Appendix A Convex analysis
The construction of
\(f_\textrm{c}\) in Theorem
2 is based on the construction of a candidate
\(f_\textrm{a}\) and a decomposition of
\(\mathbb {R}^2\) into disjoint regions with the property that
\(f_\textrm{a}\) is globally continuous and convex on the interior of the different regions. This appendix collects the results that are needed in order to prove that
\(f_\textrm{a}\) is convex on
\(\mathbb {R}^2\). The proofs follow from classical results in convex analysis, see [
41] for more information.
The subdifferential in the sense of convex analysis is denoted by \(\partial f(x)\). Then, by definition, \(\partial f(x)\subset \partial _\textrm{loc}f(x)\) and if f is differentiable, then \(\partial _\textrm{loc}f(x) \subset \{ f'(x) \}\); however, the local subdifferential may be empty if f the graph of f does not lie above the tangent plane in a sufficiently small neighborhood of x.
The construction of the convex envelope of the condensed energy relies on a local construction and the verification that the local construction leads to a convex function. The proof uses the following results.
Appendix B Proof of Theorem 2
Based on the discussion at the beginning of Sect.
5, for the construction of the convex envelope of the condensed energy
\(f^{(r)}\), we first derive pairs and triples of points
\(y^{(1)},\ldots ,y^{(q)}\in \mathbb {R}^2\) (
\(q= 2,3\)), in which
\(f^{(r)}\) is differentiable in the variable
\(y_2\) and in which
\(\partial _2 f^{(r)}\) is equal. Then, we derive a candidate for the direction
\(v\in \mathbb {R}^d\) satisfying the equation in Eq. (
5.8). However, at this point we do not know whether this
v in Eq. (
5.8) is indeed a subgradient of
f in all the points
\(y^{(1)},\ldots , y^{(q)}\). For those pairs and triples for which we expect this to be true, we do not show explicitly that the first part in Eq. (
5.8) is satisfied, rather we apply the construction described in Eqs. (
5.9) and (
5.10) and prove in Theorem
2 that this construction leads to the convex envelope. It is a remarkable result of the subsequent analysis that the explicit formulas for the relaxed energy show a very different behavior with respect to the dependence on the dissipation function
r. In fact, for small values of
b relative to
r, i.e., for
$$\begin{aligned} \sqrt{b} \cdot \frac{y_{\textrm{max}}-y_{\textrm{min}}}{2} \le r\left( \frac{y_{\textrm{min}}+y_{\textrm{max}}}{2} \right) \,, \end{aligned}$$
(B1)
which is the case of relevance in soil mechanics, the relaxed energy is in fact independent of
r. Therefore, we restrict the analysis in this paper to the case of small
b, the general case will be treated in [
41]. In fact, this assumption implies
\(r_0\le r\) by concavity of
r. In the following we consider
r as fixed, write
\(f= f^{(r)}\) and assume without loss of generality that
\(r_0<r\) on
\((y_{\textrm{min}},y_{\textrm{max}})\).
In order to implement the strategy just outlined, we construct a candidate
\(f_\textrm{a}\) for
\(f_\textrm{c}\). In a first step, we search for points with common supporting planes in
\(Y_0\) and
\(\widetilde{Y}\), see Fig.
