Conformal maps, Green’s function and Poisson
kernel.
David Gurarie
1
The re‡ection method for half-space problems
=
To get Green’s functions (for Laplaces; heat; wave) on the half-space Rn+1
+
fz = (x; y) : y > 0g we re‡ect the free-space source K0 (z ¡ w) about the boundary plane
w = (»; ´) ! w¤ = (»; ´)
and take a suitable combination of K (z; w) and K (z; w¤ ), i.e. use odd/even
extensions of K0 in the vertical variable y,
K (z; w) = K0 (z ¡ w) ¡ K0 (z ¡ w¤ )
K (z; w) = K0 (z ¡ w) + K0 (z ¡ w¤ )
Dirichlet
Neumann
In a similar way one handles the heat and wave propagators, e.g.
G (z; w; t) = G0 (z ¡ w; t) ¡ G0 (z ¡ w¤ ; t)
in terms of the free-space Gaussian G0 .
The corresponding Poisson kernels are computed from K; G; ::: via standard
relations, e. g.
P (z; w) = @n K (z; w)jw2¡ = @´ Kj´=0
K (z; w) = K (z; w)j¡ = Kj´=0
Dirichlet
Neumann
In particular for spherically-symmetric problem, like Laplacian/Helmholtz
potentials K0 = K0 (r); r = jz ¡ wj we get
q
2y
P (x ¡ »; y) = ¡ K00 (r) ; with r = (x ¡ »)2 + y2
r
In special cases 2D; 3D it yields familiar expressions, that could be also produced
by the Fourier transform
P =
P =
P =
y
¼[(x¡»)2 +y2 ]
y
3=2
2¼[(x¡»)2 +y 2 ]
2(n¡2)y
n=2
!n¡1 [(x¡»)2 +y2 ]
!n¡1 -area of the unit sphere S n¡1 in Rn .
1
2D
3D
nD
w
z
w*
Figure 1: Source and re‡ected source
Exercise 1
1. Show that P (x ¡ »; y) ! ± (x ¡ »), as y ! 1
2. Derive Poisson propagators for the heat and wave problems
1.1
Examples of Green’s functions for the Laplacian.
² The free space Green’s function in R3 is given by the Newton-Coulomb
1
,
potential (single point charge): K0 = 4¼r
² The half-space Green’s function is given by the dipole-potential - a pair
of opposite point charges:
K (x; y; a) = K0 (x ¡ a; y) ¡ K0 (x + a; y)
² The quadrant Green’s function is given by the quadrupole potentials,
KN (x; y; »; ´)
= K0 (x ¡ »; y ¡ ´) + K0 (x ¡ »; y + ´) + K0 (x + »; y ¡ ´) + K0 (x + »; y + ´)
corresponds to Dirichlet boundary conditions while NeumannKD (x; y; »; ´)
= K0 (x ¡ »; y ¡ ´) ¡ K0 (x + »; y ¡ ´) ¡ K0 (x ¡ »; y + ´) + K0 (x + »; y + ´)
2
6
0.6
5
0.5
4
0.4
3
0.3
2
0.2
1
0.1
0
-2
0
2
-2
4
2
4
Figure 2: Half-space Green
5
4
0.6
0.5
3
0.4
0.3
2
0.2
0.1
1
0
1
0
1
2
3
4
2
3
4
5
5
Figure 3: Quadrant (Dirichlet) Green
5
4
0.5
0.25
3
-0.25
-0.5
-0.75
-1
2
1
0
0
1
2
3
4
1
2
5
Figure 4: Quadrant Neumann Green
3
3
4
5
2
Conformal coordinate change in Laplacian
We take map Á : x ! y = Á(x) in Rn (or a region in Rn ), and denote by A(x) =
Á0 (x) its Jacobian matrix, and by J = det(A) - the Jacobian determinant. The
general change of variable formula for the Laplacian has the form
¡
¢¡1
rx
¢y = J1 rx ¢ J T AA
(1)
-a consequence of the grad and div-transformations
x ! y = Á (x)
¡ ¢¡1
rx ! ry = T A
rx
¡
¢
1
rx ¢ ::: ! ry ¢ ::: = J rx JA¡1 ¢ :::
The orthogonal map Á has orthogonal column-vectors in the Jacobian matrix
A = [a1~u1 ; a2~u2 ; :::; an~un ] ;
~ui ¢ ~uj = ± ij
¡T
¢
Hence diagonal product AA = diagfa21 ; :::a2n g and J = a1 :::an in (1).
