By Regner

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**Extra resources for A guide to first-passage progrecesses**

**Sample text**

We first define J from J = J (τ, u) = xτ yu − xu yτ , (156) which is the Jacobian from (x, y) to (u, τ ) coordinates, and is given explicitly by (41). From (156) we obtain Jτ = xτ τ yu + xτ yuτ − xu yτ τ − xuτ yτ (157) On the Shortest Queue Version of the Erlang Loss Model 159 while differentiating (133) with respect to x yields ∂ (xe x ) = −τx xτ τ − xτ u u x ∂x 1 = [−xτ τ yu + xτ u yτ ]. J (158) Similarly, differentiating (133) with respect to y yields ∂ (ye ∂y y − e− y ) = ∂ 1 [yτ τ xu − yτ u xτ ] .

Yao and C. Knessl which is true if 0 < a < 1/4. From (43) we see that the ray u = 1 − a is given by a − x = a(1 − e−τ ), a − y = (1 − a) (eτ − 1) (371) or x = X 1 (y) ≡ a(1 − a) , 1−y y =1+ a(a − 1) . x (372) Note that this ray passes through the points (a(1 − a), 0) and (a, a). From the above discussion it follows that if a > 1/4 the curve x = X ∗ (y) splits the domain D as D = D+ ∪ D− , (373) D+ = (x, y): y < x < X ∗ (y), 0

Knessl and then as y → Y (x) (42) becomes etˆ −16a 7/2 (2a + 1) L(u) ∼ √ J1 (x). 4a + 1 εη (292) Using (292) we then obtain from (40) the expansion of M as y → Y (x) and hence (276) becomes √ ε 4a 2 ε 2a + 1 k∗ √ a −3/2 M(x, y) ∼ k∗ π 4a + 1 2π 2 ·√ etˆ J1 (x) x (4a + 1)e−tˆ 1 , − (2a + 1) −η y → Y (x). (293) We note that√in terms of the variable η in (288), both (289) and (293) are of the order O( ε). Having obtained the local behaviors of (26) and (37) as y → Y (x)(s → a, u → u min (a), t → tˆ(x), τ → tˆ(x)) we set P(x, y) = e (x,y)/ε ¯ P(x, y) (294) and obtain from the main balance equation (8) (1 + x + y) P¯ = e− y 1 1+ ε 2 + (x + ε)e x yy 1 P¯ − ε P¯ y + ε2 P¯ yy + .