*Proof of Theorem 1* First, we will prove the uniqueness of the solution $(u,\lambda )$ of Problem 3 under the assumptions of the theorem. To this end, it is sufficient to show that the corresponding homogeneous problem has only a trivial solution. Then, by taking into account that $\stackrel{\u02c6}{L}\lambda =0$ and $F=0$, from (5) we obtain $\stackrel{\u02c6}{L}Lu=Au=0$.

Since

$u\in \mathrm{\Gamma}(A)$, there exists a sequence

$\{{u}_{k}\}\subset {C}_{\pi 0}^{3}$ such that

${u}_{k}\to u$ weakly in

${L}_{2}(\mathrm{\Omega})$ and

${(A{u}_{k},{u}_{k})}_{{L}_{2}(\mathrm{\Omega})}\to {(Au,u)}_{{L}_{2}(\mathrm{\Omega})}=0$ as

$k\to \mathrm{\infty}$. Now we want to decompose the product

$(A{u}_{k}){u}_{k}$ into the sum of a positive definite quadratic form and a divergence form. For this purpose, we have the following identities:

$\begin{array}{rcl}(A{u}_{k}){u}_{k}& =& (\stackrel{\u02c6}{L}L{u}_{k}){u}_{k}\\ =& \left(\frac{\partial}{\partial l}\left(\frac{\partial}{\partial \phi}L{u}_{k}\right)\right){u}_{k}\\ =& \frac{\partial}{\partial \phi}L{u}_{k}{\left(\frac{\partial}{\partial l}\right)}^{\ast}{u}_{k}+\frac{\partial}{\partial {x}_{1}}({u}_{k}\left(\frac{\partial}{\partial \phi}L{u}_{k}\right)sin\phi )\\ -\frac{\partial}{\partial {x}_{2}}({u}_{k}\left(\frac{\partial}{\partial \phi}L{u}_{k}\right)cos\phi )+\frac{\partial}{\partial \phi}\left({u}_{k}\left(\frac{\partial}{\partial \phi}L{u}_{k}\right)g\right)\end{array}$

(13)

and

$\begin{array}{rcl}2\frac{\partial}{\partial \phi}L{u}_{k}{\left(\frac{\partial}{\partial l}\right)}^{\ast}{u}_{k}& =& 2({u}_{k{x}_{1}\phi}cos\phi -{u}_{k{x}_{1}}sin\phi +{u}_{k{x}_{2}\phi}sin\phi +{u}_{k{x}_{2}}cos\phi \\ +{K}_{\phi}{u}_{k\phi}+K{u}_{k\phi \phi})(-{u}_{k{x}_{1}}sin\phi +{u}_{k{x}_{2}}cos\phi -g{u}_{k\phi})\\ =& {u}_{k{x}_{1}}^{2}+{u}_{k{x}_{2}}^{2}+2K{u}_{k\phi}({u}_{k{x}_{1}}cos\phi +{u}_{k{x}_{2}}sin\phi )\\ +2g{u}_{k\phi}({u}_{k{x}_{1}}sin\phi -{u}_{k{x}_{2}}cos\phi )+({g}_{{x}_{1}}cos\phi +{g}_{{x}_{2}}sin\phi \\ +{g}_{\phi}K-{K}_{{x}_{1}}sin\phi +{K}_{{x}_{2}}cos\phi -g{K}_{\phi}){u}_{k\phi}^{2}\\ +\frac{\partial}{\partial {x}_{1}}({u}_{k{x}_{2}}{u}_{k\phi}+K{u}_{k\phi}^{2}sin\phi -g{u}_{k\phi}^{2}cos\phi )\\ -\frac{\partial}{\partial {x}_{2}}({u}_{k{x}_{1}}{u}_{k\phi}+K{u}_{k\phi}^{2}cos\phi +g{u}_{k\phi}^{2}sin\phi )\\ +\frac{\partial}{\partial \phi}(-{u}_{k{x}_{1}}^{2}sin\phi cos\phi +{u}_{k{x}_{2}}^{2}sin\phi cos\phi +{u}_{k{x}_{1}}{u}_{k{x}_{2}}cos2\phi \\ -2K{u}_{k{x}_{1}}{u}_{k\phi}sin\phi +2K{u}_{k{x}_{2}}{u}_{k\phi}cos\phi -gK{u}_{k\phi}^{2}).\end{array}$

