WHD Problem 3.6b

Find the most general solution of the Black-Scholes equation that has the special form V=A(t)B(S).

(∂V/∂t) + ½ σ² S² (∂²V/∂S²) + r S (∂V/∂S) - rV = 0

V = A(t) B(s)

(∂V/∂t) = B (dA/dt)

(∂V/∂S) = A (dB/dS)

(∂²V/∂S²) = A (d²B/dS²)

B (dA/dt) + ½ σ² S² A (d²B/dS²) + r S A (dB/dS) - r A B = 0

(1/A) (dA/dt) + ½ σ² S² (1/B) (d²B/dS²) + r S (1/B) (dB/dS) - r = 0

½ σ² S² (1/B) (d²B/dS²) + r S (1/B) (dB/dS) - r = - (1/A) (dA/dt)

Since the left-hand side depends only on S and the right-hand side depends only on t, the only way the equality can hold is if both sides are equal to a constant K.

- (1/A) (dA/dt) = K

A(t) = c e^-Kt

½ σ² S² (1/B) (d²B/dS²) + r S (1/B) (dB/dS) - r = K

½ σ² S² (d²B/dS²) + r S (dB/dS) - (r + K) B = 0

S² (d²B/dS²) + (2r/σ²) S (dB/dS) - (2/σ²) (r+K) B = 0

This equation proceeds as in the previous exercise.

λ² + (2r/σ² -1)λ + (-2/σ²)(r+K) = 0

λ = { (1 - 2r/σ²) ± [ (2r/σ² - 1)² + (8/σ²)(r+K) ]^½ } / 2

λ = { (1 - 2r/σ²) ± [ (2r/σ² - 1)² + (8/σ²)(r+K) ]^½ } / 2

λ = { (1 - 2r/σ²) ± [ (2r/σ² + 1)² + (8K/σ²) ]^½ } / 2

There are three cases depending on the roots: two real roots, one real root, two complex roots. And to be explicit, we should note that the case of two real roots can come about either with a real value of K (which leaves the form of A(t) above unchanged) or with a complex value of K (which would require rewriting A as we'll see in a minute).

Case 1: λ₁ and λ₂ are distinct real roots.

B(S) = c₁ S^λ₁ + c₂ S^λ₂

V(S,t) = (c₁ S^λ₁ + c₂ S^λ₂) e^-Kt

Case 1A: K = 0
(Special scenario under case 1)

This reduces to the special case where there is no time dependence V = B(S)

V = c₁ S + c₂ S ^(-2r/σ2)

Case 2: λ₁ = λ₂ = λ is the only real root.

B(S) = S^λ (c₁ + c₂ ln S)

V(S,t) = S^λ (c₁ + c₂ ln S) e^-Kt

Case 3: λ₁ = a+ib and λ₂ = a-ib are complex roots (with K real).

B(S) = S^a [ c₁ cos (b ln S) + c₂ sin (b ln S) ]

V(S,t) = S^a [ c₁ cos (b ln S) + c₂ sin (b ln S) ] e^-Kt

Case 3A: λ₁ = a+ib and λ₂ = a-ib are complex roots (with complex K = c + id).
(This is not strictly separate from case 3; it's simply a matter of expanding e^-Kt for complex K.)

B(S) = S^a [ c₁ cos (b ln S) + c₂ sin (b ln S) ]

V(S,t) = S^a [ c₁ cos (b ln S) + c₂ sin (b ln S) ] e^-ct (cos dt - i sin dt)

Note that all of these solutions (except Case 1A) have too many degrees of freedom, due to the freedom of choice in choosing K. Normally boundary conditions would restrict the universe of valid values for K, but the problem as stated did not provide any such conditions.