# Numerical Solutions to Differential Equations

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1 Numerical Solutions to Differential Equations Lecture Notes The Finite Element Method #2 Peter Blomgren, Department of Mathematics and Statistics Dynamical Systems Group Computational Sciences Research Center San Diego State University San Diego, CA Spring 2015 Peter Blomgren, Numerical Solutions to Differential Equations (1/24)

2 Outline 1 The Finite Element Method Recap Looking for Solutions... 2 Identifying a Linear System... Properties of the Stiffness Matrix 3 Building the Toolbox Error Control 4 Initial Example: The Poisson Equation Analytical Tools: Vector Calculus Variational Formulation Minimization Formulation The Road Ahead... Peter Blomgren, Numerical Solutions to Differential Equations (2/24)

3 Recap Looking for Solutions... Quick Recap I/III Last time we formulated the Galerkin (variational, (V h )) and Ritz (minimization, (M h )) methods which will give us the finite element solutions to the differential equation (D): u = f, + Dirichlet Boundary Conditions The Function Space V h : Given the basis functions φ i (the tent functions) we can write V h = { v : v = } N η i φ i (x). i=1 Peter Blomgren, Numerical Solutions to Differential Equations (3/24)

4 Recap Looking for Solutions... Quick Recap II/III Inner Products: We defined the two (for now identical) inner products: a(u,v) = I uv dx, (u,v) = Further, we defined the energy functional: F(u) = 1 2 a(v,v ) (f,v). I uv dx. Given these definitions we formulated the Galerkin and Ritz problems... Peter Blomgren, Numerical Solutions to Differential Equations (4/24)

5 Recap Looking for Solutions... Quick Recap III/III Galerkin s Method (Variational Approach) (V h ) Find u h V h so that a(u h,v h ) = (f,v h) v h V h. Ritz Method (Minimization Approach) (M h ) Find u h V h so that F(u h ) F(v h ) v h V h. Peter Blomgren, Numerical Solutions to Differential Equations (5/24)

6 Recap Looking for Solutions... Looking for the Solution... If u h V h is a solution to (V h ), then in particular a(u h,φ j) = (f,φ j ), j = 1,2,...,N. Also, we can write u h in terms of the basis functions: u h (x) = N ξ j φ j (x), ξ j = u h (x j ). j=1 Therefore we can rewrite the above equation N ξ i a(φ i,φ j) = (f,φ j ), j = 1,2,...,N. i=1 Peter Blomgren, Numerical Solutions to Differential Equations (6/24)

7 Identifying a Linear System... Properties of the Stiffness Matrix The Stiffness Matrix and Load Vector From N ξ i a(φ i,φ j) = (f,φ j ), j = 1,2,...,N i=1 we identify the vectors ξ = {ξ 1,ξ 2,...,ξ N } T, and b = {b 1,b 2,...,b N } T, where b i = (f,φ i ), and the matrix A, where A ij = a(φ i,φ j ). We can now write our problem as A ξ = b. For historical reasons (Structural Mechanics) A is known as the stiffness matrix, and b as the load vector. Peter Blomgren, Numerical Solutions to Differential Equations (7/24)

8 Identifying a Linear System... Properties of the Stiffness Matrix Computing the Stiffness Matrix The elements of the stiffness matrix can be computed: First we notice that if the basis is the piecewise linear tent-functions, then if i j > 1, then φ i and φ j are non-overlapping which means a(φ i,φ j ) = 0. The Diagonal Entries, j = 1,2,...,N. a(φ j,φ j) = xj x j 1 1 h 2 j dx + xj+1 1 x j hj+1 2 dx = h j h j+1 The Super- and Sub-Diagonal Entries, j = 2,3,...,N. a(φ j 1,φ j) = a(φ j,φ j 1) = xj x j 1 1 h j 1 h j dx = 1 h j Peter Blomgren, Numerical Solutions to Differential Equations (8/24)

9 Identifying a Linear System... Properties of the Stiffness Matrix Illustration: Overlapping and Nonoverlapping Tent-Functions Peter Blomgren, Numerical Solutions to Differential Equations (9/24)

