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Article

# A Finite Volume Method to Solve the Ill-Posed Elliptic Problems

by
Ying Sheng
1,* and
Tie Zhang
2
1
Department of Mathematics, Northeastern University, Shenyang 110004, China
2
State Key Laboratory of Synthetical Automation for Process Industries, Department of Mathematics, Northeastern University, Shenyang 110004, China
*
Author to whom correspondence should be addressed.
Mathematics 2022, 10(22), 4220; https://doi.org/10.3390/math10224220
Submission received: 13 October 2022 / Revised: 7 November 2022 / Accepted: 8 November 2022 / Published: 11 November 2022
(This article belongs to the Topic Numerical Methods for Partial Differential Equations)

## Abstract

:
In this paper, we propose a finite volume element method of primal-dual type to solve the ill-posed elliptic problem, that is, the elliptic problem with lacking or overlapping boundary value condition. We first establish the primal-dual finite volume element scheme by introducing the Lagrange multiplier $λ$ and prove the well-posedness of the discrete scheme. Then, the error estimations of the finite volume solution are derived under some proper norms including the $H 1$-norm. Numerical experiments are provided to verify the effectiveness of the proposed finite volume element method at last.
MSC:
65N20; 65N08

## 1. Introduction

In this present paper, we consider the finite volume element method for solving the following second-order ill-posed elliptic problem
$− d i v ( a ∇ u ) = f , x ∈ Ω , u | Γ d = g 1 , ∂ u ∂ n | Γ n = g 2 ,$
where $Ω$ is a bounded polygonal domain in $R 2$ with boundary $∂ Ω$, $Γ d$ and $Γ n$ are two parts of the boundary $∂ Ω$ and $m e a s ( Γ d ) ≠ 0$, n is the unit external normal vector on $∂ Ω$, $f ∈ L 2 ( Ω )$, $g 1 ∈ H 1 ( Γ d )$ and $g 2 ∈ L 2 ( Γ n )$ are known real-valued functions, $a ( x ) ∈ W 1 , ∞ ( Ω )$ is assumed to be such that $0 < a 0 ≤ a ( x ) ≤ a 1$. When $Γ d ∪ Γ n ≠ ∂ Ω$ or $Γ d ∩ Γ n ≠ ∅$, elliptic problem (1) is called ill-posed, caused by the ill-posed boundary conditions. For example, in some engineering projects such as the dam with underground boundary, the ultra-high temperature surface of the spaceships, in these cases, part boundary conditions cannot be measured. This kind of ill-posed elliptic problem is to resolve a problem with lacking boundary conditions on part of the domain boundary [1].
Ill-posed elliptic problems have attracted many researchers since the 1960s due to their wide applications such as geophysics, steady-state inverse heat conduction, wave propagation, cardiology, electromagnetic scattering enzymatic reactions, bifurcation phenomena, phenomena in chemistry and plasma physics, and so on [2,3,4]. In 1902, Hadamard [5] first introduced the concept of the well-posed Cauchy problem, and later in 1923, he published the well-known example in [6] at Yale University. He illustrated the ill-posedness of the Cauchy problem for the Laplace equation ($Γ d = Γ n ⊂ ∂ Ω$ and $a ( x ) = 1$ in problem (1)). The work of Hadamard laid the foundation of the theory for the ill-posed Cauchy problems of elliptic equations. From then on, the theory has been further developed.
For severely ill-posed problems, the common numerical methods may not achieve optimal error estimates without some proper strategies. In [7], a stabilized finite element method was proposed in which a primal and an adjoint problem were solved simultaneously. Yagola et al. [8] discussed regularizing algorithms using a priori information. Klibanov [9] discussed a universal method of the regularization for ill-posed Cauchy problems by constructing Tikhonov functionals. Chung et al. [10] proposed a least squares formulation for the ill-posed problems and Qian [11] proposed a wavelet dual least squares method for the ill-posed elliptic problems. Cheng et al. [12] developed a numerical method for the ill-posed Cauchy problem of the Laplace equation where the essence of the method is to transform the Cauchy problem into a moment problem whose numerical method has been studied extensively. We refer the readers to [13,14,15,16] for more details on how to handle the severely ill-posed problems numerically.
Burman has conducted a series of research about ill-posed elliptic problems. In 2013, he proposed a stabilized finite element method to solve nonsymmetric indefinite problems [7] where the main idea is to solve a primal and an adjoint problem simultaneously. In 2014 [17], he further developed the analysis for the stabilized finite element method proposed in [7] where the weak continuous dependence was assumed so that the error estimate can be derived. In 2018, Burman et al. [18] showed that the stabilization methods for solving the ill-posed convection-dominated convection-diffusion problems are better than that of the Tikhonov regularization method. Wang et al. [2,19] devised a well-posed numerical scheme for a type of ill-posed elliptic problem by using a new primal-dual weak Galerkin finite element method where weak gradients were used in [19] while weak Laplacian was used in [2]. However, in their methods, the well-posedness of the discrete scheme was based on the uniqueness of the exact solution of the ill-posed problem and the error estimates were given only in some semi-norms. To the authors’ best knowledge, there is no research on the finite volume element method (FVE) for solving the ill-posed elliptic problem. The advantage of the FVE method [20,21,22,23,24,25] over some other numerical methods is that it can preserve some conservation properties of the original problem locally which are essential in some practical applications. Motivated by the work of Burman [7,17,18] and Wang [2,19], this paper aims to propose a finite volume element method for solving the ill-posed elliptic problem (1). The main difficulty of this problem is that the bilinear form of the variational equation derived from the ill-posed elliptic problem contains an individual boundary integral term because of lacking or overlap boundary conditions, so it is not positive definite on the solution space. Therefore, based on this bilinear form, the well-posedness of the discrete solution cannot be assured and it is very difficult to derive the error estimates in the $H 1$ and $L 2$ norms. To overcome this difficulty, the key technique in our paper is to introduce the primal-dual finite volume scheme and add proper stability terms on $Γ d$ and $Γ n$. Compared with the research works of Burman [7,17,18] and Wang [2,19], the advantage of our method is that our stability analysis and error estimates do not depend on some additional assumptions on the ill-posed problem. Our discrete scheme admits a unique solution and can achieve the optimal convergence order in some proper norms, including the $H 1$-norm.
According to the research in [1,26] we know that the ill-posed elliptic problem (1) has a solution $u ∈ H 1 ( Ω )$ when boundary data $g 1 × g 2$ belong to a dense subset of $H 1 2 ( Γ d ) × H 1 2 ( Γ n )$. If the solution exists, it must be unique when $Γ d ∩ Γ n$ is a nontrivial portion of the domain boundary. In this paper, we assume that the solution of the ill-posed elliptic problem (1) exists and possesses a certain smoothness needed for our analysis.
Throughout this paper, we use the notations $W m , p ( D )$ to indicate the standard Sobolev spaces on domain $D ⊆ Ω$ equipped with the norm $∥ · ∥ m , p , D$ and seminorm $| · | m , p , D$. When $p = 2$, we denote $H m ( D ) = W m , 2 ( D ) , ∥ · ∥ m , D = ∥ · ∥ m , 2 , D$ and $| · | m , D = | · | m , 2 , D$, and we omit subscript D if $D = Ω$. We also denote the letter C as a positive constant independent of the mesh size h.
This paper is organized as follows. We first establish a primal-dual finite volume scheme and prove the well-posedness of the discrete scheme in Section 2. In Section 3, we do the error analysis of the primal-dual finite volume scheme. Section 4 is devoted to the numerical experiments to illustrate the theoretical analysis.

