Lie Group of Symmetries for Differential Equation A+eB

[eljose79]

Let be a differntial equation which Lie group of symmetries is A..we will call this differential equation A(y,y´)

then suppose we perturbe the initial equation A with another differential operator B(y,y´) which lie group is B.and we form the operator C(y,y´)=A(y,y´)+eB(y,y´) where e<<1 then...

what would be the lie group of simmetries for C?...

could it be obtained by perturbation theory?..

 

[lethe]

well, the intersection of A nd B would be in the symmetry group of C, for starters...

 

[tacman]

The Lie group of symmetries for the differential equation A+eB can be obtained by considering the combined operator C(y,y´)=A(y,y´)+eB(y,y´). This operator C can be seen as a perturbation of the original operator A, with the parameter e acting as a small perturbation parameter.

The Lie group of symmetries for C can be obtained by using perturbation theory, which is a powerful mathematical tool for analyzing systems that are slightly different from a known, simpler system. In this case, we can use perturbation theory to determine the effect of the perturbation eB on the Lie group of symmetries for A.

By considering the perturbed operator C, we can determine the new Lie group of symmetries for C by finding the symmetries that leave the perturbed operator C invariant. These symmetries will be a combination of the symmetries of A and B, and can be obtained by using perturbation theory.

In summary, the Lie group of symmetries for the perturbed operator C can be obtained by using perturbation theory, and will be a combination of the symmetries of the original operator A and the perturbation operator B.