14.4: Derivation of the Infinitesimal Generators

Procedure to Find Infinitesimal Generators

We are studying differential equations, which can be written as

\[F(t, y, \dot{y}, \ldots) = 0 \nonumber \]

for some function \(F\). We are looking for continuous symmetries that can be applied to this equation such that the original equation and the transformed equation have the same solutions. The symmetries are denoted by infinitesimal generators

\[U = \xi \partial_t + \eta \partial_y \nonumber \]

that describe how the independent variable \(t\) and dependent variable \(y\) transform. Upon a symmetry transformation, the independent variable and dependent variable transform, but so do the derivatives of the dependent variable, \(\dot{y}\), \(\ddot{y}\), \(\ldots\) The prolongation of an infinitesimal generator is a generalization of the infinitesimal generator that describes the transformation of the independent variable, the dependent variable, and derivatives of the dependent variable [164, p. 94].

The \(\mathfrak{n}\)th prolongation of a generator \(U\) is defined as

\[pr^{(\mathfrak{n})}U = \xi \partial_t + \eta \partial_y + \eta^t \partial_{\dot{y}} + \eta^{tt} \partial_{\ddot{y}} + \eta^{ttt} \partial_{\dddot{y}} + \dots, \label{14.4.3} \]

and it has terms involving \(\eta^{t^{\mathfrak{n}}}\). The functions \(\eta^{t}\) and \(\eta^{tt}\) are defined [164],

\[\eta^t = \eta^t(t, y, \dot{y}) = \frac{d}{dt}(\eta - \xi\dot{y})+\xi\ddot{y} \label{14.4.4} \]

\[\eta^{tt} = \eta^{tt}(t, y, \dot{y}) = \frac{d^2}{dt^2}(\eta - \xi\dot{y})+\xi\dddot{y} \label{14.4.5} \]

The quantities \(\eta^{ttt}\), \(\eta^{tttt}\), and so on can be defined similarly, but they will not be needed for the examples below. The prolongation of the infinitesimal generator is an operator that describes the transformation of \(t\), \(y\), \(\dot{y}\), \(\ddot{y}\), and so on up to the \(\mathfrak{n}\)th derivative. Some authors [189] use the term tangential mapping instead of prolongation.

The procedure to find all possible continuous symmetries of an equation is based on the idea that the solutions of an equation remain unchanged upon a symmetry operation. For a given transformation to be a symmetry operation, not only must all the solutions remain unchanged, but so must all derivatives of the solutions. Thus, for a differential equation of the form \(F(t, y, \dot{y}, \ldots) = 0\), all symmetries \(U\) obey the symmetry condition

\[pr^{(\mathfrak{n})}UF=0. \label{14.4.6} \]

We solve this symmetry condition to find all allowed infinitesimal generators that describe continuous symmetries of the original equation.

We can use Equation \ref{14.4.4} and \ref{14.4.5} to write the symmetry condition in terms of the components of the infinitesimal generators, \(\xi\) and \(\eta\). Then, we solve the symmetry condition for \(\xi\) and \(\eta\). This step involves some algebra, but it can be accomplished with some patience and an adequate supply of ink and paper.

We can solve the symmetry condition for the allowed infinitesimal generators. By careful solution, we find all infinitesimal generators of the form \(U = \xi \partial_t + \eta \partial_y\). This procedure gives us a systematic way to find all continuous symmetries of the equation.

This technique applies to any differential equation. We are most interested in applying it to equations of motion that describe energy conversion processes. From this technique, we get information about solutions of the equation even when the equation of motion is nonlinear. Furthermore, in the Sec. 14.5 we see that we may be able to use the symmetries to find invariants of the equation, and invariants often have physical meaning. All symmetries of calculus of variations problems of the form \(\delta \int \mathcal{L}dt = 0\) are necessarily symmetries of the Euler-Lagrange equation. However, the converse is not necessarily true, so not all symmetries of the Euler-Lagrange equation are symmetries of the integral equation [164, p. 255].

Thomas-Fermi Equation Example

As an example, we apply this procedure to the Thomas-Fermi equation

\[\ddot{y} = y^{3/2}t^{-1/2}. \label{14.4.7} \]

This equation was derived in Chapter 13. From the solution of this equation \(y(t)\), the charge density \(\rho_{ch}(r)\) of electrons around an isolated atom and the voltage \(V (r)\) felt by the electrons can be calculated within the, rather severe, assumptions specified in that chapter. The independent variable of the equation is a scaled version of radial position, not time. However, \(t\) will be used as the independent variable here because the procedure applies to equations regardless of the name of the variable. Reference [190] applies this procedure to a family of equations known as the Emden-Fowler equations. The Thomas-Fermi equation is a special case of an Emden-Fowler equation, so the result of this example can be found in reference [190].

