$\Delta\chi\neq 0$, our instanton solution is new, and this term in the instanton action does not depend on the string coupling constant $g_s$. The appearance of $\Delta\chi$ in the instanton action is one of the new results in this paper. Its presence was somehow anticipated in [@BBS], and here we have computed it explicitly.

We now turn to $O_3$. The BPS equation for this case reads $$* H = \pm {\frac{\raisebox{-2pt}{$1$}}{|\tau'|}} \begin{pmatrix} -2\, {\mathrm{d}}\phi + {\mathrm{e}}^{-\phi}
  (\chi + \chi_0)\, {\mathrm{d}}\chi + \chi (\chi - \chi_0)\, {\mathrm{d}}{\mathrm{e}}^{-\phi}\,
  \\[4pt] (\chi - \chi_0)\, {\mathrm{d}}{\mathrm{e}}^{-\phi} + 2 {\mathrm{e}}^{-\phi} {\mathrm{d}}\chi
 \end{pmatrix}\ .$$ We have been unable to find the general solution[^5]. Instead, let us consider two Ansätze for which we can explicitly solve the equations. First, we set $\chi=2\chi_1{\mathrm{e}}^{\phi/2}
+\chi_0$. Then the equations simplify to $$* H = \pm 2\, {\mathrm{d}}{\mathrm{e}}^{-\phi/2} \begin{pmatrix} \sqrt{1 + \chi_1^2\,}\,
  \\[2pt] 0 \end{pmatrix}\ .$$ It follows that now ${\mathrm{e}}^{-\phi/2}$ is harmonic, with solution $${\mathrm{e}}^{-\phi/2} = {\mathrm{e}}^{-\phi_{\infty}/2} + \frac{\big|Q^{(1)}\big|}
  {2 \sqrt{1 + \chi_1^2\,}\, \Omega_D\, r^{D-2}}\ .$$ The scalar $\chi$ is then completely regular and interpolates between the boundaries as $$\Delta \chi = \chi_{\infty} - \chi_0 = \frac{2\chi_1}{g_s}\ .$$ Since $H^2=0$ we have $Q^{(2)}=0$, and for the instanton action we find $$\label{Sinst_O3_1}
  