S_3^E = \big| Q^{(1)} \big|\, \sqrt{{\frac{\raisebox{-2pt}{$4$}}{g_s^2}} + (\Delta \chi)^2}
  = \big| Q^{(1)} \big|\, \big| \tau'_{\infty} \big|\ ,$$ where $\tau'_{\infty}=(\chi_\infty-\chi_0)+2{\mathrm{i}}\,{\mathrm{e}}^{\phi_{\infty}/2}$ is the value of $\tau'$ at infinity. For $\Delta\chi=0$, a similar solution was also found in [@GS1]. Following the discussion in [@BBS], it should correspond, from a IIA point of view, to the D2-brane wrapping a three-cycle in the Calabi-Yau, or to the D1+D3+D5-branes wrapping even cycles in type IIB. Notice again consistency with the symmetries and . Observe also that for $\Delta
\chi=0$, the solution is inversely proportional to $g_s$, and is for small $g_s$ dominating over the fivebrane instanton .

As a second Ansatz, consider $\chi=2\chi_1{\mathrm{e}}^\phi+\chi_0$. This differs from the first Ansatz in the power of ${\mathrm{e}}^\phi$. The BPS condition turns into $$* H = \pm 2\, {\mathrm{d}}\sqrt{{\mathrm{e}}^{-\phi} + \chi_1^2\,} \begin{pmatrix} 1 -
  \chi_1 \chi_0 \\[2pt] -\chi_1 \end{pmatrix}\ .$$ Accordingly, the square root must be harmonic, and we find for $\chi_1
\neq 0$ (the case $\chi_1=0$ is included in the previous Ansatz), $${\mathrm{e}}^{-\phi} = (h - \chi_1) (h + \chi_1)\ ,\quad h = \sqrt{
  {\mathrm{e}}^{-\phi_\infty} + \chi_1^2\,} + \Big| {\frac{\raisebox{-2pt}{$Q^{(2)}$}}{2\chi_1}} \Big|\,
  {\frac{\raisebox{-2pt}{$1$}}{\Omega_D\, r^{D-2}}}\ .$$ The scalar field $\chi$ is regular everywhere and interpolates between zero and infinity as $$\Delta 