If (6) holds, then the quadratic form

$J(\mathrm{\nabla}{u}_{k})$ in

${u}_{k{x}_{1}}$,

${u}_{k{x}_{2}}$,

${u}_{k\phi}$ is positive definite, where

$\begin{array}{rcl}J(\mathrm{\nabla}{u}_{k})& =& {u}_{k{x}_{1}}^{2}+{u}_{k{x}_{2}}^{2}+2K{u}_{k\phi}({u}_{k{x}_{1}}cos\phi +{u}_{k{x}_{2}}sin\phi )\\ +2g{u}_{k\phi}({u}_{k{x}_{1}}sin\phi -{u}_{k{x}_{2}}cos\phi )+({g}_{{x}_{1}}cos\phi +{g}_{{x}_{2}}sin\phi \\ +{g}_{\phi}K-{K}_{{x}_{1}}sin\phi +{K}_{{x}_{2}}cos\phi -g{K}_{\phi}){u}_{k\phi}^{2}\\ =& {u}_{k{x}_{1}}^{2}+{u}_{k{x}_{2}}^{2}+2{u}_{k\phi}{u}_{k{x}_{1}}(Kcos\phi +gsin\phi )\\ +2{u}_{k\phi}{u}_{k{x}_{2}}(Ksin\phi -gcos\phi )+({g}_{{x}_{1}}cos\phi +{g}_{{x}_{2}}sin\phi \\ +{g}_{\phi}K-{K}_{{x}_{1}}sin\phi +{K}_{{x}_{2}}cos\phi -g{K}_{\phi}){u}_{k\phi}^{2}.\end{array}$

(14)

Indeed, we can estimate the terms

$2{u}_{k\phi}{u}_{k{x}_{1}}(Kcos\phi +gsin\phi )$ and

$2{u}_{k\phi}{u}_{k{x}_{2}}(Ksin\phi -gcos\phi )$ as follows:

$\begin{array}{c}2{u}_{k\phi}{u}_{k{x}_{1}}(Kcos\phi +gsin\phi )\ge -\epsilon {u}_{k{x}_{1}}^{2}-{\epsilon}^{-1}{(Kcos\phi +gsin\phi )}^{2}{u}_{k\phi}^{2},\hfill \\ 2{u}_{k\phi}{u}_{k{x}_{2}}(Ksin\phi -gcos\phi )\ge -\epsilon {u}_{k{x}_{2}}^{2}-{\epsilon}^{-1}{(Ksin\phi -gcos\phi )}^{2}{u}_{k\phi}^{2},\hfill \end{array}$

where

$0<\epsilon <1$, and we obtain

$\begin{array}{rcl}J(\mathrm{\nabla}{u}_{k})& \ge & {u}_{k{x}_{1}}^{2}+{u}_{k{x}_{2}}^{2}-\epsilon {u}_{k{x}_{1}}^{2}-\epsilon {u}_{k{x}_{2}}^{2}-{\epsilon}^{-1}{(Kcos\phi +gsin\phi )}^{2}{u}_{k\phi}^{2}\\ -{\epsilon}^{-1}{(Ksin\phi -gcos\phi )}^{2}{u}_{k\phi}^{2}+({g}_{{x}_{1}}cos\phi +{g}_{{x}_{2}}sin\phi \\ +{g}_{\phi}K-{K}_{{x}_{1}}sin\phi +{K}_{{x}_{2}}cos\phi -g{K}_{\phi}){u}_{k\phi}^{2}\\ =& (1-\epsilon )({u}_{k{x}_{1}}^{2}+{u}_{k{x}_{2}}^{2})+({g}_{{x}_{1}}cos\phi +{g}_{{x}_{2}}sin\phi +{g}_{\phi}K\\ -{K}_{{x}_{1}}sin\phi +{K}_{{x}_{2}}cos\phi -g{K}_{\phi}-{\epsilon}^{-1}({K}^{2}+{g}^{2})){u}_{k\phi}^{2}.\end{array}$

