Variational convergence of the Scharfetter–Gummel scheme to the aggregation-diffusion equation and vanishing diffusion limit

Abstract

In this paper, we explore the convergence of the semi-discrete Scharfetter–Gummel scheme for the aggregation-diffusion equation using a variational approach. Our investigation involves obtaining a novel gradient structure for the finite volume space discretization that works consistently for any non-negative diffusion constant. This allows us to study the discrete-to-continuum and zero-diffusion limits simultaneously. The zero-diffusion limit for the Scharfetter–Gummel scheme corresponds to the upwind finite volume scheme for the aggregation equation. In both cases, we establish a convergence result in terms of gradient structures, recovering the Otto gradient flow structure for the aggregation-diffusion equation based on the 2-Wasserstein distance.

Appendices

The $\cosh$-gradient structure, its tilt-dependence and its de-tilting

We link in this appendix a different possible gradient structure based on dissipation potentials based on the ‘$\cosh$’ to the gradient structure defined by the dual dissipation potential (1.7) and discussed in Sect. 3.3. We will see in Sect. A.1 that in the ‘$\cosh$’ case, the edge activity $\vartheta ^{h,\rho }$ depends on the potentials $V^h$, $W^h$ and hence also $\rho ^h$. The latter dependency would enforce stronger assumptions on the tesselation. Additionally, we point out that those dependencies are scaled by factors of $\epsilon ^{-1}$ and hence become singular in the vanishing diffusion limit $\epsilon \rightarrow 0$. Hence, one expects additional complications in proving EDP convergence for such a gradient structure.

Also from the modeling point of view, this dependence of the dissipation potential on the driving energy can be considered a drawback and unphysical, which we discuss in Sect. A.2. An in-depth discussion of tilt-dependent gradient systems, where changes in the driving energy can lead to changes in the dissipation potential, is carried out in [35]. Fortunately for the Scharfetter–Gummel scheme, it is possible to derive a tilt-independent gradient structure, which is better suited for proving EDP convergence uniform in the diffusivity $\epsilon \ll 1$. We present a de-tilting of the ‘$\cosh$’ dissipation potential towards a tilt-independent dissipation potential in Sect. A.3.

1.1 The $\cosh$-gradient structure

We show that the Scharfetter–Gummel scheme (${{{\textbf {SGE}}}_h}$) defines a random walk that possesses a ‘cosh’ gradient structure by following a similar construction of [24]. To handle the nonlocal aggregation part in the energy, we follow the strategy introduced in [16] and define a local equilibrium to arrive at a suitable gradient flow formulation incorporating, such that the scheme would indeed fit into the frame developed in [24]. From the discrete energy $\mathcal {E}_h$ given in (1.6), we identify the discrete gradient of its variational derivative as

$$\begin{aligned} \overline{\nabla }\mathcal {E}'_{\epsilon ,h}(\rho ^h) = \epsilon \overline{\nabla }\bigl (\log \rho ^h - \log \pi ^{\epsilon ,h,\rho } \bigr ), \end{aligned}$$

(A.1)

with

$$\begin{aligned} \pi ^{\epsilon ,h,\rho }_K = \frac{\vert {K}\vert e^{- \textsf {Q}_K^{h,\rho }/\epsilon }}{Z^{\epsilon ,h,\rho }},\qquad \textsf {Q}_K^{h,\rho } = V_K^h + \sum _{M\in \mathcal {T}^h} W_{KM}^h \rho _M^h, \end{aligned}$$

(A.2)

and $Z^{\epsilon ,h,\rho } = \sum _{K\in \mathcal {T}^h} \vert {K}\vert e^{-\textsf {Q}_K^{h,\rho }/\epsilon }$ is the normalization such that $\pi ^{\epsilon ,h,\rho } \in \mathcal {P}(\mathcal {T}^h)$. We note that the contribution of the variation of $Z^{\epsilon ,h,\rho }$ in (A.1) cancels due to the discrete gradient.

The ‘cosh’ dual dissipation potential is given for all $\rho ^h \in \mathcal {P}(\mathcal {T}^h)$ and $\xi ^h \in \mathcal {B}(\Sigma ^h)$ by

$$\begin{aligned} \overline{\mathcal {R}}_{\epsilon ,h}^*(\rho ^h, \xi ^h) = \frac{1}{2} \sum _{(K,L)\in \Sigma ^h} \Psi ^*_\epsilon (\xi ^h_{K|L}) \sqrt{{\bar{u}}^h_K {\bar{u}}^h_L} \, \kappa ^{\epsilon ,h,\rho }_{K|L} \pi ^{\epsilon ,h,\rho }_K, \qquad {\bar{u}}_K^h = \frac{\rho _K^h}{\pi ^{\epsilon ,h,\rho }_K},\nonumber \\ \end{aligned}$$

(A.3)

where $\Psi _\epsilon ^*(s) = 4 \epsilon ^2 (\cosh (s/2 \epsilon ) - 1)$. The idea is then to choose a jump kernel $\kappa ^{\epsilon ,h,\rho }: \Sigma ^h \rightarrow [0, \infty )$ in such a way that it satisfies the local detailed balance condition

$$\begin{aligned} \kappa ^{\epsilon ,h,\rho }_{K|L} \pi ^{\epsilon ,h,\rho }_K = \kappa ^{\epsilon ,h,\rho }_{L|K} \pi ^{\epsilon ,h,\rho }_L \qquad \text { for all } (K,L)\in \Sigma ^h \text { and all } \rho ^h \in \mathcal {P}(\mathcal {T}^h). \end{aligned}$$

(A.4)

and allows representing the flux in the gradient form (3.4).

