By Richard F. Bass

A dialogue of the interaction of diffusion tactics and partial differential equations with an emphasis on probabilistic equipment. It starts with stochastic differential equations, the probabilistic equipment had to examine PDE, and strikes directly to probabilistic representations of options for PDE, regularity of options and one dimensional diffusions. the writer discusses intensive major kinds of moment order linear differential operators: non-divergence operators and divergence operators, together with issues corresponding to the Harnack inequality of Krylov-Safonov for non-divergence operators and warmth kernel estimates for divergence shape operators, in addition to Martingale difficulties and the Malliavin calculus. whereas serving as a textbook for a graduate direction on diffusion thought with purposes to PDE, this can even be a beneficial connection with researchers in likelihood who're attracted to PDE, in addition to for analysts drawn to probabilistic tools.

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**Extra info for Bass Diffusions and elliptic operators**

**Sample text**

0 Since ∂t u = Lu, Mt is a martingale, and E x M0 = E x Mt0 . On the one hand, E x Mt0 = E x u(Xt0 , 0) = E x f (Xt0 ), while on the other, E x M0 = E x u(X0 , t0 ) = u(x, t0 ). 48 II REPRESENTATIONS OF SOLUTIONS Since t0 is arbitrary, the result follows. For bounded domains D, the Cauchy problem is to ﬁnd u such that ∂t = Lu on D, u(x, 0) = f (x) for x ∈ D, and u(x, t) = 0 for x ∈ ∂D. The solution is given by u(x, t) = E x [f (Xt ); t < τD ], where τD is the exit time of D. The proof is very similar to the case of Rd .

Let ε ∈ (0, 1), t0 > 0. There exists c1 depending only on the upper bounds of σ, b, and σ −1 such that P( sup |Xs − X0 | < ε) ≥ c1 . s≤t0 Proof. For notational simplicity assume X0 = 0. Let y = (ε/4, 0, . . , 0). Applying Itˆ o’s formula with f (z) = |z − y|2 and setting Vt = |Xt − y|2 , then V0 = (ε/4)2 and (Xti − yi ) dXti + dVt = 2 i d X i t. 3 applies and P( sup |Vs − V0 | ≤ (ε/8)2 ) = P( sup |Ys − Y0 | ≤ (ε/8)2 ) ≥ c2 . s≤t0 s≤t0 By the deﬁnition of y and Vt , this implies with probability at least c2 that Xt stays inside B(0, ε).

Dirichlet problem Let D be a ball (or other nice bounded domain) and let us consider the solution to the Dirichlet problem: given f a continuous function on ∂D, ﬁnd u ∈ C(D) such that u is C 2 in D and Lu = 0 in D, u = f on ∂D. 1) Theorem. 1) satisﬁes u(x) = E x f (XτD ). Proof. s. Let Sn = inf{t : dist (Xt , ∂D) < 1/n}. By Itˆo’s formula, t∧Sn u(Xt∧Sn ) = u(X0 ) + martingale + Lu(Xs ) ds. 0 Since Lu = 0 inside D, taking expectations shows u(x) = E x u(Xt∧Sn ). We let t → ∞ and then n → ∞. By dominated convergence, we obtain u(x) = E x u(XτD ).

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