We would like to identify continuous symmetries of Equation \ref{14.4.7}. These symmetries will be specified by infinitesimal generators of the form

\[U = \xi \partial_t + \eta \partial_y \label{14.4.8} \]

where \(\xi\) and \(\eta\) have the form \(\xi(t, y)\) and \(\eta(t, y)\). Solutions of the equation satisfy

\[(\ddot{y} - y^{3/2}t^{-1/2}) = 0. \label{14.4.9} \]

For infinitesimal generators that describe symmetries of this equation, the prolongation is also zero.

\[pr^{(\mathfrak{n})}U (\ddot{y} - y^{3/2}t^{-1/2}) = 0. \label{14.4.10} \]

Equation \ref{14.4.10} can be solved for all generators \(U\) corresponding to continuous symmetries of the Thomas-Fermi equation. Equations \ref{14.4.3} and \ref{14.4.10} can be combined.

\[\eta^{tt} + \frac{1}{2}\xi y^{3/2}t^{-3/2} - \frac{3}{2}\eta y^{1/2}t^{-1/2} = 0 \nonumber \]

Next, Equation \ref{14.4.5} is used.

\[\partial_{t t} \eta+2 \dot{y} \partial_{y t} \eta+\ddot{y} \partial_{y} \eta+\dot{y}^{2} \partial_{y y} \eta-2 \ddot{y} \partial_{t} \xi-\dot{y} \partial_{t t} \xi-2 \dot{y}^{2} \partial_{y t} \xi - \dot{y}^{3} \partial_{y y} \xi-3 \dot{y} \ddot{y} \partial_{y} \xi+\frac{1}{2} \xi y^{3 / 2} t^{-3 / 2}-\frac{3}{2} \eta y^{1 / 2} t^{-1 / 2}=0 \nonumber \]

Substitute the original equation for \(\ddot{y}\).

\[\partial_{t t} \eta+2 \dot{y} \partial_{y t} \eta+y^{3 / 2} t^{-1 / 2} \partial_{y} \eta+\dot{y}^{2} \partial_{y y} \eta-2 y^{3 / 2} t^{-1 / 2} \partial_{t} \xi-\dot{y} \partial_{t t} \xi-2 \dot{y}^{2} \partial_{y t} \xi -\dot{y}^{3} \partial_{y y} \xi-3 \dot{y} y^{3 / 2} t^{-1 / 2} \partial_{y} \xi+\frac{1}{2} \xi y^{3 / 2} t^{-3 / 2}-\frac{3}{2} \eta y^{1 / 2} t^{-1 / 2}=0 \label{14.4.13} \]

Regroup terms.

\[(\partial_{t t} \eta+y^{3 / 2} t^{-1 / 2} \partial_{y} \eta-2 y^{3 / 2} t^{-1 / 2} \partial_{t} \xi+\frac{1}{2} \xi y^{3 / 2} t^{-3 / 2}-\frac{3}{2} \eta y^{1 / 2} t^{-1 / 2}) +\dot{y}(2 \partial_{y t} \eta-\partial_{t t} \xi-3 y^{3 / 2} t^{-1 / 2} \partial_{y} \xi)+\dot{y}^{2}(\partial_{y y} \eta-2 \partial_{y t} \xi)-\dot{y}^{3}(\partial_{y y} \xi)=0 \label{14.4.14} \]

Each of the terms in parentheses in Equation \ref{14.4.14} must be zero.

\[\partial_{t t} \eta+y^{3 / 2} t^{-1 / 2} \partial_{y} \eta-2 y^{3 / 2} t^{-1 / 2} \partial_{t} \xi+\frac{1}{2} \xi y^{3 / 2} t^{-3 / 2}-\frac{3}{2} \eta y^{1 / 2} t^{-1 / 2}=0 \label{14.4.15} \]

\[2\partial_{yt}\eta - \partial_{tt}\xi - 3y^{3/2}t^{-1/2}\partial_y\xi = 0 \label{14.4.16} \]

\[\partial_{yy}\eta - 2\partial_{yt}\xi = 0 \label{14.4.17} \]

\[\partial_{yy}\xi = 0 \label{14.4.18} \]

