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; •).
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