Moreover, whenever (6) holds, for sufficiently close value of

*ε* to 1, there exists an

$\alpha \in \mathbb{R}$ such that

${g}_{{x}_{1}}cos\phi +{g}_{{x}_{2}}sin\phi +{g}_{\phi}K-{K}_{{x}_{1}}sin\phi +{K}_{{x}_{2}}cos\phi -{K}_{\phi}g-{\epsilon}^{-1}({K}^{2}+{g}^{2})\ge \alpha >0$ in Ω, and we obtain

$J(\mathrm{\nabla}{u}_{k})\ge (1-\epsilon )|{\mathrm{\nabla}}_{x}{u}_{k}{|}^{2}+\alpha {u}_{k\phi}^{2}\ge \beta |{\mathrm{\nabla}}_{x,\phi}{u}_{k}{|}^{2},$

where $\beta =min\{(1-\epsilon ),\alpha \}$.

Since the domain

*D* is bounded,

${u}_{k}=0$ on

$\partial D\times (0,2\pi )$ and

$J(\mathrm{\nabla}{u}_{k})$ is positive definite, we have

${\parallel {u}_{k}\parallel}_{{L}_{2}(\mathrm{\Omega})}^{2}\le {C}_{0}{\int}_{\mathrm{\Omega}}|{\mathrm{\nabla}}_{x}{u}_{k}{|}^{2}\phantom{\rule{0.2em}{0ex}}d\mathrm{\Omega}\le C{\int}_{\mathrm{\Omega}}J(\mathrm{\nabla}{u}_{k})\phantom{\rule{0.2em}{0ex}}d\mathrm{\Omega},$

where $C={C}_{0}{\beta}^{-1}$ and ${C}_{0}>0$ depends on the Lebesgue measure of *D* and does not depend on *k*.

Thus, since

${u}_{k}\in {C}_{\pi 0}^{3}$,

*K* and

*g* are 2

*π*-periodic with respect to

*φ*, after integrating (13) over Ω, the divergent terms disappear and we obtain

$2{(A{u}_{k},{u}_{k})}_{{L}_{2}(\mathrm{\Omega})}={\int}_{\mathrm{\Omega}}J(\mathrm{\nabla}{u}_{k})\phantom{\rule{0.2em}{0ex}}d\mathrm{\Omega}.$

(15)

Since

$u\in \mathrm{\Gamma}(A)$, from (15) we have

$\begin{array}{rl}{\parallel u\parallel}_{{L}_{2}(\mathrm{\Omega})}^{2}& \le \underset{k\to \mathrm{\infty}}{\underline{lim}}{\parallel {u}_{k}\parallel}_{{L}_{2}(\mathrm{\Omega})}^{2}\\ \le C\underset{k\to \mathrm{\infty}}{lim}{\int}_{\mathrm{\Omega}}J(\mathrm{\nabla}{u}_{k})\phantom{\rule{0.2em}{0ex}}d\mathrm{\Omega}=2C\underset{k\to \mathrm{\infty}}{lim}{(A{u}_{k},{u}_{k})}_{{L}_{2}(\mathrm{\Omega})}=0,\end{array}$

(16)

which implies that $u=0$, and since $F=0$, from (5) we get $\lambda =0$. So, the uniqueness part of the proof is completed.

Now we will prove that there exists a solution $(u,\lambda )$ of Problem 3 in $(\mathrm{\Gamma}(A)\cap {\stackrel{\u02da}{H}}_{1}^{\pi}(\mathrm{\Omega}))\times {L}_{2}(\mathrm{\Omega})$ by means of the following auxiliary problem.

*Determine* *u* *defined in* Ω

*that satisfies*
$u{|}_{\partial D\times (0,2\pi )}=0,\phantom{\rule{2em}{0ex}}u(x,0)=u(x,2\pi ),$

(18)

*where* $\mathcal{F}=\stackrel{\u02c6}{L}F$.

The solution

*u* of problem (17)-(18) will be approximated by

${u}_{N}=\sum _{i=1}^{N}{\alpha}_{{N}_{i}}{w}_{i}(x,\phi );\phantom{\rule{1em}{0ex}}{\alpha}_{N}=({\alpha}_{{N}_{1}},{\alpha}_{{N}_{2}},\dots ,{\alpha}_{{N}_{N}})\in {\mathbb{R}}^{N},$

construction of which is based on finding the vector

${\alpha}_{N}$ from the system of linear algebraic equations

${(A{u}_{N},{w}_{j})}_{{L}_{2}(\mathrm{\Omega})}={(\mathcal{F},{w}_{j})}_{{L}_{2}(\mathrm{\Omega})},\phantom{\rule{1em}{0ex}}j=1,2,\dots ,N,$

(19)

where the system of functions $\{{w}_{j}\}$ is taken as indicated in Section 2.3 and ${u}_{N}=0$ on $\partial D\times (0,2\pi )$.

