Juan Vicente Gutiérrez Santacreu Rafael Rodríguez Galván. Departamento de Matemática Aplicada I Universidad de Sevilla

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1 Doc-Course: Partial Differential Equations: Analysis, Numerics and Control Research Unit 3: Numerical Methods for PDEs Part I: Finite Element Method: Elliptic and Parabolic Equations Juan Vicente Gutiérrez Santacreu Rafael Rodríguez Galván Departamento de Matemática Aplicada I Universidad de Sevilla juanvi@us.es Departamento de Matemática Universidad de Cádiz rafael.rodriguez@uca.es

2 Outline: Parabolic Equations 1 Preliminaries Bochner Spaces An Abstract Theorem for Existence and Uniqueness 2 The Heat Equation Variational Formulation Existence and Uniqueness Positivity and Maximum Principle 3 Discretization Hypotheses Time Approximation The Euler Method Combining space and time strategies Finite Element space The Euler method 4 Second-order time approximation The Crank-Nicolson Method The BFD2 method

3 Preliminaries Bochner Spaces Let 1 Fix T > 0 and let Ω be a domain in R 2 with boundary Ω. 2 Define the space-time domain Q = Ω (0, T ) and Σ = Ω (0, T ). u : Q R (x, t) u(x, t). If one fix t [0, T ], one has that u(t, ) X with X being a Banach space, say X (= L 2 (Ω), H 1 (Ω), H 1 0 (Ω), ). That is, u : t (0, T ) u(t) u(, t) X. Definition(Bochner Spaces) Let X be a Banach space. Thus, L p (0, T ; X ) denotes the space of Bochner-measurable, X -valued functions on the interval (0, T ) such that T 0 f (s) p X ds < for 1 p < or ess sup s (0,T ) f (s) X < for p =.

4 Preliminaries An Abstract Theorem for Existence and Uniqueness Theorem (J. L. Lions) Let V L be two Hilbert spaces with dense continuous embedding. Consider a mapping a : (0, T ) V V R such that a(t,, ) is bilinear for a.e. t (0, T ) and satisfies 1 The function t a(t, u, v) is measurable for all u, v V. 2 There exists M > 0 such that a(t, u, v) M u v for a.e. t [0, T ] and for all u, v V. 3 There exist α, β > 0 such that a(t, u, u) α u V β u L for a.e. t (0, T ) and for all u V. Then, for all f L 2 (0, T, V ) and u 0 L, there exists a unique solution u L 2 (0, T ; V ) and tu L 2 (0, T ; V ) such that { < tu, v > V,V +a(t, u, v) = < f (t), v > V,V a.e.t (0, T ), v V, u(0) = u 0

5 The Heat Equation Variational Formulation The Heat Equation Find u : Ω (0, T ) R 2 R such that tu u = f in Q, u = 0 on Σ, u(0) = u 0 in Ω. (1) where f : Q R 2 R 2 and u 0 : Ω R are given function. Multiplying (1) by a test function (regular enough) such that v = 0 on Ω, integrating over Ω, and using Green s formula gives: Variacional Formulation Find u : (0, T ) H0 1 (Ω) such that tuv dx + u v dx = f v dx, Ω u(0) = u 0. Ω Ω v H 1 0 (Ω), (2) We need to make sense to the above formulation.

6 The Heat Equation Existence and Uniqueness Teorem (Weak Solution) Fix T > 0. Let Ω be a Lipschitzian domain in R 2 and let u 0 L 2 (Ω) and f L 2 (0, T ; H 1 (Ω)). Then there exists a unique (weak) solution u(t) to (1) in the sense that, a.e. t (0, T ) and for all v H 1 0 (Ω), < tu, v > H 1 (Ω),H0 1(Ω) +( u, v) = < f, v > H 1 (Ω),H0 1(Ω), u(0) = u 0. where and u L (0, T ; L 2 (Ω)) L 2 (0, T ; H 1 0 (Ω)) tu L 2 (0, T ; H 1 (Ω)). Apply Lions Theorem to a(u, v) = ( u, v) for establishing existence and uniqueness of the solution.

7 The Heat Equation Existence and Uniqueness Theorem (Strong Solution) Fix T > 0. Let Ω be a C 1,1 or convex domain in R 2 and let u 0 H 1 0 (Ω) and f L 2 (0, T ; L 2 (Ω)). Then there exists a unique (strong) solution u(t) to (1) in the sense that, a.e in Q, tu u = f a.e. in Q, u = 0 a.e. in Σ, u(0) = u 0 a.e. in Ω. where and u L (0, T ; H 1 0 (Ω)) L 2 (0, T ; H 2 (Ω) H 1 0 (Ω)) tu L 2 (0, T ; L 2 (Ω)).

8 The Heat Equation Positivity and Maximum Principle Theorem (Positivity) Let u 0 L 2 (Ω) and f L 2 (0, T ; L 2 (Ω)). Assume u 0 (x) 0 a.e. Ω and f (x, t) 0 a.e. in Q. Let u be the weak solution to (1). Then u(x, t) 0 a.e. in Q. Theorem (Maximum Principle) Let u 0 L (Ω) and assume f = 0. Let u be the weak solution to (1). Then u L (Q) u 0 L (Ω).

