Scaling Limits for the Generalized Langevin Equation

Abstract

In this paper, we study the diffusive limit of solutions to the generalized Langevin equation (GLE) in a periodic potential. Under the assumption of quasi-Markovianity, we obtain sharp longtime equilibration estimates for the GLE using techniques from the theory of hypocoercivity. We then show asymptotic results for the effective diffusion coefficient in the small correlation time regime, as well as in the overdamped and underdamped limits. Finally, we employ a recently developed numerical method (Roussel and Stoltz in ESAIM Math Model Numer Anal 52(3):1051–1083, 2018) to calculate the effective diffusion coefficient for a wide range of (effective) friction coefficients, confirming our asymptotic results.

Appendices

Confirmation of the Rate of Convergence in the Quadratic Case

To assess the sharpness of the lower bounds on the convergence rate provided by Theorem 2.2, we consider the generalized Langevin dynamics confined by the quadratic potential $V(q) = k q^2/2$ (with $k>0$) in ${\mathcal {X}}= {\mathbf {R}}$. In this case, (2.2) can be written as (2.13) with

$$\begin{aligned} {\mathbf{D}} = \begin{pmatrix} 0 &{} 1 &{} 0 \\ -k &{} 0 &{} \displaystyle \frac{\sqrt{\gamma }}{\nu } \\ 0 &{} \displaystyle -\frac{\sqrt{\gamma }}{\nu } &{} \displaystyle -\frac{1}{\nu ^2} \\ \end{pmatrix} = {\mathbf{A}} + \frac{\sqrt{\gamma }}{\nu } {\mathbf{B}} + \frac{1}{\nu ^2} {\mathbf{C}}, \end{aligned}$$

(A.1)

with

$$\begin{aligned} {\mathbf{A}} = \begin{pmatrix} 0 &{} 1 &{} 0 \\ -k &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 \\ \end{pmatrix}, \qquad {\mathbf{B}} = \begin{pmatrix} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 1 \\ 0 &{} -1 &{} 0 \\ \end{pmatrix}, \qquad {\mathbf{C}} = \begin{pmatrix} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} -1 \\ \end{pmatrix}. \end{aligned}$$

In order to determine the elements in (2.14), it suffices to compute the three eigenvalues of ${\mathbf{D}}$, and combine them linearly with positive coefficients. This means that the spectral bound corresponds to the smallest absolute value of the real parts of the eigenvalues of ${\mathbf{D}}$. For the model GL1, the eigenvalues $x_j$ (for $1 \leqslant j \leqslant 3$) are the roots of the polynomial $p(x) = \mathrm {det}(x {\mathbf{I}} - {\mathbf{D}})$, which reads

$$\begin{aligned} p(x) = x^3 + \frac{x^2}{\nu ^2} + \left( \frac{\gamma }{\nu ^2} + k \right) x + \frac{k}{\nu ^2}. \end{aligned}$$

(A.2)

The spectral gap is easily seen to be a continuous function of $(\gamma ,\nu ) \in (0,+\infty ) \times (0,+\infty )$, with positive values. To obtain the scaling of the spectral gap as a function of the parameters, it therefore suffices to consider the various limiting regimes where at least one of the parameters goes to 0 or $+\infty$. The expression (A.1) suggests that the eigenvalues can be expanded in series of (inverse or fractional) powers of $\gamma$ and $\nu$. The leading-order term depends on the asymptotic regime which is considered. We start by investigating regimes where only one of the parameters goes to 0 or $+\infty$, and then discuss regimes where both parameters are varied. The organization of this appendix is illustrated in Fig. 7.

Fig. 7

Illustration of the different limits considered in this appendix. The joint limit $\gamma \rightarrow \infty$ and $\nu \rightarrow \infty$ with $\gamma / \nu ^2$ bounded from above and below, as well as the joint limit $\gamma \rightarrow \infty$ and $\nu \rightarrow 0$ with $\gamma \nu ^2$ bounded from above and below, are omitted for conciseness

(i) Limit $\nu \rightarrow +\infty$ with $\gamma$ Fixed We denote by $\varepsilon = \nu ^{-1}$ and rewrite (A.1) as ${\mathbf{A}} + \varepsilon \sqrt{\gamma }{\mathbf{B}} + \varepsilon ^2 {\mathbf{C}}$. The matrix ${\mathbf{A}}$ is diagonalizable and its eigenvalues 0 and $\pm \mathrm {i}\sqrt{k}$ are isolated and non-degenerate. Perturbation theory (Kato 1995, Chapter II) then shows that the eigenvalues $x_j(\varepsilon )$ are analytic functions of $\varepsilon$ for $\varepsilon$ sufficiently small. We write

$$\begin{aligned} x_j(\varepsilon ) = x_j^0 + x_j^1 \varepsilon + x_j^2 \varepsilon ^2 + \dots \end{aligned}$$

(A.3)

To identify the coefficients $x_j^k$, we plug the above expansion in the characteristic polynomial, which reads

$$\begin{aligned} p(x) = x^3 + \varepsilon ^2 x^2 + \left( \gamma \varepsilon ^2 + k \right) x + k \varepsilon ^2, \end{aligned}$$

