### 3.1 Mathematical formulation of the reduced problem

Let us now introduce the Reduced Problem; essentially, its definition corresponds to setting \(\Gamma^{c}=\emptyset\) in Assumptions 2.1. Therefore, \(\Gamma^{c}\), \(\mathbf{f}^{c}\), and *g* do not take part in the formulation, and, subsequently, we have \(\Gamma =\Gamma^{r}=\partial\Omega\); moreover, for convenience, we avoid the superscript *r* for \(\mathbf{f}^{r}\), and we directly consider **f** as a given vectorial field, per unit length, on \(\Gamma=\partial \Omega\).

### Reduced Problem 1

Under the hypothesis of Assumptions 2.1, let us set \(\Gamma^{c}=\emptyset\). Find a symmetric and a.e. positive definite second-order tensor \(\boldsymbol {\sigma}=\boldsymbol {\sigma}(x,y)\) in Ω̄ such that

In the rest of this section, we show the existence and uniqueness of a solution to the Reduced Problem 1 defined in a rectangle.

### 3.2 A case of explicit resolution in a rectangle

For any \(a, c_{1}>0\) and \(n \in\mathbb{N}\) with \(n>1\), let us consider the rectangle \(\Omega=(0,a)\times(-b,0)\) with \(b=n^{1/n}\) and the function \(z(x,y)=c_{1}x^{2}-c_{2}y^{2n}\), where \(c_{2}=a^{2} c_{1}/(n(2n-1))\), which satisfies \(z_{,xx}z_{,yy}-z_{,xy}^{2}<0\) a.e. in Ω̄. Differentiating (5a) with respect to *x* and (5b) to *y* and subtracting the results from each other give

$$ \sigma_{xx,xx}-\sigma_{yy,yy} =0 \quad\text{in } \Omega. $$

(6)

On the other hand, in view of the expression of *z*, relation (5c) infers

$$ \sigma_{xx}=\frac{c_{2} n (2n-1)}{c_{1}}y^{2(n-1)} \sigma_{yy} \quad\text{in } \bar{\Omega}. $$

(7)

Hence, (6) and (7) lead to

$$ \frac{c_{2} n (2n-1)}{c_{1}}y^{2(n-1)}\sigma_{yy,xx}- \sigma_{yy,yy}=0 \quad\text{in } \Omega, $$

(8)

which degenerates for \(y=0\). To endow this equation with the desired Dirichlet conditions for the unknown \(\sigma_{yy}\), let us treat the vectorial field **f** on Γ, and let us write

$$ \mathbf{f}= \textstyle\begin{cases} (\gamma_{1}(y),\gamma_{2}(y)) \quad\text{on } x= a,\\ (-\gamma_{1}(y),\gamma_{3}(y)) \quad \text{on } x= 0, \\ (\theta_{1}(x),\theta_{2}(x)) \quad\text{on } y= 0,\\ (\theta_{3}(x),-\theta_{2}(x)) \quad \text{on } y= -b. \end{cases} $$

In the previous definition, \(\gamma_{i}(y)\) and \(\theta_{i}(x)\) (for \(i=1,2,3\)) are continuous functions for \(-b \leq y \leq0\) and \(0\leq x\leq a\), respectively, which will be chosen later on; subsequently, taking \(\mathbf{n}=(\mp1,0)\) respectively on \(x=0\) and \(x=a\) and \(\mathbf{n}=(0,\mp1)\) respectively on \(y=-b\) and \(y=0\), the boundary conditions (5d) read

$$ \textstyle\begin{cases} \sigma_{xx}(a,y)=\sigma_{xx}(0,y)=\gamma_{1}(y),\\ \sigma_{xy}(a,y)=\gamma_{2}(y) \quad\text{and}\quad \sigma_{xy}(0,y)=-\gamma _{3}(y),\\ \sigma_{xy}(x,0)=\theta_{1}(x) \quad\text{and}\quad \sigma_{xy}(x,-b)=-\theta _{3}(x),\\ \sigma_{yy}(x,0)=\sigma_{yy}(x,-b)=\theta_{2}(x). \end{cases} $$

