Systems of convection-diffusion equations Christian Reibiger, Hans-G - PowerPoint PPT Presentation

Systems of convection-diffusion equations Christian Reibiger, Hans-G¨ org Roos Technical University of Dresden hans-goerg.roos@tu-dresden.de http://math.tu-dresden.de/˜roos November 17, 2011– Dresden

Outline • motivation: optimal control • weakly coupled systems • strongly coupled systems • systems with two small parameters H.-G. Roos, TU Dresden DD-MS 60 1 • First • Prev • Next • Last • Full Screen • Close • Quit

1.Motivation Consider: � 1 0 + λ � 2 � y − y 0 � 2 2 � q � 2 min y,q J ( y, q ) := min y,q 0 subject to Ly := − ǫy ′′ + ay ′ + by = f + q in (0 , 1) , y (0) = y (1) = 0 . where a, b, y 0 , f are smooth; 0 < ε << 1 and | a ( x ) | ≥ α > 0, λ > 0. It is well-known (cf. Troeltzsch 09) that then there is an adjoint state p such that λq + p = 0 , (1a) L ∗ p = − ǫp ′′ − ap ′ + ( b − a ′ ) p = y − y 0 , (1b) p (0) = p (1) = 0 . H.-G. Roos, TU Dresden DD-MS 60 2 • First • Prev • Next • Last • Full Screen • Close • Quit

Motivation Error estimates (Becker, Vexler 07, Hinze, Yan, Zhou 09, Lube,Tews 10): � q − q h � 0 + � y − y h � 0 ≤ C ( h 3 / 2 + ε 1 / 2 h )( | y | 2 + | p | 2 ) based on 0 ≤ 1 � y − y h � 2 0 + λ � q − q h � 2 p h � 2 y h � 2 λ � p − ˜ 0 + � y − ˜ 0 (in the second step also estimates in stronger norms) Heinkenschloss, Leykekhman 10: SUPG only first order even away from the boundary layer In sharp contrast to the case of a single equation?? What about the layer structure ? H.-G. Roos, TU Dresden DD-MS 60 3 • First • Prev • Next • Last • Full Screen • Close • Quit

2.Weakly coupled systems � � � � − ǫu ′′ + a 1 u ′ + b 11 u 1 + b 12 u 2 f 1 , u 1 (0) = u 1 (1)= 0 , 1 1 = − ǫu ′′ − a 2 u ′ − b 21 u 1 + b 22 u 2 f 2 u 2 (0) = u 2 (1)= 0 . 2 2 Assuming a 1 , a 2 ≥ α > 0 , b 11 , b 22 ≥ 0 , b 12 b 21 > 0 , | b 12 | , | b 21 | ≥ β > 0 Linss 07: upwind FDM, information about first order derivatives Linss/Stynes 09: survey sp systems: CMAM, 9(2009), 165-191 Theorem : The solution of the given system can be decomposed in u 1 = S 1 + E 10 + E 11 , u 2 = S 2 + E 20 + E 21 with (for k ≤ 2) � � � � � S ( k ) � S ( k ) 0 , 0 ≤ C, � � � � 1 2 � � � � � � � E ( k ) � E ( k ) � ≤ Cǫ − k e − α 1 − x ǫ . � ≤ Cǫ 1 − k e − α x 10 ( x ) ǫ , 11 ( x ) � � � � H.-G. Roos, TU Dresden DD-MS 60 4 • First • Prev • Next • Last • Full Screen • Close • Quit

Weakly coupled systems σ 0 = 2 ǫ ln N /β σ 1 = 1 − 2 ǫ ln N/β u 1 u 2 Used Shishkin mesh Theorem : The solution of the given system is bounded by � u − u N � ǫ ≤ CN − 1 ln N Proof is based on standard results for the interpolation error. Especially the estimates 10 | 1 ≤ ˜ | E 10 − E I Ch | E 10 | 2 ≤ Cǫ − 1 / 2 N − 1 , � E 10 − E I 10 � 0 ≤ C min { ǫ 1 / 2 N − 1 , ǫ − 1 / 2 N − 2 } ≤ CN − 3 / 2 for the weak layer term H.-G. Roos, TU Dresden DD-MS 60 5 • First • Prev • Next • Last • Full Screen • Close • Quit