One possibility is to define the jump kernel as

$$\begin{aligned} \kappa ^{\epsilon ,h,\rho }_{K|L} :=\frac{1}{|K|}\frac{\tau _{K|L}^h}{\exp \big ({-\textsf {Q}_K^{h,\rho }/\epsilon }\big )} \frac{2\,q_{K|L}^h / \epsilon }{\exp (\textsf {Q}^{h,\rho }_L / \epsilon ) - \exp (\textsf {Q}^{h,\rho }_K / \epsilon )}, \qquad (K,L)\in \Sigma ^h,\nonumber \\ \end{aligned}$$

(A.5)

where we recall that $\tau _{K|L}^h:=|(K|L)| / |x_L - x_K|$ is the transmission coefficient and

$$\begin{aligned} q_{K|L}^h :=V^h_L - V^h_K + \sum _{M\in \mathcal {T}^h} \rho ^h_M (W^h_{ML} - W^h_{MK}) = \textsf {Q}_L^{h,\rho }-\textsf {Q}_K^{h,\rho },\qquad (K,L)\in \Sigma ^h.\nonumber \\ \end{aligned}$$

(A.6)

Notice that the pair $(\kappa ^{\epsilon ,h,\rho }, \pi ^{\epsilon ,h,\rho })$ satisfies the local detailed balance condition (A.4), since $\tau _{K|L}^h = \tau _{L|K}^h$ and $q_{K|L}^h = - q_{L|K}^h$. The edge conductivity is then given by

$$\begin{aligned} \vartheta ^{\epsilon ,h,\rho }_{K|L} :=\frac{\tau _{K|L}^h}{Z^{\epsilon ,h,\rho }} \frac{2\, q_{K|L}^h / \epsilon }{\exp (\textsf {Q}^{h,\rho }_L / \epsilon ) - \exp (\textsf {Q}^{h,\rho }_K / \epsilon )}. \end{aligned}$$

(A.7)

The kernel defined in (A.5) satisfies the bound

$$\begin{aligned} \sup _{h>0} \sup _{K\in \mathcal {T}^h} h^2 \sum _{L\in \mathcal {T}^h_K} \kappa ^{\epsilon ,h,\rho }_{K|L} < \infty ,\qquad \text {where}\;\;\mathcal {T}^h_K:=\bigl \{L\in \mathcal {T}^h: (K,L)\in \Sigma ^h\bigr \}, \nonumber \\ \end{aligned}$$

(A.8)

provided $\{(\mathcal {T}^h, \Sigma ^h)\}_{h>0}$ satisfy (A$_\mathcal {T}$). Indeed, for any $(K,L)\in \Sigma ^h$, it holds that

$$\begin{aligned} \kappa ^{\epsilon ,h,\rho }_{K|L} = \frac{|(K|L)|}{|K||x_K-x_L|} \frac{2\, q_{K|L}^h/\epsilon }{\exp (q_{K|L}^h/\epsilon ) - 1 } \le \frac{C_{d-1} h^{d-1}}{C_d \zeta ^{d+1} h^{d+1}} \Big (2 - \frac{q_{K|L}^h}{\epsilon } + o(h) \Big ) = O(h^{-2}). \end{aligned}$$

It is not difficult to see that the non-degeneracy assumption (A$_\mathcal {T}$) implies that [24]

$$\begin{aligned} \sup _{h>0} \sup _{K\in \mathcal {T}^h} \#\mathcal {T}^h_K < \infty , \end{aligned}$$

and thus also the asserted bound (A.8). However, we point out that the bound is non-uniform in $\epsilon \ll 1$.

To apply the strategy from [24] directly, it is left to show that the choice of $\kappa ^{\epsilon ,h,\rho }$ in (A.5) indeed gives rise to the Scharfetter–Gummel flux (1.2).

Lemma A.1

For any $\rho ^h\in \mathcal {P}(\mathcal {T}^h)$, $K\in \mathcal {T}^h$, and $(K,L)\in \Sigma ^h$, we have the identity (3.4), where $\mathcal {J}^{h,\rho }$ is the Scharfetter–Gummel flux given in (1.2) and $\overline{\mathcal {R}}_{\epsilon ,h}^*$ is the ‘cosh’ dual dissipation potential with edge conductivity $\vartheta ^{\epsilon ,h,\rho }$ defined in (A.7).

In particular, the Scharfetter–Gummel scheme (${{{\textbf {SGE}}}_h}$) possesses the ‘cosh’ gradient flow structure with (1.6) as the driving energy.

Proof

We begin by rewriting the Scharfetter–Gummel flux in (1.2) using the density ${\bar{u}}^h = \text {d} \rho ^h / \text {d} \pi ^{\epsilon ,h,\rho }$ with the reference measure $\pi ^{h,\rho }$ depending on $\textsf {Q}^{h,\rho }$:

$$\begin{aligned} \mathcal {J}_{K|L}^{h,\rho } = \frac{\epsilon \tau _{K|L}^h}{Z^{h,\rho }} \left( \mathfrak {b}( q_{K|L}^h / \epsilon ) \,{\bar{u}}^h_K e^{-\textsf {Q}^{h,\rho }_K/ \epsilon } - \mathfrak {b}(- q_{K|L}^h / \epsilon )\, {\bar{u}}^h_L e^{-\textsf {Q}^{h,\rho }_L/ \epsilon } \right) . \end{aligned}$$

(A.9)

The expression (A.9) can be simplified, since

$$\begin{aligned} \mathfrak {b}(q_{K|L}^h / \epsilon )\exp (-\textsf {Q}^{h,\rho }_K / \epsilon ) = \frac{q_{K|L}^h \exp (-\textsf {Q}^{h,\rho }_K / \epsilon )}{\epsilon \left( \exp (q_{K|L}^h / \epsilon ) - 1 \right) } = \frac{ q_{K|L}^h}{\epsilon \big ({\exp (\textsf {Q}^{h,\rho }_L / \epsilon ) - \exp (\textsf {Q}^{h,\rho }_K / \epsilon ) }\big )} \end{aligned}$$

and, similarly,

$$\begin{aligned} \mathfrak {b}(-q_{K|L}^h / \epsilon ) \exp (-\textsf {Q}^{h,\rho }_L / \epsilon ) = \frac{ q_{K|L}^h}{\epsilon \big ({\exp (\textsf {Q}^{h,\rho }_L / \epsilon ) - \exp (\textsf {Q}^{h,\rho }_K / \epsilon ) }\big )}, \end{aligned}$$

therefore

$$\begin{aligned} \mathcal {J}_{K|L}^{h,\rho } = \frac{\tau _{K|L}^h}{Z^{h,\rho }} \frac{ q_{K|L}^h}{\exp (\textsf {Q}^{h,\rho }_L / \epsilon ) - \exp (\textsf {Q}^{h,\rho }_K / \epsilon )} \left( { {\bar{u}}^h_K - {\bar{u}}^h_L }\right) = \frac{\epsilon }{2}\left( { {\bar{u}}^h_K - {\bar{u}}^h_L }\right) \vartheta _{K|L}^{\epsilon ,h,\rho }. \end{aligned}$$

