Appendix A: Proof of Lemma 1
Proof
Note that \(A^\epsilon \) is the infinitesimal generator of a \(C_0\) semigroup S(t) on H, as known in Stewart [32]. Moreover,
$$\begin{aligned} \begin{aligned}&\int _{0}^{T}\left\| B^\epsilon u^\epsilon _s(\cdot )\right\| _0+\left\| \left( u^\epsilon _s(\cdot )\right) \sigma (\frac{\cdot }{\epsilon })\right\| _0^2ds\\&\quad \le C_2\int _{0}^{T}\left\| \int _{{\mathbb {R}}}c(\frac{\cdot -y}{\epsilon })u^\epsilon _s(y)dy\right\| _0ds\\&\qquad +\, C_2\int _{0}^{T}\left\| u^\epsilon _s(\cdot )\right\| _0ds\left\| \int _{{\mathbb {R}}}c(\frac{\cdot -y}{\epsilon })dy\right\| _0\\&\qquad +\,\int _{0}^{T}\left\| (u^\epsilon _s(\cdot ))^2\sigma (\frac{\cdot }{\epsilon })^2\right\| _0ds. \end{aligned} \end{aligned}$$
(40)
Combined with the uniform estimates in Lemma 4, we conclude that
$$\begin{aligned} \begin{aligned} \left\| \int _{{\mathbb {R}}}c(\frac{\cdot -y}{\epsilon })u^\epsilon _s(y)dy\right\| _0^2&\le \int _{{\mathbb {R}}}c(y)dy\int _{{\mathbb {R}}}c(q)dq\int _{{\mathbb {R}}}u^\epsilon _s(x+\epsilon y)u^\epsilon _s(x+\epsilon q)dx\\&\le a_1^2\left\| u_s^\epsilon \right\| ^2_0<\infty . \end{aligned} \end{aligned}$$
So the right hand side of (40) is finite. Hence the equation (1) has a solution given by
$$\begin{aligned} u^\epsilon _t(x)\!=\!S(t)u_0(x)\!+\!\int _0^tS(t\!-\!s)B^\epsilon u^\epsilon _s(x)ds\!+\!\int _0^t S(t\!-\!s)u^\epsilon _s(x)\sigma \left( \frac{x}{\epsilon }\right) dW_s. \end{aligned}$$
The mild solution of the equation is unique (Theorem 3.5 of [16]). \(\square \)
Appendix B: Uniqueness of \(h_{1}(\eta ) \;\text {and} \;h_{2}(\eta )\)
Proof
We define the bilinear form:
$$\begin{aligned} \begin{aligned} a[u,v]&=\int _{{\mathbb {T}}}\left[ \int _{{\mathbb {R}}}c(\eta -q)\big (\lambda ^m(q)u(q)-\lambda ^m(\eta )u(\eta )\big ) dq\right] v(\eta )d\eta \\&\quad +\int _{{\mathbb {T}}}(a^m(\eta )u(\eta ))''v(\eta )d\eta -\int _{{\mathbb {T}}}(b^m(\eta )u(\eta ))'v(\eta )d\eta . \end{aligned} \end{aligned}$$
for every \(u, v\in H^1.\)
At first, we verify the conditions of the Fredholm alternative theorem. We want to show that there exist positive constants \(\nu ,\mu ,\) such that:
$$\begin{aligned} \left| a[u,v]\right| \le \nu \Vert u\Vert _1\Vert v\Vert _1, \end{aligned}$$
and
$$\begin{aligned} \frac{\kappa _1}{2}\Vert u\Vert _1^2\le a[u,u]+\mu \Vert u\Vert _0^2, \end{aligned}$$
for every \(u,v\in H^1({\mathbb {T}}).\) Note that
$$\begin{aligned} \begin{aligned} \left| a[u,v]\right|&\le \left| \int _{{\mathbb {T}}}\left[ \int _{{\mathbb {R}}}c(\eta -q)\left( \lambda ^m(q)u(q)-\lambda ^m(\eta )u(\eta )\right) dq \right] v(\eta )d\eta \right| \\&\quad +\left| \int _{{\mathbb {T}}}(a^m(\eta )u(\eta ))''v(\eta )d\eta \right| +\left| \int _{{\mathbb {T}}}(b^m(\eta )u(\eta ))'v(\eta )d\eta \right| . \end{aligned} \end{aligned}$$
(41)
For the first term of (41),
$$\begin{aligned} \begin{aligned}&\left| \int _{{\mathbb {T}}}\left[ \int _{{\mathbb {R}}}c(\eta -q)\lambda ^m(q)u(q)dq\right] v(\eta )d\eta -\int _{{\mathbb {T}}}\left[ \int _{{\mathbb {R}}}c(\eta -q)\lambda ^m(\eta )u(\eta ))dq\right] v(\eta )d\eta \right| \\&\quad \le \left[ \int _{{\mathbb {T}}}\left( \int _{{\mathbb {R}}}c(\eta -q)\lambda ^m(q)u(q)dq\right) ^2 d\eta \right] ^{\frac{1}{2}} \left( \int _{{\mathbb {T}}}v(\eta )^2d\eta \right) ^{\frac{1}{2}}+a_1C_2\int _{{\mathbb {T}}}u(\eta )v(\eta )d\eta .\\ \end{aligned} \end{aligned}$$
In fact,
