Analytical aspects of the Brownian motor effect in randomly flashing ratchets

Abstract

The muscle contraction, operation of ATP synthase, maintaining the shape of a cell are believed to be secured by motor proteins, which can be modelled using the Brownian ratchet mechanism. We consider the randomly flashing ratchet model of a Brownian motor, where the particles can be in two states, only one of which is sensitive the applied spatially periodic potential (the mathematical setting is a pair of weakly coupled reaction-diffusion and Fokker–Planck equations). We prove that this mechanism indeed generates unidirectional transport by showing that the amount of mass in the wells of the potential decreases/increases from left to right. The direction of transport is unambiguously determined by the location of each minimum of the potential with respect to the so-called diffusive mean of its adjacent maxima. The transport can be generated not only by an asymmetric potential, but also by a symmetric potential and asymmetric transition rates, and as a consequence of the general result we derive explicit conditions when the latter happens. When the transitions are localized on narrow active sites in the protein conformation space, we find a more explicit characterization of the bulk transport direction, and infer that some common preconditions of the motor effect are redundant.

Appendices

Appendix A: Proof of Propositions 2.1 and 3.1

Let $G(x,y)$ be Green’s function of the Sturm–Liouville operator

$$\begin{aligned} \mathcal L = - \frac{d^2}{dx^2}+\phi \end{aligned}$$

on $(A,B)$ with homogeneous Neumann boundary condition. Then $U(x)=G(x,s)$ is a solution to (9) if an only if $G(A,s)=G(B,s)$.

By the distributional maximum principle (Littman 1963, Theorem B), $G(x,y)>0$. Observe that

$$\begin{aligned} G_x(x,A)=\int \limits _{B}^x\phi (z) G(A,z) \, dz,\end{aligned}$$

(37)

$$\begin{aligned} G_x(x,B)=\int \limits _{A}^x\phi (z) G(B,z) \, dz. \end{aligned}$$

(38)

Hence, the function $G(x,A)$ is decreasing in $x$, and $G(x,B)$ is increasing. Thus, the function

$$\begin{aligned} g(x)=G(A,x)-G(B,x) \end{aligned}$$

is also (strictly) decreasing. At the ends of the segment, we have $g(A)=G(A,A)-G(B,A)> 0$ and $g(B)=G(A,B)-G(B,B) < 0$. Since $g$ is a continuous function, there is unique $s\in (A,B)$ such that $G(A,s)=G(B,s)$. The first proposition is proven.

Now, let

$$\begin{aligned} g_1(x)=G\left( \frac{A+B}{2},x\right) -G\left( \frac{A+B}{2},A+B-x\right) . \end{aligned}$$

We claim that

$$\begin{aligned} g_1(A)>0. \end{aligned}$$

(39)

Since $G$ is Green’s function,

$$\begin{aligned} (g_1)_x(A)=0, \end{aligned}$$

(40)

and

$$\begin{aligned} (g_1)_{xx}(x)&= \phi (x)G\left( \frac{A+B}{2},x\right) -\phi (A+B-x)G\left( \frac{A+B}{2},A+B-x\right) \nonumber \\&= \phi (x) g_1(x)+[\phi (x)-\tilde{\phi }(x)]G\left( \frac{A+B}{2},A+B-x\right) \nonumber \\&\le \phi (x) g_1(x), \ A\le x< \frac{A+B}{2}, \end{aligned}$$

(41)

and the inequality is strict at least at one point. In particular, $g_1$ cannot be identically zero.

Assume that $g_1(A)\le 0$. By the maximum principle, $g_1$ cannot have non-positive minima within $(A,\frac{A+B}{2})$. But

$$\begin{aligned} g_1\left( \frac{A+B}{2}\right) =0, \end{aligned}$$

(42)

so $A$ must be a minimum point. Let $g_2(x)=g_1(x)-g_1(A)$. Then $g_2$ is non-negative, and

$$\begin{aligned} (g_2)_{xx}(x) \le \phi (x) g_2(x), \ A\le x< \frac{A+B}{2}. \end{aligned}$$

(43)

Thus,

$$\begin{aligned} (g_2)_{x}(x) \le \int \limits _A^x \phi (t) g_2(t)\,dt, \ A\le x\le \frac{A+B}{2}. \end{aligned}$$

(44)

Since $g_2(A)=0$, by the mean value theorem,

$$\begin{aligned} g_2(x)&= g_2(x)-g_2(A)= (x-A)(g_2)_x(c) \nonumber \\&\le (x-A)\int \limits _A^c \phi (t) g_2(t)\,dt \le \frac{1}{2} (B-A)\int \limits _A^x \phi (t) g_2(t)\,dt \end{aligned}$$

(45)

for some $c, A<c<x\le \frac{A+B}{2}$. The Gronwall lemma implies $g_2\equiv 0$, so, by (42), $g_1\equiv 0$, and we get a contradiction.

