South China University of Technology Oscillator motions Xiaobao Yang Department of Physics
Simulation of Quasi-crystals
Annealing simulation Phase Transitions
Charged particle in electromagnetic field
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;
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;
the motion of electrons in atoms the behavior of currents and voltages in electronic circuits planetary orbits a pendulum oscillatory and periodic phenomena
Simple Harmonic Motion Euler method
Problem with Euler Method Energy not conserved with Euler method. Why? Euler-Cromer method illustration
Review of the numerical methods dynamical variable vector The accuracy of this algorithm is relatively low: Euler method
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
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 Martin Wilhelm Kutta Carl Runge
R-K method in Appendix A
Application Illustration! Rk32.m
We can also formally write the solution at t + τ as where α i (with i = 1, 2,...,m) and ν i j (with i = 2, 3,...,m and j < i ) are parameters to be determined. What is the physical meaning of the expansion?
2 nd Order Runge-Kutta method Set m=2 Now if we perform the Taylor expansion for c 2 up to the term O(τ 2 ), we have
2 nd Order Runge-Kutta method Typically, there are m Eqs and m + m(m −1)/2 unknowns. E.g., we may choose: Note:These coefficients would result in a modified Euler method …
Review of R-K method
Application
4 th Order Runge-Kutta method The well-known fourth-order Runge–Kutta algorithm is given by
Making the pendulum more interesting ►Adding dissipation ►Adding a driving force ►Nonlinear pendulum illustrations
Homework Exercise 3.1, 3.2, 3.6, Sending your home work to Both Results and source codes are required. For lecture notes, refer to 主题:学号 + 姓名 + 第?次作业