$$\begin{aligned} \begin{aligned}&\int _{{\mathbb {T}}}\left( \int _{{\mathbb {R}}}c(\eta -q)\lambda ^m(q)u(q)dq\right) ^2d\eta \\&\quad =\int _{{\mathbb {T}}}\left( \int _{{\mathbb {R}}}c(\eta -q)\lambda ^m(q)u(q)dq\right) \left( \int _{{\mathbb {R}}}c(\eta -q)\lambda ^m(q)u(q)dq\right) d\eta \\&\quad \le \frac{C^2_2}{\delta ^2}\int _{{\mathbb {R}}}c(q)dq\int _{{\mathbb {R}}}c(q)dq\int _{{\mathbb {T}}}u(\eta +q)u(\eta +q) d\eta \le a_2^2\frac{C^2_2}{\delta ^2}\left\| u\right\| _0^2. \end{aligned} \end{aligned}$$
(42)
Combining with (41), we conclude that
$$\begin{aligned} \left| a[u,v]\right|&\le C_3 \Vert u\Vert _0\Vert v\Vert _0+C_4 (\Vert u\Vert _0\Vert v'\Vert _0 +\Vert u'\Vert _0\Vert v'\Vert _0)+C_5\Vert u\Vert _0\Vert v'\Vert _0 \\&\le \nu \Vert u\Vert _1\Vert v\Vert _1. \end{aligned}$$
We now use the assumptions to infer that
$$\begin{aligned} \begin{aligned} \kappa _1\Vert u'\Vert _0&\le -\int _{{\mathbb {T}}}(a(\eta )u(\eta ))''u(\eta )d\eta =a[u,u]-\int _{{\mathbb {T}}}(b(\eta )u(\eta ))'u(\eta )d\eta \\&\quad +\int _{{\mathbb {T}}}\left[ \int _{{\mathbb {R}}}c(\eta -q)\big (\lambda (q)u(q)-\lambda (\eta )u(\eta )\big )dq\right] u(\eta )d\eta \\ {}&\le a[u,u]+\int _{{\mathbb {T}}}\left( \Vert b\Vert _{\infty }\vert u'\vert \cdot \vert u\vert +C_7\vert u\vert ^2\right) d\eta . \end{aligned} \end{aligned}$$
(43)
Now we make use of the Young’s inequality
$$\begin{aligned} ab\le \delta _1 a^2+\frac{1}{4\delta _1}b^2,\; \text {for }\;\text {every} \;\delta _1>0. \end{aligned}$$
Using this in the second term on the right hand side of (43), we obtain
$$\begin{aligned} \int _{{\mathbb {T}}}\vert u'\vert \cdot \vert u\vert d\eta \le \delta _1 \Vert u'\Vert _0^2+\frac{1}{4\delta _1} \Vert u\Vert _0^2. \end{aligned}$$
We choose \(\delta _1\), so that
$$\begin{aligned} \kappa _1-\Vert b\Vert _{\infty }\delta _1=\frac{\kappa _1}{2}. \end{aligned}$$
Thus
$$\begin{aligned} \frac{\kappa _1}{2}\Vert u'\Vert _0^2\le a[u,u]+\frac{1}{4\delta _1}\Vert b\Vert _{\infty } \Vert u\Vert _0^2+C_7\Vert u\Vert _0^2. \end{aligned}$$
We now add \(\frac{\kappa _1}{2}\Vert u\Vert _0^2\) on the both sides of the preceding inequality to obtain
$$\begin{aligned} \frac{\kappa _1}{2}\Vert u\Vert _1^2\le a[u,u]+\mu \Vert u\Vert _0^2, \end{aligned}$$
with
$$\begin{aligned} \mu =\frac{1}{4\delta _1}\Vert b\Vert _{\infty }+C_7+\frac{\kappa _1}{2}. \end{aligned}$$
Next we consider the resolvent operator
$$\begin{aligned} R_{\left( {\tilde{T}}_m\right) ^*}(\lambda )=\left( ({\tilde{T}}_m)^*+\lambda I\right) ^{-1}, \end{aligned}$$
where I stands for the identity operator and \(\lambda >0\). Note that this operator is compact. For \(\lambda \) sufficiently large, consequently, Fredholm theorem can be used for \(R_{\left( {\tilde{T}}_m\right) ^*}(\lambda )\). From the fact that the Fredholm alternative for \(R_{\left( {\tilde{T}}_m\right) ^*}(\lambda )\) implies the Fredholm alternative for \({\tilde{T}}_m^*,\) Fredholm theorem can be used for \({\tilde{T}}_m^*\) (Lemma 7.11 of [28]). Moreover, it is easy to see that \(Ker \left( {\tilde{T}}_m\right) ^*=\{C\}\), where C is a constant. Then we want to show the solvability condition:
$$\begin{aligned} \int _{{\mathbb {T}}}l(\eta )d\eta =0. \end{aligned}$$
(44)
We take \(z=\eta -q.\) Noting the fact that
$$\begin{aligned} \int _{{\mathbb {R}}}\int _{{\mathbb {T}}}q c(\eta -q)m(\eta )\lambda (\eta )d\eta dq= \int _{{\mathbb {R}}}\int _{{\mathbb {T}}}\eta c(\eta -q)m(q)\lambda (q)d\eta dq, \end{aligned}$$
