# Localization of the Black-Scholes equation using transparent by ghkgkyyt

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```									         Transparent boundary conditions
Implementation and numerical results

Localization of the Black-Scholes equation
using transparent boundary conditions

e
Ekaterina Voltchkova and S´bastien Tordeux

e
TSE – Universit´ Toulouse 1

IMT – INSA Toulouse

3d Conference on Numerical Methods in Finance
e
Marne La Vall´e, April 15–17, 2009

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions
Implementation and numerical results

Black-Scholes equation for European option prices

σ2S 2 2
∂τ V (S, τ ) +     ∂ V (S, τ ) + rS ∂S V (S, τ ) − r V (S, τ ) = 0,
2 S
V (S, T ) = (S − K )+ ,        S ∈ (0, +∞), τ ∈ [0, T )
V (S, τ )

K                    S
Asymptotic behavior:
V (S, τ )          S − K e −r (T −τ ) ,              S −→ +∞,
V (S, τ )          0,                                S −→ 0+ .
E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions
Implementation and numerical results

Change of variables

x = ln (S/S0 ),             t = T − τ,            w (x, t) = V (S, τ )

σ2 2                σ2
∂t w (x, t) =      ∂x w (x, t) + r −                       ∂x w (x, t) − r w (x, t).
2                   2
We suppress the 0-order term:

u(x, t) = exp(rt) w (x, t)

Advection diﬀusion equation with constant coeﬃcients:
σ2 2                                                                     σ2
∂t u(x, t) =      ∂ u(x, t) + µ ∂x u(x, t)                      with µ = r −
2 x                                                                      2
x ∈ (−∞, ∞),                    t ∈ (0, T ]
E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions
Implementation and numerical results

Initial condition
Put
V (S, T ) = (K − S)+                  ⇔      u(x, 0) = (K − S0 e x )+ .
Call
V (S, T ) = (S − K )+                 ⇔      u(x, 0) = (S0 e x − K )+ .
Other payoﬀ functions

V (S, T )

K1                        K2                  S

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions
Implementation and numerical results

Initial condition
Put
V (S, T ) = (K − S)+                  ⇔      u(x, 0) = (K − S0 e x )+ .
Call
V (S, T ) = (S − K )+                 ⇔      u(x, 0) = (S0 e x − K )+ .
Other payoﬀ functions

V (S, T )

K1                        K2                  S

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions
Implementation and numerical results

Initial condition
Put
V (S, T ) = (K − S)+                  ⇔      u(x, 0) = (K − S0 e x )+ .
Call
V (S, T ) = (S − K )+                 ⇔      u(x, 0) = (S0 e x − K )+ .
Other payoﬀ functions

V (S, T )

K1            K2                              S

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions
Implementation and numerical results

Localization on a bounded computational domain

x ∈ (−∞, ∞)                          −→          x ∈ (x− , x+ )
Standard approach
Dirichlet or Neumann boundary conditions based on the
asymptotics of the solution.
Example (call):

Dirichlet:      V (S− , τ ) = 0                   ⇔          u(x− , t) = 0
V (S+ , τ ) = S+ − Ke −r (T −τ )                  ⇔          u(x+ , t) = S0 e x+ +rt − K .

Neumann:              ∂S V (S− , τ ) = 0                ⇔           ∂x u(x− , t) = 0
∂S V (S+ , τ ) = 1                ⇔           ∂x u(x+ , t) = S0 e x+ +rt .

E. Voltchkova and S. Tordeux     Transparent boundary conditions for the BS equation
Transparent boundary conditions
Implementation and numerical results

Localization on a bounded computational domain

x ∈ (−∞, ∞)                          −→          x ∈ (x− , x+ )
Standard approach
Dirichlet or Neumann?
How to chose x− and x+ ?
Transparent boundary conditions
Work for any interval (x− , x+ )
(provided it contains the singularities).
Theoretically exact conditions.
Numerically almost exact.
Almost as easy to implement as Neumann or Dirichlet.

