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A singular large deviations phenomenon. (English) Zbl 0993.60025
The aim of this paper is to study the precise convergence of the law $$P_{\varepsilon}$$ (as $$\varepsilon\to 0$$) of the solution $$X_t^{\varepsilon}$$ of the stochastic differential equation $dX_t^{\varepsilon}=\varepsilon dB_t+\text{sgn}(X_t^{\varepsilon}) |X_t^{\varepsilon}|^{\gamma} dt,\quad X_0^{\varepsilon}=0,$ $$B_t$$ is one-dimensional Brownian motion, which can be regarded as a small random perturbation of the dynamical system $x_t'=\text{sgn}(x_t)|x_t|^{\gamma},\quad x_0=0. \tag{1}$
To describe the main results let us denote by $$p_t^{\varepsilon}()$$ the density of $$X_t^{\varepsilon}$$ with respect to the Lebesgue measure and observe that if $$|x |ne(t(1-\gamma))^{1/(1-\gamma)},$$ i.e. if $$(t, x)$$ does not belong to the graph of one of the extremal solutions of problem (1), then it is known that the density tends to zero. But according to the position of the point $$(t, x)$$ the authors emphasize two kinds of rate. If the point $$(t, x)$$ is such that $$|x|>(t(1-\gamma))^{1/(1-\gamma)},$$ there exists $$\lim_{\varepsilon\to 0} \varepsilon^2 \ln p_t^{\varepsilon}(x)=-k_t(|x |)$$ with some positive function $$k_t.$$ This means that the density has an exponential decay with rate $$\varepsilon^2,$$ as in large deviations theory. If the point $$(t, x)$$ lies in the domain between the two extremals, that is, if $$|x'|<(t(1-\gamma))^{1/(1-\gamma)},$$ then the density has an exponential decay with a different rate, namely $$\varepsilon^{2(1-\gamma)/(1+\gamma)}.$$ Precisely, it is shown that for such points $\lim_{\varepsilon\to 0} \varepsilon^{2(1-\gamma)/(1+\gamma)} \ln p_t^{\varepsilon}(x)=\lambda_1((x^{1-\gamma})/(1-\gamma)-t),$ where $$\lambda_1$$ is the first positive eigenvalue of the Schrödinger operator $$(\gamma|x|^{\gamma-1}+|x |^{2\gamma}-d^2/dx^2)/2$$.

##### MSC:
 60F10 Large deviations 60H10 Stochastic ordinary differential equations (aspects of stochastic analysis)
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