In this section, we show the separability properties of problem (1.5) in Sobolev spaces ${W}^{m,q}(G;E)$. The main result is the following theorem.

**Theorem 4.1** *Let the following conditions be satisfied*:

- (1)
*E* *is a* UMD *space and* *A* *is an* *R*-*positive operator in* *E*;

- (2)
*m* *is a positive integer* $q\in (1,\mathrm{\infty})$,

$0<{t}_{k}\le 1$,

*and* ${\eta}_{k}={(-1)}^{{m}_{1}}{\alpha}_{k1}{\beta}_{k2}-{(-1)}^{{m}_{2}}{\alpha}_{k2}{\beta}_{k1}\ne 0,\phantom{\rule{1em}{0ex}}k=1,2,\dots ,n.$

Then problem (3.1)-(3.2) has a unique solution

$u\in {W}^{2+m,q}(G;E(A),E)$ for

$f\in {W}^{m,q}(G;E)$,

$\lambda \in {S}_{\psi ,\varkappa}$, with sufficiently large

$\varkappa >0$, and the following coercive uniform estimate holds:

$\sum _{k=1}^{n}\sum _{i=0}^{m+2}{\epsilon}_{k}^{\frac{i}{m+2}}|\lambda {|}^{1-\frac{i}{m+2}}{\parallel \frac{{\partial}^{i}u}{\partial {x}_{k}^{i}}\parallel}_{{W}^{m,q}(G;E)}+{\parallel Au\parallel}_{{L}^{q}(G;E)}\le C{\parallel f\parallel}_{{W}^{m,q}(G;E)}$

(4.1)

with $C=C(q,A)$ independent of ${\epsilon}_{1},{\epsilon}_{2},\dots ,{\epsilon}_{n}$, *λ* and *f*.

Consider first the following nonlocal BVP for an ordinary DOE with a small parameter:

where ${\sigma}_{i}=\frac{i}{2}+\frac{1}{2p}$, ${m}_{k}\in \{0,m+1\}$, ${\alpha}_{ki}$, ${\beta}_{ki}$ are complex numbers, *t* is positive, *λ* is a complex parameter and *A* is a linear operator in *E*. Let ${A}_{\lambda}=A+\lambda I$.

To prove the main result, we need the following result in [[37], Theorem 2.1].

**Theorem A** *Let* *E* *be a* UMD

*space*,

*A* *be a* *ψ*-

*positive operator in* *E* *with bound* *M*,

$0\le \psi <\pi $.

*Let* *m* *be a positive integer*,

$1<p<\mathrm{\infty}$,

*and*$\alpha \in (\frac{1}{2p},\frac{1}{2p}+m)$.

*Then*,

*for*$\lambda \in {S}_{\phi}$,

*an operator*$-{A}_{\lambda}^{\frac{1}{2}}$*generates a semigroup*${e}^{-x{A}_{\lambda}^{\frac{1}{2}}}$*which is holomorphic for*$x>0$.

*Moreover*,

*there exists a positive constant* *C* (

*depending only on* *M*,

*ψ*,

*m*,

*α* *and* *p*)

*such that for every*$u\in {(E,E({A}^{m}))}_{\frac{\alpha}{m}-\frac{1}{2mp},p}$*and*$\lambda \in {S}_{\psi}$,

${\int}_{0}^{\mathrm{\infty}}{\parallel {A}_{\lambda}^{\alpha}{e}^{-x{A}_{\lambda}^{\frac{1}{2}}}u\parallel}^{p}\phantom{\rule{0.2em}{0ex}}dx\le C[{\parallel u\parallel}_{{(E,E({A}^{m}))}_{\frac{\alpha}{m}-\frac{1}{2mp},p}}^{p}+|\lambda {|}^{\alpha p-\frac{1}{2}}{\parallel u\parallel}_{E}^{p}].$

In a similar way as in [[43], §1.8.2, Theorem 2], we obtain the following lemma.