Hence,

$$\begin{aligned} g\left( \frac{A+B}{2}\right) =g_1(A)> 0. \end{aligned}$$

We have observed above that $g(B)<0$, so there is $s\in (\frac{A+B}{2},B)$ such that $g(s)=0$, and this number $s$ is the $\phi$-diffusive mean.

Appendix B: Proof of Proposition 3.2

Without loss of generality, $[A,B]=[0,1]$. Denoting the solutions of (9) corresponding to $\phi _n$ and $s_n$ by $U_n$, we infer that

$$\begin{aligned} \int \limits _0^1 \phi _n(x) U^2_n(x) +(U_n)_x^2\, dx = U_n(s_n). \end{aligned}$$

(46)

The function $U_n$ is positive and thus convex on $(0,s_n)$ and $(s_n,1)$. Hence, due to the boundary conditions, the derivative $(U_n)_x(x)$ is positive/negative when $x< s_n$ / $x >s_n$, and tends toward its supremum/infimum as $x$ approaches $s_n$ from the left/right, resp. Moreover, the maximum of $U_n$ is achieved at $s_n$, and the minima are reached at $0$ and $1$.

It is easy to see that

$$\begin{aligned} \min \limits _{x\in [0,1]} U_n^2(x)\le \frac{2}{K_*} \int \limits _0^1 \phi _n(x) U^2_n(x)\,dx \end{aligned}$$

(47)

for large $n$, and

$$\begin{aligned} \max \limits _{x\in [0,1]} U_n^2(x)\le 2\left[ \min \limits _{x\in [0,1]} U_n^2(x)+\int \limits _0^1 (U_n)_x^2\, dx\right] . \end{aligned}$$

(48)

Hence, (46) implies

$$\begin{aligned} \Vert U_n^2\Vert _{C[0,1]} \le C \Vert U_n\Vert _{C[0,1]}, \end{aligned}$$

(49)

whence

$$\begin{aligned} \Vert U_n\Vert _{C[0,1]} \le C. \end{aligned}$$

(50)

Without loss of generality, there exists a limit $s^*\in [0,1]$ of the sequence $\{s_n\}$. Assume that $s^*\ne s_*$, or, more particularly, $s_*>s^*$ (the opposite case may be examined in a similar way). This yields that the value of the integral $\int _0^{s_n} \phi _n(x)$ tends to zero as $n\rightarrow + \infty$. But integration of (9) implies that the left derivative $(U_n)^\prime _- (s_n)$ is equal to $\int _0^{s_n} \phi _n(x) U_n(x)\,dx$. Due to (50), this integral goes to zero as $n\rightarrow + \infty$. Using the information on the behaviour of $U_n$ summarized after equality (46), we conclude that $\Vert (U_n)_x\Vert _{C[0,s_n]}\rightarrow 0$. Due to (50), without loss of generality there exists a constant $U_*$ such that

$$\begin{aligned} \max \limits _{x\in [0,1]} U_n(x) \rightarrow U_*. \end{aligned}$$

(51)

But

$$\begin{aligned} \max \limits _{x\in [0,1]} U_n(x)- \min \limits _{x\in [0,1]} U_n(x)&= U_n(s_n)-U_n(0)\nonumber \\&\le s_n\Vert (U_n)_x\Vert _{C[0,s_n]}\rightarrow 0. \end{aligned}$$

(52)

Hence,

$$\begin{aligned} U_n(x) \rightarrow U_* \end{aligned}$$

(53)

uniformly on $[0,1]$.

Let us now test (9) with a smooth function $h$ such that $h(s_*)=0, h(s^*)=1, h_x(0)=0, h_x(1)=0$, obtaining

$$\begin{aligned} \int \limits _0^1 \phi _n(x) U_n(x) h(x)-U_n(x) h_{xx}(x)\, dx = h(s_n). \end{aligned}$$

(54)

Passing to the limit, we find that the left-hand side goes to zero and the right-hand side goes to one, arriving at a contradiction.

