Schwarz-Christoffel mapping

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In complex analysis, a discipline within mathematics, a Schwarz-Christoffel mapping is a transformation of the complex plane that maps the upper half-plane conformally to a polygon. Schwarz-Christoffel mappings are used in potential theory and some of its applications, including minimal surfaces and fluid dynamics. They are named after Elwin Bruno Christoffel and Hermann Amandus Schwarz.

1 Definition
2 Example
3 Other simple mappings
4 See also
5 External links
6 References

[edit] Definition

Consider a polygon in the complex plane. The Riemann mapping theorem implies that there is a bijective holomorphic mapping f from the upper half-plane

$\{ \zeta \in \mathbb{C}: \operatorname{Im}\,\zeta > 0 \}$

to the interior of the polygon. The function f maps the real axis to the edges of the polygon. If the polygon has interior angles $α,β,γ,...$ , then this mapping is given by

$f(\zeta) = \int^\zeta \frac{K}{(w-a)^{1-(\alpha/\pi)}(w-b)^{1-(\beta/\pi)}(w-c)^{1-(\gamma/\pi)} \cdots} \,\mbox{d}w$

where $K$ is a constant, and $a < b < c < ...$ are the values, along the real axis of the $ζ$ plane, of points corresponding to the vertices of the polygon in the $z$ plane. A transformation of this form is called a Schwarz-Christoffel mapping.

Remark it is often convenient to consider the case in which the point at infinity of the $ζ$ plane maps to one of the vertices of the $z$ plane polygon (conventionally the vertex with angle $α$ ). If this is done, the first factor in the formula is effectively a constant and may be regarded as being absorbed into the $K$ .

[edit] Example

Consider a semi-infinite strip in the $z$ plane. This may be regarded as a limiting form of a triangle with vertices $P = 0$ , $Q = π i$ , and $R$ (with $R$ real), as $R$ tends to infinity. Now $α = 0$ and $β = γ = π / 2$ in the limit. Suppose we are looking for the mapping f with f(−1) = Q, f(1) = P, and f(∞) = R. Then f is given by

$f(\zeta) = \int^\zeta \frac{K}{(w-1)^{1/2}(w+1)^{1/2}} \,\mbox{d}w. \,$

Evaluation of this integral yields

$z = f(\zeta) = C + K \operatorname{arccosh}\,\zeta,$

where $C$ is a (complex) constant of integration. Requiring that $f ( - 1) = Q$ and $f (1) = P$ gives $C = 0$ and $K = 1$ . Hence the Schwarz-Christoffel mapping is given by

$z = \operatorname{arccosh}\,\zeta.$