Theory of Functional Connections Subject to Shear-Type and Mixed Derivatives

Abstract

This study extends the functional interpolation framework, introduced by the Theory of Functional Connections, initially introduced for functions, derivatives, integrals, components, and any linear combination of them, to constraints made of shear-type and/or mixed derivatives. The main motivation comes from differential equations, often appearing in fluid dynamics and structures/materials problems that are subject to shear-type and/or mixed boundary derivatives constraints. This is performed by replacing these boundary constraints with equivalent constraints, obtained using indefinite integrals. In addition, this study also shows how to validate the constraints’ consistency when the problem involves the unknown constants of integrations generated by indefinite integrations.

1. Introduction

The linear generalization of interpolation is performed by functionals, called constrained expressions, that are analytical expressions provided by the Theory of Functional Connections (TFC) [1,2,3,4,5,6,7]. These functionals represent the whole set of functions satisfying all constraints of the interpolation problem. The TFC theory has been developed for constraints made of points, derivatives, integrals, limits, components, and any linear combination of them in univariate and multivariate dimensional spaces. This functional interpolation has been introduced by two equivalent definitions [7],

$$\begin{array}{cc}\hfill y(x,g(x\left)\right)& =g\left(x\right)+\sum _{k=1}^n{\eta }_{{}_k}(x,g\left(x\right))s_{{}_k}\left(x\right)\hfill \end{array}$$

(1)

$$\begin{array}{cc}\hfill y(x,g(x\left)\right)& =g\left(x\right)+\sum _{k=1}^n{\rho }_{{}_k}(x,g\left(x\right)){\varphi }_{{}_k}(x,s\left(x\right)).\hfill \end{array}$$

(2)

In the definition given in Equation (1), the coefficients ${\eta }_{{}_k}(x,g\left(x\right))$ are functional coefficients. They are derived by specifying the n constraints, while the $s_{{}_k}\left(x\right)$ terms are n user-defined linearly independent support functions. In the definition given in Equation (2), the terms ${\varphi }_{{}_k}(x,s\left(x\right))$ are switching functions. The k-th switching function becomes “1” when the k-th constraint is verified and “0” when a different constraint is verified. The ${\rho }_{{}_k}(x,g\left(x\right))$ terms are called projection functionals. The purpose of the k-th projection functional is to project the function $g\left(x\right)$ to the k-th constraint. Full and detailed explanation of how to use and derive constrained expressions, either with Equation (1) or with Equation (2), is found in Reference [7]. When these terms have been computed, then Equation (1) and/or Equation (2) always satisfy all the n constraint, for any expression of $g\left(x\right)$, which is called the free function.

To give an example, if the function $y\left(x\right)$ is subject to the constraints,

$${\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}}{|}_{y_{{}_1}}={\dot{y}}_{{}_1}\mathrm{and}{\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}}{|}_{y_{{}_2}}={\dot{y}}_{{}_2}.$$

(3)

then, using $s_{{}_1}\left(x\right)=x$ and $s_{{}_2}\left(x\right)=x^2$, Equation (1) becomes,

$$y(x,g\left(x\right))=g\left(x\right)+{\eta }_{{}_1}{s}_{{}_1}\left(x\right)+{\eta }_{{}_2}{s}_{{}_2}\left(x\right)=g\left(x\right)+{\eta }_{{}_1}x+{\eta }_{{}_2}{x}^2$$

(4)

where the two constants ${\eta }_{{}_1}$ and ${\eta }_{{}_2}$ are computed by imposing Equation (3) to Equation (4):

$$\left\{\begin{array}{c}{\dot{y}}_{{}_1}-{\dot{g}}_{{}_1}\\ {\dot{y}}_{{}_2}-{\dot{g}}_{{}_2}\end{array}\right\}=\left[\begin{array}{cc}{\dot{s}}_{{}_1}\left(y_{{}_1}\right)& {\dot{s}}_{{}_2}\left(y_{{}_1}\right)\\ {\dot{s}}_{{}_1}\left(y_{{}_2}\right)& {\dot{s}}_{{}_2}\left(y_{{}_2}\right)\end{array}\right]\left\{\begin{array}{c}{\eta }_{{}_1}\\ {\eta }_{{}_2}\end{array}\right\}\to \left\{\begin{array}{c}{\eta }_{{}_1}\\ {\eta }_{{}_2}\end{array}\right\}={\left[\begin{array}{cc}1& 2y_{{}_1}\\ 1& 2y_{{}_2}\end{array}\right]}^{-1}\left\{\begin{array}{c}{\dot{y}}_{{}_1}-{\dot{g}}_{{}_1}\\ {\dot{y}}_{{}_2}-{\dot{g}}_{{}_2}\end{array}\right\}$$

from which,

$$\left\{\begin{array}{c}{\eta }_{{}_1}\\ {\eta }_{{}_2}\end{array}\right\}={\displaystyle \frac{1}{y_{{}_2}-y_{{}_1}}}\left\{\begin{array}{c}{y}_{{}_2}({\dot{y}}_{{}_1}-{\dot{g}}_{{}_1})-y_{{}_1}({\dot{y}}_{{}_2}-{\dot{g}}_2)\\ -({\dot{y}}_{{}_1}-{\dot{g}}_{{}_1})+({\dot{y}}_{{}_2}-{\dot{g}}_2)\end{array}\right\}.$$

Substituting the ${\eta }_{{}_1}$ and ${\eta }_{{}_2}$ expressions into Equation (4), the following constrained expression,

$$y(x,g\left(x\right))=g\left(x\right)+\underset{{\varphi }_{{}_1}(x,s\left(x\right))}{\underbrace{{\displaystyle \frac{x(2y_{{}_2}-x)}{2(y_{{}_2}-y_{{}_1})}}}}\underset{{\rho }_{{}_1}(x,g\left(x\right))}{\underbrace{({\dot{y}}_{{}_1}-{\dot{g}}_{{}_1})}}+\underset{{\varphi }_{{}_2}(x,s\left(x\right))}{\underbrace{{\displaystyle \frac{x(x-2y_{{}_1})}{2(y_{{}_2}-y_{{}_1})}}}}\underset{{\rho }_{{}_2}(x,g\left(x\right))}{\underbrace{({\dot{y}}_{{}_2}-{\dot{g}}_2)}},$$

(5)

is obtained. This equation satisfies the constraints given in Equation (3), for any expression of the free function, $g\left(x\right)$. In this equation the expressions of the switching functions, ${\varphi }_{{}_k}(x,s\left(x\right))$, and the projection functionals, ${\rho }_{{}_k}(x,g\left(x\right))$, are also specified. It is pretty clear the mechanism of the switching functions: when the first constraint, $\dot{y}\left(y_{{}_1}\right)={\dot{y}}_{{}_1}$, is satisfied then ${\dot{\varphi }}_{{}_1}\left(y_{{}_1}\right)=1$ and ${\dot{\varphi }}_{{}_2}\left(y_{{}_1}\right)=0;$ while, if the second constraint is satisfied, $\dot{y}\left(y_{{}_2}\right)={\dot{y}}_{{}_2}$, then ${\dot{\varphi }}_{{}_1}\left(y_{{}_2}\right)=0$ and ${\dot{\varphi }}_{{}_2}\left(y_{{}_2}\right)=1$. As for the projection functionals, in this univariate case they are scalars while, as shown in Reference [7], they become functionals in the general multivariate case.

The univariate TFC framework has been extended to the multivariate case [4,7], subject to any linear (boundary and/or internal) constraints. This extension can be described by the following the expression,

$$y(x,g(x\left)\right)=h\left(c\right(x\left)\right)+g\left(x\right)-h\left(g\right(x\left)\right),$$

where $x={\left\{x_{{}_1},x_{{}_2},\cdots ,x_{{}_n}\right\}}^T$ is the vector of n independent variables, $c\left(x\right)$ is a function defining the constraints, $h\left(c\right(x\left)\right)$ is an (any) interpolating function for the constraints, and $g\left(x\right)$ is the TFC free function. References [5,6,7] contain many examples on how to derive constrained expressions. The Theory of Functional Connections is a framework for constraints made of points, derivatives, integrals, infinite, component, and any linear combination of them for univariate [1] and multivariate [4] cases on rectangular domains. However, some initial results have been proposed for domains of any shape via domain mapping [8]. The most important feature of the functionals given in Equation (1) and/or Equation (2) is that they identify the functions subspace with feasible solutions, those fully satisfying the constraints. This fact is important for a large number of constrained optimization problems because these functionals transform the constrained problems to unconstrained. This means that they can be then solved using more simple, efficient, robust, reliable, fast, and accurate methods.