Weakly coupled systems ε = 1 e − 4 ε = 1 e − 8 0 0 10 10 −1 10 −2 10 −2 10 −3 10 −4 10 error error −4 10 −6 10 −5 10 || u − u N || 0 || u − u N || 0 || u − u N || ε || u − u N || ε −6 || u N − u I || ε || u N − u I || ε 10 −8 10 ln( N ) /N ln( N ) /N (ln( N ) /N ) 2 (ln( N ) /N ) 2 −7 10 1 2 3 4 5 1 2 3 4 5 10 10 10 10 10 10 10 10 10 10 N N Error of the linear FEM on a one-sided Shishkin mesh H.-G. Roos, TU Dresden DD-MS 60 6 • First • Prev • Next • Last • Full Screen • Close • Quit

Weakly coupled systems σ 0 = 2 ǫ ln N /β σ 1 = 1 − 2 ǫ ln N/β u 1 u 2 Used two-sided Shishkin mesh ε = 1 e − 4 ε = 1 e − 8 0 0 10 10 −2 −2 10 10 −4 10 −4 10 error error −6 10 −6 10 −8 10 || u − u N || 0 || u − u N || 0 −8 10 || u − u N || ε || u − u N || ε || u N − u I || ε || u N − u I || ε −10 ln( N ) /N ln( N ) /N 10 (ln( N ) /N ) 2 (ln( N ) /N ) 2 −10 10 1 2 3 4 5 1 2 3 4 5 10 10 10 10 10 10 10 10 10 10 N N Error of the linear FEM on a two-sided Shishkin mesh H.-G. Roos, TU Dresden DD-MS 60 7 • First • Prev • Next • Last • Full Screen • Close • Quit

3.Strongly coupled systems − εu ′′ 1 − b 11 u ′ 1 − b 12 u ′ 2 + a 11 u 1 + a 12 u 2 = f 1 , (2a) − εu ′′ 2 − b 21 u ′ 1 − b 22 u ′ 2 + a 21 u 1 + a 22 u 2 = f 2 , (2b) subject to the boundary conditions u 1 (0) = u 2 (0) = u 1 (1) = u 2 (1) = 0 . Introducing the corresponding matrices B and A and vectors u and f the system takes the form Lu := − εu ′′ − Bu ′ + Au = f. O’Rordan, Stynes 09: both eigenvalues of B positive (two overlapping layers at x = 0 Linss 09: no information on pointwise derivatives, layer structure ? H.-G. Roos, TU Dresden DD-MS 60 8 • First • Prev • Next • Last • Full Screen • Close • Quit

Strongly coupled systems assumptions: ( V 1 ) B is symmetric. ( V 2 ) A + 1 / 2 B ′ is positive semidefinite. ( V 3 ) The eigenvalues of B satisfy λ 1 ( x ) > 0 , λ 2 ( x ) < 0 for all x. We have ε 3 / 2 | u | 2 + ε 1 / 2 | u | 1 ≤ � f � 0 . asymptotic approximation: � � � � α 1 α 2 exp( − ˜ exp(˜ u 0 as := w 0 + w h ˜ c 0 + d 1 λ 1 x/ε ) + d 2 λ 2 (1 − x ) /ε ) . β 1 β 2 The unknown vector ˜ c 0 and d 1 , d 2 are determined by the boundary condi- tions. Consequently: The reduced solution does not satisfies the boundary con- ditions, both solution components do have layers at x = 0 and at x = 1. H.-G. Roos, TU Dresden DD-MS 60 9 • First • Prev • Next • Last • Full Screen • Close • Quit

Strongly coupled systems Example: � � � � − 1 1 1 − εu ′′ − u ′ = 0 3 2 with � � � � � � x/ 3 + 1 / 6 1 1 u 0 as := + d 1 exp( − 3 x/ε ) + d 2 exp( − (1 − x ) /ε ) . − 2 x/ 3 + 2 / 3 4 0 Using the next term of the asymptotic approximation and the a priori esti- mate above to estimate the remainder we can prove: u allows a decomposition u = S + E 1 + E 2 with | S | 2 ≤ C, where the layer terms satisfy 1 | ≤ Cε − k exp( − ˜ 2 | ≤ Cε − k exp( − ˜ | E k | E k λ 1 x/ε ) , λ 2 (1 − x ) /ε ) . H.-G. Roos, TU Dresden DD-MS 60 10 • First • Prev • Next • Last • Full Screen • Close • Quit