9 DISCRETIZATION Hypotheses 1 Fix T > 0. 2 Let Ω be a convex, polygon domain in R 2. 3 Let u 0 H 1 0 (Ω) and f C([0, T ]; L 2 (Ω)). Remark We are assuming that the solution u to (1) has a unique strong solution. Goals 1 Develop a space-time discrete scheme. 2 Derive space and time error bounds. 3 Prove positivity and a discrete maximum principle.

10 DISCRETIZATION Time Approximation The Euler Method Given N N, choose k = T /N, t n = n k for n = 0,, N and consider a uniform partition such that [0, T ] = N 1 n=0 [tn, t n+1 ]. The Euler Method The Euler method consists in constructing a sequence of approximations {u n+1 } N 1 n=0 such that { 1 k (un+1 u n ) u n+1 = f (t n+1 ) in Ω, u n+1 = u 0 on Ω, (3) with u 0 = u 0 being known. Denote δ E t u n+1 = un+1 u n k. Variational Formulation Find u n+1 H 1 0 (Ω) such that (δ E t u n+1, v) + ( u n+1, v) = (f (t n+1 ), v) v H 1 0 (Ω). (4)

11 DISCRETIZATION Time Approximation The Euler Method Stability There holds u k n ( u k+1 u n 2 + k u n+1 2 ) u CΩ 2 k=0 Idea: Take v = u n+1 in (4) and use the identity (a b)a = 1 2 (a2 b 2 + (a b) 2 ). Error Bound Let f C([0, T ]; L 2 (Ω)) and u 0 H 1 0 (Ω). Then max n=0,...,n un+1 u(t n+1 ) + k ( N 1 k=0 n k f (t k+1 ) 2 k=0 (u n+1 u(t n+1)) 2 ) 1 2 C uk, where C u > 0 is constant depending on the strong solution s regularity.

12 DISCRETIZATION Time Approximation The Euler Method Proof For t = t n+1, we write tu(t n+1 ) u(t n+1 ) = f (t n+1 ) in Ω. A Taylor polynomial about t n+1 evaluated at t n shows that where u(t n+1 ) u(t n ) k R = 1 2k t n+1 t n u(t n+1 ) = f (t n+1 ) R, (5) (t t n+1 ) ttu(s)ds. Let e n+1 = u n+1 u(t n+1 ). Then, by subtracting (5) from (3), we have Test against e n+1 to get e n+1 e n e n+1 = R. k e n+1 2 e n 2 + e n+1 e n 2 + 2k e n+1 2 = 2k(R, e n+1 ).

13 DISCRETIZATION Time Approximation The Euler Method Let us bound the right-hand side: Now 2k(R, e n+1 ) 2k R H 1 (Ω) e n+1 k R 2 H 1 (Ω) + k en+1 2. ( k R 2 H 1 (Ω) 1 t n+1 4k 1 4k k 2 12 t n t n+1 t n t n+1 t n (s t n+1 ) ttu(s) H 1 (Ω)ds ) 2 t n+1 (s t n+1 ) 2 ds ttu(s) 2 H 1 (Ω) ds t n ttu(s) 2 H 1 (Ω) ds. Combining the above estimates and add up from n = 0 to m, we get e m k e n+1 2 e k 2 12 t m+1 0 ttu(s) 2 H 1 (Ω) ds Since e 0 = 0 and on assuming that ttu L 2 (0, T ; H 1 (Ω)), the proof is completed.

14 DISCRETIZATION Combining space and time strategies Finite Element space (H1) Let {T h } h>0 be a family of shape-regular, quasi-uniform triangulations of Ω made up of triangles, so that Ω = T Th E, where h = max K Th h T, with h K being the diameter of K. Further, let N h = {N j } j J denote the set of all the nodes of T h. (H2) A conforming finite-element space associated with T h is assumed for approximating H 1 (Ω). Let P 1(T ) be the set of linear polynomials on T ; the space of continuous, piecewise polynomial functions on T h is then denoted as } V h = {v h C 0 (Ω) : v h K P 1(K), K T h H0 1 (Ω), whose shape functions are {ϕ Ni } i I.

15 DISCRETIZATION Combining space and time strategies The Euler method Let u 0 h V h be such that u h 0 C u 0, where C > 0 is constant independent of h. The Euler Method Find u n+1 h V h such that (δ E t u n+1 h, v h ) + ( u n+1 h, v h ) = (f (t n+1 ), v h ) v h V h. Define M = (m ij ) M i,j=1 and R = (r ij ) M i,j=1 such that m ij = (ϕ Nj, ϕ Ni ) I i=0 and r ij = ( ϕ Nj, ϕ Ni ) M i=0, and b = (b j ) such that b j = (f (t n+1) M j=1, ϕ Nj ). Then where u n+1 = (u n+1 (N j )) M j=1. (M + R)u n+1 = b + Mu n,