The standard example are spherical coordinates in R2 ; R3 etc., where
µ ¶
µ
¶ µ
¶
r
x
r cos µ
Á :
!
=
µ
y
r sin µ
·
¸
·
¸
¡
¢
cos ¡r sin
1
A =
; T AA =
;J = r
sin r cos
1=r2
hence the resulting spherical Laplacian
¶
·
¸µ
¢ r
1
1¡
1
@r
@r @µ
= @r r@r + 2 @µ2
@µ
1=r
r
r
r
Conformal maps Á have conformal Jacobian matrix A = ½U - “scalar times
orthogonal”, so they make up a subclass of orthogonal transformations, and
their determinant J = ½n . The resulting Laplacian
¢=
2.1
1
½n r ¢
½n¡2 r
(2)
Radial conformal map and harmonic functions
We are looking for a change of variables (dependent and independent) u(y) !
½® (u ± Á) = v(x), that would take any harmonic function u (x) into another
harmonic v (x). Substitution into (2) yields,
½n¡2 r (½® u ± Á)
½
r½
¢ ru+
= ½¡2+® r2 u + (n ¡ 2 + 2®)
½
Ã
¯ ! )
¯
µ
¶
¯ r½ ¯2
r½
¯ u
¯
+® r ¢
+ (n ¡ 2 + ®) ¯
½
½ ¯
1
½n r ¢
4
(3)
To get the Laplace’s equation for v the …rst and 0-th order terms of (3) must
vanish. Hence, ® = ¡ n¡2
2 and the conformal factor ½ solves a nonlinear PDE,
¯
¯
µ
¶
¯ r½ ¯2
r½
¯ =0
¯
r¢
+ (n ¡ 2 + ®) ¯
½
½ ¯
We rewrite it for the log-derivative of ½ as,
¢ (ln ½) +
n¡2
jr (ln ½)j2 = 0
2
(4)
The latter could be explicitly solved when ½, hence à = ln ½ are radial
functions à = à (r). Indeed, (4) becomes a radial ODE for Ã,
¡
¢0
¡ 0 ¢2
à =0
r1¡n rn¡1 Ã0 + n¡2
(5)
2
¡ 0 ¢2
d
d
Ã
= dz
= 0, and
transformed via change of variable rn¡1 dr
into Ã00 + n¡2
2
solve by separation. The o¤-short is a general solution of (5) in the form
¡
¢2=(n¡2)
à = C1 C2 + r2¡n
(6)
A special family of solutions vanishing at 1 is given by à =
2.2
C
r2
Inversion
Let us note that ½ = r12 is precisely the conformal factor of the inversion map,
Á : x ! jxjx 2 . Indeed, its Jacobian-matrix
0
A=Á =
1
jxj2
(
± ij ¡ 2
xi xj
jxj2
)
= r¡2 U
where U is an orthogonal re‡ection-matrix about the normal hyper-plane to x
in Rn (…g.5). In fact, one could show that a radial map x ! Á (r) x is conformal,
i¤ scalar factor Á = r12 , so inversion is the only possibility.