## 2. Finite Volume Element Formulation

Let $T h$ be a quasi-regular triangulation of $Ω$ where $h = max K ∈ T h { h K }$, $h K$ is the diameter of element K. We assume that the triangulation is such that the endpoints of $Γ d$ and $Γ n$ are the nodal points of $T h$. Denote $N h$ the set of all nodal points of $T h$, $E h$ the set of all edges of $T h$ and $E h 0 = E h \ ∂ Ω$. With the triangulation $T h$, we construct a barycenter dual partition $T h *$ by connecting the midpoints of the three edges to the barycenter of each element $K ∈ T h$ by straight lines. So, for each point $P ∈ N h$, there is a polygonal element $K P * ∈ T h * = ⋃ { K P * }$ which is the dual element or the control volume of point P, see Figure 1. Let $S h$ and $S h *$ be the trail function space and the test function space, respectively,
$S h = { u h ∈ C ( Ω ¯ ) ∩ H 1 ( Ω ) : u h ∣ K ∈ P 1 ( K ) , ∀ K ∈ T h } ,$
$S h * = { v h ∈ L 2 ( Ω ) : v h ∣ K P * = c o n s t a n t , ∀ K P * ∈ T h * } ,$
where $P k ( K )$ is the set of all k-th order polynomials on K.
For function $u ∈ C ( Ω ¯ )$, define the interpolation operator $γ h * : C ( Ω ¯ ) → S h *$ by
$γ h * u ∣ K P * = u ( P ) χ P ( x ) , K P * ∈ T h * ,$
where $χ P ( x )$ is the characteristic function of the dual element $K P *$. Thus,
$γ h * u ( x ) = ∑ P ∈ N h u ( P ) χ P ( x ) , x ∈ Ω .$
It can be proved that $γ h *$ is a one to one mapping from $S h$ to $S h *$, that is, $γ h * S h = S h *$.
For smooth function $u ( x )$, we can obtain from the Green formula that
$− ∫ K P * d i v ( a ∇ u ) v d K P * = − ∫ ∂ K P * a ∂ u ∂ n v d s , ∀ v ∈ S h * .$
Consequently, when $u ∈ H 1 + s ( Ω ) ( s ≥ 1 2 )$ is the solution of problem (1), u satisfies the variational equation
$a h ( u , γ h * v h ) = ( f , γ h * v h ) + ∫ Γ n a ( x ) g 2 ( x ) γ h * v h d s , ∀ v h ∈ S h ,$
where
$a h ( u , γ h * v h ) = − ∑ K P * ∈ T h * ∫ ∂ K P * \ Γ n a ( x ) ∂ u ∂ n γ h * v h d s , ∀ v h ∈ S h .$
and
$∫ Γ n a ( x ) g 2 ( x ) γ h * v h d s = ∑ K P * ∈ T h * ∫ ∂ K P * ∩ Γ n a ( x ) g 2 ( x ) γ h * v h d s = ∑ K ∈ T h ∫ ∂ K ∩ Γ n a ( x ) g 2 ( x ) γ h * v h d s .$
In view of the ill boundary condition in general, the variational form (7) is not well-posed. In order to define a well-posed discrete scheme, we first introduce the jump $[ u ]$ of function u across element edge e by
$[ u ] | e = u ∣ K 1 n 1 + u ∣ K 2 n 2 , e ∈ E h 0 , [ u ] | e = u n , e ∈ ∂ Ω ,$
where e is the common edge of elements $K 1$ and $K 2$, $n 1 , n 2$ are the unit external normal vectors to $K 1$ and $K 2$, respectively. Furthermore, for vector function w,
$[ w ] | e = w | K 1 · n 1 + w | K 2 · n 2 , e ∈ E h 0 , [ w ] | e = w · n , e ∈ ∂ Ω .$
For any $u , v ∈ S h + H 1 + s ( Ω ) , s ≥ 1 2$, denote
$S d ( u , v ) = ∫ Γ d h μ − 1 u v d s = ∑ e ⊂ Γ d ∫ e h μ − 1 u v d s ,$
$S n , d ( u , v ) = ∫ Γ d ∪ Γ n h μ − 1 u v d s = ∑ e ⊂ Γ d ∪ Γ n ∫ e h μ − 1 u v d s ,$
$S ( u , v ) = ∫ E h 0 ∪ Γ n h μ [ ∇ u ] [ ∇ v ] d s = ∑ e ∈ E h 0 ∪ Γ n ∫ e h μ [ ∇ u ] [ ∇ v ] d s ,$
where $h μ ∣ e = h e = m e a s ( e )$.
From Equation (7), we see that when the solution of problem (1) $u ∈ H 1 + s ( Ω ) ( s ≥ 1 2 )$, u and the Lagrange multiplier $λ = 0$ satisfies the following primal-dual variation equations with stabilizers
