An Introduction to Integral Equations Adrianna Gillman Rice - PowerPoint PPT Presentation

Problem Statement Problem formulation Scattering Stokes’ flow Numerical approximations Fast direct solvers Summary An Introduction to Integral Equations Adrianna Gillman Rice University ICERM Workshop on Fast Algorithms for Generating Static and Dynamically Changing Point Configurations March 16, 2018

Problem Statement Problem formulation Scattering Stokes’ flow Numerical approximations Fast direct solvers Summary Linear boundary value problems We consider a Poisson problem Dirichlet boundary condition: Ω Γ � − ∆ u ( x ) = g ( x ) , x ∈ Ω , u ( x ) = f ( x ) , x ∈ Γ . However, the solution techniques can be extended to linear boundary value problems of the form � A u ( x ) = g ( x ) , x ∈ Ω , (BVP) B u ( x ) = f ( x ) , x ∈ Γ , where Ω is a domain in R 2 or R 3 with boundary Γ. For instance: • The equations of linear elasticity. • Stokes’ equation. • Helmholtz’ equation (at least at low and intermediate frequencies). • Time-harmonic Maxwell (at least at low and intermediate frequencies).

Problem Statement Problem formulation Scattering Stokes’ flow Numerical approximations Fast direct solvers Summary Example applications (a) (b) (c) (a) The Wraith Virginia-class submarine (b) Kayani, A., Khoshmanesh, K., Ward, S., Mitchell, A., and Kalantar-zadeh, K. Optofluidics incorporating actively controlled micro- and nano-particles. In Biomicrofluidics, vol. 6.

Problem Statement Problem formulation Scattering Stokes’ flow Numerical approximations Fast direct solvers Summary Boundary value problem We consider a Poisson problem with Dirichlet boundary condition: Ω Γ � − ∆ u ( x ) = g ( x ) , x ∈ Ω , u ( x ) = f ( x ) , x ∈ Γ .

Problem Statement Problem formulation Scattering Stokes’ flow Numerical approximations Fast direct solvers Summary Boundary value problem We consider a Poisson problem with Dirichlet boundary condition: Ω Γ � − ∆ u ( x ) = g ( x ) , x ∈ Ω , u ( x ) = f ( x ) , x ∈ Γ . Let’s write u ( x ) = v ( x ) + w ( x )

Problem Statement Problem formulation Scattering Stokes’ flow Numerical approximations Fast direct solvers Summary Boundary value problem We consider a Poisson problem with Dirichlet boundary condition: Ω Γ � − ∆ u ( x ) = g ( x ) , x ∈ Ω , u ( x ) = f ( x ) , x ∈ Γ . Let’s write u ( x ) = v ( x ) + w ( x ) where v ( x ) is the solution of x ∈ R 2 , − ∆ v ( x ) = ˆ g ( x ) , and w ( x ) is solution of � − ∆ w ( x ) = 0 , x ∈ Ω , w ( x ) = f ( x ) − v ( x ) , x ∈ Γ . The function v ( x ) is called the particular solution and w ( x ) is called the homogeneous solution.

Problem Statement Problem formulation Scattering Stokes’ flow Numerical approximations Fast direct solvers Summary The fundamental solution For a given point charge x 0 ∈ R 2 , the solution of x ∈ R 2 − ∆ u ( x ) = δ ( x − x 0 ) , is u ( x ) = − 1 2 π log | x − x 0 | .

Problem Statement Problem formulation Scattering Stokes’ flow Numerical approximations Fast direct solvers Summary The fundamental solution For a given point charge x 0 ∈ R 2 , the solution of x ∈ R 2 − ∆ u ( x ) = δ ( x − x 0 ) , is u ( x ) = − 1 2 π log | x − x 0 | . The fundamental solution G ( x , y ) is given by G ( x , y ) = − 1 2 π log | x − y | . This allows us to move the point charge around.

Problem Statement Problem formulation Scattering Stokes’ flow Numerical approximations Fast direct solvers Summary The particular solution Recall, v ( x ) satisfies Ω x ∈ R 2 , − ∆ v ( x ) = ˆ g ( x ) , Γ where � g ( x ) for x ∈ Ω ˆ g ( x ) = for x ∈ Ω c 0

Problem Statement Problem formulation Scattering Stokes’ flow Numerical approximations Fast direct solvers Summary The particular solution Recall, v ( x ) satisfies Ω x ∈ R 2 , − ∆ v ( x ) = ˆ g ( x ) , Γ where � g ( x ) for x ∈ Ω ˆ g ( x ) = for x ∈ Ω c 0 Using the fundamental solution, the particular solution is given by � v ( x ) = G ( x , y ) g ( y ) dA ( y ) . Ω

Problem Statement Problem formulation Scattering Stokes’ flow Numerical approximations Fast direct solvers Summary The homogeneous solution Recall, w ( x ) is the solution to the boundary value problem with homogeneous partial differential equation; i.e. � − ∆ w ( x ) = 0 , x ∈ Ω , w ( x ) = f ( x ) − v ( x ) = ˆ f ( x ) , x ∈ Γ . It is tempting to express w ( x ) as � w ( x ) = G ( x , y ) σ ( y ) dl ( y ) Γ where σ ( y ) is an unknown boundary charge distribution.

