mathematical methods for engineering the boundary element ...alonso/didattica/w... · ana alonso...

Mathematical methods for engineeringThe boundary element method - Collocation

Ana Alonso

University of Trento

Trento Winter School on Numerical MethodsFebruary 20th- March 3rd, 2017

Ana Alonso Mathematical methods for engineering

Introduction and notationGiven z ∈ Rd with d = 2 or 3I K (·, z) denotes the fundamental solution of the Laplace

equation:−∆K = δz .

K (x, z) =

12π log 1

|x−z| if x ∈ R2

14π

1|x−z| if x ∈ R3 .

Let us consider Ω ⊂ Rd .

I T (·, z) denotes the normal derivative of K (·, z):

T (·, z) := ∇K (·, z) · n ,being n the unit outward normal vector on ∂Ω =: Γ.For x ∈ Γ we have

T (x, z) =

−1

2π(x−z)·n(x)|x−z|2 if Ω ⊂ R2

− 14π

(x−z)·n(x)|x−z|3 if Ω ⊂ R3 .


Basic integral equations

−∆u = 0 in Ωu = ϕ in ΓD∂u∂n = g on ΓN .

Internal points: z ∈ Ω

u(z) +

∫Γ

u(x)T (x, z) ds(x)−∫

Γ

∂u

∂n(x)K (x, z) ds(x) = 0 .

Boundary points: z ∈ Γ (regular boundary)

1

2u(z) +

∫Γ


Γ

∂u

∂n(x)K (x, z) ds(x) = 0 .


Boundary element methods

I The approximate solution is piecewise-polynomial.

u ≈ uN q =∂u

∂n≈ qN .

I Also the domain Ω is approximated by a polyhedral domainΩh,

I hence the boundary Γ is approximated by Γh = ∂Ωh.

Ωh

Ω


Collocation: constant elements

I The boundary of Ω is assumed to be polygonal divided in Nelements.

I Both u and q := ∂u∂n are approximated using piecewise

constant functions: uN , qN .

I The unknowns are the values at the mid-element node, xj ,j = 1, . . . ,N.

uN(x) =N∑j=1

uN(xj)χj(x)

qN(x) =N∑j=1

qN(xj)χj(x)xi

Pi

Pi+1

Ωh

Ω


Collocation: constant elementsBoundary points: z ∈ Γ

1

2u(z)− 1

2π

∫Γ

(x− z) · n(x)

|x− z|2u(x) ds(x)− 1

2π

∫Γ

log1

|x− z|∂u

∂n(x) ds(x) = 0 .

In particular for i = 1, . . . ,N

1

2u(xi )−

1

2π

∫Γ

(x− xi ) · n(x)

|x− xi|2u(x) ds(x)− 1

2π

∫Γ

log1

|x− xi|∂u

∂n(x) ds(x) = 0 .

Now we replace

u(x) ≈ uN(x) =N∑j=1

uN(xj)χj(x) ,

q(x ≈ qN(x) =N∑j=1

qN(xj)χj(x) .



For i = 1, . . . ,N

1

2uN(xi )−

1

2π

N∑j=1

uN(xj)

∫Γj

(x− xi ) · n(x)

|x− xi |2ds(x)

− 1

2π

N∑j=1

qN(xj)

∫Γj

log1

|x− xi |ds(x) = 0

Let us set

Ti,j := − 1

2π

∫Γj

(x− xi ) · n(x)

|x− xi |2ds(x) Ki,j :=

1

2π

∫Γj

log1

|x− xi |ds(x)

uj = uN(xj) qj = qN(xj)

1

2ui +

N∑j=1

Ti,juj −N∑j=1

Ki,jqj = 0



Ti ,j := − 1

2π

∫Γj

(x− xi ) · nj|x− xi |2

ds(x) Ki ,j :=1

2π

∫Γj

log1

|x− xi |ds(x)

I For the case i 6= j integrals Ki ,j and Ti ,j can be calculatedusing, for instance, Gauss quadrature rules.

I For i = j the presence on the singularity requires a moreaccurate integration.For the particular case of constant elements Ki ,i and Ti ,i canbe computed analytically.Notice that if x ∈ Γi then (x− xi ) · ni = 0 hence Ti ,i = 0.


Evaluation of integrals

Ki ,i =1

2π

∫Γi

log1

|x− xi |ds(x) =

1

2π

Li2

∫ 1

−1log

1

|r Li/2|dr

x = Pi1− r

2+ Pi+1

1 + r

2xi =

Pi + Pi+1

2

where Li = |Pi+1 − Pi | is the length of Γi .