Appendix C: Proof of Theorem 4.1

Consider the set of functions

$$\begin{aligned} B&= \left\{ y(x)\in L_1(0,1)\left| \, \int \limits _{0}^1 \nu (x) y(x) \, dx=1, \ \widehat{(\nu y)}_k \ge M, \right. \right. \\&\quad \left. y(x)\ge y(x+1/k)+M \gamma ,\quad {\mathrm{for \, a.a.}} 0\le x\le 1-1/k.\right\} \end{aligned}$$

Inverse induction shows that for any $i=1,\dots , k$ and $y\in B$ one has

$$\begin{aligned} \widehat{(\nu y)}_i \ge M. \end{aligned}$$

(55)

Let us define a mapping $\mathcal A$ on $B$. For each $y\in B$, we let $\mathcal A (y)=Y$, where $Y$ is the solution of the problem

$$\begin{aligned} \left\{ \begin{array}{l@{\quad }l} \nu Y - \varsigma Y_{xx}= \sum \nolimits _{i=1}^k \widehat{(\nu y)}_i \delta _{a_i}, &{} x\in (0,1), \\ Y_{x}(0)=Y_{x}(1)=0. \end{array}\right. \end{aligned}$$

(56)

To put it differently,

$$\begin{aligned} Y=\sum \limits _{i=1}^k \widehat{(\nu y)}_i q_{i}. \end{aligned}$$

(57)

Then, the set $B$ is invariant for the map $\mathcal A$. In fact, let $y\in B$. Then (56) implies

$$\begin{aligned} \int \limits _{0}^1 \nu (x) Y(x) \, dx= \sum \limits _{i=1}^k \widehat{(\nu y)}_i = \int \limits _{0}^1 \nu (x) y(x) \, dx=1. \end{aligned}$$

Further,

$$\begin{aligned} \widehat{(\nu Y)}_k=\sum \limits _{i=1}^k \widehat{(\nu y)}_i \widehat{(\nu q_i)}_k\ge M\sum \limits _{i=1}^k \widehat{(\nu y)}_i =M. \end{aligned}$$

Finally, fix $x_*\in [0,1-1/k]$. Then there is a number $n$ such that $x_*\in [x_n,x_{n+1})$. Set

$$\begin{aligned} N_*=\left\{ \begin{array}{ll} \widehat{(\nu y)}_n, &{} x_*\le a_n,\\ \widehat{(\nu y)}_{n+1}, &{} x_* > a_n. \end{array}\right. \end{aligned}$$

(58)

We claim that

$$\begin{aligned} (\widehat{(\nu y)}_i - N_*)[q_i(x_*)-q_i(x_*+1/k)]\ge 0,\ i=1,\dots ,k. \end{aligned}$$

(59)

Indeed, integration of (22) gives

$$\begin{aligned}&\varsigma {q_i}_{x}(x)=\int \limits _{0}^x\nu (z) q_i(z) \, dz, \ x< a_i, \end{aligned}$$

(60)

$$\begin{aligned}&\varsigma {q_i}_{x}(x)=\int \limits _{1}^x\nu (z) q_i(z) \, dz, \ x> a_i. \end{aligned}$$

(61)

Thus, the function $q_i$ is increasing on the segment $[0,a_i]$ and decreasing on $[a_i,1]$. Assume first $x_*\le a_n$. Then, if $i<n$, we have $q_i(x_*)>q_i(x_*+1/k)$. Since $y\in B$, we also have $\widehat{(\nu y)}_i >\widehat{(\nu y)}_n= N_*$, and (59) holds true. If $i>n$, we have $q_i(x_*)<q_i(x_*+1/k)$ and $\widehat{(\nu y)}_i < N_*$, and (59) again holds. If $i=n$, (59) is trivial. Now, let $x_*> a_n$. In this case, if $i<n+1$, $q_i(x_*)>q_i(x_*+1/k)$ and $\widehat{(\nu y)}_i >\widehat{(\nu y)}_{n+1}= N_*$; if $i>n+1, q_i(x_*)<q_i(x_*+1/k)$ and $\widehat{(\nu y)}_i < N_*$; and if $i=n+1$, (59) is again trivial.

Formulas (57), (59), (26) and (55) yield

$$\begin{aligned} Y(x_*)- Y(x_*+1/k)&= \sum \limits _{i=1}^k \widehat{(\nu y)}_i [q_i(x_*)-q_i(x_*+1/k)]\\&\ge \sum \limits _{i=1}^k N_* [q_i(x_*)-q_i(x_*+1/k)]\\&= N_*[q(x_*)-q(x_*+1/k)]\ge M\gamma , \end{aligned}$$

so the invariance of $B$ is confirmed.

