18.336j/6.335j: fast methods for partial differential and...

18.336J/6.335J: Fast methods for partial differential and integral equations

Cambridge, September 21, 2017

MITMathematics Department

Instructor: Carlos Pérez Arancibia

Lecture 5

Single and double layer potentials

Single-layer potential:

Double-layer potential:

where

Sk : C(�) ! C2(Rd \ �)

Dk : C(�) ! C2(Rd \ �)

(x 2 Rd \ �)

(x 2 Rd \ �)

(� 2 C2)

(� 2 C2)

(Sk ) (x) :=

Z

�Ek(x� y) (y) dsy

(Dk ) (x) :=

Z

�

@Ek

@ny(x� y) (y) dsy

Ek(x� y) =

8>>>>>>><

>>>>>>>:

e

ik|x�y|

4⇡|x� y| if k � 0, d = 3,

� 1

2⇡log |x� y| if k = 0, d = 2,

i

4

H(1)0 (k|x� y|) if k > 0, d = 2.

where

Boundary integral operators

‣Notation:

andu|±(x) = lim✏>0,✏!0

u(x± ✏n(x))@u

@n

��±(x) = lim

✏>0,✏!0ru(x± ✏n(x)) · n(x)

where x 2 �

Dk[ ]��± =±

2+Dk[ ], Sk[ ]

��± = Sk[ ],

@Dk[ ]

@n

��±= Nk[ ],

@Sk[ ]

@n

��±=⌥

2+Kk[ ]

‣ Properties. For a densify function we have: 2 C(�)

(Sk )(x) :=

Z

�Ek(x� y) (y) dsy (Dk )(x) :=

Z

�

@Ek

@ny(x� y) (y) dsy

(Kk )(x) :=

Z

�

@Ek

@nx

(x� y) (y) dsy

(Nk )(x) :=

Z

�

@2Ek

@nx

@ny

(x� y) (y) dsy

Maue’s Integration by parts formula

The hypersingular operator has to be interpreted as a Hadamardfinite part integral that can be expressed in terms of a Cauchyprincipal value integral:

(Nk )(x) = k2Z

�Ek(x� y)n(x) · n(y) (y) ds

y

+p.v.

Z

�rx

sEk(x� y) ·ry

s (y) dsy

Calderón Identities

Assume � is of class C2. Then the single-layer (Sk), double-layer

(Dk), adjoint double-layer (Kk) and hypersingular (Nk) operators

satisfy:

DkSk = SkKk, NkDk = KkNk,

D2k � SkNk =

I

4

, K2k �NkSk =

I

4

.

Calderón Identities

Proof. Let ', 2 C1(�) and consider the function ue = �Dk'+

Sk . Evaluating ue|+ and @ue/@n|+ on � we obtain

ue|+ = �✓I

2

+Dk

◆'+Sk and

@ue

@n

��+= �Nk'+

✓�I

2

+Kk

◆ .

(1)

Since �ue + k2ue = 0 in Rd \ � and ue satisfies the radiation

condition, we have, by Green’s formula, that

ue|+ =

✓I

2

+Dk

◆ue|+ � Sk

@ue

@n

��+

and (2)

@ue

@n

��+= Nkue|+ �

✓�I

2

+Kk

◆@ue

@n

��+. (3)

Replacing (1) in (2) and (3), all four Calderon identities follow.

Weakly singular operators

Definition. A kernel K is said to be weakly singular if K is defined and

continuous for all x and y in �, x 6= y, and there exist constants ↵ 2 (0, 2] andM > 0 such that for all x,y 2 �, x 6= y, we have

|K(x,y)| M |x� y|↵�d+1

Theorem. Let � 2 C2, then there exists L > 0 such that

|n(y) · (x� y)| L|x� y|2 and

|n(x)� n(y)| L|x� y| 8x,y 2 �.

Consider the integral operator A : C(�) ! C(�), � 2 C2, defined

by

(A )(x) =

Z

�K(x,y) (y) dsy, x 2 �.

Nate that the single-layer (Sk), double-layer (Dk) and adjoint double-

layer (Kk) operators are weakly singular for a surface � 2 C2.

Compact operators

Theorem. The operators Sk : C(�) ! C(�), Dk : C(�) ! C(�), Kk : C(�) !C(�) and Nk �N0 : C(�) ! C(�) are compact.

