An Inverse Problem for a Semilinear Elliptic Equation on Conformally Transversally Anisotropic Manifolds

Abstract

Given a conformally transversally anisotropic manifold (M, g), we consider the semilinear elliptic equation

$$\begin{aligned} (-\Delta _{g}+V)u+qu^2=0\quad \hbox { on}\ M. \end{aligned}$$

We show that an a priori unknown smooth function q can be uniquely determined from the knowledge of the Dirichlet-to-Neumann map associated to the equation. This extends the previously known results of the works Feizmohammadi and Oksanen (J Differ Equ 269(6):4683–4719, 2020), Lassas et al. (J Math Pures Appl 145:44–82, 2021). Our proof is based on over-differentiating the equation: We linearize the equation to orders higher than the order two of the nonlinearity $qu^2$, and introduce non-vanishing boundary traces for the linearizations. We study interactions of two or more products of the so-called Gaussian quasimode solutions to the linearized equation. We develop an asymptotic calculus to solve Laplace equations, which have these interactions as source terms.

Appendices

Appendix A. Boundary Determination

We prove that the DN map of the semilinear elliptic equation

$$\begin{aligned} (-\Delta _g+V)u +qu^m=0 \text { in } M, \quad u=f \text { on } \partial M \end{aligned}$$

on a compact smooth Riemannian manifold with boundary determines the formal Taylor series (the jet) of the coefficient q (in the boundary normal coordinates) on the boundary. Here, $m\ge 2$ is an integer, and V and q are smooth functions on M. We assume also that zero is not a Dirichlet eigenvalue for the operator $-\Delta _g+V$ on M.

We expect this result to be well-known to experts on the field, but could not find a reference on it, so we offer detailed presentation and its proof.

Proposition A.1

(Boundary determination) For $m\ge 2$, $m\in {\mathbb {N}}$, let (M, g) be a compact Riemannian manifold with $C^\infty$ boundary $\partial M$ and consider the boundary value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} (-\Delta _g +V) u+qu^m=0 &{} \text { in } M,\\ u= f &{} \text { on } \partial M, \end{array}\right. } \end{aligned}$$

(A.1)

where $V,\hspace{0.5pt}q\in C^\infty (M)$. Assume that the DN map $\Lambda _q$ of the equation (A.1) is known for small boundary values. Then $\Lambda _q$ determines the formal Taylor series of q on the boundary $\partial M$.

In addition, if $f\in C^{\infty }(\partial M)$ is so small that (A.1) has a unique small solution, the DN map determines the formal Taylor series of the solution $u=u_f$ at any point on the boundary.

Proof

Determination of Taylor expansion of q:

We first investigate solutions of our semilinear elliptic equation could be $C^\infty$-smooth due to the following observations. Let $f\in C^{\infty }(\partial M)$. We consider boundary values $f_0,f\in C^{\infty }(\partial M)$ and $f_t=f_0 +tf$ and assume that $\Vert f_0 \Vert _{C^{2,\alpha }(\partial M)}$ and |t| are sufficiently small so that the DN maps at $f_0$ and $f_t$ are both well-defined. We denote by $u_0$ and $u_t$, the unique solutions of (A.1) with boundary data $f_0$ and $f_t$ on $\partial M$, respectively. In addition, since $V\in C^\infty (M)$ and $f,f_0\in C^\infty (\partial M)$, by elliptic regularity $u_t$ and $u_0$ are $C^\infty (M)$ functions.

By linearizing the equation (A.1) at $t=0$, we obtain

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta _g v +\left( V+mqu_0^{m-1}\right) v=0 &{} \text { in } M,\\ v= f &{} \text { on } \partial M, \end{array}\right. } \end{aligned}$$

(A.2)

where $v=\displaystyle \lim _{t\rightarrow 0}\frac{u_t-u_0}{t}$ and $u_0$ solves

$$\begin{aligned} {\left\{ \begin{array}{ll} (-\Delta _g +V) u_0+qu_0^m=0 &{} \text { in } M,\\ u_0= f_0 &{} \text { on } \partial M. \end{array}\right. } \end{aligned}$$

(A.3)

Moreover, v is the solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta _g v+ {\widetilde{q}}v=0 &{} \text { in } M,\\ v=f &{} \text { on } \partial M. \end{array}\right. } \end{aligned}$$

where

$$\begin{aligned} {\widetilde{q}}:=V+mqu_0^{m-1} \text { in }M. \end{aligned}$$

(A.4)

Note that ${\widetilde{q}}\in C^\infty (M)$, since $u_0\in C^\infty (M)$ by elliptic regularity.