10 Identifying a Linear System... Properties of the Stiffness Matrix The Stiffness Matrix is Symmetric Positive Definite (SPD) A is symmetric since a(u,v) = a(v,u), and with v(x) = N i=1 η jφ i (x) we get N N N η i a(φ i,φ j)η j = a η i φ i, η j φ j = i,j=1 i=1 j=1 I [v h (x)]2 dx 0 Equality holds if and only if v (x) 0, which implies v(x) 0 by the boundary conditions. Fact: A matrix A is SPD if x T Ax > 0, x R n \{0} Fact: An SPD matrix (i) has positive eigenvalues, (ii) is nonsingular. Peter Blomgren, Numerical Solutions to Differential Equations (10/24)

11 Identifying a Linear System... Properties of the Stiffness Matrix The Special Case h j = h, j = 1,2,...,N If we equi-partition the interval we get the linear system h ξ = If we divide by h the we recover the standard finite difference method for the problem, where the right hand-side b j = xj+1 x j 1 f(x)φ j dx is a weighted mean of f(x) over the interval [x j 1,x j+1 ]. Peter Blomgren, Numerical Solutions to Differential Equations (11/24) b 1 b 2... b N

12 Building the Toolbox Error Control Estimating the Error for the Model Problem We are now going to look at the error (u u h ) where u is the exact solution of the differential equation, and u h is the solution to the finite element problem (V h ). Since V h V, a(u,v h ) = (f,v h), v h V h a(u h,v h ) = (f,v h), v h V h a((u u h ),v h ) = 0, v h V h That means that the error is orthogonal to the function space V h (as measured by a(, ).) Peter Blomgren, Numerical Solutions to Differential Equations (12/24)

13 Building the Toolbox Error Control Definition and Tools Definition (The L 2 -norm (L 2,a?)) w 2 = a(w,w) = 1 0 w 2 dx Theorem (Cauchy s Inequality) a(v,w) v w Peter Blomgren, Numerical Solutions to Differential Equations (13/24)

14 Building the Toolbox Error Control Cauchy s Inequality: Proof Cauchy s Inequality. Given v and w, define the renormalized functions v = v v and ŵ = w w. Which means v = ŵ = 1. Now Hence, 0 ± v ŵ 2 = a(± v ŵ,± v ŵ) = a( v, v) 2a( v,ŵ)+a(ŵ,ŵ) = 2 2a( v,ŵ). a( v,ŵ) 1. Removing the linear normalization factors give a(v,w) v w. Peter Blomgren, Numerical Solutions to Differential Equations (14/24)

15 Building the Toolbox Error Control Error Control Theorem Theorem For any v h V h we have (u u h ) (u v h ). So, measured in the L 2 -norm of the derivative (also known as the H 1 -seminorm), the solution u h to the discrete problem is closer to the solution u to the original continuous ODE than any other function in the function space V h. Peter Blomgren, Numerical Solutions to Differential Equations (15/24)

16 Building the Toolbox Error Control Error Control Proof Proof. Let v h V h be arbitrary, and set w h = u h v h. Then w h V h and we get: (u u h ) 2 = a((u u h ),(u u h ) )+a((u u h ),w h) }{{} =0 = a((u u h ),(u u h +w h ) ) = a((u u h ),(u v h ) ) {Cauchy s} (u u h ) (u v h ). Dividing through by (u u h ) gives (u u h ) (u v h ). Peter Blomgren, Numerical Solutions to Differential Equations (16/24)

17 Building the Toolbox Error Control Error Control Applying the Theorem From the theorem we can get a quantitative estimate (upper bound) for the error (u u h ) by estimating (u v h ) where v h V h is a suitably chosen function. Let v h be the linear interpolant of u, i.e. v h (x j ) = u(x j ), then u (x) v h (x) hmax x [0,1] u (x) [1] u(x) v h (x) h2 8 max x [0,1] u (x) [2] The first result and the theorem gives Similarly, from [2], we can get (u u h ) C 1 h max x [0,1] u (x). u(x) u h (x) C 2 h 2 max x [0,1] u (x). Peter Blomgren, Numerical Solutions to Differential Equations (17/24)

