Erratum: A first-order dynamical transition in the displacement distribution of a driven run-and-tumble particle (2019 J. Stat. Mech. 053206)

We present here a revised version of the appendices of Gradenigo and Majumdar (2019 J. Stat. Mech. 053206). Some minor corrections are introduced and a new simplified argument to obtain the critical value of rc, the control parameter for the transition, is presented. The overall scenario and the description of the transition mechanism depicted in Gradenigo and Majumdar (2019 J. Stat. Mech. 053206) remains completely untouched, the only relevant dierence being the value of rc fixed to rc = 2 1/3 = 1.259 92 . . . rather than rc = 1.3805 . . .. This dierence also implies a small quantitative changes in figures 2 and 4; a new version of both figures is reported here. A couple of other typos discovered in the paper are pointed out and the correct version of the expressions are reported. G Gradenigo and S N Majumdar


Amendments to appendix B
In this erratum, we report a corrected version of the appendix B of [1], including dierent subsections of appendix B, i.e. B.1-B.3. In section B.1, the mechanism for choosing the correct root is pointed out, and furthermore, some algebraic errors have been corrected in section B.2. This analytically gives the correct value of r c = 2 1/3 = 1.259 92 (instead of the old value of r c = 1.3805 which was numerically obtained in the published version). Consequently the correct value of z c = 11.7771.. replaces the old numerical value z c ≈ 12.0. This change of z c appears clearly in the new figure 4 of this erratum, where the dotted vertical line is clearly shifted to the left with respect to the same figure in the published version [1]. The argument to obtain r c = 2 1/3 is presented in section B.3. In order to facilitate the comparison to the figures of the present manuscript we have given the same numbers as in [1]. Finally, we thank N Smith for pointing out the algebraic error in appendix B.2 of the published version.

B. Derivation of the rate function χ(z) in the intermediate matching regime
In this appendix we study the leading large N behavior of the integral that appears in the expression for P A (z, N ) in equation (56) of [1]: where z 0 can be thought of as a parameter and with σ 2 = 2 + 5E 2 . It is important to recall that the contour Γ (+) is along a vertical axis in the complex s-plane with its real part negative, i.e. Re(s) < 0. Thus, we can deform this contour only in the upper left quadrant in the complex s plane (Re(s) < 0 and Im(s) > 0), but we cannot cross the branch cut on the real negative axis, nor can we cross to the s-plane where Re(s) > 0. A convenient choice of the deformed contour, as we will see shortly, is the Γ (+) rotated anticlockwise by an angle π/2, so that the contour now goes along the real negative s from 0 to −∞.
To evaluate the integral in equation (1), it is natural to look for a saddle point of the integrand in the complex s plane in the left upper quadrant, with fixed z. Hence, we look for solutions for the stationary points of the function F z (s) in equation (2). They are given by the zeros of the cubic equation (3) As z 0 varies, the three roots move in the complex s plane. It turns out that for z < z l (where z l is to be determined), there is one positive real root and two complex conjugate roots. For example, when z = 0, the three roots of equation ( at s = (2Eσ 2 ) −1/2 e iφ with φ = 0, φ = 2π/3 and φ = 4π/3. However, for z > z l , all the three roots collapse on the real s axis, with s 1 < s 2 < s 3 . The roots s 1 < 0 and s 2 < 0 are negative, while s 3 > 0 is positive. For example, in figure B1, we plot the function F z (s) in equation (3) as a function of real s, for z = 12 and E = 2 (so σ 2 = 2 + 5E 2 = 22). One finds, using Mathematica, three roots at s 1 = −1/2 (the lowest root on the negative side), s 2 = −0.175 186 . . . and s 3 = 0.129 732 . . .. We can now determine z l very easily. As z decreases, the two negative roots s 1 and s 2 approach each other and become coincident at z = z l and for z < z l , they split apart in the complex s plane and become complex conjugates of each other, with their real parts identical and negative. When s 1 < s 2 , the function F z (s) has a maximum at s m with s 1 < s m < s 2 (see figure B1). As z approaches z l , s 1 and s 2 approach each other, and consequently the maximum of F z (s) between s 1 and s 2 approaches the height 0. Now, the height of the maximum of F z (s) between s 1 and s 2 can be easily evaluated. The maximum occurs at s = s m where F z (s) = 0, i.e. at s m = −(Eσ 2 ) −1/3 . Hence the height of the maximum is given by Hence, the height of the maximum becomes exactly zero when Thus we conclude that for z > z l , with z l given exactly in equation (5), the function F z (s) has three real roots at s = s 1 < 0, s 2 < 0 and s 3 > 0, with s 1 being the smallest negative root on the real axis. For z < z l , the pair of roots are complex (conjugates). However, it turns out (as will be shown below) that for our purpose, it is sucient to consider evaluating the integral in equation (1) only in the range z > z l where the roots are real and evaluating the saddle point equations is considerbaly simpler. So, focusing on z > z l , out of these three roots as possible saddle points of the integrand in equation (1), we have to discard s 3 > 0 since our saddle points have to belong to the upper left quadrant of the complex s plane. This leaves us with s 1 < 0 and s 2 < 0. Now, we deform our vertical contour Γ (+) by rotating it anticlockwise by π/2 so that it runs along the negative real axis. Between the two stationary points s 1 and s 2 , it is easy to see (see figure B1) that F z (s 1 ) > 0 (indicating that it is a minimum along real s axis) and F z (s 2 ) < 0 (indicating a local maximum). Since the integral along the deformed contour is dominated by the maximum along real negative s for large N, we should choose s 2 to be the correct root, i.e. the largest among the negative roots of the cubic equation z + σ 2 s − 1/(2Es 2 ) = 0. Thus, evaluating the integral at s * = s 2 (and discarding pre-exponential terms) we get for large N where the rate function χ(z) is given by J. Stat. Mech. (2020) 049901 The right hand side can be further simplified by using the saddle point equation (3), i.e. z + σ 2 s 2 − 1/2Es 2 2 = 0. This gives (8)