Since we know the DN map of the boundary value problem (A.1), we know the DN map of the linearized problem (A.2). This is proven in [24, Proposition 2.1], where it is shown that the DN map is $C^\infty$ in the Frechét sense. (See also the similar result [22, Theorem 2.1], which deals with local well-posedness and linearizations of (A.1) at $f_0$ not identically 0.)

It follows by [6, Theorem 8.4.] that we know the formal Taylor series of ${\widetilde{q}}$ on $\partial M$. In particular, by choosing

$$\begin{aligned} u_0 =f_0=\varepsilon _0>0 \text { on } \partial M, \end{aligned}$$

for some sufficiently small constant $\varepsilon _0>0$, and noting that

$$\begin{aligned} q=\frac{{\widetilde{q}}-V}{m\varepsilon _0^{m-1}}\text { on } \partial M, \end{aligned}$$

it follows that we know q on the boundary $\partial M$.

Next we determine first order derivatives of q on the boundary. Given a point $x_0 \in \partial M$, let $x=(x_1,\ldots , x_n)\in \partial M$ be boundary normal coordinates near $x=x_0$ in M. Differentiating (A.4) yields

$$\begin{aligned} \begin{aligned} \partial _{x_n}{\widetilde{q}}=&m\partial _{x_n}(u_0^{m-1})q+m(\partial _{x_n}q)u_0^{m-1} +\partial _{x_n} V\\ =&m(m-1)u_0^{m-2}( \partial _{x_n}u_0 ) q +m(\partial _{x_n}q)u_0^{m-1}+\partial _{x_n} V. \end{aligned} \end{aligned}$$

(A.5)

Since we have already determined the Taylor series of ${{\tilde{q}}}$ on the boundary and

$$\begin{aligned} \partial _{x_n}u_0=\Lambda _q(f_0), \end{aligned}$$

we may determine $\partial _{x_n}q$ by solving it from (A.5). Since we also know the derivatives of q in tangential directions $x_k$, where $k=1,\ldots , n-1$, we have determined all first order derivatives of q on the boundary.

To determine higher order derivatives of q on the boundary, we follow an argument similar to [28, Lemma 3.4]. On a neighborhood of $x_0$ in M we may write

$$\begin{aligned} Qu_0:=(-\Delta _g +V) u_0+qu_0^m=-\partial _{x_n}^2 u_0+Pu_0, \end{aligned}$$

where P is a non-linear partial differential operator containing derivatives in $x'$ up to order 2 and in $x_n$ up to order 1. The coefficients of P depend on pointwise values of q. By expressing

$$\begin{aligned} \partial _{x_n}^2=P-Q \end{aligned}$$

we obtain

$$\begin{aligned} \partial _{x_n}^2u_0=Pu_0-Qu_0=Pu_0. \end{aligned}$$

(A.6)

Since we already know the quantities

$$\begin{aligned} u_0,\ \partial _{x'}u_0,\ \partial _{x'}^2u_0,\ \partial _{x_n}u_0,\ \partial _{x'}\partial _{x_n}u_0,\ q,\ \partial _{x'}q \text { and }\partial _{x_n}q, \end{aligned}$$

(A.7)

it follows from (A.6) that the second derivative $\partial _{x_n}^2 u_0$ can be also determined. By using this and differentiating (A.5), we may determine second order derivatives of q on the boundary.

The higher order derivatives of q on the boundary can be determined by differentiating (A.6) and using (A.5) in succession, and by using induction.

Determination of Taylor expansions of solutions: Let then $f\in C^{\infty }(\partial M)$ be small enough so that (A.1) has a unique small solution $u=u_f$. Since we have determined the formal Taylor series of q on the boundary, the formal Taylor series of u on the boundary is determined by differentiating (A.6) with u in place of $u_0$. $\square$

Appendix B. Proof of the Carleman Estimate with Boundary Terms

In this section, we proceed to prove Lemma 4.6. Let (M, g) be a compact, smooth, transversally anisotropic Riemannian manifold with a smooth boundary and let $V\in L^{\infty }(M)$. There exists constants $\tau _0>0$ and $C>0$ depending only on (M, g) and $\Vert V\Vert _{L^{\infty }(M)}$ such that given any $|\tau |>\tau _0$ and any $v\in C^2(M)$, the following Carleman estimate holds

$$\begin{aligned}{} & {} \Vert e^{-\tau x_1}(-\Delta _g+V)(e^{\tau x_1}v)\Vert _{L^2(M)}+|\tau |^{\frac{3}{2}}\Vert v\Vert _{W^{2,\infty }(\partial M)} +|\tau |^{\frac{3}{2}}\Vert \partial _\nu v\Vert _{W^{1,\infty }(\partial M)}\nonumber \\{} & {} \quad +|\tau |^{\frac{3}{2}}\Vert \partial ^2_\nu v\Vert _{L^\infty (\partial M)} \ge C|\tau |\,\Vert v\Vert _{L^2(M)}, \end{aligned}$$

