Force-induced desorption of copolymeric comb polymers

We investigate a lattice model of comb copolymers that can adsorb at a surface and that are subject to a force causing desorption. The teeth and the backbone of the comb are chemically distinct and can interact differently with the surface. That is, the strength of the surface interaction can be different for the monomers in the teeth and in the backbone. We consider several cases including (i) the uniform case where the number of teeth is fixed and the lengths of the branches in the backbone and the lengths of the teeth are all identical, (ii) the case where the teeth are short compared to the branches in the backbone, (iii) the situation where the teeth are long compared to the backbone, and (iv) the case where the number of teeth approaches infinity. We determine the free energies in the thermodynamic limit and discuss the nature of the phase diagrams of the model.


Introduction
The introduction of experimental techniques such as atomic force microscopy [10,40] has made possible the micro-manipulation of individual polymer molecules and allowed the investigation of how individual polymer molecules respond to applied forces.This has led to renewed interest in the theoretical treatment of this topic [1,12,13,19].For a review, see for instance [30].
A particular case that has attracted attention is when the polymer is adsorbed at a surface and is desorbed by the action of the force [5,18,20,24,25,28].Several different models have been investigated [30] but we shall concentrate here on interacting models of self-avoiding walks [15,27] and related systems.For other related work, see [2,32,33].
Polymer adsorption is important in steric stabilization of colloids [29] and linear polymers terminally attached to a surface can be used as steric stabilizers.Other polymer architectures such as stars and brushes have been used for this purpose.Block copolymers are also useful for steric stabilization, where one block adsorbs on the surface and the other extends into the solution.In particular, comb polymers with the backbone and teeth having different chemical composition (see figure 1) have been investigated experimentally [39].These have several parameters that can be varied, including the length and spacing of the teeth of the comb and the chemical compositions of the teeth and backbone.
There are a few studies of the self-avoiding walk model of homopolymeric star polymers, adsorbed at a surface and subject to a desorbing force [3,4,21,22].For an investigation of the same situation for some models of homopolymeric branched polymers, including uniform combs, see reference [22].Even fewer papers [17,23] have appeared on self-avoiding walk models of copolymers adsorbed at a surface and subject to a force.Linear AB diblock copolymers and ABA triblock copolymers have been studied where the A and B monomers interact differently with the surface [17].One end monomer, ie a vertex of degree 1, of the copolymer is grafted to the surface and the force is applied either at the opposite end (the other degree 1 vertex) of the block copolymer or at the central vertex.Copolymeric stars have also been considered [23].Think of a multi-arm star with some arms (or branches) of type A and the remaining arms of type B. A vertex of degree 1 is fixed in the surface (say of an A-arm) and the force is applied either at a degree 1 vertex of type A or B, or at the central vertex.In each case the general form of the phase diagram has been established [23].
In this paper we explore the problem of copolymeric comb polymers where the backbone and the teeth have different chemical composition; see figure 1 for a schematic diagram.This is a generalization and extension of a treatment of a self-avoiding walk model of homopolymeric combs, pulled from a surface at which they adsorb, studied in [22].There it was shown that there is a critical force where the comb is pulled from an adsorbed phase into a ballistic phase.An applied force at the terminal end of the backbone may be strong enough to pull off both the last segment of the backbone and the last tooth.In this case it desorbs the entire comb.There is also the possibility that the force is strong enough to pull off the last segment (or branch) of the backbone, but not the last tooth.In this case the teeth remain adsorbed, and the comb only desorbs completely if the strength of the force is increased to also pull off the last tooth.
A lattice model of a copolymeric comb is illustrated in figure 2. The comb is embedded in the cubic lattice Z 3 (or more generically, in the d-dimensional hypercubic lattice).Attaching the coordinates (x 1 , x 2 , x 3 ) to the lattice sites (with each coordinate an integer), the comb in figure 2 is also in the half-lattice L 3 of all lattice sites with x 3 ≥ 0. The half-lattice has a boundary (hard wall) x 3 = 0 that will also be called the adsorbing surface or adsorbing plane.
A schematic diagram of a pulled adsorbing comb block copolymer.The comb consists of a backbone (blue) consisting of B-type monomers and t teeth (red) consisting of A-type monomers.An endpoint of the comb's backbone is grafted to the origin on the adsorbing surface x 3 = 0, and the other endpoint of its backbone is pulled vertically by a force f .Monomers along the backbone adsorb on the adsorbing surface with activity b = e −ǫ B /k B T where ǫ B is a binding energy, T is the absolute temperature and k B is Boltzmann's constant.Monomers in the teeth of the comb similarly adsorb on the adsorbing surface, but with activity a = e −ǫ A /k B T where ǫ A is a binding energy.Starting from the origin and moving along the backbone, the trivalent vertices (branch points) are labelled 1, 2, ..., t as indicated in circles, where the number of teeth t = 4 in this case.The teeth are similarly labelled according to the label of the trivalent vertex from which they emanate.Branches of the backbone are labelled with 1, ..., t + 1 so that the branch between the trivalent vertices i − 1 and i has label i.We assume that each backbone branch has length m b and each of the teeth has length ma.
A lattice copolymeric comb.The backbone of the comb adsorbs with activity b, and the teeth with activity a.A vertical force f pulls the comb at its endpoint, while its other endpoint is grafted at the origin.This comb has 3 teeth.
The backbone of the comb in figure 2 is composed of B-monomers, and has length 16 consisting of 4 segments or branches, each of length m b = 4.The comb has t = 3 teeth composed of Amonomers, each of length m a = 3.The total size of the comb is (t + 1) m b + t m a = 28.The backbone of the comb makes v B = 3 visits, each of weight b, to the adsorbing plane, and the teeth make v A = 2 visits, each of weight a.The last vertex of the backbone is at a height h = 3 above the adsorbing plane.Putting y = e f /kB T , where f is the applied force, k B is Boltzmann's constant and T is the absolute temperature, the partition function of the comb is where ) is the number of combs, with one terminal vertex attached to the surface, with t teeth, each of length m a , t + 1 backbone branches of length m b , having v A tooth vertices in the surface and v B + 1 backbone vertices in the surface, and with the other terminal vertex at height h above the surface.The thermodynamic limit of this model can be examined in several cases.In the first instance, putting m b = m a = m gives a uniform comb, and taking m → ∞ (with t fixed) gives a sparsely branched comb with a long backbone and long teeth.Other limits can be taken by taking m b → ∞ with m a fixed, or taking m a → ∞ with m b fixed.There is also the limit t → ∞, followed by either m b → ∞ or m a → ∞, or both.We shall investigate some of these limits.The plan of the paper is as follows.In section 2 we recall some results about self-avoiding walks adsorbing at a surface and subject to a desorbing force, and in the same section we prove some lemmas that we shall need later in the paper.
In section 3 we look at the case of uniform combs where m a = m b = m (so that all branches in the backbone and teeth are the same length) and the number of teeth, t, is fixed.We derive expressions for the free energy and give the general features of the phase diagram as slices at constant a, constant b and constant y.In section 4 we examine the situations where the backbone is long compared to the teeth, and where the teeth are long compared to the backbone, still with t fixed.These cases give rise to two limiting cases, one where the comb behaves like a self-avoiding walk and the other where it behaves like a star.In section 5 we examine the infinite t limit, and we close with a short discussion in section 6.

