Introductory lecture notes on Partial Differential Equations c Anthony Peirce.
Not to be copied, used, or revised without explicit written permission from the copyright owner.
1
( Compiled 3 March 2014 )
In this lecture we continue with the solution of Laplace’s equation on circular domains. In particular, a wedge-shaped domain with a circular cut-out, i.e., a pizza slice with a bite taken out of it, as well as the Dirichlet and Neumann problems on the interior of a circle. This leads to the well known Poisson Integral Formula and an interesting application known as Electrical Impedance Tomography (EIT) .
Key Concepts: Laplace’s equation; Circular domains; Pizza Slice-shaped regions; Dirichlet and Neumann Boundary
Conditions, Poisson Integral Formula; Applications to EIT.
Reference Section: Boyce and Di Prima Section 10.8
27 Wedges with cut-outs, circles, holes, and annuli
27.1 Wedges with cut-outs
Example 27.1
A circular wedge with a cut-out:
Figure 1.
Mixed boundary conditions on a wedge shaped domain with a cut-out (27.2) u rr
+
1 r u r
+
1 r 2 u
θθ
= 0 u
θ
( r, 0) = 0 u
θ
( r, α ) = 0 u ( b, θ ) = 0 u ( a, θ ) = f ( θ )
Let u ( r, θ ) = R ( r )Θ( θ ).
r 2 ( R 00 +
R ( r )
1 r
R )
= −
Θ 00
Θ(
( θ )
θ )
= λ 2 ⇒
½ r 2
Θ
R
00
00 + rR 0
+ λ 2
−
Θ = 0
λ 2 R = 0
(27.1)
(27.2)
(27.3)

2
Θ equation i
Θ 00 + λ 2 Θ = 0
Θ 0 (0) = 0 = Θ 0 ( α )
¾
Θ
Θ 0
= A cos λθ + B sin λθ
(0) = Bλ = 0 ⇒ B or λ = 0 ,
Θ 0
Θ 0 ( α
= − Aλ sin λθ + Bλ cos λθ
) = − Aλ sin λα = 0 , λ = nπ
α n = 0 , 1 , . . .
R equation i n = 0 : ( rR 0
0
)
0
= 0 rR 0
0
= B
0
R
0
= A
0
+ B
0 ln r . Note u
0
( b, θ ) = R
0
( b )Θ
0
( θ ) = 0 ⇒ R
0
( b ) = A
0
+ B
0 ln b = 0 , A
0
= − B
0 ln b.
(27.4)
(27.5)
(27.6)
Therefore R
0
= B
0 ln( r/b ). Choose B
0
= 1.
n ≥ 1 : r 2 R 00 n
+ rR 0 n
− λ 2 R n
= 0 R ( r ) = A n r λ n + B n r − λ n
R n
( b ) = A n b
λ n + B n b
− λ n = 0 ⇒ B n
= − A n b
2 λ n u n
R
( n
( r ) = A n r, θ ) = h r
(
[ r λ n nπ
α
) u
0
( r, θ ) = ln
³ r b
´
− b
− b
2 λ n
2
( nπ
α r
)
− λ n r −
(
] nπ
α
) i cos
µ
A
α n nπθ
· 1
Therefore
³ r
´ u ( r, θ ) = c
0 ln b u ( a, θ ) = f ( θ ) = 2
+
¡ c
0 n
¡ ln
2 c n a b
¢¢
= a
2
0
+ n =1 a n h r
( nπ
α
+
)
− c n b
( h
2 a cos n =1
µ nπθ
α
¶ nπ
α
.
(
) r nπ
α
)
−
( nπ
α
− b
(
) i cos
2 nπ
α
)
µ r −
( nπθ
α nπ
α
) i
¶ cos
µ nπθ
¶
α
Therefore
2 c
0 ln( a/b ) =
2
α
Z α
0 f ( θ ) dθ.
(27.7)
(27.8)
(27.9)
(27.10)
(27.11)
(27.12)
(27.13)
(27.14) c n
= c
0
=
α h a
( nπ
α
)
2
− b
( 2 nπ
α
) a − nπ
α i
Z α f ( θ ) cos
µ nπθ
¶
α
0
1
α ln( a/b )
Z α
0 f ( θ ) dθ.
dθ (27.15)
(27.16)
Observations: In the special case f ( θ ) = 1, c
0
1 · cos
µ nπθ
α
¶ dθ
0
= log(
1 a/b
= 0 so that the solution reduces to:
)
, and c n
= 0 n ≥ 1. By Fourier basis function orthogonality u ( r, θ ) = log( r/b ) log( a/b )
(27.17) which is purely radial i.e. has no θ dependence.