Problem Statement Problem formulation Scattering Stokes’ flow Numerical approximations Fast direct solvers Summary The homogeneous solution Recall, w ( x ) is the solution to the boundary value problem with homogeneous partial differential equation; i.e. � − ∆ w ( x ) = 0 , x ∈ Ω , w ( x ) = f ( x ) − v ( x ) = ˆ f ( x ) , x ∈ Γ . It is tempting to express w ( x ) as � w ( x ) = G ( x , y ) σ ( y ) dl ( y ) Γ where σ ( y ) is an unknown boundary charge distribution. Enforcing the boundary condition yields the following first kind Fredholm equation � G ( x , y ) σ ( y ) dl ( y ) = ˆ f ( x ) x ∈ Γ . Γ

Problem Statement Problem formulation Scattering Stokes’ flow Numerical approximations Fast direct solvers Summary The spectrum 10 0 10 -1 | λ j | 10 -2 10 -3 10 -4 0 200 400 600 800 1000 j The minimum eigenvalue in absolute value is 2 . 06 e − 04. The maximum eigenvalue in absolute value is 6 . 39 e − 1.

Problem Statement Problem formulation Scattering Stokes’ flow Numerical approximations Fast direct solvers Summary The double layer kernel For a point y on the boundary of a curve, the double layer kernel D ( x , y ) = ∂ ν y G ( x , y ) is a solution of x ∈ R 2 . − ∆ x w ( x ) = δ ( x − y ) ,

Problem Statement Problem formulation Scattering Stokes’ flow Numerical approximations Fast direct solvers Summary A second kind integral equation Ω Γ Consider the problem − ∆ w ( x ) = 0 , x ∈ Ω , w ( x ) = ˆ f ( x ) , x ∈ Γ . The solution can be represented as a double layer potential � s ( x ) = ∂ ν y G ( x , y ) σ ( y ) ds ( y ) , x ∈ Ω , Γ where ν y is the outward normal at y and G ( x , y ) is the fundamental solution G ( x , y ) = − 1 2 π log | x − y | . Then the boundary charge distribution σ satisfies the boundary integral equation 1 � ∂ ν y G ( x , y ) σ ( y ) ds ( y ) = ˆ 2 σ ( x ) + f ( x ) Γ

Problem Statement Problem formulation Scattering Stokes’ flow Numerical approximations Fast direct solvers Summary The spectrum 1 0.9 0.8 0.7 0.6 0.5 | λ j | 0.4 0.3 0 200 400 600 800 1000 j

Problem Statement Problem formulation Scattering Stokes’ flow Numerical approximations Fast direct solvers Summary Variable coefficient PDEs Consider the free space variable coefficient Poisson problem Ω for x ∈ R 2 ∇ · ( a ( x ) ∇ u ( x )) = f ( x ) where a ( x ) > 0 for x ∈ Ω and the support of f ( x ) is Ω.

Problem Statement Problem formulation Scattering Stokes’ flow Numerical approximations Fast direct solvers Summary Variable coefficient PDEs Consider the free space variable coefficient Poisson problem Ω for x ∈ R 2 ∇ · ( a ( x ) ∇ u ( x )) = f ( x ) where a ( x ) > 0 for x ∈ Ω and the support of f ( x ) is Ω. Expanding the differential operator (plus some algebra) results in the following form of the PDE; ∆ u ( x ) + ∇ a ( x ) · ∇ u ( x ) = f ( x ) for x ∈ R 2 . a ( x ) a ( x )

Problem Statement Problem formulation Scattering Stokes’ flow Numerical approximations Fast direct solvers Summary Variable coefficient PDEs Consider the free space variable coefficient Poisson problem Ω for x ∈ R 2 ∇ · ( a ( x ) ∇ u ( x )) = f ( x ) where a ( x ) > 0 for x ∈ Ω and the support of f ( x ) is Ω. Expanding the differential operator (plus some algebra) results in the following form of the PDE; ∆ u ( x ) + ∇ a ( x ) · ∇ u ( x ) = f ( x ) for x ∈ R 2 . a ( x ) a ( x ) � We let u ( x ) = Ω G ( x , y ) σ ( y ) dA ( y ) and plug this expression into the PDE. � ∇ a ( x ) · ( ∇ x G ( x , y )) σ ( y ) dA ( y ) = f ( x ) for x ∈ R 2 . σ ( x ) + a ( x ) a ( x ) Ω

Problem Statement Problem formulation Scattering Stokes’ flow Numerical approximations Fast direct solvers Summary Variable coefficient PDEs Consider the free space variable coefficient Poisson problem Ω for x ∈ R 2 ∇ · ( a ( x ) ∇ u ( x )) = f ( x ) where a ( x ) > 0 for x ∈ Ω and the support of f ( x ) is Ω. Expanding the differential operator (plus some algebra) results in the following form of the PDE; ∆ u ( x ) + ∇ a ( x ) · ∇ u ( x ) = f ( x ) for x ∈ R 2 . a ( x ) a ( x ) � We let u ( x ) = Ω G ( x , y ) σ ( y ) dA ( y ) and plug this expression into the PDE. � ∇ a ( x ) · ( ∇ x G ( x , y )) σ ( y ) dA ( y ) = f ( x ) for x ∈ R 2 . σ ( x ) + a ( x ) a ( x ) Ω 10:30 - 11:15 Mike O’Neil Integral equation methods for the Laplace-Beltrami problem