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**Additional info for Some new entire solutions of semilinear elliptic equations on R^n**

**Sample text**

3) and reasoning as in the steps after (91) we obtain I,j ∂αI,j ∂x0,l UIp−1 ∂UI ∂US = ∂zj ∂zl I=S l=1,2 ∂ ∂x0,l xS − xI · el F1 (|xS − xI |) + O e−(1+ξ)L e−η|xS | |xS − xI | . 6 implies the estimate ∂αS = ∂x0,l I=S ∂ ∂x0,l xS − xI F1 (|xS − xI |) + |xS − xI | In addition, by the symmetries of the problem we have O e−(1+ξ)L O e−(1+ξ)L e−η|xS | ∂αI,j ∂x0,l if |S| ≤ 1; if |S| ≥ 2. = 0 if j = l, so the conclusion follows. 5 We notice some further properties of the coefficients αI,j : first, still by symmetry we have ∂αI,j ∂α−I,j = ; ∂x0,l ∂x0,l I = 0.

2 a=1 This concludes the proof. 1 For a = 1, 2, 3 we set y˜a = Rθ−1 ya ; a x ˜a,i = Rθ−1 (xa,i − ya − iLa θa); a Y˜ = (˜ y1 , y˜2 , y˜3 ); α ˜ I = Rθ−1 α ˜I ; a ˜ = ((˜ X x1,i )i , (˜ x2,i )i , (˜ x3,i )i )) : our goal is to find (˜ ya )a and (xa,i )a,i such that α ˜ I = 0 for all I. We first define the operator Ta : (y, (xi )i ) ∈ R × RN → RN as xi −y, if i = 1; − 1 xi (T xi )j = (T y)i = 0 otherwise, 2 0 if j = i; if j = i ± 1; otherwise. For τ > 0, let us also introduce the weighted norm and space |(y, (xi ))|τ = |y| + sup eiτ |xi |; (103) i (104) τ = {(y, (xi )i ) : |(y, (xi ))|τ < +∞} .

Zl ∂zj I,j p−1 ∂UI We can now divide conveniently the sum I,j ∂α ∂xS,l UI ∂zj into two parts. In fact, when the index I is of the type a, h with a = 1 and h > 0, the centers xI in uL and in uX(Y ),Y coincide. Therefore, setting ∂αI,j ∂αI,j µI,j S,l = ∂xS,l − ∂xS,l for xI = x1,h with h > 0, the previous equation becomes LX,Y ∂w ∂wX,Y − ∂xS,l ∂xS,l = p up−1 − up−1 L X,Y ∂US ∂w p−1 + p up−1 X,Y − uL ∂zl ∂xS,l + p (uX,Y + wX,Y )p−1 − up−1 X,Y (98) p−1 µI,j S,l Ua,h + h>0,j To have an control ∂wX,Y ∂xS,l − ∂w ∂xS,l ∂Ua,h − ∂zj ∂wX,Y + ∂xS,l I=(1,h),h>0 i,j h≤0,j ∂U1,h ∂α U p−1 + ∂xt,s 1,h ∂zj j ∂αI,j p−1 ∂UI U ∂xS,l I ∂zj αS,j j ∂ ∂US USp−1 .