Laplace’s Equation
27.2 Problems with a complete circle as the boundary
Example 27.2
Dirichlet problem in the interior of a circle
3
Figure 2.
Dirichlet boundary condition prescribed on the on boundary of a circle ( ??
) u rr
+
1 r u r
+
1 r 2 u
θθ
= 0 0 < r < a, 0 < θ < 2 π.
BC: u ( a, θ ) = f ( θ ) u ( r, θ ) < ∞ r → 0
Periodicity u ( θ + 2 π ) = u ( θ ) periodic.
Let u ( r, θ ) = R ( r )Θ( θ ).
r 2
¡
R 00 +
R ( r )
1 r
R 0
¢
= −
Θ 00
Θ
= + λ 2 ⇒ r 2 R 00
Θ 00 +
+
λ 2 rR 0 −
Θ = 0
λ 2 R = 0 Euler Eq.
Θ equation i Θ 00 + λ 2 Θ = 0 Θ = A cos( λθ ) + B sin( λθ )
θ ( − π ) = A cos( λπ ) − B sin( λπ ) = Θ( π ) = A cos( λπ )
+ B sin( λπ ) ⇒ 2 B sin( λπ ) = 0
Θ 0 ( θ ) = − Aλ sin( λθ ) + Bλ cos( λθ )
Θ 0 ( − π ) = Aλ sin( λπ ) + Bλ cos( λπ ) = Θ 0 ( π ) = − Aλ sin( λπ )
+ Bλ cos( λπ ) ⇒ 2 Aλ sin( λπ ) = 0 .
Therefore λ n
= nπ
π
=
R equation i n = 0 : λ
0 n ; n
= 0
= 0
⇒
, 1 rR
,
00
0
2 , . . .
+ R 0
0
Θ n
( θ ) = cos( nθ ) and sin( nθ ).
= ( rR 0
0
)
0
= 0 R
0
= B
0 ln r + A
0
.
n ≥ 1 : λ n
= n : r 2
R n
R 00 n
=
+ rR 0 n r γ
− ( n ) 2 R n
= 0 Euler Eq.
⇒ γ ( γ − 1) + γ − n 2 = 0 γ = ± λ n
R n
= c n r − n + d n r n .
= ± n
(27.18)
(27.19)
(27.20)
(27.21)
(27.22)
(27.23)
(27.24)
(27.25)
(27.26)

4
Now since u ( r, θ ) < ∞ as r → 0 we must exclude solutions that blow up. Thus B
0
= 0 and c n
= 0. Therefore u ( r, θ ) = a
0
2
+
© a n cos( nθ ) + b n sin( nθ )
ª r n n =1 u ( a, θ a
) =
0
= f
1
π
( θ
Z
) =
π
− π a
0
2
+
£ a n cos( nθ ) + b n sin( nθ )
¤ a n n =1 f ( θ ) dθ a n b
= n
Z π a
=
− n
π
π f a
− π
− n
Z π
( f
θ ) cos(
( θ nθ
) sin(
) nθ dθ
) dθ.
− π
(27.27)
(27.28)
(27.29)
(27.30)
The Poisson’s Integral Formula:
Using the geometric series it is possible to sum the solution (27.27) explicitly. We proceed as follows: u ( r, θ ) =
= a
0
2
+
¡ a n cos( nθ ) + b
1
Z π n =1 f ( φ ) dφ +
2 π
− π
1
π n =1
³ n r a sin(
´ n
Z nθ
π
− π f
)
(
¢ r n
φ )
£ cos nθ cos nφ + sin( nθ ) sin( nφ )
¤ dφ
=
=
1
π
1
π


Z π
− π
1
Z π
 2
− π f ( φ f ( φ )
(
1
2
) dφ
+
+ n =1 n =1
³ r a
³ r
´ n a
´ n cos
Z n
π
− π
( θ − φ )
) dφ

 f ( φ ) cos n ( θ − φ ) dφ

Now n =1
¡ r a
¢ n cos n ( θ − φ ) =
=
Re
Re
³ =1
³ z
1 − z z n
´ z (1 − ¯ )
´
=
=
=
Re
(1 − z )(1 − ¯ )
( r a
) cos( θ − φ ) −
( r a
1 − 2 ( r a
) cos( θ − φ )+ ar cos( θ − φ ) − r
2
(
) 2 r a
) 2 a 2 − 2 ar cos( θ − φ )+ r 2
(1 − z )(1 − ¯ z =
¡
) = 1 r a
¢
− e
( i ( θ − φ ) z
¡
+ ¯ z (1 − ¯
= 1
) = r a
− e
2 r a i ( θ − φ )
¢ z ) + | z |
− r a
¢
2
2 cos( θ − φ ) + 1?
¡ r a
¢
2
Therefore u ( r, θ ) =
=
1
π
− π
1
Z π
Z π f ( φ )
½
1
2
π
− π f ( φ )
½
+
1
2 a 2 a 2
− ar cos( θ − φ ) − r ar cos( a 2
θ − φ ) +
1
2 r
2
− 2 ar cos( θ − φ ) + r
2 +
2
¾ ar dφ cos(
− 2 ar cos( θ − φ ) + r 2
θ − φ ) − r 2
¾ dφ
(27.31)
(27.32)
(27.33)
(27.34)
(27.35)
(27.36)