16 DISCRETIZATION Combining space and time strategies The Euler method Stability There holds u k+1 h 2 + n k=0 Error Bound ( u k+1 Let Ω be convex. If follows that h uh n 2 + k u n+1 2 ) uh CΩ 2 h n k f (t k+1 ) 2 N 1 max n=0,...,n un+1 h u(t n+1 ) + k (u n+1 u(t n+1)) 2 ) C u(k 2 + h 2 ) k=0 where C u > 0 is constant depending on the strong solution s regularity. k=0

17 DISCRETIZATION Combining space and time strategies The Euler method Let I h be the nodal interpolation operator from C 0 ( Ω) to V h and consider (v h, v h ) h = Ω I h (v h v h ) = M i=0 v h (N i ) v h (N i ) ϕ Ni Ω for all v h, v h N h, with the induced norm v h h = (v h, v h ) h. It is well-known that v h h v h L 2 (Ω) (d + 2) 1 2 vh h. The Euler Method Find u n+1 h V h such that (δ E t u n+1 h, v h ) h + ( u n+1 h, v h ) = 0 v h V h. Observe now that M = diag i=1,...,m m ii with m ii = (ϕ Ni, ϕ Ni ) h, because of (ϕ Ni, ϕ Nj ) h = 0 for i j.

18 DISCRETIZATION Combining space and time strategies The Euler method Definition (Nonobtuse) Let T be a triangle with α, β, γ being its internal angles. Then T is nonobtuse if 0 < α, β, γ π. Nonobtuse triangulation If T h consists of acute triangles, then if follows that ϕ Ni ϕ Nj 0 i j. Discrete Maximum Principle Let u 0 h V h be such that there exist two constant 0 m M < satisfying 0 < m u 0 h(n i ) M i I. Then there holds 0 < m u n+1 h (N i ) M i I.

19 Second-order time approximation The Crank-Nicolson Method Define t n+ 1 2 = t n+1 +t n 2 and u n+ 1 2 = u n+1 +u n 2. The Crank-Nicolson Method The Crank-Nicolson method consists in constructing a sequence of approximations {u n+1 } N 1 n=0 such that { 1 k (un+1 u n ) u n+ 2 1 = f (t n+ 1 2 ) in Ω, with u 0 = u 0 being known. Variational Formulation Find u n+1 H 1 0 (Ω) such that u n+1 = 0 on Ω, (δ tu n+1, v) + ( u n+ 1 2, v) = (f (t n+ 1 2 ), v). (6)

20 Second-order time approximation The Crank-Nicolson Method Stability There holds n n u n νk u n+1 2 u n 2 + CΩk 2 f (t n+1 ) 2. k=0 k=0 Sktech of proof Take v = u n+ 1 2 (a b)(a + b) = a 2 b 2. in (6) and use the identity Convergence Let f C([0, T ]; L 2 (Ω)) and u 0 H 1 0 (Ω). Then 1 max n=0,...,n un+ 2 u(t n+ 1 2 ) + k ( N 1 n=0 (u n+1 u(t n+1)) 2 ) 1 2 C uk 2, where C u > 0 is constant depending on the strong solution s regularity.

21 Second-order time approximation The BFD2 Method Consider the pairs of values (t n 1, u(t n 1 )), (t n, u(t n )) y (t n+1, u(t n+1 )) and construct the interpolation polynomial p 2 which goes through such points: [ ] u(t p 2(t) = u(t n+1 )+(t t n+1 n+1 ) u(t n ) ) + (t t n ) u(tn+1 ) 2u(t n ) + u(t n 1 ) k 2k 2 and is an approximation of u(t) on [t n 1, t n+1 ] and so its derivative at t = t n+1 : tu(t n+1 ) p 2(t n+1) = 1 k 3u(t n+1 ) 4u(t n ) + u(t n 1 ). 2 The BFD2 Method The BFD2 method consists in constructing a sequence of approximations {u n+1 } N n=1 such that 1 3u n+1 4u n + u n 1 ν u n+1 = f (t n+1 ) in Ω, k 2 u n+1 = 0 on Ω, with u 0 = u 0 and u 1 being known.

22 Second-order time approximation The BFD2 Method Denote δt BFD2 u n+1 = 3un+1 4u n +u n 1. 2k Variational Formulation Find u n+1 H 1 0 (Ω) such that Stability It follows that (δ BFD2 t u n+1, v) + ( u n+1, v) = (f (t n+1 ), v) (7) ( u n u n+1 u n 2 ) ( u n 2 + 2u n u n 1 2 ) n n + u n+1 2u n + u n k u n+1 2 CΩk 2 f (t n+1 ) 2. k=0 Scetch of proof Take v = u n+1 in (7) and use the identity (3a 4b + c)a = 1 2 (a2 + (2a b) 2 b 2 (2b c) + (a 2b + c) 2 ). k=0

23 Second-order time approximation The BFD2 Method Convergence It follows that N 1 max n=0,...,n un+1 u(t n+1) + k (u n+1 u(t n+1)) C uk 2, where C u > 0 is constant depending on the strong solution s regularity. n=0 The approximation u 1 should be computed by using a one-step approximation method of at least order 2.