$S d ( u , v h ) + S ( u , v h ) − a h ( v h , γ h * λ ) = ∫ Γ d h μ − 1 g 1 v h d s + ∫ Γ n h μ g 2 ∂ v h ∂ n d s , ∀ v h ∈ S h , S n , d ( λ , w h ) + a h ( u , γ h * w h ) = ( f , γ h * w h ) + ∫ Γ n a g 2 γ h * w h d s , ∀ w h ∈ S h .$
Based on the weak form (15), we define the primal-dual finite volume element approximation of problem (1): find $( u h , λ h ) ∈ S h × S h$ such that
$S d ( u h , v h ) + S ( u h , v h ) − a h ( v h , γ h * λ h ) = ∫ Γ d h μ − 1 g 1 v h d s + ∫ Γ n h μ g 2 ∂ v h ∂ n d s , ∀ v h ∈ S h , S n , d ( λ h , w h ) + a h ( u h , γ h * w h ) = ( f , γ h * w h ) + ∫ Γ n a g 2 γ h * w h d s , ∀ w h ∈ S h .$
Obviously, the discrete scheme (16) is consistent.
Next, we recall some known results which will play an important role in our analysis. Let $I h : C ( Ω ¯ ) → S h$ be the standard linear interpolation operator. The approximation property of $I h$, trace inequality and finite element inverse inequality are as follows
$∥ v − I h v ∥ m , p , K ≤ C h K 2 − m ∥ v ∥ 2 , p , K , 0 ≤ m ≤ 2 , 1 < p ≤ ∞ ,$
$∥ v ∥ 0 , p , ∂ K ≤ C h K − 1 p ( ∥ v ∥ 0 , p , K + h K ∥ ∇ v ∥ 0 , p , K ) , v ∈ W 1 , p ( K ) , 1 ≤ p ≤ ∞ ,$
$∥ v h ∥ m , p , ∂ K ≤ C h K − 1 p ∥ v h ∥ m , p , K , v h ∈ P k ( K ) , m = 0 , 1 , 1 ≤ p ≤ ∞ .$
$∥ v h ∥ m , p , K ≤ C h K 2 p − 2 q ∥ v h ∥ m , q , K , v h ∈ P k ( K ) , m = 0 , 1 , 1 ≤ q ≤ p ≤ ∞ .$
Lemma 1
([27]). Let $K ∈ T h$ and $e ⊂ ∂ K$ be an edge of K. Then, for $v h ∈ S h$, the following results hold
$∫ K ( v h − γ h * v h ) = 0 , ∫ e ( v h − γ h * v h ) d s = 0 .$
Lemma 2.
Let $K ∈ T h$. Then, for $w h ∈ S h , 1 ≤ q ≤ ∞$, the interpolation operator $γ h *$ has the following approximation properties
$∥ w h − γ h * w h ∥ 0 , q , K ≤ C h K ∥ ∇ w h ∥ 0 , q , K ,$
$∥ w h − γ h * w h ∥ 0 , q , ∂ K ≤ C h K 1 − 1 q ∥ ∇ w h ∥ 0 , q , K .$
Proof.
Let P be a vertex of element K. Denote $K ^ = K P * ⋂ K$ the one third part of element K, then $γ h * w h = w h ( P )$ on $K ^$. Following the inverse inequality, we can obtain
$∥ w h − γ h * w h ∥ 0 , q , K ^ = ∥ w h − w h ( P ) ∥ 0 , q , K ^ ≤ h K ^ | K ^ | 1 q | ∇ w h | 0 , ∞ , K ^ ≤ C h K ^ ∥ ∇ w h ∥ 0 , q , K ^ ≤ C h K ∥ ∇ w h ∥ 0 , q , K ,$
which gives estimation (22). In order to prove (23), we let $τ ^ = ∂ K ⋂ ∂ K ^$, by using trace inequality (18) and (22) we have
$∥ w h − γ h * w h ∥ 0 , q , τ ^ ≤ C h K ^ − 1 q ( ∥ w h − γ h * w h ∥ 0 , q , K ^ + h K ^ ∥ ∇ w h ∥ 0 , q , K ^ ) ≤ C h K ^ 1 − 1 q ∥ ∇ w h ∥ 0 , q , K ^ ≤ C h K 1 − 1 q ∥ ∇ w h ∥ 0 , q , K ,$
which completes the proof. □
The basic idea of our forthcoming analysis is to regard the FVE method as a disturbance of the FEM [24,27,28], so we need to give the relationship between the bilinear form of FVE and that of FEM.
Define the bilinear form of FVE and FEM, respectively,
$a V ( w h , γ h * v h ) = − ∑ K P * ∈ T h * ∫ ∂ K P * a ∂ w h ∂ n γ h * v h d s , ∀ w h , v h ∈ S h ,$
$a ( w h , v h ) = ∫ Ω a ∇ w h ∇ v h , ∀ w h , v h ∈ S h .$
Lemma 3
([27]). Let the algebraic sum space $V ( h ) = H 3 2 ( Ω ) + S h$. For all $w ∈ V ( h ) , v h ∈ S h$, we have
$a V ( w , γ h * v h ) − a ( w , v h ) = ∑ K ∈ T h ∫ ∂ K n · ( a ▽ w ) ( γ h * v h − v h ) d s + ∑ K ∈ T h ( − d i v ( a ∇ w ) , γ h * v h − v h ) K .$
From Lemma 3, we can obtain the following estimate between the bilinear form of FVE and that of FEM.
Lemma 4.
For all $w h , v h ∈ S h$, we have