we infer that
$$\begin{aligned} \begin{aligned} \int _{{\mathbb {T}}}l(\eta )d\eta&=\int _{{\mathbb {R}}}\int _{{\mathbb {T}}}qc(\eta -q)\big (m(\eta )\lambda (\eta )-m(q)\lambda (q)\big )d\eta dq\\&\quad +\,\int _{{\mathbb {T}}}b(\eta )m(\eta )d\eta -\int _{{\mathbb {T}}}2(a(\eta )m(\eta ))'d\eta \\&=\int _{{\mathbb {T}}}\eta \int _{{\mathbb {R}}}c(q-\eta )\big (m(q)\lambda (q)-m(\eta )\lambda (\eta )\big )dqd\eta \\&\quad +\,\int _{{\mathbb {T}}}b(\eta )m(\eta )d\eta -\int _{{\mathbb {T}}}2(a(\eta )m(\eta ))'d\eta \\&=\int _{{\mathbb {T}}}\eta \left[ -(a(\eta )m(\eta ))''+(b(\eta )m(\eta ))'\right] d\eta \\&\quad +\,\int _{{\mathbb {T}}}b(\eta )m(\eta )d\eta -\int _{{\mathbb {T}}}2(a(\eta )m(\eta ))'d\eta \\&=0. \end{aligned} \end{aligned}$$
The solvability condition \(\int _{{\mathbb {T}}}l_1(\eta )d\eta \) will be verified in the “Appendix C”. Thus, the solution \(h_1(\eta )\) and \(h_2(\eta )\) is existence and uniqueness. \(\square \)
Appendix C: Proof of Lemma 11
Proof
Substituting \(\xi _\epsilon \) defined in (24) into \((T^\epsilon )^*\xi _\epsilon :\)
$$\begin{aligned} \begin{aligned} (T^\epsilon )^*(\xi ^\epsilon )(x)&=\frac{1}{\epsilon ^3}\int _{{\mathbb {R}}}c\left( \frac{x-y}{\epsilon }\right) \bigg \{\lambda \left( \frac{y}{\epsilon }\right) m\left( \frac{y}{\epsilon }\right) \left[ \xi (y)\!+\!\epsilon h_{1}\left( \frac{y}{\epsilon }\right) \xi '(y)\!+\!\epsilon ^2 h_2 \left( \frac{y}{\epsilon }\right) \xi ''(y)\right] \\&\quad -\,\lambda \left( \frac{x}{\epsilon }\right) m\left( \frac{x}{\epsilon }\right) \left[ \xi (x)+\epsilon h_{1}\left( \frac{x}{\epsilon }\right) \xi '(x)+\epsilon ^2 h_2 \left( \frac{x}{\epsilon }\right) \xi ''(x)\right] \bigg \}\\&\quad +\,\left\{ a^m\left( \frac{x}{\epsilon }\right) \left[ \xi (x)+\epsilon h_{1}\left( \frac{x}{\epsilon }\right) \xi '(x)+\epsilon ^2 h_2 \left( \frac{x}{\epsilon }\right) \xi ''(x)\right] \right\} ''\\&\quad -\,\frac{1}{\epsilon }\left\{ b^m\left( \frac{x}{\epsilon }\right) \left[ \xi (x)+\epsilon h_{1}\left( \frac{x}{\epsilon }\right) \xi '(x)+\epsilon ^2 h_2 \left( \frac{x}{\epsilon }\right) \xi ''(x)\right] \right\} 'dy. \end{aligned} \end{aligned}$$
First of all, we consider the term \((B^\epsilon )^*(\xi ^\epsilon )(x)\),
$$\begin{aligned} \begin{aligned} (B^\epsilon )^*(\xi ^\epsilon )(x)&=\frac{1}{\epsilon ^2}\int _{{\mathbb {R}}}c(z)\bigg \{\lambda \left( \frac{x}{\epsilon }-z\right) m\left( \frac{x}{\epsilon }-z\right) \Big [\xi (x-\epsilon z)\\&\quad +\,\epsilon h_{1}\left( \frac{x}{\epsilon }-z\right) \xi '(x-\epsilon z) +\epsilon ^2 h_2\left( \frac{x}{\epsilon }-z\right) \xi ''(x-\epsilon z)\Big ]\\&\quad -\,\lambda \left( \frac{x}{\epsilon }\right) m\left( \frac{x}{\epsilon }\right) \left[ \xi (x)+\epsilon h_{1}\left( \frac{x}{\epsilon }\right) \xi '(x)+\epsilon ^2 h_2 \left( \frac{x}{\epsilon }\right) \xi ''(x)\right] \bigg \}dz. \end{aligned} \end{aligned}$$
Using the following identities based on the integral form of remainder term in the Taylor expansion
$$\begin{aligned} \xi (y)=\xi (x)+\int _{0}^{1}\frac{\partial }{\partial t}\xi (x+(y-x)t)dt=\xi (x)+\int _{0}^{1}\xi '(x+(y-x)t)\cdot (y-x)dt, \end{aligned}$$
and
$$\begin{aligned} \xi (y)=\xi (x)+\xi '(x)(y-x)+\int _{0}^{1}\xi ''(x+(y-x)t)(y-x)\cdot (y-x)(1-t)dt, \end{aligned}$$
which is valid for each \(x,y\in {\mathbb {R}}\), we conclude that