Initially, the Theory of Functional Connections was applied to solve linear [2] and nonlinear [3] ODEs. This was performed by expanding $g\left(x\right)$, in the constrained expression, in terms of a set of orthogonal polynomials. Then the solution (coefficients of the expansion) was obtained by least-squares. The advantages of this numerical approach has many advantages over traditional methods:

• It consists of a unified framework to solve IVP, BVP, or multi-valued problems;
• It provides an analytically approximated solution that can be used for subsequent manipulation (derivatives, integrals);
• The solution is usually obtained in millisecond and at machine error accuracy;
• The procedure is numerically robust (small condition number);
• It can solve differential equations subject to a variety of different constraint types.

In addition, constrained expressions were adopted to solve a variety of other mathematical optimization problems [9] such as: (a) homotopy continuation for control problems [10], (b) epidemiological models [11], (c) Timoshenko–Ehrenfest beam [12], (d) hybrid systems [13], (e) machine learning [14,15,16], (f) quadratic and nonlinear programming problems subject to linear equality constraints [17], (g) orbit transfer and propagation [18,19], (h) optimal control problems via indirect methods, (i) relative motion [20], and (j) landing on small and large planetary bodies [21], to mention the most important ones.

A constrained expression is the final objective of functional interpolation. Most of its applications are in solving differential equations and in optimization processes while additional applications are found in calculus of variations and in continuations (e.g., homotopy) methods. The current multivariate version of the Theory of Functional Connections includes boundary constraints specified in terms of functions or “flux derivatives” of any order, such as,

$${\displaystyle \frac{{\partial }^{n_{{}_k}}f}{\partial x_{{}_k}^{n_{{}_k}}}}{|}_{x_{{}_k}^{*}}$$

where $x_{{}_k}^{*}$ is a specific value of the $x_{{}_k}$ coordinate. This means that shear-type and/or mixed derivatives,

$${\displaystyle \frac{{\partial }^{n_{{}_i}+n_{{}_j}}f}{\partial x_{{}_i}^{n_{{}_i}}\partial x_{{}_j}^{n_{{}_j}}}}{|}_{x_{{}_k}^{*}}\mathrm{where}:k\ne (i,j)$$

are left out from the current theory. This paper shows how to incorporate shear-type and/or mixed derivatives in the current TFC framework [7], that is, how to perform perform functional interpolation with these kinds of derivative constraints. This extension is provided in a form that fits in with the previously defined theory and leverage all of the TFC theorems and proofs provided in Refs. [5,6,7] and not repeated here.

In addition, this study shows how to validate the consistency of any set of boundary conditions.

The style in which this material is presented is to provide through various examples typical cases that can be presented in practice. At the expense of a more rigorous and formal mathematical presentation, this style was chosen to increase clarity and stimulate applications.

Next section describes how to derive the constrained expressions for these partial derivatives in the two-dimensional domain defined by $[x_{{}_1},x_{{}_2}]\times [y_{{}_1},y_{{}_2}]$.

2. Two-Dimensional Domains

As a starting example, consider the constraint,

$$f_{{}_{{}_y}}(x_{{}_2},y)=p\left(y\right),$$

where $f_y(x_{{}_2},y)$ indicates the derivative of a function with respect to y, specified at coordinate $x_{{}_2}$. This shear-type constraint can be transformed into the equivalent constraint,

$${\displaystyle f(x_{{}_2},y)=\int p\left(y\right)\mathrm{d}y=\mathcal{A}\left(y\right)+c}$$

where $\mathcal{A}\left(y\right)$ is the anti-derivative of $p\left(y\right)$, that is, ${\displaystyle \frac{\mathrm{d}\mathcal{A}\left(y\right)}{\mathrm{d}y}}=p\left(y\right)$, and c is an unknown constant of integration. Then, the constrained expression is,

$$f(x,y)=g(x,y)+\mathcal{A}\left(y\right)+c-g(x_{{}_2},y).$$

(6)

The value of c may be derived by enforcing continuity conditions with neighboring constraints. For example, if $f(x,y_{{}_2})=q\left(x\right)$ is an additional constraint, then the continuity at $[x_{{}_2},y_{{}_2}]$ implies,

$$q\left(x_{{}_2}\right)=\mathcal{A}\left(y_{{}_2}\right)+c\to c=q\left(x_{{}_2}\right)-\mathcal{A}\left(y_{{}_2}\right).$$

If no continuity condition allows the computation of c, then the constant of integration will be estimated, as the free function, during the optimization process. This means that if, as done in [2,3,7], the free function is modeled as a linear combination of orthogonal polynomials, $g(x,y)={\xi }_{ij}{T}_{{}_i}\left(x\right)T_{{}_j}\left(y\right)$, then the term, ${\xi }_{{}_{00}}{T}_{{}_0}\left(x\right)T_{{}_0}\left(y\right)={\xi }_{{}_{00}}$, must be removed from the free function expansion because, in a least-squares optimization process, all the variables to estimate must be linearly independent. Removing the constants of integration instead of the associated terms of the free function expansion is a mistake because the constraint would become, $f(x_{{}_2},y)=\mathcal{A}\left(y\right)$, instead of the original, $f_y(x_{{}_2},y)=p\left(y\right)$.

The constrained example can be specified using a support function different from $s\left(x\right)=1$. For instance, using $s\left(x\right)=x$ the constrained expression will be,

$$f(x,y)=g(x,y)+{\displaystyle \frac{x}{x_{{}_2}}}\left[\mathcal{A}\left(y\right)+c-g(x_{{}_2},y)\right]$$

where the term, ${\xi }_{{}_{10}}{T}_{{}_1}\left(x\right)T_{{}_0}\left(y\right)={\xi }_{{}_{10}}x$, must be removed from the free function expansion, instead.

In general, the constrained expressions for shear-type and/or mixed constraints involve indefinite integrals and constants of integrations. The values of these constants of integration may be derived by the continuity relationships. If they cannot be derived this way, the constants of integrations become additional unknowns in the optimization process. In general, these constants of integration are terms multiplying polynomials. It is important to outline that, in optimization, the free function must be linearly independent with respect to all these multiplying polynomial terms. Therefore, the constants of integrations are not additional unknowns, because corresponding unknown terms must be removed form the free functions expansion.

Let us consider a shear-type constraint expressed by $n_{{}_y}$-derivative,

$${\displaystyle \frac{{\partial }^{n_{{}_y}}f}{\partial y^{n_{{}_y}}}}{|}_{x_{{}_2}}=p\left(y\right).$$

(7)

This constraint can be replaced with,

$${\displaystyle f(x_{{}_2},y)=\underset{n_{{}_y}}{{\displaystyle \int ...\int }}p\left(y\right){\left(\mathrm{d}y\right)}^{n_{{}_y}}=\mathcal{A}\left(y\right)+\sum _{k=0}^{n_{{}_y}-1}{c}_{{}_k}{y}^k,}$$

where $\mathcal{A}\left(y\right)$ is the $n_{{}_y}$-th anti-derivative of $p\left(y\right)$, that is, ${\displaystyle \frac{{\mathrm{d}}^{n_{{}_y}}\mathcal{A}\left(y\right)}{\mathrm{d}{y}^{n_{{}_y}}}}=p\left(y\right)$, and the $c_{{}_k}$ are $n_{{}_y}$ unknown constants of integration. Then, using $s\left(x\right)=1$, the constrained expression is,

$${\displaystyle f(x,y)=g(x,y)+\mathcal{A}\left(y\right)+\sum _{k=0}^{n_{{}_y}-1}{c}_{{}_k}{y}^k-g(x_{{}_2},y).}$$

(8)

Again, to apply Equation (8) in an optimization process, the free function must be linearly dependent with the terms generated by the $n_{{}_y}$ integrations, which must be then removed.

