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Sample polygonal meshes used
for Example 4 with (a) 21, (b) 200
and (c) 1000 elements.

This example has been solved on	meshes with polygonal and quadrilateral elements for various mesh densities and approximation orders $p=3$ up to $p=10$ on each mesh. Three polygonal meshes are shown in Fig. <ref>.	The results are presented in the form of convergence rates for various values of $p$ versus the square root of the number of degrees of freedom (see Fig. <ref>). Numerically computed convergence rates are consistent with the theoretical estimates. Note that the rate of convergence in the $L^2$ norm is $\porder+1$ except for the case when $\porder=10$ on polygonal meshes. The convergence curve for $\porder = 10$ on the polygonal case tilts for the last mesh, which is likely caused due to round-off errors when the results are close to machine precision. Such a situation is not observed for the quadrilateral mesh, which is because the Chebyshev basis functions are weight-orthogonal on square elements and so rounding errors are much smaller in this case. In  Fig. <ref>, the integrated values of the jumps of the approximate solution along the mesh skeleton are presented. Similar curves but those that pertain to the boundary conditions are depicted in Fig. <ref>. The rates of convergence of the jumps of the approximate solution and violation of boundary conditions are similar to the global error rates. It shows that the final approximate solution is not strictly continuous but the level of the discontinuity is small and tends to zero with mesh refinement.

[width=14cm, height=14cm,

xmin=10^(1.2), 	xmax=10^(2.3),
ymin=10^(-11.5), ymax=10^(-2),
legend style=at=(0.3,0.32),
xtick = 20.493902, 35.749126, 62.353829, 108.770400, 189.736660,
xticklabels = 420, 1278, 3888, 11831, 36000,

[solid, color=blue, thick, mark=*]
table [x index=3, y index=4] figures/example3/conv2_pss_polyg_p3.txt;

[dashed, color=red, thick, mark=square*]
table [x index=3, y index=4] figures/example3/conv2_pss_polyg_p5.txt;

[dashdotted, color=brown, thick, mark=triangle*, mark size=3pt]
table [x index=3, y index=4] figures/example3/conv2_pss_polyg_p7.txt;

[dotted, color=black, thick, mark=diamond*, mark size=3pt]
table [x index=3, y index=4] figures/example3/conv2_pss_polyg_p10.txt;

[width=14cm, height=14cm,

xmin=10^(1.2), 	xmax=10^(2.3),
ymin=10^(-11.5), ymax=10^(-2),
legend style=at=(0.3,0.32),
xtick = 20.493902, 35.749126, 62.353829, 108.770400, 189.736660,
xticklabels = 420, 1278, 3888, 11831, 36000,
[solid, color=blue, thick, mark=*]
table [x index=3, y index=4] figures/example3/conv2_pss_square_p3.txt;

[dashed, color=red, thick, mark=square*]
table [x index=3, y index=4] figures/example3/conv2_pss_square_p5.txt;

[dashdotted, color=brown, thick, mark=triangle*, mark size=3pt]
table [x index=3, y index=4]figures/example3/conv2_pss_square_p7.txt;

[dotted, color=black, thick, mark=diamond*, mark size=3pt]
table [x index=3, y index=4]figures/example3/conv2_pss_square_p10.txt;

Convergence study of the  method
over the domain for Example 4 on (a) polygonal meshes
and (b) quadrilateral meshes. Scale is

[width=14cm, height=14cm,

xmin=10^(1.2), 	xmax=10^(2.3),
legend style=at=(0.3,0.32),
xtick = 20.493902, 35.749126, 62.353829, 108.770400, 189.736660,
xticklabels = 420, 1278, 3888, 11831, 36000,

[solid, color=blue, thick, mark=*]
table [x index=3, y index=5] figures/example3/conv2_pss_polyg_p3.txt;

[dashed, color=red, thick, mark=square*]
table [x index=3, y index=5] figures/example3/conv2_pss_polyg_p5.txt;

[dashdotted, color=brown, thick, mark=triangle*, mark size=3pt]
table [x index=3, y index=5] figures/example3/conv2_pss_polyg_p7.txt;

[dotted, color=black, thick, mark=diamond*, mark size=3pt]
table [x index=3, y index=5] figures/example3/conv2_pss_polyg_p10.txt;

[width=14cm, height=14cm,

xmin=10^(1.2), 	xmax=10^(2.3),
legend style=at=(0.3,0.32),
xtick = 20.493902, 35.749126, 62.353829, 108.770400, 189.736660,
xticklabels = 420, 1278, 3888, 11831, 36000,
[solid, color=blue, thick, mark=*]
table [x index=3, y index=5] figures/example3/conv2_pss_square_p3.txt;

[dashed, color=red, thick, mark=square*]
table [x index=3, y index=5] figures/example3/conv2_pss_square_p5.txt;

[dashdotted, color=brown, thick, mark=triangle*, mark size=3pt]
table [x index=3, y index=5]figures/example3/conv2_pss_square_p7.txt;

[dotted, color=black, thick, mark=diamond*, mark size=3pt]
table [x index=3, y index=5]figures/example3/conv2_pss_square_p10.txt;

Convergence study of the  method
along the mesh skeleton
(discontinuity measure) for Example 4 on (a) polygonal meshes
and (b) quadrilateral meshes. Scale is

[width=14cm, height=14cm,
xmin=10^(1.2), 	xmax=10^(2.3),
legend style=at=(0.3,0.32),
xtick = 20.493902, 35.749126, 62.353829, 108.770400, 189.736660,
xticklabels = 420, 1278, 3888, 11831, 36000,
[solid, color=blue, thick, mark=*]
table [x index=3, y index=6] figures/example3/conv2_pss_polyg_p3.txt;

[dashed, color=red, thick, mark=square*]
table [x index=3, y index=6] figures/example3/conv2_pss_polyg_p5.txt;

[dashdotted, color=brown, thick, mark=triangle*, mark size=3pt]
table [x index=3, y index=6] figures/example3/conv2_pss_polyg_p7.txt;

[dotted, color=black, thick, mark=diamond*, mark size=3pt]
table [x index=3, y index=6] figures/example3/conv2_pss_polyg_p10.txt;

[width=14cm, height=14cm,
xmin=10^(1.2), 	xmax=10^(2.3),
legend style=at=(0.3,0.32),
xtick = 20.493902, 35.749126, 62.353829, 108.770400, 189.736660,
xticklabels = 420, 1278, 3888, 11831, 36000,
[solid, color=blue, thick, mark=*]
table [x index=3, y index=6] figures/example3/conv2_pss_square_p3.txt;