$$\begin{aligned} \begin{aligned} (B^\epsilon )^*(\xi ^\epsilon )(x)&= \frac{1}{\epsilon ^2}\int _{{\mathbb {R}}}c(z) \bigg \{\lambda \left( \frac{x}{\epsilon }-z\right) m\left( \frac{x}{\epsilon }-z\right) \bigg [\xi (x)-\epsilon z\xi '(x)\\&\quad +\,\epsilon ^2\int _{0}^{1}\xi ''(x-\epsilon zt)\cdot z^2 (1-t)dt+\epsilon h_{1}\left( \frac{x}{\epsilon }-z\right) \Big (\xi '(x)-\epsilon z\xi ''(x)\\&\quad +\,\epsilon ^2\int _{0}^{1}\xi '''(x-\epsilon zt)\cdot z^2 (1-t)dt\Big )+\epsilon ^2 h_2 \left( \frac{x}{\epsilon }-z\right) \xi ''(x-\epsilon z)\bigg ]\\&\quad -\,\lambda \left( \frac{x}{\epsilon }\right) m\left( \frac{x}{\epsilon }\right) \left[ \xi (x)+\epsilon h_{1}\left( \frac{x}{\epsilon }\right) \xi '(x)+\epsilon ^2 h_2 \left( \frac{x}{\epsilon }\right) \xi ''(x)\right] \bigg \}dz. \end{aligned} \end{aligned}$$
Collecting the equal power terms with \((A^\epsilon )^* \xi ^\epsilon \), we obtain
$$\begin{aligned} \begin{aligned}&(T^\epsilon )^*(\xi ^\epsilon )(x)\\&\quad = \frac{1}{\epsilon ^2}\xi (x)\bigg \{\int _{{\mathbb {R}}}c(z)\bigg [\lambda \big (\frac{x}{\epsilon }-z\big )m\big (\frac{x}{\epsilon }-z\big )-\lambda \big (\frac{x}{\epsilon }\big )m\big (\frac{x}{\epsilon }\big )\bigg ]dz+\left( am\right) ''\big (\frac{x}{\epsilon }\big )-\left( bm\right) '\big (\frac{x}{\epsilon }\big )\bigg \}\\&\qquad +\,\frac{1}{\epsilon }\xi '(x)\bigg \{\int _{{\mathbb {R}}}c(z)\bigg [\left( -z+h_1\big (\frac{x}{\epsilon }-z\big )\right) \lambda \big (\frac{x}{\epsilon }-z\big )m\big (\frac{x}{\epsilon }-z\big )-\lambda \big (\frac{x}{\epsilon }\big )m\big (\frac{x}{\epsilon }\big )h_1\big (\frac{x}{\epsilon }\big )\bigg ]dz\\&\qquad +\,2\left( am\right) '\big (\frac{x}{\epsilon }\big )+\left( amh_{1}\right) ''\big (\frac{x}{\epsilon }\big )-b\big (\frac{x}{\epsilon }\big )m\big (\frac{x}{\epsilon }\big )-\left( bmh_{1}\right) '\big (\frac{x}{\epsilon }\big )\bigg \}\\&\qquad +\,\xi ''(x)\bigg \{\int _{\mathbb {R}}c(z)\bigg [\lambda \big (\frac{x}{\epsilon }\!-\!z\big )m\big (\frac{x}{\epsilon }\!-\!z\big )\big (\frac{1}{2}z^2\!-\!zh_1\big (\frac{x}{\epsilon }\!-\!z\big )+h_2\big (\frac{x}{\epsilon }-z\big )\big )\\&\qquad -\,\lambda \big (\frac{x}{\epsilon }\big )m\big (\frac{x}{\epsilon }\big )h_2\big (\frac{x}{\epsilon }\big )\bigg ]dz +a\big (\frac{x}{\epsilon }\big )m\big (\frac{x}{\epsilon }\big )+2(amh_1)'\big (\frac{x}{\epsilon }\big )+(amh_2)''\big (\frac{x}{\epsilon }\big )-(bmh_2)'\big (\frac{x}{\epsilon }\big )\\&\qquad -\,h_1\big (\frac{x}{\epsilon }\big )b\big (\frac{x}{\epsilon }\big )m\big (\frac{x}{\epsilon }\big )\bigg \} +\phi _\epsilon (x), \end{aligned} \end{aligned}$$
(45)
with
$$\begin{aligned} \begin{aligned} \phi _\epsilon (x)&=\frac{1}{\epsilon ^2}\int _{\mathbb {R}}c(z)\bigg \{\epsilon ^2\int _0^1\lambda \big (\frac{x}{\epsilon }-z\big )m\big (\frac{x}{\epsilon }-z\big )\xi ''(x-\epsilon zt)z^2(1-t)dt\\&\quad -\, \frac{\epsilon ^2}{2}\lambda \big (\frac{x}{\epsilon }-z\big ) m\big (\frac{x}{\epsilon }-z\big )\xi ''(x)z^2 +\epsilon ^3 h_1\big (\frac{x}{\epsilon }-z\big )\int _0^1 \xi '''(x-\epsilon zt)z^2(1-t)dt\\&\quad -\, \epsilon ^3 h_2\big (\frac{x}{\epsilon }-z\big )\int _0^1\xi '''(x-\epsilon zt)zdt\bigg \}dz. \end{aligned} \end{aligned}$$
Denote \(\eta =\frac{x}{\epsilon }\) a variable on the period: \(\eta \in {\mathbb {T}}\). we collect all the terms of the order \(\epsilon ^{-2}\) in (45) and equate them to 0.