I Clearly if i = N we replace Pi+1 with P1.

By integration by parts

∫ 1

0log

(rLi2

)dr = log

Li2− 1 .

Ki ,i =Li2π

(1− log

Li2

).


Gauss quadrature rules

∫ 1

−1f (r) dr ≈

K∑k=1

ωk f (rk)

It is a quadrature rule constructed to yield an exact result forpolynomials of degree 2K − 1 or less.For instance, in the four points rule the points are

±√

1

7(3− 2

√6/5) with weight

18 +√

30

36

±√

1

7(3 + 2

√6/5) with weight

18−√

30

36.

∫ΓjF (x) ds(x) =

Lj2

∫ 1−1 F

(Pj

1−r2 + Pj+1

1+r2

)dr

≈ Lj2

∑4k=1 ωkF

(Pj

1−rk2 + Pj+1

1+rk2

).


Gauss quadrature rules

Ti,j := − 1

2π

∫Γj

(x− xi ) · nj

|x− xi |2ds(x) ≈ Lj

4π

4∑k=1

ωkT (Pj1− rk

2+ Pj+1

1 + rk2

, xi )

Ki,j :=1

2π

∫Γj

log1

|x− xi |ds(x) ≈ Lj

4π

4∑k=1

ωkK(Pj1− rk

2+ Pj+1

1 + rk2

, xi )

function [t,k]=gqcon(P1,P2,z)

r=[-0.8611363116 -0.3399810436 0.3399810436 0.8611363116];

w=[0.3478548451 0.6521451549 0.6521451549 0.3478548451];

L=norm(P2-P1);

n=[0 1; -1 0]*(P2-P1)/L;

for i=1:4

s=P1*(1-r(k))/2+P2*(1+r(k))/2;

T(k)=-(s-z)*n/norm(s-z)^2;

K(k)=-log(norm(s-z));

end

t=L/(4*pi)*w*T’;

k=L/(4*pi)*w*K’;

P1 = Pj P2 = Pj+1 z = xi


Constant elements - Boundary data

1

2ui +

N∑j=1

Ti ,juj −N∑j=1

Ki ,jqj = 0

−∆u = 0 in Ωu = ϕ in ΓD∂u∂n = g on ΓN

I If ΓD = ∂Ω the vector u = (uj)Nj=1 is given and the unknowns

are the component of the vector q = (qj)Nj=1.

I On the other hand if ΓN = ∂Ω then the vector q = (qj)Nj=1 is

given and the unknowns are the component of the vectoru = (uj)

Nj=1 (but matrix 1

2 I + T is singular!)

I In general N1 values of u corresponding to ΓD and N2 valuesof q corresponding to ΓN are known with N1 + N2 = N.


Dirichlet problem

Kq =

(1

2I + T

)ϕ

−∆u = 0 in Ωu = ϕ on Γ

function q=bemconD(coor,u)

N=size(coor,2);

T=zeros(N); K=zeros(N);

coor=[coor coor(:,1)];

L=zeros(N,1);

for i=1:N, L(i)=norm(coor(:,i+1)-coor(:,i)); end

Kdiag=L/(2*pi).*(1-log(L/2));

for i=1:N

x=0.5*(coor(:,i)+coor(:,i+1));

for j=1:N

if j~=i

[t,k]=gqcon(coor(:,j),coor(:,j+1),x);

T(i,j)=t; K(i,j)=k;

end

end

end

K=K+diag(Kdiag);

b=(1/2*eye(N)+T)*u;

q=K\b;


Dirichlet problem

Input: the geometry and the boundary conditions.

Output: the solution of the linear system.

I The geometry is given by a 2× N matrix with the coordinatesof the points Pj that define the elements.

I The boundary conditions are given in a column vectorcontaining the values of ϕ at each collocation point.

I The solution is saved in a column vector.


Example

−∆u = 0 in Ω = x ∈ R2 : |x| < 1u = 2x1 + 3x2 on ∂Ω

N=input(’Number of subintervals: ’);

x=cos(2*pi*(1:N)/N);

y=sin(2*pi*(1:N)/N);

coor=[x;y];

coorc=[coor coor(:,1)];

pm=(coorc(:,1:end-1)+coorc(:,2:end))/2;

u=2*pm(1,:)+3*pm(2,:);

qapp=bemconD(coor,u’);

Notice that in this example ∂u∂n = u!

q=u’;

errq=max(abs(q-qapp))


The solution at internal points

u(z) +

∫Γ


Γ

∂u

∂n(x)K (x, z) ds(x) = 0

u(z) ≈N∑j=1

[qj

∫Γj

K (x, z) ds(x)− uj

∫Γj

T (x, z) ds(x)

]

Input: the geometry Γ, the discrete functions on Γ uN and qN , andthe internal points z where we want to compute the value of thesolution.