(B.1)

Proof of Lemma 4.6

We may assume without loss of generality that v is real-valued and also that $\tau >0$. The proof for the case $\tau <0$ follows analogously. Throughout this proof, we use the notation C to stand for a generic positive constant that is independent of the parameter $\tau$. We also write ${\hat{v}}$ to stand for a $C^2$-extension of the function v into a slightly larger manifold ${\hat{M}}\Subset {\mathbb {R}}\times M_0$ with smooth boundary, such that $v\in C^{2}_c({\hat{M}})$ and that there holds

$$\begin{aligned} \Vert {\hat{v}}\Vert _{W^{2,\infty }({\hat{M}}\setminus M)}\le C(\Vert \partial _\nu ^2 v\Vert _{W^{2,\infty }(\partial M)}+\Vert \partial _\nu v\Vert _{W^{2,\infty }(\partial M)}+\Vert v\Vert _{W^{2,\infty }(\partial M)}), \end{aligned}$$

(B.2)

for some constant $C>0$, only depending on $({{\hat{M}}},g)$. In order to prove the latter estimate, let us consider the normal coordinate system $(y_1,\ldots ,y_n)=(y_1,y')$ near $\partial M$ in ${\mathbb {R}}\times M_0$ where we are assuming that $\partial M$ is given by $\{y_1=0\}$, and the metric g near $\partial M$ is given in these coordinates via the expression

$$\begin{aligned} g = (dy_1)^2 + g'(y_1,y'), \end{aligned}$$

where $g'(y_1,\cdot )$ can be viewed as a family of smooth Riemannian metrics on $\partial M$, smoothly depending on $y_1$ for all $|y_1|<\delta$ sufficiently small. We make the convention that $y_1>0$ on ${\hat{M}}\setminus M$. Let us now define ${\hat{v}}$ on ${\hat{M}}$ via

$$\begin{aligned} {\hat{v}}= v \quad \hbox { on}\ M,\end{aligned}$$

(B.3)

and

$$\begin{aligned} {\hat{v}}(y)= \left( v(0,y')+ y_1\, \partial _\nu v(0,y') + \frac{y_1^2}{2}\, \partial ^2_\nu v(0,y')\right) \,\eta (y_1), \quad \hbox { }\ y\in (0,\delta )\times \partial M,\nonumber \\ \end{aligned}$$

(B.4)

where $\eta$ is a smooth non-negative function such that $\eta (t)=1$ for all $|t|\le \frac{\delta }{2}$ and $\eta =0$ for all $|y_1|\ge \delta$. It is straightforward to see that ${\hat{v}}\in C^2_c({\hat{M}})$. The claimed estimate (B.2) now follows from the definition (B.4).

We return to the goal of proving (B.1) and define

$$\begin{aligned} P_\tau v=e^{-\tau x_1}\Delta _g(e^{\tau x_1}v), \end{aligned}$$

(B.5)

and note that

$$\begin{aligned} P_\tau v= \partial ^2_{x_1} v + \Delta _{g_0}v+2\tau \partial _{x_1}v + \tau ^2v. \end{aligned}$$

We claim that

$$\begin{aligned}{} & {} \left| \int _M P_\tau v\, \partial _{x_1} v\,dV_g\right| + C\tau ^2\Vert v\Vert _{W^{2,\infty }(\partial M)}^2+C\tau ^2\Vert \partial _\nu v\Vert _{W^{2,\infty }(\partial M)}^2\nonumber \\{} & {} \quad +C\tau ^2\Vert \partial ^2_\nu v\Vert _{W^{2,\infty }(\partial M)}^2 \ge 2\tau \Vert \partial _t v\Vert _{L^2(M)}^2. \end{aligned}$$

(B.6)

To show (B.6) we begin by writing

$$\begin{aligned} \begin{aligned} \int _M P_\tau v\, \partial _{x_1} v\,dV_g =&2\tau \int _M |\partial _{x_1} v|^2 \,dV_g \\&+\underbrace{\int _M \partial _{x_1}^2v\,\partial _{x_1} v\, dV_g}_{\text {I}}+\underbrace{\int _M \Delta _{g_0}v\,\partial _{x_1} v\, dV_g}_{\text {II}} +\underbrace{\int _M\tau ^2v\,\partial _{x_1} v\, dV_g}_{\text {III}}. \end{aligned} \end{aligned}$$

