This solves the problem completely. In particular if f(x) = f0 = constant
then cn sinh = (12.38)
Thus, from (12.36), we have
u(x,y) = (12.39)
Cylindrical Polar Coordinates
The three-dimensional Laplacian in the cylindrical polar coordinates
(12.40)
(Circular drum) if a circular drum is struck in the center, its vibrations are radially symmetric. We solve the boundary value problem
(12.41)
subject to the boundary conditions
u(r,0) = f(r), r< 1,
ut(r,0)=0, u(1,t)=0, t 0,(12.42)
|u(r,t)| < as r 0.
If we take u(r,t) = T(t) R(r), then Eq (12.41) reduces to the system of ordinary differential equations:
Here, again, k = - 2 yields nontrivial solutions. Then the above system gives two uncoupled ordinary differential equations:
T" + 2T =0,(12.43)
(12.44)
Eg (12.44) is the Bessel equation. The eigenvalues n are the positive zeros of J0 (), with the corresponding eigenfunctions J0 (nr). The solutions of Eq (12.43) are Tn = cos nt. Hence, the solution of the vibrating circular drum struck at the center is given by
u(x,t) = (12.45)
where the coefficients cn are given
(12.46)
Example 12.5
Solve the following boundary value problem
2u(x,y) =0 for 0 < x < l, 0 < y < k,
u(x,0)=0 for 0 x l
u(0,y) =u(l,y)= 0 for 0 y k,
u(x, k) = (l-x) sin (x) for 0 x l.
Figure 12.4 shows the region and the boundary data.
Let u(x,y) = X(x) Y(y) and substitute into Laplace's equation to obtain
Then
X" + X = 0 and Y" - Y = 0.
From the boundary conditions,
u(x,0)=X(x) Y(0)=0, so Y(0) =0.
Figure 12.4.Boundary data given on boundary sides of the rectangle
Similarly, X(0) = X(l) =0.
The problem for X(x) is a familiar one, with eigenvalues n = n22/l2 and eigenfunctions that are nonzero constant multiples of sin(nx/l).
The problem for Y is now
Y" - Y(0) =0.
Solutions of this problem are constant multiples of sinh(ny/l).
For each positive integer n=1,2,....., we now have functions
un(x,y) = bn sin ,
which are harmonic on the rectangle and satisfy the zero boundary conditions on the left, bottom, and right sides of the rectangle. To satisfy the boundary conditions on the side y = k, we must use a superposition
u(x,y) =
Choose the coefficients so that
u(x,k) = = (l – x) sin (x).
This is a Fourier sine expansion of (l –x) sin (x) on [0,l], so we must choose the entire coefficient to be the sine coefficient:
bn sinh
Then
The solution is
u(x.y) =
12.4.4. Black-scholes Model of Financial Engineering Mathematics
Black-Scholes Model represents real world system, solution of which provides the most appropriate value of options on stock, currencies and commodities. In October 1997 Myron Scholes and Robert Merton were awarded the Nobel Prize of Economics for their contribution to the study of financial problems through partial differential equations. Myron Scholes and Fischer Black in 1969, wrote the main equation, known as the Black-Scholes model, which is nothing but the diffusion equation or heat equation (see Section 12.1, (a)).
The following terminology is needed for presentation of the Black-Scholes model.
A forward contract is an agreement where one party promises to buy an asset from another party at some specified time in future and at some specified price. A future contract is very similar to a forward contract. A call option is the right to buy a particular asset for an agreed amount at a specified time in the future. A put option is the right to sell a particular asset for an agreed amount at a specified time in future. The financial instrument on which the option value depends is called underlying asset and is usually denoted by S. Date on which the option can be exercised or the date on which the option ceases to exist or gives the holder of an option any right is denoted by T and is called the expiry date. The amount for which the underlying (asset) can be bought (call) or sold (put) is called the strike or exercise price. It is denoted by E. The two most important factors having great effect on prices of options are the values of the underlying asset S and the time to expiry t. These quantities are variables, meaning that they inevitably change during the life time of the contract. This contrasts with the parameters that affect the price of options, for example volatility (volatility is considered in general, as fixed quantity). Volatility is a very important parameter denoted by and it is a measure of amount of fluctuation in the asset price; that is a measure of randomness. The technical definition of volatility is the annualized standard deviation of the asset returns. It is difficult to estimate volatility and even estimated, in practice it never stays constant and is unpredictable. In many situations it is considered as a variable but while deriving Black-Scholes model it is considered as a fixed parameter. There is another parameter denoted by called the drift rate, the, expected rate or the growth rate of the asset. It is quoted as an annualized growth rate. r is parameter associated with the currency in which the asset is quoted, for example interest rate,
Let V(S,t; ,; E,T;r) denote the option value then it is a solution of the partial differential equation
(12.47)
This equation (12.47) is known as the Black-Scholes model of option pricing. The equation was first written in 1969, but a few years elapsed, with Fisher Black and Myron Scholes justifying the model and it was finally published in 1973. The parameter was eliminated is derivation of the equation. The Black Scholes model can be written in the standard one dimensional equation
(12.48)
by transformation V(S,t) = e x+ u(x,)
where = -
S=ex and t=T -
Separation of variables-Fourier series method for solving (12.47) is given in 12.4.2. We present here elementary solutions of (12.47). For detailed discussion of Black-Scholes model we refer to Wilmott [1998,1994] and for updated literature we cite [Siddiq & Manchanda].
Elementary Solutions of the Diffusion Equation
We shall consider elementary solutions of the one-dimensional diffusion equation.
We begin by considering the expression
u = (12.49)
For this function it is readily seen that
and
showing that the function (12.49) is a solution of the diffusion equation.