On the other hand, we note that for every $(K,L)\in \Sigma ^h$ and $\xi ^h\in \mathcal {B}(\Sigma ^h)$:

$$\begin{aligned} D_2\overline{\mathcal {R}}^*_{\epsilon ,h}(\rho ^h,\xi ^h)(K,L) = \epsilon \sinh \left( {\frac{\xi ^h_{K|L}}{2\epsilon }}\right) \sqrt{{\bar{u}}^h_K {\bar{u}}^h_L} \,\vartheta ^{\epsilon ,h,\rho }_{K|L}. \end{aligned}$$

Recall from (A.1) and (A.3) that $\overline{\nabla }\mathcal {E}'_{\epsilon ,h}(\rho ^h) = \epsilon \overline{\nabla }\log ({\bar{u}}^h)$. Inserting $\xi ^h = - \overline{\nabla }\mathcal {E}'_{\epsilon ,h}(\rho ^h)$, we obtain

$$\begin{aligned} D_2 \overline{\mathcal {R}}_{\epsilon ,h}^* \big (\rho ^h, - \overline{\nabla }\mathcal {E}'_{\epsilon ,h}(\rho ^h)\bigr ) (K, L)&= \epsilon \sinh \bigg ({\frac{1}{2} \log \frac{{\bar{u}}^h_K}{{\bar{u}}^h_L}} \bigg )\sqrt{{\bar{u}}^h_K {\bar{u}}^h_L} \,\vartheta ^{\epsilon ,h,\rho }_{K|L} = \mathcal {J}_{K|L}^{h,\rho }, \end{aligned}$$

i.e. identity (3.4) holds as asserted. $\square$

Remark A.2

Since the classical Scharfetter–Gummel scheme has the ‘cosh’ gradient-flow formulation, one can ask if it is possible to use the framework of [24] to prove the convergence. The necessary assumptions on the invariant measure $\pi ^{\epsilon ,h,\rho }$ and the jump intensities $\kappa ^{\epsilon ,h,\rho }$ hold true based on the notion of local detailed balance as defined in (A.4). However, the zero-local-average assumption

$$\begin{aligned} \sum _{L\in \mathcal {T}^h_K} \vartheta ^{\epsilon ,h,\rho }_{K|L} (x_K - x_L) = 0\qquad \text {for all }K\in \mathcal {T}^h\text { with }\overline{K} \cap \partial \Omega = \emptyset \text { does not hold.}\nonumber \\ \end{aligned}$$

(A.10)

In addition, the nonlinear dependency of $\vartheta ^{\epsilon ,h,\rho }$ on $\rho$ seems to make satisfying (A.10), even only asymptotically, very hard and may require strong assumptions on the tessellations to work around.

As a last remark, we emphasize that the edge conductivity $\vartheta ^{\epsilon ,h,\rho }$ defined in (A.7) depends non-uniformly on the diffusion parameter $\epsilon >0$, which makes it difficult to pass to the limit $\epsilon \rightarrow 0$.

1.2 The tilt-dependence of the $\cosh$-gradient structure

The disadvantages of the ‘cosh’ gradient structure mentioned earlier are due to tilt-dependence as defined in [35]. To clarify this further, we decompose the free energy into entropy and potential energies by writing

$$\begin{aligned} \mathcal {E}_{\epsilon ,h}(\rho ^h) = \epsilon \mathcal {S}_h(\rho ^h) + \mathcal {V}_h^V(\rho ^h) + \mathcal {W}_h^W(\rho ^h), \end{aligned}$$

(A.11)

where $V^h:\mathcal {T}_h \rightarrow \mathbb {R}$ and $W^h:\mathcal {T}_h\times \mathcal {T}_h\rightarrow \mathbb {R}$ symmetric are given and we set

$$\begin{aligned} & \mathcal {S}_h(\rho _h) :=\sum _{K\in \mathcal {T}^h} \phi (u^h_K )|K|, \qquad \text { where } \qquad u^h_K :=\frac{\rho ^h_K}{|K|}; \\ & \mathcal {V}_h^V(\rho ^h) :=\sum _{K\in \mathcal {T}^h} V^h_K \rho ^h_K \qquad \text {and}\qquad \mathcal {W}_h^W(\rho ^h) :=\frac{1}{2} \sum _{K, L\in \mathcal {T}^h\times \mathcal {T}^h} W^h_{KL} \rho ^h_K \rho ^h_L. \end{aligned}$$

Then, we can provide a gradient structure for the Scharfetter–Gummel scheme for all possible potential energies $V^h$ and interaction energies $W^h$ altogether by introducing the set of tilts

$$\begin{aligned} \mathcal {F}_h :=\Big \{ \mathcal {V}_h^V + \mathcal {W}_h^W \,\Big |\, V^h: \mathcal {T}^h \rightarrow \mathbb {R},\; W^h: \mathcal {T}^h\times \mathcal {T}^h \rightarrow \mathbb {R}\text { symmetric}\Big \}.\nonumber \\ \end{aligned}$$

(A.12)

We can then recast Lemma A.1 as a derivation of a gradient structure with tilting [35, Definition 1.16] of the type $(\mathcal {T}^h,\Sigma ^h,\overline{\nabla },\mathcal {S}_h,\overline{\mathcal {R}}_{\epsilon ,h},\mathcal {F}_h)$. By recalling that for $\mathcal {V}_h^V + \mathcal {W}_h^W \in \mathcal {F}_h$, we find $\textsf {Q}^{h,\rho } = (\mathcal {V}_h^V)'(\rho ^h) + (\mathcal {W}_h^W)'(\rho ^h)$ as defined in (A.2) and obtain from (A.3) the dissipation potential with tilting defined by

$$\begin{aligned} & \overline{\mathcal {R}}_{\epsilon ,h}\left( \rho ^h,j^h;\mathcal {V}_h^V + \mathcal {W}_h^W\right) :=\frac{1}{2}\sum _{(K,L)\in \Sigma ^h} \Psi _\epsilon \left( \frac{j_{K|L}^h}{\sqrt{{\bar{u}}^h_K {\bar{u}}^h_L} \vartheta _{K|L}^{\epsilon ,h,\rho }}\right) \sqrt{{\bar{u}}^h_K {\bar{u}}^h_L} \vartheta _{K|L}^{\epsilon ,h,\rho },\nonumber \\ & \qquad {\bar{u}}_K^h = \frac{\rho _K^h}{\pi ^{\epsilon ,h,\rho }}. \end{aligned}$$

(A.13)

In particular, it depends on the potential energies $V^h,W^h$ through $\vartheta ^{\epsilon ,h,\rho }$ defined in (A.7) and hence is tilt-dependent. Its undesirable properties explained in Remark A.2 are a direct consequence of the dependency of the gradient structure on the potentials and the diffusivity $\epsilon >0$.