3. Assume thus that
\((y,\widetilde{y})\in Y_0\times \widetilde{Y}\) satisfy Eq. (
5.8). Since
f is differentiable in
y,
\(v= \nabla f(y)\) is the only possible choice for
v in Eq. (
5.8). If
\(\widetilde{y}\in \text {int}(\widetilde{Y})\), then
f is differentiable in
\(\widetilde{y}\) and
\(\nabla f(y)= v\in \partial f(\widetilde{y})\subset \{\nabla f(\widetilde{y})\}\) implies
\(\widetilde{y}_1= \partial _1 f(\widetilde{y})= \partial _1 f(y)= y_1\), a contradiction. Consequently we assume that
\(\widetilde{y}\in \partial \widetilde{Y}\), that is,
\(\widetilde{y}_1\in \{y_{\textrm{min}}, y_{\textrm{max}}\}\). The existence of
\(\partial _2 f(\widetilde{y})\) implies
\(\partial _2 f(\widetilde{y})= v_2= \partial _2 f(y)\) and equivalently
$$\begin{aligned} \frac{b}{b+1}\widetilde{y}_2 = y_2 \quad \Leftrightarrow \quad \widetilde{y}_2 - y_2 = \frac{1}{b}y_2\,. \end{aligned}$$
(B2)
Equation (
5.8) is equivalent to
\(f(\widetilde{y}) = (T_y f)(\widetilde{y})\). By definition of
f and since
\(f\vert _{Y_0}\) is a quadratic function, these conditions lead to
$$\begin{aligned} \frac{1}{2}\widetilde{y}_1^2 + \frac{1}{2} \frac{b}{b+1} \widetilde{y}_2^2 = f(\widetilde{y})= (T_y f)(\widetilde{y}) = \frac{1}{2}\widetilde{y}_1^2 + \frac{1}{2} \widetilde{y}_2^2 - \frac{1}{2}\nabla ^2f(y)[\widetilde{y}-y, \widetilde{y}-y]\,. \end{aligned}$$
(B3)
In view of
\(\nabla ^2 f(y) = \text {Id}\), the algebraic condition can be simplified to
$$\begin{aligned} (\widetilde{y}_1 - y_1)^2 + (\widetilde{y}_2 - y_2)^2 = \frac{1}{b+1}\widetilde{y}_2^2\,. \end{aligned}$$
(B4)
One defines
\(s= \vert \widetilde{y}_1-y_1\vert \) and substitutes the relation between
\(\widetilde{y}_2\) and
\(y_2\) to obtain
\(y_2 = \pm \sqrt{b} s\). Since
\(\widetilde{y}_1\in \{y_{\textrm{min}}, y_{\textrm{max}}\}\), one obtains four one-parameter families of pairs of points,
$$\begin{aligned} \alpha _{\textrm{min}}^\pm (s) = \left( \begin{array}{cc} y_{\textrm{min}}+ s\\ \pm \sqrt{b}\cdot s \end{array} \right)&\,,\quad \beta _{\textrm{min}}^\pm (s) = \left( \begin{array}{cc} y_{\textrm{min}}\\ \pm \frac{b+1}{\sqrt{b}} s \end{array} \right) \,, \end{aligned}$$
(B5)
$$\begin{aligned} \alpha _{\textrm{max}}^\pm (s) = \left( \begin{array}{cc} y_{\textrm{max}}- s\\ \pm \sqrt{b}\cdot s \end{array} \right)&\,,\quad \beta _{\textrm{max}}^\pm (s) = \left( \begin{array}{cc} y_{\textrm{max}}\\ \pm \frac{b+1}{\sqrt{b}} s \end{array} \right) \,. \end{aligned}$$
(B6)
where
s can take all values in
\((y_{\textrm{min}},y_{\textrm{max}})\) with
\(\alpha _{\textrm{min}}^\pm (s)\in Y_0\) (or
\(\alpha _{\textrm{max}}^\pm (s)\in Y_0\)).