Equations \ref{14.4.15}, \ref{14.4.16}, \ref{14.4.17}, and \ref{14.4.18} can be solved for \(\xi\) and \(\eta\). From Equation \ref{14.4.18}, \(\partial_{yy}\xi = 0\), so \(\xi\) must have form

\[\xi = (c_1 + c_2y) b(t). \nonumber \]

The quantities denoted \(c_{\mathfrak{n}}\) are constants. From Equation \ref{14.4.17}, \(\eta\) must have the form

\[\eta = (c_3 + c_4y + c_5y^2) g(t). \nonumber \]

Functions \(b(t)\) and \(g(t)\) only depend on \(t,\) not \(y\). The condition of Equation \ref{14.4.16} can be rewritten.

\[(2c_4\partial_tg - c_1\partial_{tt}b) + y(4c_5\partial_tg - 2c_2\partial_{tt}b) - 3y^{3/2} t ^{-1/2} c_2b = 0 \label{14.4.21} \]

To satisfy Equation \ref{14.4.21}, \(c_2\) must be zero, and either \(c_5 = 0\) or \(g(t) = 0\). From Equations \ref{14.4.16} and \ref{14.4.17}, \(\partial_y\eta\) and \(\partial_y\xi\) must be constant. Therefore, the form of \(\xi\) must be

\[\xi = c_6 + c_7t. \nonumber \]

This form can be substituted into Equation \ref{14.4.15}.

\[y^{3/2} t^{-1/2} (c_4 + 2c_5y) - 2y^{3/2} t^{-1/2} c_7 + \frac{1}{2} (c_6 + c_7t)y^{3/2} t^{-3/2} - \frac{3}{2} (c_3 + c_4y + c_5y^{2})y^{1/2} t^{-1/2} = 0 \nonumber \]

\[y^{3/2} t^{-1/2} (c_4 - 2c_7 + \frac{1}{2}c_7 - \frac{3}{2}c_4) + \frac{1}{2} c_6 y^{3/2} t^{-1/2} - \frac{3}{2} c_3 y^{1/2} t^{-1/2} + y^{5/2}t^{-1/2} (2c_5 - \frac{3}{2}c_5) = 0 \nonumber \]

The coefficients \(c_3\), \(c_5\), and \(c_6\) must be zero. Also, \(c_4 = -3c_7\). No other solutions here are possible. Thus, the symmetry condition of Equation \ref{14.4.10} can be satisfied by \(\xi = t\) and \(\eta = -3y\).

This procedure finds one regular continuous infinitesimal symmetry of the Thomas-Fermi equation, with infinitesimal symmetry generator

\[U = t\partial_t - 3y\partial_y. \nonumber \]

No other solutions can satisfy the constraints given by Equation \ref{14.4.10}. Therefore, this equation has only one continuous symmetry.

Finite transformations are related to infinitesimal transformations by Equation 14.3.8. In this case, the independent variable transforms as

\[t \rightarrow \tilde t = e^{\varepsilon (t\partial_t - 3y\partial_y)}t. \nonumber \]

\[t \rightarrow \tilde{t}=\left[1+\varepsilon\left(t \partial_{t}-3 y \partial_{y}\right)+\frac{1}{2 !} \varepsilon^{2}\left(t \partial_{t}-3 y \partial_{y}\right)^{2}+\ldots\right] t \nonumber \]

\[t \rightarrow \tilde{t}=\left[t+\varepsilon t\left(\partial_{t}t\right)+\frac{1}{2 !} \epsilon^{2}t \left(\partial_{t}t\right) \left(\partial_{t}t\right) +\ldots\right] \nonumber \]

\[t \rightarrow \tilde t = te^{\varepsilon} \nonumber \]

The dependent variable transforms as

\[y \rightarrow \tilde y = e^{\varepsilon (t\partial_t - 3y\partial_y)}y. \nonumber \]

\[y \rightarrow \tilde{y}=\left[1+\varepsilon\left(t \partial_{t}-3 y \partial_{y}\right)+\frac{1}{2 !} \varepsilon^{2}\left(t \partial_{t}-3 y \partial_{y}\right)^{2}+\ldots\right] y \nonumber \]

\[y \rightarrow \tilde y = ye^{-3\varepsilon} \nonumber \]

Defining the constant \(c_6 = e^{\varepsilon}\), the transformation can be written as

\[t \rightarrow c_6t \quad \text{ and } \quad y \rightarrow (c_6)^{-3}y. \nonumber \]

The analysis above shows that the original Thomas-Fermi equation of Equation \ref{14.4.7} and the transformed equation

\[\frac{d^2 (yc_6^{-3})}{d (tc_6)^2} = (yc_6^{-3})^{3/2}(tc_6)^{-1/2} \nonumber \]

have the same solutions. From it, we can conclude that if \(y(t)\) is a solution to the Thomas-Fermi equation, we know that \(c_6^{-3} y(\tau)\) for \(\tau = c_6t\) is also a solution.