(9)

Additionally, if \(\gamma_{1}=\gamma_{1}(y)\) is such that the function

$$ \gamma(y)=\frac{\gamma_{1}(y)c_{1}}{c_{2}n(2n-1)y^{2(n-1)}} $$

is itself continuous in \(-b \leq y \leq0\), then in light of (7), (8), and (9) we arrive at the boundary value problem

$$ \textstyle\begin{cases} \frac{c_{2} n (2n-1)}{c_{1}}y^{2(n-1)}\sigma_{yy,xx}- \sigma_{yy,yy}=0 \quad\text{in } \Omega,\\ \sigma_{yy}(x,0)=\sigma_{yy}(x,-b)=\theta_{2}(x),\\ \sigma_{yy}(0,y)=\sigma_{yy}(a,y)= \gamma(y), \end{cases} $$

(10)

for which we fix the following proper assumptions:

$$ \gamma(0)=\gamma(-b)=\theta_{2}(0)= \theta_{2}(a)=K\in \mathbb{ R}\quad\text{and}\quad \gamma_{2}(0)=H\in \mathbb{ R}. $$

(11)

Hence, let us translate the unknown \(\sigma_{yy}\) through

$$ u(x,y)=\sigma_{yy}(x,y)-\bigl[\gamma(y)+ \theta_{2}(x)-K\bigr], $$

(12)

and successively let us rescale *x* by the homogeneous dilation mapping \((0,1)\) onto \((0,a)\) by \(x(X)=aX\); the new variable \(U(X,y)=u(aX,y)\) and data \(\Theta(X)=\theta_{2}(ax)\) are so obtained. These two transformations, in conjunction with the relation \(c_{2}=a^{2} c_{1}/(n(2n-1))\), reduce (10) to

$$ \textstyle\begin{cases} y^{2(n-1)}U_{,XX}- U_{,yy}+\Theta''(X)y^{2(n-1)}-\gamma''(y)=0\quad \text{in } (0,1) \times(-b,0),\\ U(X,0)=U(X,-b)=0,\\ U(0,y)=U(1,y)=0. \end{cases} $$

(13)

Now, let us impose \(\Theta''(X)y^{2(n-1)}-\gamma''(y)=0\), that is, \(\Theta''(X)=\gamma''(y)/y^{2(n-1)}= \lambda\) for some \(\lambda\in \mathbb{ R}\). This provides, thanks to the first continuity conditions (11), which for \(-b\leq y \leq0\) and \(0\leq x \leq 1\) are \(\gamma(0)=\gamma(-b)=\Theta(0)=\Theta(1)=K\),

$$ \Theta(X)=\frac{\lambda}{2}X^{2}-\frac{\lambda}{2}X+K \quad\text{and}\quad \gamma(y)=\frac{\lambda}{2n(2n-1)}y^{2n}+\frac{\lambda b^{2n-1}}{2n(2n-1)}y+K; $$

(14)

as a consequence, problem (13) is equivalent to

$$ \textstyle\begin{cases} (-y)^{2(n-1)}U_{,XX}- U_{,yy}=0 \quad\text{in } (0,1) \times(-b,0),\\ U(X,0)=U(X,-b)=0,\\ U(0,y)=U(1,y)=0. \end{cases} $$

(15)

According to the theory of second-order linear PDE’s, and since \(b=n^{1/n}\), the characteristic curves associated with the equation \((-y)^{2(n-1)}U_{,XX}- U_{,yy}=0\) and exactly passing through the vertexes of the rectangle \((0,1) \times(-b,0)\) are (see the right side of Figure 1)

$$ X=\frac{1}{n}(-y)^{n} \quad\text{and}\quad 1-X=\frac{1}{n}(-y)^{n}. $$

Thereafter we can rely on the main statement given in [12] and apply its result to the boundary value problem (15); hence we conclude that it admits a unique solution in \((0,1)\times(-b,0)\) that is continuously differentiable everywhere in its closure, possibly except along the mentioned characteristics. As to our specific case, in view of the homogeneous boundary conditions, \(U(X,y)\equiv0\) (and hence also \(u(x,y)\equiv0\)) is the unique function with such properties solving problem (15).