We have thus shown, that any harmonic
´ function u(x) gives rise to another
³
x
.
harmonic function u¤ (x) = jxj2¡n u jxj
2
2.3
Green’s function in the ball
Clearly, both u and u¤ take on the same value on the unit sphere S = fjxj = 1g,
so inversion Á transforms harmonic functions fug in the interior of the unit ball
B = fjxj · 1g to harmonic functions fu¤g in the exterior fjxj ¸ 1g. Applying
?? to the Newton potential u = K0 (jx ¡ »j) - the free-space Green’s function,
we get the requisite harmonic correction to K0 in the unit ball B,
u¤ = jxj2¡n K0 (jx¤ ¡ »j) ; x¤ =
5
x
jxj2
(7)
Figure 5: Jacobian matrix of the inversion is the re‡ection about the normal
plane to x
1
Hence, remembering that K0 = c jx¡»j
n¡2 (constant cn is the surface area
n
n
of the unit sphere in R ) we get the Green’s function of the ball
K (x; ») = K0 (jx ¡ »j) ¡ K0 (j(^
x ¡ jxj »)j)
´¯´
³¯³
¯
¯ ^
= K0 (jx ¡ »j) ¡ K0 ¯ » ¡ j»j x ¯
(8)
»
x
^ = jxj
Here x
and ^» = j»j
denote normalized (unit) vectors in the direction
x and ». It is often convenient to rewrite (8) in the polar form, i.e. variables
r = jxj, ½ = j»j and angle µ between x and » (see …g.6),
K(r; ½; µ) = K0
2.4
2.4.1
¶
µq
³p
´
r2 + ½2 ¡ 2r½ cos µ ¡ K0
1 + (r½)2 ¡ 2r½ cos µ
(9)
Special cases.
2-D case:
1
ln jz ¡ wj, complex variables z; w 2 C being used in place of x; ».
Here K0 = 2¼
The prefactor jzj® = 1 in (7), hence (8-9) takes the form
K(z; w) =
1
2¼
¯
¯
¯ z¡w ¯
¯=
ln ¯¯
1 ¡ zw
¹¯
1
2¼
ln
s
r2 + ½2 ¡ 2r½ cos µ
1 + (r½)2 ¡ 2r½ cos µ
(10)
in polar coordinates z = reiµ ,w = ½eiÁ .Exterior disk Green’s function has the
same form (10), but this time r; ½ lie outside the unit circle
6
Figure 6:
1
0.5
0
-0.5
-1
-1
0
1
2
3
Figure 7: Interior and exterior disk Green’s function
7
2.4.2
3-D case:
Here free-space K0 =
K(x; ») =
=
2.5
1
4¼jx¡»j ,
hence
½
¾
1
1
¡
jx ¡ »j j(^
x ¡ jxj »)j
9
8
=
<
1
1
1
p
q
¡
4¼ :
;
r2 + ½2 ¡ 2r½ cos µ
1 + (r½)2 ¡ 2r½ cos µ
1
4¼
Poisson kernels
For the Dirichlet boundary condition the Poisson kernel is given by normal
@
KWe observe that
derivative of the Green’s function, P (x; ») = n» ¢ r» K = @½
on the boundary j»j = 1; n» = ». Hence,
P (x; ») =
=
(» ¡ x) ¢ »
(» ¡ x¤ ) ¢ »
n ¡
cn j» ¡ xj
cn jxjn¡2 j» ¡ x¤ jn
(11)
jxj2 ¡ x ¢ »
(» ¡ x) ¢ »
n ¡
^)jn
cn j» ¡ xj
cn j(jxj » ¡ x
It remains to note that both denominators in (11) are equal, so we get Poisson
kernel,
2
P =
1 ¡ r2
1 ¡ jxj
n =
cn j» ¡ xj
cn (1 + r2 ¡ 2r cos µ)n=2
(12)
From the unit ball one can easily pass to an arbitrary radius a,
P =
a2 ¡ r2
cn (r2 + a2 ¡ 2ra cos µ)n=2
(13)
Another way to derive (12) is via di¤erentiation of the polar form (9) in
variable ½ at ½ = 1,
¯
³p
´
@K ¯¯
1 ¡ r2
0
2 ¡ 2r cos µ p
P (r; 1; µ) =
=
K
1
+
r
= :::
@½ ¯½=1
1 + r2 ¡ 2r cos µ
In special cases we get