We must show that the solution of system (19) exists and is unique for any

$F\in {H}_{2}^{\pi}(\mathrm{\Omega})$. To demonstrate this, let us assume that the homogeneous version of (19),

*i.e.*, the system

${(A{u}_{N},{w}_{j})}_{{L}_{2}(\mathrm{\Omega})}=0,\phantom{\rule{1em}{0ex}}j=1,2,\dots ,N,$

has a nonzero solution

${\overline{\alpha}}_{N}=({\overline{\alpha}}_{{N}_{1}},{\overline{\alpha}}_{{N}_{2}},\dots ,{\overline{\alpha}}_{{N}_{N}})$. Substituting

${\overline{\alpha}}_{N}$ for

${\alpha}_{N}$, multiplying the

*j* th equation of the above homogenous system by

$2{\overline{\alpha}}_{{N}_{j}}$ and summing with respect to

*j* from 1 to

*N*, we obtain

$2{(A{\overline{u}}_{N},{\overline{u}}_{N})}_{{L}_{2}(\mathrm{\Omega})}=0,$

(20)

where

${\overline{u}}_{N}={\sum}_{i=1}^{N}{\overline{\alpha}}_{{N}_{i}}{w}_{i}$. So, from (15) and (20) we obtain

$2{(A{\overline{u}}_{N},{\overline{u}}_{N})}_{{L}_{2}(\mathrm{\Omega})}={\int}_{\mathrm{\Omega}}J(\mathrm{\nabla}{\overline{u}}_{N})\phantom{\rule{0.2em}{0ex}}d\mathrm{\Omega}=0,$

and since the quadratic form $J(\mathrm{\nabla}{\overline{u}}_{N})$ defined in (14) is positive definite and ${\overline{u}}_{N}=0$ on $\partial D\times (0,2\pi )$, we have ${\overline{u}}_{N}=0$ in Ω. But $\{{w}_{i}\}$ is linearly independent and this implies that ${\overline{\alpha}}_{{N}_{i}}=0$, $i=1,2,\dots ,N$, which contradicts with the assumption ${\overline{\alpha}}_{N}\ne 0$. So, it is shown that system (19) has a unique solution ${\alpha}_{N}$ for any $F\in {H}_{2}^{\pi}(\mathrm{\Omega})$.

Now we estimate the solution

${u}_{N}$ of system (19) in terms of

*F*. For this purpose, we multiply both sides of the

*j* th equation of (19) by

$2{\alpha}_{{N}_{j}}$ and sum the obtained equations with respect to

*j* from 1 to

*N* to obtain

$2{(A{u}_{N},{u}_{N})}_{{L}_{2}(\mathrm{\Omega})}=2{(\mathcal{F},{u}_{N})}_{{L}_{2}(\mathrm{\Omega})}=2{(\stackrel{\u02c6}{L}F,{u}_{N})}_{{L}_{2}(\mathrm{\Omega})}.$

(21)

Since

${u}_{N}\in {C}_{\pi 0}^{3}$, applying integration by parts, the right-hand side of (21) can be estimated as

$2\left|{(\stackrel{\u02c6}{L}F,{u}_{N})}_{{L}_{2}(\mathrm{\Omega})}\right|\le \gamma {\int}_{\mathrm{\Omega}}{F}_{\phi}^{2}\phantom{\rule{0.2em}{0ex}}d\mathrm{\Omega}+{\gamma}^{-1}{\int}_{\mathrm{\Omega}}{\left({\left(\frac{\partial}{\partial l}\right)}^{\ast}{u}_{N}\right)}^{2}\phantom{\rule{0.2em}{0ex}}d\mathrm{\Omega},$

for

$\gamma >0$, and from (15) and (21) we have

${\int}_{\mathrm{\Omega}}J(\mathrm{\nabla}{u}_{N})\phantom{\rule{0.2em}{0ex}}d\mathrm{\Omega}\le \gamma {\int}_{\mathrm{\Omega}}{F}_{\phi}^{2}\phantom{\rule{0.2em}{0ex}}d\mathrm{\Omega}+{\gamma}^{-1}{\int}_{\mathrm{\Omega}}{\left({\left(\frac{\partial}{\partial l}\right)}^{\ast}{u}_{N}\right)}^{2}\phantom{\rule{0.2em}{0ex}}d\mathrm{\Omega}.$