Output: the (approximated) value of the solution at the giveninternal points u(z).

The integrals∫

ΓjT (x, z) ds(x) and

∫ΓjK (x, z) ds(x) are

approximated using Gauss quadrature.



function val=SolIntcon(coor,sol,points)

m=size(points,2);

N=size(coor,2);

coor=[coor,coor(:,1)];

val=zeros(1,m);

for i=1:m

Q=points(:,i);

for j=1:N

[t,k]=gqcon(coor(:,j),coor(:,j+1),Q);

val(i)=val(i)-t*sol(j,1)+k*sol(j,2);

end

end

I The geometry is given by a 2× N matrix with the coordinates of thepoints Pj that define the elements.

I sol is a m × 2 matrix. The first column contains the coefficients ui andthe second on the coefficients qi .

I The internal points are given in a 2×m matrix being m the number ofinternal point where we will compute the solution.


Example −∆u = 0 in Ω = x ∈ R2 : |x| < 1u = 2x1 + 3x2 on ∂Ω




coor=[x;y];

coorc=[coor coor(:,1)];

pm=(coorc(:,1:end-1)+coorc(:,2:end))/2;

u=2*pm(1,:)+3*pm(2,:);

qapp=bemconD(coor,u’);

q=u’;

errq=max(abs(q-qapp))

solD=[u’,qapp];

points=[-1/2 0 1/2 0 0; 0 0 0 -1/2 1/2];

IntD=SolIntcon(coor,solD,points)

exact=2*points(1,:)+3*points(2,:)


Exercise 1

Let us set Ω =x ∈ R2 : (x1 − 1)2 + (x2 − 1)2 < 1

.

Approximate the solution of the Dirichlet boundary value problem−∆u = 0 in Ω

u = log(x21 + x2

2 ) on ∂Ω


Neumann problem

(1

2I + T

)u = Kg

∆u = 0 in Ω∂u∂n = g on Γ

I Necessary condition for the existence of solution is∫Γ g ds = 0.

I The solution is not unique because if u is a solution thenu + c is also a solution for all c ∈ R.

I In the discrete problem we choose the solution withuN(x1) = 0, namely we eliminate the first equation and thefirst unknown of the linear system.


Example

I Modify function bemconD in order to writte a functionbemconN that implements the boundary element method usingcollocation and piecewise constant approximation, for theLaplace equation with Neumann boundary conditions.

I Approximates a solution of the Neumann boundary valueproblem

−∆u = 0 in Ω = x ∈ R2 : |x| < 1∂u∂n = 2x1 + 3x2 on ∂Ω


Exercise 2

Approximate a solution of the Neumann boundary value problem

−∆u = 0 in Ω = (−1, 1)× (−1, 1)

∂u∂n =

−3 on (−1, 1)× −12 on 1 × (−1, 1)3 on (−1, 1)× 1−2 on −1 × (−1, 1) .

(Exact solution: u(x) = 2x1 + 3x2.)


Mixed problem

1

2ui +

N∑j=1

Ti ,juj −N∑j=1

Ki ,jqj = 0

−∆u = 0 in Ωu = ϕ in ΓD∂u∂n = g on ΓN

The key point is to rearrange the coefficients uj and qj in order towrite the linear system as:

Ax = b

where x is the vector of unknowns (u on ΓN , q on ΓD)and b is adata vector obtained multiplying the corresponding coefficient bythe known values of u or q.


Mixed problem

B =1

2I + T Bu− Kq = 0 Ax = b

for i=1:N

for j=1:N

if u(j) is unknown

A(i,j)=B(i,j)

b(i)=b(i)+K(i,j)*q(j)

else

A(i,j)=-K(i,j)

b(i)=b(i)-B(i,j)*u(j)

end

end

end


Linear elements

I The approximate solutions uN and qN are piecewise-linear.

I uN will be continuous while qN can be discontinuous in cornerpoints

Γj =

Pj

1− r

2+ Pj+1

1 + r

2: r ∈ [−1, 1]

.

If j = N then “j + 1 = 1”.