E. Voltchkova and S. Tordeux     Transparent boundary conditions for the BS equation
Transparent boundary conditions         Exact conditions
Implementation and numerical results        Approximate conditions

Transparent boundary conditions
t

x+                     x

Idea: the initial condition on {x > x+ } and boundary values
on {x = x+ } deﬁne completely the solution of the PDE in the
gray domain. In particular,

∂x u(x+ , t) = S[u(x+ , t)t>0 ,                        u(x, 0)x>x+ ]

Find the operator S in an explicit form and use it as a
boundary condition when solving for x ≤ x+ .
E. Voltchkova and S. Tordeux         Transparent boundary conditions for the BS equation
Transparent boundary conditions      Exact conditions
Implementation and numerical results     Approximate conditions

Laplace Transform
+∞
u(x, p) =                      u(x, t) exp(−pt)dt
0
Laplace transform of the time derivative:
∂t u(x, p) = p u(x, p) + u(x, 0)
We introduce
uH (x, t) = u(x, t) − ϕ(x, t)
where ϕ is a solution of the PDE with ϕ(x, 0) = u(x, 0).
Important: u(x, 0) has no singularity on x > x+ ,
and ϕ(x, t) has a simple explicit form, e.g.
Call: ϕ(x, t) = S0 e x+rt − K ,                          Put: ϕ(x, t) = 0.
E. Voltchkova and S. Tordeux      Transparent boundary conditions for the BS equation
Transparent boundary conditions    Exact conditions
Implementation and numerical results   Approximate conditions

Solution in the Laplace domain

In the quadrant {x > x+ } we have a homogeneous problem:

σ2 2
∂t uH (x, t) =  ∂ uH (x, t) + µ ∂x uH (x, t),                            for x > x+ ,            t≥0
2 x
uH (x, 0) = 0, for x > x+

Apply the Laplace transform:

σ2 2
puH (x, p) = ∂x uH (x, p) + µ ∂x uH (x, p),                               for x > x+ ;
2

This is an ODE on uH (·, p).

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions     Exact conditions
Implementation and numerical results    Approximate conditions

Solution in the Laplace domain (continued)

σ2 2
puH (x, p) =         ∂ uH (x, p) + µ ∂x uH (x, p),                         for x > x+ ;
2 x
Solving this ODE gives
uH (x, p) = A− e λ− x + A+ e λ+ x
with
µ    |µ|                  2pσ 2
λ± = −          2
± 2           1+
σ    σ                     µ2
To avoid exploding solutions for x → +∞, set A+ = 0 :

µ    |µ|                  2pσ 2
uH (x, p) = A− exp −                        2
+ 2           1+            x .
σ    σ                     µ2

E. Voltchkova and S. Tordeux     Transparent boundary conditions for the BS equation
Transparent boundary conditions     Exact conditions
Implementation and numerical results    Approximate conditions

The Dirichlet-to-Neumann Map
A Dirichlet-to-Neumann operator S maps the Dirichlet value
to the Neumann value of the solution of a PDE

∂x uH = SuH               at      x = x+

In the Laplace domain we have for uH

µ    |µ|                  2pσ 2
uH (x, p) = A exp −                         2
+ 2          1+             x .
σ    σ                     µ2

and consequently

µ    |µ|                  2pσ 2
∂x uH (x+ , p) = −             2
+ 2            1+           uH (x+ , p).
σ    σ                     µ2

E. Voltchkova and S. Tordeux     Transparent boundary conditions for the BS equation
Transparent boundary conditions         Exact conditions
Implementation and numerical results        Approximate conditions

Some useful Laplace transforms

function                                   Laplace transform
1                                                   1
√                                                   √
πt                                                  p
t
1                    u(s)                                        u(p)
√                    √      ds                                    √
π          0         t −s                                         p
t
1             e −a(t−s)                                          u(p)
√              √         u(s)ds                                  √
π   0           t −s                                             p+a
t
1                                e −a(t−s)                     √
√ ∂t + a                          √         u(s)ds                  p + a u(p)
π                       0          t −s

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions      Exact conditions
Implementation and numerical results     Approximate conditions

The exact DTN map: a non local operator
The Dirichlet-to-Neumann map:
a multiplication operator in Laplace domain
an integro-diﬀerential operator in time domain