Consider now the mixed derivative constraint,

$${\displaystyle \frac{{\partial }^{2}f}{\partial x\partial y}}{|}_{x_{{}_2}}=p\left(y\right).$$

This constraint can be replaced by the boundary constraint,

$${\displaystyle f_{{}_x}(x_{{}_2},y)=\int p\left(y\right)\mathrm{d}y=\mathcal{A}\left(y\right)+c.}$$

Then, using $s\left(x\right)=x^2$, the constrained expression is,

$$f(x,y)=g(x,y)+{\displaystyle \frac{x^2}{2x_{{}_2}}}\left[\mathcal{A}\left(y\right)+c-g_{{}_x}(x_{{}_2},y)\right].$$

(9)

In this example the free function must exclude the $x^2$ quadratic term.

The general $(n_{{}_x},n_{{}_y})$-th mixed derivative constraint,

$${\displaystyle \frac{{\partial }^{n_{{}_x}+n_{{}_y}}f}{\partial x^{n_{{}_x}}\partial y^{n_{{}_y}}}}{|}_{x_{{}_2}}=p\left(y\right),$$

(10)

can be transformed to the $n_{{}_x}$-th flux derivative constraint,

$${\displaystyle {\displaystyle \frac{{\partial }^{n_{{}_x}}f}{\partial x^{n_{{}_x}}}}{|}_{x_{{}_2}}=\underset{n_{{}_y}}{{\displaystyle \int ...\int }}p\left(y\right){\left(\mathrm{d}y\right)}^{n_{{}_y}}=\mathcal{A}\left(y\right)+\sum _{k=0}^{n_{{}_y}-1}{c}_{{}_k}{y}^k.}$$

Then, the constrained expression is simply,

$$f(x,y)=g(x,y)+{\displaystyle \frac{x^{n_{{}_x}+1}}{(n_{{}_x}+1)!x_{{}_2}}}\left({\displaystyle \mathcal{A}\left(y\right)+\sum _{k=0}^{n_{{}_y}-1}{c}_{{}_k}{y}^k-{\displaystyle \frac{{\partial }^{n_{{}_x}}g}{\partial x^{n_{{}_x}}}}{|}_{x_{{}_2}}}\right),$$

(11)

where the free function cannot contains the polynomial terms, $x^{n_{{}_x}+1}{y}^k$ for $k\in [1,n_{{}_y}-1]$. Note that, like the free function, the constants of integration always appear linearly in the constrained expression.

In the next section, two-dimensional examples with neighboring constraints are provided to show how to validate the consistency of the constraints.

Neighboring Constraints

Consider the example given in Figure 1, where the constraints are,

$${\displaystyle \frac{\partial f}{\partial x}}{|}_{y_{{}_2}}=p\left(x\right)\mathrm{and}{\displaystyle \frac{{\partial }^{2}f}{\partial x\partial y}}{|}_{y_{{}_2}}=q\left(y\right).$$

(12)

According to Equations (8) and (11) these constraints can be replaced as follows,

$${\displaystyle f(x,y_{{}_2})=\int p\left(x\right)\mathrm{d}x=\mathcal{P}\left(x\right)+c_p\mathrm{and}{f}_{{}_x}(x_{{}_2},y)=\int q\left(y\right)\mathrm{d}y={\mathcal{Q}}_{{}_x}\left(y\right)+c_q.}$$

The $C^1$ continuity at $[x_{{}_2},y_{{}_2}]$ implies,

$$f_{{}_x}(x_{{}_2},y_{{}_2})=p\left(x_{{}_2}\right)={\mathcal{Q}}_{{}_x}\left(y_{{}_2}\right)+c_q\to c_q=p\left(x_{{}_2}\right)-{\mathcal{Q}}_{{}_x}\left(y_{{}_2}\right),$$

Therefore, using the recursive form to derive the constrained expressions [7], we can write,

$$\left\{\begin{array}{c}{f}^x(x,y)=g(x,y)+\mathcal{P}\left(x\right)+c_p-g(x,y_{{}_2})\hfill \\ f(x,y)=f^x(x,y)+x\left[{\mathcal{Q}}_{{}_x}\left(y\right)+p\left(x_{{}_2}\right)-{\mathcal{Q}}_{{}_x}\left(y_{{}_2}\right)-f_{{}_x}^x(x_{{}_2},y)\right].\hfill \end{array}\right.$$

(13)

Then, the constrained expression is,

$$\begin{array}{cc}\hfill f(x,y)& =g(x,y)+\mathcal{P}\left(x\right)+c_p-g(x,y_{{}_2})\hfill \\ & +x\left[{\mathcal{Q}}_{{}_x}\left(y\right)-{\mathcal{Q}}_{{}_x}\left(y_{{}_2}\right)-g_{{}_x}(x_{{}_2},y)+g_{{}_x}(x_{{}_2},y_{{}_2})\right].\hfill \end{array}$$

(14)

It is easy to verify that Equation (14) satisfies the two constraints, $f_{{}_x}(x,y_{{}_2})=p\left(x\right)$ and $f_{xy}(x_{{}_2},y)=q\left(y\right)$, or the equivalent, $f(x,y_{{}_2})=\mathcal{P}\left(x\right)+c_p$ and $f_{{}_x}(x_{{}_2},y)={\mathcal{Q}}_{{}_x}\left(y\right)+c_q$, where $c_q=p\left(x_{{}_2}\right)-{\mathcal{Q}}_{{}_x}\left(y_{{}_2}\right)$, no matter what the free function is. During optimization, the term to exclude from the free function expansion is just, ${\xi }_{{}_{00}}$, which is substituted by the constant of integration, $c_p$.

The two contiguous constraints,

$$f_{xy}(x_{{}_2},y)=p\left(y\right)\mathrm{and}{f}_{xy}(x,y_{{}_2})=q\left(x\right),$$

(15)

can be replaced by,

$${\displaystyle f_{{}_x}(x_{{}_2},y)=\int p\left(y\right)\mathrm{d}y=\mathcal{A}\left(y\right)+a\mathrm{and}{f}_{{}_y}(x,y_{{}_2})=\int q\left(x\right)\mathrm{d}x=\mathcal{B}\left(x\right)+b,}$$

where the variables’ constrained expressions are,

$$\begin{array}{c}\hfill f^y(x,y)=g(x,y)+{\displaystyle \frac{y^2}{2y_{{}_2}}}\left[\mathcal{B}\left(x\right)+b-g_{{}_y}(x,y_{{}_2})\right]\\ \hfill f(x,y)=f^y(x,y)+{\displaystyle \frac{x^2}{2x_{{}_2}}}\left[\mathcal{A}\left(y\right)+a-f_{{}_x}^y(x_{{}_2},y)\right],\end{array}$$

and the global constrained expression is,

$$\begin{array}{cc}\hfill f(x,y)& =g(x,y)+{\displaystyle \frac{y^2}{2y_{{}_2}}}\left[\mathcal{B}\left(x\right)+b-g_{{}_y}(x,y_{{}_2})\right]+\hfill \\ & +{\displaystyle \frac{x^2}{2x_{{}_2}}}\left[\mathcal{A}\left(y\right)+a-g_{{}_x}(x_{{}_2},y)-{\displaystyle \frac{y^2}{2y_{{}_2}}}\left[q\left(x_{{}_2}\right)-g_{xy}(x_{{}_2},y_{{}_2})\right]\right],\hfill \end{array}$$

where the free function terms to exclude are those containing the two quadratic polynomial terms, $x^2$ and $y^2$.

Note that, in this example, no continuity equation can be specified. In general, however, continuity equations must be satisfied. This generate the consistency problem of the constraints. Next section discusses how to verify the consistency of a set of boundary constraints described by shear-type and/or mixed derivatives.

3. Consistent and Inconsistent Constraints

As already specified in Section 2, substituting shear-type and mixed derivatives constrains by indefinite integrations, a set of unknown constants of integration are introduced. Continuity conditions allow to evaluate some of them while the remaining ones will be estimated by the optimization process.