Laplace’s Equation u ( r, θ ) =
1
2 π
( a
2
− r
2
)
Z π
− π f ( φ ) a 2 − 2 ar cos( θ − φ ) + r 2 dφ
5
(27.37)
Domain exterior to a circle :
For problem exterior to a circle we require that u ( r, θ ) < ∞ as r → ∞ . In this case we require that B
0 d n
= 0 so that R
0
= A
0 and R n
= r − n ·
£ a n cos nθ + b n
= 0 and that sin( nθ )
¤
. In this case the appropriate solution to the Dirichlet problem becomes: u ( r, θ ) = a n
= a
0
2
+
£ a n cos( nθ ) + b a n
Z π n =1 f ( θ ) cos( nθ ) dθ, b n
π
− π n
= sin( nθ )
¤ r
− n
Z π a n
π
− π f ( θ ) sin( nθ ) dθ.
(27.38)
(27.39)
Example 27.3
Neumann problem on the interior of a circle
Figure 3.
Neumann boundary condition prescribed on the on boundary of a circle (27.40) u rr
∂u
∂r
(
+
1 r a, θ u r
) =
+ f ( r
1
θ
2
) u u 2 π - periodic.
θθ
= 0
(27.40)
∂u
∂r
¯ r = a u ( r, θ ) = a
0
2
= f ( θ ) =
=
+ r n
£ a n cos( nθ ) + b n sin( nθ )
¤ n =1 n =1 nr n − 1
£ a n cos( nθ ) + b n sin( nθ )
¤¯
¯ r = a na n − 1
£ a n cos( nθ ) + b n sin( nθ )
¤
.
n =1
(27.41)
(27.42)
(27.43)
Z π
A solution will not exist unless a
0
= and no steady state solution will exist.
1
2 π
− π f ( θ ) dθ = 0. Otherwise there is a net flux of heat across the boundary

6
Assume that f ( θ ) = I
0
δ
³
θ −
π
´
2
− I
0
δ
³
θ +
π
2
´
.
(a) Level sets of potential field (27.61) (b) BVP for U
1 field and voltage measurements δU
1 with σ = const . Actual voltage δV
1 with a feature (red)
(c) Field U
0 used in the sensitivity Thm
Figure 4.
Sensitivity Thm: δV
1
− δU
1
' − 1
I
R
Ω
∇ U
1
· ∇ U
0
δσdv & backprojection rule: δσ ' −
( δV
1
− δU
1
δU
1
)
σ f ( − θ ) = I
0
δ
= I
0
δ
³
−
³
θ +
³
θ +
π
´
2
π
2
´´
− I
0
δ
− I
0
³
δ
θ −
³
−
π
2
´
³
θ −
= − f
π
2
(
´´
θ )
Thus f is odd ⇒ a
0
= a n
= 0.
na n − 1 b b n n
=
= u ( r, θ ) =
2
π
2 I
0
2
2
0
πna n − 1 aI
0
X sin
³ sin( nπ
2 nθ )
´
π
Z π
I
0
δ n =1
³
θ − n
π
´ sin( nθ ) dθ sin
³ nπ
2
´ ³ r a
´ n
For enrichment ↓ =
=
=
2 aI
0 aI
π aI
π
π
0
0 n =1
"
R n =1
³ e
³ r r a a n =1
´
´ n n z n
1 n
1
−
" cos
"
2 cos n n =1 n
¡
θ −
¡
θ − n
π
2
¢ n
# z n
2 z
1
− n z
2
π
2
¢ cos n
¡
θ +
− cos n
¡
θ + n
π
2
¢ #
π
2 n
=
=
¡
¡ r a r a
¢
¢ e e i i
(
θ − π
(
θ +
2
π
2
)
)
¢ #
.
Now for | z | < 1 :
− ln(1
1
1 − z
− z
= 1 +
) = z z +
+ z z
2
2
2
+
+
· · ·
· · ·
=
= z k k =0
X z n n n =1
.
(27.44)
(27.45)
(27.46)
(27.47)
(27.48)
(27.49)
(27.50)
(27.51)
(27.52)

