Discretization of Pdes and Tools for the Parallel Solution of the Resulting Systems

Discretization of PDEs and Tools for the Parallel Solution of the Resulting Systems

Stan Tomov

Innovative Computing Laboratory Computer Science Department The University of Tennessee

Wednesday March 17, 2010

CS 594, 03-17-2010 CS 594, 03-17-2010 Outline

Part I Partial Diﬀerential Equations Part II Mesh Generation and Load Balancing Part III Tools for Numerical Solution of PDEs

CS 594, 03-17-2010 Part I

Partial Diﬀerential Equations

CS 594, 03-17-2010 Mathematical Modeling

Mathematical Model: a representation of the essential aspects of an existing system which presents knowledge of that system in usable form (Eykhoﬀ, 1974) Mathematical Modeling: Real world Model Navier-Stokes equations:

←→ ∇ · u = 0 ∂u 1 = −(u · ∇)u − ∇p + ν∇2u + f ∂t ρ B.C., etc.

CS 594, 03-17-2010 Mathematical Modeling

We are interested in models that are Dynamic i.e. account for changes in time Heterogeneous i.e. account for heterogeneous systems

Typically represented with Partial Diﬀerential Equations

CS 594, 03-17-2010 Mathematical Modeling

How can we model for e.g. Heat Transfer? Heat * a form of energy (thermal) Heat Conduction * transfer of thermal energy from a region of higher temperature to a region of lower temperature Some notations

Q : amount of heat k : material conductivity T : temperature A : area of cross-section

CS 594, 03-17-2010 Heat Transfer

The Law of Heat Conduction

4Q 4T = k A 4t 4x

Change of heat is proportional to the gradient of the temperature and the area A of the cross-section.

Q : amount of heat k : material conductivity T : temperature A : area of cross-section

CS 594, 03-17-2010 Heat Transfer

Consider 1-D heat transfer in a thin wire

so thin that T is piecewise constant along the slides, i.e. T0(t), T1(t), T2(t), etc. ideally insulated

Let us write a balance for the temperature at T1 for time t + 4t

T1(t + 4t) =?

CS 594, 03-17-2010 Heat Transfer

T1(t + 4t) ≈ T1(t) (T (t) − T (t)) + k4t 2 1 (4x)2 (T (t) − T (t)) + k4t 0 1 (4x)2 T (t) − 2T (t) + T (t) = T (t) + k4t 2 1 0 1 (4x)2

Take lim4x,4t→0

∂T ∂2T ⇒ = k (Exercise) ∂t ∂x2

CS 594, 03-17-2010 Heat Transfer

Extend to 2-D and put a source term f to easily get

∂T ∂2T ∂2T = k + + f ≡ k 4T + f ∂t ∂x2 ∂y 2

Known as the Heat equation

CS 594, 03-17-2010 Other Important PDEs

Poisson equation (elliptic)

4u = f Heat equation (parabolic)

∂T = k 4T + f ∂t Wave equation (hyperbolic)

1 ∂2u = 4u + f ν2 ∂t2

CS 594, 03-17-2010 Classiﬁcation of PDEs For a general second-order PDE in 2 variables:

Auxx + Buxy + Cuyy + ··· = 0

Elliptic: if B2 − 4AC < 0 process in equilibrium (no time dependence) easy to discretize but challenging to solve Parabolic: if B2 − 4AC = 0 processes evolving toward steady state Hyperbolic: if B2 − 4AC > 0 not evolving toward steady state diﬃcult to discretize (support discontinuoities) but easy to solve in characteristic form

CS 594, 03-17-2010 How do we solve them?

Numerical solution approaches: Finite diﬀerence method Finite element method Finite volume method Boundary element method

CS 594, 03-17-2010 Finite Diﬀerence Method

use finite differences to approximate differential operators one of the simplest and extensively used method in solving PDEs the error, called truncation error, is due to finite approximation of the Taylor series of the differential operator

CS 594, 03-17-2010 A Finite Diﬀerence Method Example

Consider the 2-D Poisson equation:

∂2u ∂2u + = f ∂x2 ∂y 2 The idea, ﬁrst in 1-D: d2u Use Taylor series to approximate dx2 (x) with u(x), u(x + h), u(x − h)

du h2 d2u h3 d3u u(x + h) = u(x) + h (x) + (x) + (x) + O(h4) dx 2 dx2 3! dx3 du h2 d2u h3 d3u u(x − h) = u(x) − h (x) + (x) − (x) + O(h4) dx 2 dx2 3! dx3

d2u 1 ⇒ (x) = (u(x + h) + u(x − h) − 2u(x)) + O(h2) dx2 h2

CS 594, 03-17-2010 A Finite Diﬀerence Method Example

Similarly in 2-D Use Taylor series to approximate 4u(x, y) with u(x, y), u(x + h, y), u(x − h, y), u(x, y + h), u(x, y − h).