For the construction of the convex envelope only the values
\(0< s< (y_{\textrm{max}}- y_{\textrm{min}})/2=: s_*\) will be relevant, since the curves
\(\alpha _{\textrm{min}}^\pm \) and
\(\alpha _{\textrm{max}}^\pm \) intersect in the point
$$\begin{aligned} y_*^\pm := \alpha _{\textrm{min}}^\pm (s_*)= \alpha _{\textrm{max}}^\pm (s_*)= (y_{\textrm{mid}},\pm \sqrt{b}(y_{\textrm{max}}- y_{\textrm{min}})/2)\in Y_0\,. \end{aligned}$$
(B7)
For each
\(y\in \mathbb {R}^2\), which lies on a connecting straight line of two corresponding points of the curves
\(\alpha _{\textrm{min}}^\pm \) and
\(\beta _{\textrm{min}}^\pm \), i.e.,
\(y= t\cdot \alpha _{\textrm{min}}^\pm (s)+ (1-t)\cdot \beta _{\textrm{min}}^\pm (s)\) with
\(s\in (0,s_*)\) and
\(t\in [0,1]\), we define
\(f_\textrm{a}(y)= t\cdot f(\alpha _{\textrm{min}}^\pm (s))+ (1-t)\cdot f(\beta _{\textrm{min}}^\pm (s))\) according to Eq. (
5.9) and we use the same construction for
\(\alpha _{\textrm{max}}^\pm \) and
\(\beta _{\textrm{max}}^\pm \). Note that
\(f_\textrm{a}\) is affine along these connecting lines and the area, which is covered by those lines consists of four triangular shaped sets denoted by
\(Y_2\) in Fig.
5. The fact that
\(\alpha _{\textrm{min}}^\pm \) and
\(\alpha _{\textrm{max}}^\pm \) intersect in
\(y_*^\pm \) means that the tangential plane of
f in
\(y_*^\pm \) touches the graph in two other points, namely
$$\begin{aligned}&y_{*\textrm{min}}^\pm := \beta _{\textrm{min}}^\pm (s_*)= \left( y_{\textrm{min}},\pm \frac{b+1}{\sqrt{b}}(y_{\textrm{max}}- y_{\textrm{min}})/2\right) \,, \end{aligned}$$
(B8)
$$\begin{aligned}&y_{*\textrm{max}}^\pm := \beta _{\textrm{max}}^\pm (s_*)= \left( y_{\textrm{max}},\pm \frac{b+1}{\sqrt{b}}(y_{\textrm{max}}- y_{\textrm{min}})/2 \right) \,. \end{aligned}$$
(B9)
Since the pairs
\((\alpha _{\textrm{min}}^\pm (s),\beta _{\textrm{min}}^\pm (s))\) and
\((\alpha _{\textrm{max}}^\pm (s),\beta _{\textrm{max}}^\pm (s))\) satisfy the equality in Eq. (
5.8) for all
\(s\in (0,s_*)\), the triples
\((y_*^\pm ,y_{*\textrm{min}}^\pm ,y_{*\textrm{max}}^\pm )\) also satisfy Eq. (
5.8). Therefore, we define
\(f_\textrm{a}\) for
y in one of the triangles
\(\text {Conv}(\{y_*^+, y_{*\textrm{min}}^+, y_{*\textrm{max}}^+\})\cup \text {Conv}(\{y_*^-, y_{*\textrm{min}}^-, y_{*\textrm{max}}^-\})\) (
\(Y_3\) in Fig.