Observe that $\mathcal A$ is a compact linear operator in $L_1(0,1)$. Indeed, let $\mathbb B$ be the unit ball of the space $L_1(0,1)$. Due to (57), its image $\mathcal A (\mathbb B )$ is a bounded subset of the linear span of $\{q_1,\dots ,q_k \}$, thus being a relatively compact subset of a finite-dimensional subspace of $L_1(0,1)$.

Let us show that (25) may have at most one solution, so $\mathcal A$ can have at most one fixed point in $B$. If not, let ${\tilde{Q}}$ be the difference of two distinct solutions. Then

$$\begin{aligned} \int \limits _{0}^1 \nu (z) \tilde{Q}(z) \, dz=0. \end{aligned}$$

(62)

Moreover,

$$\begin{aligned} {\tilde{Q}}=\sum \limits _{i=1}^k \widehat{(\nu {\tilde{Q}})}_i q_{i}, \end{aligned}$$

(63)

whence

$$\begin{aligned} \widehat{(\nu {\tilde{Q}})}_j=\sum \limits _{i=1}^k \widehat{(\nu {\tilde{Q}})}_i \widehat{(\nu q_i)}_j,\ j=1,\dots ,k. \end{aligned}$$

(64)

From (22) we deduce

$$\begin{aligned} \int \limits _{0}^1 \nu (z) q_i(z) \, dz=1. \end{aligned}$$

(65)

Therefore, the matrix $[\mathcal P _{ij}]=[\widehat{(\nu q_i)}_j ]$ is ergodic, i.e. it has positive entries, and the sum of the elements in every row is equal to one. By the Perron–Frobenius theorem, it has an eigenvector $[\xi _i]$ corresponding to the simple eigenvalue $1$, so that $\xi _j=\sum _{i=1}^k \xi _i \mathcal P _{ij}$, and all the components $\xi _i$ are positive. On the other hand, by (64), $[\Xi _i]=[\widehat{(\nu \tilde{Q})}_i ]$ is another eigenvector of $[\mathcal P _{ij}]$ corresponding to the same eigenvalue. The sum of its components is zero due to (62), so it cannot be collinear with $[\xi _i]$ unless it is a zero vector. Since $1$ is a simple eigenvalue, all $\widehat{(\nu \tilde{Q})}_i$ are zeros, so $\tilde{Q}\equiv 0$ by virtue of (63).

The set $B$ is closed, convex and bounded in $L_1(0,1)$. By Schauder’s fixed point principle, $\mathcal A$ has a fixed point $Q$ in $B$, which is automatically a solution to (25). It remains to notice that $Q$ is continuous as a linear combination of $q_i$, so (27) holds for all $0\le x\le 1-1/k$.

Appendix D: Proof of Lemma 4.1

Let us notice that

$$\begin{aligned} q=u+v, \end{aligned}$$

(66)

where $u$ and $v$ are the (unique) solutions to the following problems

$$\begin{aligned}&\left\{ \begin{array}{ll} \nu u - \varsigma {u}_{xx}= \sum \nolimits _{i=1}^k\delta _{S_i}, &{} x\in (0,1), \\ {u}_{x}(0)={u}_{x}(1)=0, \end{array}\right. \end{aligned}$$

(67)

$$\begin{aligned}&\left\{ \begin{array}{ll} \nu v - \varsigma {v}_{xx}= \sum \nolimits _{i=1}^k(\delta _{a_i}-\delta _{S_i}), &{} x\in (0,1), \\ {v}_{x}(0)={v}_{x}(1)=0, \end{array}\right. \end{aligned}$$

(68)

and

$$\begin{aligned} S_i= S+x_i,\ i=1,\dots , k. \end{aligned}$$

(69)

Since $S$ is the $\nu /\varsigma$-diffusive mean of $0$ and $1/k$, there exists a solution $U_1$ to the problem

$$\begin{aligned} \left\{ \begin{array}{ll} \nu U_1 - \varsigma {U_1}_{xx}= \delta _{S_1}, &{} x\in (0,1/k), \\ {U_1}_{x}(0)={U_1}_{x}(1/k)=0,\\ U_1(0)=U_1(1/k). \end{array}\right. \end{aligned}$$

(70)

The solution $u$ to (67) can be constructed in the following way:

$$\begin{aligned} u(x)=U_1(x-x_i),\ x_i\le x\le x_i+1/k,\ i=1,\dots , k. \end{aligned}$$

