## Abstract

In this sequel paper, we solve two boundary-value problems for the asymptotic model equation derived in the first paper. One of the purposes is to compare the analytical solutions with the experimental results obtained by a number of other authors. With the help of a phase-plane analysis on this model, first we manage to solve a force-controlled problem. We find that it is necessary to divide the external stress into seven intervals. Analytical solutions in all intervals are obtained. An interesting finding is that the value of the radius–length ratio has a great influence on the solutions. In particular, it influences the number of non-trivial solutions. In this regard, the potential application of this result to the cold drawing industry is pointed out. Our results also yield the possible nucleation stress which turns out to be dependent on the radius–length ratio *a*. A simple approximate formula for the nucleation stress is given, which shows that it is a decreasing function of *a*. This seems to capture the size effect observed in experiments: the smaller the diameter is, the larger the nucleation stress is. We also obtain the analytic solutions for a displacement-controlled problem with the help of those of a force-controlled problem. The engineering stress–strain curve plotted from the solution of the preferred configurations seems to capture the key features of the curve measured in a few experiments.

## 1. Introduction

A number of authors (e.g. Shaw & Kyriakides 1995, 1997; Sun *et al*. 2000; Favier *et al*. 2001) have carried out experimental studies on uniaxial extensions of thin structures (e.g. wires, strips and bars) made of phase-transforming materials (e.g. shape memory alloys; SMAs). Some common key features of the measured loading engineering stress–strain curves were found, including that there was a local stress maximum (nucleation stress), a sharp drop and a stress plateau. Although some comparisons have been made with numerical solutions obtained by using existing packages (which are designed for convex energy functions; e.g. Shaw & Kyriakides 1998; Favier *et al*. 2001), comparisons with analytic solutions are probably non-existent. It could be because the analytic solutions are very difficult to obtain for non-convex energy functions. One exception is the paper by Ericksen (1975), which gave analytical solutions based on a one-dimensional stress model. However, that model is probably not sophisticated enough for the purpose of making comparisons with experimental results directly. Here, we shall solve two boundary-value problems (in experiments, the specimens are always of finite length; thus, one has to deal with boundary-value problems) based on the asymptotic model equation derived in the first paper, which takes into account the influences of the radial deformation, as well as the traction-free boundary conditions. We construct all the analytical solutions and extract from them important information on the deformation configurations, the nucleation stress and the phase boundary. Comparisons with experimental results are made, which show that the analytical solutions of our model equation can capture the key features of the experimental engineering stress–strain curves. In addition, the nucleation stress obtained by us exhibits the same size effect as observed in experiments (Sun *et al*. 2000): the smaller the diameter of the specimen (wire) is, the larger the nucleation stress. These qualitative agreements with experiments give supporting evidence of the asymptotic model equation derived by us as a good model equation for phase transitions in slender elastic cylinders.

More specifically, in this sequel paper, we shall consider phase transitions in a finite slender cylinder (e.g. without loss of generality, we take the length to be 1) composed of an incompressible elastic material with two free ends. The governing equation is the asymptotic model eqn (4.26) derived in Dai & Cai (2005, hereafter referred to as part I)(1.1)By ‘free ends’, we mean that(1.2)These boundary conditions have been used by many different authors (e.g. Ericksen 1975; Tong *et al*. 2001). We consider a phase-transforming material such that the stress –strain (*U*) curve has a peak–valley combination (see figure 1). This requires that the material parameters *D*_{1} and *D*_{2} satisfy the same constraint (6.1) in part I. The critical strain values shown in figure 1 can be expressed in terms of *D*_{1} and *D*_{2}(1.3)The peak stress value *γ*_{2}, valley stress value *γ*_{1} and the Maxwell stress value *γ*_{m} are the same as given in (6.2) of part I.

In this paper, we shall consider both a force-controlled problem with a given *γ* and a displacement-controlled problem such that(1.4)where the total elongation *Δ* (which is actually the engineering strain since we take the length of the cylinder to be 1) is given and without loss of generality is taken to be zero.

## 2. Phase plane analysis

We rewrite equation (1.1) as a first-order system(2.1)To deduce the qualitative information, we conduct a phase plane analysis with *γ* as the bifurcation parameter.

The critical points of this system is given by *y*=0 and(2.2)As the bifurcation parameter *γ* varies, there are seven cases.

