The equations of motion for φ and χ are identical, so we write the generic field as Φ, and its equation of motion is We assume that the mesons will be non-interacting and plane-wave, so we make the ans¨ atz Φ = f ( ρ ) e $^{ik.x}$, with k 2 = − M 2 ( M being the mass of the meson). We then look for normalisable solutions to (3.13): these correspond to consistent field theory solutions. (3.13) can be solved exactly by: $_{2}$F$_{1}$ ( a, b ; c ; z ) is the regular hypergeometric function, defined as where ( a )$_{n}$ is the rising factorial. α is as yet undetermined. We can de- termine it by considering our requirements on f ( ρ ): the solution must be normalizable and real. To stop ( ρ $^{2}$+ L $^{2}$) − α $_{2}$F$_{1}$ ( − α, − α +1; 2; − ρ $^{2}$/L$^{2}$) run- ning off to ±∞ as ρ → ∞ we need to terminate the series at order ρ − 2 $^{α}$. A little maths shows that this can be done by setting in which case the hypergeometric function terminates at order ( ρ $^{2}$/L$^{2}$) $^{n}$. Hence φ ∼ ρ − 2 as ρ → ∞ . So the final solution to (3.13) is 53 f ( ρ ) = 1 ( ρ 2 + L $^{2}$) n +1 $_{2}$F$_{1}$ ( − n − 1 , − n ; 2; − ρ $^{2}$/L$^{2}$) (3.17) n = α − 1 (3.16) $_{2}$F$_{1}$ ( a, b ; c ; z ) ≡ ∞ ∑ n =0 ( a )$_{n}$( b )$_{n}$ ( c )$_{n}$ z n n ! (3.15) f ( ρ ) = ( ρ 2 + L 2 $_{)}$− α $_{2}$F$_{1}$ ( − α, − α + 1; 2; − ρ $^{2}$/L$^{2}$) (3.14) R 4 ( ρ 2 + L $^{2}$)$_{2}$∂ $^{µ}$∂$_{µ}$ Φ + 1 ρ $_{3}$( ρ $^{3}$∂$_{ρ}$ Φ) = 0 (3.13)