(A.4)

and identify terms with the same powers of $\varepsilon$ in $p(x_j(\varepsilon ) ) = 0$. Some straightforward computations show that $x_j^{2n+1} = 0$ (in fact, the expansion in (A.3) could be immediately restricted to even powers of $\varepsilon$ by (Reed and Simon 1978, Theorem XII.2) since the coefficients of the polynomial are analytic in $\varepsilon ^2$), and

$$\begin{aligned} x_j^2 = - \frac{k+\gamma x_j^0+ (x_j^0)^2}{3(x_j^0)^2 + k},\quad x_j^4 = - \frac{(\gamma +2x_j^0)x_j^2+ 3x_j^0 (x_j^2)^2}{3(x_j^0)^2 + k}, \end{aligned}$$

so

$$\begin{aligned} x_1(\varepsilon )= & {} -\varepsilon ^2 + {\mathcal {O}}{\left( \varepsilon ^4\right) }, \qquad \nonumber \\ x_2(\varepsilon )= & {} \mathrm {i}\sqrt{k}\left( 1 + \frac{\gamma }{2k} \varepsilon ^2 - \frac{\gamma ^2}{8k^2} \varepsilon ^4\right) - \frac{\gamma }{2k}\varepsilon ^4 + {\mathcal {O}}{\left( \varepsilon ^6\right) }, \end{aligned}$$

(A.5)

and a similar expression for $x_3(\varepsilon )$ with an imaginary part of opposite sign. This shows that the spectral gap $\lambda$ is $\gamma /(2k\nu ^4) + {\mathcal {O}}{\left( \nu ^{-6}\right) }$ as $\nu \rightarrow +\infty$ with $\gamma >0$ fixed.

(ii) Limit $\gamma \rightarrow 0$ with $\nu$ Fixed We denote by $\varepsilon = \sqrt{\gamma }$ and rewrite (A.1) as ${\mathbf{A}} + \nu ^{-2}{\mathbf{C}} + \varepsilon \nu ^{-1}{\mathbf{B}}$. The dominant part ${\mathbf{A}} + \nu ^{-2} {\mathbf{C}}$ is diagonalizable with isolated and non-degenerate eigenvalues $x_1^0 = -\nu ^{-2}$, $x_2^0 = \mathrm {i}\sqrt{k}$ and $x_3^0 = -\mathrm {i}\sqrt{k}$. The eigenvalues $x_j(\varepsilon )$ of (A.1) are therefore analytic in $\varepsilon$ for $\varepsilon$ small enough, and an expansion of the form (A.3) also holds in this case. By an analysis similar to the one leading to (A.5), we obtain $x_j^1 = x_j^3 = 0$ and $x_j^2 = -x_j^0/(3\nu ^2(x_j^0)^2+2x_j^0+k\nu ^2)$, so

$$\begin{aligned} x_1(\varepsilon ) = -\frac{1}{\nu ^2} + {\mathcal {O}}{\left( \gamma \right) }, \qquad x_2(\varepsilon ) = \mathrm {i}\sqrt{k}-\frac{1-\mathrm {i}\sqrt{k} \nu ^2}{2(1+k\nu ^4)} \gamma + {\mathcal {O}}{\left( \gamma ^2\right) }, \end{aligned}$$

(A.6)

and a similar expression for $x_3(\varepsilon )$ with an imaginary part of opposite sign. This shows that the spectral gap $\lambda$ is $\gamma /(2(1+k\nu ^4)) + {\mathcal {O}}{\left( \gamma ^{2}\right) }$ as $\gamma \rightarrow 0$ with $\nu >0$ fixed.

(iii) Limit $\gamma \rightarrow +\infty$ for $\nu >0$ Fixed In order to treat the situation when $\gamma$ diverges, we introduce $\varepsilon = \gamma ^{-1/2}$ and rewrite (A.1) as $\varepsilon ^{-1} \left[ \nu ^{-1}{\mathbf{B}} + \varepsilon \left( {\mathbf{A}} + \nu ^{-2}{\mathbf{C}}\right) \right]$. The eigenvalues $x_j(\varepsilon )$ of ${\mathbf{D}}$ are then obtained by rescaling the eigenvalues $y_j(\varepsilon )$ of $\nu ^{-1}{\mathbf{B}} + \varepsilon \left( {\mathbf{A}} + \nu ^{-2}{\mathbf{C}}\right)$ as $x_j(\varepsilon ) = y_j(\varepsilon )/\varepsilon$. The eigenproblem associated with $\nu ^{-1}{\mathbf{B}} + \varepsilon \left( {\mathbf{A}} + \nu ^{-2}{\mathbf{C}}\right)$ has the form of a standard perturbation problem, with associated characteristic polynomial

$$\begin{aligned} P(y) = y^3 + \varepsilon \nu ^{-2}y^2 + (\nu ^{-2}+k\varepsilon ^2)y + k\nu ^{-2}\varepsilon ^3. \end{aligned}$$

The dominant part $\nu ^{-1} {\mathbf{B}}$ is diagonalizable with isolated and non-degenerate eigenvalues $y_1^0 = 0$, $y_2^0 = \mathrm {i}/\nu$ and $y_3^0 = -\mathrm {i}/\nu$. The eigenvalues $y_j(\varepsilon )$ of  $\nu ^{-1}{\mathbf{B}} + \varepsilon \left( {\mathbf{A}} + \nu ^{-2}{\mathbf{C}}\right)$ are therefore again analytic in $\varepsilon$ for $\varepsilon$ small enough, and an expansion of the form (A.3) still holds, with x replaced by y. By gathering terms with the same powers of $\varepsilon$ in $P(y_j(\varepsilon )) = 0$, we find after some computations that

$$\begin{aligned} y_1(\varepsilon ) = -k \gamma ^{-3/2}+ {\mathcal {O}}{\left( \frac{1}{\gamma ^{5/2}}\right) }, \qquad y_2(\varepsilon ) = \frac{\mathrm {i}}{\nu } - \frac{1}{2\nu ^2 \sqrt{\gamma }} + {\mathcal {O}}{\left( \frac{1}{\gamma }\right) }, \end{aligned}$$

and a similar expression for $y_3(\varepsilon )$ with an imaginary part of opposite sign. This shows that the spectral gap $\lambda$ is $k/\gamma + {\mathcal {O}}{\left( \gamma ^{-2}\right) }$ as $\gamma \rightarrow +\infty$ with $\nu >0$ fixed.