5) according to Eq. (
5.10) as the unique affine function which coincides with
f in the vertices of each of these triangles,
$$\begin{aligned} f_\textrm{a}(y)= \left\{ \begin{array}{cl} (T_{y_*^+}f)(y)&{}\text { on }\text {Conv}(\{y_*^+, y_{*\textrm{min}}^+, y_{*\textrm{max}}^+\})\,,\\ (T_{y_*^-}f)(y)&{}\text { on }\text {Conv}(\{y_*^-, y_{*\textrm{min}}^-, y_{*\textrm{max}}^-\})\,. \end{array} \right. \end{aligned}$$
(B10)
Finally, consider a pair of two different points
\((\widetilde{y}^{(1)}, \widetilde{y}^{(2)})\in \partial \widetilde{Y}\) satisfying Eq. (
5.8). For the existence of a supporting hyperplane of
f, which touches the graph of
f in
\(\widetilde{y}^{(1)}\) and
\(\widetilde{y}^{(2)}\), we require equal partial derivatives in the second component, i.e.,
$$\begin{aligned} \widetilde{y}^{(1)}_2= \frac{b+1}{b} \partial _2 f(\widetilde{y}^{(1)})= \frac{b+1}{b} \partial _2 f(\widetilde{y}^{(2)})= \widetilde{y}^{(2)}_2\,. \end{aligned}$$
(B11)
With
\(\widetilde{y}^{(1)}_1, \widetilde{y}^{(2)}_1\in \{y_{\textrm{min}},y_{\textrm{max}}\}\) and
\(\widetilde{y}^{(1)}\ne \widetilde{y}^{(2)}\) we get
\(\widetilde{y}^{(1)}= y_{\textrm{min}}\) and
\(\widetilde{y}^{(2)}_1= y_{\textrm{max}}\) and obtain for
\(s> 0\) the pairs of curves
$$\begin{aligned} \alpha _\infty ^\pm (s):= (y_{\textrm{min}},\pm s)\,,\quad \beta _\infty ^\pm (s):= (y_{\textrm{max}},\pm s)\,. \end{aligned}$$
(B12)
Since we already constructed our candidate
\(f_\textrm{a}\) for the convex envelope on the set
\((y_{\textrm{min}},y_{\textrm{max}})\times [- \frac{b+1}{\sqrt{b}} s_*, \frac{b+1}{\sqrt{b}} s_*]\), we restrict
s to have values
\(s> \frac{b+1}{\sqrt{b}} s_*\). For each
\(y\in \mathbb {R}^2\), which lies on a connecting straight line of two corresponding points of the curves
\(\alpha _\infty ^\pm \) and
\(\beta _\infty ^\pm \), i.e.,
\(y= t\cdot \alpha _\infty ^\pm (s)+ (1-t)\cdot \beta _\infty ^\pm (s)\) with
\(s\in (\frac{b+1}{\sqrt{b}} s_*,\infty )\) and
\(t\in [0,1]\) (represented by the two unbounded sets forming
\(Y_4\) in Fig.
5), we define
\(f_\textrm{a}(y)= t\cdot f(\alpha _\infty ^\pm (s))+ (1-t)\cdot f(\beta _\infty ^\pm (s))\) according to Eq. (
5.9).
After these preparations, we turn to the proof of the theorem and divide it into several steps. Denote by
\(f_\textrm{a}\) the formula on the right-hand side in the assertion of the theorem in (
5.15). We are going to show in Step 1 that
\(f_\textrm{a}\) is differentiable on
\(\mathbb {R}^2{\setminus } (\{y_{\textrm{min}},y_{\textrm{max}}\}\times \mathbb {R})\) and convex. In Step 2 we verify that
\(f_\textrm{a}\) is indeed the expression we obtain from the previously described construction and
\(f_\textrm{c}\le f_\textrm{a}\). Finally in Step 3 we prove
\(f_\textrm{a}\le f\) and conclude that
\(f_\textrm{a}\) coincides with
\(f_\textrm{c}\) by
\(f_\textrm{c}\le f_\textrm{a}\le f\) and
\(f_\textrm{a}\) being convex.