Adsorbing and pulled walks
In this section we give a brief account of previous results, concentrating on self-avoiding walks.We shall need some of these results in the following sections where we construct combs from collections of self-avoiding walks and stars.We focus on the simple cubic lattice Z 3 but some of the results for self-avoiding walks can be extended to Z d for all d ≥ 2.
Consider self-avoiding walks in Z 3 where we attach the standard coordinate system (x 1 , x 2 , x 3 ) so that lattice vertices have integer coordinates.For an n-edge self-avoiding walk, number the vertices k = 0, 1, . . ., n and write x i (k) for the i-th coordinate of the k-th vertex.Write c n for the number of n-edge self-avoiding walks starting at the origin.It is known that the limit exists [6] and µ 3 is called the growth constant of the lattice (while κ 3 = log µ 3 is the connective constant of the lattice).
An n-edge self-avoiding walk is unfolded in the x i -direction (or . We shall write c † n (since the count is independent of i) for the number of these unfolded walks.It is known [8] that In general we use the superscript † to show that a walk is unfolded and †[i] when we need to emphasise that the unfolding is in the i-th direction.
If the walk is doubly unfolded, then it is unfolded in one direction and then unfolded again in a second direction, and we use the superscript ‡, or ‡[ij] if we need to specify the two unfolding directions.
If a self-avoiding walk satisfies the constraint that x 3 (k) ≥ 0 for 0 ≤ k ≤ n then the walk is a positive walk (see figure 3).A positive walk can also be unfolded.If we write c + n for the number of positive walks with n edges, and c + † [3] n for the number of positive walks with n edges, unfolded in the x 3 -direction, then lim n→∞ [8,36].If a positive walk is constrained to start and end in the plane x 3 = 0, then it is a loop.If we write ℓ n for the number of loops with n edges, and ℓ † [1] n for the number of loops, unfolded in the x 1 -direction, with n edges, then lim n→∞ [8,36].By symmetry, the same result holds for loops unfolded in the x 2 -direction.
All these results are also valid in the square lattice (or, more generally, in the hypercubic lattices).
Let c + n (v, h) be the number of n-edge positive walks, starting at the origin, having v + 1 vertices in the plane x 3 = 0 and with x 3 (n) = h.We say that the walk has v visits and that the height of the last vertex is h.Define the partition function exists for all a and y [18] and ψ(a, y) is the free energy of the model.(We use lower case letters for counts, upper case letters for partition functions, and Greek letters for free energies.)Here where ǫ is the energy associated with a vertex in the surface, k B is Boltzmann's constant, T is the absolute temperature and f is the force normal to the surface (measured in energy units).
A positive walk in the half-space x 3 ≥ 0. (Right) An x 1 -unfolded loop in the half-space x 3 ≥ 0. This loop is pulled at its last vertex by a horizontal force f parallel to the hard wall (boundary of the half-space).
We can turn off the force by setting y = 1 and we then have the pure adsorption problem where ψ(a, 1) ≡ κ(a).The free energy κ(a) is a convex function of log a [7] and is therefore continuous.
There exists a critical value of a, a c > 1, such that κ(a) = log µ 3 when a ≤ a c and κ(a) > log µ 3 when a > a c [7,14,26].
If we write ℓ n (v) for the number of loops with n edges and v + 1 vertices in the plane x 3 = 0 then [7] lim We say that ℓ n (v) is the number of adsorbing loops with n edges and v visits, and L n (a, 1) is the partition function for loops with no applied force.The free energy κ(a) is unchanged when the walks or loops are unfolded, or even doubly unfolded [7], or when the endpoints of loops are restricted to a line (for example, to the x 1 -axis), see, for example, reference [15, section 9.1].Restricting any of these walks or loops to a quarter space (for example x i ≥ 0 for i ∈ {1, 2}) also leaves κ(a) unchanged, even if endpoints are restricted to be located in a line.For example, if adsorbing loops are unfolded in the x 1 -direction (see figure 3) their partition function is L † [1] n (a, 1) and their free energy is also equal to κ(a) [7].The free energy is unchanged for the subset of these loops restricted to the quarter space x 2 , x 3 ≥ 0 and having end points in the x 1 -axis.
We can turn off the interaction between a positive walk and the surface by setting a = 1 so that the surface just acts as an impenetrable barrier (or hard wall) and we can then write ψ(1, y) ≡ λ(y).λ(y) is a convex function of log y [16], λ(y) = log µ 3 for y ≤ 1 and λ(y) > log µ 3 for y > 1 [1,12].This is the free energy of pulled positive walks.If the positive walks are unfolded, or doubly unfolded, in any direction, then the free energy is unchanged.For example, if the pulled walks are unfolded in the x 3 -direction, then the partition function is C (1, y) and the free energy is λ(y) [19].
Pulled positive walks are a special case of walks (not necessarily positive) terminally attached to the origin and then pulled at the other end by a force f in the x 3 direction; we call these x 3 -pulled walks.The partition function for these walks is denoted here by C n (y) with y being conjugate to the x 3 -span of the walk.C n (y) ≥ C + n (1, y) and its limiting free energy is also λ(y) = ψ(1, y) for y ≥ 1.This follows from unfolding arguments that give the string of inequalities For general values of a and y [18] ψ(a, y) = max[κ(a), λ(y)] so ψ(a, y) = log µ 3 when a ≤ a c and y ≤ 1.This is the free phase.There are phase boundaries in the (a, y)-plane at a = a c for y ≤ 1, at y = 1 for a ≤ a c and at the solution of κ(a) = λ(y) for a ≥ a c and y ≥ 1.The phase diagram has three phases and the phase transition for y > 1 and a > a c between the adsorbed phase and the ballistic phase is first order [5].
The definitions for positive walks and loops above are all relative to a fixed boundary surface x 3 = 0 and a corresponding half-space x 3 ≥ 0. The same results hold for negative walks and loops in the half-space x 3 ≤ 0. Similarly, the results hold for other choices for the fixed boundary surface x i = 0, i ∈ {1, 2}, and we define x i -positive (negative) walks and loops for these cases.We also refer to x i -loops as loops oriented in the x i direction.Also by symmetry, results about x 3 -pulled walks apply equally to x i -pulled walks i ∈ {1, 2} where y is conjugate to x i -span.
In some of the constructions that we use later in the paper we shall make extensive use of the properties of loops, and we shall need several new results about loops.If the endpoint of a loop (oriented in the x 3 -direction) is pulled by a force f parallel to the x 1 -direction (see the right panel of figure 3), then it is both adsorbing and pulled.We write ℓ n (v, s) for the number of loops of n edges with v visits and x 1 -span equal to s = x 1 (n) − x 1 (0) (the distance between its endpoints in the x 1 -direction), then, provided that the Boltzmann factor for adsorbed vertices is a = 1 and y ≥ 1, This is also the case if L n (1, y) is replaced by the x i -unfolded loop partition function n (1, y) for i = 1, 2, or even for the partition function L ‡ [1,2] n (1, y) of doubly unfolded loops in the x 1 and x 2 directions.The lemma also applies to these models when the walks are confined to a quarter lattice defined by x 2 , x 3 ≥ 0 and if the endpoints of the loops are restricted to the x 1 -axis and pulled parallel to the x 1 axis.We prove this result for the partition functions L † [1] n (1, y) and L n (1, y) in the lemma 1.The proofs for the other models are similar.Lemma 1.For loops with no interaction with the surface (a = 1) and pulled parallel to the surface (in the x i -direction, i ∈ {1, 2}), the free energy is λ(y) for y ≥ 1.If the loops are unfolded in the x i -direction they have free energy, λ(y), for all y > 0. Proof: Consider i = 1 and y ≥ 1. Write L n (1, y) = v,s ℓ n (v, s) y s for the partition function of loops with n edges, pulled in the x 1 -direction, and write L † [1] n (1, y) for the partition function of loops, unfolded and pulled in the x 1 -direction.Such loops are examples of x 1 -pulled walks, thus by inclusion, for y ≥ 1 Consider self-avoiding walks unfolded in the x 1 -direction and subject to a force in the same direction.These are examples of x 1 -pulled x 1 -positive walks.Suppose that their partition function is (1, y) by inclusion.Unfolding cannot decrease the distance between the end-points in the unfolding direction so, via equation (3), for y ≥ 1, C Consider n-edge self-avoiding walks pulled and unfolded in the x 1 -direction, and consider the subset of these walks unfolded in the x 3 -direction.Write C ‡ [13] n (y) for their partition function.Then C ‡ [13]   n (y) ≤ C † [1]  n (y) ≤ C ‡ [13]   n Consider the subset of these walks with the most popular span in the x 3 -direction.‡ By concatenating such a walk with most popular span in the x 3 direction and the reverse of such a walk (joined by an additional edge) a loop of length 2n + 1 edges is obtained, pulled and unfolded in the x 1 -direction.
There are at least C ‡ [13] n (y)/(n + 1) such loops obtained by this construction so For y < 1, the lemma is a corollary of lemma 1 and theorem 1 in reference [19].‡ The most popular x 3 -span is that value of the span giving a maximal contribution to the partition function.That is, the most popular span refers to an x 3 -span such that the contribution to the partition function C ‡ [13] n (y) due to walks with that span is at least as large as the contribution due to any other x 3 -span.
An unfolded walk from the origin in the slab Sw of width w defined by all vertices j with 0 ≤ x 2 (j) ≤ w.The x 3 direction is normal to the plane of the page.