(22)

It can be verified that for sufficiently large

$\gamma >0$, from (22) we obtain

${\parallel {u}_{N}\parallel}_{{\stackrel{\u02da}{H}}_{1}^{\pi}(\mathrm{\Omega})}\le C{\parallel {F}_{\phi}\parallel}_{{L}_{2}(\mathrm{\Omega})},$

(23)

where the constant

*C* is independent of

*N*. This implies that

${\{{u}_{N}\}}_{N=1}^{\mathrm{\infty}}$ is bounded in

${L}_{2}(\mathrm{\Omega})$ and

${\stackrel{\u02da}{H}}_{1}^{\pi}(\mathrm{\Omega})$, and since

${L}_{2}(\mathrm{\Omega})$ and

${\stackrel{\u02da}{H}}_{1}^{\pi}(\mathrm{\Omega})$ are Hilbert spaces, it is weakly compact in

${L}_{2}(\mathrm{\Omega})$ and

${\stackrel{\u02da}{H}}_{1}^{\pi}(\mathrm{\Omega})$. Therefore, there exists a subsequence, which we again denote by

$\{{u}_{N}\}$, such that

${u}_{N}\to u$ weakly in

${L}_{2}(\mathrm{\Omega})$ and

${\stackrel{\u02da}{H}}_{1}^{\pi}(\mathrm{\Omega})$ as

$N\to \mathrm{\infty}$ and

${\parallel u\parallel}_{{\stackrel{\u02da}{H}}_{1}^{\pi}(\mathrm{\Omega})}\le C{\parallel {F}_{\phi}\parallel}_{{L}_{2}(\mathrm{\Omega})}$

holds. Since

${u}_{N}{|}_{\partial D\times (0,2\pi )}=0$ and

${u}_{N}\to u$ weakly in

${\stackrel{\u02da}{H}}_{1}^{\pi}(\mathrm{\Omega})$, we have

$u{|}_{\partial D\times (0,2\pi )}=0$. From (23) we have also that

${\{{u}_{N{x}_{1}}\}}_{N=1}^{\mathrm{\infty}}$,

${\{{u}_{N{x}_{2}}\}}_{N=1}^{\mathrm{\infty}}$ and

${\{{u}_{N\phi}\}}_{N=1}^{\mathrm{\infty}}$ are bounded and there exists a subsequence of

$\{{u}_{N}\}$, which is again denoted by

$\{{u}_{N}\}$, such that

${u}_{N{x}_{1}}$,

${u}_{N{x}_{2}}$ and

${u}_{N\phi}$ converge weakly in

${L}_{2}(\mathrm{\Omega})$ to

${u}_{{x}_{1}}$,

${u}_{{x}_{2}}$ and

${u}_{\phi}$, respectively. Taking into account that

${u}_{N},{w}_{j}\in {C}_{\pi 0}^{3}$,

$F\in {H}_{2}^{\pi}(\mathrm{\Omega})$ and applying integration by parts in (19), for

$N\ge j$, we obtain

${(L{u}_{N}-F,{(\stackrel{\u02c6}{L})}^{\ast}{w}_{j})}_{{L}_{2}(\mathrm{\Omega})}=0.$

Since the linear span of

$\{{w}_{j}\}$ is dense on the space

${\stackrel{\u02da}{H}}_{1,2}^{\pi}(\mathrm{\Omega})$, passing to the limit as

$N\to \mathrm{\infty}$, we get

${(Lu-F,{(\stackrel{\u02c6}{L})}^{\ast}\eta )}_{{L}_{2}(\mathrm{\Omega})}=0$

(24)

for every

$\eta \in {\stackrel{\u02da}{H}}_{1,2}^{\pi}(\mathrm{\Omega})$. If we set

$\lambda =Lu-F$, since

${C}_{0}^{\mathrm{\infty}}(\mathrm{\Omega})\subset {\stackrel{\u02da}{H}}_{1,2}^{\pi}(\mathrm{\Omega})$, from (24) we have

