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Ya escribimos al respecto en este post. Aquí lo que haremos es reescribir las expresiones allí introducidas

En primer lugar, teniamos:

 \Delta X^i = 8 \pi f^{ij}S_j^* - \frac{1}{3}\mathcal{D}^i \mathcal{D}_j X^j

donde:

S_j^* := \sqrt{ \frac{\gamma}{f} } S = \psi^6 S_j,

S_j := \rho h w^2 v_j.

En el caso de estar trabajando en cartesianas y teniendo en cuenta todo el trabajo realizado en el artículo, nos queda:

\partial_{xx} X^x + \partial_{yy} X^x + \partial_{zz} X^x = 8 \pi \psi^6 \rho h w^2 v_x - \frac{1}{3} \partial_x (\partial_x X^x + \partial_y X^y + \partial_z X^z),

\partial_{xx} X^y + \partial_{yy} X^y + \partial_{zz} X^y = 8 \pi \psi^6 \rho h w^2 v_y - \frac{1}{3} \partial_y (\partial_x X^x + \partial_y X^y + \partial_z X^z),

\partial_{xx} X^z + \partial_{yy} X^z + \partial_{zz} X^z = 8 \pi \psi^6 \rho h w^2 v_z - \frac{1}{3} \partial_z (\partial_x X^x + \partial_y X^y + \partial_z X^z).

A continuación, y para la siguiente ecuación, necesitamos:

\hat{A}^{ij} = \mathcal{D}^i X^j + \mathcal{D}^j X^i - \frac{2}{3} \mathcal{D}_k X^k f^{ij}

que queda como:

\hat{A}^{xx} = 2 \partial_x X^x - \frac{2}{3} (\partial_x X^x + \partial_y X^y + \partial_z X^z),

\hat{A}^{xy} = \hat{A}^{yx}= \partial_x X^y + \partial_y X^x,

\hat{A}^{xz} = \hat{A}^{zx} = \partial_x X^z + \partial_z X^x,

\hat{A}^{yy} = 2 \partial_y X^y - \frac{2}{3} (\partial_x X^x + \partial_y X^y + \partial_z X^z),

\hat{A}^{yz} = \hat{A}^{zy} = \partial_y X^z + \partial_z X^y,

\hat{A}^{zz} = 2 \partial_z X^z - \frac{2}{3} (\partial_x X^x + \partial_y X^y + \partial_z X^z),

por lo que:

\Delta \psi = -2 \pi \psi^{-1} E^* - \psi^{-7} \frac{f_{il}f_{jm}\hat{A}^{lm}\hat{A}^{ij}}{8}

donde:

E^*:= \sqrt{ \frac{\gamma}{f} } E = \psi^6 E,

E:= D + \tau

es:

\Delta \psi = -2 \pi \psi^{-1} (D + \tau) - \psi^{-7} \frac{(\hat{A}^{xx})^2+(\hat{A}^{yy})^2+(\hat{A}^{zz})^2+2(\hat{A}^{xy})^2+2(\hat{A}^{xz})^2+2(\hat{A}^{yz})^2}{8}.

La siguiente:

\Delta (\alpha\psi) = 2 \pi (\alpha\psi)^{-1} (E^* + 2S^*) + \frac{7}{8} (\alpha\psi)^{-7} (f_{il} f{jm} \hat{A}^{lm} \hat{A}^{ij})

con:

S^*:= \sqrt{ \frac{\gamma}{f} } S = \psi^6 S,

S:= \rho h (w^2-1) + 3 p

queda:

\Delta (\alpha\psi) = 2 \pi (\alpha\psi)^{-1} ( D + \tau + 2 \rho h (w^2-1) + 6 p) +

+ \frac{7}{8}(\alpha\psi)^{-7} ((\hat{A}^{xx})^2+(\hat{A}^{yy})^2+(\hat{A}^{zz})^2+2(\hat{A}^{xy})^2+2(\hat{A}^{xz})^2+2(\hat{A}^{yz})^2)

Y la última:

\Delta \beta^i = \mathcal{D}_j (2 (\alpha\psi)^{-6} \hat{A}^{ij}) - \frac{1}{3} \mathcal{D}^i (\mathcal{D}_j \beta^j),

que escribimos como:

\Delta \beta^x = \partial_x (2 (\alpha \psi)^{-6} \hat{A}^{xx}) + \partial_y (2 (\alpha \psi)^{-6} \hat{A}^{xy}) + \partial_z (2 (\alpha \psi)^{-6} \hat{A}^{xz}) -

- \frac{1}{3} \partial_x (\partial_x \beta^x + \partial_y \beta^y + \partial_z \beta^z)

\Delta \beta^y = \partial_x (2 (\alpha \psi)^{-6} \hat{A}^{yx}) + \partial_y (2 (\alpha \psi)^{-6} \hat{A}^{yy}) + \partial_z (2 (\alpha \psi)^{-6} \hat{A}^{yz}) -

- \frac{1}{3} \partial_y (\partial_x \beta^x + \partial_y \beta^y + \partial_z \beta^z)

\Delta \beta^z = \partial_x (2 (\alpha \psi)^{-6} \hat{A}^{zx}) + \partial_y (2 (\alpha \psi)^{-6} \hat{A}^{zy}) + \partial_z (2 (\alpha \psi)^{-6} \hat{A}^{zz}) -

- \frac{1}{3} \partial_z (\partial_x \beta^x + \partial_y \beta^y + \partial_z \beta^z)

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