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.<\/p>\n
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.<\/p>\n
The Newtonian equation of motion is<\/p>\n
\\[ \\frac{d^2}{dt^2} x = F(t) \\; .\\]<\/p>\n
To get a solution, integrate twice with respect to time to arrive at<\/p>\n
\\[ x(t) = x_0 + v_0 (t – t_0) + \\int_{t_0}^{t} dt’ \\int_{t_0}^{t’} dt” F(t”) \\; , \\]<\/p>\n
where $$x_0$$ and $$v_0$$ are the initial position and velocity at time $$t_0$$.<\/p>\n
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:<\/p>\n
\\[ dU = dt’ \\]<\/p>\n
and<\/p>\n
\\[ V(t’) = \\int_{t_0}^{t’} dt” F(t”) \\; \\]<\/p>\n
Plugging these in yields<\/p>\n
\\[ 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’) \\; \\]<\/p>\n
or, upon combining the last two terms<\/p>\n
\\[ x(t) = x_0 + v_0 (t – t_0) + \\int_{t_0}^{t’} dt’ (t-t’) F(t’) \\; .\\]<\/p>\n
An alternative way to arrive at the same solution is in the state-space picture. Here the equation of motion is given by<\/p>\n
\\[ \\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] . \\]<\/p>\n
The form of the process matrix makes the construction of the state transition matrix particularly easy. To wit<\/p>\n
\\[ A = \\left[ \\begin{array}{cc} 0 & 1 \\\\ 0 & 0 \\end{array} \\right] \\; ,\\]<\/p>\n
\\[ 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]\\; ,\\]<\/p>\n
and so<\/p>\n
\\[ \\Phi(t,t_0) = \\exp((t-t_0)A) = \\left[ \\begin{array}{cc} 1 & (t-t_0) \\\\ 0 & 1 \\end{array} \\right] \\; . \\]<\/p>\n
The general solution is then<\/p>\n
\\[ \\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] \\; .\\]<\/p>\n
Using the form of $$\\Phi$$ and multiplying the terms out gives<\/p>\n
\\[ \\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] \\; , \\]<\/p>\n
which is exactly the same as the Newton method.<\/p>\n
The details for this method have actually been handled in earlier installments of Under the Hood<\/a>. It suffices to note that the one-side Green’s function in the time domai<\/a>n is obtained from the Wronskian approach and the resulting Green’s function for the operator<\/p>\n \\[ D^2 = \\frac{d^2}{dt^2} \\]<\/p>\n is<\/p>\n \\[ G(t,t’) = (t-t’) \\; , \\]<\/p>\n which leads to the exact same expression.<\/p>\n 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<\/p>\n \\[ \\frac{d}{dt} \\left[ \\begin{array}{c} x \\\\ v \\\\ t \\end{array} \\right] = \\left[ \\begin{array}{c} v \\\\ F(t) \\\\ 1 \\end{array} \\right] \\; , \\]<\/p>\n with the initial conditions of $$x_0$$, $$v_0$$, and $$t_0$$.<\/p>\n We can then read off the $$$$D$$$$ operator as<\/p>\n \\[ D = v_0 \\partial_{x_0} + F(t_0) \\partial_{v_0} + \\partial_{t_0} \\; , \\]<\/p>\n noting that the driving function must be evaluated at $$t_0$$.<\/p>\n The Lie series is then obtained by applying<\/p>\n \\[ \\exp( (t-t_0) D ) \\]<\/p>\n to the initial condition $$$$x_0$$$$.<\/p>\n As usual, it is easier to do this in steps by stacking successive applications of $$$$D$$$$:<\/p>\n \\[ D[x_0] = v_0 \\; , \\] Putting the pieces together gives<\/p>\n \\[ 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 \\; ,\\]<\/p>\n for the Lie series.<\/p>\n A Taylor’s series expansion of the integral found in the exact solution gives<\/p>\n \\[ 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 \\; ,\\]<\/p>\n which happily gives the same answer.<\/p>\n","protected":false},"excerpt":{"rendered":"Lie Series<\/h2>\n
\n\\[ D^2[x_0] = D[v_0] = F(t_0) \\; ,\\]
\n\\[ D^3[x_0] = D[F(t_0)] = F'(t_0) \\; ,\\]
\n\\[ D^4[x_0] = D[F'(t_0)] = F”(t_0) \\; ,\\]
\nand
\n\\[ D^n[x_0] = D[F^{(n-3)}(t_0)] = F^{(n-2)}(t_0) \\; ,\\]<\/p>\n