$\stackrel{\u02c6}{L}\lambda =0$ in the generalized functions sense. Moreover,

${\parallel \lambda \parallel}_{{L}_{2}(\mathrm{\Omega})}\le C{\parallel u\parallel}_{{\stackrel{\u02da}{H}}_{1}^{\pi}(\mathrm{\Omega})}+{\parallel F\parallel}_{{L}_{2}(\mathrm{\Omega})}$

holds and, by using ${\parallel u\parallel}_{{\stackrel{\u02da}{H}}_{1}^{\pi}(\mathrm{\Omega})}\le C{\parallel {F}_{\phi}\parallel}_{{L}_{2}(\mathrm{\Omega})}$, it can be seen that (7) holds. In the above expressions, by *C* we denote generic constants which depend only on the given functions and Lebesgue measure of the domain *D*.

Now it remains to show that

$u\in \mathrm{\Gamma}(A)$. Since

$u\in {L}_{2}(\mathrm{\Omega})$ and

$F\in {H}_{2}^{\pi}(\mathrm{\Omega})$, for

$\mathcal{F}=\stackrel{\u02c6}{L}F\in {L}_{2}(\mathrm{\Omega})$, from (24) we have

${(u,{A}^{\ast}\eta )}_{{L}_{2}(\mathrm{\Omega})}={(u,{L}^{\ast}{(\stackrel{\u02c6}{L})}^{\ast}\eta )}_{{L}_{2}(\mathrm{\Omega})}={(Lu,{(\stackrel{\u02c6}{L})}^{\ast}\eta )}_{{L}_{2}(\mathrm{\Omega})}={(F,{(\stackrel{\u02c6}{L})}^{\ast}\eta )}_{{L}_{2}(\mathrm{\Omega})}={(\mathcal{F},\eta )}_{{L}_{2}(\mathrm{\Omega})}$

for any $\eta \in {C}_{0}^{\mathrm{\infty}}(\mathrm{\Omega})$, which implies that $\mathcal{F}=Au$ in the generalized functions sense, *i.e.*, $u\in {\mathrm{\Gamma}}^{\u2033}(A)$.

Moreover, from (19) we have

${\mathcal{P}}_{N}A{u}_{N}={\mathcal{P}}_{N}\mathcal{F}$, where

${\mathcal{P}}_{N}$ is the orthogonal projector. Since the system

$\{{w}_{1},{w}_{2},\dots \}$ is orthogonal and complete in

${L}_{2}(\mathrm{\Omega})$,

${\mathcal{P}}_{N}\mathcal{F}\to \mathcal{F}$ strongly,

*i.e.*,

${\mathcal{P}}_{N}A{u}_{N}\to \mathcal{F}=Au$ strongly in

${L}_{2}(\mathrm{\Omega})$ as

$N\to \mathrm{\infty}$. Then, since

${u}_{N}\to u$ weakly in

${L}_{2}(\mathrm{\Omega})$ as

$N\to \mathrm{\infty}$, we have

${({\mathcal{P}}_{N}A{u}_{N},{u}_{N})}_{{L}_{2}(\mathrm{\Omega})}\to {(Au,u)}_{{L}_{2}(\mathrm{\Omega})}$ as

$N\to \mathrm{\infty}$. On the other hand, since the projection operator

${\mathcal{P}}_{N}$ is self-adjoint in

${L}_{2}(\mathrm{\Omega})$,

${({\mathcal{P}}_{N}A{u}_{N},{u}_{N})}_{{L}_{2}(\mathrm{\Omega})}={(A{u}_{N},{\mathcal{P}}_{N}^{\ast}{u}_{N})}_{{L}_{2}(\mathrm{\Omega})}={(A{u}_{N},{\mathcal{P}}_{N}{u}_{N})}_{{L}_{2}(\mathrm{\Omega})}={(A{u}_{N},{u}_{N})}_{{L}_{2}(\mathrm{\Omega})},$

thus ${(A{u}_{N},{u}_{N})}_{{L}_{2}(\mathrm{\Omega})}\to {(Au,u)}_{{L}_{2}(\mathrm{\Omega})}$ as $N\to \mathrm{\infty}$, and so $u\in \mathrm{\Gamma}(A)$. The proof of Theorem 1 is complete. □