**Lemma 4.1** *Let* *m* *and* *j* *be integer numbers*,

$0\le j\le m-1$,

${\theta}_{j}=\frac{pj+1}{pm}$,

$0<t\le 1$,

${x}_{0}\in [0,b]$.

*Then*,

*for*$u\in {W}_{p}^{m}(0,b;{E}_{0},E)$,

*the transformation*$u\to {u}^{(j)}({x}_{0})$*is bounded linear from*${W}_{p}^{m}(0,b;{E}_{0},E)$*onto*${({E}_{0},E)}_{{\theta}_{j},p}$*and the following inequality holds*:

${t}^{{\theta}_{j}}{\parallel {u}^{(j)}({x}_{0})\parallel}_{{({E}_{0},E)}_{{\theta}_{j},p}}\le C({\parallel t{u}^{(m)}\parallel}_{{L}_{p}(0,b;E)}+{\parallel u\parallel}_{{L}_{p}(0,b;{E}_{0})}).$

*Consider at first the homogeneous problem of* (4.2)

*Let*
${E}_{k}={(E(A),E)}_{{\theta}_{k},p}.$

**Lemma 4.2** *Let* *A* *be an* *R*-

*positive operator in a* UMD

*space* *E* *and*$0<t\le 1,\phantom{\rule{1em}{0ex}}\eta ={(-1)}^{{m}_{1}}{\alpha}_{1}{\beta}_{2}-{(-1)}^{{m}_{2}}{\alpha}_{2}{\beta}_{1}\ne 0.$

*Then problem* (4.3)

*has a unique solution*$u\in {W}^{m+2,p}(0,1;E(A),E)$*for*${f}_{k}\in {E}_{k}$,

$p\in (1,\mathrm{\infty})$,

$\lambda \in {S}_{\psi}$,

*and the coercive uniform estimate holds*$\sum _{i=0}^{m+2}{t}^{\frac{i}{m+2}}|\lambda {|}^{1-\frac{i}{m+2}}{\parallel {u}^{(i)}\parallel}_{{L}^{p}(0,1;E)}+{\parallel Au\parallel}_{{L}^{p}(0,1;E)}\le M\sum _{k=1}^{2}({\parallel {f}_{k}\parallel}_{{E}_{k}}+|\lambda {|}^{1-{\theta}_{k}}{\parallel {f}_{k}\parallel}_{E}).$

(4.4)

*Proof* In a similar way as in [[

40], Theorem 3.1], we obtain the representation of the solution of (4.3)

$\begin{array}{rcl}u(x)& =& {t}^{-\frac{1}{2p}}\{{e}^{-x{t}^{-\frac{1}{2}}{A}_{\lambda}^{\frac{1}{2}}}[{C}_{11}+{d}_{11}(\lambda ,t)]+{e}^{-(1-x){t}^{-\frac{1}{2}}{A}_{\lambda}^{\frac{1}{2}}}[{C}_{12}+{d}_{12}(\lambda ,t)]\}{A}_{\lambda}^{-\frac{{m}_{1}}{2}}{f}_{1}\\ +{t}^{-\frac{1}{2p}}\{{e}^{-x{t}^{-\frac{1}{2}}{A}_{\lambda}^{\frac{1}{2}}}[{C}_{21}+{d}_{21}(\lambda ,t)]+{e}^{-(1-x){t}^{-\frac{1}{2}}{A}_{\lambda}^{\frac{1}{2}}}[{C}_{22}+{d}_{22}(\lambda ,t)]\}{A}_{\lambda}^{-\frac{{m}_{2}}{2}}{f}_{2},\end{array}$

(4.5)

where

${C}_{ij}$ and

${d}_{ij}$ are uniformly bounded operators. Then, in view of positivity of

*A*, we obtain from (4.5)

By changing the variable

$x{t}^{-\frac{1}{2}}=y$ and in view of Theorem A, we obtain

By using the estimate (4.8), by virtue of Theorem A, we get the uniform estimate

Then from (4.6)-(4.9) we obtain (4.4). □

Now we can represent a more general result for nonhomogeneous problem (4.2).

