Abstract

Models of Fisher's runaway process show that if there is a cost to female preference, no preference or male trait exaggeration will evolve. Surprisingly, this is true no matter how small the cost, which reveals that these models of Fisher's process are structurally unstable (Bulmer 1989). Here a model of Fisher's runaway process is presented to demonstrate that costly female preference evolves very easily when space is explicitly included in the model. The only requirement is that the optimal male phenotype changes across the species' range. The model shows that the spatial average of the female preference and male trait reach an evolutionary equilibrium that is identical to those of nonspatial models, but that the preference and male trait can deviate greatly from these averages at any point in space. For example, if random mating results in the lowest cost to females, then at equilibrium the spatial average preference will be zero. Nevertheless, there will be some locations at which females prefer males with larger ornaments and others where they prefer males with smaller ornaments. Results also show that the structural instability of nonspatial models of Fisher's process is less of a problem in spatial models. In particular, many of the main qualitative features of cost-free spatial models of Fisher's process remain valid even when there are small costs of female preference. Finally, the model shows that abrupt changes in the optimal male phenotype across space can result in an amplification of this pattern when preference has a small cost, but it can also result in a pattern similar to reproductive character displacement. Which of these occurs depends on the magnitude of the cost of female preference. This suggests that some patterns of reproductive character displacement in nature might be explained simply by sexual selection rather than by hybrid dysgenesis and reinforcement.Corresponding Editor: J. Mallet

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