*Proof of Lemma 1* Since

$K=\frac{1}{r}$ is constant, and hence

${K}_{{x}_{1}}={K}_{{x}_{2}}={K}_{\phi}=0,$

to prove that condition (6) holds, we only need to show that

*g* satisfies

${g}_{{x}_{1}}cos\phi +{g}_{{x}_{2}}sin\phi +{g}_{\phi}\frac{1}{r}>\frac{1}{{r}^{2}}+{g}^{2}.$

By taking into account that

$M={sup}_{({x}_{1},{x}_{2})\in D}({x}_{1}^{2}+{x}_{2}^{2})$ and

$r>3\sqrt{M}$, for the function

$g({x}_{1},{x}_{2},\phi )=\frac{1}{3M}({x}_{1}cos\phi +{x}_{2}sin\phi )$

given in (8) defined on

$D\times (0,2\pi )$, we obtain

$\begin{array}{r}{g}_{{x}_{1}}cos\phi +{g}_{{x}_{2}}sin\phi +{g}_{\phi}\frac{1}{r}-\frac{1}{{r}^{2}}-{g}^{2}\\ \phantom{\rule{1em}{0ex}}=\frac{1}{3M}(1+\frac{1}{r}(-{x}_{1}sin\phi +{x}_{2}cos\phi ))\\ \phantom{\rule{2em}{0ex}}-\frac{1}{{r}^{2}}-\frac{1}{9{M}^{2}}{({x}_{1}cos\phi +{x}_{2}sin\phi )}^{2}\\ \phantom{\rule{1em}{0ex}}\ge \frac{1}{3M}(1-\frac{1}{r}\sqrt{M})-\frac{1}{{r}^{2}}-\frac{1}{9M}\\ \phantom{\rule{1em}{0ex}}>\frac{1}{3M}(1-\frac{1}{3})-\frac{2}{9M}\\ \phantom{\rule{1em}{0ex}}=0,\end{array}$

and the proof is complete. □

*Proof of Lemma 2* For

$\begin{array}{rcl}K({x}_{1},{x}_{2},\phi )& =& {({x}_{1}sin\phi -{x}_{2}cos\phi +{({R}^{2}-{({x}_{1}cos\phi +{x}_{2}sin\phi )}^{2})}^{1/2})}^{-1}\\ =& \frac{{({R}^{2}-{({x}_{1}cos\phi +{x}_{2}sin\phi )}^{2})}^{1/2}-{x}_{1}sin\phi +{x}_{2}cos\phi}{{R}^{2}-{x}_{1}^{2}-{x}_{2}^{2}},\end{array}$

we have

$\begin{array}{c}\begin{array}{rl}{K}_{{x}_{1}}=& \frac{1}{{({R}^{2}-{x}_{1}^{2}-{x}_{2}^{2})}^{2}}((\frac{-({x}_{1}cos\phi +{x}_{2}sin\phi )cos\phi}{{({R}^{2}-{({x}_{1}cos\phi +{x}_{2}sin\phi )}^{2})}^{1/2}}-sin\phi )({R}^{2}-{x}_{1}^{2}-{x}_{2}^{2})\\ +2{x}_{1}({({R}^{2}-{({x}_{1}cos\phi +{x}_{2}sin\phi )}^{2})}^{1/2}-{x}_{1}sin\phi +{x}_{2}cos\phi )),\end{array}\hfill \\ \begin{array}{rl}{K}_{{x}_{2}}=& \frac{1}{{({R}^{2}-{x}_{1}^{2}-{x}_{2}^{2})}^{2}}((\frac{-({x}_{1}cos\phi +{x}_{2}sin\phi )sin\phi}{{({R}^{2}-{({x}_{1}cos\phi +{x}_{2}sin\phi )}^{2})}^{1/2}}+cos\phi )({R}^{2}-{x}_{1}^{2}-{x}_{2}^{2})\\ +2{x}_{2}({({R}^{2}-{({x}_{1}cos\phi +{x}_{2}sin\phi )}^{2})}^{1/2}-{x}_{1}sin\phi +{x}_{2}cos\phi )),\end{array}\hfill \\ {K}_{\phi}=\frac{-({x}_{1}cos\phi +{x}_{2}sin\phi )}{({R}^{2}-{x}_{1}^{2}-{x}_{2}^{2})}(\frac{(-{x}_{1}sin\phi +{x}_{2}cos\phi )}{{({R}^{2}-{({x}_{1}cos\phi +{x}_{2}sin\phi )}^{2})}^{1/2}}+1)\hfill \end{array}$

