Another place that free-boundary problems appear is in option pricing,
particularly American options.
An American call option is the right to buy a stock at the exercise price
K any time up to expiry date t = T . In contrast, a European call option is
the right to buy a stock at the exercise price K at the expiry date t = T .
The option price C satisfies the Black-Scholes PDE
−Ct =
S 2 σ2
Css + S(r − D)Cs − rS
2
where S is the stock price (which follows geometric brownian motion), σ
is the volatility, r is the interest rate, and D is the (continuous) dividend
rate. In the European case, C satisfies this equation for all S > 0, with the
additional conditions:
C(S, T ) =
(
0
S<K
S−K S >K
This condition essentially is a statement that the option is worthless if S <
K; that is, we would lose money by buying the stock for price K which is
larger than the market value S. For S − K > 0, the price of the option is
set at S − K to eliminate chance of arbitrage, that is, profit with no risk.
For the American option, the price of the option C(S, t) satisfies the
Black-Scholes PDE only for values of S < S f (t) where Sf (t) is a free
boundary in the problem that must be determined. For S > S f (t), we
set C(S, t) = S − K. If we allowed C(S, t) to take the value given by the
Black-Scholes PDE, then there would be an arbitrage opportunity. That
is, we could buy the option for price C(S, t) < S − K, as follows from the
behavior of the solution for the Black-Scholes PDE. Then we could exercise
the option for and buy the stock for price K and sell it for the market value
of S. Then our profit is
Profit = S − K − C(S, t) > 0
taking C(S, t) as the solution fot the Black-Scholes PDE. Then we have an
opportunity to make a profit at no risk.
In order to eliminate this possibility, the price of the American option
satisfies the following equation:
S 2 σ2
Css + S(r − D)Cs − rS for S < Sf (t)
2
C = S − K for S > Sf (t)
−Ct =
11
Here Sf (t) is a free boundary which will have to be determined as part of the
problems. In addition, we have the same boundary and initial conditions as
the European option. We also need conditions at the free boundary.
C(Sf , t) = Sf − K,
CS (Sf , t) = 1
and an initial condition for the free boundary. This follows from the balance
of the yield rate from dividends and the rate of interest from the stock’s
worth:
DSf (T ) = rK. ⇒ Sf (T ) = rK/D
In order to solve the equation for C we first convert it into a convectiondiffusion equation with constant coefficients.
x = log(S/K), Kex = S
σ 2 (T − t)
= τ
2
C(S, t) = S − K + Ku(x, τ )
Then the derivatives are
SCS = (
K
ux + 1)S
S
S 2 CSS = (
K
K
uxx − 2 ux )S 2
2
S
S
Substituting in the equation for u yields:
σ2
Kuτ
2
=
σ2
Kuxx − Kux + (r − D)Kux − rKu + S(r − D) − r(S − K)
2
If we divide by K and σ 2 /2 we get
uτ
= uxx − (ρ − ν − 1)ux + −ρu + ρ − νex
ρ =
2r
,
σ2
ν=
2
D
σ2
| {z }
−f (x)
In addition, the conditions for u are then:
C(Sf , t) = Sf − K = Sf − K + Ku(xf , t) ⇒ u(xf , τ ) = 0
∂xf
CS (Sf , t) = 1 = 1 + Kux
⇒ ux (xf , τ ) = 0
∂S
12
As τ → 0, xf = log(r/D) = x0 and the initial conditions on u are:
u(x, 0) =
(
(−S + K)/K = 1 − ex x < 0
0
x>0
We can’t solve this problem exactly, but we can look at asymptotic
behavior.
√
Near expiry: i.e. near t = T , which is τ near 0 we might expect τ
behavior for xf , similar to the Stefan problem.
Then we would expect something like
√
xf = x 0 + c τ
In order to confirm this, we need to consider local variables in the region
near expiry:
x = x0 + y
τ = 2 τ
u(x, τ ) = v(y, τ
Note the relative scalings and 2 for space and time respectively. Then
the equation for v(y, τ ) is
−2 vτ
vτ
= −2 vyy + (ρ − ν − 1)−1 vy − ρv − f (x0 + y)
= vyy + (ρ − ν − 1)vy − 2 ρv − f (x0 + y)
If we write out f (x0 + y)
νex − ρ = νey+x0 − ρ
= νex0 (1 + y + 2 y 2 /2 + . . .) − ρ
= νeln ρ/ν (1 + y + 2 y 2 /2 + . . .) − ρ
= ρ(1 + y + 2 y 2 /2 + . . .) − ρ = yρ + . . .
