Complex Analysis - Maths KU

v
B′
C′
w 0
∇Φ
N
Φ = c0 u
Figure 7.52 Figure for the proof of
Theorem 7.8
7.5 Applications 435
Proof Assume that f and h satisfythe hypothesis of the theorem. Let
z0 = x0 + iy0 be a point on C and let w0 = u0 + iv0 = f(z0) be its image on
C ′ . Recall from calculus that if N is a normal vector to C ′ at w0, then the
normal derivative at w0 is given bythe dot product
dΦ
dN
= ∇Φ · N,
where ∇Φ is the gradient vector Φu(u0, v0)i +Φv(u0, v0)j. The condition
dΦ/dN = 0 implies that ∇Φ and N are orthogonal, or, equivalently, that
∇Φ is a tangent vector to C ′ at w0. Let B ′ be the level curve Φ(u, v) =c0
containing (u0, v0). In multivariable calculus, you learned that the gradient
vector ∇Φ is orthogonal to the level curve B ′ . Thus, since the gradient is
tangent to C ′ and orthogonal to B ′ , we conclude that C ′ is orthogonal to B ′
at w0. See Figure 7.52.
Now consider the level curve B in the z-plane given by
φ(x, y) =Φ(u(x, y),v(x, y)) = c0.
The point (x0, y0) isonB and the gradient vector ∇φ is orthogonal to B at
this point. Moreover, given anypoint (x, y) onB in the z-plane, we have
that the point (u(x, y), v(x, y)) is on B ′ in the w-plane. That is, the image
of B under w = f(z) isB ′ . The curve C intersects B at z0, and because f
is conformal at z0, it follows that the angle between C and B at z0 is the
same as the angle between C ′ and B ′ at w0. In the preceding paragraph we
found that this angle is π/2, and so C and B are orthogonal at z0. Since
∇φ is orthogonal to B, we must have that ∇φ is tangent to C. If n is a
normal vector to C at z0, then we have shown that ∇φ and n are orthogonal.
Therefore,
dφ
= ∇φ · n =0.
dn
Theorem 7.8 gives us a procedure for solving Nuemann problems associated
with boundaryconditions of the form dφ/dn = 0. Namely, we follow
the same four steps given on page 429 to solve a Dirichlet problem. In Step
1, however, we find a conformal mapping from D onto D ′ . Since conformal
mappings preserve boundaryconditions of the form dφ/dn = 0, solving the
associated Nuemann problem in D ′ will give us a solution of the original
Nuemann problem. Because analytic mappings are conformal at noncritical
points, this approach also works with mixed boundaryconditions. Roughly,
these are boundaryconditions where values of φ are specified on some boundarycurves,
whereas the normal derivative is required to satisfydφ/dn =0on
other boundarycurves.
EXAMPLE 3 A Heat Flow Application
Find the steady-state temperature φ in the first quadrant, shown in color in
Figure 7.53(a), which satisfies the indicated mixed boundaryconditions.
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