Let us start with some calculus facts that you should be familiar with.3 A functionfWR!Risdifferentiableif
df.x/
dx WD lim
h!0
f.xCh/f.x/
h
exists and is finite. We callf0.x/WDdf.x/=dxthederivativeoff (with respect to the variablex). Iff0is continuous, we callf continuously differentiable.
A functionfWR!RisHửlder continuous with Hửlder exponent˛atxif there is a constantCand an exponent˛ 2 .0; 1such thatjf.xCh/f.x/j Ch˛ for allh> 0. If˛D1, we say that the function isLipschitz continuous. The following result is non-trivial and very useful:
Theorem A.11 (Rademacher Theorem) Every Lipschitz continuous function is a.e. differentiable.
2Indeed, this is the form in which the inequality is presented in mathematical text books. In finance we more frequently deal with concave functions.
3If this is all new for you, we advise you to study your favorite calculus textbook first.
Next we consider functions with more than one variable. As an example take the elevationh of a hilly area that depends on two variables, say, latitude x and longitudey. We can take partial derivativesof the elevation functionh, e.g., we can consider its slope in the direction ofxwhich we denote by@h.x;y/=@x. This means that we cut out a thin slice of the hills along thex-direction and look at the derivative of the elevation on this slice (which is just a function in one variable). Of course we can generalize this idea to arbitrarily many dimensions. An example for a vector composed of partial derivatives is thegradientof a functionhW.x1; : : : ;xn/! h.x1; : : : ;xn/2Rwhich is defined by
rh.x/WD 0 BB
@
@h.x/
@x1
@h:::.x/
@xn
1 CC A;
wherexD.x1; : : : ;xn/and sometimesDhis written instead ofrh.
In the following we need to extend the real numbersRtocomplex numbersC that can be written as a linear combination of a real and animaginary number, i.e., zDxCiy2Cwithx;y2Randi2 D 1. For complex numbers particularly the Euler formula holds:
ei DcosCisin:
We can now define aFourier transformation: it maps a functionfWR!Rto its Fourier transformFfWR!Cand is defined by
.Ff/./WD p1 2
Z C1
1 eitf.t/dt:
The Fourier transformation describes in a certain way the frequency distribution off. The most natural application is in acoustics wherefdescribes the oscillation (e.g., of the air) and its Fourier transform corresponds to the distribution of frequencies, i.e., its sound. A sine function would lead to a Dirac distribution as Fourier transform:
in other words there is only one frequency, the sound is a pure tone. Such a tone would sound very harsh and artificial, since natural tones (like the sound of a piano) are composed of many different tones, i.e., they correspond to a weighted sum of sine functions. Their Fourier transform is therefore a weighted sum of Dirac distributions. Another example is a normal distribution: their Fourier transform is again a normal distribution.
The Fourier transformation has a couple of interesting properties. We collect the most important in the following lemma:
Lemma A.12 (Properties of the Fourier transformation) LetF be the Fourier transformation, then:
342 A Mathematics (i) Fis a linear map.
(ii) The inverse of the Fourier transformation is given by .F1f/.x/D p1
2 Z C1
1 eixtOf.t/dt:
(iii) The Fourier transform of a derivative is a polynomial, more precisely:
F @n
@xnf.x/
./D.i/n.Ff/./:
(iv) It is possible to define an appropriate space (of functions or distributions) such thatFis a bijective map on this space.
Property (iii) can be summarized by saying that the Fourier transformation turns derivatives into a product – although this has nothing to do with the marketing of options. . . The property is often the main reason to use the Fourier transformation and could even be used as an alternative definition of differentiation. While this is certainly a complicated way for reaching this goal, it enables us to generalize the definition of derivatives: if we take the Fourier transform of a multiplication off with, e.g.,ijj1=2, we get something like a “half” derivative. If that sounds esoteric to you, then be ensured that it is not: very much to the contrary it is surprisingly useful.