**Theorem 4.2** *Assume that the following conditions are satisfied*:

- (1)
*E* *is a* UMD *space and* *A* *is an* *R*-*positive operator in* *E*;

- (2)
$\eta ={(-1)}^{{m}_{1}}{\alpha}_{1}{\beta}_{2}-{(-1)}^{{m}_{2}}{\alpha}_{2}{\beta}_{1}\ne 0$, ${\theta}_{k}=\frac{{m}_{k}}{m+2}+\frac{1}{2p}$, $k=1,2$, $0<t\le 1$, $p\in (1,\mathrm{\infty})$.

*Then the operator*$u\to \{({L}_{t}+\lambda )u,{L}_{1t}u,{L}_{2t}u\}$*is an isomorphism from*${W}^{m+2,p}(0,1;E(A),E)$*onto*${W}^{m,p}(0,1;E)\times {E}_{1}\times {E}_{2}$*for*$\lambda \in {S}_{\psi ,\varkappa}$*with large enough*$\varkappa >0$.

*Moreover*,

*the uniform coercive estimate holds**Proof* The uniqueness of a solution of problem (4.2) is obtained from Lemma 4.2. Let us define

$\overline{f}(x)=\{\begin{array}{cc}f(x),\hfill & \text{if}x\in [0,1],\hfill \\ 0,\hfill & \text{if}x\notin [0,1].\hfill \end{array}$

We will show that problem (4.2) has a solution

$u\in {W}^{m+2,p}(0,1;E(A),E)$ for

$f\in {W}^{m,p}(0,1;E)$,

${f}_{k}\in {E}_{k}$ and

$u={u}_{1}+{u}_{2}$, where

${u}_{1}$ is the restriction on

$[0,1]$ of the solution of the equation

$({L}_{t}+\lambda )u=\overline{f}(x),\phantom{\rule{1em}{0ex}}x\in R=(-\mathrm{\infty},\mathrm{\infty}),$

(4.11)

and

${u}_{2}$ is a solution of the problem

$({L}_{t}+\lambda )u=0,\phantom{\rule{2em}{0ex}}{L}_{kt}u={f}_{k}-{L}_{kt}{u}_{1}.$

(4.12)

A solution of equation (

4.11) is given by

$u(x)={F}^{-1}{L}^{-1}(\lambda ,t,\xi )F\overline{f}=\frac{1}{2\pi}{\int}_{\mathrm{\infty}}^{\mathrm{\infty}}{e}^{i\xi x}{L}^{-1}(\lambda ,t,\xi )(F\overline{f})(\xi )\phantom{\rule{0.2em}{0ex}}d\xi ,$

where

$L(\lambda ,t,\xi )=A+t{\xi}^{2}+\lambda $. It follows from the above expression that

It is sufficient to show that the operator-functions

are uniform Fourier multipliers in

${L}^{p}(R;E)$. Actually, due to the positivity of

*A*, we have

It is clear to observe that

$\xi \frac{d}{d\xi}{\mathrm{\Psi}}_{t\lambda}(\xi )=-2t{\xi}^{2}A{L}^{-2}(\lambda ,t,\xi )=[-2t{\xi}^{2}{L}^{-1}(\lambda ,t,\xi )]A{L}^{-1}(\lambda ,t,\xi ).$

Due to

*R*-positivity of the operator

*A*, the sets

$\{-2t{\xi}^{2}{[A+t{\xi}^{2}+\lambda ]}^{-1}\},\phantom{\rule{2em}{0ex}}\left\{A{[A+t{\xi}^{2}+\lambda ]}^{-1}\right\},\phantom{\rule{1em}{0ex}}\xi \in \mathbb{R}\mathrm{\setminus}\{0\}$

are

*R*-bounded. Then, in view of the Kahane contraction principle, from the product properties of the collection of