Confined walks
In addition to the above results, we shall also need a result on walks confined to slabs.
The situation we consider is illustrated in figure 4. A positive walk from the origin in the half space x 3 ≥ 0 and adsorbing in the x 3 = 0 plane is also confined to a slab S w with boundaries being the two planes x 2 = 0 and x 2 = w.That is, this walk is adsorbing in the plane x 3 = 0 (namely the (x 1 , x 2 )-plane shown in the figure) and is confined in the slab with x 3 ≥ 0 between the two planes x 2 = 0 and x 2 = w normal to the x 2 -direction and a distance w apart (these two planes project to the horizontal lines bounding the slabe in figure 4).Denote the number of these walks of length n by c (w) n .Using the methods in references [9,35,37,38] it can be shown that the limit exists, and that µ (w) < µ (w+1) < µ 3 and lim w→∞ µ (w) = µ 3 .
We extend these results to walks adsorbing in the x 3 = 0 plane below.The partition function of these walks is then n (v) is the number of walks with v vertices in the plane x 3 = 0 (apart from the origin), and confined to the slab S w .
Next, consider the walks in Denote the number of these unfolded walks of length n in S w having v vertices (apart from the origin) in the adsorbing plane (a) which are also loops with both endpoints in the x 1 -axis.That is, x 1 -unfolded walks in the slab with x j (0) = x j (n) = 0, j = 2, 3.In this case, the double dagger superscript does not denote double unfolding but instead denotes that the endpoints of the loops are restricted to the surface x 3 = 0 and also to the line x 2 = x 3 = 0 (namely the x 1 -axis).This subset of unfolded loops will be used in constructions later in the paper.
We have the following lemma.

Lemma 2. The limit
The same results hold if the walks in S w are restricted to the quarter space x 1 , x 3 ≥ 0 and/or restricted to loops or x 1 -unfolded loops with or without the endpoints restricted to the x 1 -axis.
Proof: Unfolded loops with both endpoints in the x 1 -axis contributing to the partition function L ‡ [1],(w) n (a) can be concatenated with unfolded loops in the partition function L ‡ [1],(w) m (a) by placing the first vertex of the second loop on the last vertex of the first loop.This shows that and this establishes the existence of the limit lim (a) and the partition functions for walks or loops in S w and restricted to the quarter space x 1 , x 3 ≥ 0 fit between these two extremes.
Consider the subset of walks from the origin, confined to S w , unfolded in the x 1 -direction, with partition function There is a subset of these walks with most popular (x 2 , x 3 )coordinates of their end-point.Such walks can be concatenated with a reverse of such a walk (and an additional edge in the x 1 -direction), to form a loop with endpoints in the x 1 -axis.This gives the bound . exist.
Since any subwalk with x 2 -span greater than w cannot occur as a subwalk of any walk in the slab S w , it follows from the pattern theorem for adsorbing half-space walks (see [15, To prove that lim w→∞ κ (w) (a) = κ(a), consider, L ‡ [1] m (a), the partition function for unconfined x 1 -unfolded loops with end points in the x 1 -axis.Its limiting free energy is κ(a).Thus for any ǫ > 0 there exists m (a) > κ(a) − ǫ for m ≥ N .These loops will fit into a slab with w > m.Fix w > m ≥ N and then put n = mp + q with n ≥ N , non-negative integers p and q, and 0 ≤ q < m.By concatenating p unfolded loops of size m and a final loop of size q, Take n → ∞ by taking p → ∞ to obtain κ(a) − ǫ ≤ κ (w) (a).This completes the proof.