[dashed, color=red, thick, mark=square*]
table [x index=3, y index=6] figures/example3/conv2_pss_square_p5.txt;

[dashdotted, color=brown, thick, mark=triangle*, mark size=3pt]
table [x index=3, y index=6]figures/example3/conv2_pss_square_p7.txt;

[dotted, color=black, thick, mark=diamond*, mark size=3pt]
table [x index=3, y index=6]figures/example3/conv2_pss_square_p10.txt;

Convergence study of the  method
along the Dirichlet boundary
for Example 4 on (a) polygonal meshes
and (b) quadrilateral meshes. Scale is

§.§ Example 5: Cook’s membrane problem for hyperelasticity

In this example, we solve the well-known Cook's membrane problem, [23] which combines bending and shear deformation. The cantilever is clamped on the left-hand side and is subjected to a constant shear loading on the right-hand side. The geometry, boundary conditions, and material parameters for this problem are shown in Fig. <ref>.

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[-latex] (48+1.5,44) – ++(0,16/5);
[-latex] (48+1.5,44+16/5) – ++(0,16/5);
[-latex] (48+1.5,44+2*16/5) – ++(0,16/5);
[-latex] (48+1.5,44+3*16/5) – ++(0,16/5);
[-latex] (48+1.5,44+4*16/5) – ++(0,16/5);
(48+1.5,44+16/2) node[right]$p_0$;
[black] (48,44+16) circle (15pt) node[anchor=south west]$A$;
[] (0,0)–++(-2,-4) ++(2,4+4) – ++(-2,-4)
++(2,4+4) – ++(-2,-4)  ++(2,4+4) – ++(-2,-4)  ++(2,4+4) – ++(-2,-4)
++(2,4+4) – ++(-2,-4)  ++(2,4+4) – ++(-2,-4)  ++(2,4+4) – ++(-2,-4)
++(2,4+4) – ++(-2,-4)  ++(2,4+4) – ++(-2,-4)  ++(2,4+4) – ++(-2,-4)
++(2,4+4) – ++(-2,-4)  ++(2,4+4)

[] (0,0-5)
– ++(-1,-1) – ++(2,2) ++(-1,-1)
– ++(48,0)
– ++(-1,-1) – ++(2,2) ++(-1,-1)
++(-48/2,0) node[anchor = north]48 mm

[] (44+12,0)
– ++(-1,-1) – ++(2,2) ++(-1,-1)
– ++(0,44)
– ++(-1,-1) – ++(2,2) ++(-1,-1)
– ++(0,16)
– ++(-1,-1) – ++(2,2) ++(-1,-1)
++ (0,-16/2) node[anchor = north, rotate=90]16 mm
++ (0,-16/2-44/2) node[anchor = north, rotate=90]44 mm
[](10,40) node[]$\times$ 1 mm;
[] (21,17) node[anchor=west]$E = 500$ MPa
(21,17-5) node[anchor=west]$\nu = 0.35$
(21,17-10) node[anchor=west]$p_0 = 20$ MPa

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Cook's membrane problem (Example 5)
with (a) geometry, material parameters and boundary conditions, and (b) polygonal mesh
discretization of the domain.

The Cook's membrane is analyzed under plane strain conditions. The hyperelastic model for the material is defined by the strain energy density function[57]

\begin{align}
\psi = \frac{\mu}{2} \left(\Tr \RightCG -3 \right) + \frac{\lambda}{4} \left(J^2 -1 \right)
- \left( \frac{\lambda}{2} + \mu \right) \ln J  \eqcomma
\end{align}

where $\lambda$ and $\mu$ are the Lamé parameters, $J$ is the determinant of the deformation gradient $\Jacob=\gradoper \displu + \unitmatrix$, and $\bm C = \Jacob\tran \bm \Jacob$ is the right Cauchy-Green tensor. The second Piola Kirchhoff stress tensor and the material tangent tensor are

\begin{align}
\spk = 2\frac{\partial \psi}{\partial \RightCG} =
\frac{\lambda}{2} \left(J^2-1 \right) \bm C^{-1} + \mu \left( \unitmatrix - \RightCG^{-1} \right)  \eqcomma
\\&
\mattangent = 2\frac{\partial \spk}{\partial \RightCG} \eqcomma
\end{align}

where symbolic computations are used to compute the tensor $\mattangent$.

The momentum balance, in the absence of the body forces, in the initial configuration can be expressed in terms of the first Piola-Kirchhoff stress tensor as:

\begin{align}
\diverg \fpk = \bm 0 \quad \text{in } \Domain \eqcomma
\\&
\fpk\scalarprod \VVnormal = \boundarysign{\spow} \quad \text{on } \Boundaryneum \eqcomma
\\&
\displu= \bm 0  \quad \text{on } \Boundarydirich \eqcomma
\end{align}

where $\VVnormal$ denotes the unit outward normal.

The weak form of (<ref>) is: Find $\displu \in \funspacecz$ such that

\begin{align}\label{eq:momentum_pk_weak}
\omint \gradoper \fvu   : \fpk  \dOm  - \gaintt{\Boundaryneum} \fvu\tran \boundarysign{\spow} \dGa = 0 \ \
\forall \fvu \in \funspacecz
\eqdot
\end{align}

The Green-Lagrange strain tensor and its variation are

\begin{align}
\GLstrain = \frac{1}{2} \left( \Jacob\tran\Jacob  - \unitmatrix \right) \eqcomma
\\&
\delta\GLstrain = \frac{1}{2} \left(\delta\Jacob\tran\Jacob + \Jacob\tran\delta\Jacob \right)
\delta \Jacob = \frac{\partial \delta \displu}{\partial \bm X} = \frac{\partial \fvu }{\partial \bm X} = \gradoper \fvu
\end{align}

Using the relation between the first and the second Piola-Kirchhoff stress tensors, $\fpk = \Jacob \spk$, the inner product (<ref>) is rewritten as follows:

\begin{align}
\gradoper \fvu   : \fpk =
\gradoper \fvu  : \left(\Jacob\cdot\spk\right) = \left( \gradoper\fvu\tran \cdot \Jacob \right) : \spk
\\&=
\frac{1}{2}  \left( \gradoper\fvu\tran \cdot \Jacob  + \Jacob\tran \gradoper\fvu \right) : \spk
+ \cancelto{0}{\frac{1}{2}  \left( \gradoper\fvu\tran \cdot \Jacob  - \Jacob\tran \gradoper\fvu \right): \spk}
\delta\GLstrain(\fvu) : \spk \eqdot
\end{align}

