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I am reading Rodrigues, Henrion, and Cantwell - Symmetries and analytical solutions of the Hamilton–Jacobi–Bellman equation for a class of optimal control problems, p.753.

Consider the following PDE: $$q_1x_2^2+q_2x_2^2+V_{x_1}x_2-\frac{V^2_{x_2}b^2}{4r}=0.$$

This PDE has the following dilation symmetry: $$\tilde{x}_1=e^sx_1,\, \tilde{x}_2=e^sx_2, \, \tilde{V}=e^{2s}V.$$ Note that $\tilde{V}_{\tilde{x}_1}=e^sV_{x_1}.$ So the above PDE in the tilde variables becomes $$e^{2s}\bigg(q_1x_2^2+q_2x_2^2+V_{x_1}x_2-\frac{V^2_{x_2}b^2}{4r}\bigg)=0.$$

So we can form the following characteristic equations $$ \frac{d x_1}{x_1}=\frac{d x_2}{x_2}=\frac{dV}{2V}.$$

my question is from the following statement,

Integrating and rearranging terms, the PDE is invariant under the change of variables: $\alpha= \frac{x_2}{x_1}$, $V=x_1^2 G(\alpha).$

How to understand the above statement? Does the dilation symmetry implies the above? How to see "being invariant"?

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  • $\begingroup$ I cannot find the quoted statement in the cited paper; instead, I read on page 753 that $\alpha=x_2/x_1$ and $V=x_1^2 F(\alpha)$ are integrals of the characteristic equations, which indeed they are. Are we talking about the same paper?? $\endgroup$ – Carlo Beenakker May 22 at 21:07
  • $\begingroup$ @CarloBeenakker Yes, that is what I say. The "invariant" statement is from its extension paper. ieeexplore.ieee.org/document/8443024 (p.206, same example) $\endgroup$ – sleeve chen May 22 at 21:09
  • $\begingroup$ What is G here? $\endgroup$ – Piyush Grover May 23 at 0:16
  • $\begingroup$ @PiyushGrover $G$ is a function of $\alpha$ to be calculated (and is calculated in tha paper). $\endgroup$ – sleeve chen May 23 at 1:24
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Here's an answer, at least as I understand your question.

Geometrically, you can understand the solutions to your PDE as graphs of surfaces in $\mathbb{R}^3$ given by $(x_1,x_2,V(x_1,x_2))$ (at least locally). From this viewpoint, to say that a PDE has a symmetry, is to say that a solution surface "moved" in the direction of the symmetry will also be a solution surface to your PDE. This may mean that a solution surface moves along itself.

The dilation symmetry you give is precisely the 1-parameter family of diffeomorphisms (or flow) generated by the vector field $X=x_1\partial_{x_1}+x_2\partial_{x_2}+2V\partial_V$. The change of variables you mention come from the invariant surfaces of this flow. That is, the surfaces in $\mathbb{R}^3$ that flow along themselves. These correspond to level sets of functions $f=f(x_1,x_2,V)$ such that $X(f)=0$ (these are the invariant functions, or first integrals of $X$). In this case, all invariant functions are generated by the two independent invariant functions $\alpha=x_2/x_1$ and $G=V/x_1^2$ (I have note yet prescribed $G$ as a function of $\alpha$ here).

As it seems we are concerned with solutions to the PDE that are invariant under the symmetry, we want to understand the PDE purely in terms of the invariants $(\alpha,G)$. This means we want a differential equation involving the two invariants. Since $V=V(x_1,x_2)$, then using the second invariant function, we conclude that $G=G(\alpha)$, so that $V(x_1,x_2)=x_1^2 G(\alpha)$. Throwing this into the PDE produces an ODE which is the reduction.

I've thrown details and rigor under the bus here (for example, there are issues concerning group actions and quotient manifolds). For a rigorous treatment of this approach I suggest looking at Peter Olver's book "Applications of Lie Groups to Differential Equations." Specifically, you can learn about symmetry invariant solutions in Chapter 3.

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