Step 1: Differentiability and convexity of \(f_\textrm{a}\). The differentiability of
\(f_\textrm{a}\) on
\(\text {int}(\widetilde{Y})\cup (y_{\textrm{min}},y_{\textrm{max}})\times \mathbb {R}\) follows from
$$\begin{aligned} \nabla f_\textrm{a}(y) = \left\{ \begin{array}{cl} \displaystyle (y_1, \frac{b}{b+1}y_2) &{} \text { on }\text {int}(\widetilde{Y})\,,\\ [.1in] \displaystyle (y_1, y_2) &{} \text { on } Y_1\,, \\ [.1in] \displaystyle \left( \begin{array}{c} y_1+ \frac{\text {sign}(y_{\textrm{mid}}- y_1) \sqrt{b}}{b+1}(\vert y_2\vert - r_0(y_1))\\ \frac{b}{b+1} y_2+ \frac{\text {sign}(y_2) r_0(y_1)}{b+1} \end{array} \right) ^\textrm{T} &{} \text { on }Y_2\,,\\ [.1in] \displaystyle \left( y_{\textrm{mid}}, \text {sign}(y_2) \sqrt{b}\cdot \frac{y_{\textrm{max}}-y_{\textrm{min}}}{2}\right) &{} \text { on }Y_3\,,\\ [.1in] \displaystyle \left( y_{\textrm{mid}}, \frac{b}{b+1}y_2\right) &{} \text { on }Y_4\,,\end{array} \right. \end{aligned}$$
(B13)
and verifying the continuity of
\(\nabla f_\textrm{a}\) along the boundaries of the different regions. With the notation
\(e_1\odot e_2:= \frac{1}{2} (e_1\otimes e_2+ e_2\otimes e_1)\), the second derivatives are given by
$$\begin{aligned} \nabla ^2 f_\textrm{a}(y) = \left\{ \begin{array}{cl} \displaystyle e_1\otimes e_1+ \frac{b}{b+1}e_2\otimes e_2 &{} \text { on }\text {int}(\widetilde{Y})\,,\\ [.1in] \displaystyle e_1\otimes e_1+ e_2\otimes e_2 &{} \text { on } \text {int}(Y_1)\,, \\ [.1in] \displaystyle \frac{1}{b+1}\cdot \left( e_1\otimes e_1+ \frac{2 \sqrt{b} \text {sign}(y)}{\text {sign}(y_{\textrm{mid}}- y_1)}e_1\odot e_2 + b e_2\otimes e_2 \right) &{} \text { on }\text {int}(Y_2)\,,\\ [.1in] \displaystyle 0 &{} \text { on }\text {int}(Y_3)\,,\\ [.1in] \displaystyle \frac{b}{b+1}e_2\otimes e_2 &{} \text { on }\text {int}(Y_4)\,. \end{array} \right. \end{aligned}$$
(B14)
In fact,
\(\nabla ^2 f_\textrm{a}\) is positively semidefinite on the interior of each domain
\(\widetilde{Y}, Y_1, Y_2, Y_3, Y_4\), consequently
\(f_\textrm{a}\) is locally convex on those domains. Since
\(f_\textrm{a}\) is differentiable on
\((y_{\textrm{min}},y_{\textrm{max}})\times \mathbb {R}\), an iterative application of Lemma
7 gives us the local convexity (and hence the convexity by Lemma
10) of
\(f_\textrm{a}\) on
\((y_{\textrm{min}},y_{\textrm{max}})\times \mathbb {R}\). A short calculation shows that
\(g: \mathbb {R}^2\rightarrow \mathbb {R},\ g(y)= \frac{1}{2}y_1^2+ \frac{1}{2}\frac{b}{b+1}y_2^2\) is a convex lower bound of
\(f_\textrm{a}\), and since
\(f_\textrm{a}\) and
g coincide on
\(\widetilde{Y}\) and
\(f_\textrm{a}\) is convex on
\(\mathbb {R}{\setminus } \widetilde{Y}= (y_{\textrm{min}},y_{\textrm{max}})\times \mathbb {R}\), the convexity of
\(f_\textrm{a}\) follows by Lemma
11.
Step 2: \(f_\textrm{c}\le f_\textrm{a}\). Since the calculation of the explicit formula of the candidate
\(f_\textrm{a}\) derived from the previously described construction is extensive, we just prove that the given formula is the one, which is obtained by this construction. Refer to Fig.
5 for a sketch of the various domains.
On \(\widetilde{Y}\cup Y_1\): Since by assumption \(r_0\le r\) and hence \(Y_1\subset Y_0\), \(f_\textrm{a}\) coincides with f in \(Y_1\) and in \(\widetilde{Y}\) and \(f_\textrm{c}\le f= f_\textrm{a}\), as asserted.
