# Reduce PDE to ODE by dilation symmetry

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"?

• 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?? – Carlo Beenakker May 22 '20 at 21:07
• @CarloBeenakker Yes, that is what I say. The "invariant" statement is from its extension paper. ieeexplore.ieee.org/document/8443024 (p.206, same example) – sleeve chen May 22 '20 at 21:09
• What is G here? – Piyush Grover May 23 '20 at 0:16
• @PiyushGrover $G$ is a function of $\alpha$ to be calculated (and is calculated in tha paper). – sleeve chen May 23 '20 at 1:24

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.