µ
∂x uH (x+ , t) = −         uH (x+ , t)−
σ2
µ2
2     1
2        µ2                 t
e − 2σ2 (t−s) uH (x+ , s)
− 2          ∂t + 2                               √                ds          =: S+ uH (x+ , t)
σ π            2σ              0                  t −s
Coming back to the non-homogeneous solution we obtain

∂x u(x+ , t) = S+ u(x+ , t) + (∂x ϕ(x+ , t) − S+ ϕ(x+ , t))

E. Voltchkova and S. Tordeux      Transparent boundary conditions for the BS equation
Transparent boundary conditions            Exact conditions
Implementation and numerical results           Approximate conditions

The Dirichlet-to-Neumann map at x = x−
t

x−                                   x

Using the same arguments for {x < x− }, we obtain

∂x u(x− , t) = S− u(x− , t) + (∂x ψ(x− , t) − S− ψ(x− , t))

with
µ2
µ         2                            1
2            µ2               t
e − 2σ2 (t−s) u(s)
S− u(t) := − 2 u(t)+ 2                                     ∂t + 2                         √             ds.
σ        σ π                                       2σ            0               t −s

E. Voltchkova and S. Tordeux            Transparent boundary conditions for the BS equation
Transparent boundary conditions    Exact conditions
Implementation and numerical results   Approximate conditions

Implementation of the exact transparent conditions

PDE: usual Finite Diﬀerence scheme (Cranck-Nicolson).
Boundary: integration by parts to remove the singularity
+ trapezoidal rule.
Numerical experiments

Very good precision. The error is only due to the
discretization of the integral which may be improved.
Drawbacks
Messy formulas to implement.
At each time iteration, boundary conditions involve all
previous values of u on the boundary (not only u n−1 ).

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions        Exact conditions
Implementation and numerical results       Approximate conditions

Approximation of the transparent conditions
First idea: approximate the integro-diﬀerential
Dirichlet-to-Neumann operator by a diﬀerential operator
by approximating its Laplace symbol by a polynomial.

µ   |µ|                      2pσ 2
∂x uH (x+ , p) = −               + 2               1+            uH (x+ , p).
σ2  σ                         µ2

⇓
∂x uH (x+ , p) = − a0 + a1 p + · · · + am p m uH (x+ , p).
⇓
m
∂x uH (x+ , t) = − a0 + a1 ∂t + · · · + am ∂t uH (x+ , t).

E. Voltchkova and S. Tordeux        Transparent boundary conditions for the BS equation
Transparent boundary conditions    Exact conditions
Implementation and numerical results   Approximate conditions

Numerical results for polynomial approximation

We have performed numerical experiments using Taylor
expansion of the symbol at p = 0.

Unfortunately, this approximation does not work well:
The precision is not better than for Dirichlet or Neumann.
Increasing m does not improve the precision.

(see next slide)

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions    Exact conditions
Implementation and numerical results   Approximate conditions

r = 0.05
σ = 0.2
S0 = 100

Put option
K = 100
T =1
Nx = 101
Nt = 100

Dirichlet
Neumann
Taylor

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions    Exact conditions
Implementation and numerical results   Approximate conditions

r = 0.05
σ = 0.2
S0 = 100

Put option
K = 100
T =1
Nx = 101
Nt = 100

Dirichlet
Neumann
Taylor

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions    Exact conditions
Implementation and numerical results   Approximate conditions

The polynomial approximation does not give good results.
The reason is that polynomials are not a good approximation
for 1 + (2σ 2 /µ2 )p ( see graphics ):
the radius of convergence is small
⇒ bad approximation for p > rconv = µ2 /2σ 2 ;
Remark: small p (low frequencies) correspond to large t,
large p (high frequencies) correspond to small t.

Example: if r = 0.05, σ = 0.2, then rconv ≈ 0.01.
Heuristically, the approximation is bad for t < 100!
We have also tried Taylor expansion at a diﬀerent point p > 0.
The results are globally the same.

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions    Exact conditions
Implementation and numerical results   Approximate conditions

Approximation by rational functions
Alternative approach: approximate by rational functions.