On \(Y_2\): The set
\(Y_2\) is exactly the set of points, which are covered by the straight line segments connecting the corresponding pairs of points on the four pairs of curves
\((\alpha _{\textrm{min}}^\pm , \beta _{\textrm{min}}^\pm )\) and
\((\alpha _{\textrm{max}}^\pm ,\beta _{\textrm{max}}^\pm )\). Assume
\(y\in Y_2\) with
\(y_1< y_{\textrm{mid}}\) and
\(y_2 > 0\) (the other cases are analogous). There exists
\(s\in (0,(y_{\textrm{max}}- y_{\textrm{min}})/2)\) and
\(t\in (0,1)\) with
\(y= t\cdot \alpha _{\textrm{min}}^+(s)+ (1-t)\cdot \beta _{\textrm{min}}^+(s)\). To show that the given expression for
\(f_\textrm{a}(y)\) results from the construction described above, we notice that
\(f_\textrm{a}(\alpha _{\textrm{min}}^+(s))= f(\alpha _{\textrm{min}}^+(s))\) and
\(f_\textrm{a}(\beta _{\textrm{min}}^+(s))= f(\beta _{\textrm{min}}^+(s))\) and that
\(f_\textrm{a}\) is affine along the connecting line between
\(\alpha _{\textrm{min}}^+(s)\) and
\(\beta _{\textrm{min}}^+(s)\). This can be seen by
\(\beta _{\textrm{min}}^+(s)- \alpha _{\textrm{min}}^+(s)=(s-y_{\textrm{min}})\cdot (-1,\sqrt{b}^{-1})^T\) and using
\(y_2\),
\(y_{\textrm{mid}}- y_1> 0\) to obtain
$$\begin{aligned} (-1,\sqrt{b}^{-1}) \nabla ^2 f_\textrm{a}(y) \begin{pmatrix} -1 \\ \sqrt{b}^{-1} \end{pmatrix}= \frac{1}{b+1} (-1,\sqrt{b}^{-1}) \begin{pmatrix} 1 &{} \sqrt{b} \\ \sqrt{b} &{} b \end{pmatrix} \begin{pmatrix} -1 \\ \sqrt{b}^{-1}\end{pmatrix}= 0\,. \end{aligned}$$
(B15)
Since
\(f_\textrm{c}\) must lie below the linear interpolation, we obtain
\(f_\textrm{c}\le f_\textrm{a}\).
On \(Y_3\): By construction,
\(f_\textrm{a}\) and
f have the same values in
\(y_*^\pm , y_{*\textrm{min}}^\pm , y_{*\textrm{max}}^\pm \) and on
\(\text {Conv}(y_*^+, y_{*\textrm{min}}^+, y_{*\textrm{max}}^+)\) and
\(\text {Conv}(y_*^-, y_{*\textrm{min}}^-, y_{*\textrm{max}}^-)\) the function
\(f_\textrm{a}\) is affine due to
\(\nabla ^2 f_\textrm{a}= 0\) on
\(\text {int}(Y_3)\). By
\(y_*^\pm \in Y_0\),
f is differentiable in
\(y_*^\pm \) with
\(\nabla f(y_*^\pm )= \nabla f_\textrm{a}(y_*^\pm )\). Now it follows in view of Eq. (
B10) that
\(f_\textrm{a}\) is the above constructed function.
On \(Y_4\): Similar to the argument on \(Y_2\), we recognize that for any \(y\in Y_4\) with \(\pm y_2> 0\) there is an \(s\in (0,\frac{b+1}{\sqrt{b}}(y_{\textrm{max}}- y_{\textrm{min}})/2)\) and \(t\in (0,1)\) with \(y= t\cdot \alpha ^\pm (s)+ (1-t)\cdot \beta ^\pm (s)\). We have \(f_\textrm{a}(\alpha _\infty ^\pm (s))= f(\alpha _\infty ^\pm (s))\) and \(f_\textrm{a}(\beta _\infty ^\pm (s))= f(\beta _\infty ^\pm (s))\), while \(f_\textrm{a}\) is affine along the connecting line between \(\alpha _\infty ^\pm (s)\) and \(\beta _\infty ^\pm (s)\) since \(\beta _\infty ^\pm (s)- \alpha _\infty ^\pm (s)= (y_{\textrm{max}}-y_{\textrm{min}},0)\) and in the formula of \(f_\textrm{a}\) on \(Y_4\) the quadratic term in \(y_1\) cancels out. Finally, we get \(f_\textrm{c}\le f_\textrm{a}\) since any value of \(f_\textrm{a}\) is a convex combination of two or three function values of f.
