Solving a second order differential equation by fourth order Runge-Kutta







Numerical Methods

Solving a second order differential equation by fourth order Runge-Kutta

Any second order differential equation can be written as two coupled first order equations,

\[ \begin{equation} \frac{dx_1}{dt} =f_1(x_1,x_2,t)\qquad\frac{dx_2}{dt} =f_2(x_1,x_2,t). \end{equation} \]

These coupled equations can be solved numerically using a fourth order Runge-Kutta routine. The equations for a damped driven pendulum, $f_1(x_1,x_2,t) = x2$, $f_2(x_1,x_2,t) = -x_2 - \sin (x_1) +\sin (t)$ is coded below for the intial conditions $x_1(t=0)=0$, $x_2(t=0)=1$.

<script> var x1=0; var x2=1; var t=0; var dt=0.01; function f1(x1,x2,t) { dx1dt = x2; return dx1dt; } function f2(x1,x2,t) { dx2dt = -x2 - Math.sin(x1) + Math.sin(t); return dx2dt; } for (j=0; j<20; j++){ document.write("t="+t+" x1="+x1+" x2="+x2+"<\/br>"); k11 = dtf1(x1,x2,t); k21 = dtf2(x1,x2,t); k12 = dtf1(x1+0.5k11,x2+0.5k21,t+0.5dt); k22 = dtf2(x1+0.5k11,x2+0.5k21,t+0.5dt); k13 = dtf1(x1+0.5k12,x2+0.5k22,t+0.5dt); k23 = dtf2(x1+0.5k12,x2+0.5k22,t+0.5dt); k14 = dtf1(x1+k13,x2+k23,t+dt); k24 = dtf2(x1+k13,x2+k23,t+dt); x1 += (k11+2k12+2k13+k14)/6; x2 += (k21+2k22+2k23+k24)/6; t += dt; } </script>