2D
3D
Green’s functions and Poisson kernels in 2D and 3D balls
K0
P
q 2K 2
r +½ ¡2r½ cos µ
1
1
1¡r2
¡ 2¼ ln r
2¼ ln
2¼(1+r2 ¡2r cos µ)
1+(r½)2 ¡2r½ cos µ
½
¾
1
1
1¡r2
1
p 2 21
¡p
3=2
2
2
4¼r
4¼
r +½ ¡2r½ cos µ
1+(r½) ¡2r½ cos µ
8
4¼(1+r ¡2r cos µ)
2.6
Other examples
2.6.1
Half-disk
Green’s function and the Poisson kernel in the upper half-disk are obtained by
a combination of the re‡ection and inversion:
K (z; ³) =
1
2¼
¯
¯
½ ¯
¹ ¯¯¾
¯
¯ z¡³ ¯
¯ ¡ ln ¯ z ¡ ³ ¯
ln ¯¯
¯ 1 ¡ z³ ¯
1 ¡ z¹
³¯
or in polar form z = reiµ , ³ = ½eiÁ
s
( s
)
r2 + ½2 ¡ 2r½ cos (µ ¡ Á)
r2 + ½2 ¡ 2r½ cos (µ + Á)
1
K = 2¼ ln
¡ ln
1 + (r½)2 ¡ 2r½ cos (µ ¡ Á)
1 + (r½)2 ¡ 2r½ cos (µ + Á)
The corresponding Poisson kernel is also obtained by re‡ection of (12)
½
¾
1 ¡ r2
1 ¡ r2
1
¡
P (r; Á; µ) = 2¼
1 + r2 ¡ 2r cos (µ ¡ Á) 1 + r2 ¡ 2r cos (µ + Á)
on the upper (semicircular) boundary and
(
)
1
1
r sin µ
¡
P (r; ½; µ) = ¼
r2 + ½2 ¡ 2r½ cos µ 1 + (r½)2 ¡ 2r½ cos µ
on the lower (‡at) boundary.
2.6.2
General 2-D regions in complex plane.
By the general Riemann mapping Theorem of complex analysis any planar region D ½ C can be conformally mapped onto any other such region, e.g. onto
the unit disk D = fjzj < 1g or a half-plane H = fIm(z) > 0g.
Example 2 Map Á : z ! w = z¡i
fIm(z) < 0g onto the
z+i takes a half-plane
¯
¯
2
unit disk fjwj < 1g, and has a conformal factor ½ = ¯Á0 (z)¯ = jz+ij
2z
4
1
3
0.5
2
-1 -0.8-0.6-0.4-0.2
1
-3
-2
-1
0
0 0.2 0.4 0.6 0.8
-0.5
1
2
-1
3
9
Example 3 Map Á : z ! ez takes a complex strip f0 · Im(z) · 2¼g onto the
entire plane C.
4
4
3
2
2
-4
-2
0
2
4
-2
1
-4
-3
-2
0
-1
1
2
3
Example 4 Map w = cosh(z) takes a coordinate rectangle f0 · x · a; 0 · y ·
2¼g in the z-plane onto an ellipse in the w-plane (elliptical coordinates).
6
1
5
0.5
4
3
-1.5
-1
2
-0.5
0
0.5
1
1.5
-0.5
1
-1
0
0.2
0.4
0.6
0.8
1
Conformal maps allow to transfer Green’s functions from one region to another. Namely, we let map z ! w = Á (z) take region D (with the known
Green’s function KD (z; ³)) onto region M, whose Green’s function KM is to be
computed. Assuming the boundary of D mapped onto the boundary of M , two
Laplacians are related via ¢M [u] ± Á = J1 ¢D [u ± Á] , where J = ½2 denotes the
Jacobian determinant. It follows then that the Green’s functions of two regions
are related by a change of variables
¡
¢
KM (w; ´) = KD Á¡1 (w) ; Á¡1 (´)
In the above examples map w = z¡i
z+i from the half-plane H onto the disk
D. The half-space Green’s function,¯ computed
by the re‡ection (complex con¯
¯ z¡³ ¯
1
jugation) has the form: K = 2¼ ln ¯ z¡¹³ ¯ Substitution of the inverse map z =
¡1
Á¡1 (w) = i w+1
(´) = ::: yields
w¡1 and ³ = Á
¯
¯
¯ w¡´ ¯
1
¯
¯
KD (w; ´) = 2¼ ln ¯
1 ¡ w¹́ ¯
that was produced earlier by the inversion method.