(iv) Limit $\nu \rightarrow 0$ for $\gamma$ Fixed We introduce here $\varepsilon = \nu$ and rewrite (A.1) as $\varepsilon ^{-2} \left[ {\mathbf{C}} + \varepsilon \sqrt{\gamma } {\mathbf{B}} + \varepsilon ^2 {\mathbf{A}} \right]$. As in the previous situation, we look for the eigenvalues $y_j(\varepsilon )$ of ${\mathbf{C}} + \varepsilon \sqrt{\gamma } {\mathbf{B}} + \varepsilon ^2 {\mathbf{A}}$, whose characteristic polynomial reads

$$\begin{aligned} P(y) = y^3 + y^2 + (\gamma \varepsilon ^2 + k\varepsilon ^4)y + k\varepsilon ^4. \end{aligned}$$

(A.7)

The dominant part ${\mathbf{C}}$ of the eigenvalue problem is diagonalizable with eigenvalues $y_1^0 = -1$ and $y_2^0 = y_3^0 = 0$. Since the latter eigenvalue is doubly degenerate, the results of (Kato 1995, Chapter II) show, by reducing the eigenvalue problem to the subspace associated with $y_2(\varepsilon )$ and $y_3(\varepsilon )$, that $y_1(\varepsilon )$ is analytic in $\varepsilon$ for $\varepsilon$ sufficiently small, while $y_2(\varepsilon )$ and $y_3(\varepsilon )$ are analytic in $\sqrt{\varepsilon }$:

$$\begin{aligned} y_2(\varepsilon ) = y_2^{1/2} \sqrt{\varepsilon } + y_2^1 \varepsilon + y_2^{3/2} \varepsilon ^{3/2} + \dots , \end{aligned}$$

(A.8)

and a similar expansion for $y_3(\varepsilon )$. Note that $y_1(\varepsilon ) = -1 + {\mathcal {O}}{\left( \varepsilon \right) }$, so the spectral gap is determined by the lowest order terms in the expansions of $\lambda _2(\varepsilon )$ and $\lambda _3(\varepsilon )$. By plugging (A.8) into (A.7), we find that the first nonzero terms in the expansions satisfy

$$\begin{aligned} (y_j^2)^2 + \gamma y_j^2 + k = 0, \qquad y_j^{5/2}(2y_j^2+\gamma ) = 0. \end{aligned}$$

We need to distinguish three situations at this stage:

• (i) When $\gamma ^2 > 4k$, one finds $y_2^2 = -\frac{\gamma }{2} \left( 1+\sqrt{1-\frac{4k}{\gamma ^2}}\right)$ and $y_3^2 = -\frac{\gamma }{2} \left( 1-\sqrt{1-\frac{4k}{\gamma ^2}}\right)$, so $y_j^{5/2} = 0$. In this case, the spectral gap of the rescaled matrix ${\mathbf{C}} + \varepsilon \sqrt{\gamma } {\mathbf{B}} + \varepsilon ^2 {\mathbf{A}}$ scales as $\frac{\gamma }{2} \left( 1-\sqrt{1-\frac{4k}{\gamma ^2}}\right) + {\mathcal {O}}{\left( \nu \right) }$ as $\nu \rightarrow 0$ with $\gamma >0$ fixed.

• (ii) When $\gamma ^2 < 4k$, one finds $y_2^2 = -\frac{\gamma }{2} \left( 1-\mathrm {i}\sqrt{\frac{4k}{\gamma ^2}-1}\right)$ and $y_3^2 = -\frac{\gamma }{2} \left( 1+\mathrm {i}\sqrt{\frac{4k}{\gamma ^2}-1}\right)$, so $y_j^{5/2} = 0$. In this case, the spectral gap of the rescaled matrix scales as $\frac{\gamma }{2} + {\mathcal {O}}{\left( \nu \right) }$ as $\nu \rightarrow 0$ with $\gamma >0$ fixed.

• (iii) When $\gamma ^2 = 4k$, one finds $y_2^2 = y_3^2 = -\frac{\gamma }{2}$, and actually $y_j^{5/2} = 0$ (because of the next order condition in $\varepsilon$ which reads $(y_j^{5/2})^2+(\gamma +2y_j^2)y_j^3 = 0$). The spectral gap of the rescaled matrix scales as in the previous case as $\frac{\gamma }{2} + {\mathcal {O}}{\left( \nu \right) }$ as $\nu \rightarrow 0$.

In conclusion, the spectral gap $\lambda$ of ${\mathbf{D}}$ scales as $\frac{\gamma }{2}\left( 1-{\mathbb {1}}_{\gamma ^2 > 4k}\sqrt{1-\frac{4k}{\gamma ^2}}\right) + {\mathcal {O}}{\left( \nu ^{3}\right) }$.