µ |µ|                  2pσ 2                              β1           βmr
−     +            1+                   ≈ η0 + η1 p +              +···+
σ2 σ2                   µ2                              p + γ1       p + γmr

This leads to the boundary condition

∂x uH (x+ , t) = η0 uH (x+ , t) + η1 ρ0 (t) + β1 ρ1 (t) + · · · + βmr ρmr (t)

where the auxiliary functions ρi satisfy the following ODEs:

ρ0 (t) = ∂t uH (x+ , t),
∂t ρi (t) + γi ρi (t) = uH (x+ , t),               ρi (0) = 0,                i = 1, . . . , mr

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions    Exact conditions
Implementation and numerical results   Approximate conditions

Approximation by rational functions

√
z
Approximation with 2 interpolation points
3

2

1                                                                                 8

4

z                                             z
1   2          3      4       5      6       7     8     9                            20      40         60

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions    Exact conditions
Implementation and numerical results   Approximate conditions

Approximation by rational functions

√
z
Approximation with 3 interpolation points
3

2

1                                                                                 8

4

z                                             z
1   2          3      4       5      6       7     8     9                            20      40         60

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions    Exact conditions
Implementation and numerical results   Approximate conditions

Approximation by rational functions

√
z
Approximation with 4 interpolation points
3

2

1                                                                                 8

4

z                                             z
1   2          3      4       5      6       7     8     9                            20      40         60

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions    Exact conditions
Implementation and numerical results   Approximate conditions

Approximation by rational functions

√
z
Approximation with 5 interpolation points
3

2

1                                                                                 8

4

z                                             z
1   2          3      4       5      6       7     8     9                            20      40         60

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions    Exact conditions
Implementation and numerical results   Approximate conditions

Approximation by rational functions

√
z
Approximation with 6 interpolation points
3

2

1                                                                                 8

4

z                                             z
1   2          3      4       5      6       7     8     9                            20      40         60

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions
Implementation and numerical results

r = 0.05
σ = 0.2
S0 = 100

Put option
K = 100
T =1
Nx = 101
Nt = 100

Dirichlet
Neumann
Rational

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions
Implementation and numerical results

r = 0.05
σ = 0.2
S0 = 100

Put option
K = 100
T =1
Nx = 101
Nt = 100

Transparent
Rational

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions
Implementation and numerical results

r = 0.05
σ = 0.2
S0 = 100

Put option
K = 100
T =1
Nx = 101
Nt = 100

Transparent
Rational

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions
Implementation and numerical results

r = 0.05
σ = 0.2
S0 = 100

Put option
K = 100
T =1
Nx = 101
Nt = 100

Transparent
Rational

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions
Implementation and numerical results

The implementation is easy due to the introduction of
auxiliary functions. The vector of unknowns is
u n = (λn l , . . . , λn , λn , u0 , u1 , . . . , uN , uN+1 , ρn , ρn , . . . , ρn r )
˜       m              1    0
n    n            n    n
0    1            m

Iterations: A˜n+1 = b(˜n ) with constant matrix A
u        u                                                              show matrix   .
...
n+1              n+1
aui−1 + buin+1 + cui+1 = dui−1 + euin + fui+1 ,
n              n
i = 1, . . . , N
n+1       n+1
uN+1 −    uN−1
˜
= η0 u n+1 + η1 ρn+1 + β1 ρn+1 + · · · + βmr ρn+1 + (Sφ)n+1
mr
0         1                            N
2∆x
n+1    n
uN − uN     1
= (ρn+1 + ρn )0
∆t       2 0
ρn+1 − ρn  1 n+1
i      i
= (uN − γi ρn+1 + uN − γi ρn )
i
n
i
∆t      2

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions
Implementation and numerical results

Conclusion

Transparent conditions are:
suitable for any small computational domain
(containing the singularities)
easy to implement
apply to a wide range of initial conditions
Remark: apply without changes to σ(S) provided that
σ = const outside of (S− , S+ ).
⇒ Should be used instead of Dirichlet and Neumann
Future work
PDE with time-dependent coeﬃcients, PIDE.