Bounds on the free energy of uniform combs
In this section we are concerned with uniform combs.See figure 2 for a sketch.A comb can be thought of as a self-avoiding walk making up the backbone of the comb, with t teeth attached at regular intervals along the backbone.The teeth are also self-avoiding walks, and the teeth and backbone are mutually avoiding.The comb is placed in the half lattice L 3 which is that part of the cubic lattice with non-negative x 3 coordinates.The first vertex of the backbone of the comb is grafted at the origin, and all other vertices are in L 3 .
A comb with t teeth has t vertices of degree 3 and t + 2 vertices of degree 1.The number of edges in each subwalk between two adjacent vertices of degree 3 and between the end vertices of degree 3 and the initial and terminal vertices of degree 1 is m b .The number of edges in each tooth is m a .The total number of edges in the comb is n = (t + 1) m b + t m a .
In the uniform case we take m a = m b = m so that n = (2t + 1) m.In this section we consider t-combs where the number of teeth t ≥ 1 is a fixed integer and define to be the number of uniform t-combs with a total of n edges, having v A A-visits and v B B-visits, and with the x 3 -coordinate of the terminal vertex of the backbone equal to h.The corresponding partition function in terms of the general partition function ( 1) is In order to prove existence of a unique limiting free energy ζ(a, b, y) = lim n→∞ 1 n log K n (a, b, y) in this model, we shall consider the schematic conformations of adsorbed combs shown in figures 5 and 6.The proof proceeds by establishing lower and upper bounds on K n (a, b, y).
We proceed by first considering x 1 -unfolded adsorbing B-loops and denote their partition function by L † [1] n (b).In figure 5 we are interested in such loops along the backbone of the comb with x 1 -spans greater than a constant value w, and we denote the partition function of these by L † [1] n (>w, b).The remaining loops of x 1 -span less than or equal to w have partition function A corresponding equation arises for the partition functions for subsets of these loops that are restricted further either to x 2 ≥ 0 and/or to having end points in the x 1 -axis.We need the following lemma in order to find a lower bound on the free energy of uniform combs.
Lemma 3.For any fixed w ≥ 0, lim The same result holds for the subsets of these loops with x 2 ≥ 0 and/or with the endpoints restricted to the x 1 -axis.
Proof: Take logarithms on the left hand side of equation ( 13), divide by n, and let n → ∞.This gives for this case (and for the other restrictions considered) The walks contributing to L † [1] n (≤w, b) correspond to subsets of walks that fit in the slab of width w, S w , as in figure 4 except now the restricting planes are constant x 1 planes instead of constant x 2 Figure 5. Top view schematic projected onto the x 1 x 2 -plane of an adsorbed comb in the halfspace x 3 ≥ 0 grafted at the origin (on the left) and pulled by a vertical force at its endpoint (on the right) from the adsorbing plane.The backbone consists of unfolded loops with endpoints in the x 1 -axis (dotted line) in the half-space x 2 ≥ 0. The teeth are confined in slabs of width w normal to the x 1 -axis in the half-space x 2 < 0.
x 2 planes.By symmetry, the results of lemma 2 also hold for walks in S w (or with the additional restriction x 2 ≥ 0).It is thus a corollary of lemma 2 that lim sup Existence of the limits above then implies that lim n→∞ n (>w, b) = κ(b).Next, consider figure 6.On the left hand side is a negative x 1 -loop (that is, a loop oriented in the x 1 direction in the half-space x 1 ≤ 0).It is x 3 -unfolded and pulled by a force in the x 3 -direction.These loops have the same partition function and free energy, λ(y) for y > 0, as the unfolded loops of Lemma 1. Suppose that the length of the loop is m b .Then there is a most popular (vertical) distance or height between its endpoints, denoted by h y as in the figure (this is a function of the pulling force f = k B T log y and of the length m b ).The number of unfolded loops of length m b and height h y is denoted by ℓ † [3] m b (h), and it follows that Since, by Lemma 1, the free energy of these pulled unfolded loops exists and is equal to λ(y), this shows that lim It is the case that h y is not bounded, but grows with m b .This is shown in the next lemma.
Lemma 4. lim inf A schematic drawing of a pulled unfolded loop.The loop is oriented in the half-space x 1 ≤ 0 and its endpoints are in the plane x 1 = 0.A force f is applied in the vertical direction.If the length of the loop is m b , then there is a most popular vertical separation (or height) hy between its endpoints.This most popular height is a function of both m b and y. (Right) Constructing a pulled comb by combining pulled loops with teeth confined to horizontal slits or slabs of (constant) height w.In this case, each loop has length m b and height hy (the most popular height at an applied force f ).Each tooth has length ma and is joined to the backbone by a u horizontal edges.
Proof: Suppose that there exists a w ≥ 0 such that h y < w for all m b ≥ 0. Then (that is, each loop of length m b and height h y is also a self-avoiding walk confined in a slab of width w, bounded by the planes x 3 = 0 and x 3 = w).Taking logarithms, dividing by m b and then taking m b → ∞ gives via equation ( 9), for y > 0, This is a contradiction (since λ(y) ≥ log µ 3 ), and thus h y is not bounded.Since the limit on the left exists, this is valid if the limits are taken along any subsequence, with the result that h y is unbounded along any subsequence.Thus, for any N > 0 there is an M such that h y > N for all m b > M .

Proof:
Lower bound: First establish a lower bound by considering t-combs in conformations shown in figures 5 and 6.We label the trivalent nodes in the combs, starting in the node closest to the origin by 1, 2,. . ., t, as shown.A lower bound is constructed by considering three cases, based on the schematic diagrams: 1) the height of the pulled vertex is 0 (a special case of figure 5); 2) the height of trivalent vertex labeled t is 0 and all the vertices in the (t + 1)-st branch are out of the surface (see figure 5); 3) all vertices of the comb are above the surface except for the origin (see figure 6).
Consider case (1) and figure 5 but with the last (pulled) vertex at height zero.The backbone of the comb is a sequence of t loops unfolded in the x 1 -direction, each of length m b and confined to the sublattice x 2 ≥ 0 and x 3 ≥ 0, with endpoints in the x 1 -axis.The last segment along the backbone is also assumed to be an unfolded loop with endpoints in the x 1 -axis.
Next, assume that the x 1 -span of each loop along the backbone exceeds w.We simplify the presentation by denoting the partition function for each of these loops by L † [1] n (>w, b) (this includes the restrictions that x 2 ≥ 0 and the loop endpoints are in the x 1 -axis).Thus by lemma 3 their limiting free energies are κ(b).
The teeth in figure 5 are quarantined in slabs of width w (bounded by constant x 1 planes) in the sublattice x 2 < 0 and x 3 ≥ 0, each of length m a , and interacting with the x 3 -plane.Each tooth is attached to the backbone by a sequence of u edges in the x 2 direction, and one may assume that u = 1.By symmetry, these walks have the same partition function as the quarter space walks of lemma 2. For simplicity, we denote that partition function here by C (w) ma−u (a) (this includes the quarter space restriction).By lemma 2 its limiting free energy is κ (w) (a).
Thus, the comb is now put together by assuming each tooth is quarantined in a slab of width fixed at w, and the loops along the backbones have x 1 -spans exceeding w.This shows that for case (1) above, Next, put m a = m b = m, take logarithms, divide by (2t + 1) m, and then take m → ∞.By lemmas 2 and 3 and equation (7), Since w is arbitrary, one may take w → ∞ on the right hand side.By lemma 2 κ (w) (a) → κ(a).This gives the lower bound t 2t+1 κ(a) + t+1 2t+1 κ(b) from case (1).Case (2) is treated in the same way, except that the last segment of the backbone is anchored by the t-th trivalent vertex to the adsorbing plane x 3 = 0 but is otherwise disjoint from it, and is pulled in its endpoint.This gives the lower bound lim inf Taking w → ∞ gives the lower bound 1 2t+1 λ(y) + t 2t+1 (κ(a) + κ(b)) from case (2).Consider case (3) next.The proof is again similar to that of case (1), but now loops are stacked in the x 3 -direction as shown in figure 6.Let y > 0, fix m b and choose w < h y , with h y as defined before equation (14).Put u = 1.Then Next, put m a = m b = m, take logarithms, divide by (2t+ 1) m, and then take m → ∞.By equation ( 15) and lemma 2, By lemma 4, h y increases to infinity as m → ∞, thus w < h y can be made arbitrarily large.Since µ (w) → µ 3 , this completes case (3).This completes the lower bound.
. An adsorbing and pulled uniform comb.A force f is pulling at the end of the backbone, desorbing the comb from the surface.Cutting the comb in its trivalent nodes gives a sequence of copolymeric 2-stars and a final branch where the comb is pulled.The trivalent nodes of the comb are labelled starting at the origin by 0, 1, . .., t, t + 1.

