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Diffstat (limited to 'Gallavotti_Jauslin_2015.tex')
| -rw-r--r-- | Gallavotti_Jauslin_2015.tex | 4 |
1 files changed, 2 insertions, 2 deletions
diff --git a/Gallavotti_Jauslin_2015.tex b/Gallavotti_Jauslin_2015.tex index e39a636..c0eb89c 100644 --- a/Gallavotti_Jauslin_2015.tex +++ b/Gallavotti_Jauslin_2015.tex @@ -112,7 +112,7 @@ and find that with \begin{equation}\begin{array}{r@{\quad}l} O_{0,\eta}^{[\le 0]}(\Delta):=\frac12\mathbf A^{[\le 0]}_\eta(\Delta)\cdot\bm\tau,& O_{1,\eta}^{[\le 0]}(\Delta):=\frac12\mathbf A^{[\le 0]}_\eta(\Delta)^2,\\[0.3cm] -O_{4,\eta}^{[\le 0]}(\Delta):=\frac12\mathbf A^{[\le 0]}_\eta(\Delta)\cdot\bm\omega,& O_{5,\eta}^{[\le 0]}(\Delta):=\frac12\mathbf \bm\tau\cdot\bm\omega,\\[0.3cm] +O_{4,\eta}^{[\le 0]}(\Delta):=\frac12\mathbf A^{[\le 0]}_\eta(\Delta)\cdot\bm\omega,& O_{5,\eta}^{[\le 0]}(\Delta):=\frac12 \bm\tau\cdot\bm\omega,\\[0.3cm] O_{6,\eta}^{[\le 0]}(\Delta):=\frac12(\mathbf A^{[\le 0]}_\eta(\Delta)\cdot\bm\omega)(\bm\tau\cdot\bm\omega),& O_{7,\eta}^{[\le 0]}(\Delta):=\frac12(\mathbf A^{[\le 0]}_\eta(\Delta)^2)(\bm\tau\cdot\bm\omega) \end{array}\label{eqOdef}\end{equation} (the numbering is meant to recall that in [\cite{BGJ15}]) in which $\bm\tau=(\tau^1,\tau^2,\tau^3)$ and $\mathbf A_\eta^{[\le 0]}(\Delta)$ is a vector of polynomials in the fields whose $j$-th component for $j\in\{1,2,3\}$ is @@ -123,7 +123,7 @@ $\psi_\alpha^{[\le 0]\pm}:=\sum_{m\le0}2^{\frac m2}\psi_\alpha^{[m]\pm}$, and \begin{equation}\begin{array}{r@{\ }>{\displaystyle}l} C=&\cosh(\tilde h),\quad \ell_0^{[0]}=\frac1C\frac{\lambda_0}{\tilde h}\sinh(\tilde h),\quad \ell_1^{[0]}=\frac1C\frac{\lambda_0^2}{12\tilde h}(\tilde h\cosh(\tilde h)+2\sinh(\tilde h))\\[0.3cm] -\ell_4^{[0]}=&\frac1C\lambda_0\sinh(\tilde h),\quad \ell_5^{[0]}=\frac1C\sinh(\tilde h),\quad +\ell_4^{[0]}=&\frac1C\lambda_0\sinh(\tilde h),\quad \ell_5^{[0]}=\frac2C\sinh(\tilde h),\quad \ell_6^{[0]}=\frac1C\frac{\lambda_0}{\tilde h}(\tilde h\cosh(\tilde h)-\sinh(\tilde h))\\[0.3cm] \ell_7^{[0]}=&\frac1C\frac{\lambda_0^2}{12\tilde h^2}(\tilde h^2\sinh(\tilde h)+2\tilde h\cosh(\tilde h)-2\sinh(\tilde h)) \end{array}\label{eqinitcd}\end{equation} |