# What does the time-reversibility of Verlet (or other) integration mean?

What does the time-reversibility of Verlet or any other integration method mean? The wikipedia article about it is very complex, unclear and confusing. And how can I determine whether a method is time reversible or not?

For example the classical Störmer-Verlet method

$$x_{n+1} = 2x_n - x_{n-1} + a(x_n) · dt^2$$

How is this time-reversible? If I change the sign of the timestep dt, because of the square nothing changes.

• For a mathematical analysis, you should better write it not as computer instruction but as the recursion formula $$x_{n+1}=2x_n-x_{n-1}+a(x_n)·dt^2.$$ Sep 23, 2015 at 18:59

Yes, it is exactly that.

The time reverse of the explicit Euler method is the implicit Euler method. $y_{n+1}=y_n+f(y_n)\,dt$ gets reversed to $y_{n-1}=y_n+f(y_n)\,(-dt)$ and after index shift $y_{n+1}=y_n+f(y_{n+1})\,dt$.

The same for the symplectic Euler methods. \begin{array}{lll} forward:&x_{n+1}=x_n+v_n\,dt,& v_{n+1}=v_n+a(x_{n+1})\\ reverse:&x_{n-1}=x_n-v_n\,dt,& v_{n-1}=v_n+a(x_{n-1})\\ shifted:&v_{n+1}=v_n+a(x_n)\,dt,& x_{n+1}=x_n+v_{n+1}\,dt \end{array} (Velocity) Verlet is a combination of an explicit and implicit symplectic Euler step, thus invariant under time reversal. \begin{align} v_{n+1/2}&=v_n+a(x_{n})\,dt/2\\ x_{n+1/2}&=x_n+v_{n+1/2}\,dt/2\\ x_n&=x_{n+1/2}+v_{n+1/2}\,dt/2\\ v_n&=v_{n+1/2}+a(x_{n+1})\,dt/2 \end{align} Elimination of $x_{n+1/2}$ gives velocity Verlet, elimination of the integer-indexed velocities gives the Leapfrog method, elimination of all velocities gives the basic Stoermer-Verlet method.

The first reversion-invariant Runge-Kutta methods are the (necessarily implicit) trapezoidal and midpoint methods. \begin{align} &trapezoid:&y_{n+1}&=y_n+\tfrac12(f(y_n)+f(y_{n+1}))\,dt\\ &midpoint:& y_{n+1}&=y_n+f(\tfrac12(y_n)+y_{n+1}))\,dt \end{align}

The advantage of reversion-invariant methods is that the error in scalar invariants is also reversion invariant and thus an even function. For the second-order Verlet method one would expect from the order alone that the first term of the local error in energy and momentum is $O(dt^3)$, but since the odd powers are missing the actual error is $O(dt^4)$.

Of course, the same or better can be achieved with the classical RK4, but symplectic Euler and Verlet are better suited for real-time simulations and provide, because of "symplectic", global (very slowly eroding) error bounds on conserved quantities of the dynamical system.

• "Yes, it is exactly that" - exactly what? I'm thankful for taking the time, but you wrote down a bunch of information without answering the question. What does time-reversibility mean? How Verlet method is time-reversible? If I use negative timestep, the simulation doesn't go backward, since the (dt*dt) part just make the sign positive, making unchanged everything. So what does time-reversibility mean? Sep 23, 2015 at 17:05
• Yes to your conclusion that changing the sign of the timestep (and rearranging the indices to reflect the reverse order) does not change the method. Sep 23, 2015 at 18:36
• Yeah, but if the method (and the signs of the polynomials) doesn't change how is this reversing the time? I mean the trajectories will still advance forward, just like nothing happened. It is possible that I misunderstand the concept of time-reversibility. I tought when I use negative timesteps, then the particles should go backward, backtracking their trajectories, just like when the time is reversed. Sep 23, 2015 at 18:43
• That is another valid way to view it, that integrating first forward and then backwards with the same method (and constant time-step length) will pass through the same points of the numerical trajectory in the reverse order. I've expanded on my view of the reversal in the answer. Sep 23, 2015 at 18:57