This shows that the function f (x) = O(). Then the leading order equation
is vτ = vyy + homogeneous initial condition and boundary conditions. This
gives v = 0 to leading order, which suggests that we also need to scale v
with . The behavior of the forcing term in the equation suggests what this
scaling should be, i.e. from the term 2 f (x) we expect
u(x, τ ) = 3 v(y, τ )
xf − x 0
also
= yf
13
Substituting yields:
3 vτ
= 3 vyy + 4 (ρ − ν − 1)vy − ρ5 v − 3 ρy
v(yf , τ ) = 0
vy (yf , τ ) = 0
yf (0) = 0
Now we have a PDE for which, to leading order, we can find a similarity
solution:
√
v ∼ τ λ V (ξ) ξ = cy/ τ
Substituting in the leading order equation for v:
λτ λ−1 V + τ λ
−cy
2τ 3/2
Vξ = τ λ Vξξ
c2
− ρτ 1/2 ξ/c
τ
In order to have a similarity solution of this form we need
λ − 1 = 1/2
So v ∼ τ 3/2 V (ξ) with
ξ
3
V + − Vξ = Vξξ c2 − ρξ/c
2
2
So we take c = 1 for convenience and solve the ODE for V . Particular solution: Vp = −ρξ. One can look for polynomial solutions for the homogeneous
solution, and get the second solution by the reduction of order.
w1 = ξ 3 + 6ξ
w2 = (ξ 2 + 4)e−ξ
2 /4
1
+ (ξ 2 + 6ξ)
2
Z
ξ
e−ξ
2 /4
dξ
−∞
Then
V = Vp + Aw1 + Bw2
(1)
We set A = 0 to eliminate terms that violate the condition of the asymptotic
behavior as ξ → −∞. This follows from considering
lim C = 0 ⇒ lim u = 1 − ex
S→0
x→−∞
14
Consider the behavior of this solution near x = 0, which we can view as the
“outer” behavior of the solution (away from the free boundary). This outer
solution has linear behavior, so we can conclude that the solution V should
not grow faster than linearly as ξ → −∞, that is, as τ → 0 for x < x 0 .
Now we use the boundary conditions (i.e. on the free boundary), to get
Bw20 (y0 ) = ρ
Bw2 (y0 ) = ρy0
(2)
√
Here we have taken ξ = y0 for y = y0 τ . Note that since the conditions at
the free boundary are homogeneous, this allows us to write things directly
in terms of y0 rather than y(τ ). Then we get
Bw2 (y0 ) = ρy0
(3)
which can then be used in the second condition to get a transcendental
equation for y0 :
2
y03 ey0 /4
Z
ξ0
e−ξ
2 /4
−∞
dξ = 2(2 − y02 )
Solving for y0 we then have the expression for the optimal exercise boundary:
Sf
Sf
√
∼ Kexf ∼ Kex0 +y0 τ
√
√
2
2
∼ Kex0 +y0 (T −t)σ /2 ∼ Kr/Dey0 (T −t)σ /2
Note that as D → 0, Sf → 0, the location of the free boundary goes
to ∞, that is, there is no free boundary. This could have been guessed
from looking at the behavior of the European option price (solution of the
Black-Scholes PDE).
What about r < D?
In that case we find that the solution to the transcendental equation
y0 → ∞. That is, we can not find a finite solution for the transcendental
equation for y0 .
What is different about this case? Recall the initial condition for x f :
Sf = Kr/D
xf = log r/D
end condition
initial conditon
Notes that for r < D this gives a negative value for x f , or a value less than
K for Sf when r < D. This gives a contradiction to the end condition, so
for r < D, we take the initial condition:
Sf = K ⇒ x f = 0
15
(4)
Even when the correct initial condition is applied for the free boundary
we still have a transcendental equation for y 0 which does not have a finite
solution:
What happens? When we reconsider the rescaled equation, we see that
the inhomogeneous term has a different form. Recall that this term comes
from the transformation C = S − K + Ku, which eventuallly leads to the
inhomogeneous term in the equation for u which is f (x) = ρ − νe x . When
we look for √
an expansion near expiry, we look near the initial condition
x ∼ x0 + y0 τ but in this case x0 = 0 so that
f (x) ∼ ρ − ν
(5)
Thus the scaling of u in terms of v has to be different and it leads to a
different transcendental equation.