The tensor $\spk$ is an unknown tensor and is updated using its increments, which can be expressed in the recurrence form:

\begin{align}
\spk = \spk  + \Delta \spk \eqdot
\end{align}

The tensor increment $\Delta\spk$ is expressed using a linearized relation.  There is no unique technique for linearization and in this paper, the following one is used:

\begin{align}
\Delta\spk \approx \mattangent \strain(\Delta\displu),
\quad
\strain(\Delta\displu) = \frac{1}{2} \left( \gradoper \Delta\displu +
\gradoper \Delta\displu\tran \right) \eqdot
\end{align}

The nonlinear problem in (<ref>) is solved using an incremental iterative procedure, which is written in the following form:

\begin{align}
\omint 	\delta\GLstrain(\fvu) : \mattangent \strain(\Delta\displu) \dOm
= \Lambda \gaintt{\Boundaryneum} \fvu\tran \boundarysign{\spow} \dGa - \omint \gradoper \fvu   : \fpk  \dOm
\eqdot
\end{align}

In this procedure, the external force multiplier $0< \Lambda <1$ is increased at every step. At every incremental step, equilibrium is found by updating the tensor $\fpk$ at every iterative step. The tangent tensor $\mattangent$ is updated at every incremental step.

Meshes with polygonal elements are used in the analysis (see Fig. <ref>). The calculations are performed using approximation orders in the elements $p=$ 3, 5, 7, and 10. Numerical results are shown in Fig. <ref> in the form of convergence curves for the displacement $u_y$ at point $A$ for various $p$ and in Fig. (<ref>) the maps of the stress components, smoothed using the least-square technique. Convergence of the tip displacement to the value around $u_y=10.59$ is observed, which agrees with the results shown in Schröder at al.[57] 	In this example, the calculations are performed on the initial configuration, so the boundary and continuity condition components $\ttransfmatrix$ and $\tilde{\vm u}_b$ from (<ref>) are computed only once, just before the incremental-iterative procedure.

ylabel=Displacement $u_y$ at point $A$,

legend style=at=(0.7,0.03),anchor=south west,
xmode = log,

[solid, color=blue, thick, mark=*]
table [x index=3, y index=4] figures/example6_cook/cook2_p3.txt;

[dashed, color=red, thick, mark=square*]
table [x index=3, y index=4] figures/example6_cook/cook2_p5.txt;

[dashdotted, color=brown, thick, mark=triangle*, mark size=3pt]
table [x index=3, y index=4] figures/example6_cook/cook2_p7.txt;

[dotted, color=black, thick, mark=diamond*, mark size=3pt]
table [x index=3, y index=4] figures/example6_cook/cook2_p10.txt;

Convergence curves of
displacement $u_y$ at point $A$ for Example 5.

Stress components for the Cook's membrane problem (a) $\sigma_{xx}$, (b) $\sig_{yy}$, (c) $\sig_{xy}$  for Example 5.

§.§ Example 6: Fourth-order biharmonic problem

In this example, we consider the fourth-order biharmonic boundary-value problem. The formulation for this problem is presented in Section <ref>. The exact solution is chosen to be trigonometric, the same as in (<ref>). The biharmonic problem presented in eq:4problem is chosen with the parameters defined as  trigonometric or polynomial functions:

\begin{align}
\paramone =  \sin(1.3\pi\,x)\,\cos(1.8\pi\,y)+2 \eqcomma
\\&
\paramtwo =  4\,x^2\,y^2 + 2\,x y^2 + y +  \eqcomma
\\&
\paramthree = 3\,x y^2 - 2\,x,y+2  \eqdot
\end{align}

The domain $\Domain=(-1,\,1)^2$  is covered by polygonal elements. Convergence curves for various approximation orders are presented in Fig. <ref>. In all cases, optimal convergence rates are obtained, confirming the effectiveness of the proposed method.

[width=14cm, height=14cm,
xtick = 36.331804, 52.829916, 76.830983, 111.731822, 162.480768,
xticklabels = 1320, 2791, 5903, 12484, 26400,

[solid, color=blue, thick, mark=*]
table [x index=3, y index=4] figures/example4/ex3_poly_p4_vsin.txt;

[dashed, color=red, thick, mark=square*]
table [x index=3, y index=4] figures/example4/ex3_poly_p6_vsin.txt;

[dashdotted, color=brown, thick, mark=triangle*, mark size=3pt]
table [x index=3, y index=4] figures/example4/ex3_poly_p8_vsin.txt;

[dotted, color=black, thick, mark=diamond*, mark size=3pt]
table [x index=3, y index=4] figures/example4/ex3_poly_p10_vsin.txt;

Convergence study of the
method for Example 6 on polygonal meshes. Scale is

§.§ Example 7: 3D benchmark problem

A Poisson problem in $\Domain=(-1,1)^3$ is solved with an exponential exact solution of the form:

\begin{align}\label{eq:exp3d_example}
u(x,y,z) = \sum_{i=1}^{50}  \exp \left(-a_i (x - x_i)^2 - b_i (y-y_i)^2 - c_i (y-z_i)^2\right) \eqcomma
\end{align}

where the positive values of the parameters $a$, $b$ and $c$  are chosen randomly and the points $(x_i, y_i, z_i)$ are randomly generated in the domain. This benchmark problem is the three-dimensional extension of the one presented in Section <ref>.

The calculations are performed using structured hexahedral meshes for $p = 3, \ 4,\ 6,\ 10$. Convergence curves in the $L^2$ norm for different $p$ are presented in Fig. <ref>. The convergence plots once again reveal that the  method converges optimally for different approximation orders.

