Abstract
In this paper, we investigate a mixed discontinuous Galerkin approximation of time dependent convection diffusion optimal control problem with control constraints based on the combination of a mixed finite element method for the elliptic part and a discontinuous Galerkin method for the hyperbolic part of the state equation. The control variable is approximated by variational discretization approach. A priori error estimates of the state, adjoint state, and control are derived for both semidiscrete scheme and fully discrete scheme. Numerical example is given to show the effectiveness of the numerical scheme.
1. Introduction
The objective of this paper mainly focuses on developing a mixed discontinuous Galerkin scheme for the following control constrained optimal control problem governed by a transient convection diffusion equation:More details will be specified later.
This kind of problem plays an important role in many fields, such as air pollution ([1]) and waste water treatment ([2]). In recent years the research of numerical method for this kind of problem forms a hot topic. Lots of literatures are devoted to developing effective numerical methods for this kind of problem. In [3–6] the stabilization method such as local projection stabilization, SUPG, and continuous interior penalty method are investigated. The discontinuous Galerkin approximation including primary discontinuous Galerkin method and local discontinuous Galerkin method is addressed in [7–12]. In [13] the authors discuss the characteristic finite element approximation of transient convection diffusion optimal control problem. For more references, one can refer to [14].
In this paper, we investigate a mixed discontinuous Galerkin approximation of transient convection diffusion optimal control problem with control constraints. This scheme is based on the combination of a discontinuous Galerkin method for the hyperbolic part and a mixed finite element method for the elliptic part of the state equation. Variational discretization approach is utilized to approximate the control variable. The work of this paper is motivated by [8, 15] where a similar scheme was proposed for convection diffusion equations and stationary convection diffusion optimal control problems, respectively. Similar to other discontinuous Galerkin methods, this scheme is also locally conservative, which makes it much suitable for problems where conservation is important, for example, for time dependent convection diffusion problems. Moreover, when the diffusion coefficient tends to zero, this scheme reduces to the classical discontinuous Galerkin method. Thus it inherits the stabilizing features of discontinuous Galerkin methods. We derive a priori error estimates of the state, adjoint state, and control for both semidiscrete scheme and fully discrete scheme. Numerical experiment is carried out to show the performance of our scheme.
The rest of the paper is organized as follows: In Section 2, semidiscrete and fully discrete mixed discontinuous Galerkin scheme are defined for control constrained optimal control problems governed by the time dependent convection diffusion equations. In Section 3, a priori error estimates of the semidiscrete scheme are derived. In Section 4, we derive a priori error estimates of the fully discrete scheme. Finally numerical example is given to illustrate the theoretical findings.
2. Mixed Discontinuous Galerkin Scheme
Consider the following optimal control problem with control constraints:where is a convex polygon with piecewise smooth boundary . Here is the admissible set defined by and are given functions, is a constant, and is a given vector valued function, which satisfies with a constant .
To define a mixed discontinuous Galerkin scheme for (2), we introduce a new variable:Then the optimal control problem (2) can be rewritten asLet be a regular triangulation of with element and . Let , where denotes the diameter of the element . We define the following spaces:For simplicity, we setwhere is the outward norm direction on ,and on when . Then the weak formulation for the optimal control problem (5) reads as follows: finding such that
Standard arguments techniques imply that optimal control problem (10)-(11) admits a unique solution and the following first-order optimality condition holds:
Let denote the Raviart-Thomas element space of the lowest order associated with a triangular or rectangular mesh of (see [16] for details). Then the approximation scheme of (10)-(11) is as follows: finding such thatHere is an approximation of the initial value . Variational discretization approach is used for the control .
Similar to continuous case we derive the following semidiscrete optimality conditions:
To define a fully discrete scheme we introduce a time partition. Let be a time grid with . Let be a half-open interval. We write for a smooth .
We use backward Euler scheme for time discretization. Then a fully discrete scheme of (10)-(11) is characterized as follows: finding such that
To obtain the fully discrete first-order optimality condition we define a Lagrange functional as follows:
Here , , , , and . Then we derive the discrete first-order optimality condition:
3. Semidiscrete Error Estimate
The goal of this section is to prove the semidiscrete error estimates for the state, the adjoint state, and the control. We firstly decompose and aswhere and satisfy the following auxiliary problems:For the following analysis we introduce a new norm:Following [8] we have .
Let and be the projections defined byAccording to [8] we have the following.
Lemma 1. If and satisfy (21)-(22), respectively, then we have
Lemma 2. If and satisfy (21)-(22), respectively, then the following estimates hold:
Proof. Differentiating (21) about , we haveSet and to be the interpolation operators for the standard RT element space (see, e.g., [17]) such thatThen we have the following approximation properties:LetThen we deriveBy the definition of interpolation operators and we obtainSetting and leads tonamely,Since the function in is piecewise constant, using the definition of , we have for all and Therefore, we get the estimate as follows:Note thatand according to [18] we have Then we deriveBy setting small enough we obtainUsing triangle inequality we havewhich implies the first theorem result. Similar to the above proof we haveUsing the above estimates we can derive the following.