∂u h2 ∂2u h3 ∂3u u(x + h, y) = u(x, y) + h (x, y) + (x, y) + (x, y) + O(h4) ∂x 2 ∂x2 3! ∂x3 ∂u h2 ∂2u h3 ∂3u u(x − h, y) = u(x, y) − h (x, y) + (x, y) − (x, y) + O(h4) ∂x 2 ∂x2 3! ∂x3 ∂u h2 ∂2u h3 ∂3u u(x, y + h) = u(x, y) + h (x, y) + (x, y) + (x, y) + O(h4) ∂y 2 ∂y 2 3! ∂y 3 ∂u h2 ∂2u h3 ∂3u u(x, y − h) = u(x, y) − h (x, y) + (x, y) − (x, y) + O(h4) ∂y 2 ∂y 2 3! ∂y 3

1 ⇒ ∆u(x, y) = (u(x + h, y) + u(x − h, y) + u(x, y + h) + u(x, y − h) − 4u(x)) + O(h2) h2

CS 594, 03-17-2010 A Finite Diﬀerence Method Example Consider the 1-D equation:

d2u (x) = f (x), for x ∈ (0, 1) dx2 and the Dirichlet boundary condition we obtain a linear system of the form

u(0) = u(1) = 0 Ax = b

The interval [0, 1] is discretized uniformly with where b = (fi )i=1,n and x = (ui )i=1,n and n + 2 points x0 x1 xn xn+1 0 2 −1 1 −1 2 −1 -- -- B C B −1 2 −1 C h h h h 1 B C b b b b b b b b A = B . . . C 2 B . . . C h B . . . C At any point xi we are looking for ui , an B C @ −1 2 −1 A approxmation of the exact solution u(xi ), using −1 2 the approximation

2 −ui−1 + 2ui − ui+1 = h fi , and the fact that u0 = un+1 = 0 ,

(slide used material from Julien Langou’s presentation)

CS 594, 03-17-2010 A Finite Diﬀerence Method Example

The interval [0, 1] × [0, 1] is discretized uniformly with (n + 2) × (n + 2) points

Consider the 2-D Poisson equation: b b b b b b ∆u = f b b b b b b and the Dirichlet boundary condition b b b b b b

u(x, y) = 0 for (x, y) ∈ ∂Ω b b b b b b h 6 ?b- b b b b b h b b b b b b 0 B −I 1 0 4 −1 1 −IB −1 −1 4 −1 B C B C B −IB −1 C B −1 4 −1 C 1 B C B C B C B C A = B . . . C where B = B . . . C h2 B . . . C B . . . C B . . . C B . . . C @ −IB −I A @ −1 4 −1 A −IB −1 4

(slide used material from Julien Langou’s presentation)

CS 594, 03-17-2010 Finite Element Method

Remember the slides from lecture 2 http://www.cs.utk.edu/∼dongarra/WEB-PAGES/SPRING-2010/Lect02-2010.pdf

Main pluses/minuses of FEM vs FDM FEM can handle complex geometries FDM is easy to implement

CS 594, 03-17-2010 A Finite Element Method Example

Consider the 1-D Dirichlet problem:

(1) u00(x) = f (x), for x ∈ (0, 1)

and the Dirichlet boundary condition u(0) = u(1) = 0

Weak or Variational formulation:

Multiply (1) by smooth v and integrate over (0,1)

Z 1 Z 1 f (x)v(x)dx = u00(x)v(x)dx 0 0

Integrate by parts the above RHS

Z 1 Z 1 00 0 1 0 0 u (x)v(x)dx = u (x)v(x)|0 − u (x)v (x)dx 0 0 Z 1 = − u0(x)v 0(x)dx ≡ −a(u, v) 0

1 Variational formulation: Find u ∈ H0 (0, 1) such that

Z 1 1 f (x)v(x)dx = −a(u, v) for ∀v ∈ H0 (0, 1) 0

CS 594, 03-17-2010 A Finite Element Method Example

Discretization (Galerkin FE problem): 1 Replace H0 (0, 1) with ﬁnite dimensional subspace V

Shown is a 4 dimensional space V (basis in blue) and a linear combination (in red)

What is the matrix form of the problem (Exercise)

CS 594, 03-17-2010 Part II

Mesh Generation and Load Balancing slides at: http://www.cs.utk.edu/∼dongarra/WEB-PAGES/SPRING-2010/Lect09-p2.pdf

CS 594, 03-17-2010 Part III

Tools for Numerical Solution of PDEs

CS 594, 03-17-2010 Parallel PDE Computations

Challenges:

Software Complexity Data Distribution and Access Portability, Algorithms, and Data Redistribution