$| a ( w h , v h ) − a V ( w h , γ h * v h ) | ≤ C h ∥ ∇ w h ∥ ∥ ∇ v h ∥ .$
Proof.
From Lemma 3, Lemma 1, Lemma 2 and noting that $w h$ is a piecewise linear function on $T h$, we can derive
$| a ( w h , v h ) − a V ( w h , γ h * v h ) | ≤ | ∑ K ∈ T h ∫ ∂ K a ∂ w h ∂ n ( γ h * v h − v h ) d s | + | ∑ K ∈ T h ∫ K d i v ( a ∇ w h ) ( γ h * v h − v h ) | = | ∑ K ∈ T h ∫ ∂ K ( a − a c ) ∂ w h ∂ n ( γ h * v h − v h ) d s | + | ∑ K ∈ T h ∫ K ∇ a · ∇ w h ( γ h * v h − v h ) | ≤ | a − a c | ∞ ∑ K ∈ T h ( ∫ ∂ K | ∂ w h ∂ n | 2 ) 1 2 ( ∫ ∂ K | γ h * v h − v h | 2 ) 1 2 + C h ∥ ∇ w h ∥ ∥ ∇ v h ∥ ≤ C h | a | 1 , ∞ ∑ K ∈ T h ∥ ∇ w h ∥ ∂ K h K 1 2 ∥ ∇ v h ∥ K + C h ∥ ∇ w h ∥ ∥ ∇ v h ∥ ≤ C h | a | 1 , ∞ ∑ K ∈ T h h K − 1 2 ∥ ∇ w h ∥ K h K 1 2 ∥ ∇ v h ∥ K + C h ∥ ∇ w h ∥ ∥ ∇ v h ∥ ≤ C h ∥ ∇ w h ∥ ∥ ∇ v h ∥ .$
where $a c | K = 1 | K | ∫ K a d x$ is the mean of $a ( x )$ on element K. The proof is completed. □
From Lemma 4 and $a ( u , v ) = ( a ∇ u , ∇ v )$, we see that when h is sufficiently small, the following inequality holds
$a V ( u h , γ h * u h ) ≥ C 0 ∥ ∇ u h ∥ 2 , ∀ u h ∈ S h .$
By the definitions of $a h ( · , · )$ and $a V ( · , · )$, we have
$a h ( u , γ h * v ) = a V ( u , γ h * v ) + ∫ Γ n a ∂ u ∂ n γ h * v d s , ∀ u ∈ V ( h ) , v ∈ S h .$
Lemma 5.
For sufficiently small h, we have
$a h ( w h , γ h * w h ) ≥ γ 0 ∥ ∇ w h ∥ 2 − C 1 ∫ Γ n h μ − 1 | w h | 2 d s , ∀ w h ∈ S h ,$
where $γ 0$ is a constant.
Proof.
By using Lemma 1, Lemma 2 and the inverse inequality, we obtain
$∫ Γ n a ∂ w h ∂ n γ h * v h d s = ∫ Γ n a ∂ w h ∂ n ( γ h * v h − v h ) d s + ∫ Γ n a ∂ w h ∂ n v h d s = ∑ K ∈ T h ∫ Γ n ∩ ∂ K ( a − a c ) ∂ w h ∂ n ( γ h * v h − v h ) d s + ∑ K ∈ T h ∫ Γ n ∩ ∂ K a ∂ w h ∂ n v h d s ≤ C h ∑ K ∈ T h ∥ ∇ w h ∥ ∂ K ∥ γ h * v h − v h ∥ ∂ K + C ∑ K ∈ T h ∥ h μ 1 2 ∇ w h ∥ ∂ K ∩ Γ n ( ∫ Γ n ∩ ∂ K h μ − 1 v h 2 d s ) 1 2 ≤ C h ∥ ∇ w h ∥ ∥ ∇ v h ∥ + C ∥ ∇ w h ∥ ( ∫ Γ n h μ − 1 v h 2 d s ) 1 2 ,$
We complete the proof by combining (31) and (32). □
Theorem 1.
Assume that either $Γ n ≠ ∅$ or $Γ d$ is not a straight line, then the solution $( u h , λ h ) ∈ S h × S h$ of finite volume element scheme (16) exists uniquely.
Proof.
We just need to prove that the discrete linear homogeneous equations (16) only have the trivial solutions. Let $f ( x ) = g 1 ( x ) = g 2 ( x ) = 0$ in (16) and take $v h = u h , w h = λ h$, we can derive by adding the two equations
$S d ( u h , u h ) + S ( u h , u h ) + S n , d ( λ h , λ h ) = 0 .$
With the definitions of $S d ( u , v ) , S n , d ( u , v )$ and $S ( u , v )$, we obtain
$u h | Γ d = 0 , ∂ u h ∂ n | Γ n = 0 , [ ∇ u h ] | e = 0 , e ∈ E h 0 , λ h | Γ d = λ h | Γ n = 0 .$
Notice that $∇ u h$ is piecewise constant on $T h$, then by $[ ∇ u h ] | e = 0$ we know that $∇ u h$ is a constant on the whole domain $Ω$ and $u h$ is a single linear polynomial on $Ω$. Thus, by $∂ u h ∂ n | Γ n = 0$, $u h | Γ d = 0$ and the condition of Theorem 1, we can assert that $u h = 0$ on $Ω$. Furthermore, from the first homogeneous Equation in (16) we can obtain
$a h ( v h , γ h * λ h ) = 0 , ∀ v h ∈ S h .$
Taking $v h = λ h$ in (37), using Lemma 5 and $λ h | Γ n = 0$ we can deduce $∥ ∇ λ h ∥ = 0$, that is, $λ h$ is a constant. Furthermore, $λ h | Γ d = 0$ implies that $λ h ≡ 0$ on $Ω$. □
Theorem 1 gives the well-posedness of the primal-dual finite volume element scheme (16).