Note that the continuity conditions do not involve free functions and the constants of integrations appear linearly in the continuity equations. Therefore, a set of m continuity equations constitutes a linear system,

$$Ax=\left[\begin{array}{ccc}{a}_{11}& \cdots & a_{1n}\\ ⋮& \ddots & ⋮\\ a_{m1}& \cdots & a_{mn}\end{array}\right]\left\{\begin{array}{c}{c}_{{}_1}\\ ⋮\\ c_n\end{array}\right\}=\left\{\begin{array}{c}{b}_{{}_1}\\ ⋮\\ b_m\end{array}\right\}=b$$

where the vector $x$ contains the set of the n unknown constants of integration, while matrix A and vector $b$ contain functions, derivatives, and anti-derivatives evaluated where the continuity equations are specified.

In general, the consistency of a linear system requires computing the rank of the coefficients matrix, A, and the rank of the augmented matrix, $[A,b]$, obtained by adding the vector $b$ as a new column to matrix A. Specifically,

$$\begin{array}{ccc}\hfill \mathrm{if}\mathrm{rank}\left(A\right)=\mathrm{rank}\left(\right[A,b\left]\right)& \to & \mathrm{consistent}\mathrm{system}\hfill \\ \hfill \mathrm{if}\mathrm{rank}\left(A\right)<\mathrm{rank}\left(\right[A,b\left]\right)& \to & \mathrm{inconsistent}\mathrm{system}\hfill \end{array}$$

If the continuity conditions lead to an inconsistent linear system then the set boundary constraints under analysis is inconsistent. No method can be then used to obtain a solution: the problem is unsolvable and no associated constrained expression can be derived. Otherwise, if the continuity conditions lead to a redundant system, then $n-\mathrm{rank}\left(A\right)$ variables can be considered independent while the remaining variables, rank$\left(A\right)$, can be expressed in terms of the independent ones.

In the case the continuity equations do not involve constants of integration as, for instance, for the constraints, $p(x_{{}_2},y)$, $q_{{}_y}(x,y_{{}_2})$ and $r_{{}_y}(x,y_{{}_1})$, then the continuity equations, can be written with a dummy unknown variable (c) as,

$$Ax=\left[\begin{array}{c}0\\ 0\end{array}\right]\left\{\begin{array}{c}c\end{array}\right\}=\left\{\begin{array}{c}{p}_{{}_y}(x_{{}_2},y_{{}_2})-q_{{}_y}(x_{{}_2},y_{{}_2})\\ p_{{}_y}(x_{{}_2},y_{{}_1})-r_{{}_y}(x_{{}_2},y_{{}_1})\end{array}\right\}=b$$

Since in this example rank$\left(A\right)=0$, then in order to obtain rank$\left(\right[A,b\left]\right)=0$, the conditions, $p_{{}_y}(x_{{}_2},y_{{}_2})=q_{{}_y}(x_{{}_2},y_{{}_2})$ and $p_{{}_y}(x_{{}_2},y_{{}_1})=r_{{}_y}(x_{{}_2},y_{{}_1})$, must be satisfied.

Appendix A provides a robust method to compute the rank of a matrix via QR decomposition and to derive an equivalent and smaller redundant linear system.

To make clear this important aspect, the following sections provide a few examples of consistent and inconsistent boundary constraints.

3.1. Continuous Shear-Type Constraints Example

Consider the boundary conditions scenario depicted on the left of Figure 2,

$$f_{{}_x}(x,y_{{}_2})=p\left(x\right)\mathrm{and}{f}_{{}_y}(x_{{}_2},y)=q\left(y\right).$$

These constraints can be replaced by integration with the two new constraints,

$${\displaystyle f(x,y_{{}_2})=\int p\left(x\right)\mathrm{d}x=\mathcal{P}\left(x\right)+c_p\mathrm{and}f(x_{{}_2},y)=\int q\left(y\right)\mathrm{d}y=\mathcal{Q}\left(y\right)+c_q.}$$

The continuity condition implies, $\mathcal{P}\left(x_{{}_2}\right)+c_p=\mathcal{Q}\left(y_{{}_2}\right)+c_q$. This equation can be written in matrix form,

$$Ac=\begin{array}{cc}\hfill [1,& -1]\end{array}\left\{\begin{array}{c}{c}_p\\ c_q\end{array}\right\}=\left\{\begin{array}{c}\mathcal{Q}\left(y_{{}_2}\right)-\mathcal{P}\left(x_{{}_2}\right)\end{array}\right\}=b.$$

In this example, rank$\left(A\right)$ = rank$\left(\right[A,b\left]\right)=1$ for any value of $\mathcal{Q}\left(y_{{}_2}\right)$ and $\mathcal{P}\left(x_{{}_2}\right)$. Therefore, the two shear-type constraints are always consistent. Since rank$\left(A\right)=r=1$, then $n-r=1$ unknown is independent (e.g., $c_p$) while $r=1$ is dependent (e.g., $c_q$).

Consider now the boundary conditions scenario depicted on the center of Figure 2,

$$f_{{}_x}(x,y_{{}_2})=p\left(x\right),f_{{}_y}(x_{{}_2},y)=q\left(y\right),\mathrm{and}{f}_{{}_x}(x,y_{{}_1})=r\left(x\right).$$

These shear-type constraints can be replaced by $f(x,y_{{}_2})=\mathcal{P}\left(x\right)+c_p,f(x_{{}_2},y)=\mathcal{Q}\left(y\right)+c_q,$ and $f(x,y_{{}_1})=\mathcal{R}\left(x\right)+c_r$. The two continuity conditions are,

$$\mathcal{P}\left(x_{{}_2}\right)+c_p=\mathcal{Q}\left(y_{{}_2}\right)+c_q\mathrm{and}\mathcal{Q}\left(y_{{}_1}\right)+c_q=\mathcal{R}\left(x_{{}_2}\right)+c_r,$$

in matrix form,

$$Ax=\left[\begin{array}{ccc}\hfill 1& \hfill -1& \hfill 0\\ \hfill 0& \hfill 1& \hfill -1\end{array}\right]\left\{\begin{array}{c}{c}_p\\ c_q\\ c_r\end{array}\right\}=\left\{\begin{array}{c}\mathcal{Q}\left(y_{{}_2}\right)-\mathcal{P}\left(x_{{}_2}\right)\\ \mathcal{R}\left(x_{{}_2}\right)-\mathcal{Q}\left(y_{{}_1}\right)\end{array}\right\}=b.$$

In this example, rank$\left(A\right)$ = rank$\left(\right[A,b\left]\right)=2=r$, no matter the values of the anti-derivatives. Therefore: the constraints are always consistent, $n-r=1$ unknown is independent, and $r=2$ unknowns are dependent.

The four shear-type constraints (right example in Figure 2),

$$f_{{}_x}(x,y_{{}_2})=p\left(x\right),f_{{}_y}(x_{{}_2},y)=q\left(y\right),f_{{}_x}(x,y_{{}_1})=r\left(x\right),\mathrm{and}{f}_{{}_y}(x_{{}_1},y)=s\left(y\right),$$

give the linear system,

$$Ax=\left[\begin{array}{cccc}\hfill 1& \hfill -1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 1& \hfill -1& \hfill 0\\ \hfill 0& \hfill 0& \hfill 1& \hfill -1\\ \hfill -1& \hfill 0& \hfill 0& \hfill 1\end{array}\right]\left\{\begin{array}{c}{c}_p\\ c_q\\ c_r\\ c_s\end{array}\right\}=\left\{\begin{array}{c}\mathcal{Q}\left(y_{{}_2}\right)-\mathcal{P}\left(x_{{}_2}\right)\\ \mathcal{R}\left(x_{{}_2}\right)-\mathcal{Q}\left(y_{{}_1}\right)\\ \mathcal{S}\left(y_{{}_1}\right)-\mathcal{R}\left(x_{{}_1}\right)\\ \mathcal{P}\left(x_{{}_1}\right)-\mathcal{S}\left(y_{{}_2}\right)\end{array}\right\}=b.$$

In this case, the coefficients matrix is singular, with rank$\left(A\right)=3$. This means that the rows (or columns) of the coefficient matrix are not independent. The dependency relationship implies finding the coefficients of a linear combination of rows summing to zero. Let $\xi$ be the coefficients vector of this linear combination. This implies, $A^T\xi =0$. This means that $\xi$ can be seen as the eigenvector associated with the zero eigenvalue of the $A^T$ matrix.