Laplace’s Equation
Therefore u ( r, θ ) = − aI
π
0
R e
· ln
µ
1 − z
1
1 − z
2
¶¸
.
Now if (1 − z ) = A e iφ then
R e
£ ln(1 − z )
¤
= R e
£ ln
¡
A e iφ
¢¤
= R e[ln A + iφ ] = ln A.
Therefore u ( r, θ ) = − aI
2 π
0 ln
1 − z
1
1 − z
2
¯ 2
.
Now z
1 k 1 − z
1 k
2
=
³ r a
´ e i
(
θ −
π
2
) =
ρ e
= (1 − z
1
= 1 − ρ
¡ e
)(1 − z
1 iφ
1 + e
) =
− iφ
1 iφ
1
¢
¡
1 − ρ e iφ
1
+ ρ 2
¢¡
1 − ρ e
− iφ
1
¢
Similarly z
2
=
³ r
´ a e i
(
φ +
π
2
) =
ρ e iφ
2 and | 1 − z
2
| 2 u u
(
( r, θ r, θ
= 1 − 2 ρ cos φ
1
) =
) = aI
2 π aI
2 π
0
0
= 1 −
+
2 ρ
ρ
2
.
cos φ
2
+ ρ 2 . Therefore ln ln
"
1
· a 2
− 2(
1 − 2(
+ 2 r a r a
) cos(
) cos( ar sin θ
θ
θ
+
−
+ r
π
2
π
2
2
) + (
) + (
¸ r a r a
)
2
)
2 a 2 − 2 ar sin θ + r 2
#
Electrical Impedance Tomography - from the US Patent
7
(27.53)
(27.54)
(27.55)
(27.56)
(27.57)
(27.58)
(27.59)
(27.60)
(27.61)
(a) Level sets of potential field used in the US Patent
(b) Medical imaging device (c) Actual image obtained from EIT
Figure 5.
Figures taken from a US Patent for a medical imaging device based in EIT and the level sets of the solution given in (27.61)

3/10/2012 Monitor: We’ve got it covered | The Ec…
Technology Quarterly: Q1 2012
In this Technology Quarterly
What happened to the flying car?
We’ve got it covered
Reviving autopsy
An open-source robo-surgeon
Pumping ions
Pivoting pixels
No more whirly-splat
Starting from scratch
Meaningful gestures
This is not a video game
Unblinking eyes in the sky
Computing with soup
Packing some power
Can the scientists keep up?
Counting every moment
Taking the long view
Reprints
Mar 3rd 2012 | from the print edition
IT WOULD be much easier to locate and repair damage to bridges, wind turbines and other dumb objects if those objects could tell you what the problem was. Researchers at the University of Strathclyde, in
Britain, led by Mohamed Saafi, are therefore trying to give them a voice, by devising a new sort of smart paint.
It is composed of what sounds like a bizarre mixture: fly-ash, a finegrained waste product from coal-fired power stations; carbon nanotubes, cylindrical molecules made of elemental carbon; and two binding agents, sodium silicate and sodium hydroxide. The result is a economist.com/node/21548497/print 1/2

3/10/2012 Monitor: We’ve got it covered | The Ec… material similar to cement, which makes a suitably tough paint. When it dries, the fly-ash acts as a coating, able to withstand the elements in exposed places. The carbon nanotubes are there to conduct electricity.
The smart bit is that the tubes’ conductivity is affected by cracks in, or corrosion of, the painted surface. When put under stress, for example, the nanotubes bend and become less conductive. If inundated by chloride ions, as a result of corrosion by salt water, their conductivity increases. This makes it possible to monitor damage.
The voltage running through any part of the painted area can be measured remotely, with an array of electrodes distributed across its surface, and data for the entire structure dispatched, via a central transmitter, to a computer. Using a medical-imaging technique called electrical-impedance tomography, Dr Safi and his colleague David
McGahon are devising software that can draw a conductivity map of an entire painted structure.
Nanotechnology has been used in paints before. Sometimes the goal is to bind the paint tightly to the material it has been applied to.
Sometimes it is to channel water molecules efficiently, thus keeping a surface clean. Some paints incorporate tiny silver particles, which capture atmospheric pollutants. But Dr Saafi’s smart paint is new in several ways.
It is cheap, so it is possible to imagine whole structures being built out of it, instead of cement. It is also versatile, theoretically able to detect a broad range of stresses and pollutants. If it performs well, there are currently 3,500 wind turbines—and counting—in Britain alone that could do with a lick of it.
from the print edition | Technology Quarterly
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