(v) Joint Limits $\gamma \rightarrow 0$ and $\nu \rightarrow +\infty$, or $\nu \rightarrow +\infty$ with $\gamma /\nu ^2 \rightarrow 0$ We denote by $\eta = \frac{\sqrt{\gamma }}{\nu }$ and $\varepsilon = \frac{1}{\nu ^2}$. These two parameters are small in the asymptotic regime which is considered. The eigenvalues of the leading-order matrix ${\mathbf{A}}$ in (A.1) are isolated, so, by perturbation theory (upon writing the spectral projector using contour integrals and expanding the resolvents which appear), we can write the eigenvalues of ${\mathbf{D}} = {\mathbf{A}} + \eta {\mathbf{B}} + \varepsilon {\mathbf{C}}$ as

$$\begin{aligned} x_j(\eta ,\varepsilon )= & {} x_j^0 + x_j^{1,0} \eta + x_j^{0,1} \varepsilon + x_j^{2,0} \eta ^2 + x_j^{1,1} \eta \varepsilon + x_j^{0,2} \varepsilon ^2 + \dots , \end{aligned}$$

(A.9)

with $x_1^0 = 0$, $x_2^0 = \mathrm {i}\sqrt{k}$ and $x_3^0 = -\mathrm {i}\sqrt{k}$. We next identify terms with the same powers of $\eta ,\varepsilon$ in $p(x_j(\eta ,\varepsilon ) )=0$ with $p(x) = x^3 + \varepsilon x^2 + (\eta ^2+k)x+k\varepsilon$. Some straightforward computations show that

$$\begin{aligned} x_1(\eta ,\varepsilon )= & {} -\varepsilon + \frac{1}{k} \eta ^2 \varepsilon + {\mathcal {O}}{\left( \eta ^2\left( \varepsilon ^2 + \eta ^2\right) \right) }, \qquad \\ x_2(\eta ,\varepsilon )= & {} \mathrm {i}\left( \sqrt{k} + \frac{\eta ^2}{2\sqrt{k}}\right) - \frac{1}{2k} \eta ^2\varepsilon + {\mathcal {O}}{\left( \varepsilon ^4 + \eta ^4\right) }, \end{aligned}$$

and a similar expansion for $x_3(\eta ,\varepsilon )$ with imaginary parts of opposite signs. Note that the expressions of $x_1,x_2$ coincide with (A.5), as well with (A.6) in the limit $\nu \rightarrow +\infty$. It can be shown that $x_j^{2m,0} \in \mathrm {i}{\mathbb {R}}$ and $x_j^{2m+1,0} = 0$ for any $m \geqslant 1$ and $1 \leqslant j \leqslant 3$ (it suffices in fact to set $\varepsilon =0$ to identify these coefficients, which amounts to considering the polynomial $x(x^2+\eta ^2+k)$, whose roots are 0 and $\pm \mathrm {i}\sqrt{k+\eta ^2}$). Similarly, $x_j^{0,m} = 0$ for $m \geqslant 2$ and $1 \leqslant j \leqslant 3$, as seen by setting $\eta = 0$ and factorizing p(x) as $(x+\varepsilon )(x^2+k)$. The spectral gap is therefore $\displaystyle \frac{\gamma }{2k \nu ^4} + {\mathcal {O}}{\left( \varepsilon ^2 \eta ^2\right) }$.

(vi) Joint Limits $\gamma \rightarrow +\infty$ with $\gamma /\nu ^2 \rightarrow +\infty$ and $\gamma \nu ^2 \rightarrow +\infty$ We denote by $\eta = \nu /\sqrt{\gamma }$ and $\varepsilon = 1/(\nu \sqrt{\gamma })$ the two small parameters in the asymptotic regime considered here. We write ${\mathbf{D}}$ as $\eta ^{-1}\left( {\mathbf{B}} + \eta {\mathbf{A}} + \varepsilon {\mathbf{C}}\right)$, the characteristic polynomial associated with ${\mathbf{B}} + \eta {\mathbf{A}} + \varepsilon {\mathbf{C}}$ being $P(y) = y^3 + \varepsilon y^2 + (k\eta ^2 +1)y + k\varepsilon \eta ^2$. We can use an argument similar to the one used to write (A.9) with $y_1^0 = 0$, $y_2^0 = \mathrm {i}$ and $y_3^0 = -\mathrm {i}$ to obtain the following expansion for the zeros of the polynomial P(y):

$$\begin{aligned} y_1(\eta ,\varepsilon ) = - k\eta ^2\varepsilon + {\mathcal {O}}{\left( \eta ^4 + \eta ^4\right) }, \qquad y_2(\eta ,\varepsilon ) = \mathrm {i}- \frac{\varepsilon }{2} + {\mathcal {O}}{\left( \varepsilon ^2 + \eta ^2\right) }, \end{aligned}$$

and a similar expansion for $y_3(\eta ,\varepsilon )$ with imaginary parts of opposite signs. Note that it can be shown that the remainders for $y_1(\eta ,\varepsilon )$ involve only powers of $\eta ^2$ (since the polynomial itself is analytic in $\eta ^2$, see (Reed and Simon 1978, Theorem XII.2)) and there are no remainders of the form $\varepsilon ^n$ or $\eta ^n$ (as seen by setting, respectively, $\eta =0$ and $\varepsilon =0$ in the expression of P(y)). For $y_2$, it can similarly be shown that $y_2^{2m,0} \in \mathrm {i}{\mathbb {R}}$ and $y_2^{2m+1,0} = 0$ for any $m \geqslant 1$. The spectral gap is therefore $\displaystyle k\eta \varepsilon + {\mathcal {O}}{\left( \eta \varepsilon ^2\right) }$ with $k\eta \varepsilon = k/\gamma$, which coincides with the limit obtained for $\gamma \rightarrow +\infty$ with $\nu$ fixed.