Note that $M \Subset {\mathbb {R}}\times M_0$ and $dV_g= dx_1\,dV_{g_0}$. We can use integration by parts to bound each of the terms I–III as follows. For I, we first note that

$$\begin{aligned} \int _{{\hat{M}}} \partial _{x_1}^2{{\hat{v}}}\,\partial _{x_1} {{\hat{v}}}\, dV_g=\frac{1}{2}\int _{{{\hat{M}}}} \partial _{x_1}(|\partial _{x_1} {{\hat{v}}}|^2)\,dV_g =0. \end{aligned}$$

Together with the estimate (B.2), we obtain

$$\begin{aligned}{} & {} |\text {I}|= \left| \int _{{{\hat{M}}}\setminus M}\partial _{x_1}^2{{\hat{v}}}\,\partial _{x_1} {{\hat{v}}}\, dV_g\right| \\{} & {} \quad \le C\left( \Vert \partial _\nu ^2 v\Vert ^2_{W^{2,\infty }(\partial M)}+\Vert \partial _\nu v\Vert ^2_{W^{2,\infty }(\partial M)}+\Vert v\Vert ^2_{W^{2,\infty }(\partial M)}\right) . \end{aligned}$$

For II, since $\left[ \partial _{x_1}, \Delta _g\right] =0$ on $({{\hat{M}}},g)$, we may apply integration by parts again to deduce that

$$\begin{aligned} \int _{{{\hat{M}}}} \Delta _{g_0}{{\hat{v}}}\,\partial _{x_1} {{\hat{v}}}\, dV_g=0. \end{aligned}$$

Thus, using (B.2), we can show analogously to term I that

$$\begin{aligned} |\text {II}| \le C\left( \Vert \partial _\nu ^2 v\Vert ^2_{W^{2,\infty }(\partial M)}+\Vert \partial _\nu v\Vert ^2_{W^{2,\infty }(\partial M)}+\Vert v\Vert ^2_{W^{2,\infty }(\partial M)}\right) . \end{aligned}$$

Finally for the term III we first note that

$$\begin{aligned} \tau ^2\int _M {{\hat{v}}}\,\partial _{x_1} {{\hat{v}}}\, dV_g=\frac{\tau ^2}{2}\int _M \partial _{x_1}({\hat{v}}^2)\,dV_g=0. \end{aligned}$$

Thus, using (B.2), we have

$$\begin{aligned} |\text {III}|\le C\tau ^2\left( \Vert \partial _\nu ^2 v\Vert ^2_{W^{2,\infty }(\partial M)}+\Vert \partial _\nu v\Vert ^2_{W^{2,\infty }(\partial M)}+\Vert v\Vert ^2_{W^{2,\infty }(\partial M)}\right) . \end{aligned}$$

Combining the previous three bounds yields the claimed estimate (B.6). Using (B.6) and applying the Cauchy-Schwarz inequality

$$\begin{aligned} \left| \int _M P_\tau v\, \partial _{x_1} v\,dV_g\right| \le \frac{1}{4\tau }\Vert P_\tau v\Vert _{L^2(M)}^2 + \tau \Vert \partial _{x_1} v\Vert _{L^2(M)}^2, \end{aligned}$$

we deduce that

$$\begin{aligned}{} & {} \Vert P_\tau v\Vert _{L^2(M)}^2+ C\tau ^3\Vert v\Vert _{W^{2,\infty }(\partial M)}^2+C\tau ^3\Vert \partial _\nu v\Vert _{W^{2,\infty }(\partial M)}^2\nonumber \\{} & {} \quad +C\tau ^3\Vert \partial ^2_\nu v\Vert _{W^{2,\infty }(\partial M)}^2 \ge \tau ^2 \Vert \partial _{x_1} v\Vert _{L^2(M)}^2, \end{aligned}$$

(B.7)

We recall that by the standard Poincaré inequality on ${\hat{M}}$, there exists $C>0$ such that

$$\begin{aligned} \Vert \partial _{x_1} w\Vert _{L^2({\hat{M}})} \ge C\Vert w\Vert _{L^2({\hat{M}})}\qquad \forall w \in C^1_0({\hat{M}}). \end{aligned}$$

Also, analogously to the proof of the estimate (B.2), given any $r\in C^1(M)$, there is a $C^1$-extension of r into ${\hat{M}}$ such that ${\hat{r}}\in C^1_c({\hat{M}})$ and there holds

$$\begin{aligned} \Vert {\hat{r}}\Vert _{W^{1,\infty }({\hat{M}}\setminus M)} \le C(\Vert \partial _\nu r\Vert _{W^{1,\infty }(\partial M)}+\Vert r\Vert _{W^{1,\infty }(\partial M)}), \end{aligned}$$

(B.8)

for some constant $C>0$ only depending on $({\hat{M}},g)$. Combining the latter two bounds, we deduce that given any $v\in C^1(M)$ there holds

$$\begin{aligned} \Vert \partial _{x_1} v\Vert _{L^2(M)} \ge C_1\Vert v\Vert _{L^2(M)} - C_2\Vert v\Vert _{W^{1,\infty }(\partial M)}-C_3\Vert \partial _\nu v\Vert _{W^{1,\infty }(\partial M)}, \end{aligned}$$

(B.9)

for all $v\in C^1(M)$, where the positive constants $C_1$, $C_2$ and $C_3$ only depend on (M, g).