$$\begin{aligned} \int _{{\mathbb {R}}}c(z)\big [\lambda (\eta -z)m(\eta -z)-\lambda (\eta )m(\eta )\big ]dz +(a(\eta )m(\eta ))''-(b(\eta )m(\eta ))'=({\tilde{T}})^*m(\eta )=0. \end{aligned}$$
From the fact that \(({\tilde{T}}_m)^*(h_1)(\eta )=l(\eta )\), for the terms of order \(\epsilon ^{-1}\), we have
$$\begin{aligned} \begin{array}{rl} 0&{}=\displaystyle \int _{{\mathbb {R}}}c(z)\Big [\big (-z+h_1(\eta -z)\big )\lambda (\eta -z)m(\eta -z)-\lambda (\eta )m(\eta )h_1(\eta ))\Big ]dz\\ &{}\quad +\,2(a(\eta )m(\eta ))' +\left( a(\eta )m(\eta )h_{1}(\eta )\right) ''-b(\eta )m(\eta )-(b(\eta )m(\eta )h_{1}(\eta ))'. \end{array} \end{aligned}$$
(46)
At last, we collect the term of the order \(\varepsilon ^0.\) Our goal is to find the function \(h_2\), such that the sum of these terms will be equal to \(T^0\xi =Q \xi ''\) with \(Q>0\). Then we have
$$\begin{aligned} \begin{aligned}&({\tilde{T}}_m)^*(h_2)(\eta )\\&\quad =-Q+\int _{{\mathbb {R}}}c(z)\lambda (\eta -z)m(\eta -z)\left[ \frac{1}{2}z^2-zh_1(\eta -z)\right] dz +a(\eta )m(\eta )\\&\qquad +\,2(a(\eta )m(\eta )h_1(\eta ))'-b(\eta )m(\eta )h_1(\eta ). \end{aligned} \end{aligned}$$
(47)
Similar to the equality (44). In order to ensure the uniqueness of the function \(h_2\), we see that Q is determined from the following solvability condition for Eq. (44)
$$\begin{aligned} \begin{aligned} Q&=\int _{{\mathbb {T}}}\int _{{\mathbb {R}}}c(z)\lambda (\eta -z)m(\eta -z)\left[ \frac{1}{2}z^2-zh_1(\eta -z)\right] dzd\eta \\&\quad +\,\int _{{\mathbb {T}}}a(\eta )m(\eta )d\eta -\int _{{\mathbb {T}}}b(\eta )m(\eta )h_1(\eta )d\eta . \end{aligned} \end{aligned}$$
(48)
Next, let’s simplify the expression of Q. A short calculation revealed that
$$\begin{aligned} \begin{aligned}&\int _{{\mathbb {T}}}\int _{{\mathbb {R}}}c(z)\lambda (\eta -z)m(\eta -z)zh_1(\eta -z)dzd\eta \\&\quad =\int _{{\mathbb {T}}}\int _{{\mathbb {R}}}(\eta -q)c(\eta -q)\lambda (q)m(q)h_1(q)dqd\eta \\&\quad =\int _{{\mathbb {T}}}\int _{{\mathbb {R}}}c(q-\eta )(q-\eta )\lambda (\eta )m(\eta )h_1(\eta )dqd\eta \\&\quad =\int _{{\mathbb {T}}}\left[ \int _{{\mathbb {R}}}c(\eta -q)(q-\eta )dq\right] \lambda (\eta )m(\eta )h_1(\eta )d\eta \\&\quad =0 \end{aligned} \end{aligned}$$
For the third term of Q,
$$\begin{aligned} \begin{aligned}&\int _{{\mathbb {T}}}-b^m(\eta )h_1(\eta )d\eta =\int _{{\mathbb {T}}}{\tilde{T}}_m\chi (\eta )h_1(\eta )d\eta =\int _{{\mathbb {T}}}\chi (\eta )({\tilde{T}}_m)^*h_1(\eta )d\eta \\&\quad =\int _{{\mathbb {T}}}\chi (\eta )\left[ \int _{{\mathbb {R}}}zc(z)m(\eta -z)\lambda (\eta -z)dz+b^m(\eta )-2(a^m(\eta ))'\right] d\eta , \end{aligned} \end{aligned}$$
and
$$\begin{aligned} \begin{aligned}&\int _{{\mathbb {T}}}\chi (\eta )b^m(\eta )d\eta =-\int _{{\mathbb {T}}}\chi (\eta ){\tilde{T}}_m\chi (\eta )d\eta \\&\quad =\int _{{\mathbb {T}}}a^m(\eta )(\chi '(\eta ))^2d\eta +\frac{1}{2}\int _{{\mathbb {T}}}\int _{{\mathbb {R}}}\lambda ^m(\eta )c(z)\left[ \chi (\eta -z)-\chi (\eta )\right] ^2dzd\eta \\&\quad =\int _{{\mathbb {T}}}a^m(\eta )(\chi '(\eta ))^2d\eta +\frac{1}{2}\int _{{\mathbb {T}}}\int _{{\mathbb {R}}}\lambda ^m(q)c(\eta -q)\left[ \chi (q)-\chi (\eta )\right] ^2dqd\eta ,\\ \end{aligned} \end{aligned}$$
we conclude that
$$\begin{aligned} Q\!=\!\int _{{\mathbb {T}}}a(\eta )m(\eta )(\chi '(\eta )\!+\!1)^2d\eta \!+\!\frac{1}{2}\int _{{\mathbb {T}}}\int _{{\mathbb {R}}}c(\eta \!-\!q)\lambda (q)m(q)\left[ (\eta \!-\!q)\!+\!(\chi (\eta )\!-\!\chi (q))\right] ^2d\eta dq. \end{aligned}$$