1.3 Detilting of the $\cosh$-gradient structure

In this section, we deduce the dual dissipation potential (1.7) from the one including the ‘$\cosh$’ in (A.3) by a de-tilting construction explained in [35, Remark 1.17]. In this way, we prove the following Lemma.

Lemma A.3

The Scharfetter–Gummel with flux-force relation (1.2) is induced by a gradient structure with tilting $(\mathcal {T}^h,\Sigma ^h,\overline{\nabla }, \mathcal {S}_h,\mathcal {R}_{\epsilon ,h},\mathcal {F}_h)$ with tilt set $\mathcal {F}_h$ given in (A.12). Moreover, the dissipation potential $\mathcal {R}_{\epsilon ,h}$ is tilt-independent and given by

$$\begin{aligned} \mathcal {R}_{\epsilon ,h}(\rho ^h, j^h) = 2\sum _{(K,L)\in \Sigma ^h} \tau _{K|L}^h \,\alpha _\epsilon \left( { u^h_K,u^h_L, \frac{j^h_{K|L}}{\tau _{K|L}^h}}\right) , \qquad u_K^h:=\frac{\rho _K^h}{|K|}, \end{aligned}$$

(A.14)

where $\alpha _\epsilon$ is the Legendre dual of $\alpha _\epsilon ^*$ given in (1.8) with respect to the third variable.

Proof

First, we make the tilt-dependence of the ‘$\cosh$’-dual dissipation potential $\overline{\mathcal {R}}_h^*$ from (A.3) explicit, for which use the primal dissipation potential defined in (A.13) and can rewrite (A.3) as

$$\begin{aligned} \overline{\mathcal {R}}^*_{\epsilon ,h}(\rho ^h,\xi ^h; \mathcal {V}_h^V+\mathcal {W}_h^W) = \frac{1}{2} \sum _{(K,L)\in \Sigma ^h} \Psi ^*_\epsilon (\xi ^h_{K|L})\, \sqrt{{\bar{u}}^h_K {\bar{u}}^h_L} \vartheta _{K|L}^{\epsilon ,h,\rho },\qquad {\bar{u}}_K^h = \frac{\rho _K^h}{\pi _K^{\epsilon ,h,\rho }}. \end{aligned}$$

Note, that the tilt-dependence comes through $\vartheta ^{\epsilon ,h,\rho }$ in terms of $\textsf {Q}^{h,\rho }$ defined in (A.7) and  (A.2), respectively. Specifically, we fix $(K,L)\in \Sigma ^h$ and identify $q_{K|L}^h = \overline{\nabla }\textsf {Q}^{h,\rho }$ in $\vartheta ^{\epsilon ,h,\rho }_{K|L}$ from (A.7) to obtain

$$\begin{aligned} \sqrt{{\bar{u}}^h_K {\bar{u}}^h_L} \vartheta _{K|L}^{\epsilon ,h,\rho } = \tau _{K|L}^h \sqrt{u_K^h u_L^h} \frac{\overline{\nabla }\textsf {Q}^{h,\rho }_{K|L}}{\exp \big ({\overline{\nabla }\textsf {Q}^{h,\rho }_{K|L}/(2\epsilon )}\big )-\exp \big ({-\overline{\nabla }\textsf {Q}^{h,\rho }_{K|L}/(2\epsilon )}\big )} . \end{aligned}$$

In this way, we can write

$$\begin{aligned} \overline{\mathcal {R}}^*_{\epsilon ,h}\left( \rho ^h,\xi ^h; \mathcal {V}_h^V+\mathcal {W}_h^W\right) = \frac{1}{2} \sum _{(K,L)\in \Sigma ^h} \overline{\textsf {R}}^*_{\epsilon ,h}\left( K|L,\rho ^h_K,\rho ^h_L,\xi ^h_{K|L};\textsf {Q}^{h,\rho }_K - \textsf {Q}^{h,\rho }_L\right) , \end{aligned}$$

with

$$\begin{aligned} \overline{\textsf {R}}^*_{\epsilon ,h}\left( K|L;\rho ^h_K,\rho ^h_L;\xi ^h_{K|L};q_{K|L}^h\right) :=\Psi ^*_\epsilon (\xi ^h_{K|L})\, \tau _{K|L}^h \sqrt{ u_K^h u_L} \frac{q_{K|L}^h}{e^{q_{K|L}^h/(2\epsilon )}- e^{-q_{K|L}^h/(2\epsilon )}}.\nonumber \\ \end{aligned}$$

(A.15)

Following the construction from [35, Remark 1.17], we can make use of the fact that the solution is given for the specific force $\xi _{K|L}^h=- \epsilon \overline{\nabla }\log \rho ^h(K,L)-q_{K|L}^h$, which is the negative discrete gradient of the discrete free energy from (A.11). With this, one can define an evolution-equivalent dissipation potential $\tilde{\textsf {R}}^*_{\epsilon ,h}$ (cf. [35, Eqn. (1.64)]) by first differentiating the dual dissipation potential $\overline{\textsf {R}}^*_{\epsilon ,h}$ in (A.15) w.r.t. the $\xi$-variable denoted with $D_3$, then use the substitution for $q_{K|L}^h$ along solutions of the scheme and finally integrate w.r.t. the $\xi$-variable again to obtain

$$\begin{aligned} \tilde{\textsf {R}}^*_{\epsilon ,h}(K|L;\rho ^h_K,\rho ^h_L;\Xi ^h_{K|L}):= & \int \limits _0^{\Xi ^h_{K|L}} D_3 \overline{\textsf {R}}^*_{\epsilon ,h}(K|L;\rho ^h_K,\rho ^h_L;\xi ^h_{K|L};-\xi _{K|L}^h\\ & - \epsilon \overline{\nabla }\log \rho ^h(K,L)) \text {d} \xi _{K|L}^h. \end{aligned}$$

Instead of calculating the integral, we check that $\tilde{\textsf {R}}^*_{\epsilon ,h}$ agrees with $D_2 \mathcal {R}_{\epsilon ,h}^*(\rho ^h,\xi ^h)$ from (1.7), which amounts to verifying the identity

$$\begin{aligned} D_2 \mathcal {R}_{\epsilon ,h}^*(\rho ^h,\xi ^h)(K,L) {\mathop {=}\limits ^{!}} D_2\overline{\mathcal {R}}_{\epsilon ,h}^*\left( {\rho ^h,\xi ^h; -\xi ^h-\epsilon \overline{\nabla }\mathcal {S}_h(\rho )}\right) (K,L). \end{aligned}$$