Upper bound:
The case y ≥ 1: The bound is constructed by viewing the comb as composed of a sequence of copolymer 2-stars, followed by a final pulled branch on the backbone as shown schematically in figure 7. The pulling force f is transmitted along the backbone to the origin, or along the backbone via an absorbed trivalent vertex or a tooth, or an adsorbed branch along the backbone, into the adsorbing plane.
We construct an upper bound by considering the 2-stars to be independent of each other, and bound each from above.
As before, assume that K (t) (m a , m b , a, b, y) is the partition function of the comb, where the backbone branches have lengths m b , the teeth have lengths m a , and (a, b, y) are the parameters as defined before.
Next, consider one of the 2-stars along the comb (as shown in figure 7).A schematic diagram of such a 2-star is shown in figure 8.The 2-star has arms of lengths equal to m a and m b .Each arm is partitioned into a part from the central trivalent node (with label denoted by i in figure 8 and preceded by the trivalent node i 0 = i − 1) to its first visit in the adsorbing surface, and then a remaining part.The remaining part intersects the adsorbing surface, and those visits are weighted as shown.
The trivalent node i 0 = i − 1 in figure 8 is the central node of the preceding 2-star, and it has height h 0 above the adsorbing plane.h 0 is bounded by 0 ≤ h 0 ≤ (t + 1) m b = H.The trivalent node i in figure 8 has height h above the adsorbing surface.Clearly, 0 ≤ h ≤ H.The pulling force on the comb is transmitted to i and then passed to the adsorbing plane at the visits a or b, or to the node i 0 .Denote the height of the i-th trivalent node by h (i) and the height of the first vertex of the i-the star by h 1) .The height of the entire comb is h (t+1) and the height of the final pulled walk is h (t+1) − h (t) .Then which is the height of the comb.In the partition function of the i-th 2-star the height is defined as 1) and this is conjugate to y.The arm of the 2-star which is also a branch along the backbone of the comb (the blue arm in figure 8) has an adsorbing part of length ⌊β 2 m b ⌋ + σ b and a pulled part of length ⌊β 1 m b ⌋.Here, Similarly, the tooth (the red arm in figure 8) in the 2-star has a pulled part of length ⌊α 1 m a ⌋ and an adsorbing part of length ⌊α 2 m a ⌋ + σ a , where α 1 + α 2 = 1 and σ a is a function of (α 1 , m a ) such that ⌊α m b , then h = 0, and thus α 1 < 1/m a .Similarly, if α 1 < 1/m a , then h = 0 and thus A similar approach to the above with α 1 and β 1 taking discrete values such that α 1 m a and β 1 m b are integers can be followed instead.
Proceed by excising the 2-star in figure 8 and considering it in isolation, independent from the rest of the comb.These 2-stars have a fixed end point in one arm at height h 0 above the adsorbing surface, and are pulled in the midpoint by a vertical force, while the arms adsorb with activities a and b.
Copolymer 2-stars terminally attached to the surface and pulled from their midpoint have been studied previously in reference [23].The 2-star in figure 8 is not terminally attached but rather the height of the initial vertex is fixed at h 0 where h 0 can range from 0 to (i − 1) m b ≤ (t + 1) m b = H in the comb.
Denote by S α2α1β1β2 ma,m b (h 0 ; a, b, y) the partition function of 2-stars as in figure 8 with 0 ≤ h 0 ≤ H and where y is conjugate to h − h 0 .If the central vertex of a 2-star is under tension due to the pulling force, then its contribution to the comb partition function is bounded above by Denote the partition function of a pulled and adsorbing self-avoiding walk of length m b starting in a vertex at height h 0 above the adsorbing surface by C m b (h 0 ; b, y).This corresponds to the 2star partition function with m a = 0.The partition function of the comb is bounded above by the product of the partition functions of the contributing 2-stars and the partition function of the final pulled walk.In the upper bound these contributing factors are treated as independent.Since y ≥ 1, it follows that The maximum over α 1 and β 1 is over all values of α 1 and β 1 between 0 and 1 and the maxima fix the values of {α 2 , β 2 , σ a , σ b }.We continue by putting m a = m b and by examining the factors in equation (22).
Claim: Given ǫ > 0, there exists an M ≥ 0 such that for all m ≥ M , Proof of the claim: Each 2-star in figure 8 is partitioned in 4 (possibly empty) subwalks.Ignore intersections between these subwalks.We now construct bounds on the contributions of the arms of the 2-star to the partition function.Assume first that both arms intersect the adsorbing surface.
In figure 8 the tooth of the 2-star is partitioned into two walks by the first vertex in the adsorbing surface.The second part is an adsorbing walk of length ⌊α 2 m a ⌋ + σ a , while the first part is a walk of length ⌊α 1 m a ⌋ from the trivalent node i to the adsorbing surface.
The other arm of the 2-star is similarly partitioned into an adsorbing walk from the trivalent node i 0 to the adsorbing surface, of length ⌊β 2 m b ⌋ + σ b , and a walk from the trivalent node i to the adsorbing surface, of length ⌊β 1 m b ⌋ (see figure 8).
Denote the number of self-avoiding walks of length n from a vertex at height h 0 to a vertex at height h by e n (h 0 , h).Note that the partition function introduced in section 2.1 for x 3 -pulled (not necessarily positive) walks C n (y) = h e n (0, h)y h = h e n (h 0 , h)y h−h0 , independent of h 0 .Also, denote the number of positive self-avoiding walks from the origin making v visits, including the origin, to the adsorbing surface by c + n (v).Denote their partition function by ).Then counting the walks in figure 8 in terms of e n (h 0 , h) and c + n (v) and noting that y ≥ 1 while h 0 ≥ 0, the following bound is obtained: Observe that the right hand side of equation ( 23) is independent of h 0 .For the three cases when at least one arm of the 2-star does not intersect the adsorbing surface (here, α 2 = 0 or β 2 = 0, or both), it can be verified that the same final upper bound in equation ( 23) applies.
In addition, [7] and for y ≥ 1 lim Thus, for an arbitrary and fixed ǫ > 0 there exists an M > 0 such that for all n > M , Using these in equation (23) shows that there are large, but finite values of m > M (a function of (h 0 , a, b)), such that This completes the proof of the claim.
Proceed by taking logarithms of equation (22).Divide by (2t+1) m, and take the limit superior as m → ∞ on the left hand side.Since H = O(m), this gives lim sup m,m (h 0 ; a, b, y) By equation ( 26), lim sup It remains to examine the last term above.We notice that it cannot exceed a linear combination of κ(b) and λ(y).Thus, if MAX = λ(y) Evaluating the maximum gives the upper bound (t κ(a) + (t + 1)κ(b))/(2t + 1).This completes the proof of the theorem.

