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

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Zhen Zhen 2019 年 3 月 25 日

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)
Cheers,
Jane
##### 2 件のコメント表示非表示 1 件の古いコメント
Walter Roberson 2019 年 3 月 26 日
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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### 採用された回答

Teja Muppirala 2019 年 3 月 26 日

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
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)) ];
end
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;
end
end
end
##### 1 件のコメント表示非表示 なし
Zhen Zhen 2019 年 3 月 28 日
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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### その他の回答 (1 件)

Lewis Fer 2021 年 6 月 10 日

Hello, I am having troubles solving a system of second order nonlinear equations with boundary conditions using MATALB
Here is the equations:
f''(t)=3*f(t)*g(t) +5;
g''(t)=-4f(t)*g(t)+7;
the boundary conditions are: f'(0)=0 et g'(o)=5;
g(0)=3 et f'(2)=f(2)

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