Theorem 3. Let and be the solutions of (12) and (19), respectively. Then we haveHereand the constant depends on the norms of and .
Proof. Using (12) and (19) we derive the following error equations:We split the errors and intowhere is defined in (21). Then we haveSetting and givesChoosing and in the above equation leads toSince , then we deduce by Young inequality thatBy setting small enough we deriveNote that . Integrating (49) from leads toSimilarly, integrating (49) from yieldswhich impliesCombining (50) and (52) we deduce thatBy Lemma 2 we further deriveThen by triangle inequality and Lemma 1 we obtain This implies the theorem result.
Theorem 4. Let and be the solutions of (12) and (19), respectively. Then we haveHere the constant depends on the norms of and .
Proof. Using (12) and (19) we derive the following error equations:We rewrite the errors and aswhere is defined in (22). Then we haveSetting and we haveTaking and yieldsSince , thenUsing inequality and Young inequality we deriveChoosing small enough we obtainIntegrating (64) with respect to from yieldswhere was used. In an analogue way, integrating (64) from results inWe further haveCombining (65) and (67) we deduce thatThen we can derive the theorem result by using Lemmas 1 and 2 and Theorem 8 and triangle inequality.
Lemma 5. Let and be the solutions of (14) and (19), respectively. Then we have
Proof. Using (19) along with (14) leads toSetting and and choosing and lead toBy Young inequality we haveSimilar to the proof of Lemmas 1 and 2 we haveBy (17) and (19) we deriveSetting and and choosing and yieldFurthermore, we derive by Young inequalitySimilar to the proof of Lemmas 1 and 2 we haveThen the theorem result follows from (73) and (77).
Lemma 6. Assume that and be the solutions of (11) and (14), respectively. Let where is the solution of (19). Then the following estimate holds:
Proof. By the definitions of , we haveNote thatBy setting , and , in the above equations, respectively, we can proveSince , , andwe can prove thatTherefore, we derive
Lemma 7. Let and be the solutions of (11) and (14), respectively. Then we have
Proof. By Lemma 6 we haveUsing the result of Theorem 4 yields
Theorem 8. Let and be the solutions of (12) and (14), respectively. Then we have
Proof. By Lemma 2 and (88) we deriveCombining Lemma 5 and (90) yieldsCombining (88)–(91) we can derive the theorem result.
4. Fully Discrete Error Estimate
In this section we will prove the error estimates for the fully discrete scheme. For this purpose we firstly introduce the following auxiliary problems:
Lemma 9. Let and be the solutions of (12) and (92), respectively. Then we havewhere
Proof. Using (12) and (92) we derive the following error equations:Setting , , and and combining the definition of and we haveTesting (96) with and yieldsBy Taylor expansion with integral reminder we deduce thatBy inequality we obtainThen by Young inequality we deriveChoosing to be small enough leads toMultiplying (114) by and summing up with respect to from 1 to yieldMultiplying (101) by and summing up with respect to from 1 to lead toCollecting above estimates we haveUsing above estimates, Lemmas 1 and 2, and triangle inequality we deduce that
Lemma 10. Let and be the solutions of (12) and (92), respectively. Then we have
Proof. Using (12) and (92) we derive the following error equations:Setting , , and and using the definitions of and giveChoosing and in (108) yields Using inequality and Young inequality we obtainThen by setting small enough we haveMultiplying (111) by and summing up with respect to from to yieldMultiplying (111) by and summing up with respect to from to giveCollecting (112) and (113) we obtainBy Lemmas 1 and 2 and triangle inequality we derive
Lemma 11. Let and be the solutions of (12) and (92), respectively. Then we have
Proof. Using (92) along with (17) leads toChoosing and in the above equations and adding the resulting equations together yieldSetting and in the above equation leads toand thenFurthermore, we haveSimilar to the proof of Lemmas 9 and 10 we haveSimilarly, by (17) and (92) we deriveTaking and in the above equation, setting and , and using similar argument to (124) giveThen the theorem result follows from (122) and (124).
Lemma 12. Let and be the solutions of (12) and (92), respectively. Letwhere is the solution of (92). Then the following estimate holds:Here .
Proof. Note thatSetting , , , and , by the definition of and we haveChoosing , and , in the above equations, respectively, we can easily prove thatTherefore, we derive
Lemma 13. Let and be the solutions of (12) and (92), respectively. Then we have
Proof. Following Lemma 12 we deriveUsing Young inequality we obtain
Theorem 14. Let and be the solutions of (11) and (17), respectively. Then we have
Proof. By Lemma 9 and using triangle inequality we deriveBy Lemma 10 we deduce thatCombining Lemma 13, (135), and (136) we arrive atwhich completes the theorem.
Competing Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgments
The authors would like to acknowledge the support of Natural Science Foundation of Shandong Province (no. ZR2016JL004).