Theorem. The space of complex-valued continuous functions C(�), � 2 C2,

equipped with the norm k�k = max

x2� |�(x)|, is a Banach space (i.e, a complete

normed vector space).

Definition. A linear operator A : X ! Y (X and Y are Banach spaces) is

compact if for each bounded sequence {�n} ⇢ X the sequence {A�n} contains

a convergent subsequence.

Note. The eigenvalues � of A (A compact) can only accumulate at 0.

Theorem. An integral operator A with continuous or weakly singular kernel is

a compact operator on C(�).

Spectral properties�(Sk) �(Dk)

�(Kk) �(Nk)

Interior problems

Interior Neumann. Find u : ⌦ ! C, u 2 C2(⌦) \ C1

(⌦), � = @⌦ 2 C2, such

that:

(IN)

8<

:

�u+ k2u = 0 in ⌦,

@u

@n= g on �.

Interior Dirichlet. Find u : ⌦ ! C, u 2 C2(⌦) \ C1

(⌦), � = @⌦ 2 C2, such

that:

(ID)

(�u+ k2u = 0 in ⌦,

u = f on �.

Theorem. The interior Dirichlet problem has at most one solution

for all k > 0 such that k 6=p�D where �D is a Laplace Dirichet

eigenvalue, while the interior Neumann problem has at most one

solution for all k > 0 such that k 6=p�N , where �N is a Laplace

Neumann eigenvalue.

Exterior problems

Exterior Dirichlet. Find u : Rd \ ⌦ ! C, u 2 C2(Rd \ ⌦) \ C1

(Rd \ ⌦),

� = @⌦ 2 C2, such that:

(ED)

8>>>><

>>>>:

�u+ k2u = 0 in Rd \ ⌦,

u = f on �,

lim

|x|!1|x|(d�1)/2

⇢@u

@|x| � iku

�= 0

Exterior Neumann. Find u : Rd \ ⌦ ! C, u 2 C2(Rd \ ⌦) \ C1

(Rd \ ⌦),

� = @⌦ 2 C2, such that:

(EN)

8>>>>><

>>>>>:

�u+ k2u = 0 in Rd \ ⌦,@u

@n= g on �,

lim

|x|!1|x|(d�1)/2

⇢@u

@|x| � iku

�= 0

Theorem. The exterior Dirichlet and Neumann problems have at

most one solution.

Green formula:

Interior Dirichlet problem

• Direct Integral equation formulations

second-kind integral equation for the unknown function

first-kind integral equation for the unknown function

Integral equation formulations: Interior problems

Interior Neumann problem:

=@u

@n

��

= u��

Sk[ ] =f

2

+Dk[f ] on �

2

+Dk[ ] = Sk[g] on �

(Dku) (x)�✓Sk

@u

@n

◆(x) =

⇢�u(x), x 2 ⌦,0, x 2 Rd \ ⌦.


• Direct Integral equation formulations

“first”-kind integral equation for the unknown function



Interior Neumann problem:

=@u

@n

��

= u��

Taking normal derivative of the Green’s representation formula we

obtain:

(Nku)(x)�✓Kk

@u

@n

◆(x) = �1

2

@u(x)

@n, x 2 �

� 2

+Kk[ ] = Nk[f ] on �

Nk[ ] = �g

2

+Kk[g] on �

Green formula:

Exterior Dirichlet problem:

• Direct integral equation formulations


first-kind integral equation for the unknown function

Integral equation formulations: Exterior problems

Exterior Neumann problem:

= u��+

=@u

@n

��+

Sk[ ] = �f

2

+Dk[f ] on �

� 2

+Dk[ ] = Sk[g] on �

(Dku) (x)�✓Sk

@u

@n

◆(x) =

⇢0, x 2 ⌦,

u(x), x 2 Rd \ ⌦.