(A.16)

By substituting once more $\overline{\nabla }\textsf {Q}^{h,\rho }_{K|L} = q_{K|L}^h = -\xi _{K|L}^h- \epsilon \overline{\nabla }\log \rho ^h(K,L)$, which amounts in using the identity (3.6), we observe that

$$\begin{aligned}&D_2\overline{\mathcal {R}}_{\epsilon ,h}^*\left( {\rho ^h,\xi ^h; - \xi ^h -\overline{\nabla }\mathcal {S}_h(\rho )}\right) (K,L)\\&\quad = \epsilon \tau _{K|L}^h \sinh \bigg ({\frac{\xi ^h_{K|L}}{2\epsilon }} \bigg )\sqrt{u_K^h u_L^h}\frac{\log \big ({u^h_K e^{-\xi ^h_{K|L} / 2\epsilon }}\big ) - \log \big ({u^h_L e^{-\xi ^h_{K|L} / 2\epsilon } }\big )}{e^{-\xi ^h_{K|L}/2\epsilon -\overline{\nabla }\log \sqrt{u^h}(K,L) }-e^{\xi _{K|L}/2\epsilon +\overline{\nabla }\log \sqrt{u^h}(K,L)}} \\&\quad = \alpha _\epsilon \left( {u^h_K,u^h_L, \xi ^h_{K|L}/2}\right) = D_2\mathcal {R}_{\epsilon ,h}^*(\rho ^h,\xi ^h)(K,L), \end{aligned}$$

which verifies the claimed identity (A.16). Therefore, the solution property is a consequence of (3.7) and the remaining statements about the tilt-independence in Lemma A.3 follow by construction as argued in [35, Remark 1.17]. $\square$

Properties of the tilted dual dissipation potential

The following lemma contains some properties and an integral representation of the harmonic-logarithm mean $\Lambda _H$ introduced in (1.9).

Definition B.1

(Mean) A function $M:\mathbb {R}_+ \times \mathbb {R}_+ \rightarrow \mathbb {R}_+$ is a mean if it is

• (1) positively one-homogeneous: $M(\lambda s,\lambda t) = \lambda M(s,t)$ for all $s,t\in \mathbb {R}_+$ and $\lambda >0$;

• (2) bounded by $\min \{s,t\}\le M(s,t)\le \max \{s,t\}$ for all $s,t\in \mathbb {R}_+$;

• (3) jointly concave.

Lemma B.2

(Harmonic-logarithmic mean) The logarithmic mean $\Lambda : \mathbb {R}_+ \times \mathbb {R}_+ \rightarrow \mathbb {R}_+$,

$$\begin{aligned} \Lambda (s,t) = \int \limits _0^1 s^\tau t^{1-\tau } \text {d} \tau = {\left\{ \begin{array}{ll} \frac{s-t}{\log s - \log t}, & s\ne t;\\ s, & s=t. \end{array}\right. } \end{aligned}$$

is a mean between the geometric and arithmetic mean

$$\begin{aligned} \sqrt{st } \le \Lambda (s,t) \le \frac{s+t}{2}, \end{aligned}$$

with derivatives bounded

$$\begin{aligned} \partial _1 \Lambda (s,t) = \partial _2\Lambda (t,s) \qquad \text {and}\qquad \partial _1 \Lambda (s,t) = \frac{\Lambda (s,t)(s-\Lambda (s,t))}{s(s-t)} \,. \end{aligned}$$

The harmonic-logarithmic mean $\Lambda _H: \mathbb {R}_+ \times \mathbb {R}_+ \rightarrow \mathbb {R}_+$ defined by

$$\begin{aligned} \Lambda _H(s,t) = \frac{1}{\Lambda \left( {1/s, 1/t}\right) } = \frac{st}{\Lambda (s, t)} \end{aligned}$$

is a mean between the harmonic and geometric mean

$$\begin{aligned} \frac{2}{\frac{1}{s}+ \frac{1}{t}} \le \Lambda _H(s,t) \le \sqrt{st} \end{aligned}$$

with the integral representations

$$\begin{aligned} \Lambda _H(a,b) = \int \limits _0^1 \frac{\text {d} \tau }{\tau /s+ (1-\tau )/t} = \int \limits _0^\infty \frac{s\, t\text {d} \tau }{(\tau +s) (\tau +t)} \end{aligned}$$

and derivatives

$$\begin{aligned} \partial _1 \Lambda _H(s,t)=\partial _2 \Lambda _H(t,s) = \frac{t\left( \Lambda (s, t) - t \right) }{\Lambda (s, t)}. \end{aligned}$$

Proof

See, for instance [8] for many properties of the logarithmic mean, from which the analogous ones of the harmonic-logarithmic mean follow. $\square$

The tilt-independent dual dissipation potential $\mathcal {R}_{\epsilon ,h}^*$ in (1.7) is given in terms of the function $\alpha ^*_\epsilon$ defined in (1.8), which we recall here for convenience

$$\begin{aligned} \alpha _\epsilon ^*(a, b, \xi ) = \epsilon \int \limits _0^\xi \sinh \left( \frac{x}{\epsilon } \right) \Lambda _H (a e^{-x/\epsilon }, b e^{x/\epsilon }) \text {d} x= \epsilon ^2 \alpha _1 \left( a, b, \frac{\xi }{\epsilon } \right) . \end{aligned}$$

Below we prove useful properties of $\alpha _\epsilon ^*$.

