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for example given the differential operator $ \partial _{x} $ we know that it belong to the translation $ y=x+a $ for some 'a' given the differential operator $ x\partial _{x} $ we know that we are dealing with dilations $ y=ax$ however for any particular first order differential operator $ a(x) \partial _{x} +b(x) $ for some polynomials a(x) and b(x) can we obtain the symmetry ?? |
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An infinitesimal generator $X$ is a vector field, which satisfies Leibniz rule $$X[fg]~=~ fX[g]+g X[f].$$ In the $1$-dimensional case, the generator is of the form $X=p(x)\frac{\partial}{\partial x}$, where $p=p(x)$ is some function. Assume furthermore that there exists a bijective smooth function $h=h(x)$, such that $$p(x) h'(x)~=~1.$$ In other words, that $$h(x)~=~ \int^x \frac{dx^{\prime}}{p(x^{\prime})}.$$ Then the symmetry flow becomes $$ e^{a X} f(x) ~=~ e^{a p(x) \frac{\partial}{\partial x}} f(x) ~=~ \left. e^{a \frac{\partial}{\partial y}} f(h^{-1}(y)) \right|_{y=h(x)}$$ $$~=~ \left.f(h^{-1}(y+a)) \right|_{y=h(x)}~=~ f(h^{-1}(h(x)+a)). $$ Example 1. Translation: $$p(x)~=~1, \qquad h(x)~=~x, \qquad e^{a X} f(x)~=~f(x+a).$$ Example 2. Dilation: $$p(x)~=~x, \qquad h(x)~=~\ln x, \qquad e^{a X} f(x)~=~f(e^ax).$$ |
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