B.1. Asymptotic behavior of χ(z)
We now determine the asymptotic behavior of the rate function χ(z) in the range z l < z < ∞, where z l is given in equation (5). Essentially, we need to determine s 2 (the largest among the negative roots) as a function of z by solving equation (3), and substitute it into equation (8) to determine χ(z). We first consider the limit z → z l from above, where z l is given in equation (5). As z → z l from above, we have already mentioned that the two negative roots s 1 and s 2 approach each other. Finally at z = z l , we have s 1 = s 2 = s m where s m = −(Eσ 2 ) −1/3 is the location of the maximum between s 1 and s 2 . Hence as z → z l from above, s 2 → s m = −(Eσ 2 ) −1/3 . Substituting this value of s 2 in equation (8) gives the limiting behavior as announced in the first line of equation (24) in [1].
To derive the large z → ∞ behavior of χ(z) as announced in the second line of equation (24) in [1], it is first convenient to re-parametrize s 2 and define Substituting this into equation (3), it is easy to see that θ z satisfies the cubic equation Note that due to the change of sign in going from s 2 to θ z , we now need to determine the smallest positive root of θ z in equation (11). In terms of θ z , χ(z) in equation (8) reads The formulae in equations (11)-(13) are now particularly suited for the large z analysis of χ(z). From equations (11) and (12) it follows that in the limit z → ∞ we have that b(z) → 0, so that θ z → 1. Hence, for large z or equivalently small b(z), we can obtain a perturbative solution of equation (11). To leading order, it is easy to see that with b(z) given in equation (12). Substituting this into equation (13) gives the large z behavior of χ(z) The dependence on N of the prefactor in the right hand side of both equations (56) and (85) in [1] is wrong, 1/ √ N 1/3 must be replaced with N 5/6 . In fact, the correct expression to be considered in place of equation (56) whereas the correct expression to be considered in place of equation (85)