[width=14cm, height=14cm,
legend style=at=(0.03,0.03),anchor=south west,
legend style=font=,
xtick = 13.177065, 16.938649, 21.774557, 27.991069, 35.981729,
xticklabels = 2288, 4860, 10324, 21931, 46585,

[solid, color=blue, thick, mark=*]
table [x index=3, y index=4] figures/example5_3d/ex_exp3d_p3.txt;

[dashed, color=red, thick, mark=square*]
table [x index=3, y index=4] figures/example5_3d/ex_exp3d_p4.txt;

[dashdotted, color=brown, thick, mark=triangle*, mark size=3pt]
table [x index=3, y index=4] figures/example5_3d/ex_exp3d_p6.txt;

[dotted, color=black, thick, mark=diamond*, mark size=3pt]
table [x index=3, y index=4] figures/example5_3d/ex_exp3d_p10.txt;

Convergence study in the $L^2$ norm
of the
method for Example 7 on hexahedral meshes. Scale is

§ CONCLUSIONS

In this paper, a new DG method is presented, in which no penalty or stability parameter is used, and hence the method is referred to as the penalty-free DG method. Generally, in DG methods, the global approximations are not continuous a priori on the mesh skeleton and continuity has to be enforced by penalty-like parameters. A similar scenario arises when imposing  Dirichlet boundary conditions. An alternative approach is proposed in this paper where the consistency of the approximation field, i.e., continuity on the mesh skeleton and satisfaction of the Dirichlet boundary conditions, is obtained by adding least-squares constraint relations. In the discrete version, the constraints require solving a singular system of equations. Such an approach allows for the construction of the discrete, continuous approximation fields over the whole domain, independent of the basis functions within the elements, and at the same time, the Dirichlet boundary conditions are satisfied. In the implementation, the consistent approximation field is augmented (especially needed on polygonal meshes), which allows for small levels of discontinuities on the mesh skeleton.

The augmented consistent approximation field consists of basis functions in all finite elements, and the constraints are applied to the degrees of freedom, which guarantee the consistency of the approximation field. The constraints are expressed via a rectangular system of equations that is singular. In this paper, the method for solving such a singular system of equations is proposed, which can be solved globally, or the solution is constructed using a special assembly-free approach. The constraint relations are similar to the ones used in the DG method with Lagrange multipliers. However, in the case of the  method, there is no need to construct an additional space for the Lagrange multipliers.

The  method is verified on several benchmark problems where various types of boundary-value problems are analyzed in 2D and for one problem in 3D. In each example, the convergence plots are presented for the approximation orders up to $\porder=10$, and optimal convergence rates in the $L^2$ norm and the energy seminorm are obtained. In finite-dimensional space, small jumps           (discontinuities) are observed in the approximate solution, but it has been shown via the examples that the jumps converge to zero with similar rates as the global error. The method can be especially effective in the nonlinear regime since the constraint problem is solved only once before the incremental iterative procedure. It is shown that the method works well for standard elliptic problems, such as the Poisson equation or linear and nonlinear elasticity. It was also tested for the biharmonic problem, where the approximation in a standard Galerkin method must be $C^1$ continuous. As part of future research, application of the  method to other classes of boundary-value problems will be pursued.

§ DATA AVAILABILITY STATEMENT

The data that support the findings of this study are available from the corresponding author upon reasonable request.

§ SOLVER FOR THE  METHOD

The following  algebraic system of equations is considered:

\begin{align}\label{eq:sq_basic_equation}
\thematrix \rarg = \rvector \eqcomma
\end{align}

where $\thematrix$ is an $\matrows\times \matcols$ matrix, $\rvector$ is a $\matrows$-component column vector and $\rarg$ is an $\matcols$-component column vector. The vector $\rarg$ belongs to the set that consists of all possible solutions:

\begin{align}
\rarg \in \solsset = \left\{ \bm y:  \norm{\thematrix \bm y - \rvector} =  0 \right \} \eqdot
\end{align}

It is assumed that the system of equations (<ref>) is consistent, which means that $\solsset \neq \emptyset$. Since we have a linear, consistent system of equations, there exists one or infinite number of solutions. The matrix $\thematrix$ can be a square or rectangular matrix, i.e., $\matrows=\matcols$ or $\matrows<\matcols$. The matrix $\thematrix$ can be singular in the sense that the rows in the matrix are linearly dependent. The solution of the system of (<ref>) can be unique, i.e., there exists only one $\rarg$, otherwise the set $\solsset$ consists of an infinite number of certain vectors.

In this paper, such a general linear system of equations as defined above is solved using the Gauss-Jordan approach as presented in Valco and Vajda.[62] Here the approach is presented in a more general way so that it can be applied to an arbitrarily large system of equations. For the sake of clarity, the nonsingular matrices are firstly considered,  which is extended to singular matrices in <ref>.

The procedure for solving the square (consistent) singular system of equations is presented in Farhat and Roux,[31] where the solver is applied to the floating subdomains as a result of domain partitioning in parallel calculations. This approach is referred to as the null-space method, [9, 51] which is used to solve the saddle-point problem. [58]. This method has successfully been applied to solve problems in incompressible linear and nonlinear elasticity. [46] In this approach, the singular stiffness matrix is partitioned concerning the principal and redundant quantities, which is performed during the factorization process. Then the null space of the stiffness matrix can be identified, which contains the rigid-body modes of the subdomain. Then the final solution is presented as the linear combination of the null space basis vectors plus the vector constructed as the right-hand side vector multiplied by the pseudo-inverse stiffness matrix. However, the procedure presented here is more general, since it can also handle rectangular and singular matrices.

The rectangular system of equations can be solved using the $LQ$ factorization, which is presented in Slobodkins and Tausch.[60] In this method, the matrix is decomposed as $\thematrix = \begin{bmatrix} \vm L & \vm 0\end{bmatrix} \begin{bmatrix}\vm Q^1 & \vm Q^2\end{bmatrix}\tran$, where the matrix $\vm L$ is lower triangular and matrix $\vm Q$ consists of orthogonal vectors where the vectors in $\vm Q^2$ form the basis of the null space of matrix $\thematrix$. Another approach that can be applied to solve the linear equation with a rectangular matrix is the singular value decomposition (SVD),[2] and after some modifications, the SVD can also handle the solution of the singular system of equations.