End-ball A-ads
End-ball AB-ads which is a fully ballistic phase.For a > a c , κ(a) > log µ 3 and we need to compare λ(y) and κ(a).If λ(y) < κ(a), then the middle term is larger than both the first and the last term since λ(y) > log µ 3 .

B-ads AB-ads
Hence ζ(a, b, y) = 1 2t+1 λ(y) + t 2t+1 (κ(a) + log µ 3 ) and we have an end-ballistic/A-adsorbed phase.For λ(y) > κ(a), then the last term is larger than both the first term and the middle term since λ(y) > log µ 3 .Hence ζ(a, b, y) = t+1 2t+1 λ(y) + t 2t+1 log µ 3 and we have a fully ballistic phase.It follows that for b < b c and y < 1, there is a phase boundary from a free phase to an Aadsorbed phase at a = a c (this phase boundary is illustrated in figure 9 (top left)).If b < b c and y > 1, there is a phase boundary at λ(y) = κ(a) (or y = λ −1 (κ(a))) which divides a fully ballistic and an end-ballistic (and A-adsorbed) phase (the case a < a c lies in the fully ballistic phase).Along y = 1 and a > a c runs a phase boundary separating the A-adsorbed and end-ballistic phases.Thus the boundary y = 1 bounds four different phases (free, ballistic, A-adsorbed and end-ballistic); see figure 9 (top left).
The phase boundary between the ballistic and end-ballistic phases is first order, while the other phase boundaries are presumably continuous.Since λ(y) is asymptotic to log y [18] and κ(a) is asymptotic to log a + log µ 2 [18,31], the phase boundary between the ballistic and end-ballistic phases is asymptotic to log y = log a + log µ 2 .The phase boundary between the ballistic and B-adsorbed phases is first order, while the other phase boundaries are presumably continuous.The phase boundary between the ballistic and B-adsorbed phases is asymptotic to log y = log b + log µ 2 .The free energy is equal to ((t+1)λ(y)+t log µ 3 )/(2t+1) at both these points.Since the free energy is a convex function, the free energy is less than or equal to ((t + 1)λ(y) + t log µ 3 )/(2t + 1) at the mid-point of the chord joining these points, so that this point is on or below the phase boundary.

The case
The concavity of the boundary then follows from the mid-point theorem.

Ballistic
End-ball A-ads

Non-uniform combs
In this section we examine the limits that the teeth are short compared to the backbone, or the backbone is short compared to the teeth.

Short teeth and long backbone
As before, the number of teeth is fixed at t and we are considering combs with initial vertex at the origin, in the half-space x 3 ≥ 0, with m a edges in each tooth, m b edges in each segment of the backbone, v A A-visits, v B B-visits and having the terminal vertex at height h, and with partition function: With t and m a fixed, we have the following lemma as m b → ∞: Proof: Define a structure S to be 2m a edges in the x 1 -direction and m a edges starting at the centre vertex of this subwalk and extending in the x 3 -direction.This is a 3-star with m a edges in each arm, and each arm being a straight line.Construct a comb with t teeth, each tooth with m a edges, and a backbone of total length (t + 1) m b , as follows (see figure 11).Take a walk unfolded in the x 1 -direction of length m b − m a , concatenated with a structure S, then recursively concatenated with a walk unfolded in the x 1 -direction of length m b − 2m a , concatenated with an S, a total of t times, and ending with a walk unfolded in the x 1 -direction of length m b − m a .Consider the subset of these combs with the first s nodes of degree 3 in x 3 = 0 plane (that is, the first s x 1 -unfolded walks are loops) and the remaining t − s nodes of degree 3 located in the half-space This completes the proof.This can be extended to the situation where t is fixed and m b goes to infinity with m a = o(m b ).We give this result in the next theorem.This result is expected, since the backbone dominates the partition function in the limit; this is the walk limit in this model.

Long teeth and short backbone
Assume that t is fixed, m b = o(m a ) and that if the limit m a → ∞ is taken, then m b → ∞ as well.In this case the teeth are long while the backbone is short, and the partition function is again given by equation (29).An upper bound is found by considering the teeth to be self-avoiding walks independent of one another and of the backbone.Denoting the partition function of pulled adsorbing positive walks by C + n (b, y) (see equation ( 4)), and the partition function of adsorbing self-avoiding walks by C n (a), this shows that lim sup For a lower bound, equation ( 16) holds here and gives Since w is arbitrary, one may take w → ∞ on the right hand side.By lemma 2, κ (w) (a) → κ(a), and comparing this result to equation (30), the following theorem is obtained.This result is expected, since the adsorbing teeth dominate the partition function in the limit.Since the backbone has length o(m a ), the pulling force cannot pull any teeth from the adsorbing plane, and if the teeth are adsorbed, then so is the backbone.This is the star limit in this model.

The limit t approaches infinity
In this section the limit as t → ∞ examined.In this limit the comb becomes of infinite length, but with the lengths of teeth and of the segments between teeth, fixed at m a and m b respectively.
Denote by g(m a , m b , t) the number of combs from the origin, in the bulk lattice Z 3 , with t teeth of length m a and t + 1 backbone segments of length m b .Then a comb from the origin of backbone length equal to (s + t) m b with s + t − 1 teeth of length m a , can be cut, at the node along the backbone where the s-th tooth is attached, into two subcombs with s − 1 teeth and t − 1 teeth respectively, and a tooth which is a self-avoiding walk of length m a .This shows that g(m a , m b , s + t − 1) ≤ c ma g(m a , m b , s − 1) g(m a , m b , t − 1), (33) where c ma accounts for the conformations of the orphaned tooth and is the number of self-avoiding walks of length m a from the origin in the lattice.This shows that c ma g(m a , m b , t) satisfies a submultiplicative inequality, and by reference [11] the connective constant [6] of lattice combs is defined by Notice that ζ 0,m b = m b log µ 3 where µ 3 is the growth constant of the self-avoiding walk.

