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Learning about Differential Equations from Their Symmetries
Application of MathSym to Analyzing a Partial Differential Equation As a final example, I will discuss the computation of the symmetries of the cubic nonlinear Schrödinger equation, a complex valued partial differential equation. Using the symmetries of the equation it is possible to generate exact solutions by a couple of different methods. First since symmetries of an equation transform its solutions to other solutions, I will demonstrate a family of solutions that are the transforms of a spatially invariant solution. Also, I will show that the symmetries can be used to reduce the partial differential equation to an ordinary differential equation in a couple of different ways. For ordinary differential equations, it is theoretically possible to reduce the order of the equation by one for every symmetry of the equation. So, a second-order equation can be reduced to two integrals if two symmetries are known. Of course, as we saw above, in practice it can be difficult to perform the necessary computations. For a partial differential equation, it is generally not possible to get the full solution set from just the knowledge of several symmetries. However, by looking for fixed points of a given symmetry, we can find reductions of the equation and often special solutions. The cubic nonlinear Schrödinger equation is derived in descriptions of nonlinear optics, water waves, and plasma physics [11]. Usually, the equation is written in terms of a complex valued function
When computing the symmetries, we set
which we denote as NLS. Again, MathSym returns a system of linear coupled partial differential equations for the generators of the symmetries. MathSym uses the convention that the independent variables
MathSym has been successful in generating and reducing these equations. The solution of the determining equations is
where the Recall that symmetries map solutions to solutions so knowledge of a solution and a symmetry allows us to generate a family of new solutions. As an example, let us use the Galilean boost, which is represented above by the constant
The point in this exercise was to generate transformations of the variables appearing in the nonlinear Schrödinger equation that sent solutions of the equation to other solutions. This means that any solution to NLS can be used to generate a family of solutions. It is straightforward to check that a solution of NLS is
where The transformation in equations (13) to (16) gives new solutions to NLS when applied to equations (17) and (18). If we carry out these substitutions, we find
which can be shown to satisfy NLS for any choices of
Figure 2. The real part of a spatially invariant solution of the nonlinear Schrödinger equation.
Figure 3. The real part of a transformed solution of the nonlinear Schrödinger equation. Finally, we will use the symmetries to derive ordinary differential equations. In order to compute the reductions, we pose the question: "What are the solutions of NLS that are invariant under a given symmetry?" Notice that this question is similar to the one that we asked to define the symmetries. A symmetry is a mapping of the solution set of an equation to itself. Reductions arise from looking at the invariant subset of that transformation. To find these invariant solutions we look for the intersection of the solution sets of the original equation, here the NLS equation, and a pair of first-order, quasi-linear partial differential equations, which are
These equations are known as the invariant surface condition equations. We will solve this new pair of equations using the method of characteristics and then substitute the solutions into the NLS equation. We start with Substituting
We can approximate solutions to this pair of ordinary differential equations by changing to polar coordinates
with
If we let
Our last reduction will come from letting
We can solve these two equations to get
In terms of the original variables this is
Here it is nice to have Mathematica check our computations. First we will derive the equations. Notice that we are using a little
Before we substitute for
We apply these two functions.
We can now substitute our solution in and run the result through Simplify.
So it works. Let us see what some pictures look like. The constant
Figure 4. The real part of a boost invariant solution of the nonlinear Schrödinger equation.
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