(vii)Joint Limits $\nu \rightarrow 0$ and $\gamma \rightarrow +\infty$ with $\gamma \nu ^2 \rightarrow 0$ It is difficult to rely on a perturbative approach here. Indeed, denoting by $\eta = \nu \sqrt{\gamma }$ and $\varepsilon = \nu ^2$ the two small parameters in the asymptotic regime considered here, we write ${\mathbf{D}}$ as $\varepsilon ^{-1}\left( {\mathbf{C}} + \eta {\mathbf{B}} + \varepsilon {\mathbf{A}}\right)$. The dominant term ${\mathbf{C}}$, however, has degenerate eigenvalues, so it is not clear whether the eigenvalues can be expanded as in (A.9) (in fact, it is in general not true that this is the case, see (Kato 1995, Section II.5.7)). We use for this case another argument, based on a localization of the zeros of the characteristic polynomial $P(y) = y^3 + y^2 + (k\varepsilon ^2 + \eta ^2)y + k\varepsilon ^2$ associated with ${\mathbf{C}} + \eta {\mathbf{B}} + \varepsilon {\mathbf{A}}$. We first note that the discriminant of the third-order polynomial P is

$$\begin{aligned}&\left( k\varepsilon ^2 + \eta ^2\right) ^2 \left[ 1- 4 \left( k\varepsilon ^2 + \eta ^2\right) \right] - \left( 4 + 9 k\varepsilon ^2 - 18 \eta ^2\right) k\varepsilon ^2 \\&\qquad = -4k\varepsilon ^2 + \eta ^4 + {\mathcal {O}}{\left( \varepsilon ^4 + \eta ^2 \varepsilon ^2 + \eta ^6\right) }. \end{aligned}$$

Since $\eta ^4/\varepsilon ^2 = \gamma ^2 \rightarrow +\infty$, the discriminant is positive in the limiting regime we consider, so that P admits three real zeros. The polynomial p in (A.4) therefore also admits three real zeros. We next compute

$$\begin{aligned} p\left( -\frac{1}{\nu ^2} + \delta \gamma \right) = \frac{(\delta -1)\gamma }{\nu ^4} + \delta \gamma \left( \frac{(1-2\delta )\gamma }{\nu ^2} + \delta ^2\gamma ^2 + k\right) . \end{aligned}$$

When $\delta = 1$, the right-hand side of the above equality is $-\gamma ^2/\nu ^2<0$ at dominant order, while, for $\delta = 1+\varepsilon$ with a fixed small parameter $\varepsilon > 0$, the right-hand side scales as $\varepsilon \gamma /\nu ^4 > 0$. This shows that one of the roots is in the interval $[-\nu ^{-2}+\gamma ,-\nu ^{-2}+(1+\varepsilon )\gamma ]$ for $\gamma$ large enough and $\nu ,\gamma \nu ^2$ sufficiently small. To localize the second root, we consider

$$\begin{aligned} \frac{\nu ^2}{\gamma ^2} p\left( -(1 + \delta )\gamma \right) = \delta (1+\delta ) - \gamma \nu ^2(1+\delta )^3-\frac{k\nu ^2}{\gamma }(1+\delta ) + \frac{k}{\gamma ^2}. \end{aligned}$$

The right-hand side converges to $\delta (1+\delta )$ as $\gamma \rightarrow +\infty$ and $\nu \rightarrow 0$ with $\gamma \nu ^2\rightarrow 0$, which shows that, for any $\varepsilon \in (0,1]$, the second root is in the interval $[-(1+\varepsilon )\gamma ,-(1-\varepsilon )\gamma ]$ for $\gamma$ large enough and $\nu ,\gamma \nu ^2$ sufficiently small. Finally,

$$\begin{aligned} \begin{aligned} \nu ^2 p\left( -\frac{k}{\gamma }-\frac{(1+\delta )k^2}{\gamma ^3}\right) = -\delta \frac{k^2}{\gamma ^2}&- \frac{k\nu ^2}{\gamma }\left( k +\frac{(1+\delta )k^2}{\gamma ^2}\right) - \frac{\nu ^2}{\gamma ^3}\left( k +\frac{(1+\delta )k^2}{\gamma ^2}\right) ^3 \\&+ \frac{(1+\delta )k^3}{\gamma ^4}\left( 2 +\frac{(1+\delta )k}{\gamma ^2}\right) . \end{aligned} \end{aligned}$$

The right-hand side scales as $-\delta k^2/\gamma ^2$ as $\gamma \rightarrow +\infty$ and $\nu \rightarrow 0$ with $\gamma \nu ^2\rightarrow 0$, which shows that, for any $\varepsilon \in (0,1]$, the third root is in the interval $[-k/\gamma -(1+\varepsilon )k^2/\gamma ^2,-k/\gamma -(1-\varepsilon )k^2/\gamma ^2]$ for $\gamma$ large enough and $\nu ,\gamma \nu ^2$ sufficiently small. The spectral gap is dictated by the location of this eigenvalue and scales therefore as $k/\gamma$ in the limit which is considered here.