18 Initial Example: The Poisson Equation Analytical Tools: Vector Calculus Variational Formulation Minimization Formulation The Road Ahead... FEM for PDEs the Poisson Equation Consider the following boundary value problem u = f u = 0 in Ω on Γ = Ω, where Ω is a bounded domain in R 2 = {(x 1,x 2 ) : x 1 R,x 2 R}. Γ is the boundary of Ω, f(x 1,x 2 ) a given function, and [ 2 u = x 2 1 ] + 2 x2 2 u(x 1,x 2 ) = u x1 x 1 (x 1,x 2 )+u x2 x 2 (x 1,x 2 ). Physical problems: Our simple 1-D model problems from last lecture carry over to this 2-D model: heat distribution in a plate, the displacement of an elastic membrane fixed at the boundary under transverse load, etc... Peter Blomgren, Numerical Solutions to Differential Equations (18/24)

19 Initial Example: The Poisson Equation Analytical Tools: Vector Calculus Variational Formulation Minimization Formulation The Road Ahead... Vector Calculus Review (or Crash Course) I/III We need to generalize integration by parts to higher dimensions... We start with the divergence theorem: ũd x = ũ ñds, Ω where ũ = {u 1,u 2 } T, and ũ = u 1 x 1 + u 2 x 2. ñ = {n 1,n 2 } T is the unit (length-one, outward) normal to Γ. d x denotes area-integration, and ds integration along the boundary. c.f. in 1-D: du [a,b] dx dx = u( 1)ds = u(a)+u(b) {a,b} Γ Peter Blomgren, Numerical Solutions to Differential Equations (19/24)

20 Initial Example: The Poisson Equation Analytical Tools: Vector Calculus Variational Formulation Minimization Formulation The Road Ahead... Vector Calculus Review (or Crash Course) II/III If we apply the divergence theorem to the vector functions ũ 1 = (vw,0) and ũ 2 = (0,vw), we find v w +v w d x = vwn i ds, i = 1,2. x i x i Ω Let v denote the gradient v = Γ { v, v } T, x 1 x 2 and use the result above to write Green s Formula Ω v w d x [ ] v w Ω x 1 x 1 + v w x 2 x 2 d x = [ ] Γ v w x 1 n 1 +v w x 2 n 2 ds [ Ω v 2 w x1 2 = Γ v w n ds Ωv w d x. ] + 2 w d x x2 2 Peter Blomgren, Numerical Solutions to Differential Equations (20/24)

21 Initial Example: The Poisson Equation Analytical Tools: Vector Calculus Variational Formulation Minimization Formulation The Road Ahead... Vector Calculus Review (or Crash Course) III/III The normal derivative w n = w n 1 + w n 2 = w ñ x 1 x 2 is the derivative in the outward normal direction to the boundary Γ. With this notation we are ready to state the variational formulation corresponding to the differential equation... Peter Blomgren, Numerical Solutions to Differential Equations (21/24)

22 Initial Example: The Poisson Equation Analytical Tools: Vector Calculus Variational Formulation Minimization Formulation The Road Ahead... Variational Formulation where (V) Find u V so that a(u,v) = (f,v), v V. V = a(u,v) = Ω (f,v) = u v d x Ω fv d x v : v C(Ω) v x1 and v x2 are piecewise continuous in Ω v = 0 on Γ Note: We have slightly changed the notation of the a(, ) inner-product. This is very common in the PDE framework. The a(, ) innerproduct generically involves integrating the product(s) of derivative(s) of the two arguments over the domain. Peter Blomgren, Numerical Solutions to Differential Equations (22/24)

23 Initial Example: The Poisson Equation Analytical Tools: Vector Calculus Variational Formulation Minimization Formulation The Road Ahead... Minimization Formulation u is a solution to (V) if and only if it is also a solution to the following minimization problem. (M) Find u V such that F(u) F(v) v V, where F(v) is the potential energy F(v) = 1 2 a(v,v) (f,v). Peter Blomgren, Numerical Solutions to Differential Equations (23/24)

24 Initial Example: The Poisson Equation Analytical Tools: Vector Calculus Variational Formulation Minimization Formulation The Road Ahead... Next... Constructing a Finite Element Method for this problem... Peter Blomgren, Numerical Solutions to Differential Equations (24/24)

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