Lemma B.3

The function $\alpha _\epsilon ^*:\mathbb {R}_+\times \mathbb {R}_+\times \mathbb {R}\rightarrow \mathbb {R}_+$ in (1.8) has the following useful properties:

• (a) $\alpha _\epsilon ^* (a, b, \xi )$ is convex in $\xi$ for fixed $a,b>0$, with $\min \{a,b\} \le \partial _{\xi }^2 \alpha _\epsilon ^* (a, b, \xi ) \le \max \{a,b\}$;

• (b) $\alpha _\epsilon ^* (a, b, \xi )$ is positively one-homogeneous and jointly concave in (a, b) for fixed $\xi$;

• (c) $\alpha _\epsilon ^*$ satisfies the following bound:

  $$\begin{aligned} \alpha _\epsilon ^* (a, b, \xi ) \le \epsilon ^2 \sqrt{ab} \bigg ( \cosh \bigg ( \bigg | \frac{\xi }{\epsilon } \bigg | \bigg ) - 1 \bigg )= \frac{1}{4}\sqrt{ab}\,\Psi ^*(2\xi ). \end{aligned}$$

  Moreover, the expansion for $\vert {\xi }\vert \ll 1$ is given by

  $$\begin{aligned} \alpha _\epsilon ^*(a,b,\xi ) = \Lambda _H(a,b) \frac{\xi ^2}{2} + O\left( {\frac{\vert {\xi }\vert ^3}{\epsilon }}\right) ; \end{aligned}$$

• (d) It holds that

  $$\begin{aligned} \alpha _\epsilon ^*(a,b,\xi ) \rightarrow \frac{1}{2} \big ( a (\xi ^+)^2 + b (\xi ^-)^2 \big ) =:\alpha _0^*(a,b,\xi ) \qquad \text {as } \epsilon \rightarrow 0 \,, \end{aligned}$$

  where $\xi ^\pm$ is the positive and negative part of $\xi$, respectively. Moreover,

  $$\begin{aligned} | \alpha _\epsilon ^*(a,b,\xi ) - \alpha _0^* (a,b,\xi ) | = O(C_{a, b, \xi } \, \epsilon ), \end{aligned}$$

  where the constant $C_{a, b, \xi } < \infty$ depends on $a, b, \xi$.

• (e) The function $\beta _\epsilon : \mathbb {R}_+\times \mathbb {R}_+\rightarrow \mathbb {R}_+$ defined for the argument $\xi = - \epsilon \log \sqrt{b/a}$ in $\alpha _\epsilon ^*$ has the representation

  $$\begin{aligned} \beta _\epsilon (a, b) :=\alpha _\epsilon ^* (a, b, -\epsilon \log \sqrt{b/a})&= \frac{\epsilon ^2}{4} \int \limits _a^b \frac{ab}{z} \left[ \frac{1}{\Lambda (z,a)} - \frac{1}{\Lambda (z,b)} \right] \text {d} z; \end{aligned}$$

• (f) The function $\beta _\epsilon :\mathbb {R}_+\times \mathbb {R}_+\rightarrow \mathbb {R}_+$ defined in (e) is jointly convex, continuous with

  $$\begin{aligned} \beta _\epsilon (a, 0) :=\frac{\epsilon ^2}{4} \frac{\pi ^2}{6} a \quad \text {and, symmetrically, } \quad \beta _\epsilon (0, b) :=\frac{\epsilon ^2}{4} \frac{\pi ^2}{6} b, \end{aligned}$$

  and satisfies the following bounds:

  $$\begin{aligned} \frac{\epsilon ^2}{4} (\sqrt{a}-\sqrt{b})^2 \le \frac{\epsilon ^2}{4} \frac{\left( {a-b}\right) ^2}{a+b} \le \beta _\epsilon (a, b) \le \frac{\epsilon ^2}{2} (\sqrt{a}-\sqrt{b})^2; \end{aligned}$$

  Moreover, the function $\mathbb {R}_+ \times \mathbb {R}_+ \ni (a, b) \mapsto \beta _\epsilon (a^2, b^2)$ is differentiable.

• (g) The function $\alpha _\epsilon ^*\bigl (a,b,-\epsilon \log \sqrt{b/a} + q / 2 \bigr )$ has the expansion

  $$\begin{aligned} \alpha _\epsilon ^*\left( a,b,-\epsilon \log \sqrt{b/a} + q/2 \right)&= \beta _\epsilon (a, b) + \frac{\epsilon }{4}(a-b)\,q + \frac{q^2}{4} \mathbb {h}_\epsilon (a, b, q) \end{aligned}$$

  with

  $$\begin{aligned} \mathbb {h}_\epsilon (a, b, q) :=\int \limits _0^1 \Bigl [a\, \mathfrak {h}\left( \lambda q/\epsilon \right) + b\,\mathfrak {h}\left( -\lambda q/\epsilon \right) \Bigr ](1-\lambda )\text {d} \lambda ,\qquad \mathfrak {h}(s) = \frac{1}{4}\frac{e^s-1-s}{\sinh ^2(s/2)}. \end{aligned}$$

Proof

(a) From the representation of $\alpha ^*_1$ in terms of the harmonic-logarithmic mean, it follows that

$$\begin{aligned} \partial _\xi \alpha ^*_1(a, b, \xi ) = \sinh (\xi ) \Lambda _H (ae^{-\xi }, b e^\xi ) = \sinh (\xi ) \frac{ab}{\Lambda (ae^{-\xi }, b e^\xi )}. \end{aligned}$$

It also holds

$$\begin{aligned} \partial _\xi ^2 \alpha ^*_1(a,b,\xi )= & \frac{a\, b}{\left( {a e^{-\xi } - b e^{\xi }}\right) ^2} \\ & \left( { a (e^{-2\xi }-1)+ b (e^{2\xi } -1) + (a-b)\left( { \log \frac{e^{-\xi }}{b} - \log \frac{e^{\xi }}{a}}\right) }\right) \,, \end{aligned}$$

which can be rewritten with the help of the function

$$\begin{aligned} g(x) = \frac{x \log x - x +1}{(x-1)^2} \end{aligned}$$

as

$$\begin{aligned} \partial _\xi ^2 \alpha ^*_1(a,b,\xi ) = a \, g\left( {\frac{a}{b}e^{-2\xi }}\right) + b \, g\left( { \frac{b}{a} e^{2\xi }}\right) \,. \end{aligned}$$

The convexity follows now by observing that

$$\begin{aligned} \forall x\in [0,1]: 0\le g(x) \le 1 \quad \text {and}\quad g(x) + g(x^{-1}) = 1 \end{aligned}$$

and hence the bound

$$\begin{aligned} \min \{a,b\} \le \partial _\xi ^2 \alpha ^*_1(a,b,\xi ) \le \max \{ a,b \} \,, \end{aligned}$$

implying the convexity in $\xi$ for fixed $a,b>0$.

(b) The positive one-homogeneity and joint concavity follow from the properties of $\Lambda _H$.