*R*-bounded operators (see,

*e.g.*, [

36] Lemma 3.5, Proposition 3.4), we obtain

$\begin{array}{r}R\{{\xi}^{i}\frac{{d}^{i}}{d{\xi}^{i}}{\mathrm{\Psi}}_{t\lambda}(\xi ):\xi \in \mathbb{R}\mathrm{\setminus}\{0\}\}\le {C}_{1},\\ R\{{\xi}^{i}\frac{{d}^{i}}{d{\xi}^{i}}{\sigma}_{t\lambda}(\xi ):\xi \in \mathbb{R}\mathrm{\setminus}\{0\}\}\le {C}_{2},\phantom{\rule{1em}{0ex}}i=0,1.\end{array}$

(4.15)

By [[

33], Theorem 3.4] it follows that

${\mathrm{\Psi}}_{t,\lambda}(\xi )$ and

${\sigma}_{t\lambda}(\xi )$ are the uniform collection of multipliers in

${L}^{p}(R;E)$. Then in view of (4.13) we obtain that problem (4.11) has a solution

$u\in {W}^{m+2,p}(R;E(A),E)$ and the uniform coercive estimate holds

$\sum _{j=0}^{m+2}{t}^{\frac{j}{m+2}}|\lambda {|}^{1-\frac{j}{m+2}}{\parallel {u}^{(j)}\parallel}_{{L}^{p}(R;E)}+{\parallel Au\parallel}_{{L}^{p}(R;E)}\le C{\parallel \overline{f}\parallel}_{{L}^{p}(R;E)}.$

(4.16)

Let

${u}_{1}$ be the restriction of

*u* on

$(0,1)$. The estimate (4.16) implies that

${u}_{1}\in {W}^{m+2,p}(0,1;E(A),E)$. By virtue of Lemma 4.1, we get

${u}_{1}^{({m}_{k})}(\cdot )\in {(E(A);E)}_{{\theta}_{k},p},\phantom{\rule{1em}{0ex}}k=1,2.$

Hence,

${L}_{kt}{u}_{1}\in {E}_{k}$. Thus, by virtue of Lemma 4.2, problem (4.12) has a unique solution

${u}_{2}(x)$ that belongs to the space

${W}^{m+2,p}(0,1;E(A),E)$ and

Moreover, from (4.16) we obtain

$\sum _{j=0}^{m+2}{t}^{\frac{j}{m+2}}|\lambda {|}^{1-\frac{j}{m+2}}{\parallel {u}_{1}^{(j)}\parallel}_{{L}^{p}(0,1;E)}+{\parallel A{u}_{1}\parallel}_{{L}^{p}(0,1;E)}\le C{\parallel f\parallel}_{{W}^{m,p}(0,1;E)}.$

(4.18)

Therefore, by Lemma 4.1 and by estimate (4.17), we obtain

${t}^{{\theta}_{k}}{\parallel {u}_{1}^{({m}_{k})}(\cdot )\parallel}_{{E}_{k}}\le C{\parallel {u}_{1}\parallel}_{{W}_{t}^{m+2,p}(0,1;E(A),E)}\le C{\parallel f\parallel}_{{W}^{m,p}(0,1;E)}.$

(4.19)

So, in view of Lemma 4.1 and estimates (4.17)-(4.19), we get

Finally, from (4.18) and (4.20) we obtain (4.10). □

Now, we can prove the main result of this section.

*Proof of Theorem 4.1* Let

${G}_{2}=(0,{b}_{1})\times (0,{b}_{2})$. It is clear to see that

${W}^{m,q}({G}_{2};E)={W}^{m,q}(0,{b}_{1};{X}_{0},X)={W}^{m,q}(0,{b}_{1};X)\cap {L}^{q}(0,{b}_{1};{X}_{0}),$

where ${X}_{0}={W}^{m,q}(0,{b}_{2};E)$ and $X={L}^{q}(0,{b}_{2};E)$.

