Finding Lie point symmetries of ODE $\frac{du}{dx}=xG(\frac{u}{x^2})$

Question

I have been tasked with finding all the Lie point symmetries of the following ODE: $$\frac{du}{dx}=xG\left(\frac{u}{x^2}\right)$$ I have found by applying the prolongation formula that a symmetry generated by the vector field $V=a(x,u)\partial_x+b(x,u)\partial_u$ must satisfy $$0=b_x+(b_u-a_x-xG)xG-aG-2a\frac{u}{x^2}G'-\frac{b}{x}G'$$ But I am not sure how to proceed to solve this fully and the best I can do is guess at solutions. Any tips would be greatly appreaciated.

Answer 1

Not sure if this is helpful, but too long for a comment.

Consider $$\frac{du}{dx}=xG\left( \frac{u}{{{x}^{2}}} \right)=w\left( x,u \right)$$ This is a first order ODE and so either via the prolongation formula or just expanding in a Taylor series, etc, we find for a generator $X=\xi \left( x,u \right)\frac{\partial }{\partial x}+\eta \left( x,u \right)\frac{\partial }{\partial u}$ the requirement $${{\eta }_{x}}+\left( {{\eta }_{u}}-{{\xi }_{x}} \right)w-{{\xi }_{u}}{{w}^{2}}=\xi {{w}_{x}}+\eta {{w}_{u}}$$ Which in terms of the reduced characteristic $K=\eta -w\xi $ becomes $${{K}_{x}}+w{{K}_{y}}={{w}_{y}}K$$ In principle we could determine all symmetries from this equation by using the method of characteristics but upon doing so we find that we need to solve the original equation we started with! (this is something that typically doesn’t happen for equations of order 2 or above). Hence we must start with an ansatz. So for example suppose $$\xi \left( x,u \right)=\alpha x,\,\,\eta \left( x,u \right)=\gamma u$$
Note in this instance the requirement, as you have stated, $${{\eta }_{x}}+\left( {{\eta }_{u}}-{{\xi }_{x}} \right)xG-{{\xi }_{u}}{{x}^{2}}{{G}^{2}}-\xi \left( G-\frac{2u}{{{x}^{2}}}G' \right)-\frac{\eta }{x}G'=0$$ is reduced under the assumption to

$$\left( \gamma -2\alpha \right)\left( xG-\frac{uG'}{x} \right)=0$$ which implies for example $\alpha =1,\,\,\gamma =2$ and ${{X}_{1}}=x\frac{\partial }{\partial x}+2u\frac{\partial }{\partial u}$