$$A^T\xi =0\to \xi ={\left\{\begin{array}{cccc}1,& 1,& 1,& 1\end{array}\right\}}^T$$

Therefore, in order for the system to admit solutions (be consistent) also rank$\left(\right[A,b\left]\right)=3$. Therefore, the dependency relationship found for the rows of matrix A must also be valid for the elements of the $b$ vector. This means, $b^T\xi ={\xi }^{T}b=0$. Therefore, in order for the constraints to be consistent, their anti-derivatives must satisfy,

$$\mathcal{Q}\left(y_{{}_2}\right)-\mathcal{P}\left(x_{{}_2}\right)+\mathcal{R}\left(x_{{}_2}\right)-\mathcal{Q}\left(y_{{}_1}\right)+\mathcal{S}\left(y_{{}_1}\right)-\mathcal{R}\left(x_{{}_1}\right)+\mathcal{P}\left(x_{{}_1}\right)-\mathcal{S}\left(y_{{}_2}\right)=0$$

If this relationship is not satisfied, then rank$\left(\right[A,b\left]\right)=4$, and the constraints are inconsistent, unable to satisfy the continuity conditions.

3.2. Example of Inconsistent Constraints

Consider the three constraints,

$$T(0,t)=0,T(L,t)=0,\mathrm{and}{\displaystyle \frac{{\partial }^{2}T}{\partial x\partial t}}{|}_{t=0}=sin\left({\displaystyle \frac{5\pi x}{L}}\right),$$

(16)

the third constraint can be replaced by

$${\displaystyle T_t(x,0)=\int sin\left({\displaystyle \frac{5\pi x}{L}}\right)\mathrm{d}x+c=-{\displaystyle \frac{L}{5\pi }}cos\left({\displaystyle \frac{5\pi x}{L}}\right)+c.}$$

The constraints consistency check must be performed for $T_t(0,0)$ and $T_t(L,0)$. This gives,

$$Ax=\left[\begin{array}{c}+1\\ +1\end{array}\right]\left\{\begin{array}{c}c\end{array}\right\}={\displaystyle \frac{L}{5\pi }}\left\{\begin{array}{c}+1\\ -1\end{array}\right\}=b.$$

In this example, rank$\left(A\right)=1$ while rank$\left(\right[A,b\left]\right)=2$. Therefore, the constraints are inconsistent. The consistency can be obtained by changing the third constraint to ${\displaystyle \frac{{\partial }^{2}T}{\partial x\partial t}}{|}_{t=0}=cos\left({\displaystyle \frac{5\pi x}{L}}\right)$, constraint that can be replaced by,

$${\displaystyle T_t(x,0)=\int cos\left({\displaystyle \frac{5\pi x}{L}}\right)\mathrm{d}x+c={\displaystyle \frac{L}{5\pi }}sin\left({\displaystyle \frac{5\pi x}{L}}\right)+c.}$$

The consistency equations give, $\left[\begin{array}{c}+1\\ +1\end{array}\right]\left\{\begin{array}{c}c\end{array}\right\}=\left\{\begin{array}{c}0\\ 0\end{array}\right\}$, with rank$\left(A\right)=$ rank$\left(\right[A,b\left]\right)=1$. In this case the consistency equations also provide the value of the integration constant, $c=0$. Then, the two variables’ constrained expressions are,

$$\begin{array}{cc}\hfill T^x(x,t)& =g(x,t)+{\displaystyle \frac{x-L}{L}}g(0,t)-{\displaystyle \frac{x}{L}}g(L,t)\hfill \\ \hfill T(x,t)& =T^x(x,t)+t(t+1)\left[{\displaystyle \frac{L}{5\pi }}sin\left({\displaystyle \frac{5\pi x}{L}}\right)-T_t^x(x,0)\right],\hfill \end{array}$$

and the final constrained expression is,

$$\begin{array}{cc}\hfill T(x,t)& =g(x,t)+{\displaystyle \frac{x-L}{L}}g(0,t)-{\displaystyle \frac{x}{L}}g(L,t)+\hfill \\ \hfill & +t(t+1)\left[{\displaystyle \frac{L}{5\pi }}sin\left({\displaystyle \frac{5\pi x}{L}}\right)-g_t(x,0)-{\displaystyle \frac{x-L}{L}}{g}_t(0,0)+{\displaystyle \frac{x}{L}}{g}_t(L,0)\right].\hfill \end{array}$$

It is easy to verify that this equation satisfies the differential equation constraints for any $g(x,t)$ function. A necessary (but not sufficient) condition to validate a constrained expression is setting $g(x,t)=0$. Then, the constrained expression becomes just a function interpolating the constraints, $T(x,t)=t(t+1){\displaystyle \frac{L}{5\pi }}sin\left({\displaystyle \frac{5\pi x}{L}}\right)$.

3.3. Example: 2D Plane Strain Flow

Let us consider the following boundary constraints,

$$\left\{\begin{array}{c}{f}_r(r,0)=0\hfill \\ f_r(r,\pi /2)=0\hfill \\ f_r(b,\vartheta )=c_{{}_0}sin\left(2\vartheta \right)\hfill \\ f_{\vartheta }(b,\vartheta )=c_{{}_1}{cos}^2\vartheta +c_{{}_2}{sin}^2\vartheta ,\hfill \end{array}\right.$$

where $c_{{}_0}$, $c_{{}_1}$ and $c_{{}_2}$ are known parameters. This example come from a material problem of plastic flow in a two dimensional porous medium, expressed in cylindrical coordinates, and under the assumptions of perfectly plastic material with yield stress $\sigma$ that obeys $J_{{}_2}$ plasticity and in-compressible. These constraint can be replaced by,

$$\left\{\begin{array}{c}{\displaystyle f(r,0)=\int f_r(r,0)\mathrm{d}r=d_{{}_1}}\hfill \\ {\displaystyle f(r,\pi /2)=\int f_r(r,\pi /2)\mathrm{d}r=d_{{}_2}}\hfill \\ f_r(b,\vartheta )=c_{{}_0}sin\left(2\vartheta \right)\hfill \\ {\displaystyle f(b,\vartheta )=\int f_{\vartheta }(b,\vartheta )\mathrm{d}\vartheta ={\displaystyle \frac{\vartheta }{2}}(c_{{}_1}+c_{{}_2})+{\displaystyle \frac{sin\vartheta }{4}}(c_{{}_1}-c_{{}_2})+d_{{}_3},}\hfill \end{array}\right.$$

by adding three constants of integration, $d_{{}_1}$, $d_{{}_2}$, and $d_{{}_3}$. Four continuity conditions can be specified,

$$\begin{array}{cc}\hfill f(b,0)& =d_{{}_1}=d_{{}_3}\hfill \\ \hfill f(b,\pi /2)& =d_{{}_2}={\displaystyle \frac{\pi }{4}}(c_{{}_1}+c_{{}_2})+{\displaystyle \frac{1}{4}}(c_{{}_1}-c_{{}_2})+d_{{}_3}\hfill \\ \hfill f_r(b,0)& =0=0\hfill \\ \hfill f_r(b,\pi /2)& =0=0.\hfill \end{array}$$

In matrix form,

$$Ax=\left[\begin{array}{ccc}1& 0& -1\\ 0& 1& -1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\left\{\begin{array}{c}{d}_{{}_1}\\ d_{{}_2}\\ d_{{}_3}\end{array}\right\}=\left\{\begin{array}{c}0\\ {\displaystyle \frac{\pi }{4}}(c_{{}_1}+c_{{}_2})+{\displaystyle \frac{1}{4}}(c_{{}_1}-c_{{}_2})\\ 0\\ 0\end{array}\right\}=b.$$

This linear system has, rank$\left(A\right)=$ rank$\left(\right[A,b\left]\right)=2$. Two variables are dependent (i.e., $d_{{}_1}$ and $d_{{}_2}$) and the dependent variable ($d_{{}_3}$) must be estimated by the optimization process,

$$d_{{}_1}=d_{{}_3}\mathrm{and}{d}_{{}_2}={\displaystyle \frac{\pi }{4}}(c_{{}_1}+c_{{}_2})+{\displaystyle \frac{1}{4}}(c_{{}_1}-c_{{}_2})+d_{{}_3}.$$