Grafted combs pulled at an endpoint
In this section we consider models of adsorbing combs in the half-lattice where, as above, the backbone of the comb is taken to infinity by letting t → ∞ (but with (m a , m b ) fixed).For a lower bound, equation ( 19) holds here for y ≥ 1, that is for w < h y and a, b ≥ 0: where K (t) (m a , m b , 0, 0, y) denotes the partition function for combs with the backbone and the teeth disjoint from the adsorbing surface (except at the origin).Take logarithms, divide by (t + 1) m b + t m a and let t → ∞: ≤ lim inf t→∞ 1 (t + 1) m b + t m a log K (t) (m a , m b , 0, 0, y) = Λ inf (m a , m b , 0, 0, y).
To find an upper bound, use a construction similar to figure 6. Replace the backbone with a positive self-avoiding walk of length (t + 1) m b from the origin, and pulled at its endpoint.The teeth are replaced by self-avoiding walks, each of length m a − u.This gives Take logarithms, divide by (t + 1) m b + t m a and let t → ∞: Λ sup (m a , m b , 1, 1, y) = lim sup The bounds in equations (36) and (38) are valid in the square and in the cubic lattices.The bounds above are independent of a and b, but may be generalised by noting that for 0 ≤ a, b ≤ 1 and y ≥ 1, K (t) (m a , m b , 0, 0, y) ≤ K (t) (m a , m b , a, b, y) ≤ K (t) (m a , m b , 1, 1, y). (39)

Adsorbed pulled combs
In this section consider a fully adsorbed comb grafted at its first vertex, and being pulled at its endpoint by a vertical force.For a lower bound, equation ( 16) holds here and gives Taking logarithms, dividing by (t + 1) m b + t m a and then taking t → ∞ gives for any choice of w, where C + n (a) is the partition function of adsorbing positive walks.These bounds are valid in the cubic lattice.

Limits in the cubic lattice
The limit t → ∞ is an infinite comb with finite segments of length m b along the backbone, and teeth each of finite length m a .If m a → ∞ with m b = o(m a ) and diverging, then the limiting object will be a star with an infinite number of arms (this will be the star limit ), and, if instead, m b → ∞ with m a = o(m b ) and diverging, then the limit should be a self-avoiding walk limit.
One may also calculate limits of intermediate combs by putting m b = ⌊α n⌋ and m a = ⌊(1 − α) n⌋ and then taking n → ∞.These limits are given in the following theorems.
Consider the star limit first.for all a, b ≥ 0 and y ≥ 0.

Discussion
We have investigated a self-avoiding walk model of the adsorption of comb copolymers at a surface.These are interesting because of their use as steric stabilizers of dispersions [39].The teeth of the comb are chemically different from the backbone of the comb and can interact differently with the surface.In addition, we have considered the effect of a force pulling the adsorbed comb off the surface.We have concentrated on the case where one end vertex of the backbone is fixed in the surface and where the force is applied at the other end of the backbone.
We have considered several cases: (i) a uniform comb where the lengths of the teeth and the backbone branches (or segments) are all equal and where the number of teeth (t) is fixed, (ii) a comb with t teeth where the teeth are short compared to the backbone branches, (iii) a comb with t teeth where the backbone branches are short compared to the teeth, and (iv) the case where the number of teeth goes to infinity.In each case we have determined the free energy rigorously and investigated the forms of the phase diagrams.In particular, for the uniform case (where the lengths of the teeth and the backbone branches are equal and where the number of teeth is fixed) we have established the form of the phase diagram by taking various slices through the three dimensional diagram.When the teeth do not adsorb, the form of the (two dimensional) diagram is similar to that of self-avoiding walks adsorbing in a surface and being desorbed by a force [18].When both the teeth and the backbone adsorb, there are strong similarities to the phase diagram for copolymeric stars [23].
Comparing the infinite number of teeth case (iv) to the other cases considered, our results establish that some limits can be interchanged.For case (ii), when the teeth are short compared to the backbone branches, then the walk limit is obtained whether one first lets the number of teeth (t) go to infinity and then lets the backbone length go to infinity (as in section 5.3, theorem 6) or if the limits are in the reverse order (see section 4, theorem 2).Similarly, for case (iii), when the the backbone branches are short compared to the teeth, no matter the order of the limits, the star limit is obtained (see theorems 4 and 3).For the uniform case (i), comparing theorems 5 (α = 1/2) and 1, no matter the order of the limits (t → ∞, length of the branches goes to infinity) the free energy is the same.The resulting comb free energy is either equal to that of the adsorbed case (y = 1) or that of the pulled/non-adsorbed case (a = b = 1).
Although we have concentrated on the case of the simple cubic lattice our results can be extended to the d-dimensional hypercubic lattice for all d ≥ 3, but our methods do not extend to the d = 2 case.The two dimensional case requires further work.In three dimensions, our methods could be extended to apply to other lattices such as the body centred and face centred cubic lattices, although we have not worked out the details.
The model can be extended in other ways, for instance to a brush copolymer where we have a backbone with t vertices of degree k and with k − 2 side chains (the analogue of teeth of a comb) at each of these t vertices.It would also be interesting to consider cases where the force is applied at vertices other than the terminal vertex of degree 1.

Figure 8 .
Figure 8.A block copolymeric 2-star in the comb.The trivalent node labelled i is the central node of the star, and i 0 = i − 1 is the central node of the preceding 2-star.The arm of the 2-star from i 0 to i has length m b and is in the backbone of the comb.The pulling force on the comb is transmitted along this arm from i to i 0 .The i-th tooth of the comb is shown on the right, and it is the other arm of the 2-star.It has length ma and adsorbs in the adsorbing plane with activity a. Vertices in the backbone adsorb with activity b.The lengths of parts of the arms are shown.Here, σa, σ b ∈ {0, 1}.

