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

Lecture 27: Wedges with cut-outs, Dirichlet and Neumannproblems on Circular domains

(Compiled 3 March 2014)

In this lecture we continue with the solution of Laplaces 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: Laplaces equation; Circular domains; Pizza Slice-shaped regions; Dirichlet and Neumann BoundaryConditions, 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)

urr +1rur +

1r2u = 0 (27.1)

u(r, 0) = 0 u(r, ) = 0u(b, ) = 0 u(a, ) = f()

(27.2)

Let u(r, ) = R(r)().

r2(R + 1rR)R(r)

= ()()

= 2 {

r2R + rR 2R = 0 + 2 = 0

(27.3)

2 equation + 2 = 0(0) = 0 = ()

} = A cos +B sin(0) = B = 0 B or = 0, (27.4)

= A sin +B cos() = A sin = 0, = npi n = 0, 1, . . .

(27.5)

R equation n = 0: (rR0) = 0 rR0 = B0 R0 = A0 +B0 ln r. Note

u0(b, ) = R0(b)0() = 0 R0(b) = A0 +B0 ln b = 0, A0 = B0 ln b. (27.6)

Therefore R0 = B0 ln(r/b). Choose B0 = 1.

n 1: r2Rn + rRn 2Rn = 0 R(r) = Anrn +Bnrn

Rn(b) = Anbn +Bnbn = 0 Bn = Anb2n (27.7)Rn(r) = An[rn b2nrn ] Choose An = 1. (27.8)

un(r, ) =[r(

npi ) b2(npi )r(npi )

]cos(npi

)(27.9)

u0(r, ) = ln(rb

) 1 (27.10)

Therefore

u(r, ) = c0 ln(rb

)+

n=1

cn

[r(

npi ) b( 2npi )r(npi )

]cos(npi

)(27.11)

u(a, ) = f() = 2

(c0 ln

(ab

))2

+n=1

cn

[a(

npi ) b( 2npi )r(npi )

]cos(npi

)(27.12)

=a02+

n=1

an cos(npi

). (27.13)

Therefore

2c0 ln(a/b) =2

0

f() d. (27.14)

cn =2

[a(

npi ) b( 2npi )anpi

] 0

f() cos(npi

)d (27.15)

c0 =1

ln(a/b)

0

f() d. (27.16)

Observations: In the special case f() = 1, c0 =1

log(a/b), and cn = 0 n 1. By Fourier basis function orthogonality

0

1 cos(npi

)d = 0 so that the solution reduces to:

u(r, ) =log(r/b)log(a/b)

(27.17)

which is purely radial i.e. has no dependence.

Laplaces Equation 3

27.2 Problems with a complete circle as the boundary

Example 27.2 Dirichlet problem in the interior of a circle

Figure 2. Dirichlet boundary condition prescribed on the on boundary of a circle (??)

urr +1rur +

1r2u = 0 0 < r < a, 0 < < 2pi. (27.18)

BC: u(a, ) = f() u(r, )

4Now since u(r, )

Laplaces Equation 5

u(r, ) =12pi

(a2 r2)pi

pi

f()a2 2ar cos( ) + r2 d (27.37)

Domain exterior to a circle:

For problem exterior to a circle we require that u(r, )

6Special Case - Electrical Impedance Tomography (EIT)

Assume that f() = I0( pi

2

) I0

( +

pi

2

).

(a) Level sets of potential field (27.61) (b) BVP for U1 field and voltage mea-surements U1 with = const. Actualvoltage V1 with a feature (red)

(c) Field U0 used in the sensitivity Thm

Figure 4. Sensitivity Thm: V1 U1 ' 1I

U1 U0dv & backprojection rule: ' (V1U1)U1

f() = I0(( +

pi

2

)) I0

(( pi

2

))(27.44)

= I0( +

pi

2

) I0

( pi

2

)= f() (27.45)

Thus f is odd a0 = an = 0.

nan1bn =2pi

pi0

I0( pi

2

)sin(n) d (27.46)

bn =2I0

pinan1sin(npi2

)(27.47)

u(r, ) =2aI0pi

n=1

sin(n)n

sin(npi2

)( ra

)n(27.48)

For enrichment = 2aI0pi

n=1

( ra

)n 12

[cosn

( pi2

)n

cosn( + pi2

)n

](27.49)

=aI0pi

n=1

( ra

)n [cosn ( pi2 )n

cosn( + pi2

)n

](27.50)

=aI0piRe

[ n=1

zn1n

n=1

zn2n

]z1 =

(ra

)ei(

pi2 )

z2 =(ra

)ei(+

pi2 )

. (27.51)

Now for |z| < 1:1

1 z = 1 + z + z2 + =

k=0

zk

ln(1 z) = z + z2

2+ =

n=1

zn

n

. (27.52)

Laplaces Equation 7

Therefore

u(r, ) = aI0piRe[ln(1 z11 z2

)]. (27.53)

Now if (1 z) = Aei then

Re[ln(1 z)] = Re [ln (Aei)] = Re[lnA+ i] = lnA. (27.54)

Therefore

u(r, ) = aI02pi

ln1 z11 z2

2. (27.55)Now

z1 =( ra

)ei(

pi2 ) = ei1 (27.56)

1 z12 = (1 z1)(1 z1) =(1 ei1)(1 ei1) (27.57)

= 1 (ei1 + ei1)+ 2 (27.58)= 1 2 cos1 + 2. (27.59)

Similarly z2 =( ra

)ei(+

pi2 ) = ei2 and |1 z2|2 = 1 2 cos2 + 2. Therefore

u(r, ) =aI02pi

ln

[1 2( ra ) cos( + pi2 ) + ( ra )21 2( ra ) cos( pi2 ) + ( ra )2

](27.60)

u(r, ) =aI02pi

ln[a2 + 2ar sin + r2

a2 2ar sin + r2]

(27.61)

Electrical Impedance Tomography - from the US Patent

(a) Level sets of potential field used inthe 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 solutiongiven in (27.61)

Mar 3rd 2012 | from the print edition

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M257_316_2012_Lecture_27.pdfEIT_ The EconomistMonitor_ Weve got it covered _ The Economist.pdf