Monday Math 51

Last week, we looked at solving Laplace’s equation, in two dimensions. In particular, I demonstrated the technique of separation of variables for both Cartesian and polar coordinates.

Now, let us consider Laplace’s equation in three dimensions. In Cartesian coordinates, this is . If we use separation of variables by trying f(x,y,z)=X(x)Y(y)Z(z), then we get

As before, each of the three terms on the left hand side is a function of a single, different independent variable, each of these three terms is equal to a constant, the sum of which is zero:
, , ,
and so
, ,
Note that we have either two of the three constants positive, and the third negative, or we have one positive and two negative. In the first case, the resulting function is periodic in one dimension and exponential in the other two; in the second, it is periodic in two dimensions and exponential in the third.

Consider the function over the cube 0≤x≤1, 0≤y≤1, 0≤z≤1 with (Dirichlet) boundary conditions , , , . Seeing the x and y boundary conditions, it must be periodic in x and y: the x=0 and y=0 conditions give , , and our x=1 and y=1 conditions give and , where m and n are positive integers. The original equation tells us then that . The condition that tells us that our Am,n are all zero, and we obtain:
.
For z=1, we get

Using the orthogonality relationships of circular functions, we get:
,
and so
.

Now, let us instead consider cylindrical coordinates (r,φ,z). Drawing upon our plane polar coordinate laplacian, we get
. Trying a solution of the form f(r,φ,z)=R(r)Φ(φ)Z(z), and then dividing the resulting equation by R(r)Φ(φ)Z(z), we get

We see that the third term depends only on z, while the other terms depend only on the other two coordinates; thus we have that the third term must be a constant:

,

and also

Here, we have that the last term on the left hand side is a function of φ only, while the remaining terms depend on r only. As we usually want periodic solutions in φ, we set its term at –m2, so


.
To have a continuous function on all φ, 0≤φ<2π, we require that m be a non-negative integer (it can be zero for the constant case ).
The differential equation for R is now


Making the change in variables to the dimensionless variable , this becomes:

which is known as Bessel’s differential equation.
It’s solutions are known as the Bessel functions of the first and second kind, and . Bessel functions of the first kind, , are finite at u=0 for non-negative integer m, while the are singular at u=0. Thus, most problems will have solutions involving Bessel functions of the first kind.

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2 Responses to “Monday Math 51”

  1. What are the odds of landing a pair with a A-K pre-flop in texas hold’em? – Smarter Poker Says:

    […] Monday Math 51 « Twisted One 151’s Weblog […]

  2. Monday Math 52 « Twisted One 151’s Weblog Says:

    […] Math 52 By twistedone151 Last week, we looked at solving Laplace’s equation, in three dimensions, specifically Cartesian and […]

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