## 3. Error Estimation

In this section, we do the error analysis of the primal-dual finite volume element method (16).
Introduce the semi-norm notations
$| u | Γ d = S d 1 2 ( u , u ) , | u | Γ n , d = S n , d 1 2 ( u , u ) , | u | S = S 1 2 ( u , u ) .$
Lemma 6.
Let $e u = u h − I h u ∈ S h$, then the following estimate holds
$| a h ( u − I h u , γ h * v h ) | ≤ C h ∥ u ∥ 2 ( ∥ ∇ v h ∥ + | v h | Γ n , d ) , ∀ v h ∈ S h .$
Proof.
From (32), we first have
$a h ( u − I h u , γ h * v h ) = a V ( u − I h u , γ h * v h ) + ∫ Γ n a ∂ ( u − I h u ) ∂ n γ h * v h d s .$
Using Lemma 3, Lemma 2, trace inequality (18) and the approximation property of $I h u$, we obtain
$| a V ( u − I h u , γ h * v h ) | ≤ | a ( u − I h u , v h ) | + ∑ K ∈ T h | a | ∞ ∥ ∇ ( u − I h u ) ∥ 0 , ∂ K ∥ γ h * v h − v h ∥ 0 , ∂ K + ∑ K ∈ T h [ | ∇ a | ∞ ∥ ∇ ( u − I h u ) ∥ 0 , K + | a | ∞ ∥ Δ u ∥ 0 , K ] ∥ γ h * v h − v h ∥ 0 , K ≤ C h ∥ u ∥ 2 ∥ ∇ v h ∥ .$
It follows from Lemma 2 and trace inequality (18) that
$| ∫ Γ n a ∂ ( u − I h u ) ∂ n γ h * v h d s | ≤ ∑ e ∈ Γ n | ∫ e a ∂ ( u − I h u ) ∂ n ( γ h * v h − v h ) d s | + ∑ e ∈ Γ n | ∫ e a ∂ ( u − I h u ) ∂ n v h d s | ≤ C h ∥ u ∥ 2 ∥ ∇ v h ∥ + ∑ e ∈ Γ n | ∫ e a h μ 1 2 ∂ ( u − I h u ) ∂ n h μ − 1 2 v h d s | ≤ C h ∥ u ∥ 2 ∥ ∇ v h ∥ + C h ∥ u ∥ 2 | v h | Γ n , d .$
Combining (40), (41) and (42), we complete the proof. □
Lemma 7.
Let $( u , λ = 0 )$ be the solution of problem (15) and $( u h , λ h ) ∈ S h × S h$ be the solution of problem (16). Denote error functions $e u = u h − I h u , e λ = λ h − λ = λ h$. Then, the following error estimate holds
$∥ ∇ e λ ∥ ≤ C ( | e u | Γ d + | e u | S + | e λ | Γ n , d ) + C h ∥ u ∥ 2 .$
Proof.
Subtracting (15) from (16), it leads to
$S d ( e u , v h ) + S ( e u , v h ) − a h ( v h , γ h * e λ ) = S d ( u − I h u , v h ) + S ( u − I h u , v h ) , ∀ v h ∈ S h ,$
$S n , d ( e λ , w h ) + a h ( e u , γ h * w h ) = a h ( u − I h u , γ h * w h ) , ∀ w h ∈ S h .$
Taking $v h = e λ$ in (44), we obtain from (17)–(19) that
$| a h ( e λ , γ h * e λ ) | = | S d ( e u , e λ ) + S ( e u , e λ ) − S d ( u − I h u , e λ ) − S ( u − I h u , e λ ) | ≤ | e u | Γ d | e λ | Γ d + | e u | S | e λ | S + C h ∥ u ∥ 2 | e λ | Γ d + C h ∥ u ∥ 2 ∥ ∇ e λ ∥ ≤ | e u | Γ d | e λ | Γ d + C | e u | S ∥ ∇ e λ ∥ + C h ∥ u ∥ 2 ( | e λ | Γ d + ∥ ∇ e λ ∥ ) .$
Hence, by using Lemma 5 we have
$γ 0 ∥ ∇ e λ ∥ 2 ≤ | e u | Γ d | e λ | Γ d + C | e u | S ∥ ∇ e λ ∥ + C h ∥ u ∥ 2 ( | e λ | Γ d + ∥ ∇ e λ ∥ ) + C 1 ∫ Γ n h μ − 1 | e λ | 2 d s .$
By using the Cauchy inequality it yields that
$∥ ∇ e λ ∥ 2 ≤ C ( | e u | Γ d 2 + | e u | S 2 + | e λ | Γ n , d 2 ) + C h 2 ∥ u ∥ 2 2 ,$
which completes the proof. □
Theorem 2.
Under the assumptions of Lemma 7, the following error estimates hold
$| e u | Γ d + | e u | S ≤ C h ∥ u ∥ 2 ,$
$∥ ∇ e λ ∥ + | e λ | Γ n , d ≤ C h ∥ u ∥ 2 .$
Proof.
Setting $v h = e u , w h = e λ$ in error Equations (44) and (45), respectively, and adding these two equations, we obtain
