Page 104 - 6099
P. 104
time. Assume that the fluid is incompressible (e.g.,water) and that there are no sources
or sinks. Then div v = 0. Hence v = − grad φ for some φ (called the velocity potential)
and ∆φ = − div v = 0, which is Laplace’s equation.
Analytic functions of a complex variable. Write z = x + iy and f(z) = u(z) +
iv(z) = u(x + iy) + iv(x + iy), where u and v are real-valued functions. An analytic
function is one that is expressible as a power series in z. This means that the powers are
n
m n
n
not x y but z = (x + iy) . Thus
∞
X
f(z) = a n z n
n=0
(a n complex constants). That is,
∞
X
n
u(x + iy) + iv(x + iy) = a n (x + iy) .
n=0
Formal differentiation of this series shows that
∂u ∂c ∂u ∂v
= and = − .
∂x ∂y ∂y ∂x
These are the Cauchy-Riemann equations. If we differentiate them, we find that
u xx = v yx = v xy = −u yy ,
so that ∆u = 0. Similarly ∆v = 0, where ∆ is the two-dimensional laplacian.Thus the
real and imaginary parts of an analytic function are harmonic.
Brownian motion. Imagine brownian motion in a container D. This means that
particles inside D move completely randomly until they hit the boundary, when they stop.
Divide the boundary arbitrarily into two pieces, C 1 and C 2 . Let u(x, y, z) be the probability
that a particle which begins at the point (x, y, z) stops at some point of C 1 . Then it can
be deduced that
∆u = 0 in D
u = 1 on C 1 u = 0 on C 2 .
Thus u is the solution of a Dirichlet problem.
The basic mathematical problem is to solve Laplace’s or Poisson’s equation in a given
domain D with a condition on boundary D :
∆u = f in D,
∂u ∂u
u = h or = h or + au = h on boundary D.
∂n ∂n
In one dimension the only connected domain is an interval {a ≤ x ≤ b}. We will see that
what is interesting about the two- and three-dimensional cases is the geometry.
We begin our analysis with the maximum principle, which is easier for Laplace’s equa-
tion than for the diffusion equation. By an open set we mean a set that includes none of
its boundary points.
97