Coming back to the tensorial unknown *σ* in Ω̄, expression (14) produces through the relations \(X=x/a\), (12), and (7)

$$ \textstyle\begin{cases} \sigma_{yy}(x,y)=\frac{\lambda}{2a^{2}}x^{2}-\frac{\lambda}{2a}x+\frac {\lambda}{2n(2n-1)}y^{2n}+\frac{\lambda b^{2n-1}}{2n(2n-1)}y+K & \text{in } \bar{\Omega},\\ \sigma_{xx}(x,y)=a^{2}y^{2(n-1)}\sigma_{yy}(x,y) & \text{in } \bar {\Omega}. \end{cases} $$

(16)

As to \(\sigma_{xy}=\sigma_{yx}\), from (5b) we deduce

$$ \sigma_{xy}(x,y)=- \int\sigma_{yy,y}\,dx=-\frac{2n\lambda y^{2n-1}+\lambda b^{2n-1}}{2n(2n-1)}x+ h(y), $$

so that imposing (5a), we get

$$ h'(y)=\frac{a}{2}\lambda y^{2(n-1)}\quad\Leftrightarrow \quad h(y) =\frac{a \lambda y^{2n-1}}{2(2n-1)} +h_{0}, \quad h_{0}\in \mathbb{ R}. $$

Now, taking into account the second position in (11) and the boundary conditions (9), the last two expressions yield

$$\begin{aligned} \sigma_{xy}(x,y)&= \sigma_{yx}(x,y) \\ &=-\frac{2n\lambda y^{2n-1}+\lambda b^{2n-1}}{2n(2n-1)}x+ \frac{a \lambda y^{2n-1}}{2(2n-1)} +H+\frac{a\lambda b^{2n-1}}{2n(2n-1)} \quad\text{in } \bar{ \Omega}. \end{aligned}$$

(17)

Lately, to guarantee the positive definiteness of *σ* a.e. in Ω̄ in the sense of (H2), we have to impose, inter alia, that \(\sigma_{xx}\sigma_{yy}-(\sigma_{xy})^{2}> 0\) a.e. in Ω̄. From (16) and (17) we obtain that \(\sigma_{xx}(x,0)=0\) for \(x\in [0,a]\), whereas \(\sigma_{xy}(x,0)=-\lambda b^{2n-1}x/(2n(2n-1))+H+a\lambda b^{2n-1}/(2n(2n-1))\) for \(x\in[0,a]\); therefore, without specific assumptions and relations on \(a,b,n,\lambda \), and *H*, generally, \(\sigma_{xx}(x,0)\sigma_{yy}(x,0)-(\sigma _{xy}(x,0))^{2}<0\) in \([0,a]\). Under these conditions, since for continuity arguments, there would exist a constant \(\varepsilon>0\) such that \(\sigma_{xx}\sigma_{yy}-(\sigma_{xy})^{2}< 0\) in \((0,a)\times (-\varepsilon,0)\), in (17) we have to impose \(H=\lambda=0\) obtaining \(\sigma_{xy}(x,y)=\sigma_{yx}(x,y)\equiv0\) in Ω̄. In addition, to avoid the nil solution \(\boldsymbol {\sigma}\equiv{\mathbf{0}}\), we choose a strictly positive value for *K*, and from (16) we explicitly write \(\sigma_{yy}(x,y)=K\) and \(\sigma_{xx}(x,y)=Ka^{2} y^{2(n-1)}\) in Ω̄ and also obtain the following formulas for the functions \(\gamma_{i}(y)\) and \(\theta_{i}(x)\) defining the boundary conditions (9):

$$ \gamma_{1}(y)=a^{2}K y^{2(n-1)},\qquad \theta_{2}(x)=K,\qquad \theta_{1}(x)=\theta _{3}(x)=\gamma_{2}(y)= \gamma_{3}(y)=0. $$

So we have proved our main result.

