Daniel J. Duffy - Numerical Methods in Computational Finance

exists and is equal to f ( a , b ). We now need definitions for the derivatives of f in the x and y directions.

In general, we calculate the partial derivatives by keeping one variable fixed and differentiating with respect to the other variable; for example:

We now discuss the situation when we introduce a change of variables into some problem and then wish to calculate the new partial derivatives. To this end, we start with the variables ( x , y ), and we define new variables ( u , ). We can think of these as ‘original’ and ‘transformed’ coordinate axes, respectively. Now define the function z ( u , ) as follows:

This can be seen as a function of a function . The result that we are interested in is the following: if z is a differentiable function of ( u , ) and u , are themselves continuous functions of x , y , with partial derivatives, then the following rule holds:

(1.11)

This is a fundamental result that we shall apply in this chapter. We take a simple example of Equation (1.11)to show how things work. To this end, consider the Laplace equation in Cartesian geometry:

We now wish to transform this equation into an equation in a circular region defined by the polar coordinates:

The derivative in r is given by:

and you can check that the derivative with respect to is:

hence:

and:

Combining these results allows us to write Laplace's equation in polar coordinates as follows:

Thus, the original heat equation in Cartesian coordinates is transformed to a PDE of convection-diffusion type in polar coordinates.

We can find a solution to this problem using the Separation of Variables method , for example.

Some problems use functions of two variables that are written in the implicit form :

In this case we have an implicit relationship between the variables x and y . We assume that y is a function of x . The basic result for the differentiation of this implicit function is:

(1.12a)

or:

(1.12b)

We now use this result by posing the following problem. Consider the transformation:

and suppose we wish to transform back:

To this end, we examine the following differentials :

(1.13)

Let us assume that we wish to find dx and dy , given that all other quantities are known. Some arithmetic applied to Equation (1.13)(two equations in two unknowns!) results in:

where J is the Jacobian determinant defined by:

We can thus conclude the following result.

Theorem 1.1The functions and exist if :

are continuous at ( a , b ) and if the Jacobian determinant is non-zero at ( a , b ).