Line Equation Example

Consider another example of this procedure applied to the equation \(\ddot{y} = 0\). The solution of this equation can be found by inspection

\[y{t}=c_0t + c_1 \nonumber \]

because this is the equation of a straight line. The coefficients \(c_{\mathfrak{n}}\) are constants, and they are different from the previous example. In this example, we will identify the infinitesimal generators for continuous symmetries of this equation, and we will find eight infinitesimal generators. The result of this problem appears in [191], and it is a modified version of problem 2.26 of reference [164, p. 180].

Solutions of the original equation must be the same as solutions of an equation transformed by a continuous symmetry, and this idea is contained in the symmetry condition of Equation \ref{14.4.6}. In this case, the original equation is \(\ddot{y} = 0\), so the prolongation of an infinitesimal generator acting on this equation must also be zero for an infinitesimal generator \(U\) that describes a continuous symmetry.

\[pr^{(\mathfrak{n})}U (\ddot{y}) = 0 \nonumber \]

Using Equations \ref{14.4.3}, \ref{14.4.4}, and \ref{14.4.5}, we can write this symmetry condition in terms of \(\xi\) and \(\eta\).

\[\eta^{tt} = 0 \nonumber \]

\[\eta^{tt} = 0 = \partial_{tt}\eta + 2 \dot{y}\partial_{yt}\eta + \ddot{y}\partial_{y}\eta + \dot{y}^2\partial_{yy}\eta - 2\ddot{y}\partial_{t}\xi - \dot{y}\partial_{tt}\xi - 2\dot{y}^2\partial_{yt}\xi - \dot{y}^3\partial_{yy}\xi - 3\dot{y}\ddot{y}\partial_{y}\xi \nonumber \]

Use \(\ddot{y} = 0\), and regroup the terms.

\[(\partial_{tt}\eta) + \dot{y}(2\partial_{yt}\eta - \partial_{tt}\xi) + \dot{y}^2 (\partial_{yy}\eta - 2\partial_{yt}\xi) - \dot{y}^3(\partial_{yy}\xi) = 0 \nonumber \]

The above equation is true for all \(y\) only if all of the quantities in parentheses are zero.

\[\partial_{tt}\eta = 0 \label{14.4.40} \]

\[2\partial_{yt}\eta - \partial_{tt}\xi = 0 = 0 \label{14.4.41} \]

\[\partial_{yy}\eta - 2\partial_{yt}\xi = 0 \label{14.4.42} \]

\[\partial_{yy}\xi = 0 \label{14.4.43} \]

The next step is to solve the above set of equations for all possible solutions of \(\xi\) and \(\eta\) which will determine the infinitesimal generators of all possible continuous symmetry transformations.

We will consider three cases: case 1 with \(\eta = 0\), case 2 with \(\xi = 0\), and case 3 with both \(\xi\) and \(\eta\) nonzero.

Case 1 with \(\eta = 0\): Assume \(\eta = 0\). What solutions can be found for \(\xi\)? Equation \ref{14.4.40} to Equation \ref{14.4.43} can be reduced.

\[\partial_{tt}\xi = 0 \nonumber \]

\[\partial_{yy}\xi = 0 \nonumber \]

\[\partial_{yt}\xi = 0 \nonumber \]

There are three possible independent solutions for \(\xi\). They are \(\xi = 1\), \(\xi = t\), and \(\xi = y\). So, we found three infinitesimal generators.

\[U_1 = \partial_t \nonumber \]

\[U_2 = t\partial_t \nonumber \]

\[U_3 = y\partial_t \nonumber \]

Case 2 with \(\xi = 0\): Suppose \(\xi = 0\). What solutions can be found for \(\eta\)? Equation \ref{14.4.40} to Equation \ref{14.4.43} simplify.