### Theorem 3.1

*Let*
\(a, c_{1}>0\)
*and*
\(n\in\mathbb{N}\)
*with*
\(n>1\). *Moreover*, *for*
\(b=n^{1/n}\)
*and*
\(c_{2}=a^{2} c_{1}/(n(2n-1))\), *the rectangle*
\(\Omega =(0,a)\times(-b,0)\)
*and the function*
\(z(x,y)=c_{1}x^{2}-c_{2}y^{2n}\)
*are given*. *Then*, *for any fixed*
\(K>0\)
*and vectorial field* (*per unit length*) *on*
\(\Gamma=\partial\Omega\)

$$\mathbf{f}= \textstyle\begin{cases} (a^{2}Ky^{2(n-1)},0) \quad\textit{on } x= a,\\ (-a^{2}Ky^{2(n-1)},0) \quad\textit{on } x= 0, \\ (0,K) \quad\textit{on } y= 0,\\ (0,-K) \quad \textit{on } y= -b, \end{cases} $$

*the symmetric and a*.*e*. *positive definite tensor*

$$ \boldsymbol {\sigma}(x,y)= \begin{pmatrix} a^{2}Ky^{2(n-1)} & 0 \\ 0 & K \end{pmatrix} \quad\textit{in } \bar{\Omega} $$

(18)

*is the unique classical solution of the Reduced Problem*
1.

### 3.3 The case of no restriction on the sign definiteness of *σ*

By retracing the proof of Theorem 3.1 we observe that, behind other technical reasons, the final expression of the solution *σ* derived in (18) is deeply tied to the requirement of the a.e. positivity definiteness of such a tensor; conversely, as announced in the introductory comments of Section 1, if this restriction is omitted, then for the same function \(z(x,y)=c_{1}x^{2}-c_{2}y^{2n}\), the unique solution in Ω̄ exhibits a more general representation, precisely given by (16) and (17). Subsequently, we have this further result, which we state without further comments.

### Theorem 3.2

*Let*
\(a, c_{1}>0\)
*and*
\(n\in\mathbb{N}\)
*with*
\(n>1\). *Moreover*, *for*
\(b=n^{1/n}\)
*and*
\(c_{2}=a^{2} c_{1}/(n(2n-1))\), *the rectangle*
\(\Omega =(0,a)\times(-b,0)\)
*and the function*
\(z(x,y)=c_{1}x^{2}-c_{2}y^{2n}\)
*are given*. *Then for any fixed*
\(H,K,\lambda\in \mathbb{ R}\)
*and vectorial field* (*per unit length*) *on*
\(\Gamma=\partial\Omega\)

$$ \mathbf{f}= \textstyle\begin{cases} (\gamma_{1}(y),\gamma_{2}(y)) \quad\textit{on } x= a,\\ (-\gamma_{1}(y),\gamma_{3}(y)) \quad \textit{on } x= 0, \\ (\theta_{1}(x),\theta_{2}(x)) \quad \textit{on } y= 0,\\ (\theta_{3}(x),-\theta_{2}(x)) \quad \textit{on } y= -b, \end{cases} $$

