Let $\mathfrak{b}_{0}\equiv\frac{H^{\alpha}y_{\alpha}}{4r}dy_{1}\wedge dy_{2}\wedge dy_{3}$, then Item (2) of Lemma 3.19 tells us that

(3.193)

$$\Delta\mathfrak{b}_{0}=\frac{H^{\alpha}y_{\alpha}}{2r^{3}}dy_{1}\wedge dy_{2}\wedge dy_{3}+r^{-5}\Gamma_{3}^{(4)}+O^{\prime}(1).$$

Let $\Pi^{(4)}$ be the $3$-form in the expansion of $\Delta\phi_{1}$ given by (3.119) in Proposition 3.15.

Next, Lemma 3.18 and Lemma 3.19 tell us that there are $3$-forms $\widehat{\Gamma}_{3}^{(4)}$ and $\widehat{\Pi}_{3}^{(4)}$ which are also of the form as in (3.45) such that

(3.194)		$\displaystyle\Delta(r^{-3}\widehat{\Gamma}_{3}^{(4)})=-r^{-5}\Gamma_{3}^{(4)}+O^{\prime}(1),$
(3.195)		$\displaystyle\Delta(r^{-3}\widehat{\Pi}_{3}^{(4)})=-r^{-5}\Pi_{3}^{(4)}+O^{\prime}(1).$

Now let

(3.196)

$$\mathfrak{b}_{1}\equiv r^{-3}\widehat{\Gamma}_{3}^{(4)}+r^{-3}\widehat{\Pi}_{3}^{(4)},$$

then the correction term $\mathfrak{b}_{0}+\mathfrak{b}_{1}$ is chosen as the above such that $\Delta(\mathfrak{b}_{0}+\mathfrak{b}_{1})$ in fact eliminates the $O^{\prime}(r^{-2})$-term and implicit $O^{\prime}(r^{-1})$-terms in the expansion of $\Delta\phi_{1}$ (see Proposition 3.15).

In the following, we will make a further correction such that those explicit $O^{\prime}(r^{-1})$-terms will be cancelled out as well. In fact, we define

(3.197)

$$\mathfrak{b}_{2}\equiv\Big(\Omega_{ij\alpha\beta}(\frac{1}{16}y_{\widehat{\alpha\beta}}dr-\frac{3}{16}rdy_{\widehat{\alpha\beta}})-\frac{1}{16}A_{ij\alpha\beta}(y_{\widehat{\alpha\beta}}dr+rdy_{\widehat{\alpha\beta}})\Big)\wedge dx_{i}\wedge dx_{j},$$

applying Lemma 3.18 and Lemma 3.19 again, then

(3.198)

$$\Delta\mathfrak{b}_{2}=\Omega_{ij\alpha\beta}\Big(\frac{1}{4r}dy_{\widehat{\beta\alpha}}+\frac{y_{\widehat{\alpha\beta}}}{4r^{2}}dr\Big)\wedge dx_{i}\wedge dx_{j}-A_{ij\alpha\beta}\Big(\frac{y_{\widehat{\alpha\beta}}}{4r^{2}}dr-\frac{1}{4r}dy_{\widehat{\alpha\beta}}\Big)\wedge dx_{i}\wedge dx_{j},$$

and hence

(3.199)

$$\Delta(\phi_{1}+\mathfrak{b}_{0}+\mathfrak{b}_{1}+\mathfrak{b}_{2})=O^{\prime}(1).$$

Therefore, it suffices to choose the correction term

(3.200)

$$\mathfrak{B}_{0}\equiv\mathfrak{b}_{0}+\mathfrak{b}_{1}+\mathfrak{b}_{2},$$

which gives $\Delta(\phi_{1}+\mathfrak{B}_{0})=O^{\prime}(1)$.

Notice that, $\mathfrak{b}_{2}$ has a further cancellation,

	$\displaystyle\mathfrak{b}_{2}=$	$\displaystyle\Big(\Omega_{ij\alpha\beta}(\frac{1}{16}y_{\widehat{\alpha\beta}}dr-\frac{3}{16}rdy_{\widehat{\alpha\beta}})-\frac{1}{16}A_{ij\alpha\beta}(y_{\widehat{\alpha\beta}}dr+rdy_{\widehat{\alpha\beta}})\Big)\wedge dx_{i}\wedge dx_{j},$
	$\displaystyle=$	$\displaystyle-\frac{1}{4}A_{ij\alpha\beta}rdy_{\widehat{\alpha\beta}}\wedge dx_{i}\wedge dx_{j}-\frac{1}{16}(A_{i,\alpha,\alpha+1}A_{j,\alpha,\alpha+2}-A_{i,\alpha,\alpha+2}A_{j,\alpha,\alpha+1})y_{\alpha}dr\wedge dx_{i}\wedge dx_{j}$
(3.201)			$\displaystyle+\frac{3}{16}(A_{i,\alpha,\alpha+1}A_{j,\alpha,\alpha+2}-A_{i,\alpha,\alpha+2}A_{j,\alpha,\alpha+1})rdy_{\alpha}\wedge dx_{i}\wedge dx_{j}.$

Therefore,

	$\displaystyle\phi_{2}=$	$\displaystyle\phi_{1}+\mathfrak{B}_{0}$
	$\displaystyle=$	$\displaystyle\frac{1}{2r}(1-\frac{H^{\alpha}y_{\alpha}}{2})dy_{1}\wedge dy_{2}\wedge dy_{3}+\frac{1}{2r}y_{\beta}A_{i\alpha\beta}dx_{i}\wedge dy_{\widehat{\alpha}}-\frac{1}{4}A_{ij\alpha\beta}r\cdot dy_{\widehat{\alpha\beta}}\wedge dx_{i}\wedge dx_{j}$
(3.202)			$\displaystyle+\frac{3}{16}(A_{i\alpha,\alpha+1}A_{j\alpha,\alpha+2}-A_{i\alpha,\alpha+2}A_{j\alpha,\alpha+1})d(ry_{\alpha})\wedge dx_{i}\wedge dx_{j}+r^{-3}\Pi_{3}^{(4)}+O^{\prime}(r^{2}),$

and

(3.203)

$$\Delta\phi_{2}=O^{\prime}(1).$$

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