It follows immediately that
(12.50)
where is an arbitrary real constant, is also a solution. Furthermore, if the function (x) is bounded for all real values of x, then it is possible that the integral
d(12.51)
is also, in some sense, a solution of the diffusion equation.
It may readily be proved that the integral (12.51) is convergent if t>0 and that the integrals obtained from it by differentiating under the integral sign with respect to x and t are uniformly convergent in the neighbourhood of the point (x,t). The function u(x,t) and its derivatives of all orders therefore exist for t>0, and since the integrand satisfies the one-dimensional diffusion equation, it follows that u (x,t) itself satisfies that equation for t > 0.
=| 1 + 2 + 3 - 4 |
where 1 =
2 =
3 =
4 =
If the function (x) is bounded, we can make each of the integrals 2, 3, 4 as small as we please by taking N to be sufficiently large, and by the continuity of the function we can make the integral 1 as small as we please by taking t sufficiently small. Thus as t 0, u(x,t) (x).
u(x,t) = (12.52)
is the solution of the initial value problem
(12.53)
u(x,0) = (x)
It will be observed that by a simple change of variable we can express the solution (12.52) in the form
u(x,t) = (12.54)
We shall now show how this solution may be modified to obtain the solution of the boundary value problem
(12.55)
u(x,0) = f(x) x > 0 u(0,t) =0 t > 0
If we write (x) =
then the Poisson integral (12.51) assumes the form
u (x,t) = (12.56)
and it is readily verified that this is the solution of the boundary value problem (12.55). We may express the solution (12.56) in the form
u (x,t) =
- (12.57)
The solution of (12.48) can be obtained by taking =1.
12.5 Exercises
1.Derive the one-dimensional heat equation
where k is the diffusivity of the bar under consideration
- Derive the one-dimensional wave equation
where c2 is a constant.
3.Derive the two-dimensional Laplace equation
Find the Fourier series of the following functions
- f(x) = |x|, - x
- f(x) =
- f (x) = x – x2 from x = - to x =
- f(x) = e-x , 0 < x < 2
- f (x) =
Write down the Fourier cosine series and the Fourier sine series of functions given in problems 9 to 10.
- f(x) = x in 0 < x < 2
- f (x) =
- Find half range cosine series for the function
f (x) = x2 , 0 x
12.Express sinx as a cosine series in 0 < x < .
- Find half range sine series for
f (x) =
14.Find the Fourier series of the function
f (x) =
- Find the Fourier series to represent
f (x) = x2 -2, -2 x 2.
16.A homogeneous isotropic rod of length 1 with curved side insulated is initially and uniformly at temperature 1. Its ends are suddenly brought in contact with sinks at temperature 0. How will the temperature profile of the rod relax to 0?
Mathematical formulation is
subject to
(i)u(x,0)=1, 0 < x < 1 (initial condition)
(ii)u(0,t) = 0, t 0 (left boundary condition)
(iii)u(1,t) = 0, t 0 (right boundary condition)
17.Solve the boundary value problem
, 0 < x < 1,
subject to
(iv)u(x,0)=1, 0 < x < 1
(v)u(0,t) = 0, t 0
(vi)ux(1,t) = -u(1,t), t 0
18.A long rod is bent and its ends welded to form a ring. Given an initial temperature profile around the ring, how will this temperature profile evolve over time?
In symbols, solve
, - < x < ,
subject to the initial profile :
(i)u(x,0) = f(x)
(ii)u(-,t) = u(,t)
(iii) (-,t) = (,t).
19.Solve
, 0 < x < 1,
subject to
(i)u(x,0) = f(x)
(ii)ut(x,0) = 0, 0 x 1
(iii)u(0,t) = 0, t 0
(iv)u(1,t) = 0, t 0
20.Solve
2u = 0, 0 <x, y < 1
subject to
(i)u(0,y) = 0, 0 y 1
(ii)u(1,y) = 1, 0 < y < 1
(iii)u(x,0) = 0, 0 x 1
(iv)u(x,1) = 0, 0 x 1
(see Figure 12.4).
Figure 12.4 Steady problem.
In each of problems 21 through 30, write a solution of the boundary value problem.
21. 0 < x < 9, t > 0
u (0,t) = T1, u(9, t) = T2 for t > 0
u (x,0) = x2, 0 x 9
22.
(i) u is finite when t
(ii) = 0 when x=0 and u=0 when x=l for all t
(iii) u=u0 for t=0 and all values of x between 0 and l.
23. 0 < x < 2, t > 0
u (0,t) =u(2,t) =0 , t 0
u(x,0)= 0, (x,0) = g(x), 0 x 2.
24. 0 < x < 3, t 0
u(0,t) = u (3,t) = 0, t > 0
u (x,0) = 0, (x, 0) = x (3-x), 0 x 3
25.0 < x < 2, t > 0
u (0,t) = u (2,t) = 0, t 0
u(x,0) = 0, (x, 0) = 0, 0 x 2
26.; 0 < x < 2, t > 0
u (0,t) = u (2,t) = 0, t 0
u (x,0) =0, (x, 0) = 0, 0 x 2
27., 0 < < l, t > 0
u (0, t) = u (, t) = 0, t 0
u (x, 0) = f(x), (x, 0) = 0,
28.2 u (x, y) = 0, 0 < x < l, 0 < y < m
u (x, 0) = 0, 0 x l
u (0,y) = u(l,y) = 0, 0 y m
u (x, m) = (l – x) sin x, 0 x l
29.2 u (r, ) = 0, 0 r < h, - ,
u (h, ) = f(), -
30.2u(x, y) = 0, x2 + y2 < 16
u(x, y) = x2, x2+y2 = 16
31.Write down the partial differential equation
-r V = 0, S > 0, 0 t T
subject to boundary conditions
V(0, t) = 0,
in the form
subject to appropriate conditions.
1