Therefore, the constraints become,

$$\left\{\begin{array}{c}f(r,0)=d_{{}_3}\hfill \\ f(r,\pi /2)={\displaystyle \frac{\pi }{4}}s+{\displaystyle \frac{1}{4}}d+d_{{}_3}\hfill \\ f_r(b,\vartheta )=c_{{}_0}sin\left(2\vartheta \right)\hfill \\ f(b,\vartheta )={\displaystyle \frac{\vartheta }{2}}s+{\displaystyle \frac{sin\vartheta }{4}}d+d_{{}_3}.\hfill \end{array}\right.$$

(17)

The variables’ constrained expressions are,

$$\begin{array}{cc}\hfill f^r(r,\vartheta )& =g(r,\vartheta )+\left(1-{\displaystyle \frac{2\vartheta }{\pi }}\right)\left[d_{{}_3}-g(r,0)\right]+{\displaystyle \frac{2\vartheta }{\pi }}\left[{\displaystyle \frac{\pi }{4}}s+{\displaystyle \frac{1}{4}}d+d_{{}_3}-g(r,\pi /2)\right]\hfill \\ \hfill f(r,\vartheta )& =f^r(r,\vartheta )+{\displaystyle \frac{\vartheta }{2}}s+{\displaystyle \frac{sin\vartheta }{4}}d+d_{{}_3}-f^r(b,\vartheta )+(r-b)\left[c_{{}_0}sin\left(2\vartheta \right)-f_r^r(b,\vartheta )\right],\hfill \end{array}$$

and the global constrained expression is,

$$\begin{array}{cc}\hfill f(r,\vartheta )& =g(r,\vartheta )-\left(1-{\displaystyle \frac{2\vartheta }{\pi }}\right)g(r,0)-{\displaystyle \frac{2\vartheta }{\pi }}g(r,\pi /2)\hfill \\ \hfill & +{\displaystyle \frac{\vartheta }{2}}s+{\displaystyle \frac{sin\vartheta }{4}}d+d_{{}_3}-g(b,\vartheta )+\left(1-{\displaystyle \frac{2\vartheta }{\pi }}\right)g(b,0)+{\displaystyle \frac{2\vartheta }{\pi }}g(b,\pi /2)\hfill \\ \hfill & +(r-b)\left[c_{{}_0}sin\left(2\vartheta \right)-g_r(b,\vartheta )+\left(1-{\displaystyle \frac{2\vartheta }{\pi }}\right)g_r(b,0)+{\displaystyle \frac{2\vartheta }{\pi }}{g}_r(b,\pi /2)\right].\hfill \end{array}$$

It is easy to verify that this constrained expression satisfies the new constraints given in Equation (17). In particular, setting $g(r,\vartheta )=0$, the simplest function interpolating the constraints,

$$\begin{array}{cc}\hfill f(r,\vartheta )& ={\displaystyle \frac{\vartheta }{2}}s+{\displaystyle \frac{sin\vartheta }{4}}d+d_{{}_3}+(r-b)c_{{}_0}sin\left(2\vartheta \right),\hfill \end{array}$$

is obtained.

3.4. Constrained Expression for $f_{xx}(x_{{}_b},y)$ and $f_{xy}(x_{{}_b},y)$

For these two constraints the projection functionals [7] are,

$${\displaystyle {\rho }_{{}_1}\left(y\right)=f_{xx}(x_{{}_b},y)-g_{xx}(x_{{}_b},y)\mathrm{and}{\rho }_{{}_2}\left(y\right)=\int f_{xy}(x_{{}_b},y)\mathrm{d}y+c-g_{{}_x}(x_{{}_b},y).}$$

Using $s_{{}_k}\left(x\right)=\{x^2,x^3\}$ as support functions (1 and x are excluded because the constraints involves the second derivative) the switching matrix satisfies,

$$\left[\begin{array}{cc}2& 6x_{{}_b}\\ 2x_{{}_b}& 3x_{{}_b}^2\end{array}\right]\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\to \left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]={\displaystyle \frac{1}{6x_{{}_b}^2}}\left[\begin{array}{cc}-3x_{{}_b}^2& 6x_{{}_b}\\ 2x_{{}_b}& -2\end{array}\right].$$

This provides the coefficients of the switching functions, ${\varphi }_{{}_k}\left(x\right)=a_{1k}{x}^2+a_{2k}{x}^3$, and the constrained expression,

$$\begin{array}{cc}\hfill f(x,y,g(x,y\left)\right)=g(x,y)& +x^2\left({\displaystyle \frac{x}{3x_{{}_b}}}-{\displaystyle \frac{1}{2}}\right)\left[f_{xx}(x_{{}_b},y)-g_{xx}(x_{{}_b},y)\right]\hfill \\ & +x^2\left({\displaystyle \frac{1}{x_{{}_b}}}-{\displaystyle \frac{x}{3x_{{}_b}^2}}\right)\left[{\displaystyle \int f_{xy}(x_{{}_b},y)\mathrm{d}y+c-g_{{}_x}(x_{{}_b},y)}\right].\hfill \end{array}$$

No continuity relationships can be written for this example.

3.5. Continuous Shear-Type Constraints: Second Derivative

Consider the two shear-type constraints,

$$f_{yy}(x_{{}_2},y)=p\left(y\right)\mathrm{and}{f}_{xx}(x,y_{{}_2})=q\left(x\right).$$

(18)

Using anti-derivatives these two constraints can be replaced by,

$$\begin{array}{c}{\displaystyle f(x_{{}_2},y)=\int \int p\left(y\right)\mathrm{d}y=\mathcal{A}\left(y\right)+a_{{}_1}y+a_{{}_2}}\\ {\displaystyle f(x,y_{{}_2})=\int \int q\left(x\right)\mathrm{d}x=\mathcal{B}\left(x\right)+b_{{}_1}x+b_{{}_2}.}\end{array}$$

The continuity equation at $[x_{{}_2},y_{{}_2}]$ is,

$$Ax=\left[\begin{array}{cccc}{y}_{{}_2},& 1,& -x_{{}_2},& -1\end{array}\right]\left\{\begin{array}{c}{a}_{{}_1}\\ a_{{}_2}\\ b_{{}_1}\\ b_{{}_2}\end{array}\right\}=\left\{\begin{array}{c}\mathcal{B}\left(x_{{}_2}\right)-\mathcal{A}\left(y_{{}_2}\right)\end{array}\right\}=b,$$

which is always satisfied. This equation allows to eliminate one constant of integration. For instance,

$$a_{{}_2}=b_{{}_2}+\mathcal{B}\left(x_{{}_2}\right)+b_{{}_1}{x}_{{}_2}-\mathcal{A}\left(y_{{}_2}\right)-a_{{}_1}{y}_{{}_2}.$$

This constant of integration can be then substituted in the constrained expression. Using the recursive approach,

$$\begin{array}{cc}\hfill f^y(x,y)& =g(x,y)+{\displaystyle \frac{y}{y_2}}\left(\mathcal{B}\left(x\right)+b_{{}_1}x+b_{{}_2}-g(x,y_{{}_2}\right)\hfill \\ \hfill f(x,y)& =f^y(x,y)+{\displaystyle \frac{x}{x_2}}\left(\mathcal{A}\left(y\right)+a_{{}_1}y+a_{{}_2}-f^y(x_{{}_2},y)\right),\hfill \end{array}$$

the global constrained expression is,

$$\begin{array}{cc}\hfill f(x,y,g(x,y\left)\right)& =g(x,y)+{\displaystyle \frac{y}{y_{{}_2}}}\left(\mathcal{B}\left(x\right)+b_{{}_1}x+b_{{}_2}-g(x,y_{{}_2})\right)\hfill \\ \hfill & +{\displaystyle \frac{x}{x_{{}_2}}}[\mathcal{A}\left(y\right)+a_{{}_1}y+b_{{}_2}+\mathcal{B}\left(x_{{}_2}\right)+b_{{}_1}{x}_{{}_2}-\mathcal{A}\left(y_{{}_2}\right)-a_{{}_1}{y}_{{}_2}\hfill \\ \hfill & -g(x_{{}_2},y)-{\displaystyle \frac{y}{y_{{}_2}}}\left(\mathcal{B}\left(x_{{}_2}\right)+b_{{}_1}{x}_{{}_2}+b_{{}_2}-g(x_{{}_2},y_{{}_2})\right)].\hfill \end{array}$$