What is the implication of finding that there is no finite solution for the
transcendental equation? Something wrong with a previous ansatz - scaling
for u? Similarity solution? Ansatz for y f ?
We reconsider this problem in a similar framework, but in this case it is
convenient to use a different transformation from C → u
C = Ke−ρτ u(x, τ ) + S − K
This has the effect of changing the form of f (x) in the equation for u:
∂u
∂2u
∂u
+ (ρ − ν − 1)
=
+ eρτ (ρ − νex )
∂τ
∂x2
∂x
We will perform the analysis for the put option (right to sell, rather than
right to buy) with price P , which satisfies a similar problem:
∂P
1
∂2P
∂P
+ σ 2 S 2 2 + (r − D) S
− rP = 0 ,
∂t
2
∂S
∂S
∂P
= −1 ,
P = K − Sf ,
∂S
P ∼0,
P (S, TF ) = max (K − S, 0) ,
in 0 < t < TF , S > Sf (t)
at S = Sf (t)
as S → +∞
(
K
if D ≤ r ,
−
r
Sf (TF ) =
K if D ≥ r .
D
Again we use dimensionless variables:
S = Kex ,
t = TF −
2
τ,
σ2
P = Ke−ρτ u(x, τ )+K−S ,
Sf = Kexf .
(6)
16
So that the equation for u is:
∂2u
∂p
∂u
=
+ (ρ − ν − 1)
+ eρτ (νex − ρ)
2
∂τ
∂x
∂x
∂u
u=
=0
∂x
u ∼ eρτ (ex − 1)
x
xf (0) = x0 =
(
u = max (e − 1, 0)
in 0 < τ <
σ2
TF , xf (τ ) < x
2
at x = xf (τ ) ,
as x → +∞ ,
at τ = 0 ,
0
if ν ≤ ρ ,
− log (ν/ρ) if ν > ρ .
To analyze the small-time behavior (near expiry) of the system (7)–(7)
we set
τ = θT .
(7)
Here T = O(1) and θ is an artificial small parameter. Then the problem for
u(x, t) becomes for D < r:
"
#
∂2u
∂u
= θ
+ (ρ − ν − 1)
+ eθρT (νex − ρ)
2
∂x
∂x
∂
u(xf , T ) =
u(xf , T ) = 0, u(x, T ) = eθρT (ex − 1), as x → ∞
∂x
(
0,
x < 0,
u(x, 0) =
ex − 1 , x ≥ 0 .
∂u
∂T
In the limit θ → 0 we obtain the following three layer structure:
An expansion in regular powers of θ gives the outer expansion valid for
x > 0 and x = O(1),
u = ex − 1 + θρ(ex − 1)T + O(θ 2 ) , x > 0.
(8)
To satisfy the boundary conditions we introduce a local expansion in the
region x = xf (τ )+θz with z = O(1). There the problem for p(x, τ ) = p(z, T )
is given by
θ
∂u
dxf ∂u
−
∂T
dτ ∂z
=
u(xf , T ) =
∂u
∂2u
+ θ(ρ − ν − 1)
+ θ 2 eθρT (νexf +θz − ρ)
2
∂z
∂z
∂
u(xf , T ) = 0 ,
∂z
17
which yields
u = O(θ 2 ) .
Since the outer expansion breaks down when x = O(θ 1/2 ), we require
an inn er region with the scaling x = θ 1/2 X that bridges between the outer
region and a region near xf . Where X = O(1) we write
u(x, τ ) = θ 1/2 P0 (X, T ) + θP1 (X, T ) + θ 3/2 P2 (X, T ) + O(θ 2 ).
(9)
This leads to the following sequence of problems:
(i) First for P0 we have
∂ 2 P0
∂P0
=
in − ∞ < X < +∞ , T > 0 ,
∂T
∂X 2
P0 (X, 0) = max(X, 0) , as X → −∞, P0 → 0
, as X → +∞, P0 ∼ X .