Pn

n

PP

j−1

j

j+1

j−1

j

uqq

jjj−+

uN|Γj= uN(Pj)N1(r) + uN(Pj+1)N2(r)

qN|Γj= q+

N (Pj)N1(r) + q−N (Pj+1)N2(r)

N1(r) = 1−r2 , N2(r) = 1+r

2 , with r ∈ [−1, 1].


Linear elements

Corner points: z ∈ Γ

C (z)u(z) +

∫Γ


Γ

∂u

∂n(x)K (x, z) ds(x) = 0

C (z) ≡ free term ≡ α(z)

2π. α

z(z) Ω


Linear elements

C (Pi )uN(Pi )+

∫∂Ω

uN(x)T (x,Pi ) ds(x)−∫∂Ω

qN(x)K (x,Pi ) ds(x) = 0

∫ΓjuN(x)T (x,Pi ) ds(x) =

ujLj2

∫ 1−1 N1(r)T (PjN1(r) + Pj+1N2(r),Pi ) dr

+uj+1Lj2

∫ 1−1 N2(r)T (PjN1(r) + Pj+1N2(r),Pi ) dr

For k = 1, 2

Ak(i , j) :=Lj2

∫ 1

−1Nk(r)T (PjN1(r) + Pj+1N2(r),Pi ) dr


Linear elements

Analogously∫ΓjqN(x)K (x,Pi ) =

q+j

Lj2

∫ 1−1 N1(r)K (PjN1(r) + Pj+1N2(r),Pi ) dr

+q−j+1Lj2

∫ 1−1 N2(r)K (PjN1(r) + Pj+1N2(r),Pi ) dr

For k = 1, 2

Bk(i , j) :=Lj2

∫ 1

−1Nk(r)K (PjN1(r) + Pj+1N2(r),Pi ) dr

C (Pi )ui+N∑j=1

[ujA1(i , j) + uj+1A2(i , j)] =N∑j=1

[q+j B1(i , j) + q−j+1B2(i , j)

]


Linear elements

C (Pi )ui+N∑j=1

[A1(i , j)uj + A2(i , j)us ] =N∑j=1

[B1(i , j)q+

j + B2(i , j)q−s

]

s = s(j) =

j + 1 if j = 1, . . . ,N − 11 if j = N

for i=1:Nfor j=1:N

compute A1(i , j), A2(i , j), B1(i , j), B2(i , j).For each one of this four coefficients

if it multiplies an unknown then the coefficient is added to thecorresponding entry of the matrix (ai,j or ai,s),

if it multiplies a data the product is added to the right hand term.


Boundary conditions

1. Neumann - Neumann conditionuj unknown, q−j given, q+

j given.

2. Dirichlet - Neumann conditionuj given, q−j unknown, q+

j given.

3. Neumann - Dirichlet conditionuj given, q−j given, q+

j unknown.

4. Dirichlet - Dirichlet condition (regular point)uj given, q−j unknown, q+

j unknown.

but q−j = q+j

5. Dirichlet - Dirichlet condition (non regular point)uj given, q−j unknown, q+

j unknown.

and q−j 6= q+j The gradient approach


The gradient approach

q−j = ∇u(Pj) · nj−1

q+j = ∇u(Pj) · nj

∇u(Pj) = |∇u(Pj)|e(Pj)

Pn

n

PP

j−1

j

j+1

j−1

j

uqq

jjj−+

Assuming linear evolution along the ele-ments adjacent to the corner in Pj a linearuj(P) can be calculated.

I e(Pj) is approximated using ∇uj (thatis a constant vector).

e(Pj) ∇uj|∇uj |

I |∇u(Pj)| is the unknown vj .

q−j vj∇uj|∇uj |

· nj−1 q+j vj

∇uj|∇uj |

· nj .


Evaluation of the coefficients

I If Pi 6∈ Γj Gauss quadratureI If Pi ∈ Γj

Ak(i , j) = 0 for k = 1, 2 because ∀P ∈ Γj (P− Pi ) · n = 0 .Bk(i , j) are improper integrals that can be computed exactly.

B1(j , j) = −Lj2

∫ 1−1

12π

1−r2 log

∣∣Pj1−r

2 + Pj+11+r

2 − Pj

∣∣ dr= −Lj

2

∫ 1−1

12π

1−r2 log

∣∣(Pj+1 − Pj)1+r

2

∣∣ dr(1+r

2 = s , dr = 2 ds , 1−r2 = 1− s

)= − Lj

2π

∫ 10 (1− s) log(Ljs) ds =

Lj2π

(34 −

12 log Lj

)


Evaluation of the coefficients

B1(j , j) =Lj2π

(34 −

12 log Lj

)B2(j , j) =

Lj2π

(14 −

12 log Lj

)B1(j + 1, j) =

Lj2π

(14 −

12 log Lj

)B2(j + 1, j) =

Lj2π

(34 −

12 log Lj

)I Free terms: for a constant solution u = M

C (Pi )M +N∑j=1

[A1(i , j)M + A2(i , j)M

]= 0

C (Pi ) = −N∑j=1

[A1(i , j) + A2(i , j)

]Ana Alonso Mathematical methods for engineering

Input data

The geometry and the boundary conditions.