Our last step is to show that \(\left\| \phi _\varepsilon (x)\right\| _0\) is vanishing as \(\epsilon \rightarrow 0.\)
Choose the term of order \(\epsilon ^0\) of \(\phi _\epsilon (x)\), and denote it by \(\phi _{\epsilon }^{(1)}(x)\). For an arbitrary positive constant M, we infer that
$$\begin{aligned} \begin{aligned} \phi _{\epsilon }^{(1)}(x)&\!=\!\frac{1}{\epsilon ^2}\left[ \int \limits _{\left\{ |z|\!\le \!M\cup |z|\!>\!M\right\} }c(z)\epsilon ^2\lambda (\frac{x}{\epsilon }-z)m(\frac{x}{\epsilon }-z) \int _0^1( \xi ''\left( x-\epsilon zt)-\xi ''(x)\right) z^2(1-t)dt\right] dz\\&:=\phi _{\epsilon }^{(2)}(x)+\phi _{\epsilon }^{(3)}(x). \end{aligned} \end{aligned}$$
Then
$$\begin{aligned} \left\| \phi _{\epsilon }^{(2)}\right\| _0\le & {} \frac{C_2}{2\delta }\sup _{|z|\le M}\left\| \xi ''(x-\epsilon zt)-\xi ''(x)\right\| _0\int _{{\mathbb {R}}}z^2c(z)dz, \\ \Vert \phi _{\epsilon }^{(3)}\Vert _0\le & {} \frac{2C_2}{\delta }\left\| \xi ''(x)\right\| _0 \int _{|z|>M}z^2c(z)dz. \end{aligned}$$
If we take \(M=\frac{1}{\sqrt{\epsilon }}\), then
$$\begin{aligned} \left\| \phi _{\epsilon }^{(2)}\right\| _0\rightarrow 0 \quad and \quad \left\| \phi _{\epsilon }^{(3)}\right\| _0\rightarrow 0, \quad as\quad \epsilon \rightarrow 0. \end{aligned}$$
This implies that
$$\begin{aligned} \left\| \phi _{\epsilon }^{(1)}\right\| _0\rightarrow 0, \quad \epsilon \rightarrow 0. \end{aligned}$$
For the second term of \(\phi _\epsilon (x),\)
$$\begin{aligned} \phi _{\epsilon }^{(4)}(x)=\epsilon \int _{{\mathbb {R}}}c(z)h_1\left( \frac{x}{\epsilon }-z\right) \left[ \int _0^1 \xi '''(x-\epsilon zt)z^2(1-t)dt\right] dz, \end{aligned}$$
we have
$$\begin{aligned} \left\| \phi _{\epsilon }^{(4)}(x)\right\| _0\le \frac{\epsilon C_2}{2\delta }\sup _{z,q\in {\mathbb {R}}}\left\| h_1\left( \frac{x}{\epsilon }-z\right) \xi '''(x-\epsilon z+q)\right\| _0\int _{{\mathbb {R}}}z^2c(z)dz. \end{aligned}$$
(49)
Next, we estimate
$$\begin{aligned} \sup _{z,q\in {\mathbb {R}}}\left\| h_1\left( \frac{x}{\epsilon }-z\right) \xi '''(x-\epsilon z+q)\right\| _0. \end{aligned}$$
Taking \(y=x-\epsilon z\), we deduce that
$$\begin{aligned} \begin{aligned} \sup _{q\in {\mathbb {R}}}\left\| h_1\left( \frac{y}{\epsilon }\right) \xi '''(y+q)\right\| _0&=\sup _{q\in \epsilon {\mathbb {T}}}\left\| h_1\left( \frac{y}{\epsilon }\right) \xi '''(y+q)\right\| _0\\&\le \sup _{q\in \epsilon {\mathbb {T}}} \displaystyle \sum _{k\in {\mathbb {Z}}}\int _{\epsilon k}^{\epsilon k+\epsilon }h_1\left( \frac{y}{\epsilon }\right) ^2 [\xi '''(y+q)]^2dy\\&\le \left\| h_1\right\| ^2_0\sum _{k\in {\mathbb {Z}}}\max _{y\in [\epsilon k,\epsilon k+\epsilon ],q\in \epsilon {\mathbb {T}}}[\xi '''(y+q)]^2dy\\&\rightarrow \left\| h_1\right\| ^2_0\Vert \xi '''\Vert ^2_0, \end{aligned} \end{aligned}$$
as \(\epsilon \rightarrow 0\). Thus from (49), it follows that \(\left\| \phi _{\epsilon }^{(4)}\right\| _0\rightarrow 0\), as \(\epsilon \rightarrow 0.\)
Similarly, for the third term, we have
$$\begin{aligned} \begin{aligned} \left\| \phi _{\epsilon }^{(5)}(x)\right\| _0&=\left\| \epsilon \int _{{\mathbb {R}}}dz c(z) h_2(\frac{x}{\epsilon }-z)\int _0^1\xi '''(x-\epsilon zt)zdt\right\| _0\\&\rightarrow 0. \end{aligned} \end{aligned}$$
In summary, we have \(\left\| \phi _\epsilon (x)\right\| _0 \rightarrow 0,\) as \(\epsilon \rightarrow 0.\) \(\square \)
Appendix D: Convergence of \((V^\epsilon )^*\xi ^\epsilon \)
Here we will show the convergence of \((V^\epsilon )^*\xi ^\epsilon ,\) as \(\epsilon \) goes to 0.