Via the bounds (B.7)–(B.9), we deduce that

$$\begin{aligned}{} & {} \Vert (P_\tau -V) v\Vert _{L^2(M)}^2+ C\tau ^3\Vert v\Vert _{W^{2,\infty }(\partial M)}^2+C\tau ^3\Vert \partial _\nu v\Vert _{W^{2,\infty }(\partial M)}^2\nonumber \\{} & {} \quad +C\tau ^3\Vert \partial ^2_\nu v\Vert _{W^{2,\infty }(\partial M)}^2 \ge \tau ^2 \Vert v\Vert _{L^2(M)}^2. \end{aligned}$$

(B.10)

This proves the assertion.

Appendix C. Computations of ${\textbf{D}}_{ik}$

In the end of this paper, we compute the values ${\textbf{D}}_{ik}$, for different sub-indices $i,k\in \{1,2,3,4,5\}$. Recalling that

$$\begin{aligned}&\zeta _1=\xi _1, \qquad \zeta _2=\xi _2, \qquad \\&\quad \zeta _3= \left( 1+\sqrt{\frac{2}{2-\delta }}\right) \xi _3 , \quad \zeta _4=\left( 1+\sqrt{\frac{2}{2-\delta }}\right) \xi _4, \\&\quad \zeta _5=\sqrt{\frac{2}{2-\delta }}(\xi _1+\xi _2), \end{aligned}$$

and

$$\begin{aligned}&\overline{\zeta }_1=|\zeta _1 |e_1+ {\textbf{i}}\zeta _1, \quad \overline{\zeta }_2=|\zeta _2 |e_1+{\textbf{i}}\zeta _2, \\&\quad \overline{\zeta }_3=|\zeta _3 |e_1+{\textbf{i}}\zeta _3, \quad \overline{\zeta }_4=-|\zeta _4 |e_1+{\textbf{i}}\zeta _4,\\&\quad \overline{\zeta }_5=-|\zeta _5 |e_1+{\textbf{i}}\zeta _5, \end{aligned}$$

where

$$\begin{aligned}&|\xi _1 |=|\xi _2 |=1, \quad \langle \xi _1 , \xi _2 \rangle =1-\delta ,\\&\quad \xi _3=-\frac{1}{1+\delta }(\xi _1+\delta \xi _2), \quad \xi _4=-\frac{1}{1+\delta }(\delta \xi _1 + \xi _2). \end{aligned}$$

Via straightforward computations, we have

$$\begin{aligned}&\langle \xi _1 , \xi _2\rangle = 1- \delta , \quad \langle \xi _1,\xi _3 \rangle =-\frac{1+\delta -\delta ^2}{1+\delta }, \quad \langle \xi _1, \xi _4\rangle =-\frac{1}{1+\delta }, \\&\quad \langle \xi _2 , \xi _3\rangle = -\frac{1}{1+\delta }, \quad \langle \xi _2,\xi _4 \rangle =-\frac{1+\delta -\delta ^2}{1+\delta }, \text { and } \langle \xi _3,\xi _4 \rangle =\frac{1+\delta +\delta ^2-\delta ^3}{(1+\delta )^2}. \end{aligned}$$

By

$$\begin{aligned} {\textbf{D}}_{ij}=\langle \overline{\zeta }_i+\overline{\zeta }_j, \overline{\zeta }_i+\overline{\zeta }_j \rangle , \end{aligned}$$

for different $i,k\in \{1,2,3,4,5\}$, direct computations yield that

$$\begin{aligned} \begin{aligned} {\textbf{D}}_{12}=&\left( |\zeta _1 |e_1+ {\textbf{i}}\zeta _1+|\zeta _2 |e_1+{\textbf{i}}\zeta _2\right) \cdot \left( |\zeta _1 |e_1+ {\textbf{i}}\zeta _1+|\zeta _2 |e_1+{\textbf{i}}\zeta _2\right) \\ =&2 \left( |\zeta _1||\zeta _2|-\langle \zeta _1, \zeta _2 \rangle \right) \\ =&2 \left( |\xi _1||\xi _2|-\langle \xi _1, \xi _2 \rangle \right) = 2\delta , \end{aligned} \end{aligned}$$