Let us consider the BVP

$-{\epsilon}_{1}\frac{{\partial}^{2}u}{\partial {x}_{1}^{2}}-{\epsilon}_{2}\frac{{\partial}^{2}u}{\partial {x}_{2}^{2}}+(A+\lambda )u({x}_{1},{x}_{2})=f({x}_{1},{x}_{2}),\phantom{\rule{2em}{0ex}}{L}_{kj\epsilon}u=0,\phantom{\rule{1em}{0ex}}k,j=1,2,$

(4.21)

where

${L}_{kj\epsilon}$ are defined by equalities (1.5). Problem (4.21) can be expressed as the following BVP for an ordinary DOE:

$Lu=-{\epsilon}_{1}\frac{{d}^{2}u}{d{x}_{1}^{2}}+({B}_{{\epsilon}_{2}}+\lambda )u({x}_{2})=f({x}_{2}),\phantom{\rule{1em}{0ex}}{x}_{1}\in (0,{b}_{1}),\phantom{\rule{2em}{0ex}}{L}_{k1{\epsilon}_{1}}u=0,$

(4.22)

where

${L}_{k1{\epsilon}_{1}}$ are boundary conditions of type (3.2),

${B}_{{\epsilon}_{2}}$ is the operator acting in

${X}_{0}$ and

*X* defined by

Since

${X}_{0}$ and

*X* are UMD spaces, (see,

*e.g.*, [[

35], Theorem 4.5.2]) by virtue of Theorem 4.2, we obtain that problem (4.22) has a unique solution

$u\in {W}^{m+2,q}(0,{b}_{1};D({B}_{{\epsilon}_{2}}),X)$ for

$f\in {W}^{m,q}(0,{b}_{1};X)$ and

$\lambda \in {S}_{\psi ,\varkappa}$ with sufficiently large

$\varkappa >0$. Moreover, the coercive uniform estimates holds

$\begin{array}{r}\sum _{i=0}^{m+2}{\epsilon}_{1}^{\frac{i}{m+2}}|\lambda {|}^{1-\frac{i}{m+2}}{\parallel {u}^{(i)}\parallel}_{{L}^{q}(0,{b}_{1};X)}+{\parallel {B}_{{\epsilon}_{2}}u\parallel}_{{L}^{q}(0,{b}_{1};X)}\le C{\parallel f\parallel}_{{W}^{m,q}(0,{b}_{1};X)},\\ \sum _{i=0}^{2}{\epsilon}_{1}^{\frac{i}{2}}|\lambda {|}^{1-\frac{i}{2}}{\parallel {u}^{(i)}\parallel}_{{L}^{q}(0,{b}_{1};{X}_{0})}+{\parallel {B}_{{\epsilon}_{2}}u\parallel}_{{L}^{q}(0,{b}_{1};{X}_{0})}\le C{\parallel f\parallel}_{{L}^{q}(0,{b}_{1};{X}_{0})}.\end{array}$

(4.23)

From (4.23) we obtain that problem (4.22) has a unique solution

$u\in {W}^{m+2,q}({G}_{2};E(A),E)\phantom{\rule{1em}{0ex}}\text{for}{W}^{m,q}({G}_{2};E).$

Moreover, the uniform coercive estimates hold

$\sum _{i=0}^{m+2}{\epsilon}_{1}^{\frac{i}{m+2}}|\lambda {|}^{1-\frac{i}{m+2}}{\parallel {u}^{(i)}\parallel}_{{L}^{q}(0,{b}_{1};X)}+{\parallel {B}_{{\epsilon}_{2}}u\parallel}_{{L}^{q}(0,{b}_{1};X)}\le C{\parallel f\parallel}_{{W}^{m,q}({G}_{2};E)}.$

(4.24)

By applying Theorem 4.2 for

${f}_{k}=0$ and

$E=X$, we get the following uniform estimate:

$\sum _{j=0}^{m+2}{\epsilon}_{2}^{\frac{i}{m+2}}|\lambda {|}^{1-\frac{i}{m+2}}{\parallel {u}^{(i)}\parallel}_{X}+{\parallel Au\parallel}_{X}\le C{\parallel {B}_{{\epsilon}_{2}}u\parallel}_{{W}^{m,q}(0,{b}_{2};E)}.$

(4.25)

From estimates (4.24)-(4.25) we conclude the corresponding claim for problem (4.21). Then, by continuing this process *n*-times, we obtain the assertion. □