(c) Let $\xi >0$. Using the inequality between the harmonic-logarithmic and geometric mean, we obtain

$$\begin{aligned} \alpha _1 (a, b, \xi ) = \int \limits _0^\xi \sinh (x) \Lambda _H (a e^{-x}, b e^x) \text {d} x \le \int \limits _0^\xi \sinh (x) \sqrt{ab} \text {d} x = \sqrt{ab} \big ( \cosh (\xi ) - 1 \big ). \end{aligned}$$

If $\xi < 0$, then

$$\begin{aligned} \alpha _1 (a, b, \xi ) = \int \limits _0^{|\xi |} \sinh (x) \Lambda _H(ae^x, be^{-x}) \text {d} x \le \sqrt{ab} \big ( \cosh (|\xi |) - 1 \big ). \end{aligned}$$

Combining the two cases and considering $\alpha _\epsilon$, we get

$$\begin{aligned} \alpha _\epsilon (a, b, \xi ) \le \epsilon ^2 \sqrt{ab} \bigg ( \cosh \bigg ( \bigg | \frac{\xi }{\epsilon } \bigg | \bigg ) - 1 \bigg ). \end{aligned}$$

As for the asymptotic expansion, we obtain, by definition of $\alpha ^*_1$,

$$\begin{aligned} \alpha ^*_1(a, b, \xi ) = \partial _\xi ^2 \alpha ^*_1(a, b, \xi ) |_{\xi =0} \frac{\xi ^2}{2} + O\left( |\xi |^3 \right) = \Lambda _H (a, b) \frac{\xi ^2}{2} + O\left( |\xi |^3 \right) . \end{aligned}$$

Then it follows directly that

$$\begin{aligned} \alpha ^*_\epsilon (a, b, \xi ) = \epsilon ^2 \alpha ^*_1 \left( a, b, \frac{\xi }{\epsilon } \right) = \Lambda _H (a, b) \frac{\xi ^2}{2} + O\left( \frac{|\xi |^3}{\epsilon } \right) . \end{aligned}$$

(d) We rewrite $\alpha ^*_\epsilon$ as

$$\begin{aligned} \alpha ^*_\epsilon (a, b, \xi )&=\epsilon ^2 \int \limits _0^{\xi /\epsilon } \sinh (x) \Lambda _H (ae^{-x}, be^x) \text {d} x\\&= \frac{\epsilon ^2}{2} \int \limits _0^{\xi /\epsilon } \big ( \Lambda _H (a, be^{2x}) - \Lambda _H (ae^{-2x}, b) \big ) \text {d} x \\&= \frac{\epsilon }{2} \int \limits _0^\xi \big ( \Lambda _H (a, be^{2x/\epsilon }) - \Lambda _H (ae^{-2x/\epsilon }, b) \big )\text {d} x. \end{aligned}$$

For $x > 0$, it holds that

$$\begin{aligned} \frac{\epsilon }{2} \Lambda _H (a, be^{2x/\epsilon }) = \frac{\epsilon }{2} \frac{ab e^{2x/\epsilon }}{a - be^{2x/\epsilon }} \Big ( \log \frac{a}{b} - \frac{2x}{\epsilon } \Big ) = \frac{ab}{a e^{-2x/\epsilon }- b} \Big ( \frac{\epsilon }{2} \log \frac{a}{b} - x\Big ) \xrightarrow {\epsilon \rightarrow 0} ax, \end{aligned}$$

and

$$\begin{aligned} - \frac{\epsilon }{2} \Lambda _H (ae^{-2x/\epsilon }, b) = - \frac{ab}{a - b e^{2x/\epsilon }} \Big ( \frac{\epsilon }{2} \log \frac{a}{b} - x\Big ) \xrightarrow {\epsilon \rightarrow 0} 0. \end{aligned}$$

For $x < 0$, similarly, we obtain

$$\begin{aligned} \frac{\epsilon }{2} \Big ( \Lambda _H (a, be^{2x/\epsilon }) - \Lambda _H (ae^{-2x/\epsilon }, b) \Big ) \xrightarrow {\epsilon \rightarrow 0} bx. \end{aligned}$$

Combining the two cases yields

$$\begin{aligned} \lim _{\epsilon \rightarrow 0} \alpha ^*_\epsilon (a, b, \xi ) = \mathbbm {1}_{\xi > 0} \int \limits _0^\xi ax \text {d} x + \mathbbm {1}_{\xi < 0} \int \limits _0^\xi bx \text {d} x = \frac{1}{2} \big ( a (\xi ^+)^2 + b (\xi ^-)^2 \big ). \end{aligned}$$

(e) Direct calculation shows

$$\begin{aligned} \beta _\epsilon (a, b)&= \alpha ^*_\epsilon \left( a, b, -\epsilon \log \sqrt{\frac{b}{a}} \right) = \epsilon ^2 \alpha ^*_1 \left( a, b, \log \sqrt{\frac{a}{b}} \right) \\&= \epsilon ^2 \int \limits _0^{\log \sqrt{a/b}} \sinh (x) \Lambda _H(ae^{-x}, be^x) \text {d} x\\&= \frac{\epsilon ^2}{4} \int \limits _1^{a/b} \left( \sqrt{y} - \frac{1}{\sqrt{y}} \right) \frac{ab}{\Lambda \left( \frac{a}{\sqrt{y}}, b\sqrt{y} \right) } \frac{1}{y} \text {d} y \\&= \frac{\epsilon ^2}{4} \int \limits _1^{a/b} \frac{ab}{y} \left[ \frac{1}{\Lambda \left( a/y, b \right) } - \frac{1}{\Lambda \left( a, b y \right) } \right] \text {d} y \\&= \frac{\epsilon ^2}{4} \int \limits _a^b \frac{ab}{z} \left[ \frac{1}{\Lambda (z, a)} - \frac{1}{\Lambda (z, b)} \right] \text {d} y. \end{aligned}$$

(f) The joint convexity of $\beta _\epsilon$ follows from (a) and (b). It is clear that $\beta _\epsilon$ is continuously differentiable in $\mathbb {R}_+ \times \mathbb {R}_+$ since it is defined as an integral of a bounded continuous function. However, on the boundary $\{0\}\times [0, +\infty ) \cup [0, +\infty ) \times \{0\}$ some partial derivatives become $-\infty$. In the case of $(a, b) \mapsto \beta _\epsilon (a^2, b^2)$, the directional derivatives are continuous and bounded:

$$\begin{aligned} 0 \ge \partial _1 \beta _1 (a^2, 1)&= - 2a \int \limits _{a^2}^1 \frac{\log z}{z (z - 1)} \text {d} z \ge - 2a \int \limits _{a^2}^1 \frac{1}{z \sqrt{z}} \text {d} z = 4a \frac{1}{\sqrt{z}}\Big |_{a^2}^1 \\&= 4a \left( 1 - \frac{1}{a} \right) = 4 (a - 1) > -\infty . \end{aligned}$$