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions
Implementation and numerical results

Bibliography

Halpern Laurence, Artiﬁcial boundary conditions for the
linear advection diﬀusion equation. Math. Comp. (86)
Joly Patrick, Pseudo-transparent boundary conditions for
the diﬀusion equation. M2AS (89)
Dubach Eric, Artiﬁcial boundary conditions for diﬀusion
equations: numerical study. JCAM (95)
Halpern L. and Rauch J., Absorbing boundary conditions
for diﬀusion equation, Numer. math. (95)

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions
Implementation and numerical results

Bibliography

Halpern Laurence, Artiﬁcial boundary conditions for the
linear advection diﬀusion equation. Math. Comp. (86)
Joly Patrick, Pseudo-transparent boundary conditions for
the diﬀusion equation. M2AS (89)
Dubach Eric, Artiﬁcial boundary conditions for diﬀusion
equations: numerical study. JCAM (95)
Halpern L. and Rauch J., Absorbing boundary conditions
for diﬀusion equation, Numer. math. (95)

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
Transparent boundary conditions
Implementation and numerical results

Bibliography

Halpern Laurence, Artiﬁcial boundary conditions for the
linear advection diﬀusion equation. Math. Comp. (86)
Joly Patrick, Pseudo-transparent boundary conditions for
the diﬀusion equation. M2AS (89)
Dubach Eric, Artiﬁcial boundary conditions for diﬀusion
equations: numerical study. JCAM (95)
Halpern L. and Rauch J., Absorbing boundary conditions
for diﬀusion equation, Numer. math. (95)

E. Voltchkova and S. Tordeux    Transparent boundary conditions for the BS equation
r = 0.05
σ = 0.2
S0 = 100

Put option
K = 100
T =1
Nx = 101
Nt = 100

Dirichlet
Neumann
Transparent
next

E. Voltchkova and S. Tordeux   Transparent boundary conditions for the BS equation
r = 0.05
σ = 0.2
S0 = 100

Put option
K = 100
T = 10
Nx = 101
Nt = 100

Dirichlet
Neumann
Transparent
back

E. Voltchkova and S. Tordeux   Transparent boundary conditions for the BS equation
Taylor approximation of the symbol
1
μ   |μ|  2pσ 2       2
+ 2 1+ 2
6     σ2  σ     μ
Taylor order 0
5

4

3

2

1

0
p
.1                  .2                .3
−1

−2

−3

E. Voltchkova and S. Tordeux   Transparent boundary conditions for the BS equation
Taylor approximation of the symbol
1
μ   |μ|  2pσ 2       2
+ 2 1+ 2
6     σ2  σ     μ
Taylor order 1
5

4

3

2

1

0
p
.1                  .2                .3
−1

−2

−3

E. Voltchkova and S. Tordeux   Transparent boundary conditions for the BS equation
Taylor approximation of the symbol
1
μ   |μ|  2pσ 2       2
+ 2 1+ 2
6     σ2  σ     μ
Taylor order 2
5

4

3

2

1

0
p
.1                  .2                .3
−1

−2

−3

E. Voltchkova and S. Tordeux   Transparent boundary conditions for the BS equation
Taylor approximation of the symbol
1
μ   |μ|  2pσ 2       2
+ 2 1+ 2
6     σ2  σ     μ
Taylor order 3
5

4

3

2

1

0
p
.1                  .2                .3
−1

−2

−3

E. Voltchkova and S. Tordeux   Transparent boundary conditions for the BS equation
Taylor approximation of the symbol
1
μ   |μ|  2pσ 2       2
+ 2 1+ 2
6     σ2  σ     μ
Taylor order 4
5

4

3

2
back
1

0
p
.1                  .2                .3
−1

−2

−3

E. Voltchkova and S. Tordeux   Transparent boundary conditions for the BS equation
Matrix A of the linear system A˜n+1 = b n
u

1 + Δt γ1
2                          − Δt
2
1+   Δt
2 γ2               − Δt
2
1 + Δt γ3
2            − Δt
2
− Δt
2      1
−2Δxα1 −2Δxα2 −2Δxα3 −2Δxξ1 −1 −2Δxξ0 1
a b      c

a b c
−1 −2Δxη0 1 −2Δxη1 −2Δxβ1 −2Δxβ2 −2Δxβ3
1     − Δt
2
−2Δt
1 + Δt δ1
2
− Δt
2                     1 + Δt δ2
2
−2Δt
1 + Δt δ3
2

E. Voltchkova and S. Tordeux     Transparent boundary conditions for the BS equation

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