$S d ( e u , e u ) + S ( e u , e u ) + S n , d ( e λ , e λ ) = S d ( u − I h u , e u ) + S ( u − I h u , e u ) + a h ( u − I h u , γ h * e λ ) .$
By the definitions of $S d ( u , v )$ and $S ( u , v )$, estimates (17) and (18), we have
$| S d ( u − I h u , e u ) | ≤ | u − I h u | Γ d | e u | Γ d ≤ C h ∥ u ∥ 2 | e u | Γ d ,$
$| S ( u − I h u , e u ) | ≤ | u − I h u | S | e u | S ≤ C h ∥ u ∥ 2 | e u | S .$
From Lemma 6 and Lemma 7, it yields that
$| a h ( u − I h u , γ h * e λ ) | ≤ C h ∥ u ∥ 2 ( | e λ | Γ n , d + ∥ ∇ e λ ∥ ) ≤ C h ∥ u ∥ 2 ( h ∥ u ∥ 2 + | e u | Γ d + | e u | S + | e λ | Γ n , d ) .$
Taking these estimates into (51) together with the Cauchy inequality we obtain
$| e u | Γ d 2 + | e u | S 2 + | e λ | Γ n , d 2 ≤ C h 2 ∥ u ∥ 2 2 .$
Then, the conclusions of Theorem 2 follow from (55) and Lemma 7. □
Introduce the notations
$| | | u h | | | Γ 2 = | | ∇ u h | | 2 + | u h | Γ d 2 , | | | u h | | | S 2 = | u h | Γ d 2 + | u h | S 2 .$
Obviously, $| | | u h | | | Γ$ defines a norm on $H 1 ( Ω )$ and $| | | u h | | | S$ is also a norm on $S h$ under the condition of Theorem 1. Then, Theorem 2 implies the following error estimates
$| | | u h − I h u | | | S + | | | λ − λ h | | | Γ ≤ C h ∥ u ∥ 2 .$
Theorem 3.
Suppose the boundary condition satisfies $Γ n ⊂ Γ d$. Let u and $u h$ be the solutions of problem (1) and (16), respectively, then the following gradient error estimate holds
$∥ ∇ ( u h − I h u ) ∥ ≤ C h ∥ u ∥ 2 .$
Proof.
When $Γ n ⊂ Γ d$, $S n , d ( u , v ) = S d ( u , v )$ and $| u | Γ d = | u | Γ n , d$. Take $w h = e u$ in (45) we have from Lemma 6
$| a h ( e u , γ h * e u ) | ≤ | S d ( e λ , e u ) | + | a h ( u − I h u , γ h * e u ) | ≤ | e λ | Γ d | e u | Γ d + C h ∥ u ∥ 2 ( ∥ ∇ e u ∥ + | e u | Γ d ) .$
It follows from Lemma 5 that
$γ 0 ∥ ∇ e u ∥ 2 ≤ | e λ | Γ d 2 + | e u | Γ d 2 + C | e u | Γ d 2 + C h 2 ∥ u ∥ 2 2 + γ 0 2 ∥ ∇ e u ∥ 2 + C | e u | Γ d 2 ,$
Combining with Theorem 2, it yields the desired result.
Lemma 8.
On space $H 1 ( Ω )$, the norm $| | | w | | | Γ = ( | w | Γ d 2 + ∥ ∇ w ∥ 2 ) 1 2$ is equivalent to the $H 1$-norm $∥ w ∥ 1$.
Proof.
Let $w ∈ H 1 ( Ω )$. We only need to prove $∥ w ∥ ≤ C ( | w | Γ d + ∥ ∇ w ∥ )$. Introduce the following auxiliary problem
$− ▵ ψ = w , ψ ∈ H 2 ( Ω ) , ψ | Γ d = 0 , ∂ ψ ∂ n | ( Γ d ) c = 0 , ∥ ψ ∥ 2 ≤ C ∥ w ∥ ,$
where $( Γ d ) c$ is the supplementary set of $Γ d$. By Green formula, we arrive at
$∥ w ∥ 2 = ( ∇ w , ∇ ψ ) − ∫ Γ d ∂ ψ ∂ n w d s ≤ ∥ ∇ w ∥ ∥ ∇ ψ ∥ + | ∫ Γ d h μ 1 2 ∂ ψ ∂ n h μ − 1 2 w d s | ≤ ∥ ∇ w ∥ ∥ ∇ ψ ∥ + C t r a ∥ ψ ∥ 2 | w | Γ d ≤ C ( | w | Γ d + ∥ ∇ w ∥ ) ∥ w ∥ ,$
where $C t r a$ is a constant related to the trace inequality. This implies that $∥ w ∥ ≤ C ( | w | Γ d + ∥ ∇ w ∥ )$. □
Theorem 4.
Under the assumption of Theorem 3, the following error estimate holds
$∥ u − u h ∥ 1 ≤ C h ∥ u ∥ 2 .$
Proof.
With Lemma 8, Theorem 2, Theorem 3 and the triangle inequality, we obtain the desired result. □