(viii) Limit $\gamma \rightarrow 0$ and $\nu \rightarrow 0$ Here also, it is difficult to rely on a perturbative approach, so we use an alternative method. Employing the same reasoning as in §(vii), it is possible to show that the characteristic polynomial p admits only one real root in this limit, denoted by $x_1$, and that this root is, for $\varepsilon \in (0,1]$ fixed, in the interval $[- \nu ^{-2} + (1-\varepsilon )\gamma , - \nu ^{-2} + (1+\varepsilon ) \gamma ]$ provided that $\gamma \nu ^2$ is sufficiently small. Inspired by (A.6), we show that the other two, complex conjugate roots, which we denote by $x^{\pm }$, with the superscript indicating the sign of the imaginary part, are close to ${\hat{x}}^{\pm } := - \gamma /2 \pm \mathrm {i}\sqrt{k}$. To this end, we calculate

$$\begin{aligned} p({\hat{x}}^{\pm }) = k \gamma \pm \frac{3}{4} \mathrm {i}\sqrt{k} \gamma ^{2} - \frac{\gamma ^{3}}{8} - \frac{\gamma ^{2}}{4 \nu ^{2}}. \end{aligned}$$

Given the factorization $p(x) = (x - x_1)(x - x^+)(x - x^-)$, this implies

$$\begin{aligned} ({\hat{x}}^{\pm } - x^{+}) ({\hat{x}}^{\pm } - x^{-}) = \frac{\displaystyle k \gamma \pm \frac{3}{4} \mathrm {i}\sqrt{k} \gamma ^{2} - \frac{\gamma ^{3}}{8} - \frac{\gamma ^{2}}{4 \nu ^{2}}}{{\hat{x}}^{\pm } - x_1}. \end{aligned}$$

(A.10)

Since $|{\hat{x}}^{\pm } - x_1|$ scales as $\nu ^{-2}$, the modulus of the right-hand side scales as $k \gamma \nu ^2 - \frac{\gamma ^2}{4}$ in the limit as $\gamma \rightarrow 0$ and $\nu \rightarrow 0$. This implies

$$\begin{aligned} |{\hat{x}}^{\pm } - x^{\pm }|^2 \leqslant |{\hat{x}}^{\pm } - x^{+}| \, |{\hat{x}}^{\pm } - x^{-}| = {\mathcal {O}}{\left( \gamma \nu ^2 + \gamma ^2\right) }, \end{aligned}$$

so $|{\hat{x}}^{\pm } - x^{\pm }| = {\mathcal {O}}{\left( \sqrt{\gamma } \nu + \gamma \right) }$. Therefore, $|{\hat{x}}^{\pm } - x^{\mp }|$ converges to $2\sqrt{k}$ in the limit $\gamma ,\nu \rightarrow 0$, so (A.10) implies that in fact $|{\hat{x}}^{\pm } - x^{\pm }| = {\mathcal {O}}{\left( \gamma \nu ^2 + \gamma ^2\right) }$, which is negligible in front of $\gamma /2 = -\mathrm {Re}({\hat{x}}^{\pm })$. In conclusion, the spectral gap scales as $\gamma /2$ in the limit considered here.

Remark A.1

The scaling of the exponential decay rate can also be obtained from explicit expressions for the roots of the characteristic polynomial, as we illustrate below in the particular case of the limit $\nu \rightarrow \infty$ with fixed $\gamma$. For simplicity, we consider $k = \gamma = 1$. In this case, the characteristic polynomial is

$$\begin{aligned} p(x) = \frac{1}{\nu ^2} \left( \nu ^2 x^3 + x^2 + \left( 1 +\nu ^2 \right) x + 1 \right) . \end{aligned}$$

Letting $\lambda = \nu ^2$, we define

$$\begin{aligned} q(x) = \lambda x^3 + x^2 + \left( 1 + \lambda \right) x + 1 =: ax^3 + b x^2 + cx + d. \end{aligned}$$

Using the standard notation for expressing the roots of a cubic polynomial, we define

$$\begin{aligned} \Delta _0&= b^2 - 3 ac = 1 - 3 \lambda ( 1 + \lambda ) = -3 \lambda ^2 - 3 \lambda + 1, \\ \Delta _1&= 2 b^3 - 9 abc + 27 a^2 d = 2 - 9 \lambda (1 + \lambda ) + 27 \lambda ^2 = 18 \lambda ^2 - 9 \lambda + 2, \end{aligned}$$

and

$$\begin{aligned} C&= \root 3 \of {\frac{\Delta _1 + \sqrt{\Delta _1^2 - 4 \Delta _0^3}}{2}}\\&= \root 3 \of {\frac{18 \lambda ^2 - 9 \lambda + 2 + \sqrt{108 \lambda ^{6} + 324 \lambda ^{5} + 540 \lambda ^{4} - 432 \lambda ^{3} + 81 \lambda ^{2}}}{2}}. \end{aligned}$$

By a Taylor expansion, obtained by symbolic calculation, we calculate

$$\begin{aligned} \frac{C}{\lambda }&= \sqrt{3} + \left( \frac{\sqrt{3}}{2} + 1\right) \frac{1}{\lambda } - \left( 3 + \frac{\sqrt{3}}{4} \right) \frac{1}{2\lambda ^2} + {\mathcal {O}}{\left( \frac{1}{\lambda ^3}\right) } \\&=: {\hat{C}}(\lambda ) + {\mathcal {O}}{\left( \frac{1}{\lambda ^3}\right) }. \end{aligned}$$

The roots of q (and therefore also of p) are given by

$$\begin{aligned} x_k&= - \frac{1}{3a}\left( b+\xi ^kC+\frac{\Delta _0}{\xi ^kC}\right) , \\&= - \frac{1}{3}\left( \frac{1}{\lambda }+\xi ^k {\hat{C}}+\frac{\Delta _0}{\xi ^k {\hat{C}} \lambda ^2} + {\mathcal {O}}{\left( \frac{1}{\lambda ^3}\right) }\right) , \qquad \xi = \frac{-1 + i \sqrt{3}}{2}, \qquad k \in \{0,1,2\}. \end{aligned}$$

Carrying out the calculation of the leading-order terms using symbolic calculations, we obtain

$$\begin{aligned} \mathrm {Re} (x_0)&= - \frac{1}{\lambda } + {\mathcal {O}}{\left( \frac{1}{\lambda ^2}\right) }, \quad \mathrm {Re} (x_1) = - \frac{1}{2\lambda ^2} + {\mathcal {O}}{\left( \frac{1}{\lambda ^3}\right) }, \quad \\ \mathrm {Re} (x_2)&= - \frac{1}{2\lambda ^2} + {\mathcal {O}}{\left( \frac{1}{\lambda ^3}\right) }, \end{aligned}$$

so the decay rate scales as $-1/2 \nu ^4$ in the limit, which coincides with the result found in §(i). This example shows that, while feasible, obtaining the scaling of the spectral gap from the explicit expressions for the roots of the characteristic polynomial presents a level of difficulty similar to the one encountered above. One advantage of our approach is that it can easily be extended to higher dimensions, which is relevant when more auxiliary processes are considered in the GLE.

