236
13 Solutions for Selected Exercises
We have to keep in mind that the year is divided into two segments and the annual
growth rate ρ has to be divided by two and the volatility σ by
√
2. Using these
values, we calculate the stock values after two segments, as shown in the Fig. 13.1.
We then compare the values with the strike price K and find that only the upper-most
branch contributes, because f 1 = 1/ f 2 and the payoff function for a call option is
max(S T − K , 0). Thus we find that, after applying the discount factor e
−ρT
, the value
of the call option at the start of the contract is c 0 = e
−ρT p
2
( f
2
1 − 1)S 0 = 0.129S 0 ,
where T is one year, such that ρT = 0.05. See ex4_1.m from the ESM for the
numerical details.
Exercise 4.2
Since G(r ) has radial symmetry, we only need the radial part of the Laplace operator
in three dimensions r . Using spherical coordinates in three dimensions, it is given
by
r G(r ) =
1
r 2
∂
∂r
r
2 ∂G(r )
∂r
(13.5)
For non-zero r , inserting G(r ) = 1/4πr immediately shows that bracket becomes a
constant, such that the “outer” derivative results in zero. In order to justify the 4π
in the denominator, we employ Gauss’ theorem and write the Laplace operator as
the scalar product of the divergence ∇ and the gradient ∇ applied to the function
G, which is related to the field E = −∇G. Integrating the field over a small sphere
with radius ε around the origin yields
V (ε)
(∇ · ∇)GdV = −
∂ V (ε)
Ed S =
∂ V (ε)
1
4πr 2 d S
(13.6)
where V (ε) is the volume of the small sphere and ∂ V (ε) its surface. In the second
equality we insert the radial gradient of E = −∇G = 1/4πr
2 . Since we have r = ε
on the surface of the sphere and d S = 4πε
2 we see that the integral over G is
unity. Summarily, the value is zero, except at the origin, where the integral amounts
to unity.
Exercise 4.3
The response of the system to an initial velocity perturbation v 0 can be calculated by
matching the initial values of the homogeneous solution of the differential equation
to the initial velocity. The trial solution x = Ae
iωt leads to the following equation to
determine the eigenfrequencies ω
ω
2
− 2iαω − ω
2
0 = 0
(13.7)
which has the solutions ω ± = iα ±
ω
2
0 − α 2 . We assume that the damping is weak
with α < ω 0 . The general solution is then given by
13 Solutions for Selected Exercises
We have to keep in mind that the year is divided into two segments and the annual
growth rate ρ has to be divided by two and the volatility σ by
√
2. Using these
values, we calculate the stock values after two segments, as shown in the Fig. 13.1.
We then compare the values with the strike price K and find that only the upper-most
branch contributes, because f 1 = 1/ f 2 and the payoff function for a call option is
max(S T − K , 0). Thus we find that, after applying the discount factor e
−ρT
, the value
of the call option at the start of the contract is c 0 = e
−ρT p
2
( f
2
1 − 1)S 0 = 0.129S 0 ,
where T is one year, such that ρT = 0.05. See ex4_1.m from the ESM for the
numerical details.
Exercise 4.2
Since G(r ) has radial symmetry, we only need the radial part of the Laplace operator
in three dimensions r . Using spherical coordinates in three dimensions, it is given
by
r G(r ) =
1
r 2
∂
∂r
r
2 ∂G(r )
∂r
(13.5)
For non-zero r , inserting G(r ) = 1/4πr immediately shows that bracket becomes a
constant, such that the “outer” derivative results in zero. In order to justify the 4π
in the denominator, we employ Gauss’ theorem and write the Laplace operator as
the scalar product of the divergence ∇ and the gradient ∇ applied to the function
G, which is related to the field E = −∇G. Integrating the field over a small sphere
with radius ε around the origin yields
V (ε)
(∇ · ∇)GdV = −
∂ V (ε)
Ed S =
∂ V (ε)
1
4πr 2 d S
(13.6)
where V (ε) is the volume of the small sphere and ∂ V (ε) its surface. In the second
equality we insert the radial gradient of E = −∇G = 1/4πr
2 . Since we have r = ε
on the surface of the sphere and d S = 4πε
2 we see that the integral over G is
unity. Summarily, the value is zero, except at the origin, where the integral amounts
to unity.
Exercise 4.3
The response of the system to an initial velocity perturbation v 0 can be calculated by
matching the initial values of the homogeneous solution of the differential equation
to the initial velocity. The trial solution x = Ae
iωt leads to the following equation to
determine the eigenfrequencies ω
ω
2
− 2iαω − ω
2
0 = 0
(13.7)
which has the solutions ω ± = iα ±
ω
2
0 − α 2 . We assume that the damping is weak
with α < ω 0 . The general solution is then given by