§.§ Global approach

The variables in $\rarg$ are divided into basic variables vector $\rarg_b$ and free variables vector $\rarg_f$.  For the sake of clarity, it can be assumed without loss of generality, that basic and free variables are contiguous in the $\rarg$ vector. The matrix $\thematrix$ is decomposed on:

\begin{align}\label{eq:seteq_divided}
\rarg = \begin{bmatrix} \rarg_b \\ \rarg_f \end{bmatrix}
\quad \Rightarrow \quad
\begin{bmatrix}	\thematrix_b & \thematrix_f \end{bmatrix}
\begin{bmatrix} \rarg_b \\ \rarg_f \end{bmatrix} = \rvector \eqdot
\end{align}

The division into the basic and free parts is done in such a way that the matrix $\thematrix_b$ is square and well-conditioned. On multiplication both sides of (<ref>)  by $\thematrix_b^{-1}$, we get

\begin{align}\label{eq:seteq_divided2}
%	\begin{bmatrix}	\thematrix_b & \thematrix_f \end{bmatrix}
%	\begin{bmatrix} \rarg_b \\ \rarg_f \end{bmatrix} = \rvector \,\Bigr|\, \thematrix_b^{-1}
%	\quad \rightarrow \quad
\begin{bmatrix}	\eye & \tilde{\thematrix}_f \end{bmatrix}
\begin{bmatrix} \rarg_b \\ \rarg_f \end{bmatrix} = \tilde{\rvector} \eqcomma
\end{align}

\begin{align}\label{eq:tilde_defs}
\tilde{\thematrix}_f = \thematrix_b^{-1} \thematrix_f \quad \text{and} \quad
\tilde{\rvector} = \thematrix_b^{-1} \rvector \eqdot
\end{align}

The system of equations shown in (<ref>), is derived by multiplication  both sides of (<ref>) by $\thematrix_b^{-1}$. However, the same results can be obtained using the Gauss-Jordan (GJ) elimination procedure. The GJ approach can be applied when the leading components of the matrix $\thematrix_b$ are located on the diagonal. In other cases the equations in (<ref>) can be rearranged to place the leading components on the diagonal or the leading component can be searched in every row of the matrix $\thematrix_b$. In this approach, the matrix $\thematrix_b$ is substituted by the matrix $\bar{\eye}$ which is the zero-one matrix, but the ones are not located on the diagonal. In this situation, it can be observed that

\begin{align}
\begin{bmatrix}	\bar\eye & \bar{\thematrix}_f \end{bmatrix}
\begin{bmatrix} \rarg_b \\ \rarg_f \end{bmatrix} = \bar{\rvector}  \,\Bigr|\, 	\bar\eye\tran
\quad \rightarrow \quad
\begin{bmatrix}	\eye & \tilde{\thematrix}_f \end{bmatrix}
\begin{bmatrix} \rarg_b \\ \rarg_f \end{bmatrix} = \tilde{\rvector} \eqdot
\end{align}

Using (<ref>) the basic variables can be expressed by the free variables:

\begin{align}\label{eq:rbasic_rfree}
\rarg_b = -\tilde{\thematrix_f}\rarg_f + \tilde{\rvector} \eqdot
\end{align}

Using the relation in (<ref>) the vector $\rarg$ can be expressed by the free variable vector $\freevar$:

\begin{align}\label{eq:solution}
&\rarg = \transfmatrix\freevar + \transfvector \eqcomma
\intertext{where}
\label{eq:sols_defs}
&\transfmatrix = \begin{bmatrix} -\tilde{\thematrix_f} \\ \eye \end{bmatrix}
\transfvector = \begin{bmatrix} \tilde{\rvector} \\ \vm 0 \end{bmatrix}
\freevar = \rarg_f \eqdot
\end{align}

If a  vector $\vm y$ is defined as $\vm y = \transfmatrix\vm z + \transfvector$, then

\begin{align*}
\vm y \in \solsset \ \ \forall \vm z \eqdot
\end{align*}

When $\vm y \in \solsset$ that means that the vector satisfies the equation 	 (<ref>). Let us suppose that there exists a vector $\vm z$ for which 	$\vm y \notin \solsset$, then there exists a nonzero vector $\residvector$  such that

\begin{align}
\residvector = \thematrix \vm y - \rvector =
\thematrix \transfmatrix \vm z + \thematrix\transfvector - \rvector \eqdot
\end{align}

Using the definitions of the vector $\vm y$ and in the (<ref>) and (<ref>) and applying the matrix division as in (<ref>), we have

\begin{align}
\residvector = \begin{bmatrix} \thematrix_b & \thematrix_f \end{bmatrix}
\begin{bmatrix}
-\thematrix_b^{-1} \thematrix_f \\ \eye
\end{bmatrix} \vm z
+ \begin{bmatrix} \thematrix_b & \thematrix_f \end{bmatrix}
\begin{bmatrix} \thematrix_b^{-1} \rvector \\ \bm 0 \end{bmatrix} - \rvector \eqcomma
\end{align}

\begin{align}
\residvector = (- \thematrix_f + \thematrix_f) \vm z + \rvector - \rvector
\quad \Rightarrow \quad \residvector = \vm 0 \eqdot
\end{align}

It was shown that for every vector $\vm z$ the residual vector is zero.

The solution in (<ref>) represents the hypersurface in which $\transfvector$ represents the point on that surface and the vectors of the matrix $\transfmatrix$ are the basis vectors for that surface.  But in the case when $\matrows=\matcols$, the number of columns of the matrix $\transfmatrix$ is reduced to zero and the vector $\transfvector$ is the unique solution of the problem in (<ref>). When $\matrows<\matcols$ the vector $\rarg$ represents the points belonging to the hypersurface defined by the pair $(\transfmatrix,\transfvector)$. In this hypersurface, a vector $\rarg_m$ can be distinguished that is the shortest among all vectors in $\solsset$

\begin{align}
\norm{\rarg_m} = \min_{\rarg\in\solsset} \norm{\rarg} \eqdot
\end{align}

In order to identify the $\rarg_m$ vector the relevant $\vm z_m$ has to be found by solving the following minimization problem:

\begin{align}
\vm z_m \,: f(\vm z_m) = \min_{\vm z} f(\vm z) \eqcomma
\intertext{where}
f(\vm z) = \left(\transfmatrix\vm z + \transfvector \right)\tran \left(\transfmatrix\vm z + \transfvector \right) \eqcomma
\end{align}

which leads to the following algebraic system of equations:

\begin{align}
\transfmatrix\tran\transfmatrix \vm z_m = -\transfmatrix\tran \transfvector \eqdot
\end{align}

On solving for $\vm z_m$ the vector $\rarg_m$ is then given by

\begin{align}
\rarg_m = \transfmatrix \vm z_m + \transfvector \eqdot
\end{align}

§.§ Recursive approach

The approach presented in the previous section can be applied to obtain the recursive solution of the set of equations (<ref>). For the recursive approach, this set of equations is divided into $\nsubdivision$ subproblems. For the sake of clarity, firstly let $\nsubdivision=2$:

\begin{align}
\begin{bmatrix}	\thematrix_1 \\ \thematrix_2 \end{bmatrix}	\rarg =
\begin{bmatrix} \rvector_1 \\ \rvector_2 \end{bmatrix} \eqcomma
\end{align}

where $\thematrix_i$ and $\rvector_i$ are the appropriate parts of the matrix $\thematrix$ and $\rvector$, respectively. Now, the problem defined in (<ref>) is reformulated: find the vector $\rarg$ that satisfy the following equations:

\begin{align}
\label{eq:basic_equation_2split_a}
\thematrix_1 \rarg = \rvector_1 \eqcomma
\\&
\label{eq:basic_equation_2split_b}
\thematrix_2 \rarg = \rvector_2 \eqdot
\end{align}