## 4. Numerical Experiment

In this section, we provide two numerical examples to illustrate the effectiveness of the proposed primal-dual finite volume element method (16). In the numerical experiment, all computations are carried out by using Matlab R2019a.
We consider problem (1) on the domain $[ 0 , 1 ] 2$ with $a ( x ) = 1$. The source term $f = f ( x , y )$ is determined by the exact solution $u ( x , y )$ with the formula $f = − d i v ( a ∇ u )$. The boundary data $g 1 = u ( x , y )$ on $Γ d$ and $g 2 = ∂ u ∂ n$ on $Γ n$ depends on the given exact solution $u ( x , y )$.
In the numerical examples, we first partition the domain $Ω$ into the squares with mesh size $1 / N$. Then, each square is divided into two triangles by the diagonal line connection which results in a uniform triangulation $T h$ whose mesh size is $h = 2 / N$. Based on the uniform triangulation, we construct the barycenter dual partition $T h *$ by connecting the midpoints of the three edges to the barycenter of each element in $T h$ with straight lines.
The error estimation of the primal-dual finite volume element solution is computed in the following three norms
$∥ ∇ ( u − u h ) ∥ = ( ∑ K ∈ T h ( ∇ ( u − u h ) , ∇ ( u − u h ) ) ) 1 2 ,$
$| | | u − u h | | | S = ( | u − u h | Γ d 2 + | u − u h | S 2 ) 1 2 = ( S d ( u − u h , u − u h ) + S ( u − u h , u − u h ) ) 1 2 ,$
$∥ u − u h ∥ = ( ∑ K ∈ T h ( u − u h , u − u h ) K ) 1 2 .$
Example 1.
In this experiment, the exact solution is $u ( x ) = x 2 + y 2$. We consider three boundary conditions as follows
B.C. I: Lacking boundary condition (See Figure 2.)
$Γ d : = { y = 0 ; x ∈ ( 0 , 1 ) } ∪ { x = 1 ; y ∈ ( 0 , 1 ) } ∪ { y = 1 ; x ∈ ( 0 , 1 ) } .$
B.C. II: Lacking and overlap boundary condition (See Figure 3.)
$Γ d : = { y = 0 ; x ∈ ( 0 , 1 ) } ∪ { x = 1 ; y ∈ ( 0 , 1 ) } ∪ { y = 1 ; x ∈ ( 0 , 1 ) } , Γ n : = { y = 0 ; x ∈ ( 0 , 1 ) } .$
B.C. III: Overlap boundary condition (See Figure 4.)
$Γ d : = { y = 0 ; x ∈ ( 0 , 1 ) } ∪ { x = 1 ; y ∈ ( 0 , 1 ) } ∪ { y = 1 ; x ∈ ( 0 , 1 ) ∪ { x = 0 ; y ∈ ( 0 , 1 ) } , Γ n : = { x = 0 ; y ∈ ( 0 , 1 ) } ∪ { y = 0 ; x ∈ ( 0 , 1 ) } .$
The errors and convergence rates of the numerical solution for the above three kinds of boundary conditions are shown in Table 1, Table 2 and Table 3, respectively.
The exact solution u and the numerical solution $u h$ of B.C. I are shown in Figure 5 and Figure 6, respectively.
Example 2.
In this experiment, the exact solution is $u ( x ) = x 2 e y$. We consider three boundary conditions as follows
B.C. I: Lacking boundary condition (See Figure 2).
$Γ d : = { y = 0 ; x ∈ ( 0 , 1 ) } ∪ { x = 1 ; y ∈ ( 0 , 1 ) } ∪ { y = 1 ; x ∈ ( 0 , 1 ) } .$
B.C. IV: Lacking and overlap boundary condition (See Figure 7).
$Γ d : = { y = 0 ; x ∈ ( 0 , 1 ) } ∪ { x = 1 ; y ∈ ( 0 , 1 ) } ∪ { y = 1 ; x ∈ ( 0 , 1 ) } , Γ n : = { y = 0 ; x ∈ ( 0 , 1 ) } ∪ { x = 1 ; y ∈ ( 0 , 1 ) } .$
B.C. V: Overlap boundary condition (See Figure 8).
$Γ d : = { y = 0 ; x ∈ ( 0 , 1 ) } ∪ { x = 1 ; y ∈ ( 0 , 1 ) } ∪ { y = 1 ; x ∈ ( 0 , 1 ) ∪ { x = 0 ; y ∈ ( 0 , 1 ) } , Γ n : = { y = 1 ; x ∈ ( 0 , 1 ) } .$
The errors and convergence rates of the numerical solution for the above three kinds of boundary conditions are shown in Table 4, Table 5 and Table 6, respectively.
The exact solution u and the numerical solution $u h$ of B.C. V. are shown in Figure 9 and Figure 10, respectively.
Example 3.
In this experiment, the exact solution is $u ( x ) = x 2 + e y$. We consider three boundary conditions as follows:
B.C. I: Lacking boundary condition (See Figure 2).
$Γ d : = { y = 0 ; x ∈ ( 0 , 1 ) } ∪ { x = 1 ; y ∈ ( 0 , 1 ) } ∪ { y = 1 ; x ∈ ( 0 , 1 ) } .$
B.C. II: Lacking and overlap boundary condition (See Figure 3).
$Γ d : = { y = 0 ; x ∈ ( 0 , 1 ) } ∪ { x = 1 ; y ∈ ( 0 , 1 ) } ∪ { y = 1 ; x ∈ ( 0 , 1 ) } , Γ n : = { y = 0 ; x ∈ ( 0 , 1 ) } .$
B.C. V: Overlap boundary condition (See Figure 8).
$Γ d : = { y = 0 ; x ∈ ( 0 , 1 ) } ∪ { x = 1 ; y ∈ ( 0 , 1 ) } ∪ { y = 1 ; x ∈ ( 0 , 1 ) ∪ { x = 0 ; y ∈ ( 0 , 1 ) } , Γ n : = { y = 1 ; x ∈ ( 0 , 1 ) } .$
The errors and convergence rates of the numerical solution for the above three kinds of boundary conditions are shown in Table 7, Table 8 and Table 9, respectively.
The exact solution u and the numerical solution $u h$ of B.C. V are shown in Figure 11 and Figure 12, respectively.

