one point in the parameter space. We will now show how this works in two simple examples. In both cases, we will consider a kink field φ in addition to just one stabilization field χ , with no interaction terms among them in the scalar potential. In both examples there will be regions of parameter space in which two distinct static configurations are possible for the same action. In both of the following examples we consider a Lagrangian for two scalar fields L matter = 1 2 g $^{MN}$( ∂$_{M}$ φ ) ∂$_{N}$ φ − V ( φ ) − ∑ i =1 , 2 β$_{i}$ ( φ ) δ ( y − y$_{i}$ ) + 1 2 g $^{MN}$( ∂$_{M}$ χ ) ∂$_{N}$ χ − U ( χ ) − ∑ i =1 , 2 λ$_{i}$ ( χ ) δ ( y − y$_{i}$ ) , (38) U ( χ ) = 1 2 m 2 $_{χ}$χ 2 (39) λ$_{i}$ ( χ ) = M − 1 ∗ ( 1 2 µ 2 $_{i}$χ 2 + Σ$_{i}$ ) . (40) The fact that the second brane potential for χ is parameterized in terms of two parameters, µ 2 and Σ$_{2}$, will allow us to find unique solutions to the boundary conditions on the second brane. The junction conditions (35)-(36) become where φ is the kink field and χ is the stabilization field with potentials On the second brane ( i = 2) these can be inverted to give so that once we determine the fields on the second brane, we can extract the boundary conditions (and therefore Lagrangian) to which those fields are a solution. The only things left to specify are the bulk potential for the kink field and the initial- boundary conditions. In this first example, we take the kink potential to be V ( φ ) = − 1 2 m 2 $_{φ}$φ 2 + 1 4 λφ 4 . (45) µ 2 = − 2 χ $^{′}$( y$_{2}$ ) χ ( y$_{2}$ ) (43) Σ$_{2}$ = − 6 M 4 $_{∗}$σ ( y$_{2}$ ) + χ $^{′}$( y$_{2}$ ) χ ( y$_{2}$ ) (44) σ $^{′}$( y$_{i}$ ) = ( − 1) i − 1 1 6 M 4 ∗ ( 1 2 µ 2 $_{i}$χ $^{2}$( y$_{i}$ ) + Σ$_{i}$ ) (41) χ $^{′}$( y$_{i}$ ) = ( − 1) i − $_{1}$1 2 µ 2 $_{i}$χ ( y$_{i}$ ) . (42) B. Example 1: Quartic Potential 16