The solution of (<ref>) leads to the following

\begin{align}\label{eq:split2_sol1}
\rarg =\transfmatrix_1 \freevar_1 + \transfvector_1 \eqdot
\end{align}

When applying this solution to (<ref>), we have

\begin{align}\label{eq:split2_sol2}
\thematrix\transfmatrix_1 \freevar_1 = \rvector_2 - \thematrix_2\transfvector_2 \eqdot
\end{align}

The solution of (<ref>) leads to

\begin{align}
\freevar_1 = \transfmatrix_2 \freevar_2 + \transfvector_2 \eqdot
\end{align}

When applying this equation to (<ref>) we have the final solution

\begin{align}
\rarg = \transfmatrix_1\transfmatrix_2\freevar_2 + \transfmatrix_1\transfvector_2 + \transfvector_1 \eqdot
\end{align}

In a case when there are multiple divisions of the set of equations (<ref>) then it is convenient to show the construction of the matrix $\transfmatrix$ and the vector $\transfvector$ using the recursive procedure:

\begin{align}
\begin{aligned}
&\transfvector = \transfmatrix \transfvector_i + \transfvector \eqcomma
\\
&\transfmatrix = \transfmatrix\transfmatrix_i \eqcomma
\end{aligned}
\end{align}

where $\transfvector_i$ and $\transfmatrix_i$ come from the $i$-th block of equations and the initial values are $\transfvector=\vm 0$ and $\transfmatrix=\eye$.

The recursive procedure presented in this section has the same complexity as the global approach in (<ref>), i.e., the number of multiplications in these two approaches is the same. However, the recursive approach can be applied when the system of equations is constructed by segments. Then the solution can be applied to each of the segments sequentially without the need to construct the full set of equations.

§.§ Singular system of equations

When the matrix $\thematrix$ in the problem (<ref>), then the dependent equations should be removed from the system of equations, and then the solution shown in (<ref>) can be achieved. The linearly dependent equations can be identified when applying the Gauss-Jordan elimination procedure to calculate the quantities in (<ref>).  The Gauss-Jordan procedure is performed using the leading position in each row to locate the pivot variable. When reaching a linear dependent equation all the elements in that record become zero, so this equation is deleted from the set of equations. There may be multiple such equations, and each time these equations are eliminated.

§  IN ONE DIMENSION

We provide the step-by-step procedure for the  method on two Poisson problems in 1D. In the first case, $\porder=1$ and $\bporder=1$ are chosen; in the second case, $\porder=2$ and $\bporder=1$ are selected. To clarify the exposition, in both cases, we divide the unit interval into two finite elements $[0,\,\frac{1}{2}]$ and $[\frac{1}{2},\,1]$.

Monomial basis functions are chosen to construct the approximation, which is defined in the element's local coordinates scaled by the element length. For first- and second-order finite element approximations, the $e$-th element basis are:

\begin{align}
\Gshapev^e = \begin{bmatrix} 1 & \frac{2}{L^e} \left(x - x_m^e \right)
\end{bmatrix}
\,,\quad
\Gshapev^e = \begin{bmatrix} 1 & \frac{2}{L^e} \left(x - x_m^e \right)
& \frac{4}{{L^e}^2} \left(x - x_m^e \right)^2 \end{bmatrix} \eqcomma
\end{align}

where $L^e=\frac{1}{2}$ is the length of the element and $x_m^e$ is the mid-point of the element, i.e., $x_m^1=\frac{1}{4}$, $x_m^2 = \frac{3}{4}$.

§.§ Poisson problem with affine solution

We consider the following 1D Poisson problem:

\begin{align}
\label{eq:1d_poisson_a}
\temperature'' = 0 \quad \text{in } \Domain=(0,\,1),
\\&
\label{eq:1d_poisson_b}
u(0)=2,\quad u(1)=5 \eqcomma
\end{align}

which has the exact solution $u(x) = 3x+2$. In this case $\porder=\bporder=1$ so the subscript in the approximation matrix $\Gshapev$ is omitted for clarity. The approximation of $\temperature$ and its derivative in  $[0,\,1]$  is:

\begin{align}\label{eq:1d_ex_approx}
\temperature = \Gshapev \dftemp,\quad
\temperature = \begin{cases}
\begin{bmatrix} 1 & 4x-1 \end{bmatrix} \begin{bmatrix} \dftempsc_1 \\ \dftempsc_2\end{bmatrix}
& \text{if } 0 \le x < 0.5
\\
\begin{bmatrix} 1 & 4x-3 \end{bmatrix} \begin{bmatrix} \dftempsc_3 \\ \dftempsc_4\end{bmatrix}
& \text{if } 0.5 \le x \le 1
\end{cases} \eqcomma
\end{align}

\begin{align}\label{eq:ex_1d_approx}
\temperature' = \Gshapev' \dftemp,\quad
\temperature' = \begin{cases}
\begin{bmatrix} 0 & 4 \end{bmatrix} \begin{bmatrix} \dftempsc_1 \\ \dftempsc_2\end{bmatrix}
& \text{if } 0 \le x < 0.5
\\
\begin{bmatrix} 0 & 4 \end{bmatrix} \begin{bmatrix} \dftempsc_3 \\ \dftempsc_4\end{bmatrix}
& \text{if } 0.5 \le x \le 1
\end{cases}  \eqdot
\end{align}