3.6. Example with Constraints: $f_{{}_x}(x_{{}_b},y)=p\left(y\right)$, $f_{{}_y}(x_{{}_b},y)=q\left(y\right)$, and $f(x,y_{{}_b})=r\left(x\right)$

The shear-type constraint can be replaced by integration,

$${\displaystyle f(x_{{}_b},y)=\int q\left(y\right)\mathrm{d}y=\mathcal{A}\left(y\right)+c.}$$

The constraints consistency implies,

$$Ax=\left[\begin{array}{c}0\\ 1\end{array}\right]\left\{\begin{array}{c}c\end{array}\right\}=\left\{\begin{array}{c}p\left(y_b\right)-r_{{}_y}\left(x_b\right)\\ \mathcal{A}\left(y_b\right)-r_{{}_y}\left(x_b\right)\end{array}\right\}=b.$$

This system is consistent if rank$\left(\right[A,b\left]\right)=$ rank$\left(A\right)=1$. This implies that, $r_{{}_x}\left(x_{{}_f}\right)=p\left(y_{{}_f}\right)$, must be satisfied to obtain consistency between constraints. It also provides the value of the constant of integration, $c=\mathcal{A}\left(y_b\right)-r_{{}_y}\left(x_b\right)$.

Using the recursive approach [7] the overall constrained expression is,

$$\begin{array}{cc}\hfill f(x,y)& =g(x,y)+\left({\displaystyle \frac{x^2}{x_{{}_f}}}-x\right)\left[p\left(y\right)-g_{{}_x}(x_{{}_f},y)\right]+\left({\displaystyle \frac{2x}{x_{{}_f}}}-{\displaystyle \frac{x^2}{x_{{}_f}^2}}\right)[\mathcal{A}\left(y\right)+r\left(x_{{}_f}\right)-\mathcal{A}\left(y_{{}_f}\right)\hfill \\ \hfill & -g(x_{{}_f},y)]+{\displaystyle \frac{y}{y_{{}_f}}}\{r\left(x\right)-g(x,y_{{}_f})-\left({\displaystyle \frac{x^2}{x_{{}_f}}}-x\right)\left[r_{{}_x}\left(x_{{}_f}\right)-g_{{}_x}(x_{{}_f},y_{{}_f})\right]\hfill \\ \hfill & -\left({\displaystyle \frac{2x}{x_{{}_f}}}-{\displaystyle \frac{x^2}{x_{{}_f}^2}}\right)\left[r\left(x_{{}_f}\right)-g(x_{{}_f},y_{{}_f})\right]\}.\hfill \end{array}$$

3.7. No Analytic Expression for the Anti-Derivative

Consider just the single shear-type constraint, $f_{{}_y}(x_{{}_b},y)$. This constraints can be replaced by the integral, $\int f_{{}_y}(x_{{}_b},y)\mathrm{d}y=\mathcal{A}\left(y\right)+c$, where $\mathcal{A}\left(y\right)$ is the anti-derivative of $f_{{}_y}(x_{{}_b},y)$. In the case the anti-derivative expression cannot be obtained analytically, then the anti-derivative can be numerically approximated by least-squares using a linear expansion model, $\mathcal{A}\left(y\right)\approx {\xi }^{T}h\left(y\right)$, where $h\left(y\right)$ is a set of m basis functions (e.g., orthogonal polynomials). The unknown vector of coefficients, $\xi$, can be obtained by discretizing the y-domain range in $n>m$ points and least-squares,

$${\displaystyle min\sum _{k=1}^n{\left[f_{{}_y}(x_{{}_b},y_{{}_k})-{\xi }^T\dot{h}\left(y_{{}_k}\right)\right]}^2\to \xi ={\left({\dot{H}}^T\dot{H}\right)}^{-1}{\dot{H}}^T{f}_{{}_y},}$$

where, ${\dot{H}}^T=\left[\dot{h}\left(y_{{}_1}\right),\dot{h}\left(y_{{}_2}\right),\cdots ,\dot{h}\left(y_n\right)\right]$ and $f_{{}_y}=\left\{f_{{}_y}(x_{{}_b},y_{{}_1}),f_{{}_y}(x_{{}_b},y_{{}_2}),\cdots ,f_{{}_y}(x_{{}_b},y_n)\right\}$.

Example of Inconsistent Constraints

Consider the 8 constraints, defined in the $(x,y)\in [0,L]\times [0,L]$ domain with their equivalent constraints,

$$\begin{array}{ccc}{f}_{xx}(x,0)=0\hfill & \to & f(x,0)=a_{{}_0}^{x}x+b_{{}_0}^x\hfill \\ f_{xx}(x,L)=0\hfill & \to & f(x,L)=a_{{}_L}^{x}x+b_{{}_L}^x\hfill \\ f_{yy}(0,y)=0\hfill & \to & f(0,y)=a_{{}_0}^{y}y+b_{{}_0}^y\hfill \\ f_{yy}(L,y)={\displaystyle \frac{P}{L}}(y-L)\hfill & \to & f(L,y)={\displaystyle \frac{P}{L}}\left({\displaystyle \frac{y^3}{6}}-{\displaystyle \frac{L{y}^2}{2}}\right)+a_{{}_L}^{y}y+b_{{}_L}^y\hfill \\ f_{xy}(x,0)=0\hfill & \to & f_{{}_y}(x,0)=c_{{}_0}^x\hfill \\ f_{xy}(x,L)=0\hfill & \to & f_{{}_y}(x,L)=c_{{}_L}^x\hfill \\ f_{xy}(0,y)=0\hfill & \to & f_{{}_x}(0,y)=c_{{}_0}^y\hfill \\ f_{xy}(L,y)=0\hfill & \to & f_{{}_x}(L,y)=c_{{}_L}^y,\hfill \end{array}$$

(19)

where 12 unknown constants of integration are appearing. These constants are derived by the 12 continuities conditions,

$$\begin{array}{ccc}f(0,0)\hfill & \to & b_{{}_0}^x=b_{{}_0}^y\hfill \\ f(0,L)\hfill & \to & b_{{}_L}^x=a_{{}_0}^{y}L+b_{{}_0}^y\hfill \\ f(L,0)\hfill & \to & a_{{}_0}^{x}L+b_{{}_0}^x=b_{{}_L}^y\hfill \\ f(L,L)\hfill & \to & a_{{}_L}^{x}L+b_{{}_L}^x=-{\displaystyle \frac{P{L}^2}{3}}+a_{{}_L}^{y}L+b_{{}_L}^y\hfill \\ f_{{}_x}(0,0)\hfill & \to & c_{{}_0}^y=a_{{}_0}^x\hfill \\ f_{{}_x}(0,L)\hfill & \to & c_{{}_0}^y=a_{{}_L}^x\hfill \\ f_{{}_x}(L,0)\hfill & \to & c_{{}_L}^y=a_{{}_0}^x\hfill \\ f_{{}_x}(L,L)\hfill & \to & c_{{}_L}^y=a_{{}_L}^x\hfill \\ f_{{}_y}(0,0)\hfill & \to & c_{{}_0}^x=a_{{}_0}^y\hfill \\ f_{{}_y}(0,L)\hfill & \to & c_{{}_L}^x=a_{{}_0}^y\hfill \\ f_{{}_y}(L,0)\hfill & \to & c_{{}_0}^x=a_{{}_L}^y\hfill \\ f_{{}_y}(L,L)\hfill & \to & c_{{}_L}^x=-{\displaystyle \frac{pL}{2}}+a_{{}_L}^y.\hfill \end{array}$$