Exterior Dirichlet problem:

• Direct integral equation formulations

“first”-kind integral equation for the unknown function



Exterior Neumann problem:

= u��+

=@u

@n

��+

Taking normal derivative of the Green’s representation formula we

obtain:

(Nku)(x)�✓Kk

@u

@n

◆(x) =

1

2

@u(x)

@n, x 2 �

2

+Kk[ ] = Nk[f ] on �

Nk[ ] =g

2

+Kk[g] on �

We look for a solution of the form:

and imposing the corresponding boundary condition we obtain:

• Indirect integral equation formulation



Interior Neumann problem:second-kind integral equation for the unknown function

first-kind integral equation for the unknown functionSk[ ] = f on �

2

+Kk[ ] = g on �

u(x) = (Sk ) (x), x 2 Rd \ �






Interior Neumann problem:“first”-kind integral equation for the unknown function


u(x) = (Dk ) (x), x 2 Rd \ �

Nk[ ] = g on �

� 2

+Dk[ ] = f on �





Exterior Dirichlet problem

Exterior Neumann problem:second-kind integral equation for the unknown function

first-kind integral equation for the unknown functionSk[ ] = f on �

� 2

+Kk[ ] = g on �

u(x) = (Sk ) (x), x 2 Rd \ �





Exterior Dirichlet problem

Exterior Neumann problem:“first”-kind integral equation for the unknown function


u(x) = (Dk ) (x), x 2 Rd \ �

Nk[ ] = g on �

2

+Dk[ ] = f on �

Calderón Preconditioners



satisfy:


D2k � SkNk =

I

4

, K2k �NkSk =

I

4

.

Consider the first kind integral equation Sk[ ] = f on �

Problem: Sk has eigenvalues clustered at 0.

second-kind integral equation

Solution: Precondition the integral equation from the left by Nk

and use the fact that NkSk = �I/4 +K2k :

NkSN [ ] =

✓�I

4

+K2k

◆[ ] = Nk[f ] on �


�(Sk)

... GMRES convergence ...

krnk ✓1� �2

min

(1/2(A+AT))

�max

(ATA)

◆n/2

kr0

k

�

✓�I

4+K2

k

◆




satisfy:


D2k � SkNk =

I

4

, K2k �NkSk =

I

4

.

Consider the “first” kind integral equation

second-kind integral equation

Nk[ ] = g on �

Problem: Nk has eigenvalues that tend to infinity.

Solution: Precondition the integral equation from the left by Sk

and use the fact that SkNk = �I/4 +D2k :

SNNk[ ] =

✓�I

4

+D2k

◆[ ] = Sk[g] on �


... GMRES convergence ...

krnk ✓1� �2

min

(1/2(A+AT))

�max

(ATA)

◆n/2

kr0

k

�(Nk) �

✓�I

4+D2

k

◆

Uniqueness of solutionsThe first and second kind integral equations

Sk[ ] = f and

✓�I

2

+Kk

◆[ ] = g

have at most one solution for all k > 0 except when �k2 is an

eigenvalue of the interior Dirichlet problem for the Laplacian, i.e.,

when there exist uD 6= 0 and �D > 0 such that ��uD = �DuD in

⌦ and uD = 0 on � = @⌦.

The “first” and second kind integral equations

Nk[ ] = g and

✓I

2

+Dk

◆[ ] = f

have at most one solution for all k > 0 except when �k2 is an

eigenvalue of the interior Neumann problem for the Laplacian, i.e.,

when there exist uN 6= 0 and �N � 0 such that ��uN = �NuN in

⌦ and

@uN

@n= 0 on � = @⌦.

Discretization methods: Collocation Method

Say we want to solve a second-kind integral equation:

(x) +

Z

�K(x,y) (y) dsy = f(x) on �.

1. Approximate using certain basis functions {vn}Nn=1 (vn :

� ! C):

(x) ⇡ N (x) =

NX

n=1

cnvn(x).

2. Replace the approximation in the integral equation:

NX

n=1

cn

✓vn(x) +

Z

�K(x,y)vn(y) dsy

◆= f(x), x 2 �.

3. Evaluate the expressions on both sides of the identity above

at points {xm}Nm=1 on �:

NX

n=1

cn

✓vn(xm) +

Z

�K(xm,y)vn(y) dsy

◆= f(xm), m = 1, . . . , N.

4. Solve the linear system Ac = f where c = [c1, . . . , cN ]

T 2 CN,

f = [f(x1), . . . , f(xN )]

T 2 CNand

Am,n = vn(xm) +

Z

�K(xm, y)vn(y) dsy, n,m = 1, . . . , N.

Discretization methods: Collocation Method

Say we want to solve a second-kind integral equation:

(x) +

Z

�K(x,y) (y) dsy = f(x) on �.

1. Approximate using certain basis functions {vn}Nn=1 (vn :

� ! C):

(x) ⇡ N (x) =

NX

n=1

cnvn(x).