4.Strongly coupled systems: two small pa- rameters − ε 1 u ′′ 1 + b 11 u ′ 1 − b 12 u ′ 2 + a 11 u 1 + a 12 u 2 = f 1 , (3a) − ε 2 u ′′ 2 − b 21 u ′ 1 − b 22 u ′ 2 + a 21 u 1 + a 22 u 2 = f 2 , (3b) subject to the boundary conditions u 1 (0) = u 2 (0) = u 1 (1) = u 2 (1) = 0 . Introducing the corresponding matrices B and A and vectors u and f the system takes the form Lu := − diag( ε 1 , ε 2 ) u ′′ − Bu ′ + Au = f. assumptions: • 0 < ε 1 < ε 2 << 1 • B constant, b 11 > 0 , b 22 > 0 H.-G. Roos, TU Dresden DD-MS 60 11 • First • Prev • Next • Last • Full Screen • Close • Quit

Strongly coupled systems: two small parameters Example (Linß SINUM 09): − ε 1 u ′′ 1 + u ′ 1 − u ′ 2 = 1 − ε 2 u ′′ 2 − 3 u ′ 2 = 2 , u 1 (0) = u 1 (1) = 0 , u 2 (0) = u 2 (1) = 0 asymptotic approximation: � x 3 + 2 1 � − 2 1 ε 1 � � 3 ε 2 − (1 3+2 1 � � 1 1 e − 1 − x 1+3 ε 1 3 e − x ε 1 . u as ( x ) = ˜ ) ε 2 1 + 3 ε 1 1 + 3 ε 1 0 − 2 3 x + 2 3 1 + 3 ε 2 3 ε 2 ε 2 3 Thus, the ”reduced” solution � x 3 + 2 1 � 1+3 ε 1 3 u 0 ( x ) = ε 2 − 2 3 x + 2 3 varies with ε 1 /ε 2 . Questions: ′′ reduced ′′ solution ? layer structure ? H.-G. Roos, TU Dresden DD-MS 60 12 • First • Prev • Next • Last • Full Screen • Close • Quit

Strongly coupled systems: two small parameters asymptotic approximation: u as ( x ) = u p ( x ) + u h ( x ) c + d 1 V 1 e − λ 1 x + d 2 V 2 e λ 2 (1 − x ) • u p ( x ) + u h ( x ) c general solution of the reduced system • λ 1 , λ 2 eigenvalues of � � − b 11 /ε 1 b 12 /ε 1 ˜ B = . b 21 /ε 2 b 22 /ε 2 • V 1 , V 2 scaled eigenvectors: ε 1 � � � ε 1 λ 2 − b 2 � b 12 ε 2 V 1 := , V 2 := ε 1 b 11 + ε 1 λ 1 b 21 ε 2 . Assuming ε 1 << ε 2 < 1 we have ε 1 λ 2 = − b 11 + 0( ε 1 D ) , ε 2 λ 1 = b 11 + 0( ε 1 ε 2 ε 2 ) H.-G. Roos, TU Dresden DD-MS 60 13 • First • Prev • Next • Last • Full Screen • Close • Quit

Strongly coupled systems: two small parameters main result: solution decomposition d 1 e − λ 1 x + ˜ u 1 = S 1 + ˜ d 2 e λ 2 (1 − x ) with � S ′ � S 1 � ∞ ≤ C, 1 � ∞ ≤ C. steps of the proof: • study the bvp for the remainder u − u as • apply the strong stability result from Linß 09: ||| u ||| ∞ ≤ C || f || L 1 ≤ C || f || − 1 , ∞ with ||| v ||| i = max ( ε i � v ′ i � ∞ , � v i � ∞ ) • insert the solution decomposition obtained for u 2 into the first equation and use Kelloggs’s technique H.-G. Roos, TU Dresden DD-MS 60 14 • First • Prev • Next • Last • Full Screen • Close • Quit