By Beattie J. A.

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Additional info for A New Equation of State for Fluids IV. An Equation Expressing the volume as an Explicit Function of the Pressure and Temperature

Example text

The solution is p∗ (t) = e−tM h; and hence T p∗ (t)T = hT X−1 (t), since (e−tM )T = e−tM = X−1 (t). T 2. 3 that hT X−1 (t)N α∗ (t) = max{hT X−1 (t)N a} a∈A Since p∗ (t)T = hT X−1 (t), this means that p∗ (t)T (M x∗ (t) + N α∗ (t)) = max{p∗ (t)T (M x∗ (t) + N a)}. a∈A 3. Finally, we observe that according to the deﬁnition of the Hamiltonian H, the dynamical equations for x∗ (·), p∗ (·) take the form (ODE) and (ADJ), as stated in the Theorem. 3 EXAMPLES 35 EXAMPLE 1: ROCKET RAILROAD CAR. 2. We have ˙ x(t) = (ODE) 0 0 1 x(t) + 0 0 α(t) 1 M for x(t) = N x1 (t) x2 (t) , A = [−1, 1].

We will as above suppose that R has the explicit representation R = {x ∈ Rn | g(x) ≤ 0} for a given function g(·) : Rn → R. DEFINITION. It will be convenient to introduce the quantity c(x, a) := ∇g(x) · f (x, a). Notice that if x(t) ∈ ∂R for times s0 ≤ t ≤ s1 , then c(x(t), α(t)) ≡ 0 (s0 ≤ t ≤ s1 ). This is so since f is then tangent to ∂R, whereas ∇g is perpendicular. 6 (MAXIMUM PRINCIPLE FOR STATE CONSTRAINTS). Let α∗ (·), x∗ (·) solve the control theory problem above. Suppose also that x∗ (t) ∈ ∂R for s0 ≤ t ≤ s1 .

We modify the problem above by introducing the region R := {x ∈ Rn | g(x) ≤ 0}, determined by some given function g : Rn → R. Suppose x∗ ∈ R and f (x∗ ) = maxx∈R f (x). We would like a characterization of x∗ in terms of the gradients of f and g. Case 1: x∗ lies in the interior of R. Then the constraint is inactive, and so ∇f (x∗ ) = 0. 3) gradient of f X* R figure 1 Case 2: x∗ lies on ∂R. We look at the direction of the vector ∇f (x∗ ). A geometric picture like Figure 1 is impossible; for if it were so, then f (y ∗ ) would be greater that f (x∗ ) for some other point y ∗ ∈ ∂R.