2. Replace the approximation in the integral equation:

NX

n=1

cn

✓vn(x) +

Z

�K(x,y)vn(y)

◆= f(x), x 2 �.

3. Evaluate the expressions on both sides of the identity above

at points {xm}Nm=1 on �:

NX

n=1

cn

✓vn(xm) +

Z

�K(xm,y)vn(y) dsy

◆= f(xm), m = 1, . . . , N.

4. Solve the linear system Ac = f where c = [c1, . . . , cN ]

T 2 CN,

f = [f(x1), . . . , f(xN )]

T 2 CNand

Am,n = vn(xm) +

Z

�K(xm, y)vn(y) dsy, n,m = 1, . . . , N.

Discretization methods

(y) ⇡NX

j=1

jvj(y)

• Boundary element method (Galerkin):i) Replace the curve by its discretization and expand the unknown function in terms of basis functions:

ii) Replace the approximate density in the integral operator

iii) Test against the same basis functions

� �h

y 2 �h

Z

�K(x,y) (y) dsy ⇡

NX

j=1

j

Z

�h

K(x,y)vj(y) dsy

NX

j=1

j

Z

�h

vi(x)

Z

�h

K(xi,y)vj(y) dsy dsx

=

Z

�h

f(x)vi(x) dsx

i = 1, . . . , Nfor

Discretization methods

• Nyström method:i) Parametrize and express the integral equation as

ii) Use a quadrature rule to approximate the integral operator:

iii) Evaluate the integral equation at the quadrature points:

�

i = 1, . . . , Nfor

(t) +

Z T

0K(t, ⌧) (⌧) d⌧ = g(t), t 2 [0, T )

Z T

0K(t, ⌧) (⌧) d⌧ ⇡

NX

j=1

K(t, tj) (tj)wj , t 2 [0, T )

i +NX

j=1

K(ti, tj) (tj)wj = g(ti)

In parametric form a second-kind integral equation is given by:

K(t, ⌧) = K1(t, ⌧) ln

✓4 sin2

t� ⌧

2

◆+K2(t, ⌧)where (K1 and K2 analytic)

Use the trapezoidal rule:

(t)�Z 2⇡

0K(t, ⌧) (⌧) d⌧ = g(t), 0 t 2⇡ ( is 2⇡-periodic)

Discretized integral equation:

It can be shown that for an analytic right-hand side it holds that:

tj =j⇡

n

Z 2⇡

0ln

✓4 sin2

t� ⌧

2

◆f(⌧) d⌧ ⇡

2n�1X

j=0

R(n)j (t)f(tj),

Z 2⇡

0f(⌧) d⌧ ⇡ ⇡

n

2n�1X

j=0

f(tj),

(n)(ti)�2n�1X

j=0

n

R(n)j (ti)K1(ti, tj) +

⇡

nK2(ti, tj)

o

(n)(tj) = g(ti), i = 0, . . . , 2n� 1

| (n)(t)� (t)| C e�n�, 0 t 2⇡, � > 0.

Martensen-Kussmaul Nyström method

tj =j⇡

n

Z 2⇡

0ln

✓4 sin2

t� ⌧

2

◆f(⌧) d⌧ ⇡

2n�1X

j=0

R(n)j (t)f(tj),

Nyström method

t 2 [0, 2⇡]

R(n)j (t) = �2⇡

n

n�1X

m=1

1

mcos(m(t� tj))�

⇡

n2cos(n(t� tj)), j = 0, . . . , 2n� 1

where

For the exterior Neumann problem, for example, we have:

K(t, ⌧) =ik

2{x0

2(⌧)[x1(⌧)� x1(t)]� x01(⌧)[x2(⌧)� x2(t)]}

H(1)1 (k|x(t)� x(⌧)|)|x(t)� x(⌧)|

K1(t, ⌧) =k

2⇡{x0

2(⌧)[x1(t)� x1(⌧)]� x01(⌧)[x2(t)� x2(⌧)]}

J1(k|x(t)� x(⌧)|)|x(t)� x(⌧)|

K2(t, ⌧) = K(t, ⌧)�K1(t, ⌧) ln

✓4 sin2

t� ⌧

2

◆

� = {(x1(t), x2(t)) 2 R2 : 0 t 2⇡}where

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18.336j/6.335j: fast methods for partial differential and...

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