(71)

Thus, $u$ is $1/k$-periodic, i.e.

$$\begin{aligned} u(x)- u(x+1/k)=0, \ 0\le x\le x_k. \end{aligned}$$

(72)

Set

$$\begin{aligned} d(x)=v(x)- v(x+1/k), \ 0\le x\le x_k. \end{aligned}$$

(73)

Then it suffices to show that

$$\begin{aligned} \gamma =\min _{0\le x\le x_k} d(x)>0. \end{aligned}$$

(74)

Note that

$$\begin{aligned} \nu d - \varsigma {d}_{xx}= 0, \ x\in (0,x_k). \end{aligned}$$

(75)

By the maximum principle, if the minimum of $d$ is non-positive, it is attained at $0$ or $x_k$. To ascertain that this cannot happen, we are going to prove that

$$\begin{aligned} d_x(0)<0, \ d_x(x_k)>0. \end{aligned}$$

(76)

Set

$$\begin{aligned} V(x)=\int \limits _{0}^x\nu (z) v(z) \, dz, \end{aligned}$$

(77)

and let $\theta$ be the solution of the Cauchy problem

$$\begin{aligned} \left\{ \begin{array}{ll} \theta _x= \sum \nolimits _{i=1}^k(\delta _{a_i}-\delta _{S_i}), &{} x\in (0,1), \\ \theta (0)=0. \end{array}\right. \end{aligned}$$

(78)

Note that $\theta$ is non-negative and $1/k$-periodic.

Integration of (68) gives

$$\begin{aligned} V-\varsigma v_x= \theta . \end{aligned}$$

(79)

Therefore

$$\begin{aligned} d_x=\frac{V(x)-V(x+1/k)}{\varsigma }. \end{aligned}$$

(80)

From (79) we deduce

$$\begin{aligned} V(0)=V(1)=0, \end{aligned}$$

(81)

and

$$\begin{aligned} V-\varsigma \left( \frac{V_x}{\nu }\right) _x= \theta \ge 0. \end{aligned}$$

(82)

Using the distributional maximum principle (Littman 1963, Theorem B), we conclude that

$$\begin{aligned} V(x)> 0, \ 0<x<1, \end{aligned}$$

(83)

so

$$\begin{aligned} d_x(0)=-\frac{V(1/k)}{\varsigma }<0, \ d_x(x_k)=\frac{V(1-1/k)}{\varsigma }>0. \end{aligned}$$

(84)

Appendix E: Proof of Lemma 5.1

It suffices to prove that the set

$$\begin{aligned} O=\left\{ \varphi +\kappa b\varphi _x\Big | \varphi \in C^2[0,1], \varphi _x(0)=\varphi _x(1)=0\right\} \end{aligned}$$

is dense in $C[0,1]$.

Let $h\in C^2[0,1]$ be an arbitrary function which is locally constant near the zeros of $b$. These functions constitute a dense subset $O_1$ of $C[0,1]$. Let

$$\begin{aligned} \varphi (x)&= h(x)+ \int \limits _{x}^{a_i} \exp \left( \int \limits _x^y\frac{1}{\kappa b(t)}\,dt \right) h_y(y)\,dy,\nonumber \\&\qquad \,\,\,\,\,\,x_i< x< x_{i+1}, \ i=1,\dots , k, \nonumber \\&\qquad \,\,\,\,\,\,\varphi (x_i)=h(x_i),\ i=1,\dots , k+1. \end{aligned}$$

(85)

Clearly, $\varphi$ is equal to a constant $c^-_i$ (resp. $c^+_i$) in a left (resp. right) neighbourhood of the point $x_i$. But

$$\begin{aligned} h=\varphi +\kappa b\varphi _x, \end{aligned}$$

(86)

so $c^-_i=c^+_i=h(x_i)$. Thus, $\varphi$ is $C^2$-smooth and $\varphi _x(0)=\varphi _x(1)=0$. By virtue of (86), $O_1$ is contained in $O$.