*where*

$$ \textstyle\begin{cases} \gamma_{1}(y)=a^{2}y^{2(n-1)}(\frac{\lambda}{2n(2n-1)}y^{2n}+\frac {\lambda b^{2n-1}}{2n(2n-1)}y+K),\\ \gamma_{2}(y)=H +a \lambda\frac{ n y^{2 n-1}-b^{2 n-1} (n-1)}{2n (2n-1) },\\ \gamma_{3}(y)=-H-a\lambda\frac{ b^{2 n-1} + y^{2 n-1}}{2(2n-1)},\\ \theta_{1}(x)= \frac{\lambda b^{2n-1}}{2n(2n-1)}(an-x) +H,\\ \theta_{2}(x)=\frac{\lambda}{2a^{2}}x^{2}-\frac{\lambda}{2a}x+K,\\ \theta_{3}(x)=-\frac{\lambda b^{2n-1}}{2n}x -H, \end{cases} $$

*the symmetric tensor*

$$ \textstyle\begin{cases} \sigma_{yy}(x,y)=\frac{\lambda}{2a^{2}}x^{2}-\frac{\lambda}{2a}x+\frac {\lambda}{2n(2n-1)}y^{2n}+\frac{\lambda b^{2n-1}}{2n(2n-1)}y+K \quad \textit{in } \bar{\Omega},\\ \sigma_{xx}(x,y)=a^{2}y^{2(n-1)}\sigma_{yy}(x,y) \quad\textit{in } \bar {\Omega},\\ \sigma_{xy}(x,y)=\sigma_{yx}(x,y)=\frac{\lambda y^{2n-1}}{2n-1} (\frac{a}{2}-x )+\frac{\lambda b^{2n-1}}{2n(2n-1)}(an-x) +H \quad \textit{in } \bar{\Omega} \end{cases} $$

(19)

*is the unique classical solution of the Reduced Problem*
1.

### 3.4 Two specific examples: representation of the solution

To give an explicit example to each one of the results claimed in Theorems 3.1 and 3.2, we analyze Figures 2 and 3. They graphically show the behavior of the tensor *σ*, which solves the Reduced Problem 1, once in the hypothesis of such theorems the same surface \(z=c_{1}x^{2}-c_{2}y^{2n}\) and the rectangle \(\Omega=(0,a)\times(-b,0)\) are fixed by means of the values \(n=3\), \(c_{1}=4\), and \(a=5\) (the surface and the domain are shown at the top-left corners of Figures 2 and 3).

More precisely, for \(K=5\) in expression (18), Figure 2 represents the case of the equilibrium between the stress and shape of a membrane structure. We can realize that the component \(\sigma_{xx}\) is positive a.e. in Ω and increases as \(y\rightarrow-b\) and constant values of *x* (see the below part of the top-right corner of Figure 2); in the limit, it exactly corresponds to a zone on the membrane with major tension, along the *x*-direction, with respect to others (Figure 2, above part, top-right). As to \(\sigma_{yy}\), it is constant and positive in Ω, so that the corresponding tension along the *y*-direction is uniformly distributed on the surface (see the lower-left corner of Figure 2); finally, the lower-right corner of Figure 2 highlights the nil contribution of \(\sigma _{yx}=0\) in Ω, that is, the absence of shear stress on the membrane.

Conversely, if in (19) we set \(\lambda =4\), \(K=0.5\), and \(H=2\), the features of the solution *σ* are summarized in Figure 3, which models the balance between the stress and shape for a shell structure. Relaxing the assumption on the sign definiteness of *σ*, we obtain not only positive expressions for the components \(\sigma_{xx}\) and \(\sigma_{yy}\) on the whole Ω, but also regions of the rectangle where they are negative (see the below part of the top-right and lower-left corner of Figure 3, respectively); this aspect identifies zones of the shell where tensions or compressions are present along both the *x*- and *y*-directions (same corners of Figure 3, but the above part).

By the above we stress again that the general solution for the tensor *σ* given by relation (18) represents a very particular and simplified case of solution (19). Such a leap has not to appear surprising since, indeed, it is intimately linked to the different natures of the problems: in particular, when a membrane is considered, a strong limitation on the state of its stress tensor that exactly balances its shape is naturally expected and absolutely consistent with the mechanical problem.