(C.1)

$$\begin{aligned} \begin{aligned} {\textbf{D}}_{13}=&\left( |\zeta _1 |e_1+ {\textbf{i}}\zeta _1 +|\zeta _3 |e_1+{\textbf{i}}\zeta _3\right) \cdot \left( |\zeta _1 |e_1+ {\textbf{i}}\zeta _1 +|\zeta _3 |e_1+{\textbf{i}}\zeta _3\right) \\ =&2 \left( |\zeta _1 ||\zeta _3 |-\langle \zeta _1, \zeta _3 \rangle \right) \\ =&2 \left( 1+\sqrt{\frac{2}{2-\delta }}\right) \left( |\xi _1 | |\xi _3 | -\langle \xi _1, \xi _3 \rangle \right) \\ =&2 \left( 1+\sqrt{\frac{2}{2-\delta }}\right) \left( 2+2\delta +{\mathcal {O}}(\delta ^2)\right) , \end{aligned} \end{aligned}$$

(C.2)

$$\begin{aligned} \begin{aligned} {\textbf{D}}_{14}=&\left( |\zeta _1 |e_1+ {\textbf{i}}\zeta _1 -|\zeta _4 |e_1+{\textbf{i}}\zeta _4\right) \cdot \left( |\zeta _1 |e_1+ {\textbf{i}}\zeta _1 -|\zeta _4 |e_1+{\textbf{i}}\zeta _4\right) \\ =&-2\left( 1+\sqrt{\frac{2}{2-\delta }}\right) \left( |\xi _1 ||\xi _4 |+\langle \xi _1 ,\xi _4 \rangle \right) \\ =&-2\left( 1+\sqrt{\frac{2}{2-\delta }}\right) \frac{\delta +{\mathcal {O}}(\delta ^2)}{1+\delta }, \end{aligned} \end{aligned}$$

(C.3)

$$\begin{aligned} \begin{aligned} {\textbf{D}}_{15}=&\left( |\zeta _1 |e_1+ {\textbf{i}}\zeta _1-|\zeta _5 |e_1+{\textbf{i}}\zeta _5 \right) \cdot \left( |\zeta _1 |e_1+ {\textbf{i}}\zeta _1-|\zeta _5 |e_1+{\textbf{i}}\zeta _5 \right) \\ =&-2 \left( |\zeta _1 ||\zeta _5 | + \langle \zeta _1, \zeta _5 \rangle \right) \\ =&-2 \sqrt{\frac{2}{2-\delta }} \left( |\xi _1 ||\xi _1 +\xi _2 | +\langle \xi _1,\xi _1+\xi _2\rangle \right) \\ =&-8+\frac{\delta }{2}+{\mathcal {O}}(\delta ^2), \end{aligned} \end{aligned}$$

(C.4)

$$\begin{aligned} \begin{aligned} {\textbf{D}}_{23}=&\left( |\zeta _2 |e_1+{\textbf{i}}\zeta _2+|\zeta _3 |e_1+{\textbf{i}}\zeta _3 \right) \cdot \left( |\zeta _2 |e_1+{\textbf{i}}\zeta _2+|\zeta _3 |e_1+{\textbf{i}}\zeta _3 \right) \\ =&2 \left( 1+\sqrt{\frac{2}{2-\delta }}\right) \left( |\xi _2 ||\xi _3 | -\langle \xi _2 ,\xi _3\rangle \right) \\ =&2 \left( 1+\sqrt{\frac{2}{2-\delta }}\right) \left( \frac{2+\delta +{\mathcal {O}}(\delta ^2)}{1+\delta } \right) . \end{aligned} \end{aligned}$$

(C.5)

In order to compute ${\textbf{D}}_{24}$ more carefully, let us recall the Taylor expansion of $\sqrt{1+\delta }=1+\frac{\delta }{2}-\frac{\delta ^2}{8}+{\mathcal {O}}(\delta ^3)$, then we have