As for the bounds, we begin with the upper bound. Using the inequality that the harmonic-logarithmic mean is less or equal to the geometric mean yields

$$\begin{aligned} \beta _\epsilon (a, b) \le \epsilon ^2 \sqrt{ab} \int \limits _0^{-\log \sqrt{b/a}} \sinh (x) \text {d} x&= \epsilon ^2 \sqrt{ab} \left( \cosh \left( -\log \sqrt{b/a} \right) - 1 \right) \\&= \frac{\epsilon ^2}{2}\left( \sqrt{a} - \sqrt{b} \right) ^2. \end{aligned}$$

Tight lower bound. Since $\beta _1$ is positively one-homogeneous it is enough to prove that

$$\begin{aligned} \beta _1(a, 1) \ge \gamma (a) :=\frac{1}{4} \frac{(a-1)^2}{a+1} \qquad \forall a\ge 0. \end{aligned}$$

For $a = 0$ the inequality holds, since $\beta _1(0, 1) = \frac{1}{4} \frac{\pi ^2}{6} \ge \frac{1}{4} = \gamma (0)$. It is left to consider $a > 0$.

We notice that $\beta _1(1, 1) = 0 = \gamma (1)$. Now we aim to compare the derivatives $\partial _a \beta _1(a,1)$ and $\partial _a \gamma (a)$ for $a\in (0,1)$ and $a\in (1,\infty )$. The derivative of $\gamma$ is

$$\begin{aligned} \partial _a \gamma (a) = \frac{1}{4} \frac{(a-1)(a+3)}{(a+1)^2} = \int \limits _1^a \frac{2}{(z+1)^3} \text {d} z \end{aligned}$$

We use the representation of $\beta _1$ from (e) and apply the change of variables $y = z/a$ in the first part of the integral

$$\begin{aligned} \partial _a \beta _1(a, 1)&= \frac{1}{4} \partial _a \left[ \int \limits _1^{1/a} \frac{1}{y \Lambda (y,1)} \text {d} y - a \int \limits _a^1 \frac{1}{z \Lambda (z,1)} \text {d} z \right] \\&= \frac{a}{\Lambda (1/a, 1)} \left( -\frac{1}{a^2} \right) - \int \limits _a^1 \frac{1}{z \Lambda (z,1)} \text {d} z + \frac{1}{\Lambda (a, 1)} \\&= \int \limits _1^a \frac{1}{z \Lambda (z,1)} \text {d} z = \int \limits _1^a \frac{\log z}{z (z - 1)} \text {d} z . \end{aligned}$$

Therefore,

$$\begin{aligned} \partial _a \left( \beta _1(a,1) - \gamma (a) \right) = \int \limits _1^a \left[ \frac{\log z}{z (z - 1)} - \frac{2}{(z+1)^3} \right] \text {d} z. \end{aligned}$$

We are left to show that the integrand is positive, and then the bound follows. For $z>1$, the integrand is positive, if and only if

$$\begin{aligned} \log z \ge \frac{8z(z-1)}{(z+1)^3}, \end{aligned}$$

which can be shown again by comparing the derivatives

$$\begin{aligned} \frac{1}{z} - 8\frac{-z^2 + 4z -1}{(z+1)^4} = \frac{(z-1)^2 (z^2 + 14z + 1)}{z (z+1)^4}> 0 \qquad \forall z >1. \end{aligned}$$

Rough lower bound. This lower bound follows from the inequality between the geometric and arithmetic means

$$\begin{aligned} \frac{(a-b)^2}{a+b} = \left( \sqrt{a} - \sqrt{b} \right) ^2 \left( 1 + \frac{2\sqrt{ab}}{a +b} \right) \ge 2 \left( \sqrt{a} - \sqrt{b} \right) ^2. \end{aligned}$$

(g) We apply the second-order Taylor expansion for a function f:

$$\begin{aligned} f(y) = f(x) + f'(x)(y-x) + (y-x)^2\int \limits _0^1 f''((1-\lambda )x + \lambda y)(1-\lambda )\text {d} \lambda \end{aligned}$$

to expand the function $\alpha ^*_\epsilon$, obtaining

$$\begin{aligned}&\alpha ^*_\epsilon \Big (a,b,-\epsilon \log \sqrt{\frac{b}{a}} + \frac{q}{2} \Big ) = \alpha ^*_\epsilon \Big (a,b,-\epsilon \log \sqrt{\frac{b}{a}} \Big ) + \frac{q}{2} \,\partial _\xi (\alpha ^*_\epsilon ) \Big (a,b,-\epsilon \log \sqrt{\frac{b}{a}} \Big ) \\&\qquad + \frac{q^2}{4} \int \limits _0^1 (\partial _\xi ^2\alpha ^*_\epsilon ) \Big (a,b,-\epsilon \log \sqrt{\frac{b}{a}} + \lambda \frac{q}{2} \Big )(1-\lambda )\text {d} \lambda . \end{aligned}$$

After some manipulation, we find that

$$\begin{aligned} (\partial _\xi \alpha ^*_\epsilon )\Big (a,b,-\epsilon \log \sqrt{\frac{b}{a}} \Big )&= \frac{\epsilon }{2}(a-b),\\ (\partial _\xi ^2\alpha ^*_\epsilon ) \Big (a,b,-\epsilon \log \sqrt{\frac{b}{a}} + \frac{q}{2} \Big )&= a\, \mathfrak {h} \Big (\frac{q}{\epsilon } \Big ) + b\,\mathfrak {h}\left( -\frac{q}{\epsilon }\right) , \end{aligned}$$

with

$$\begin{aligned} \mathfrak {h}(s) = \frac{1}{4}\frac{e^s-1-s}{\sinh ^2(s/2)}. \end{aligned}$$

Hence,

$$\begin{aligned} \alpha ^*_\epsilon \Big (a,b,-\epsilon \log \sqrt{\frac{b}{a}} + \frac{q}{2} \Big )&= \beta _\epsilon (a,b) + \frac{\epsilon }{4}(a-b)\,q \\&\quad + \frac{q^2}{4} \int \limits _0^1 \Bigl [a\, \mathfrak {h}\left( \lambda q/\epsilon \right) + b\,\mathfrak {h}\left( -\lambda q/\epsilon \right) \Bigr ](1-\lambda )\text {d} \lambda , \end{aligned}$$

therewith concluding the proof. $\square$