Appendix F: Proof of Lemma 5.2

The solution $r$ to (30) can be written explicitly:

$$\begin{aligned} r(x)&= -\frac{1}{\kappa b(x)}\int \limits _{x_i}^x \exp \left( \int \limits _s^x\frac{1}{\kappa b(t)}\,dt \right) R(s)\,ds,\quad x_i<x < a_i,\end{aligned}$$

(87)

$$\begin{aligned} r(x)&= \frac{1}{\kappa b(x)}\int \limits _{x}^{x_{i+1}} \exp \left( \int \limits _s^x\frac{1}{\kappa b(t)}\,dt \right) R(s)\,ds,\quad a_i<x < x_{i+1}. \end{aligned}$$

(88)

Note that it is unique in $L_1(0,1)$. Indeed, if $r_1$ is another solution, then $\omega =r-r_1$ satisfies the conditions of Lemma 5.1.

We need to show that $r\rightarrow \sum \limits _{i=1}^k\widehat{R}_i\delta _{a_i}$ weakly-$*$, uniformly with respect to $R\in \mathcal R$. It suffices to prove that, for each $i, r\rightarrow \widehat{R}_i\delta _{a_i}$ weakly-$*$ on the interval $(x_i,x_{i+1})$, uniformly in $R\in \mathcal R$. We restrict ourselves to the case $i=1$, and the others are analogous.

We calculate, integrating by parts,

$$\begin{aligned} \widehat{r}_1&= \int \limits _0^{1/k} r(x)\,dx \nonumber \\&= -\int \limits _0^{a} \frac{1}{\kappa b(x)}\int \limits _{0}^x \exp \left( \int \limits _s^x\frac{1}{\kappa b(t)}\,dt \right) R(s)\,ds\,dx \nonumber \\&+\int \limits _{a}^{1/k}\frac{1}{\kappa b(x)}\int \limits _{x}^{1/k} \exp \left( \int \limits _s^x\frac{1}{\kappa b(t)}\,dt \right) R(s)\,ds\,dx\nonumber \\&= \left[ \int \limits _{0}^x \exp \left( \int \limits _s^x\frac{1}{\kappa b(t)}\,dt \right) R(s)\,ds\right] _a^0+\int \limits _0^{a} R(x)\,dx\nonumber \\&+\left[ \int \limits _{1/k}^x \exp \left( \int \limits _s^x\frac{1}{\kappa b(t)}\,dt \right) R(s)\,ds\right] _{1/k}^a+\int \limits _a^{1/k} R(x)\,dx\nonumber \\&= \widehat{R}_1-\int \limits _{0}^{1/k} \exp \left( \int \limits _s^a\frac{1}{\kappa b(t)}\,dt \right) R(s)\,ds= \widehat{R}_1. \end{aligned}$$

(89)

Let us show that for every $x_*\in (0,a)$

$$\begin{aligned} \lim \limits _{\kappa \rightarrow +\infty } \int \limits _0^{x_*} r(x)\,dx= 0, \end{aligned}$$

(90)

uniformly in $R\in \mathcal R$. Indeed, let $s_\kappa <x_*$ be such that

$$\begin{aligned} \int \limits _{x_*}^{s_\kappa }\frac{1}{ b(t)}\,dt= \sqrt{\kappa }. \end{aligned}$$

(91)

Observe that $s_\kappa \rightarrow 0$ as $\kappa \rightarrow +\infty$. We have

$$\begin{aligned} \int \limits _0^{x_*} r(x)\,dx&= -\int \limits _0^{x_*} \frac{1}{\kappa b(x)}\int \limits _{0}^x \exp \left( \int \limits _s^x\frac{1}{\kappa b(t)}\,dt \right) R(s)\,ds\,dx\nonumber \\&= \left[ \int \limits _{0}^x \exp \left( \int \limits _s^x\frac{1}{\kappa b(t)}\,dt \right) R(s)\,ds\right] _{x_*}^0+\int \limits _0^{x_*} R(x)\,dx\nonumber \\&= \int \limits _{0}^{x_*} \left[ 1-\exp \left( \int \limits _s^{x_*}\frac{1}{\kappa b(t)}\,dt \right) \right] R(s)\,ds\nonumber \\&\le \int \limits _{0}^{s_\kappa } \left[ 1-\exp \left( \int \limits _s^{x_*}\frac{1}{\kappa b(t)}\,dt \right) \right] R(s)\,ds \nonumber \\&+ \int \limits _{s_\kappa }^{x_*} \left[ 1-\exp \left( \int \limits _{s_\kappa }^{x_*}\frac{1}{\kappa b(t)}\,dt \right) \right] R(s)\,ds\nonumber \\&\le \int \limits _{0}^{s_\kappa } R(s)\,ds + [1-\exp (-\kappa ^{-1/2})]\int \limits _{s_\kappa }^{x_*}R(s)\,ds\rightarrow 0 \end{aligned}$$

(92)

as $\kappa \rightarrow +\infty$.

