*7.3. Proof*

This section demonstrates that the term *<sup>A</sup>*(*x*) from Equation (20) generates a surface satisfying the boundary constraints defined by the function *<sup>c</sup>*(*x*). First, it is shown that *<sup>A</sup>*(*x*) satisfies boundary constraints on the value, and then that *<sup>A</sup>*(*x*) satisfies boundary constraints on arbitrary-order derivatives.

Equation (23) for *d* = 0 allows us to write,

$${}^{k}b\_{p\_{q-1}}^{0}[A(\mathbf{x})] = \,^{k}b\_{p\_{q-1}}^{0}[\mathcal{M}\_{i\_1i\_2\ldots i\_k\ldots i\_n}]\boldsymbol{\upsilon}\_{i\_1}\boldsymbol{\upsilon}\_{i\_2}\ldots \,^{k}b\_{p\_{q-1}}^{0}[\boldsymbol{\upsilon}\_{i\_k}]\ldots \boldsymbol{\upsilon}\_{i\_n}.\tag{26}$$

The boundary constraint operator applied to *vk* yields,

$$^k b\_{p\_{q-1}}^0 [v\_{i\_k}] = \begin{cases} = 1, & i\_k = 1, q \\ = 0, & i\_k \neq 1, q. \end{cases} \tag{27}$$

Since the only nonzero terms are associated with *ik* = 1, *q*, we have,

$${}^{k}b\_{p\_{q-1}}^{0}[A(\mathbf{x})] = \left( \begin{array}{c} {}^{k}b\_{p\_{q-1}}^{0} \left[ \mathcal{M}\_{i\_{1}i\_{2}...1...i\_{n}} \right] + {}^{k}b\_{p\_{q-1}}^{0} \left[ \mathcal{M}\_{i\_{1}i\_{2}...q...i\_{n}} \right] \right) \boldsymbol{\uppi}\_{i\_{1}} \boldsymbol{\uppi}\_{i\_{2}} \ldots \boldsymbol \uppi}\_{i\_{n}}.\tag{28}$$

Applying the boundary constraint operator to the *n* − 1-dimensional M tensor where index *ik* = *q* has no effect, because all of the functions already have coordinate *xk* substituted for the value *pq*−<sup>1</sup> (see Equation (24)). Moreover, applying the boundary constraint operator to the M tensor where index *ik* = 1 causes all terms in the sum within the parenthesis in Equation (28) to cancel each other, except when all of the non-*ik* indices are equal to one. This leads to Equation (29).

$$\mathbf{a}^{k} b\_{p\_{q-1}}^{0} [A(\mathbf{x})] = \left( \mathcal{M}\_{11\dots1\dots1} + \mathcal{M}\_{11\dots q\dots1} \right) \mathbf{v}\_{1} \mathbf{v}\_{1} \dots \mathbf{v}\_{1} \tag{29}$$

Since *vj* = 1 when *j* = 1 and M11...1 = 0 by definition, then,

$${}^{k}b\_{p\_{q-1}}^{0}[A(\mathfrak{x})] = \mathcal{M}\_{11\ldots q\ldots 1} = c(\mathfrak{x}\_1, \mathfrak{x}\_2, \ldots, p\_{q-1}, \ldots, \mathfrak{x}\_{\mathfrak{n}})\_{\mathfrak{n}}$$

which proves Equation (20) works for boundary constraints on the value.

Now, we show that Equation (20) holds for arbitrary-order derivative type boundary constraints. Equation (23) for *d* > 0 allows us to write,

$${}^{k}b\_{p\_{q-1}}^{d\_{q-1}}[A(\mathbf{x})] = {}^{k}b\_{p\_{q-1}}^{d\_{q-1}}[\mathcal{M}\_{i\_1i\_2\ldots i\_k\ldots i\_\pi}] {}\_{\overline{v}\_1}\overline{v}\_{i\_1}\ldots\overline{v}\_{i\_k}\ldots\overline{v}\_{i\_k} + \mathcal{M}\_{i\_1i\_2\ldots i\_k\ldots i\_\pi}{}^{k}\_{\overline{v}\_1}\overline{v}\_{i\_1}\ldots\ldots\overline{v}\_{i\_\pi}\ldots\tag{30}$$

