Download Differential equations methods for the Monge-Kantorovich by Lawrence C. Evans, Wilfrid Gangbo PDF

April 5, 2017 | Mathematics | By admin | 0 Comments

By Lawrence C. Evans, Wilfrid Gangbo

During this quantity, the authors display below a few assumptions on $f^+$, $f^-$ technique to the classical Monge-Kantorovich challenge of optimally rearranging the degree $\mu{^+}=f^+dx$ onto $\mu^-=f^-dy$ might be built through learning the $p$-Laplacian equation $- \mathrm{div}(\vert DU_p\vert^{p-2}Du_p)=f^+-f^-$ within the restrict as $p\rightarrow\infty$. the belief is to teach $u_p\rightarrow u$, the place $u$ satisfies $\vert Du\vert\leq 1,-\mathrm{div}(aDu)=f^+-f^-$ for a few density $a\geq0$, and then to construct a move via fixing a nonautonomous ODE concerning $a, Du, f^+$ and $f^-$.

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Ig of P . We have i1 = 1 and ig ≤ 2g − 1 . Proof. 6, and 0 is a pole number. We have the following obvious characterization of gap numbers: i is a gap number of P ⇐⇒ L ((i − 1)P ) = L (iP ) . 17. 8. 39) exactly g − 1 numbers 1 ≤ i ≤ 2g − 1 with L ((i − 1)P ) L (iP ). The remaining g numbers are gaps of P . Finally we must show that 1 is a gap. Suppose the converse, so 1 is a pole number of P . Since the pole numbers form an additive semigroup, every n ∈ IN is a pole number, and there are no gaps at all.

CΩ (D, G) is an [n, k , d ] code with parameters k = i(G − D) − i(G) and d ≥ deg G − (2g − 2) . Under the additional hypothesis deg G > 2g − 2, we have k = i(G − D) ≥ n + g − 1 − deg G. If moreover 2g − 2 < deg G < n then k = n + g − 1 − deg G . Proof. Let P ∈ IPF be a place of degree one and let ω be a Weil differential with vP (ω) ≥ −1. We claim that ωP (1) = 0 ⇐⇒ vP (ω) ≥ 0 . 3 which states that for an integer r ∈ ZZ, 52 2 Algebraic Geometry Codes vP (ω) ≥ r ⇐⇒ ωP (x) = 0 for all x ∈ F with vP (x) ≥ −r .

One checks easily that ϕ is surjective and that the kernel of ϕ is AF (A1 ). Consequently deg A2 − deg A1 = deg P = [FP : K] = dim(AF (A2 )/AF (A1 )) . Step 2. Let A1 , A2 ∈ Div(F ) and A1 ≤ A2 as before. Then dim((AF (A2 ) + F )/(AF (A1 ) + F ) = (deg A2 − (A2 )) − (deg A1 − (A1 )) . 25) Proof of Step 2. 26) where σ1 and σ2 are defined in the obvious manner. In fact, the only non-trivial assertion is that the kernel of σ2 is contained in the image of σ1 . In order to prove this, let α ∈ AF (A2 ) with σ2 (α+AF (A1 )) = 0.

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