√
√
This has the solution P0 = T h0 (ζ) where ζ = X/2 T and h0 (ζ) is given
by
1
2
h0 (ζ) = √ e−ζ + ζerfc(−ζ) ,
π
(ii) For P1 we obtain
∂P0
∂P1
∂ 2 P1
+ (ρ − ν − 1)
=
+ (ν − ρ)
2
∂T
∂X
∂X
(
1 2
2X , X ≥ 0 ,
P1 (X, 0) =
0,
X < 0,
∂P1
→ 0,
∂X
in − ∞ < X < +∞ , T > 0 ,
as X → −∞,
1
as X → +∞, P1 ∼ X 2 .
2
This has the solution P1 = T h1 (ζ) where h1 (ζ) satisfies,
in − ∞ < ζ < +∞ ,
as ζ → −∞, h01 → 0,
h001 + 2ζh01 − 4h1 = 2(1 − ν − ρ)h00 + 4(ν − ρ) ,
as ζ → +∞, h1 ∼ 2ζ .
(iii) Finally for P2 we have
∂ 2 P2
∂P1
∂P2
=
+ (ρ − ν − 1)
+νX
2
∂X
∂X
∂T
1 3
∂P2
P2 (X, 0) = max
X , 0 , as X → −∞,
→ 0,
6
∂X
1
as X → +∞, P2 ∼ X 3 + ρXT .
6
18
in − ∞ < X < +∞ , T > 0 ,
This has the solution P2 = T 3/2 h2 (ζ) where h2 (ζ) satisfies
h002 + 2ηh02 − 6h2 = 2(1 − ν − ρ)h01 − 8νζ ,
4
as ζ → +∞, h2 ∼ ζ 3 + 2ρζ ,
3
in − ∞ < ζ < +∞ ,
as ζ → −∞, h02 → 0,
The forms of h1 and h2 as X → −∞, needed for matching purposes, are
1
e−ζ
h1 ∼ ν − ρ + √ (1 + ν − ρ)
2 π
ζ
2
1
h2 ∼ 2νζ + √ (1 + ν − ρ)2 e−ζ
4 π
2
as ζ → −∞ ,
as ζ → −∞ .
This expansion breaks down as x → 0+ , and it begins to fail when
√
x = O( τ ). Then u is represented instead by the inner expansion
u(x, τ ) ∼ τ 1/2 h0
x
√
2 τ
+ τ h1
x
√
2 τ
+ τ 3/2 h2
x
√
2 τ
+ O(τ 2 ),
√
x = O( τ ).
The terms in the expansion have the following asymptotic behaviors as ζ →
−∞:
h0 (ζ) ∼
2
1
√
e−ζ
2 π
h
1
ζ2
i
+ O(ζ −4 ) ,
h2 (ζ) ∼ 2νζ + O e−ζ
2
.
h1 (ζ) ∼ (ν − ρ) + O ζ −1 e−ζ
2
,
(10)
Near the moving boundary, where x = x f (τ ) + O(τ ), u has the behavior
u(x, τ ) ∼ O(τ 2 ) ,
x − xf (τ ) = O(τ ).
(11)
Now we match the values of uτ by using the relationship at x = xf :
du
= uτ + x0f (τ )ux
dτ
and the fact that u = O(τ 2 ) near the moving boundary. Then we must
have that uτ = 0. This gives us a condition for the matching in the limit
√
xf (τ )/ τ → −∞. We use the asymptotic forms for the h j ’s and we obtain
the transcendental equation
2
1
√
e−xf /4τ + ν − ρ ∼ 0,
2 πτ
2
1
√
e−xf /4τ + νxf ∼ 0,
2 πτ
19
ν < ρ,
ν = ρ.
This gives
2
e−α (τ )
√
π
∼ 2(ρ − ν)τ 1/2 ,
ν < ρ,
2
e−α (τ )
√
π
∼ 4ντ α(τ ),
ν = ρ.
which gives the solution for α(τ ) is
α2 (τ ) ∼ ln
1
√ ,
2(ρ − ν) πτ
ν < ρ.
Then Sf (t) is given by
√ 1/2
,
Sf (t) = Kexf (τ ) ∼ K + Kxf (τ ) ∼ K − 2Kτ 1/2 ln 1/2(ρ − ν) πτ
1
Sf (t) = Kexf (τ ) ∼ K + Kxf (τ ) ∼ K − 2Kτ 1/2 ln √
4 πντ
20
1/2
,
D = r.
D < r,