From Equation (23), we note that boundary constraint operators that take a derivative follow the usual product rule when applied to a product. Moreover, we note that all of the *v* vectors except *<sup>v</sup>ik* do not depend on *xk*, thus applying the boundary constraint operator to them results in a vector of zeros. Applying the boundary constraint operator to *<sup>v</sup>ik*yields,

$$\begin{aligned} {}^k b\_{p\_{q-1}}^{d\_{q-1}} [\mathbf{z}\_{i\_k}] &= \begin{cases} = \mathbf{1}, & i\_k = q \\ = 0, & i\_k \neq q\_k \end{cases} \end{aligned}$$

and applying the boundary constraint operator to M yields,

$${}^{k}b\_{p\_{q-1}}^{d\_{q-1}}[\mathcal{M}\_{i\_1i\_2\ldots 1\ldots i\_n}] = \begin{cases} = {}^{k}b\_{p\_{q-1}}^{d\_{q-1}}[\mathcal{M}\_{i\_1i\_2\ldots 1\ldots i\_n}], & i\_k = 1\\ = 0, & i\_k \neq 1 \ldots k \end{cases}$$

Substituting these simplifications into *<sup>A</sup>*(*x*) = M*<sup>i</sup>*1*i*2...*ik* ...*in <sup>v</sup>i*1*vi*2 ... *<sup>v</sup>ik* ... *<sup>v</sup>in* , after applying the boundary constraint operator, results in Equation (31).

$${}^{k}b\_{p\_{q-1}}^{d\_{q-1}}[A(\mathbf{x})] = \left( {}^{k}b\_{p\_{q-1}}^{d\_{q-1}}[\mathcal{M}\_{i\_1i\_2\dots 1\dots i\_n}] + \mathcal{M}\_{i\_1i\_2\dots q\dots i\_n} \right) \mathbf{v}\_{i\_1}\mathbf{v}\_{i\_2}\dots\mathbf{v}\_{i\_n} \tag{31}$$

Similar to the proof for value-based boundary constraints, based on Equation (24), all terms in the sum within the parenthesis in Equation (31) cancel each other, except when all of the non-*ik* indices are equal to one. Thus, Equation (31) can be simplified to Equation (32).

$${}^{k}b\_{p\_{q-1}}^{d\_{q-1}}[A(\mathbf{x})] = \left( {}^{k}b\_{p\_{q-1}}^{d\_{q-1}}[\mathcal{M}\_{11\dots1\dots1}] + \mathcal{M}\_{11\dots q\dots1} \right) \mathbf{v}\_{1}\mathbf{v}\_{1}\dots\mathbf{v}\_{1} \tag{32}$$

Again, all of the vectors *v* were designed such that their first component is 1, and the value of the element of M for all indices equal to 1 is 0. Therefore, Equation (32) simplifies to,

$$\prescript{k}{}{b}\_{p\_{q-1}}^{d\_{q-1}}[A(\mathfrak{x})] = \mathcal{M}\_{11\ldots q\ldots 1} = \frac{\partial^d c(\mathfrak{x})}{\partial \mathfrak{x}\_k^d} \Big|\_{\mathfrak{x}\_k = p\_{q-1}}.$$

which proves Equation (20) works for arbitrary-order derivative boundary constraints.

In conclusion, the term *<sup>A</sup>*(*x*) from Equation (20) generates a manifold satisfying the boundary constraints given in terms of arbitrary-order derivative in *n*-dimensional space. The term *<sup>B</sup>*(*x*) from Equation (20) projects any free function *g*(*x*) onto the space of functions that are vanishing at the specified boundary constraints. As a result, Equation (20) can be used to produce the family of *all* possible functions satisfying assigned boundary constraints (functions or derivatives) in rectangular domains in *n*-dimensional space.