Step 3: The inequality \(f_\textrm{a}\le f\). We show
\(f_\textrm{a}\le f^{(r_0)}\le f\), where the second inequality is an immediate consequence of Lemma
1. For the first inequality, recognize that
\(f_\textrm{a}= f^{(r_0)}\) holds on
\(\widetilde{Y}\cup Y_1\cup Y_2\) and the inequality only has to be shown on
\(Y_3\) and
\(Y_4\).
On
\(Y_3\), the estimate follows by
$$\begin{aligned} f_\textrm{a}(y)= f^{(r_0)}(y)- \frac{1}{2} \frac{b}{b+1} \left( \vert y_2\vert - \frac{b+1}{\sqrt{b}} \frac{y_{\textrm{max}}-y_{\textrm{min}}}{2}+ \frac{r_0(y_1)}{b} \right) ^2\,. \end{aligned}$$
(B16)
On
\(Y_4\), we can calculate
$$\begin{aligned} f^{(r_0)}(y)- \left( \frac{1}{2}y_1^2+ \frac{1}{2}\frac{b}{b+1}y_2^2 \right)&= \frac{y_2^2}{2(b+1)}- \frac{(\vert y_2\vert - r_0(y_1))_+^2}{2(b+1)} \end{aligned}$$
(B17)
$$\begin{aligned}&\ge - \frac{r_0(y_1)^2}{2(b+1)}+ \frac{r_0(y_1)}{b+1}\vert y_2\vert \end{aligned}$$
(B18)
$$\begin{aligned}&\ge - \frac{r_0(y_1)^2}{2b}+ \frac{r_0(y_1)}{\sqrt{b}}\frac{y_{\textrm{max}}-y_{\textrm{min}}}{2}\,. \end{aligned}$$
(B19)
For
\(y_1\in (y_{\textrm{min}},y_{\textrm{mid}}]\), we have
$$\begin{aligned} - \frac{r_0(y_1)^2}{2b}+ \frac{r_0(y_1)}{\sqrt{b}}\frac{y_{\textrm{max}}-y_{\textrm{min}}}{2}&= - \frac{b(y_1-y_{\textrm{min}})^2}{2b}+ \frac{\sqrt{b}(y_1-y_{\textrm{min}})}{\sqrt{b}}\cdot \frac{y_{\textrm{max}}-y_{\textrm{min}}}{2} \end{aligned}$$
(B20)
$$\begin{aligned}&= \frac{1}{2}(y_1-y_{\textrm{min}})(y_{\textrm{max}}-y_1) \end{aligned}$$
(B21)
and for
\(y_1\in (y_{\textrm{mid}},y_{\textrm{max}})\) we have
$$\begin{aligned} - \frac{r_0(y_1)^2}{2b}+ \frac{r_0(y_1)}{\sqrt{b}}\frac{y_{\textrm{max}}-y_{\textrm{min}}}{2}&= - \frac{b(y_{\textrm{max}}-y_1)^2}{2b}+ \frac{\sqrt{b}(y_{\textrm{max}}-y_1)}{\sqrt{b}}\cdot \frac{y_{\textrm{max}}-y_{\textrm{min}}}{2} \end{aligned}$$
(B22)
$$\begin{aligned}&= \frac{1}{2}(y_1-y_{\textrm{min}})(y_{\textrm{max}}-y_1)\,, \end{aligned}$$
(B23)
which gives us the desired estimate on
\(Y_4\).