I've tried dsolve and ode45 yet there always seems to be some problems. I think ode45 might work better because apparently it would be easier to plot the graph by using some numerical method?
Here's my failed attempt to solve it: (I just set some variables to be 1 to make the problem easier here)
syms theta(t) phi(t) psi(t) C
dtheta = diff(theta , t);
d2theta = diff(theta , t , 2);
dphi = diff(phi , t);
%dpsi = diff(psi , t);
%alpha = C * (dpsi + dphi * cos(theta));
%beta = alpha * cos(theta) + dphi * (sin(theta))^2;
alpha = 1;
beta = 1;
dpsi = 1;
eqn1 = dphi == (beta - alpha * cos(theta)) / (sin(theta))^2 ; %equations of motion
eqn2 = d2theta == (dphi*(dphi * cos(theta) - alpha)+1)*sin(theta) ; %equations of motion
eqns = [eqn1 , eqn2];
%cond = [dpsi == 1];
[thetaSol(t) phiSol(t)] = dsolve(eqns)
Thanks a lot for your help and time in advance!


There are at least four solutions to that.
Two of them are pretty abstract, involving integrals of roots of an equation -- not closed form solutions but rather a description of what properties the solution function would have to have.
The other two solutions are closed form:
theta(t) = arctan(sqrt(2*diff(phi(t), t) - 1)/diff(phi(t), t), (-diff(phi(t), t) + 1)/diff(phi(t), t))
and the same except the negative of the first parameter.
MATLAB is not powerful enough to arrive at these solutions.
Have a look at odeFunction() and in particular the first example. It shows you the functions you need to use in order to convert the symbolic forms into something you can call with ode45.
Use the options structure created by odeset to designate an OutputFcn . You might want to use @odephas2 to construct a 2D phase plot, perhaps.

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1 Answer

Answer by Teja Muppirala
on 26 Mar 2019
Edited by Teja Muppirala
on 26 Mar 2019
 Accepted Answer

If you just need a plot and not a closed-form solution, then I'd recommend just using ODE45 without worrying about symbolic stuff. This is an example of how to solve this using ODE45 for initial conditions psi(0) = 0, theta(0) = 0, thetadot(0) = 1 over the time span [0 10].
function doODE
a = 1; % alpha
b = 1; % beta
% Define variables as:
% Y(1): phi
% Y(2): theta
% Y(3): thetadot
ic = [0;0;1]; % Pick some initial Conditions
tspan = [0 10];
[tout, yout] = ode45(@deriv,tspan,ic);
subplot(1,3,1), plot(tout,yout(:,1)); title('psi(t)')
subplot(1,3,2), plot(tout,yout(:,2)); title('theta(t)')
subplot(1,3,3), plot(yout(:,1),yout(:,2)); title('theta vs psi')
function dY = deriv(t,Y)
dY = [dpsi(Y(2)); ...
Y(3); ...
[dpsi(Y(2))*(dpsi(Y(2))*cos(Y(2))-a)+1]*sin(Y(2)) ];
function dp = dpsi(th)
dp = (b - a*cos(th))./(sin(th)).^2 ;
if isnan(dp) % Need to take care at theta = 0
dp = 0.5;

  1 Comment

Thank you very much! it worked well!
I was wondering, if now I need to use the real alpha and beta (alpha = C * (dpsi + dphi * cos(theta)); beta = alpha * cos(theta) + dphi * (sin(theta))^2;) I know they are constants. Where should I put them in the code?
And if I need to print one more constant (say, E = 0.5 * (Y(3))^2 + (Y(1))^2 * (sin(Y(2)))^2) +0.5* a^2 + cos(Y(2)) ) where should I put it?
Thanks so much for your help!

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