Remark 2 Sample Clauses

Remark 2. Disregarding the right hand side of (2.1.0.19a) for now, we see that in terms of derivatives of η, the left hand side looks like a set of wave equations. However, the presence of powers of w in front of both the velocity and pressure terms, as well as the way A and J linearise, will present additional difficulties. This is discussed in more detail in Section 2.2.2.

Remark 2. 2.1 Integrating (2.6) and setting u = t we see that t = ∫ t κ t −s √V dW 0( ) + 0 ( )

Remark 2. One may ask why we choose to privilege the renormalised enthalpy w over the initial density ˚ρ in our formulation of the system. The reason is that w, as shown in (2.1.0.8), is a quantity that exhibits the same behaviour as dictated by the physical vacuum condition on the initial sound speed ˚cs, and therefore allows us to more clearly investigate the nature of the degeneracy at the boundary. Moving on to (2.0.0.1d), note that since ρ has compact support, we can solve the Poisson’s equation using the fundamental solution: φ(t, x) = −c ∫ Ω(t) ρ(t, z) dz |x − z| By applying the flow map η to both the x and z variables, we can obtain an explicit solution for ψ: ψ(t, x) = −c ∫ Ω = −c Ω ρ(t, η(t, z)) J (t, z)dz = c ∫− |η(t, x) − η(t, z)|

Remark 2. 1. If χ = 0 this is the situation where there is no offsetting (current situation). In the following, we assume that χ ‰ 1, in coherence with the regulation14. 14The revised ETS state aid guidelines for the period after 2021 include hydrogen as a sub-sector at risk of carbon leakage. As such, it can benefit from a unit offsetting of up to 75% of the indirect costs of its emissions.

Remark 2. We defined T^ = argmin||B − U^ Λ^ 2 U^ T ||2, with B that satisfies all the constraints in Eq.

Remark 2. ■ The par swap rate appears as the coupon rate (or yield to maturity) of a bond quoting at par. ■ This is directly deduced from equation (1) above: