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This is the Einstein-Maxwell-Dilaton Gravity action: \begin{eqnarray*} S_{EM} = -\frac{1}{16 \pi G_5} \int \mathrm{d^5}x \sqrt{-g} \ [R - \frac{f(\phi)}{4}F_{MN}F^{MN} -\frac{1}{2}D_{M}\phi D^{M}\phi -V(\phi)]\,, \end{eqnarray*} where $R$ is the Ricci scalar, $G_5$ is the Newton's constant in five dimensions, $F_{MN}$ is the field strength tensor of $U(1)$ gauge field $A_M$, and $L$ is the AdS radius. Moreover, $f(\phi)$ is the gauge kinetic function that acts as coupling between $U(1)$ gauge field and dilaton field $\phi$, and $V(\phi)$ is the potential of the dilaton field.

To introduce the magnetic field, we're taking the following ansatz: \begin{eqnarray*} & & ds^2=\frac{L^2 e^{2A(z)}}{z^2}\biggl[-g(z)dt^2 + \frac{dz^2}{g(z)} + dx_{1}^2+ e^{B^2 z^2} \biggl( dx_{2}^2 + dx_{3}^2 \biggr) \biggr]\,, \nonumber \\ & & \phi=\phi(z), \ \ F_{MN}=B dx_{2}\wedge dx_{3}\,,A(z)=-a z^2 \end{eqnarray*} By solving the equation of motions, we obtained the metric function $g(z)$ and the gauge kinetic function $f(\phi)$: \begin{eqnarray*} g(z) &=& 1 - \frac{ \int_0^z , d\xi \ \xi^3 e^{-B^2 \xi^2 -3A(\xi) } }{\int_0^{z_h} , d\xi \ \xi^3 e^{-B^2 \xi^2 -3A(\xi) }}\\ f(z) &=& g(z)e^{2 A(z)+2 B^2 z^2} \left(-\frac{6 A'(z)}{z}-4 B^2+\frac{4}{z^2}\right)-\frac{2 e^{2 A(z)+2 B^2 z^2} g'(z)}{z} \end{eqnarray*} In our model, everything is working fine except that $f(z)$ diverges at the boundary $(z \rightarrow 0)$.

Will the divergence of the gauge kinetic function at the boundary affect the completeness of our model? If it does, how to rectify this problem?

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It is not clear to me how you can deduce that $f(\phi)$ is divergent at $z\to0$, if the other functions involved are not known, but since the fundamental quantity of your theory is $f(\phi)F_{MN}F^{MN}$ as long as this is finite your theory is fine.

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  • $\begingroup$ My mistake, I forgot to state the other functions. I will edit my question to explain the divergence. $\endgroup$
    – codebpr
    Commented Aug 1, 2023 at 6:43
  • $\begingroup$ @codebpr just check every scalar in your action. If any of them is singular at $z\to0$ then your theory does not hold at this point. $\endgroup$
    – Noone
    Commented Aug 1, 2023 at 6:45

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