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Remark 3.3 . [022Z]

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Remark 3.3.

The symmetry of the ODE (cf. Remark 2.6) translates into the following. Let

w⁡(t)=t​w~​(1/t),w(t)=t\tilde{w}(1/t),

then

w′=w~−t−1​w~′,w′′=t−3​w~′′,w^{\prime}=\tilde{w}-t^{-1}\tilde{w}^{\prime},\quad w^{\prime\prime}=t^{-3}\tilde{w}^{\prime\prime},
w+(1−t)​w′=w~+(1−t−1)​w~′,w′′​w3=w~′′​w~3.w+(1-t)w^{\prime}=\tilde{w}+(1-t^{-1})\tilde{w}^{\prime},\quad w^{\prime\prime}w^{3}=\tilde{w}^{\prime\prime}\tilde{w}^{3}.

Thus if ww solves (10), then so does w~\tilde{w}.

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