This is a system of linear equations than can be written in matrix form,

$$Ax=\left[\begin{array}{cccccccccccc}\hfill 0& \hfill 1& 0\hfill & \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill -1& \hfill 0& \hfill 0& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& 0\hfill & \hfill 0& \hfill 1& \hfill 0& \hfill -L& \hfill -1& \hfill 0& \hfill 0& \hfill 0& \hfill 0\\ \hfill L& \hfill 1& 0\hfill & \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill -1& \hfill 0\\ \hfill 0& \hfill 0& 0\hfill & \hfill L& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill -L& \hfill -1& \hfill 0\\ \hfill -1& \hfill 0& 0\hfill & \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 1& \hfill 0& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& 0\hfill & \hfill -1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 1& \hfill 0& \hfill 0& \hfill 0\\ \hfill -1& \hfill 0& 0\hfill & \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 1\\ \hfill 0& \hfill 0& 0\hfill & \hfill -1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 1\\ \hfill 0& \hfill 0& 1\hfill & \hfill 0& \hfill 0& \hfill 0& \hfill -1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& 0\hfill & \hfill 0& \hfill 0& \hfill 1& \hfill -1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& 1\hfill & \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill -1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& 0\hfill & \hfill 0& \hfill 0& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill -1& \hfill 0& \hfill 0\end{array}\right]\left\{\begin{array}{c}\hfill a_{{}_0}^x\\ \hfill b_{{}_0}^x\\ \hfill c_{{}_{{}_0}}^x\\ \hfill a_{{}_L}^x\\ \hfill b_{{}_L}^x\\ \hfill c_{{}_L}^x\\ \hfill a_{{}_0}^y\\ \hfill b_{{}_0}^y\\ \hfill c_{{}_0}^y\\ \hfill a_{{}_L}^y\\ \hfill b_{{}_L}^y\\ \hfill c_{{}_L}^y\end{array}\right\}=-{\displaystyle \frac{PL}{6}}\left\{\begin{array}{c}\hfill 0\\ \hfill 0\\ \hfill 0\\ \hfill 2L\\ \hfill 0\\ \hfill 0\\ \hfill 0\\ \hfill 0\\ \hfill 0\\ \hfill 0\\ \hfill 0\\ \hfill 3\end{array}\right\}=b.$$

The consistency check for this linear system gives ($P,L\ne 0$),

$$\mathrm{rank}\left(A\right)=9<10=\mathrm{rank}\left([A,b]\right).$$

Therefore, the constraints given in Equation (19) are inconsistent.

4. Higher Dimensional Domains

The extension to higher dimensional space follows the same logic. For instance, in three dimensions, let us consider the generic constraint acting at $z=z_{{}_2}$,

$${\displaystyle \frac{{\partial }^{(n_{{}_x}+n_{{}_y}+n_{{}_z})}f}{\partial x^{n_{{}_x}}\partial y^{n_{{}_y}}\partial z^{n_{{}_{{}_x}}}}}{|}_{z_{{}_2}}=p(x,y).$$

(20)

By first performing $n_{{}_x}$ indefinite integrals in x and then $n_{{}_y}$ indefinite integrals in y, this constraint can be substituted by,

$${\displaystyle \underset{n_{{}_y}}{{\displaystyle \int ...\int }}\left[\underset{n_{{}_x}}{{\displaystyle \int ...\int }}p(x,y){\mathrm{d}}^{n_{{}_x}}x\right]{\mathrm{d}}^{n_{{}_y}}y=\mathcal{A}(x,y)+y^{n_{{}_y}}\sum _{k=0}^{n_{{}_x}}{c}_{{}_k}{x}^k+\sum _{k=0}^{n_{{}_y}-1}{d}_{{}_k}{y}^k.}$$

Because of the “symmetry of second derivatives” (Clairaut’s theorem), the anti-derivative $\mathcal{A}(x,y)$, satisfies,

$${\displaystyle \frac{{\partial }^{(n_{{}_x}+n_{{}_y})}\mathcal{A}}{\partial x^{n_{{}_x}}\partial y^{n_{{}_y}}}}=p(x,y),$$

(21)

no matter the sequence of derivations. Therefore, the expression of $\mathcal{A}(x,y)$ is independent from the integration order. However, the terms involving the constants of integration dependent from the integration order. The constrained expression of Equation (20) has the form,

$$f(x,y,z)=g(x,y,z)+{\displaystyle \frac{z^{n_{{}_z}}}{n_{{}_z}!}}\left[\mathcal{A}(x,y)+p(x,y)-{\displaystyle \frac{{\partial }^{n_{{}_z}}g}{\partial z^{n_z}}}{|}_{z_{{}_2}}\right],$$

where $p(x,y)$ is a polynomial of degree $n_{{}_x}{n}_{{}_y}$ whose expression depends on the integral order selected. For instance, by first performing $n_{{}_x}$ indefinite integrals in x and then $n_{{}_y}$ indefinite integrals in y, the $p(x,y)$ polynomial is,

$${\displaystyle p(x,y)=y^{n_{{}_y}}\sum _{k=0}^{n_{{}_x}}{c}_{{}_k}{x}^k+\sum _{k=0}^{n_{{}_y}-1}{d}_{{}_k}{y}^k.}$$

For example, using $n_{{}_x}=2$, $n_{{}_y}=1$, with the function, $f(x,y)=ysinx-xcosy$, any sequence of integration provide the anti-derivative, $\mathcal{A}(x,y)=-{\displaystyle \frac{y^2}{2}}sinx-{\displaystyle \frac{x^3}{6}}siny$, while the polynomial terms associated with the constants of integration are,

$$\begin{array}{cc}\hfill {\displaystyle \int \int \int f(x,y)\mathrm{d}x\mathrm{d}x\mathrm{d}y}& =\mathcal{A}(x,y)+a_{{}_1}xy+a_{{}_2}y+a_{{}_{{}_3}}\hfill \\ \hfill {\displaystyle \int \int \int f(x,y)\mathrm{d}x\mathrm{d}y\mathrm{d}x}& =\mathcal{A}(x,y)+b_{{}_1}xy+b_{{}_2}x+b_{{}_{{}_3}}\hfill \\ \hfill {\displaystyle \int \int \int f(x,y)\mathrm{d}y\mathrm{d}x\mathrm{d}x}& =\mathcal{A}(x,y)+c_{{}_1}{x}^2+c_{{}_2}x+c_{{}_{{}_3}}.\hfill \end{array}$$

This means that the sequence of integrations, when substituting shear-type and mixed derivative constraints with new constraints, is important in optimization. The sequence specifies what are the polynomial terms to exclude in the free function expansion.

Discussion

This study extends the Theory of Functional Connections framework to perform functional interpolation by including constraints expressed in terms of shear-type and mixed derivative constraints. This has been achieved by replacing these derivative constraints by equivalent constraints, obtained using indefinite integrals. At first look, this approach appears at the cost of additional unknowns, the constants of integration. However, this is not correct. In fact, when using constrained expressions in optimization, these additional constants just replace the terms in the free function expansion that are linearly dependent with the additional constants terms. It is important to outline that, the unknowns constants of integration, as well as the coefficients of the free function expansion, always appear linear in any constrained expression. This fact implies that linear differential equations can be solved by the Theory of Functional Connections just using least-squares.

This study outlines that, before proceeding with the derivation of constrained expressions, the consistency of the boundary constraints must be validated. This consistency problem is a linear problem, $Ax=b$. This system admits a solution if the rank of the coefficients matrix, A, is equal to the rank of the augmented matrix, $[A,b]$. In some example, enforcing the consistency allows to derive the expressions of some constants of integrations. This reduces the number of terms to exclude in the free function expansion and, in some cases, can make the final expression of the constrained expression be independent from the specific functions representing the constraints. It is important to outline that the constraints’ consistency problem is an interpolation problem, which can be obtained just by setting the free function to zero. This is because the constrained expression obtained by functional interpolation must satisfy all constraints for any expression of the free function, and by setting the free function to zero the functional interpolation problem becomes an interpolation problem for the support functions selected.

The constraints’ consistency validates the constraints’ continuity. The constraints’ consistency problem does not involve the free functions. Constraints can be interpolated (or functionally interpolated) if the constraints are consistent, only. This means that, in the case the a constrained expression is derived before validating the constraints’ consistency, the constraints’ continuity can be validated by setting the free function equal to zero, because the continuity problem is independent from free function.