Varying $\Phi$ and integrating by parts gives
$$
\boxed{\nabla_a\nabla^a\Phi
-\alpha e^{\alpha\Phi}F_{ab}F^{ab}=0}.
$$
Since $\delta F_{ab}=2\nabla_{[a}\delta A_{b]}$, variation of the potential and another integration by parts give
$$
\boxed{\nabla_a(e^{\alpha\Phi}F^{ab})=0}.
$$
These are the scalar and Maxwell equations of <Einstein-Maxwell-dilaton theory>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 309 / 4 / b / i / Solution

{c}

Einstein-Maxwell-dilaton theory couples the metric to a scalar field and a Maxwell field with scalar-dependent gauge coupling, typically through $e^{\alpha\Phi}F_{ab}F^{ab}$. Its Maxwell equation is $\nabla_a(e^{\alpha\Phi}F^{ab})=0$.


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 Incoming links: Einstein-Maxwell-dilaton theory