First, Let’s do a simple calculation for fractional Laplace operator. For every functions \(f,g,\psi \in H^{\alpha /2}\),
$$\begin{aligned} \begin{aligned}&\big \langle (-\varDelta )^{\alpha /2}(f\cdot g)(x),\psi (x)\big \rangle =\int _{{\mathbb {R}}}\int _{{\mathbb {R}}}\big (f(x)g(x)-f(y)g(y)\big )\psi (x)\gamma ^2(x,y)dxdy\\&\quad =\frac{1}{2}\int _{{\mathbb {R}}}\int _{{\mathbb {R}}}\big (f(x)g(x)-f(y)g(y)\big )\big (\psi (x)-\psi (y)\big )\gamma ^2(x,y)dxdy\\&\quad =\frac{1}{2}\big (\mathcal {D^{*}}(fg)(x,y),\mathcal {D^{*}}\psi (x,y)\big )_{L^2{({\mathbb {R}}\times {\mathbb {R}})}}\\&\quad =\small {\frac{1}{2}\int _{{\mathbb {R}}}\int _{{\mathbb {R}}}\big [(f(x)-f(y)g(x)+f(x)(g(x)-g(y))\big ](\psi (x)-\psi (y))\gamma ^2(x,y)dxdy}\\&\quad =\frac{1}{2}\Big (\mathcal {D^{*}}(f)(x,y)g(x)+\mathcal {D^{*}}(g)(x,y)f(y),\mathcal {D^{*}}\psi (x,y)\Big )_{L^2{({\mathbb {R}}\times {\mathbb {R}})}}.\\ \end{aligned} \end{aligned}$$
For the operator \(L^\epsilon ,\) we have
$$\begin{aligned} \begin{aligned}&\big \langle (L^\epsilon )^*\xi ^\epsilon ,\psi \big \rangle =\bigg (-(-\varDelta )^{\alpha /2}(\delta _1^\epsilon \xi ^\epsilon )(x),\psi (x)\bigg ) -\frac{1}{\epsilon ^{\alpha -1}}\bigg ((p^\epsilon (x)\xi ^\epsilon (x))',\psi (x)\bigg )\\&\quad =-\frac{1}{2}\bigg (\mathcal {D^{*}}(\delta _1^\epsilon \xi ^\epsilon )(x,y),\mathcal {D^{*}}\psi (x,y)\bigg )_{L^2{({\mathbb {R}}\times {\mathbb {R}})}} -\frac{1}{\epsilon ^{\alpha -1}}\bigg ((p^\epsilon (x)\xi ^\epsilon (x))',\psi (x)\bigg )\\&\quad =-\frac{1}{2}\bigg (\mathcal {D^{*}}(\delta _1^{m_1,\epsilon }\xi )(x,y),\mathcal {D^{*}}\psi (x,y)\bigg )_{L^2{({\mathbb {R}}\times {\mathbb {R}})}} -\frac{\epsilon }{2}\bigg (\mathcal {D^{*}}(\delta _1^{m_1,\epsilon } h_{3}^\epsilon \xi ^{'})(x,y),\mathcal {D^{*}}\psi (x,y)\bigg )_{L^2{({\mathbb {R}}\times {\mathbb {R}})}}\\&\qquad -\,\bigg (\epsilon ^{-\alpha }(pm_1)'\left( \frac{x}{\epsilon }\right) +\epsilon ^{1-\alpha }p^{m_1,\epsilon }(x)\xi '(x)+\epsilon ^{2-\alpha }\Big (p^{m_1,\epsilon }(x)h_3^{\epsilon }(x)\Big )'\xi '(x),\psi (x)\bigg )\\&\qquad -\,\bigg (\epsilon ^{2-\alpha }p^{m_1,\epsilon }(x)h_3^{\epsilon }(x)\xi ''(x),\psi (x)\bigg )\\&\quad :=G_1+G_2-\bigg (\epsilon ^{-\alpha }(pm_1)'\left( \frac{x}{\epsilon }\right) \xi (x)+\epsilon ^{1-\alpha }p^{m_1,\epsilon }(x)\xi '(x)+\epsilon ^{1-\alpha }(pm_1h_3)'\left( \frac{x}{\epsilon }\right) \xi '(x),\psi (x)\bigg )\\&\qquad -\,\bigg (\phi ^1_{\epsilon },\psi (x)\bigg ), \end{aligned} \end{aligned}$$
where \((\phi ^1_{\epsilon },\psi (x))\rightarrow 0,\) as \(\epsilon \) goes to 0. Furthermore, we can show that,
$$\begin{aligned} \begin{aligned} G_1&=-\frac{1}{2}\bigg (\mathcal {D^{*}}(\delta _1^{m_1,\epsilon }\xi )(x,y),\mathcal {D^{*}}\psi (x,y)\bigg )_{L^2{({\mathbb {R}}\times {\mathbb {R}})}}\\&=-\frac{1}{2}\bigg (\mathcal {D^{*}}(\delta _1^{m_1,\epsilon })(x,y)\xi (x)+\mathcal {D^{*}}(\xi )(x,y)\delta _1^{m_1,\epsilon }(y),\mathcal {D^{*}}\psi (x,y)\bigg )_{L^2{ ({\mathbb {R}}\times {\mathbb {R}})}}\\&=-\frac{1}{2}\bigg (\mathcal {D^{*}}(\delta _1^{m_1,\epsilon })(x,y),\mathcal {D^{*}}(\psi \xi )(x,y)-\mathcal {D^{*}}(\xi )(x,y)\psi (y)\bigg )_{L^2{({\mathbb {R}}\times {\mathbb {R}})}}\\&\quad -\,\frac{1}{2}\bigg (\mathcal {D^{*}}(\xi )(x,y),\delta _1^{m_1,\epsilon }(y)\mathcal {D^{*}}\psi (x,y)\bigg )_{L^2{({\mathbb {R}}\times {\mathbb {R}})}} =\bigg (-(-\varDelta )^{\alpha /2}\delta _1^{m_1,\epsilon }(x),\psi (x)\xi (x)\bigg )\\&\quad -\,\frac{1}{2}\bigg (\mathcal {D^{*}}(\xi )(x,y), \delta _1^{m_1,\epsilon }(y)\mathcal {D^{*}}\psi (x,y)-\mathcal {D^{*}}\delta _1^{m_1,\epsilon }(x,y)\psi (y)\bigg )_{L^2{({\mathbb {R}}\times {\mathbb {R}})}}\\&:=I_1+I_2. \end{aligned} \end{aligned}$$