\[\partial_{tt}\eta = 0 \nonumber \]

\[\partial_{yt}\eta = 0 \nonumber \]

\[\partial_{yy}\eta = 0 \nonumber \]

There are three possible independent solutions for \(\eta\). They are \(\eta = 1\), \(\eta = y\), and \(\eta = t\). So, we found three more infinitesimal generators.

\[U_4 = \partial_y \nonumber \]

\[U_5 = y\partial_y \nonumber \]

\[U_6 = t\partial_y \nonumber \]

Case 3 where both \(\xi\) and \(\eta\) are nonzero: From Equation \ref{14.4.40}, we can write

\[\eta = (c_1 + c_2t) b(y). \nonumber \]

Here, \(b\) is a function of only \(y\), not \(t\). Therefore,

\[\partial_{yt}\eta = c_2\partial_y b(y) \nonumber \]

which is not a function of \(t\). From Equation \ref{14.4.43}, we can write

\[\xi = (c_3 + c_4y) g(t). \nonumber \]

Here, \(g\) is a function of only \(t\), not \(y\). Therefore,

\[\partial_{yt}\xi = c_4\partial_t g(t) \nonumber \]

which is not a function of \(y\). Now use Equation \ref{14.4.41}.

\[2c_2\partial_y b(y) - (c_3 + c_4y) \partial_{tt}g = 0 \nonumber \]

The first term is not a function of \(t\). Therefore, \(\xi\) is at most quadratic in \(t\). So, \(\xi\) has the form

\[\xi = (c_3 + c_4y) (c_5 + c_6t + c_7t^2). \nonumber \]

Distribute out the multiplication.

\[\xi = c_3c_5 + c_3c_6t + c_3c_7t^2 + c_4c_5y + c_4c_6yt + c_4c_7yt^2 \nonumber \]

\[\partial_{yt}\xi = c_4c_6 + 2c_4c_7t \label{14.4.63} \]

Next, use Equation \ref{14.4.42}.

\[\partial_{yy}\eta = 2c_4\partial_t g = 0 \nonumber \]

The second term is not a function of \(y\). Therefore, \(\eta\) is at most quadratic in \(y\). So, \(\eta\) has the form

\[\eta = (c_1 + c_2t) (c_8 + c_9y + c_{10}y^2). \nonumber \]

Distribute out the multiplication.

\[\eta = c_1c_8 + c_1c_9y + c_1c_{10}y^2 + c_2c_8t + c_2c_9yt + c_2c_{10}ty^2 \nonumber \]

\[\partial_{yt}\eta = c_2c_9 + 2c_2c_{10}y \label{14.4.67} \]

Now use Equations \ref{14.4.41} and \ref{14.4.67}.

\[2 (c_2c_9 + 2c_2c_{10}y) - 2 (2c_3c_7 + 2yc_4c_7) = 0 \nonumber \]

\[(2c_2c_9 - 4c_3c_7) + y (4c_2c_{10} - 4c_4c_7) = 0 \nonumber \]

We end up with the pair of equations

\[c_2c_9 = 2c_3c_7 \label{14.4.70} \]

\[c_2c_{10} = 2c_4c_7 \label{14.4.71} \]

Next use Equations \ref{14.4.42} and \ref{14.4.63}.

\[(2c_1c_{10} + 2tc_2c_{10}) - 2 (c_4c_6 + 2c_4c_7t) = 0 \nonumber \]

\[(2c_1c_{10} - 2c_4c_6) + t(2c_2c_{10} - 4c_4c_7) = 0 \nonumber \]

and we end up with a pair of equations.

\[c_1c_{10} = c_4c_6 \label{14.4.74} \]

\[c_2c_{10} = 2c_4c_7 \label{14.4.75} \]

These are the only possible solution of Equations \ref{14.4.71} and \ref{14.4.75}.

Finally, there are two possible solutions which are independent from the previously found solutions. We can set the coefficients of Equation \ref{14.4.74} to 1. The first solution is \(\eta = y^2\) and \(\xi = yt\) corresponding to

\[U_7 = yt\partial_t + y^2\partial_y. \nonumber \]

For the second solution, we can set the coefficients of Equation \ref{14.4.70} to 1. The second solution is \(\eta = yt\) and \(\xi = t^2\) corresponding to

\[U_8 = t^2 \partial_t + yt\partial_y. \nonumber \]

At this point, we have found eight infinitesimal generators. These are all possible generators of continuous regular nongeometrical symmetries.