One of the topics touched on in some of the recent posts about Lie series was the idea that a non-autonomous equation can be handled relatively easily in the Lie series framework by enlarging the state space from $$2N$$-dimensions to $$2N+1$$-dimensions. I thought it would be beneficial to see how this is done in an explicit example where the actual equations of motion can be solved analytically.

The candidate for this is treatment is an otherwise free particle, moving in one dimension, subjected to a time varying force. The reason is for this is that the second-order Newtonian, first-order state space, and Green’s functions methods (both time and frequency space) all provide the same answer with relatively little work.

The Newton Method

The Newtonian equation of motion is

\[ \frac{d^2}{dt^2} x = F(t) \; .\]

To get a solution, integrate twice with respect to time to arrive at

\[ x(t) = x_0 + v_0 (t – t_0) + \int_{t_0}^{t} dt’ \int_{t_0}^{t’} dt” F(t”) \; , \]

where $$x_0$$ and $$v_0$$ are the initial position and velocity at time $$t_0$$.

This solution isn’t particularly useful and a much better form results after an integration-by-parts. The usual form of integration-by-parts $$ \int dU \, V = U V – \int U dV $$ is carried out by identifying:

\[ dU = dt’ \]

and

\[ V(t’) = \int_{t_0}^{t’} dt” F(t”) \; \]

Plugging these in yields

\[ x(t) = x_0 + v_0 (t – t_0) + (t-t_0) \int_{t_0}^{t} dt” F(t”) – \int_{t_0}^{t} dt’ (t’-t_0) F(t’) \; \]

or, upon combining the last two terms

\[ x(t) = x_0 + v_0 (t – t_0) + \int_{t_0}^{t’} dt’ (t-t’) F(t’) \; .\]

State Space Method

An alternative way to arrive at the same solution is in the state-space picture. Here the equation of motion is given by

\[ \frac{d}{dt} \left[ \begin{array}{c} x \\ v \end{array} \right] = \left[ \begin{array}{c} x \\ v \end{array} \right] = \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right] \left[ \begin{array}{c} x \\ v \end{array} \right] + \left[ \begin{array}{c} 0 \\ F(t) \end{array} \right] . \]

The form of the process matrix makes the construction of the state transition matrix particularly easy. To wit

\[ A = \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right] \; ,\]

\[ A^2 = \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right] \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right] = \left[ \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right]\; ,\]

and so

\[ \Phi(t,t_0) = \exp((t-t_0)A) = \left[ \begin{array}{cc} 1 & (t-t_0) \\ 0 & 1 \end{array} \right] \; . \]

The general solution is then

\[ \left[ \begin{array}{c} x \\v \end{array} \right] = \Phi(t,t_0) \left[ \begin{array}{c} x_0 \\ v_0 \end{array} \right] + \int_{t_0}^t dt’ \Phi(t,t’) \left[ \begin{array}{c} 0 \\ F(t’) \end{array} \right] \; .\]

Using the form of $$\Phi$$ and multiplying the terms out gives

\[ \left[ \begin{array}{c} x \\ v \end{array} \right] = \left[ \begin{array}{c} x_0 + v_0 (t – t_0) + \int_{t_0}^t dt’ (t-t’) F(t’) \\ v_0 + \int_{t_0}^t dt’ F(t’) \end{array} \right] \; , \]

which is exactly the same as the Newton method.

Green’s Function Method

The details for this method have actually been handled in earlier installments of Under the Hood. It suffices to note that the one-side Green’s function in the time domain is obtained from the Wronskian approach and the resulting Green’s function for the operator

\[ D^2 = \frac{d^2}{dt^2} \]

is

\[ G(t,t’) = (t-t’) \; , \]

which leads to the exact same expression.

Lie Series

With that preliminary work under our belt, let’s turn to the application of the Lie series. Since the equation is non-autonomous, the dimensionality of the system has to be expanded. The resulting structure is

\[ \frac{d}{dt} \left[ \begin{array}{c} x \\ v \\ t \end{array} \right] = \left[ \begin{array}{c} v \\ F(t) \\ 1 \end{array} \right] \; , \]

with the initial conditions of $$x_0$$, $$v_0$$, and $$t_0$$.

We can then read off the $$$$D$$$$ operator as

\[ D = v_0 \partial_{x_0} + F(t_0) \partial_{v_0} + \partial_{t_0} \; , \]

noting that the driving function must be evaluated at $$t_0$$.

The Lie series is then obtained by applying

\[ \exp( (t-t_0) D ) \]

to the initial condition $$$$x_0$$$$.

As usual, it is easier to do this in steps by stacking successive applications of $$$$D$$$$:

\[ D[x_0] = v_0 \; , \]
\[ D^2[x_0] = D[v_0] = F(t_0) \; ,\]
\[ D^3[x_0] = D[F(t_0)] = F'(t_0) \; ,\]
\[ D^4[x_0] = D[F'(t_0)] = F”(t_0) \; ,\]
and
\[ D^n[x_0] = D[F^{(n-3)}(t_0)] = F^{(n-2)}(t_0) \; ,\]

Putting the pieces together gives

\[ x(t) = L[x_0] = x_0 + v_0 (t – t_0) + F(t_0) \frac{ (t-t_0)^2 }{2} \\ + F'(t_0) \frac{ (t-t_0)^3 }{3!} + \ldots + F^{(n-2)}(t_0) \frac{ (t-t_0)^n }{n!} + \ldots \; ,\]

for the Lie series.

A Taylor’s series expansion of the integral found in the exact solution gives

\[ x(t) = x_0 + v_0 (t – t_0) + F(t_0) \frac{ (t-t_0)^2 }{2} \\ + F'(t_0) \frac{ (t-t_0)^3 }{3!} + \ldots + F^{(n-2)}(t_0) \frac{ (t-t_0)^n }{n!} + \ldots \; ,\]

which happily gives the same answer.