Let the trial solution be:
φ (β) = φ « I, C 25 st, vt, τ ) = exp [p (τ ; ζ 1, ζ 2) + q (τ ; ζ 1 ,ζ 2) vt + ιζ 1 st],
where p (τ; ζ 1, ζ2) and q (τ; ζ 1, ζ2) are complex-valued functions.
Obviously:
Φτ∕Φ = pτ + qτvt, Φs∕Φ = zζ 1, Ψ∙∕Φ = q,
Φss∕Φ = (iC 1)2 , Φυυ∕Φ = q2, Φsυ∕Φ = iζ χq.
Plugging these into (7), factoring out Φ, and simplifying:
0 = [-pτ + iζ 1 (r - δ) + aq] + vt
-qτ- Iiζ 1 - βq +1(iC 1)2 +172q2 + iC 1p7q
Since p.d.e. (7) holds for all values of vt, it must be the case that functions p and q are the solution to
the system of ordinary differential equations:
qτ = ∣ [(iC 1)2 - iC 1] + [iC 1P7 - β∖ q +|72q2i,
Pτ = iC 1 (r - δ) + aq.
These equations are similar to the ones in Epps (2004b). However, the initial conditions are determined
by:
Φ (C 1, C25 sτ, vτ, 0) = exp [iC2vτ + iC 1^τ].
τherefore, q (0; c 1, c2) = iC2 and P (0; c 1, c2) = 0.
Solutions were obtained with Maple.3
Case 1. 7 = 0.
Let:
A ≡ A (C 1)= 72 (1 - P2) C2 + (72 - ∣P7β)iC 1 + β2,
r - r ∕λ Λ ʌ _ P7iC 1 - β - √a (c 1) + 72iC2
R ≡ R (Ci, C ) . .
P7iC 1 - β + √a (c 1) + 72iC2
3In the most interesting case, 7 ≠ 0, Maple gives solution either in terms of trigonometric and inverse trigonometric functions
or in terms of hyperbolic functions with a non-closed-form expression for p (τ; ∙). The “trigonometric” solution is perfectly
acceptable, but is not convenient to program. I started with the “hyperbolic” solution and derived a closed-form expression for
p (T; •).
More intriguing information
1. A Pure Test for the Elasticity of Yield Spreads2. The name is absent
3. L'organisation en réseau comme forme « indéterminée »
4. The name is absent
5. Intertemporal Risk Management Decisions of Farmers under Preference, Market, and Policy Dynamics
6. Regional dynamics in mountain areas and the need for integrated policies
7. Modellgestützte Politikberatung im Naturschutz: Zur „optimalen“ Flächennutzung in der Agrarlandschaft des Biosphärenreservates „Mittlere Elbe“
8. The name is absent
9. The name is absent
10. The name is absent