Longtime Behavior for Model GL2

We present here elements on how to obtain results for GL2 similar to the ones proved for GL1 in Sect. 3, with choices of coefficients providing a modified $H^1(\mu )$ inner product and guaranteeing hypocoercivity and hypoelliptic regularization. We omit the details of the calculations, which are more cumbersome than for GL1. Defining the generator associated with GL2 as

where $\partial ^*_{z_2} := \beta z_2 - \partial _{z_2}$, the approach outlined below leads to the resolvent bound

$$\begin{aligned} \left\| {\mathcal {L}}^{-1}\right\| _{{\mathcal {B}}\left( L^2_0(\mu )\right) } \leqslant C \max \left( \gamma , \ \frac{1}{\gamma }, \ \frac{\nu ^{4}}{\gamma }, \ \frac{\nu ^{8}}{\alpha ^{4} \gamma }, \ \frac{\gamma }{\alpha ^{2}}, \ \frac{\gamma ^{2} \nu ^{2}}{\alpha ^{4}}\right) . \end{aligned}$$

(B.1)

In contrast to our observations in the case of model GL1, the scaling on the right-hand side of this equation appears not to be sharp. Indeed, an explicit calculation of the scaling of the spectral bound in the case of a quadratic potential, which is possible by using the same approach as in Appendix A, shows that the resolvent bound scales in fact as $\gamma$ in the limit as $\gamma \rightarrow \infty$ for fixed $\alpha$ and $\nu$, and not as $\gamma ^2$ as suggested by (B.1).

Let us now make precise how we obtain (B.1). Define the coefficients

$$\begin{aligned} a_0&= 2 A^{4} \min \left( 1, \alpha , \frac{\alpha ^{2}}{\nu ^{2}}, \frac{\alpha ^{2}}{\sqrt{\gamma } \nu }\right) , \\ a_1&= 2 A^{10} \min \left( 1, \alpha , \frac{1}{\nu ^{2}}, \frac{\alpha ^{4}}{\nu ^{6}}, \frac{1}{\sqrt{\gamma } \nu }, \frac{\alpha ^ {4}}{\gamma ^{\frac{3}{2}} \nu ^{3}}\right) , \\ a_2&= 2 A^{14} \min \left( 1, \alpha , \gamma , \frac{\gamma }{\nu ^{4}}, \frac{\alpha ^{4} \gamma }{\nu ^{8}}, \frac{1}{\sqrt{\gamma } \nu }, \frac{\alpha ^{4}}{\gamma ^{\frac{3}{2}} \nu ^{3}}\right) , \\ a_3&= 2 A^{16} \min \left( \gamma , \frac{1}{\gamma }, \frac{\gamma }{\nu ^{4}}, \frac{\alpha ^{2}}{\gamma }, \frac{\alpha ^{4} \gamma }{\nu ^{8}},\frac{\alpha ^{4}}{\gamma ^{2} \nu ^{2}}\right) , \end{aligned}$$

and

$$\begin{aligned} b_0&= A^{7} \min \left( \alpha , \frac{1}{\alpha }, \frac{\alpha ^{3}}{\nu ^{4}}, \frac{\alpha ^{3}}{\gamma \nu ^{2}}\right) , \\ b_1&= A^{12} \min \left( \sqrt{\gamma } \nu ,\frac{1}{\sqrt{\gamma } \nu }, \frac{\sqrt{\gamma }}{\nu ^{3}}, \frac{\alpha ^{4} \sqrt{\gamma }}{\nu ^{7}}, \frac{\alpha ^{4}}{\gamma ^{\frac{3}{2}} \nu ^{3}}\right) , \\ b_2&= A^{15} \min \left( \gamma , \frac{1}{\gamma }, \frac{\gamma }{\nu ^{4}}, \frac{\alpha ^{2}}{\gamma }, \frac{\alpha ^{4} \gamma }{\nu ^{8}},\frac{\alpha ^{4}}{\gamma ^{2} \nu ^{2}}\right) . \end{aligned}$$

It is possible to show that, for any sufficiently small A and for fixed $\beta$, the following conditions are satisfied for any values of the parameters $\gamma$, $\nu$ and $\alpha$:

• The following matrix is positive definite:

  $$\begin{aligned} \begin{pmatrix} a_0 &{} -b_0 &{} 0 &{} 0 \\ -b_0 &{} a_1 &{} -b_1 &{} 0 \\ 0 &{} -b_1 &{} a_2 &{} -b_2 \\ 0 &{} 0 &{} -b_2 &{} a_3 \end{pmatrix}. \end{aligned}$$

  Consequently, the inner product defined by polarization from the norm

  

  is equivalent to the usual $H^{1}(\mu )$ inner product, by an inequality similar to (3.5).