$$\begin{aligned} \begin{aligned} {\textbf{D}}_{24} =&\left( |\zeta _2 |e_1+{\textbf{i}}\zeta _2 -|\zeta _4 |e_1+{\textbf{i}}\zeta _4\right) \cdot \left( |\zeta _2 |e_1+{\textbf{i}}\zeta _2 -|\zeta _4 |e_1+{\textbf{i}}\zeta _4\right) \\ =&-2 \left( |\zeta _2 ||\zeta _4 | +\langle \zeta _2, \zeta _4 \rangle \right) \\ =&-2\left( 1+\sqrt{\frac{2}{2-\delta }}\right) \left( |\xi _2 ||\xi _4 | +\langle \xi _2, \xi _4 \rangle \right) \\ =&-2\left( 1+\sqrt{\frac{2}{2-\delta }}\right) \left( \sqrt{1+2\delta -\delta ^2}-(1+\delta -\delta ^2) \right) \\ =&-2\left( 1+\sqrt{\frac{2}{2-\delta }}\right) \left( 1 +\delta -\delta ^2 +{\mathcal {O}}(\delta ^3)-(1+\delta -\delta ^2) \right) \\ =&-2 \left( 1+\sqrt{\frac{2}{2-\delta }}\right) \frac{{\mathcal {O}}(\delta ^3)}{1+\delta }, \end{aligned} \end{aligned}$$

(C.6)

$$\begin{aligned} \begin{aligned} {\textbf{D}}_{25} =&\left( |\zeta _2 |e_1+{\textbf{i}}\zeta _2-|\zeta _5 |e_1+{\textbf{i}}\zeta _5 \right) \cdot \left( |\zeta _2 |e_1+{\textbf{i}}\zeta _2-|\zeta _5 |e_1+{\textbf{i}}\zeta _5 \right) \\ =&-2 \left( |\zeta _2 ||\zeta _5 | +\langle \zeta _2,\zeta _5 \rangle \right) \\ =&-2\sqrt{\frac{2}{2-\delta }} \left( \sqrt{4-2\delta }+2-\delta \right) \\ =&-8+\delta +{\mathcal {O}}(\delta ^2), \end{aligned} \end{aligned}$$

(C.7)

$$\begin{aligned} \begin{aligned} {\textbf{D}}_{34} =&\left( |\zeta _3 |e_1+{\textbf{i}}\zeta _3-|\zeta _4 |e_1+{\textbf{i}}\zeta _4 \right) \cdot \left( |\zeta _3 |e_1+{\textbf{i}}\zeta _3-|\zeta _4 |e_1+{\textbf{i}}\zeta _4 \right) \\ =&-2 \left( |\zeta _3 ||\zeta _4 | + \langle \zeta _3,\zeta _4 \rangle \right) \\ =&-2\left( 1+\sqrt{\frac{2}{2-\delta }}\right) ^2\left( |\xi _3 ||\xi _4 | + \langle \xi _3,\xi _4 \rangle \right) \\ =&-2\left( 1+\sqrt{\frac{2}{2-\delta }}\right) ^2 \frac{1}{(1+\delta )^2} \left( 2+3\delta -\delta ^3 \right) , \end{aligned} \end{aligned}$$

(C.8)

$$\begin{aligned} \begin{aligned} {\textbf{D}}_{35}=&\left( |\zeta _3 |e_1+{\textbf{i}}\zeta _3-|\zeta _5 |e_1+{\textbf{i}}\zeta _5 \right) \cdot \left( |\zeta _3 |e_1+{\textbf{i}}\zeta _3-|\zeta _5 |e_1+{\textbf{i}}\zeta _5 \right) \\ =&-2 \left( |\zeta _3 ||\zeta _5 |+\langle \zeta _3 , \zeta _5\rangle \right) \\ =&-\frac{2}{1+\delta } \left( 1+\sqrt{\frac{2}{2-\delta }}\right) \sqrt{\frac{2}{2-\delta }} \left( \sqrt{4-2\delta }(1+\delta +{\mathcal {O}}(\delta ^2)) - 2-\delta +\delta ^2\right) \\ =&-\frac{2}{1+\delta } \left( 1+\sqrt{\frac{2}{2-\delta }}\right) \sqrt{\frac{2}{2-\delta }} \\&\qquad \times \left( (2-\frac{\delta }{2}+{\mathcal {O}}(\delta ^2))(1+\delta +{\mathcal {O}}(\delta ^2)) - 2-\delta +\delta ^2\right) \\ =&-\frac{1}{1+\delta } \left( 1+\sqrt{\frac{2}{2-\delta }}\right) \sqrt{\frac{2}{2-\delta }} \left( \delta +{\mathcal {O}}(\delta ^2) \right) \end{aligned} \end{aligned}$$