Due to (90), for any $f_0\in C[0,1/k]$,

$$\begin{aligned} \lim \limits _{\kappa \rightarrow +\infty } \int \limits _0^{x_*} r(x)f_0(x)\,dx= 0, \end{aligned}$$

(93)

uniformly in $R\in \mathcal R$.

Similarly, for all $x^*\in (a,1/k)$ and $f_0\in C[0,1/k]$,

$$\begin{aligned} \lim \limits _{\kappa \rightarrow +\infty } \int \limits _{x^*}^{1/k} r(x)f_0(x)v= 0, \end{aligned}$$

(94)

uniformly in $R\in \mathcal R$.

Fix $\varepsilon >0$ and $f\in C[0,1/k]$. Let $x_*$ and $x^*$ be so close to $a$ that $|f(x)-f(a)|\le \varepsilon /2$ provided $x_*\le x \le x^*$. Then

$$\begin{aligned} \left| \,\,\int \limits _{x_*}^{x^*} r(x)[f(x)-f(a)]\,dx\right| \le \varepsilon /2. \end{aligned}$$

(95)

Due to (93) and (94) with $f_0=f-f(a)$,

$$\begin{aligned} \left| \int \limits _{0}^{1/k} r(x)[f(x)-f(a)]\,dx-\int \limits _{x_*}^{x^*} r(x)[f(x)-f(a)]\,dx\right| \le \varepsilon /2 \end{aligned}$$

(96)

for sufficiently large $\kappa$. Thus,

$$\begin{aligned} \langle r- \widehat{R}_1\delta _{a_1},f\rangle =\langle r- \widehat{r}_1\delta _{a},f\rangle =\int \limits _{0}^{1/k} r(x)[f(x)-f(a)]\,dx\le \varepsilon . \end{aligned}$$

(97)

Appendix G: Proof of Lemma 5.3

Let $r_{\sigma ,\kappa }$ be the solution of the system

$$\begin{aligned} \left\{ \begin{array}{ll} r_{\sigma ,\kappa } -\kappa (br_{\sigma ,\kappa })_x= R_{\sigma ,\kappa }, &{} x\in (0,1)\\ r_{\sigma ,\kappa }\ge 0. \end{array}\right. \end{aligned}$$

(98)

Then, by Lemma 5.2,

$$\begin{aligned} \lim \limits _{\kappa \rightarrow +\infty } d(r_{\sigma ,\kappa }, \sum \limits _{i=1}^k\widehat{(R_{\sigma ,\kappa })}_i\delta _{a_i}) =0, \end{aligned}$$

(99)

uniformly in $\sigma$. Thus, it suffices to prove that for every $\kappa$ there is $\epsilon _\kappa >0$ such that

$$\begin{aligned} \lim \limits _{\kappa \rightarrow +\infty ,\ \sigma \le \epsilon _\kappa } d(w_{\sigma ,\kappa }, r_{\sigma ,\kappa }) =0. \end{aligned}$$

This would follow from the claim that for every $\kappa$ there is $\epsilon _\kappa >0$ so that for $\sigma \le \epsilon _\kappa$ we have $d(w_{\sigma ,\kappa }, r_{\sigma ,\kappa }) < 1/\kappa$. If it is not true, then for some $\kappa$ there exists a sequence $\sigma _n\rightarrow 0$ such that

$$\begin{aligned} d(w_{\sigma _n,\kappa }, r_{\sigma _n,\kappa }) \ge 1/\kappa . \end{aligned}$$

Since $w_{\sigma _n,\kappa }$ and $r_{\sigma _n,\kappa }$ are solutions of the problems (32) and (98), we have

$$\begin{aligned} -&\sigma _n \langle w_{\sigma _n,\kappa },\varphi _{xx}/\eta \rangle +\langle w_{\sigma _n,\kappa },\varphi +\kappa b\varphi _x\rangle = \langle R_{\sigma _n,\kappa },\varphi \rangle ,\end{aligned}$$

(100)

$$\begin{aligned}&\langle r_{\sigma _n,\kappa },\varphi +\kappa b\varphi _x\rangle = \langle R_{\sigma _n,\kappa },\varphi \rangle , \end{aligned}$$

(101)

for any $\varphi \in C^2[0,1], \varphi _x(0)=\varphi _x(1)=0$. Since the sequences $w_{\sigma _n,\kappa }$ and $r_{\sigma _n,\kappa }$ lie in the space of probability measures, which is weakly-* compact, without loss of generality there exist their weak-* limits $w_\kappa$ and $r_\kappa$. Clearly,

$$\begin{aligned} d(w_{\kappa }, r_{\kappa }) \ge 1/\kappa . \end{aligned}$$

(102)

On the other hand, taking the difference of (100) and (101), and passing to the limit, we find $\langle w_{\kappa }-r_{\kappa },\varphi +\kappa b\varphi _x\rangle =0,$ so $w_{\kappa }=r_{\kappa }$ by Lemma 5.1, and we arrive at a contradiction.