I The geometry is given by a 2× N matrix with the coordinatesof the points Pj that define the elements (as in the case ofconstant elements).

I The type of boundary conditions is given in a vector with Ncomponents. Each entry is an integer from 1 to 5.

I The boundary data are given in a matrix 3× N matrix. Thefirst line contains the values of u, the second one q− and thelast one q+.Notice that in this matrix there are unknowns coefficients.The same matrix is used to save the solution, filling theunknowns coefficients



u(z) ≈ 1

2π

∫∂Ω

uN(x)(x− z) · n(x)

|x− z|2ds(x)− 1

2π

∫∂Ω

qN(x) log |x− z| ds(x)

The integrals are calculated using Gauss quadrature.

function s=SolIntlin(coor,sol,data)

K=length(data);

N=length(coor);

s=zeros(1,K);

for i=1:K

Q=data(:,i);

for j=1:N-1

[a1,a2, b1,b2]=gqlin(coor(:,j),coor(:,j+1),Q);

s(i)=s(i)-a1*sol(1,j)-a2*sol(1,j+1)+b1*sol(3,j)+b2*sol(2,j+1);

end

[a1,a2, b1,b2]=gqlin(coor(:,N),coor(:,1),Q);

s(i)=s(i)-a1*sol(1,N)-a2*sol(1,1)+b1*sol(3,N)+b2*sol(2,1);

end


gqlin

function [a1,a2,b1,b2]=gqlin(P1,P2,Q)

r=[-0.8611363116 -0.3399810436 0.3399810436 0.8611363116];

w=[0.3478548451 0.6521451549 0.6521451549 0.3478548451];

l=norm(P2-P1);

n=[0 1; -1 0]*(P2-P1)/l;

for i=1:4

x=P1*(1-r(i))/2+P2*(1+r(i))/2;

A1(i)=(1-r(i))*(x-Q)’*n/norm(x-Q)^2;

A2(i)=(1+r(i))*(x-Q)’*n/norm(x-Q)^2;

B1(i)=(1-r(i))*log(norm(x-Q));

B2(i)=(1+r(i))*log(norm(x-Q));

end

a1=-l/(8*pi)*w*A1’;

a2=-l/(8*pi)*w*A2’;

b1=-l/(8*pi)*w*B1’;

b2=-l/(8*pi)*w*B2’;


Example

The following script uses function bemlin for the approximation ofthe solution of the Dirichlet boundary value problem

−∆u = 0 in Ω

u = log(x21 + x2

2 ) on ∂Ω

being Ω =x ∈ R2 : (x1 − 1)2 + (x2 − 1)2 < 1

.


Example




coor=[x;y]+1;

u=log(coor(1,:).^2+coor(2,:).^2);

q=2*(1-(coor(1,:)+coor(2,:))./(coor(1,:).^2+coor(2,:).^2));

bccode=4*ones(1,N);

bcdata=zeros(3,N);

bcdata(1,:)=u;

solD=bemlin(coor,bccode,bcdata);

errq=max(abs(q-solD(3,:)))

points=[1 1/2 1 3/2 3/2; 1/2 1 1 1 3/2];

IntvalD=SolIntlin(coor,solD,points)

solex=log(points(1,:).^2+points(2,:).^2)


Exercises

Using linear boundary elements approximate the solution of the followingboundary value problems:

1.

−∆u = 0 in Ω∂u∂n = 2

(1− x1+x2

x21 +x2

2

)on ∂Ω ,

being Ω =x ∈ R2 : (x1 − 1)2 + (x2 − 1)2 < 1

.

2.

−∆u = 0 in Ω = (−1, 1)× (−1, 1)u = 2x1 + 3x2 on ∂Ω .

3.

−∆u = 0 in Ω = (−1, 1)× (−1, 1)∂u∂n = −3 on (−1, 1)× −1u = 2 + 3x2 on 1 × (−1, 1)∂u∂n = 3 on (−1, 1)× 1u = −2 + 3x2 on −1 × (−1, 1) .


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