## 5. Conclusions

We consider how to solve numerically an ill-posed elliptic problem with missing or overlap boundary conditions. A primal-dual finite volume scheme with proper stabilized terms is proposed and analyzed. We first prove the well-posedness of the discrete scheme. Then, we derive the gradient error estimate of $O ( h )$-order and some other error estimates under proper norms. Numerical experiments verify the effectiveness of the proposed primal-dual finite volume element method.

## Author Contributions

All authors have contributed equally to the article. All authors have read and agreed to the published version of the manuscript.

## Funding

This research was funded by the State Key Laboratory of Synthetical Automation for Process Industries Fundamental Research Funds, No. 2013ZCX02.

Not applicable.

## Conflicts of Interest

The authors declare no conflict of interest.

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Figure 1. Dual element $K P *$ at node P.
Figure 1. Dual element $K P *$ at node P.
Figure 2. B.C. I.
Figure 2. B.C. I.
Figure 3. B.C. II.
Figure 3. B.C. II.
Figure 4. B.C. III.
Figure 4. B.C. III.
Figure 5. Exact solution $u = x 2 + y 2$.
Figure 5. Exact solution $u = x 2 + y 2$.
Figure 6. Numerical solution $u h$ of B.C. I.
Figure 6. Numerical solution $u h$ of B.C. I.
Figure 7. B.C. IV.
Figure 7. B.C. IV.
Figure 8. B.C. V.
Figure 8. B.C. V.
Figure 9. Exact solution $u = x 2 e y$.
Figure 9. Exact solution $u = x 2 e y$.
Figure 10. Numerical solution $u h$ of B.C. V.
Figure 10. Numerical solution $u h$ of B.C. V.
Figure 11. Exact solution $u = x 2 + e y$.
Figure 11. Exact solution $u = x 2 + e y$.
Figure 12. Numerical solution $u h$ of B.C. V.
Figure 12. Numerical solution $u h$ of B.C. V.
Table 1. Experiment results for the exact solution $u = x 2 + y 2$ with B.C. I.
Table 1. Experiment results for the exact solution $u = x 2 + y 2$ with B.C. I.
N$∥ ∇ ( u − u h ) ∥$$| | | u − u h | | | S$$∥ u − u h ∥$
ErrorOrderErrorOrderErrorOrder
40.2757N/A0.4144N/A6.6300 × 10$− 2$N/A
80.13011.08350.20321.02812.2700 × 10$− 2$1.5463
160.05971.12380.09671.07136.4000 × 10$− 3$1.8265
320.02911.03670.04751.02561.8000 × 10$− 3$1.8301
640.01441.01500.02361.00914.7860 × 10$− 4$1.9111
1280.00721.00000.01181.00001.2875 × 10$− 4$1.8942
Table 2. Experiment results for the exact solution $u = x 2 + y 2$ with B.C. II.
Table 2. Experiment results for the exact solution $u = x 2 + y 2$ with B.C. II.
N$∥ ∇ ( u − u h ) ∥$$| | | u − u h | | | S$$∥ u − u h ∥$
ErrorOrderErrorOrderErrorOrder
40.2392N/A0.3898N/A5.5900 × 10$− 2$N/A
80.11551.05030.19211.02092.1300 × 10$− 2$1.3920
160.04861.24890.08731.13786.2000 × 10$− 3$1.7805
320.02231.12390.04101.09041.7000 × 10$− 3$1.8667
640.01081.04600.02001.03564.5130 × 10$− 4$1.9134
1280.00531.02700.00991.01451.2153 × 10$− 4$1.8928
Table 3. Experiment results for the exact solution $u = x 2 + y 2$ with B.C III.
Table 3. Experiment results for the exact solution $u = x 2 + y 2$ with B.C III.
N$∥ ∇ ( u − u h ) ∥$$| | | u − u h | | | S$$∥ u − u h ∥$
ErrorOrderErrorOrderErrorOrder
40.2482N/A0.4199N/A5.8400 × 10$− 2$N/A
80.12760.95990.20701.02042.0000 × 10$− 2$1.5460
160.06281.02280.10480.98206.5000 × 10$− 3$1.6215
320.03051.04200.05320.97811.9000 × 10$− 3$1.7744
640.01491.03350.02680.98925.5340 × 10$− 4$1.7796
1280.00741.00970.01350.98931.5403 × 10$− 4$1.8451
Table 4. Experiment results for the exact solution $u = x 2 e y$ with B.C. I.
Table 4. Experiment results for the exact solution $u = x 2 e y$ with B.C. I.
N$∥ ∇ ( u − u h ) ∥$$| | | u − u h | | | S$$∥ u − u h ∥$
ErrorOrderErrorOrderErrorOrder
40.5723N/A0.8975N/A1.2870 × 10$− 1$N/A
80.25411.17140.39111.19843.9100 × 10$− 2$1.7188
160.11591.13250.18171.10601.1000 × 10$− 2$1.8297
320.05731.01630.09081.00083.1000 × 10$− 3$1.8272
640.02870.99750.04620.97488.7486 × 10$− 4$1.8251
1280.01440.99500.02340.98142.4102 × 10$− 4$1.8599
Table 5. Experiment results for the exact solution $u = x 2 e y$ with B.C. IV.
Table 5. Experiment results for the exact solution $u = x 2 e y$ with B.C. IV.
N$∥ ∇ ( u − u h ) ∥$$| | | u − u h | | | S$$∥ u − u h ∥$
ErrorOrderErrorOrderErrorOrder
40.5365N/A0.9491N/A1.2670 × 10$− 1$N/A
80.30920.79500.52650.85015.0000 × 10$− 2$1.3414
160.17020.86130.29780.82211.8000 × 10$− 2$1.4739
320.08910.93370.16170.88105.9000 × 10$− 3$1.6092
640.04580.96010.08510.92611.8000 × 10$− 3$1.7127
1280.02330.97500.04380.95825.0591 × 10$− 4$1.8310
Table 6. Experiment results for the exact solution $u = x 2 e y$ with B.C. V.
Table 6. Experiment results for the exact solution $u = x 2 e y$ with B.C. V.
N$∥ ∇ ( u − u h ) ∥$$| | | u − u h | | | S$$∥ u − u h ∥$
ErrorOrderErrorOrderErrorOrder
40.5050N/A0.8853N/A1.1150 × 10$− 1$N/A
80.20011.33560.35801.30623.2100 × 10$− 2$1.7964
160.09011.15110.15371.21988.9000 × 10$− 3$1.8507
320.04351.05050.07221.09002.5000 × 10$− 3$1.8319
640.02161.01000.03551.02427.0990 × 10$− 4$1.8162
1280.01081.00000.01771.00411.9578 × 10$− 4$1.8584
Table 7. Experiment results for the exact solution $u = x 2 e y$ with B.C. I.
Table 7. Experiment results for the exact solution $u = x 2 e y$ with B.C. I.
N$∥ ∇ ( u − u h ) ∥$$| | | u − u h | | | S$$∥ u − u h ∥$
ErrorOrderErrorOrderErrorOrder
40.4205N/A0.6184N/A1.0720 × 10$− 1$N/A
80.19721.09240.31120.99073.5800 × 10$− 2$1.5823
160.09201.10000.14951.05771.0300 × 10$− 2$1.7973
320.04461.04460.07291.03622.8000 × 10$− 3$1.8791
640.02211.01300.03601.01797.5851 × 10$− 4$1.8842
1280.01101.00650.01791.00802.0323 × 10$− 4$1.9001
Table 8. Experiment results for the exact solution $u = x 2 e y$ with B.C. II.
Table 8. Experiment results for the exact solution $u = x 2 e y$ with B.C. II.
N$∥ ∇ ( u − u h ) ∥$$| | | u − u h | | | S$$∥ u − u h ∥$
ErrorOrderErrorOrderErrorOrder
40.3758N/A0.5988N/A9.5300 × 10$− 2$N/A
80.19620.93760.33030.85833.8300 × 10$− 2$1.3151
160.08981.12750.16171.03181.1900 × 10$− 2$1.6864
320.04261.07590.07831.04623.4000 × 10$− 3$1.8074
640.02091.02740.03851.02429.2049 × 10$− 4$1.8851
1280.01041.00690.01911.01132.4911 × 10$− 4$1.8856
Table 9. Experiment results for the exact solution $u = x 2 e y$ with B.C. V.
Table 9. Experiment results for the exact solution $u = x 2 e y$ with B.C. V.
N$∥ ∇ ( u − u h ) ∥$$| | | u − u h | | | S$$∥ u − u h ∥$
ErrorOrderErrorOrderErrorOrder
40.3638N/A0.5879N/A8.8300 × 10$− 2$N/A
80.25260.52630.38640.60553.6600 × 10$− 2$1.2706
160.14890.76250.23810.69851.4600 × 10$− 2$1.3259
320.07620.96650.12920.88204.7000 × 10$− 3$1.6352
640.03840.98870.06690.94951.4000 × 10$− 3$1.7472
1280.01921.00000.03400.97653.8797 × 10$− 4$1.8514
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Sheng, Y.; Zhang, T. A Finite Volume Method to Solve the Ill-Posed Elliptic Problems. Mathematics 2022, 10, 4220. https://doi.org/10.3390/math10224220

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Sheng Y, Zhang T. A Finite Volume Method to Solve the Ill-Posed Elliptic Problems. Mathematics. 2022; 10(22):4220. https://doi.org/10.3390/math10224220

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Sheng, Ying, and Tie Zhang. 2022. "A Finite Volume Method to Solve the Ill-Posed Elliptic Problems" Mathematics 10, no. 22: 4220. https://doi.org/10.3390/math10224220

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