10
2.6.3
Green and Poisson kernels on strip
1. . We use conformal map: z ! ez from strip f0 · Im z · ¼g onto halfspace fIm w ¸ 0g. Then
¯
¯
¯ z
¯
¯ sinh ¡ z¡w ¢ ¯
¯ e ¡ ew ¯
1
1
¯
2
¯ = ¡ ln ¯
¢ ¯¯
¡
G (z; w) = ¡ ln ¯¯ z
2¼
e ¡ ew¹ ¯
2¼ ¯ sinh z¡2w¹ ¯
³
´
8
¡
¢9
=
< cosh2 x¡»
¡ cos2 y¡´
2
2
1
³
´
= ¡ ln
¡
¢
4¼ : cosh2 x¡» ¡ cos2 y+´ ;
2
2
½
¾
cosh (x ¡ ») ¡ cos (y ¡ ´)
1
= ¡ ln
4¼
cosh (x ¡ ») ¡ cos (y + ´)
1
sin y
P (z; ») = @´ G (z; w)j´=0 =
2¼ cosh (x ¡ ») ¡ cos (y)
3
2.5
2
1.5
1
0.5
0
-3
-2
-1
0
1
2
3
Function P (x; y) along with its contour map near zero, in strip 0 < y < ¼
2.6.4
Problems
1. Show that complex map w = cosh(z) takes strip P =f0 · Im(z) · 2¼g
onto the upper half-plane H, so that the boundaries of P:fIm(z) = 0g and
fIm(z) = 2¼g go into the boundary of H. Use this map and the known
Green’s function of H to construct the Green’s function and the Poisson
kernel of P.
2. Do the same exercise with the fractional power map Á (z) = z ®=¼ that
takes the upper half-plane onto a sector of angle ®, and derive the Green’s
function and the Poisson kernel of the sector.
3
Applications to ‡uid ‡ows
A 2D ideal (incompressible, inviscid) Euler ‡ow is described by its stream-…eld
Ã. When viewed as a function of complex variable z = x + iy, Ã (z; z¹), the
11
¡
¢
Eulerian velocity u = ¡Ãy ; Ãx is given by a complex derivative u = i@z¹Ã.
Irrotational ‡ow r £ u = 0 corresponds to harmonic stream-…eld: ¢Ã = 0.
Equivalently u has velocity potential u = rÁ. In fact, it could be described
by a complex analytic potential w (z) = Á + iÃ, and complex velocity w0 (z) =
u ¡ iv. Indeed, the relations u = rÁ = rÃ? are nothing but Cauchy-Riemann
equations for w.
Let us also remark that steady Euler ‡ow obeys the Bernoulli relation:
½
2
ju j2 + p = Const - the hydrostatic pressure
(14)
where ½ an p are ‡uid density and pressure. The latter follows from the momentum conservation
³
´
u t + r ½2 ju j2 + p ¡ u £ ! = 0
where ! = r £ u denotes vorticity of u . We consider a few examples and
basic problems of ideal ‡uid ‡ows.
3.1
Potential ‡ow passed an obstacle.
We shall study a incompressible potential ‡ow passed an obstacle D, e.g. a
cylinder, or sphere of radius a. It produced by a moving parallel ‡ow passed the
body, or the motion of the body in a quiescent ‡uid. In either case potential,
laminar ‡ow has stream-…eld à and potential Á - both harmonic functions in
the exterior of D, ¢Ã = ¢Á = 0 in Rn nD; n = 2; 3.