For \(I_2,\) we deduce that,
$$\begin{aligned} I_2= & {} -\frac{1}{2}\bigg (\mathcal {D^{*}}(\xi )(x,y),\delta _1^{m_1,\epsilon }(y)\mathcal {D^{*}}\psi (x,y)-\mathcal {D^{*}}\delta _1^{m_1,\epsilon }(x,y)\psi (y)\bigg ) _{L^2{({\mathbb {R}}\times {\mathbb {R}})}}\\= & {} -\frac{1}{2}\bigg (\mathcal {D^{*}}(\xi )(x,y),\mathcal {D^{*}}(\delta _1^{m_1,\epsilon }\psi )(x,y)-\mathcal {D^{*}}\delta _1^{m_1,\epsilon }(x,y)\left[ \psi (x)+\psi (y)\right] \bigg ) _{L^2{({\mathbb {R}}\times {\mathbb {R}})}}\\:= & {} \bigg (-(-\varDelta )^{\alpha /2}\xi (x),\delta _1^{m_1,\epsilon }(x)\psi (x)\bigg )+I_3,\\ \end{aligned}$$
where we have \(I_3\rightarrow 0\). In fact,
$$\begin{aligned} \begin{aligned} I_3&=\frac{1}{2}\bigg (\mathcal {D^{*}}(\xi )(x,y),\mathcal {D^{*}}\delta _1^{m_1,\epsilon }(x,y)\left[ \psi (x)+\psi (y)\right] \bigg )_{L^2{({\mathbb {R}}\times {\mathbb {R}})}}\\&=\int _{{\mathbb {R}}}\int _{{\mathbb {R}}}\left[ \xi (x)-\xi (y)\right] \delta _1^{m_1,\epsilon }(x)\left[ \psi (x)+\psi (y)\right] \gamma ^2(x,y)dxdy\\&=\bigg (\int _{{\mathbb {R}}}\left[ \xi (x)-\xi (y)\right] \left[ \psi (x)+\psi (y)\right] \gamma ^2(x,y)dy,\delta _1^{m_1,\epsilon }(x)\bigg ) \rightarrow 0. \end{aligned} \end{aligned}$$
From the calculation above, we can obtain that,
$$\begin{aligned} \begin{aligned} G_2=\epsilon \bigg (-(-\varDelta )^{\alpha /2}(\delta _1^{m_1,\epsilon } h_3^\epsilon )(x),\psi (x)\xi '(x)\bigg )-\bigg (\phi ^2_\epsilon ,\psi (x)\xi '(x)\bigg ), \end{aligned} \end{aligned}$$
where \(\bigg (\phi _2^\epsilon ,\psi (x)\xi '(x)\bigg )\rightarrow 0,\) as \(\epsilon \) goes to 0. Then, we have,
$$\begin{aligned} \begin{aligned} \big \langle (L^\epsilon )^*\xi ^\epsilon ,\psi \big \rangle&=\bigg (-(-\varDelta )^{\alpha /2}\delta _1^{m_1,\epsilon }(x)-\epsilon ^{-\alpha }(pm_1)'\left( \frac{x}{\epsilon }\right) ,\psi (x)\xi (x)\bigg )\\&\quad +\,\bigg (-\epsilon (-\varDelta )^{\alpha /2}(\delta _1^{m_1,\epsilon } h_3^\epsilon )(x)-\epsilon ^{1-\alpha }p^{m_1,\epsilon }(x)-\epsilon ^{1-\alpha }(pm_1h_3)'\left( \frac{x}{\epsilon }\right) ,\psi (x)\xi '(x)\bigg )\\&\quad +\,\bigg (-(-\varDelta )^{\alpha /2}\xi (x), \delta _1^{m_1,\epsilon }(x)\psi (x)\bigg )+\phi _\epsilon ^3\\ \end{aligned} \end{aligned}$$
where \(\phi _\epsilon ^3\) goes to 0. Using the Eqs. (12), (29), as \(\epsilon \rightarrow 0\), we have
$$\begin{aligned} \left\langle (L^\epsilon )^*\xi ^\epsilon ,\psi \right\rangle \rightarrow \left( \int _{\mathbb {T}}\delta ^\alpha (\eta )m_1(\eta )d\eta \right) \cdot \bigg (-(-\varDelta )^{\alpha /2}\xi (x),\psi (x)\bigg ). \end{aligned}$$
From the fact that
$$\begin{aligned} \left\langle F^\epsilon )^*\xi ^\epsilon ,\psi \right\rangle \rightarrow \left( -\xi '(x)\int _{{\mathbb {T}}}g(\eta )m_1(\eta )d\eta +\xi (x)\int _{{\mathbb {T}}}f(\eta )m_1(\eta )d\eta ,\psi (x)\right) \end{aligned}$$
We can infer that
$$\begin{aligned} \begin{aligned} \left\langle (V^\epsilon )^*\xi ^\epsilon ,\psi \right\rangle \rightarrow&\bigg (\int _{{\mathbb {T}}}\delta ^\alpha (\eta )m_1(\eta )d\eta \cdot \Big (-(-\varDelta )^{\alpha /2}\Big )\xi (x)+\xi '(x)\int _{{\mathbb {T}}}g(\eta )m_1(\eta )d\eta \\&+\xi (x)\int _{{\mathbb {T}}}f(\eta )m_1(\eta )d\eta ,\psi (x)\bigg ). \end{aligned} \end{aligned}$$
