How to solve a second order nonlinear differential equations with a step function
显示 更早的评论
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
采纳的回答
Edit: The 
is not signum-based, and the constraint for the velocity 
is defined in an Event function called velocityEventsFcn.
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 个评论
Sudipta Mukherjee
2022-6-15
Walter Roberson
2022-6-15
ode45 and related functions are not compatible with discontinuous functions. You would need to stop at the boundaries and restart.
Effective clarification will improve the communication. I have two queries.
Query #1:
Perhaps I interpreted your u(x) incorrectly. It is true that ode45 does not work well with jump discontinuities. Would you enlighten exactly which part of the mathematics that incorrectly describes 
? Or, the signum function is disallowed by your Professor? Or, a fancy continuous function is preferred?
? Or, the signum function is disallowed by your Professor? Or, a fancy continuous function is preferred?x = 0:0.01:3;
u = max(0, min(min(1000*x - 999, 1), min(1, -1000*x + 2001)));
plot(x, u, 'linewidth', 1.5)
grid on
xlabel('x')
ylabel('u')
ylim([-0.5 1.5])
Query #2:
You have mentioned that 
cannot be negative for the entire simulation. Does your ODE correctly reflect this? The ODE solver only integrates the differential equations that are given. If you can provide a little more critical information about the system that you want to simulate, then it will be helpful in the coding process.
cannot be negative for the entire simulation. Does your ODE correctly reflect this? The ODE solver only integrates the differential equations that are given. If you can provide a little more critical information about the system that you want to simulate, then it will be helpful in the coding process.
Sudipta Mukherjee
2022-6-17
At u(1) and u(2) , It's not mandatory to use ode45 only.
Actually 
is the velocity then how to put a condition that 
is the velocity then how to put a condition that 
Sam Chak
2022-6-17
@Sudipta Mukherjee, Still waiting for your technical reply on Query #1. Why would you want to avoid using signum function to describe 
? Is the newly proposed function acceptable to describe 
?
? Is the newly proposed function acceptable to describe ?On 
, we know nothing about the physical dynamics of your system. Either the differential equations are modeled by you, or they come from a reference (your supervisor, books, journal paper, etc.).
, we know nothing about the physical dynamics of your system. Either the differential equations are modeled by you, or they come from a reference (your supervisor, books, journal paper, etc.).
Sudipta Mukherjee
2022-6-17
then u(1)=u(2)=
at boundary as u=(sign(x(1)-1)-sign(x(1)-2))/2;Thanks for your clarification. Now I see...
If 
is strictly desired, and the proposed signum-based boxcar function gives 
, then the requirement 
is not met.
is strictly desired, and the proposed signum-based boxcar function gives , then the requirement is not met. However, thinking deeper, would x (position) reach 
or 
in finite time?
or in finite time?I'm just trying to help and I know that 
is a piecewise function. What mathematical differentiable function do you suggest for ? Anyhow, I have edited the Answer to run the sim for 
.
is a piecewise function. What mathematical differentiable function do you suggest for ? Anyhow, I have edited the Answer to run the sim for .
Sudipta Mukherjee
2022-7-5
if 
then what will be the u(x) function in the matlab code?
then what will be the u(x) function in the matlab code?
Sam Chak
2022-7-5
I did't try, but you can use the non-signum template to design for the new 
.
.
Sudipta Mukherjee
2022-7-5
编辑：Sudipta Mukherjee
2022-7-5
Walter Roberson
2022-7-5
Yes, go ahead. It can't hurt, at least not more than what you are already doing.
Walter Roberson
2022-7-5
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?
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
更多回答（0 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Programming 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
