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South China University of Technology

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Oscillator motions

Xiaobao Yang

Department of Physics

http://www.compphys.cn/~xbyang/lectures.html

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Simulation of Quasi-crystals

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Annealing simulation

Phase Transitions

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Charged particle in electromagnetic field

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One particle fx1(ii)=q1*E0+q1*B*vy1(ii); fy1(ii)=-q1*B*vx1(ii); vx1(ii+1)=vx1(ii)+fx1(ii)/m1*dt; vy1(ii+1)=vy1(ii)+fy1(ii)/m1*dt; x1(ii+1)=x1(ii)+vx1(ii)*dt; y1(ii+1)=y1(ii)+vy1(ii)*dt; fx1(ii+1)=q1*E0+q1*B*vy1(ii+1); fy1(ii+1)=-q1*B*vx1(ii+1); vx1(ii+1)=vx1(ii)+0.5*(fx1(ii)+fx1(ii+1))/m1*dt; vy1(ii+1)=vy1(ii)+0.5*(fy1(ii)+fy1(ii+1))/m1*dt; x1(ii+1)=x1(ii)+0.5*(vx1(ii)+vx1(ii+1))*dt; y1(ii+1)=y1(ii)+0.5*(vy1(ii)+vy1(ii+1))*dt;

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Two particles fx1(ii)=q1*E0+q1*B*vy1(ii)-k*q1*q2*(x2(ii)-x1(ii))/norm([x2(ii)-x1(ii) y2(ii)-y1(ii)])^3; fy1(ii)=-q1*B*vx1(ii)-k*q1*q2*(y2(ii)-y1(ii))/norm([x2(ii)-x1(ii) y2(ii)-y1(ii)])^3; fx2(ii)=q2*E0+q2*B*vy2(ii)+k*q1*q2*(x2(ii)-x1(ii))/norm([x2(ii)-x1(ii) y2(ii)-y1(ii)])^3; fy2(ii)=-q2*B*vx2(ii)+k*q1*q2*(y2(ii)-y1(ii))/norm([x2(ii)-x1(ii) y2(ii)-y1(ii)])^3;

vx1(ii+1)=vx1(ii)+fx1(ii)/m1*dt; vy1(ii+1)=vy1(ii)+fy1(ii)/m1*dt; x1(ii+1)=x1(ii)+vx1(ii)*dt; y1(ii+1)=y1(ii)+vy1(ii)*dt; vx2(ii+1)=vx2(ii)+fx2(ii)/m1*dt; vy2(ii+1)=vy2(ii)+fy2(ii)/m1*dt; x2(ii+1)=x2(ii)+vx2(ii)*dt; y2(ii+1)=y2(ii)+vy2(ii)*dt;

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the motion of electrons in atoms

the behavior of currents and voltages in electronic circuits

planetary orbits

a pendulum

oscillatory and periodic phenomena

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Simple Harmonic Motion

Euler method

1

1

1

( / )i i i

i i i

i i

g l t

t

t t t

2

2

d g

dt l

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Problem with Euler Method

Energy not conserved with Euler method. Why?

Euler-Cromer method

1

1 1

1

( / )i i i

i i i

i i

g l t

t

t t t

illustration

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Review of the numerical methods

dynamical variable vector

The accuracy of this algorithm is relatively low:

Euler method

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If we carry out the integration with g(y, t) given from this equation, we obtain a new algorithm

with preparation of

We can always include more points in the integral to obtain algorithms with apparently higher accuracy, but we will need the values of more points in order to start the algorithm. This becomes impractical if we need more than two points in order to start the algorithm.

Is there a more practical method?

Alternative way to improve the accuracy

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The Runge-Kutta method

A more practical method that requires only the first point in order to start or to improve the algorithm is the Runge–Kutta method.

Remember

( ') ( , )''

'''

t y

d y dg y t g g yy g gg

dt dt t y t

y

Martin Wilhelm Kutta

Carl Runge

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R-K method in Appendix A

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Application

Illustration! Rk32.m

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We can also formally write the solution at t +τ as

where αi (with i = 1, 2, . . . ,m) and νi j (with i = 2, 3, . . . ,m and j < i ) are parameters to be determined.

What is the physical meaning of the expansion?

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2nd Order Runge-Kutta method

Set m=2

Now if we perform the Taylor expansion for c2 up to the term O(τ2), we have

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2nd Order Runge-Kutta method

Typically, there are m Eqs and m + m(m −1)/2 unknowns.

E.g., we may choose:1 2 21

1 1, , 1

2 2v

1 2 21

1 2 3, ,

3 3 4v

Note: These coefficients would result in a modified Euler method …

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Review of R-K method

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Application

1 i i

2 i 1 i

i 1 i 1 2

,

,

1

2

c g y t

c g y c t

y y c c

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4th Order Runge-Kutta method

The well-known fourth-order Runge–Kutta algorithm is given by

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Making the pendulum more interesting

2

2

d g dq

dt l dt

2

2sin( )D D

d g dq F t

dt l dt

►Adding dissipation

►Adding a driving force

►Nonlinear pendulum2

2sin

d g

dt l

illustrations

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Homework

Exercise 3.1, 3.2, 3.6,

Sending your home work to 17273799@qq.comBoth Results and source codes are required.

For lecture notes, refer to http://www.compphys.cn/~xbyang/

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