Figure 9 .
Figure 9. Phase diagrams of pulled and adsorbing uniform combs at constant a or constant b.The top left diagram is for fixed b < bc and the top right for fixed a < ac.The top right diagram is qualitatively similar to that of a pulled adsorbing self-avoiding walk.On the bottom left b > bc and the backbone is adsorbed or partially adsorbed in all but the ballistic phase.The bottom right phase diagram has five phases, similar to what was established in pulled adsorbing copolymeric stars in reference [23, see figure 10(b)].
a < a c : If a < a c then κ(a) = log µ 3 and ζ(a, b, y) = max t 2t+1 log µ 3 + t+1 2t+1 κ(b), 1 2t+1 λ(y) + t 2t+1 (κ(b) + log µ 3 ), t+1 2t+1 λ(y) + t 2t+1 log µ 3 .The phase diagram of this case is illustrated in figure 9 (top right).For y < 1, λ(y) = log µ 3 and κ(b) ≥ log µ 3 thus the first term is always greater than the other two.Thus for a < a c and y < 1, the free energy ζ(a, b, y) = t+1 2t+1 κ(b) + t 2t+1 log µ 3 which has a critical point at b = b c and there is a phase boundary from B-adsorbed to B-desorbed and no ballistic phases.For y > 1, λ(y) > log µ 3 and κ(b) ≥ log µ 3 .For b < b c , κ(b) = log µ 3 and the last term is always larger than the other two.Thus for a < a c , y > 1, b < b c , the free energy ζ(a, b, y) = t+1 2t+1 λ(y) + t 2t+1 log µ 3 which is a fully ballistic phase.For b > b c , κ(b) > log µ 3 and we need to compare λ(y) and κ(b).If λ(y) < κ(b), then the first term is greater than the other two since λ(y) > log µ 3 .Hence ζ(a, b, y) = t 2t+1 log µ 3 + t+1 2t+1 κ(b) which corresponds to a B-adsorbed/toothfree phase (not end ballistic).For λ(y) > κ(b), then the last term is greater than both the first term and the middle term since λ(y) > log µ 3 .Hence ζ(a, b, y) = t+1 2t+1 λ(y) + t 2t+1 log µ 3 and we have a fully ballistic phase.Thus for a < a c and y < 1, there is a phase boundary from a free phase to a B-adsorbed phase at b = b c .For y > 1, there is a phase boundary at λ(y) = κ(b) (y = λ −1 (κ(b))) which divides a ballistic phase for large y from the B-adsorbed phase at large b (the region b < b c and y > 1 is in the fully ballistic phase).There is no end-ballistic phase in this phase diagram.At the point b = b c and y = 1 there are three phase boundaries meeting, as shown in figure 9 (top right).
b > b c : The phase diagram is illustrated in figure 9 (bottom left).If b > b c then κ(b) > log µ 3 .In the case that a < a c , then, as discussed above, for y < 1 or y > 1 and λ(y) < κ(b), the free energy ζ(a, b, y) = t+1 2t+1 κ(b) + t 2t+1 log µ 3 which corresponds to a B-adsorbed/tooth-free phases is asymptotic to log y = log b + log µ 2 .Notice that these curves are parallel in the (log y)-(log b) plane.3.2.5.The (a, b)-plane for y > 1: In figure 10 we show a slice through the three dimensional phase diagram at fixed y > 1.At small values of a and b the system is ballistic.If b increases at fixed a < a c we enter a phase where the backbone is adsorbed for b > b * , where b * = κ −1 (λ(y)).(The inverse function exists since κ(b) is continuous and strictly monotone for b > b c [7].)If a increases at fixed b < b c the teeth adsorb when a > a * where a * = κ −1 (λ(y)).In this phase the last branch of the backbone is ballistic and the teeth are adsorbed.If a > a c and b > b * the backbone and the teeth are adsorbed and we have an AB-adsorbed phase.There is also a curved phase boundary between the ballistic phase and a phase in which the teeth are adsorbed, all except the last branch of the backbone are adsorbed, and the last branch is ballistic.This boundary is given by the solution of the equation λ(y) + log µ 3 = κ(a) + κ(b), and the boundary is concave down in the (log a, log b)plane.To see this, take two points on the phase boundary, say (log a 1 , log b 1 ) and (log a 2 , log b 2 ).

Figure 11 .Lemma 5 .b →∞ 1 (
Figure 11.Concatenating unfolded walks with the structure S recursively to form a pulled adsorbing comb.The branches along the backbone each have length m b and the straight teeth are each of length ma.Each of the three branches of the structure S has length ma.

inf m b →∞ 1 (sup m b →∞ 1 (
x 3 > 0. If s = 0 this gives the lower bound log K (t) (m a , m b , a, b, y) ≥ (m b − m a )ψ(b, y) + ((t − 1)(m b − 2m a )λ(y) + (m b − m a )λ(y) + o(m b ).If s = t then the lower bound log K (t) (m a , m b , a, b, y) ≥ (m b −m a )κ(b)+2tm a log b+(t−1)(m b −2m a )κ(b)+(m b −m a )ψ(b, y)+o(m b ) is obtained instead.Also, if 0 < s < t we have the lower bound log K (t) (m a , m b , a, b, y) ≥ (m b − m a )κ(b) + 2sm a log b + (s − 1)(m b − 2m a )κ(b) + (m b − 2m a )ψ(b, y) + (t − s − 1)(m b − 2m a )λ(y) + (m b − m a )λ(y) + o(m b ).Since s is arbitrary we can choose its value to optimize the bound.When κ(b) > λ(y) the bound is most effective when s = t and when λ(y) > κ(b) it is most effective when s = 0. Dividing by (t + 1) m b + t m a and letting m b → ∞ with t and m a fixed gives lim t+1) m b +t ma log K (t) (m a , m b , a, b, y) ≥ max[κ(b), λ(y)] = ψ(b, y).To obtain an upper bound we can regard the comb as being composed of a backbone with (t + 1) m b edges and t teeth, and treat these components as being independent.This gives the bound logK (t) (m a ,m b , a, b, y) ≤ (t + 1) m b ψ(b, y) + tm a κ(a) + o(m b ).Dividing by (t + 1) m b + t m a and letting m b → ∞ with t and m a fixed gives lim t+1) m b +t ma log K (t) (m a , m b , a, b, y) ≤ ψ(b, y).

Theorem 2 . 1 ( 1 ( 1 (
If t is fixed and m b goes to infinity with m a = o(m b ) the free energy is given by lim m b →∞,ma=o(m b ) t+1) m b +t ma log K (t) (m a , m b , a, b, y) = max[κ(b), λ(y)] = ψ(b, y).Proof: Since m a being fixed is a special case of m a = o(m b ) the result of the previous lemma gives the lower bound lim inf m b →∞,ma=o(m b ) t+1) m b +t ma log K (t) (m a , m b , a, b, y) ≥ ψ(b, y).We can obtain an upper bound by considering the case where the backbone and teeth behave independently.Again we have log K (t) (m a , m b , a, b, y) ≤ (t + 1) m b ψ(b, y) + tm a κ(a) + o(m b ).Dividing by (t + 1) m b + t m a and letting m b → ∞ with t fixed and m a = o(m b ) gives lim sup m b →∞,ma=o(m b ) t+1) m b +t ma log K (t) (m a , m b , a, b, y) ≤ ψ(b, y), which completes the proof.

ζ ma,m b = lim t→∞ 1 t 1 t
log g(m a , m b , t) = inf t≥0 log (c ma g(m a , m b , t)) .

1 m 1 m 1 ( 1 ( 1 m 1 m
b +ma log L [ †,1] m b (>w, b) + b +ma log C t+1) m b +t ma log K (t) (m a , m b , a, b, y) = Λ inf (m a , m b , a, b, y).An upper bound is obtained by replacing the loops and walks in figure5with adsorbing positive self-avoiding walks.This shows that, for a, b ≥ 0 and y ≥ 0, Λ sup (m a , m b , a, b, y) = lim sup t→∞ t+1) m b +t ma log K (t) (m a , m b , a, b, y) ≤ b +ma log C + m b (b) + b +ma log C + ma (a),