Download The flow lines are curves that meet each isotherm

MATH 4/5/7880, SPRING 2013
ISOTHERMS AND FLOW LINES
Suppose we want to know the steady state temperature distribution across the
upper half plane h when the temperature is held to a constant T0 along the segment
from −1 to 1 and a constant 0 elsewhere along the real line. Our text argues from
Fourier’s Equation that the temperature T must satisfy the equation
Txx + Tyy = 0.
That is, T is harmonic. One way to solve this problem is to exploit the fact that the
composition of a harmonic function H and an analytic function f is harmonic on the
image under f of the domain of H. Thus we wish to find a conformal transformation
f : h → D to a domain on which the temperature distribution problem is easy to
solve.
One possibility is to take D to be the infinite strip {w | 0 ≤ Im w ≤ π} with
constant temperature T0 along the upper edge and 0 along the lower edge. We
can see by inspection that a solution is T (u, v) = T0 v/π. The isotherms {(u, v) |
T (u, v) = c} for this temperature distribution are simply horizontal lines, pictured
(in red) below.
The flow lines are curves that meet each isotherm orthogonally. Clearly in our
example these are vertical line segments, pictured in green above.
1
2
ISOTHERMS AND FLOW LINES
More generally, if F (x, y) is a smooth function then dF = Fx dx + Fy dy. Along
its level curves dF = 0 and so these have tangent slopes given by the formula
dy
Fx
=− .
dx
Fy
In particular if F is harmonic and G is a harmonic conjugate then the level curves
of F and G meet orthogonally whenever they meet. This is proved using the
Cauchy-Riemann equations:
Fx
Gx
Fx Gx
−
−
=
·
= 1 · (−1).
Fy
Gy
Gy Fy
Back to our example on the infinite strip D we see that the harmonic conjugate
of T is S = −T0 u/π whose level curves are indeed the vertical line segments. When
we find our conformal transformation f : h → D then the isotherms on h will be
the level curves of Im f and the flow lines will be the level curves of Re f .
The logarithm w = log ζ — say the real branch — maps h onto D. It maps the
negative real axis onto the upper edge of D and the positive real axis onto the lower
edge. Hence it yields the flow lines and isotherms of the temperature distribution
T0
ξ
T0
Im log ζ =
arccot
π
π
η
T0
T0
S(ξ, η) = − Re log ζ = −
ln(ξ 2 + η 2 ).
π
2π
T (ξ, η) =
Semicircles centered at the origin are the flow lines of this solution half lines through
the origin are the isotherms. This is a valid solution of the steady state heat
equation in the upper half plane but it does not match the give boundary conditions.
We can rectify this by a linear fraction transformation of h to itself that maps
the negative real axis to the segment [−1, 1]. The transformation we seek is
ζ=
z−1
.
z+1
ISOTHERMS AND FLOW LINES
3
We obtain the steady-state temperature distribution
T0
z−1
T0
x2 + y 2 − 1
Im log
=
arccot
π
z+1
π
2y
which equals T0 along the interval −1 ≤ x ≤ 1 and 0 elsewhere along the real axis.
The isotherms are the arcs of the circles through ±1 that lie in the upper half plane.
(1)
T (x, y) =
Exercises.
(1) Find explicit equations for the isotherms and flow lines for our final result (1) above and verify directly that these meet orthogonally.
(2) Suppose that G(u, v) is a smooth real-valued function and that f (z) =
u(x, y) + iv(x, y) is analytic. Set H = G ◦ f . Show that
Hxx + Hyy = |f 0 |2 · (Guu + Gvv ) ◦ f.
Conclude that if G is harmonic then so is H, and that the converse is true
provided that f is conformal.