Appendix H: Proof of Lemma 5.4

Multiplying the second equation in (16) by $P$ and integrating, we find

$$\begin{aligned} \int \limits _0^1\nu (x) P^2(x) -\varsigma P_{xx}(x)P(x)\,dx=\int \limits _0^1\eta (x) p(x)P(x)\,dx, \end{aligned}$$

(103)

whence

$$\begin{aligned} \inf \limits _{0\le x\le 1}\nu (x)\int \limits _0^1 P^2(x)\,dx +\varsigma \int \limits _0^1 P_{x}^2(x)\,dx\le \sup \limits _{0\le x\le 1} P(x). \end{aligned}$$

(104)

Hence,

$$\begin{aligned} \Vert P\Vert ^2_{W_2^1(0,1)} \le C\Vert P\Vert _{C[0,1]}\le C\Vert P\Vert _{W_2^1(0,1)}\le C. \end{aligned}$$

(105)

The pair $(w_{\sigma ,\kappa }, R_{\sigma ,\kappa })=(\eta p,\nu P)$ satisfies (32). Due to (105), the set $\mathcal R =\{\nu P\,|\, \sigma >0,\kappa >0\}$ is uniformly bounded and thus uniformly integrable. By Lemma 5.3, for every $\kappa$ there exists $\epsilon _\kappa >0$ such that

$$\begin{aligned} \lim \limits _{\kappa \rightarrow +\infty ,\ \sigma \le \epsilon _\kappa } d\left( \eta p, \sum \limits _{i=1}^k\widehat{(\nu P)}_i\delta _{a_i}\right) =0. \end{aligned}$$

(106)

Assume that (34) is not true, i.e. there exist $\delta >0$ and sequences $\kappa _n\rightarrow \infty$ and $\sigma _n\le \epsilon _{\kappa _n}$ such that for the corresponding solutions $(p_n, P_n)=(p_{\sigma _n,\kappa _n},P_{\sigma _n,\kappa _n})$ to (16) we have $\Vert P_{n}-Q\Vert _{C[0,1]}>\delta$. Since the embedding $W_2^1(0,1)\subset C[0,1]$ is compact, without loss of generality we may assume that $P_n$ converges to some limit $P_0$ in $C[0,1]$. Obviously,

$$\begin{aligned} \Vert P_{0}-Q\Vert _{C[0,1]}\ge \delta . \end{aligned}$$

(107)

Passing to the limit in the second, forth and the last equations in (16)—the combination of the first two is understood in the weak sense (8)—and remembering (106), we find

$$\begin{aligned} \nu P_0 - \varsigma (P_0)_{xx}&= \sum \limits _{i=1}^k \widehat{(\nu P_0)}_i \delta _{a_i},\\ (P_0)_x(0)=(P_0)_x(1)&= 0,\\ \int \limits _0^1 \nu (x)P_0(x)\, dx&= 1. \end{aligned}$$

By Theorem 4.1, $P_0$ coincides with $Q$, which contradicts (107).

From (106) we deduce

$$\begin{aligned} \lim \limits _{\kappa \rightarrow +\infty ,\ \sigma \le \epsilon _\kappa } d\left( p, \sum \limits _{i=1}^k\frac{\widehat{(\nu P)}_i\delta _{a_i}}{\eta (a_i)} \right) =0. \end{aligned}$$

(108)

Due to (34) and $1/k$-periodicity of $\eta$, (108) implies that

$$\begin{aligned} p \rightarrow \sum \limits _{i=1}^k\frac{\widehat{(\nu Q)}_i\delta _{a_i}}{\eta (a)}. \end{aligned}$$

(109)

weakly-* as $\kappa \rightarrow +\infty ,\ \sigma \le \epsilon _\kappa$. Taking test functions which are equal to $1$ in one of the wells and are zero at the minima of the potential located outside of that well, we derive (35) from (109).