Depending on the reference frame one either …xes ¡ and certain asymptotic
‡ow at 1 e.g. u ¼ (U; 0) - (body frame), or allows moving ¡ and zero 1-‡ow
(‡uid frame). We shall use the latter as it permits non-stationary motion of D.
The body velocity U creates a boundary potential Á0 = ¡ U ¢ ~rj¡ , to maintain the tangential ‡ow along the boundary, i.e. Ãj¡ = Const. The latter is
extended through the entire Rn nD via the exterior Poisson kernel Á = P¡ [Á0 ].
Any harmonic function Á could be expanded at 1 as
A0 A1 ¢ ~r
An (~r)
+
+ ::: + n+1 + :::
(15)
r
r3
r
with spherical harmonics An (~r) of degree n. For velocity potential Á coe¢cient
A0 = 0, since the net outward ‡ux (at large ~r) must be 0 (no sources).
Clearly, potential Á, and velocity u = rÁ depend linearly on U
Á¼
Á = A (~r) ¢ U
Ai = P¡ [xi ]
(16)
the components Ai - represent harmonic extensions of linear coordinate functions
on ¡.
To get the ‡uid reaction force on the body we compute the total (kinetic)
energy of the ‡uid
ZZ
X
½
u 2 = ½2
mij Ui Uj
E=2
(17)
12
½ - ‡uid density. The coe¢cients of the quadratic form E = E (U ) make up the
so-called virtual mass-tensor
ZZ
I
rAi (~r) ¢ rAj (~r) = ½ Ai (@n Aj )
mij = ½
(18)
= ½
I
Rn nD
xi @n P¡ [xj ] = ½
¡
I
¡
xj @n P¡ [xi ]
¡
De…nition (18) resembles the inertia tensor for a rotating
of massH xjsolid
xi
density ½r distributed over surface ¡. It’s not however, equal ½
,
as
Poisson
r
¡
P¡ makes important di¤erence. Though Ai coinsides with a linear function
xi j¡ , its gradient rAi and normal derivative @n Aj on ¡are not e i and Ni
(the
component
of n ) respectively. A simple illustartion would be a disk
© 2 i-the
ª
x
x + y 2 · 1 , where A1 = x2 +y
2 has rA1 j¡ = (cos 2µ; sin 2µ) as opposed to
rxj¡ = (cos µ; 0). Let us also remark that tensor mij is a geometric invariant
of surface ¡, independent of its position in Rn , as one could easily verify, using
translational/ rotational symmetries of solution (16), and volume-integral1 form
(18) .
In a similar fashion one computes the total momentum of the induced ‡ow
1
0
1 0
I
ZZ
X
P :::
mij Uj A
u =½ Ni @
Uj xj A = @
~p = ½
(19)
Rn nD
::::
j
¡
The rate of the change of the momentum gives the ‡uid reaction force,
exerted on the body. So the resulting equation of motion take the
M
dU
d~
p
+
= F -external force
dt
dt
RR
½0 is the body mass. Remembering the exact form of ‡uid
where M =
DP
dU
momentum (19) j (M ± ij + mij ) dtj = Fi . Thus the e¤ect of ‡uid on the
moving body is to replace the standard mass by the virtual mass-tensor in the
p
kinetic energy. The components of the reaction force ¡ d~
dt along U , and its
?
normal U gives the drag and buoyancy (lift) forces
X
X
Fd = ¡
mij U_ i Uj ; Fb = ¡
mij U_ i Uj?
j
j
In the 2D-case and horizontal U = (U; 0) we get Fd = ¡m11 U U_ ; Fb = ¡m12 U U_ .
Exercise 5 Compute virtual mass-tensor and the drag and buoyancy forces for
the cylinder (disk in 2D), half-cylinder (0 · µ · ¼) , and quarter-cylinder
(0 · µ · ¼=2). Do the same exercise for solid sphere, hemi-sphere and quartersphere in 3D.