and

$\begin{array}{r}-{K}_{{x}_{1}}sin\phi +{K}_{{x}_{2}}cos\phi \\ \phantom{\rule{1em}{0ex}}=\frac{1}{{({R}^{2}-{x}_{1}^{2}-{{x}_{2}}^{2})}^{2}}({R}^{2}-{({x}_{1}cos\phi +{x}_{2}sin\phi )}^{2}+{({x}_{1}sin\phi -{x}_{2}cos\phi )}^{2}\\ \phantom{\rule{2em}{0ex}}-2({x}_{1}sin\phi -{x}_{2}cos\phi ){({R}^{2}-{({x}_{1}cos\phi +{x}_{2}sin\phi )}^{2})}^{1/2})\\ \phantom{\rule{1em}{0ex}}={K}^{2}.\end{array}$

Since

$M={sup}_{({x}_{1},{x}_{2})\in D}({x}_{1}^{2}+{x}_{2}^{2})$ and

$R>\sqrt{M}$, for

$g({x}_{1},{x}_{2},\phi )=\frac{1}{2M}({x}_{1}cos\phi +{x}_{2}sin\phi )$

given in (12) defined on

$D\times (0,2\pi )$, we have

$\begin{array}{rl}-{K}_{\phi}g=& \frac{({x}_{1}cos\phi +{x}_{2}sin\phi )}{({R}^{2}-{x}_{1}^{2}-{x}_{2}^{2})}(\frac{(-{x}_{1}sin\phi +{x}_{2}cos\phi )}{{({R}^{2}-{({x}_{1}cos\phi +{x}_{2}sin\phi )}^{2})}^{1/2}}+1)\\ \times \frac{1}{2M}({x}_{1}cos\phi +{x}_{2}sin\phi )\\ =& \frac{{({x}_{1}cos\phi +{x}_{2}sin\phi )}^{2}(-{x}_{1}sin\phi +{x}_{2}cos\phi +{({R}^{2}-{({x}_{1}cos\phi +{x}_{2}sin\phi )}^{2})}^{1/2})}{2M({R}^{2}-{x}_{1}^{2}-{x}_{2}^{2}){({R}^{2}-{({x}_{1}cos\phi +{x}_{2}sin\phi )}^{2})}^{1/2}}\\ \ge & 0.\end{array}$

Hence, to prove that condition (6) holds, we only need to show that

${g}_{{x}_{1}}cos\phi +{g}_{{x}_{2}}sin\phi +{g}_{\phi}K>{g}^{2}$

holds. If we take into account again that

$M={sup}_{({x}_{1},{x}_{2})\in D}({x}_{1}^{2}+{x}_{2}^{2})$ and

$R>\sqrt{M}$, then we obtain

$\begin{array}{r}{g}_{{x}_{1}}cos\phi +{g}_{{x}_{2}}sin\phi +{g}_{\phi}K-{g}^{2}\\ \phantom{\rule{1em}{0ex}}=\frac{1}{2M}(1+\frac{-{x}_{1}sin\phi +{x}_{2}cos\phi}{{x}_{1}sin\phi -{x}_{2}cos\phi +{({R}^{2}-{({x}_{1}cos\phi +{x}_{2}sin\phi )}^{2})}^{1/2}})\\ \phantom{\rule{2em}{0ex}}-\frac{1}{4{M}^{2}}{({x}_{1}cos\phi +{x}_{2}sin\phi )}^{2}\\ \phantom{\rule{1em}{0ex}}\ge \frac{1}{2M}\frac{{({R}^{2}-{({x}_{1}cos\phi +{x}_{2}sin\phi )}^{2})}^{1/2}}{{x}_{1}sin\phi -{x}_{2}cos\phi +{({R}^{2}-{({x}_{1}cos\phi +{x}_{2}sin\phi )}^{2})}^{1/2}}-\frac{1}{4M}\\ \phantom{\rule{1em}{0ex}}>0.\end{array}$

The proof is complete. □