The constrained problem in (<ref>) requires that (<ref>) has to be firstly solved to yield $\ttransfmatrix$ and $\ttransfvectorb$. The values of the approximation matrix $\Gshapev$ on the boundaries and its jump on the skeleton are:

\begin{align}
&\Gshapev(0) = \begin{bmatrix}1 & -1 & 0 & 0 \end{bmatrix},\quad
\Gshapev'(0) = \begin{bmatrix}0& 4 & 0 & 0 \end{bmatrix} \eqcomma
\\&
\Gshapev(1) = \begin{bmatrix}0 & 0 & 1 & 1 \end{bmatrix},\quad
\Gshapev'(1) = \begin{bmatrix}0& 0 & 0 & 4 \end{bmatrix} \eqcomma
\\&
\inbbr{\Gshapev(0.5)} = \begin{bmatrix} 0 & 0 & 1 & -1\end{bmatrix} - \begin{bmatrix} 1 & 1 & 0 & 0\end{bmatrix} = \begin{bmatrix}-1 & -1 & 1 & -1 \end{bmatrix}  \eqcomma
\\&
\iinbbr{\Gshapev'(0.5)} =  \frac{1}{2}\left(\begin{bmatrix} 0 & 4 & 0 & 0\end{bmatrix} + \begin{bmatrix} 0 & 0 & 0 & 4\end{bmatrix}\right) = \begin{bmatrix}0 & 2 & 0 & 2 \end{bmatrix}  \eqdot
\end{align}

The ${\tcmtx}$ matrix and $\tcmtvb$ vector in this case are:

\begin{align}\label{eq:1d_example_constr_a}
\begin{aligned}
\tcmtx =& 	\inbbr{\Gshapev(0.5)}\tran\inbbr{\Gshapev(0.5)} + \Gshapev(0)\tran\Gshapev(0) + \Gshapev(1)\tran\Gshapev(1)
\\=& \begin{bmatrix}
1 &    1 &   -1  &   1\\
1  &   1 &   -1  &   1\\
-1  &  -1 &    1  &  -1\\
1  &   1  &  -1   &  1
\end{bmatrix}
\begin{bmatrix}
1 & -1 & 0 & 0\\
-1 & 1 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 1 & 1\\
0 & 0 & 1 & 1
\end{bmatrix}
\begin{bmatrix}
2 & 0 & -1 & 1\\
0 & 2 & -1 & 1\\
-1 &-1 &  2 & 0\\
1 & 1 &  0 & 2
\end{bmatrix}
\end{aligned} \eqcomma
\end{align}

\begin{align}\label{eq:1d_example_constr_b}
\tcmtvb =  \Gshapev(0)\tran a + \Gshapev(1)\tran b = \begin{bmatrix} 2 \\-2\\5\\5 \end{bmatrix} \eqdot
\end{align}

The solution of the system of equations $(\tcmtx,\, \tcmtvb)$ gives the pair $(\ttransfmatrix,\, \ttransfvectorb)$, which are

\begin{align}
\ttransfmatrix = \begin{bmatrix} -1\\-1\\-1\\1\end{bmatrix} ,
\quad
\ttransfvectorb =  \begin{bmatrix} \frac{7}{2}\\\frac{3}{2}\\5\\0\end{bmatrix} \eqdot
\end{align}

The step-by-step solution procedure of the constraints problem is presented in Appendix <ref>.

The matrix $\vm{K}$, vector $\vm{f}$ in (<ref>)  are defined as:

\begin{align}
\begin{aligned}
\vm{K} =& \mint_0^1 {\Gshapev'}\tran\Gshapev' \intend x  + \inbbr{\Gshapev(0.5)} \iinbbr{\Gshapev'(0.5)}
+ \Gshapev(0)\Gshapev'(0) - \Gshapev(1)\Gshapev'(1)
\\=&
\begin{bmatrix}
0 & 0 & 0 & 0\\
0 & 8 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 8\\
\end{bmatrix}
\begin{bmatrix}
0 & -2 & 0 & -2\\
0 & -2 & 0 & -2\\
0 & 2 & 0 & 2\\
0 & -2 & 0 & -2\\
\end{bmatrix}
\begin{bmatrix}
0 & 4 & 0 & 0\\
0 & -4 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
\end{bmatrix}
\begin{bmatrix}
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 4\\
0 & 0 & 0 & 4\\
\end{bmatrix}
\\=&
\begin{bmatrix}
0 & 2 & 0 & -2\\
0 & 2 & 0 & -2\\
0 & 2 & 0 & -2\\
0 & -2 & 0 & 2\\
\end{bmatrix} \eqcomma
\end{aligned}
\end{align}

\begin{align}
\vm{f} = -\mint_0^1 \Gshapev\tran 0 \intend x = \begin{bmatrix}0 \\ 0 \\ 0 \\0\end{bmatrix} \eqdot
\end{align}

Now, the linear system of equations given in (<ref>) has the following form:

\begin{align}
\begin{bmatrix} 16 \end{bmatrix} \dftempfree = \begin{bmatrix} 12 \end{bmatrix}
\quad\rightarrow\quad \dftempfree = \begin{bmatrix} 0.75\end{bmatrix} \eqdot
\end{align}

Finally, on using (<ref>) we have the vector $\dftemp$:

\begin{align}
\dftemp =  \begin{bmatrix} -1\\-1\\-1\\1\end{bmatrix} \begin{bmatrix} \frac{3}{4} \end{bmatrix} +
\begin{bmatrix}  \frac{7}{2}\\\frac{3}{2}\\5\\0\end{bmatrix} = \begin{bmatrix} \frac{11}{4}\\\frac{3}{4}\\\frac{17}{4}\\\frac{3}{4}\end{bmatrix} \eqdot
\end{align}

On applying the vector $\dftemp$ to the approximation in (<ref>) we recover the exact solution (affine function) of the problem.

§.§ Poisson problem with quadratic solution

In this case, we consider the following 1D Poisson problem:

\begin{align}
\label{eq:1d2p_poisson_a}
\temperature'' = 2 \quad \text{in } \Domain=(0,\,1),
\\&
\label{eq:1d2p_poisson_b}
u(0)=2,\quad u(1)=5 \eqcomma
\end{align}

which has the exact solution $u(x) = x^2 + 2x+2$. The approximation of $\temperature$ and its derivative in  $[0,\,1]$  is:

\begin{align}\label{eq:1d2p_ex_approx}
\temperature = \Gshapev_\porder \dftemp,\quad
\temperature = \begin{cases}
\begin{bmatrix} 1 & 4x-1  & 16x^2 - 8x + 1
\end{bmatrix}
\begin{bmatrix} \dftempsc_1 \\ \dftempsc_2 \\ \dftempsc_3\end{bmatrix}
& \text{if } 0 \le x < 0.5
\\
\begin{bmatrix} 1 & 4x-3  & 16x^2 - 24x + 9	\end{bmatrix}
\begin{bmatrix} \dftempsc_4 \\ \dftempsc_5  \\ \dftempsc_6\end{bmatrix}
& \text{if } 0.5 \le x \le 1
\end{cases} \eqcomma
\end{align}

\begin{align}\label{eq:ex_1d2p_approx}
\temperature' = \Gshapev_\porder' \dftemp,\quad
\temperature' = \begin{cases}
\begin{bmatrix} 0 & 4 & 32x-8 \end{bmatrix}
\begin{bmatrix} \dftempsc_1 \\ \dftempsc_2 \\ \dftempsc_3\end{bmatrix}
& \text{if } 0 \le x < 0.5
\\
\begin{bmatrix} 0 & 4& 32x-24  \end{bmatrix}
\begin{bmatrix} \dftempsc_4 \\ \dftempsc_5  \\ \dftempsc_6\end{bmatrix}
& \text{if } 0.5 \le x \le 1
\end{cases}  \eqdot
\end{align}