1 Observe, that coe¢cients m , hence the total energy of the induced ‡ow is …nite, as
ij
¡
¢
r) ¼ O r¡2 decay su¢ciently fast at 1.
functions Ai (~
13
_ so there
Remark 6 Notice that the reaction force depends on acceleration U,
is no net drag or lift in a uniform ‡ow (d’Alembert paradox). This is not
surprising, particularly for the drag-force, as its presence would require either
energy dissipation or a non-zero energy ‡ux to 1. Both are absent in the ideal
‡uid.
We shall demonstrate the general principles with two speci…c examples.
3.1.1
Potential ‡ow passed the cylinder.
of parallel ‡ow passed the cylinder r · a. Its stream-…eld is the sum of the
principal (parallel) ‡ow-component ª = ¡Uy and a perturbation Ã0 , that would
make the sum à = ª + Ã0 -constant (e.g. 0) on the boundary. So we solve the
Poisson equation for Ã0 (r; µ) in the exterior of disk with boundary value ¡Uy
¢Ã0
¯
Ã0 ¯
= 0 for r < a
= Ua sin µ = ¡ ªjr=a
r=a
(20)
The solution is obtained via exterior Poisson kernel P (r; µ) =
0
à =
Z2¼
P (r; µ ¡ ¿ ) U a sin ¿
r2 ¡a2
2¼(r2 ¡2ar cos µ+a2 )
(21)
0
ikµ
imµ
= 2¼± km yields
Direct evaluation of (21) using
¡ a ¢convolution identity e ¤e
Ã0 = ¡Ua rsin µ ¢= Uray
=
U
Im
.
The
latter
is
also
an
obvious
consequence
2
z
z
a
of Im a + z = 0 on the
³ circle2 ´of radius jzj = a. We plot stream-lines of
¡z a¢
à = U a Im a + z = U y 1 ¡ ar2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
-1
0
1
2
Flow passed cylinder
From the explicit Ã, one computes the velocity u = @z¹Ã and the hydrostatic
2
pressure (14) p = p1 ¡ 12 jrÃj . The reaction forces could be obtained by
14
evaluating pressure gradient along the circle F =
H
C
rpds. Thus we get
µ
¶
a2
à (r; µ) = U r ¡
sin µ -stream
r
(µ
µ
¶2
¶2 )
@Ã (r; µ)
@Ã (r; µ)
2
1
+r
p (r; µ) = ¡ 2
-pressure
@r
@µ
µ
¶
µ
¶
@p (r; µ)
sin µ @p (r; µ)
f (r; µ) = r cos µ
¡
-horizontal force
@r
r
@µ
3.1.2
Point-vortex ‡ow passed cylinder
The point-vortex at ³ of strength ¡ creates a ‡ow passed an obstacle D whose
stream-…eld is given by the exterior (Dirichlet) Green’s function
¡¢Ã
Ãj@D
= ¡± (z ¡ ³)
= Const
For the cylinder D = fjzj < ag we have à (z; ³) =
sponding velocity
u = i@z¹Ã =
i¡ n 1
+
4¼ z¹¡³¹
³
a2 ¡¹
z³
o
=
¡
2¼
¯
¯
¯
¯
ln ¯ a(z¡³)
and the corre2
¹
a ¡z ³ ¯,
i¡
a2 ¡ j³j2
¡
¢
4¼ z¹ ¡ ¹
³ (a2 ¡ z¹³)
Due to rotational symmetry we could place source on the real axis ³ = b > 0
call z = reiµ and compute pressure
¡ 2
¢
a ¡ b2
¡2
p=¡
(22)
2 (4¼)2 jr ¡ beiµ j2 jbr ¡ a2 eiµ j2
Clearly, the net force is exerted by the vortex on the cylinder is directed toward the vortex. Its precise form is obtained
by evaluating the pressure gradient
H
¡2
px of (22) along the circle f = 2(4¼)2 C px ds.
15