How to solve a second order nonlinear differential equations with a step function
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Sudipta Mukherjee
on 14 Jun 2022
Commented: Walter Roberson
on 5 Jul 2022
for 0<=t<=10 solve
5 x'' + x' + t x^3 = t + 3 u(x) where u(x) = 1 for 1<=x<=2
=0 otherwise
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Accepted Answer
Sam Chak
on 14 Jun 2022
Edited: Sam Chak
on 18 Jun 2022
Edit: The is not signum-based, and the constraint for the velocity is defined in an Event function called velocityEventsFcn.
options = odeset('Events', @velocityEventsFcn);
[t, x, te, xe, ie] = ode23s(@odefcn, [0 10], [0.5 1.5], options);
plot(t, x, 'linewidth', 1.5)
function dxdt = odefcn(t, x)
dxdt = zeros(2, 1);
u = max(0, min(min(100000*x(1) - 99999, 1), min(1, -100000*x(1) + 200001)));
dxdt(1) = x(2);
dxdt(2) = (t + 3*u - (x(2) + t*x(1)^3))/5; % 5*x" + x' + t*x^3 = t + 3*u(x)
end
function [position, isterminal, direction] = velocityEventsFcn(t, x)
position = x(2); % When velocity x(2) = 0,
isterminal = 1; % the integration stops,
direction = 0; % and the velocity cannot go into negative no matter what
end
13 Comments
Walter Roberson
on 5 Jul 2022
Are you still working on the 5 x'' + x' + t x^3 = t + 3 u(x) just with a different range over which u(x) is 1?
Walter Roberson
on 5 Jul 2022
Could you remind us what the boundary conditions are?
options = odeset('Events', @velocityEventsFcn);
[t, x, te, xe, ie] = ode23s(@odefcn, [0 10], [-0.5 1e-5], options);
plot(t, x, 'linewidth', 1.5)
legend({"x'", "x''"})
function dxdt = odefcn(t, x)
dxdt = zeros(2, 1);
u = x(1)>=0 && x(1)<=10^-9;
dxdt(1) = x(2);
dxdt(2) = (t + 3*u - (x(2) + t*x(1)^3))/5; % 5*x" + x' + t*x^3 = t + 3*u(x)
end
function [position, isterminal, direction] = velocityEventsFcn(t, x)
position = x(2); % When velocity x(2) = 0,
isterminal = 1; % the integration stops,
direction = 0; % and the velocity cannot go into negative no matter what
end
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