How can I solve the periodic function with ode45 solver?

Question

Doruk Can Ogan 2018-2-8

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回答： Walter Roberson 2018-2-9

hello everyone,

alpha=t*(2*pi/T) (angular velocity and T is the period=0.02s) (alpha [0 2*pi])

V=5*alpha^2

dVdt=10*alpha*(2*pi/T)

I have a periodic function, the volume changes with alpha and also its derivative with time. I want to solve basically this differential equation dVdt=10*alpha*(2*pi/T) with ode45 solver and at the end of the integration, I should get a solution like below. I know I already have the V=5*alpha^2 solution and I can use module operator. But here I want to know how to do this with ode solver for more complicated equations. Thank you

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Answer 1

JK 2018-2-8

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Hi,

use the mod-command to account for ther periodicality of t.

tspan=linspace(0,0.08,1000);
[t,y]=ode45(@odefun,tspan,0);
figure
plot(t,y)
ylabel ('V')
xlabel ('t')
function dVdt=odefun(t,~)
    T=0.02;
    alpha=mod(t,T)*2*pi;
    V=5*alpha^2;
    dVdt=10*alpha*2*pi/T;
end

though I am not getting V = 200 at t=T. however, this will still produce an ongoing increase of V. If you want to set V back to 0 every T, use a boudary value solver. See example 4 in the tutorial for BV-problems.

example 4 tutorial

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Answer 2

Walter Roberson 2018-2-9

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No, you cannot do this. The ode*() functions require that the function be continuous, and to have two further derivatives that are continuous (it will usually detect if the first derivative is not continuous, but if the second derivative is not continuous it will probably just give the wrong answer.)

You have to stop the ode at the point of discontinuity and restart it. If your function is periodic in time, then that can be done by setting the tspan to cover just one period at a time. If your boundaries are not easily computed as time, then you need to create an event function that signals termination.