The concept leads to the definition of pseudo differential operatorsand they are needed, e.g., in solving certain asset pricing problems when the underlying process is of a more complicated form. We mention this point and generally use Fourier transformations when we discuss Lévy processes in Sect.8.8.
Apartial differential equation(short:PDE) is an equation that contains different partial derivatives of an unknown function and is used to determine this function. In contrast, anordinary differential equation(short:ODE) only involves one kind of derivative (e.g., only derivatives with respect tot). As an example for a simple PDE we consider theheat equation4which is needed to solve the Black-Scholes equation (see Sect.8):
@u.x;t/
@t D @2u.x;t/
@x2 ;
wheret 2 Œ0;Tandx2 R. (In the original physics model for heat transport,tis the time andxthe space variable, whereasu.x;t/is the temperature at timet and positionx.) The above equation therefore means that the partial derivative ofuwith respect totequals its second partial derivative with respect tox.
4This PDE was originally used to describe the transport of heat in a material. There are, however, various other applications for this equation, therefore it is sometimes also called, e.g.,diffusion equation.
Typically, PDEs can only be solved uniquely when we have additional condi- tions. In the case above this would be an initial condition (specifyinguatt D 0) plus some boundary conditions (e.g., specifying the behavior ofuforx! ˙1).
PDEs are a central modeling tool in all scientific disciplines that rely on sophisticated mathematical models (like physics, chemistry, biology, engineering – and some areas in finance). Therefore their analytical and numerical investigation is very important. But how can we solve such a PDE? First, we need to stress that there is no general method that works for all kinds of PDEs. Very much to the contrary, specific methods need to be developed for different situations, and there is a whole research area in mathematics dedicated to this. In the case of the simple linear heat equation given above, things look better, of course: there are in fact several methods that can be applied. In the following we sketch one particularly simple method (the separation of variables). Other methods that could be used are the Fourier transformation,5variational methods or the finite element method. The latter is the standard way for numerical computations and works for a large class of PDEs. We refer the reader to [Eva98] and [RR04] for in-depth introductions to PDEs.
The key idea of the separation of variables is to look not for all possible solutions, but only for solutions of a special form, namelyu.x;t/ D a.x/b.t/. Once we have found such a solution, we only need to prove uniqueness, and we know that the solution we have found is notanysolution, butthe onlysolution. In fact, the uniqueness proof will be omitted here, but can be found in most mathematical textbooks on PDEs.
Using the ansatzu.x;t/Da.x/b.t/we can rewrite the PDE as follows:
a.x/b0.t/Da00.x/b.t/:
Sorting terms (and assuming that non of them vanishes, which is another point that would have to be justified later) we obtain
b0.t/
b.t/ D a00.x/
a.x/:
The central observation is now the following: the left side only depends ont and the right side only onx. Since both sides agree for all xand t, both terms have to be constant. Let us call this constant, then we get two ordinary differential
5Here we can exploit that the Fourier transform of a derivative is a simple multiplication. Thus after taking the Fourier transform of a PDE some of the derivatives become multiplications (compare LemmaA.12). The result is typically an ODE that can be solved much easier, either analytically or numerically. Finally the solution needs to be Fourier transformed again to return to the original formulation.
344 A Mathematics
equations:
b0.t/D b.t/;
a00.x/D a.x/:
The first of these equations can be solved by an exponential function:
b.t/Db.0/et;
the second can be solved by a combination of sine and cosine functions, e.g.
a.x/Dsin.x=p /:
To determine the precise form ofaandbwe need to take into account the initial and boundary conditions of the heat equation: we superimpose solutions forasuch that a.0/corresponds to the initial condition . For instance, if the boundary condition is given byu.0;t/D u.1;t/D 0, then anyak.x/Dsin.kx=/fork2Nsatisfies the boundary condition, since sin.k/D0for allk2N. Denoteuk.x/Dak.x/bk.x/, wherebk.x/Db.0/ekt=. Then a weighted sum of theukcan be constructed to fit the initial condition. This sum still solves the heat equation and the boundary and initial condition, since the heat equation is linear, i.e., weighted sums of solutions are also solutions.