(C.9)

and similarly,

$$\begin{aligned} \begin{aligned} {\textbf{D}}_{45}=&\left( -|\zeta _4 |e_1+{\textbf{i}}\zeta _4-|\zeta _5 |e_1+{\textbf{i}}\zeta _5 \right) \cdot \left( -|\zeta _4 |e_1+{\textbf{i}}\zeta _4-|\zeta _5 |e_1+{\textbf{i}}\zeta _5 \right) \\ =&2 \left( |\zeta _4 ||\zeta _5 |-\langle \zeta _4, \zeta _5 \rangle \right) \\ =&2 \left( 1+\sqrt{\frac{2}{2-\delta }}\right) \sqrt{\frac{2}{2-\delta }} \left( |\xi _4 ||\xi _1 +\xi _2 |-\langle \xi _4, \xi _1+\xi _2 \rangle \right) \\ =&\frac{2}{1+\delta } \left( 1+\sqrt{\frac{2}{2-\delta }}\right) \sqrt{\frac{2}{2-\delta }} \left( \sqrt{4-2\delta }(1+\delta +{\mathcal {O}}(\delta ^2))+2+\delta -\delta ^2\right) \\ =&\frac{2}{1+\delta } \left( 1+\sqrt{\frac{2}{2-\delta }}\right) \sqrt{\frac{2}{2-\delta }} \left( 4+\frac{5}{2}\delta + {\mathcal {O}}(\delta ^2)\right) . \end{aligned} \end{aligned}$$

(C.10)

Proof of Lemma 5.2

With (C.1)–(C.10) at hand, let us split the analysis into two cases.

(1) By using (C.5), (C.6) and (C.8), we have that $\frac{1}{{\textbf{D}}_{23}+{\textbf{D}}_{24}+{\textbf{D}}_{34}}$ is a bounded as $\delta \rightarrow 0$. Similarly, (C.2), (C.3) and (C.8) imply that $\frac{1}{{\textbf{D}}_{13}+{\textbf{D}}_{14}+{\textbf{D}}_{34}}$ is also bounded as $\delta \rightarrow 0$. Similarly $\frac{1}{{\textbf{D}}_{12}+{\textbf{D}}_{13}+{\textbf{D}}_{23}}$ is bounded as $\delta \rightarrow 0$. On the other hand, by (C.1), (C.3) and (C.6), we observe that $\frac{1}{{\textbf{D}}_{12}+{\textbf{D}}_{14}+{\textbf{D}}_{24}}={\mathcal {O}}(\delta ^{-1})$. Meanwhile, ${\textbf{D}}_{15}^{-1}$, ${\textbf{D}}_{25}^{-1}$ and ${\textbf{D}}_{45}^{-1}$ are bounded as $\delta \rightarrow 0$, but ${\textbf{D}}_{35}^{-1}={\mathcal {O}}(\delta ^{-1})$.

(2) Similarly, $\frac{1}{{\textbf{D}}_{12}}\frac{1}{{\textbf{D}}_{34}}={\mathcal {O}}(\delta ^{-1})$, $\frac{1}{{\textbf{D}}_{13}}\frac{1}{{\textbf{D}}_{24}}={\mathcal {O}}(\delta ^{-3})$ and $\frac{1}{{\textbf{D}}_{14}}\frac{1}{{\textbf{D}}_{23}}={\mathcal {O}}(\delta ^{-1})$.

Therefore, combining the above, we conclude that

$$\begin{aligned} \begin{aligned} {\textbf{E}}_\delta =&\left| \frac{1}{{\textbf{D}}_{15}}\left( \frac{1}{{\textbf{D}}_{23}+{\textbf{D}}_{24}+{\textbf{D}}_{34}}\right) + \frac{1}{{\textbf{D}}_{25}}\left( \frac{1}{{\textbf{D}}_{13}+{\textbf{D}}_{14}+{\textbf{D}}_{34}}\right) \right. \\&\qquad + \frac{1}{{\textbf{D}}_{35}}\left( \frac{1}{{\textbf{D}}_{12}+{\textbf{D}}_{14}+{\textbf{D}}_{24}}\right) + \frac{1}{{\textbf{D}}_{45}}\left( \frac{1}{{\textbf{D}}_{12}+{\textbf{D}}_{13}+{\textbf{D}}_{23}}\right) \\&\qquad \left. + \frac{1}{{\textbf{D}}_{12}}\frac{1}{{\textbf{D}}_{34}}+ \frac{1}{{\textbf{D}}_{13}}\frac{1}{{\textbf{D}}_{24}} + \frac{1}{{\textbf{D}}_{14}}\frac{1}{{\textbf{D}}_{23}}\right| \\ \ge&\frac{C_0}{\delta ^3} -\frac{C_1}{\delta ^2}-C_2>0, \end{aligned} \end{aligned}$$

for all sufficiently small $\delta >0$, where $C_0$, $C_1$ and $C_2$ are some positive constants independent of $\delta$. Hence, the coefficient ${\textbf{E}}_\delta ={\mathcal {O}}(\delta ^{-3})\ne 0$

for all sufficiently small $\delta >0$.