The constrained problem in (<ref>) require that (<ref>) has to be firstly solved to yield $\ttransfmatrix$ and $\ttransfvectorb$. The values of the approximation matrix $\Gshapev_\porder$ and $\Gshapev_\bporder$ on the boundaries and its jump on the skeleton are:

\begin{align}
&\Gshapev_\porder(0) = \begin{bmatrix}1 & -1 & 1 & 0 & 0 & 0 \end{bmatrix},\quad
\Gshapev_\porder'(0) = \begin{bmatrix}0& 4 & -8 & 0 & 0 &0 \end{bmatrix} \eqcomma
\\&
\Gshapev_\porder(1) = \begin{bmatrix}0 & 0 & 0 & 1 & 1 & -1 \end{bmatrix},\quad
\Gshapev_\porder'(1) = \begin{bmatrix}0& 0 & 0 & 0 & 4 & -24\end{bmatrix} \eqcomma
\\&
\inbbr{\Gshapev_\porder(0.5)} = \begin{bmatrix} 0 & 0 & 0 & 1 & -1 & 1\end{bmatrix} - \begin{bmatrix} 1 & 1 & 1 & 0 & 0 & 0\end{bmatrix} = \begin{bmatrix}-1 & -1 & -1 & 1 & -1 & 1 \end{bmatrix}  \eqcomma
\\&
\iinbbr{\Gshapev_\porder'(0.5)} =  \frac{1}{2}\left(\begin{bmatrix} 0 & 4 & 8 & 0 & 0 & 0\end{bmatrix} + \begin{bmatrix} 0 & 0 & 0 & 4 & -8\end{bmatrix}\right) = \begin{bmatrix}0 & 2 & 4 & 0 & 2 & -4 \end{bmatrix}  \eqdot
\end{align}

\begin{align}
&\Gshapev_\bporder(0) = \begin{bmatrix}1 & -1 & 0 & 0 \end{bmatrix},\quad
\Gshapev_\bporder'(0) = \begin{bmatrix}0& 4 & 0 & 0 \end{bmatrix} \eqcomma
\\&
\Gshapev_\bporder(1) = \begin{bmatrix}0 & 0 & 1 & 1 \end{bmatrix},\quad
\Gshapev_\bporder'(1) = \begin{bmatrix}0& 0 & 0 & 4 \end{bmatrix} \eqcomma
\\&
\inbbr{\Gshapev_\bporder(0.5)} = \begin{bmatrix} 0 & 0 & 1 & -1\end{bmatrix} - \begin{bmatrix} 1 & 1 & 0 & 0\end{bmatrix} = \begin{bmatrix}-1 & -1 & 1 & -1 \end{bmatrix}  \eqcomma
\\&
\iinbbr{\Gshapev_\bporder'(0.5)} =  \frac{1}{2}\left(\begin{bmatrix} 0 & 4 & 0 & 0\end{bmatrix} + \begin{bmatrix} 0 & 0 & 0 & 4\end{bmatrix}\right) = \begin{bmatrix}0 & 2 & 0 & 2 \end{bmatrix}  \eqdot
\end{align}

The ${\tcmtx}$ matrix and $\tcmtvb$ vector in this case are:

\begin{align}\label{eq:1d2p_example_constr_a}
\begin{aligned}
\tcmtx =& 	\inbbr{\Gshapev_\bporder(0.5)}\tran\inbbr{\Gshapev_\porder(0.5)} + \Gshapev_\bporder(0)\tran\Gshapev_\porder(0) + \Gshapev_\bporder(1)\tran\Gshapev_\porder(1)
\\=& \begin{bmatrix}
1  &    1  & 1  &   -1  &   1 & -1\\
1   &   1  & 1  &   -1  &   1 & -1\\
-1  &  -1  & -1 & 1 &  -1  &  1\\
1  &   1 & 1  &  -1 &  1 & -1
\end{bmatrix}
\begin{bmatrix}
1 & -1 & 1 & 0 & 0 & 0\\
-1 & 1 & -1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
0 & 0 & 0 & 0& 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 1& 1 & 1\\
0 & 0 &0 & 1& 1 & 1
\end{bmatrix}
\begin{bmatrix}
2 & 0 & 2 & -1 & 1 & -1\\
0 & 2 &0 &  -1 & 1 & -1\\
-1 &-1 &-1&  2 & 0& 2\\
1 & 1 & 1&  0 & 2& 0
\end{bmatrix}
\end{aligned} \eqcomma
\end{align}

\begin{align}\label{eq:1d2p_example_constr_b}
\tcmtvb =  \Gshapev_\bporder(0)\tran a + \Gshapev_\bporder(1)\tran b = \begin{bmatrix} 2 \\-2\\5\\5 \end{bmatrix} \eqdot
\end{align}

The solution of the system of equations $(\tcmtx,\, \tcmtvb)$ gives the pair $(\ttransfmatrix,\, \ttransfvectorb)$, which are:

\begin{align}
\ttransfmatrix =
\begin{bmatrix}
-1 & -1 & 0 \\
0 & -1 & 0\\
1 & 0 & 0\\
0 & -1 & -1\\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix} ,
\quad
\ttransfvectorb =  \begin{bmatrix} \frac{7}{2}\\\frac{3}{2}\\0\\5\\0\\0\end{bmatrix} \eqdot
\end{align}

The matrix $\vm{K}$, vector $\vm{f}$ in (<ref>)  are defined as:

\begin{align}
\begin{aligned}
\vm{K} =& \mint_0^1 {\Gshapev_\porder'}\tran\Gshapev_\porder' \intend x  + \inbbr{\Gshapev_\porder(0.5)} \iinbbr{\Gshapev_\porder'(0.5)}