• The operator $- {\mathcal {L}}$ is coercive in $H^1_0(\mu )$ endowed with the inner product $\left( \!\left( \,\cdot \,, \,\cdot \,\right) \!\right)$. More precisely, it holds for any $h \in H^1_0(\mu )$ that

  $$\begin{aligned} - \left( \!\left( h, \mathcal Lh\right) \!\right) \geqslant C(A) \lambda (\mu , \nu , \alpha ) \left( \!\left( h, h\right) \!\right) , \end{aligned}$$

  where

  $$\begin{aligned} \lambda (\mu , \nu , \alpha ) = \min \left( \gamma , \ \frac{1}{\gamma }, \ \frac{\gamma }{\nu ^{4}}, \ \frac{\alpha ^{4} \gamma }{\nu ^{8}}, \ \frac{\alpha ^{2}}{\gamma }, \ \frac{\alpha ^{4}}{\gamma ^{2} \nu ^{2}}\right) . \end{aligned}$$

  (B.2)

• For any $h \in L^2_0(\mu )$, it holds for $t \in [0, 1]$, where $N_h(t)$ is defined by

  

This leads finally to (B.1).

Technical Results Used in Sect. 4.3

In this section, we present technical results that are used in Sect. 4.3. The following estimate is useful in motivating (4.19).

Lemma C.1

Assume that $c > 0$ is a constant, that h is a smooth function on the torus ${\mathbf {T}}$ and that g is a smooth, everywhere positive function on ${\mathbf {T}}$. Then there exists a unique smooth solution f to the equation

(C.1)

In addition,

(C.2)

Proof

Let $f_a(q)$ denote the solution to

Since g is smooth and everywhere positive, this equation admits a unique smooth solution on the interval $[-\pi , \pi ]$, by the standard theory of ordinary differential equations. Rewriting the equation for $f_a$ as

it is clear that

$$\begin{aligned} f_a(q) = \exp \left( \int _{-\pi }^{q} \frac{c}{g(x)} \, {\mathrm {d}}x \right) \left[ a + \int _{-\pi }^{q} \, \exp \left( - \int _{-\pi }^{y} \frac{c}{g(x)} \, {\mathrm {d}}x \right) \, \frac{h(y)}{g(y)} \, {\mathrm {d}}y \right] .\qquad \end{aligned}$$

(C.3)

The unique value $a^*$ for which the function is periodic (namely $f_{a^*}(\pi ) = a^*$) is

$$\begin{aligned} a^* = -\left[ 1- \exp \left( -\int _{-\pi }^{\pi } \frac{c}{g(x)} \, {\mathrm {d}}x\right) \right] ^{-1}\int _{-\pi }^{\pi } \, \exp \left( - \int _{-\pi }^{y} \frac{c}{g(x)} \, {\mathrm {d}}x \right) \, \frac{h(y)}{g(y)} \, {\mathrm {d}}y. \end{aligned}$$

It is easily checked that all derivatives of $f_{a^*}$ are also continuous, so that $f_{a^*} \in C^{\infty }({\mathbf {T}})$.

To obtain the estimate (C.2), we introduce and note by that, by (C.1),

The term vanishes at the extrema of r(q), which leads then to (C.2). $\square$

We next prove another technical result, which we then employ to show that the leading-order term ${\overline{\phi }}_0$ in the underdamped limit, defined in (4.20), indeed belongs to $H^1(\mu )$ when V is not constant.

Lemma C.2

If $q \mapsto V(q)$ is not constant, then there exists $M > 0$ such that

$$\begin{aligned} \forall E \in (E_0, \infty ), \qquad S_{\nu }(E) \geqslant \frac{\nu ^2}{1 + \nu ^2 M} \int _{\mathbf {T}}\sqrt{2(E_0-V(q))} \, {\mathrm {d}}q. \end{aligned}$$

(C.4)

In particular, $S_{\nu }(E)$ is bounded below by a positive constant uniformly in $E\in (E_0,\infty )$.

Proof

Fix $E > E_0$. We integrate (4.15) over ${\mathbf {T}}$ and use integration by parts for the first term, which gives

(C.5)

We now show that that there exists a positive constant M such that

(C.6)

To this end, notice that is a decreasing function of E for fixed q, so it is sufficient to show that

Since V is smooth, it holds that for any q such that $V(q) = E_0$, so

By using L’Hôpital’s rule, we notice that, for values of q in a neighborhood of any $q^* \in V^{-1}\{ E_0 \}$ with $V'(q) \ne 0$,

so L(q) is bounded uniformly from above, implying that (C.6) holds. This concludes the proof because, by (C.5), and recalling that $s_\nu \geqslant 0$ by (4.16),

and the last integral is positive because V is not constant. $\square$

Lemma C.3

If $q \mapsto V(q)$ is not constant, then the function ${\overline{\phi }}_0$ defined in (4.19) belongs to $H^1(\mu )$.

Proof

It is easy to see that the function ${\overline{\phi }}_0$ defined in (4.19) belongs to $L^2(\mu )$. We next consider the distributional derivatives for $y \in \{q,p\}$, since derivatives in z vanish: For $H(q,p) < E_0$, these derivatives vanish; while for $H(q,p) > E_0$, it holds and . Note that all these derivatives belong to $L^{\infty }_{\mathrm{loc}}({\mathbf {T}}\times {\mathbf {R}}\times {\mathbf {R}})$ in view of the lower bound provided by (C.4). Moreover, $\left| \nabla {{\overline{\phi }}}_0(q, p, z)\right|$ grows sufficiently slowly as $|p| \rightarrow \infty$. This allows to conclude that ${\overline{\phi }}_0 \in H^1(\mu )$, as claimed. $\square$