*Case* (*a*) 0≤*γ*<*γ*_{1}

In this case, equation (2.2) has a unique positive root, say, denoted by *U*_{a}, and further *U*_{a}<*U*_{1} (see figure 1). The system (2.1) has only one critical point (*U*_{a}, 0), which is a saddle point.

*Case* (*b*) *γ*=*γ*_{1}

In this case, there are two critical points (*U*_{1}, 0) and (*U*_{5}, 0), the first one being a saddle point and the second one being a cusp point.

*Case* (*c*) *γ*_{1}<*γ*<*γ*_{m}

In this case, equation (2.2) has three positive roots *U*_{a}<*U*_{b}<*U*_{c}, which satisfy(2.3)Correspondingly, (*U*_{a}, 0) and (*U*_{c}, 0) are saddle points and (*U*_{b}, 0) is a centre point. In the phase plane, the centre point is surrounded by a homoclinic orbit connecting the right saddle point (*U*_{c}, 0) (this actually corresponds to the solitary-wave solution (6.7) in part I).

*Case* (*d*) *γ*=*γ*_{m}

There are three critical points (*U*_{2}, 0), (*U*_{4}, 0) and (*U*_{6}, 0), the first and the third being saddle points and the second being a centre point. Different from case (*c*), in this case, two saddle points are connected by two heteroclinic orbits (corresponds to the kink/antikink solution (6.8) in part I).

*Case* (*e*) *γ*_{m}<*γ*<*γ*_{2}

Similar to case (*c*), but *U*_{a}, *U*_{b}, *U*_{c} satisfy the following inequalities:(2.4)Different from case (*c*), now the centre point (*U*_{b}, 0) is surrounded by a homoclinic orbit connecting the left saddle point (*U*_{a}, 0) (this corresponds to the anti-solitary-wave solution (6.9) in part I).

*Case* (*f*) *γ*=*γ*_{2}

In this case, there are two critical points (*U*_{3}, 0) and (*U*_{7}, 0), the first one being a cusp point and the second one being a saddle point.

*Case* (*g*) *γ*>*γ*_{2}

In this case, equation (2.2) has a unique positive root *U*_{c}. Correspondingly, the system (2.1) has only a saddle point (*U*_{c}, 0).

The phase planes corresponding to these seven cases are shown in figure 2.

We may call the segment of the curve between 0 and *U*_{3} in figure 1 a ‘low-strain phase’ and the segment to the right of (*U*_{5}, *γ*_{1}) a ‘high-strain phase’. Then, the physical problem of the phase transition from the low-strain phase to the high-strain phase is equivalent to the mathematical problem how the saddle point (*U*_{a}, 0) in figure 2*a* transforms into the saddle point (*U*_{c}, 0) in figure 2*g*.

## 3. Solutions in a force-controlled problem

We now consider the solutions for a given external stress *γ* under the boundary conditions (1.2).

Every trajectory in the phase planes in figure 2 is a solution of the system (2.1). However, only the trajectories that satisfy (1.2) represent the physical solutions for the present problem.

It is easy to see that to satisfy (1.2), we require that a trajectory is in contact with the *U*-axis at *Z*=0 and 1. Based on this observation, one can obtain all solutions from these phase planes.

*Case* (*a*) 0≤*γ*<*γ*_{1}

In this case, from figure 2*a*, it can be seen that the only possible solution is *U*=*U*_{a} (*U*_{Z}≡0, so that (1.2) is satisfied).

*Case* (*b*) *γ*=*γ*_{1}

From figure 2*b*, it can be seen that there are two constant solutions only: *U*=*U*_{1} and *U*=*U*_{5}. The total potential energy values are *P*_{1}=−5.1123*μ*×10^{−4} and *P*_{2}=−3.6723*μ*×10^{−4}, respectively. Thus, *U*=*U*_{1} represents a preferred configuration and *U*=*U*_{5} is an unstable solution.

*Case* (*c*) *γ*_{1}<*γ*<*γ*_{m}

From figure 2*c*, it can be seen that there are three constant solutions: *U*=*U*_{a}, *U*=*U*_{b} and *U*=*U*_{c}.

There are also non-trivial solutions, which are represented by certain closed orbits in figure 2*c*, as they can contact the *U*-axis at least twice. To deduce the solution expression, we integrate (1.1) once to obtain(3.1)where *H*, the integration constant, is actually the Hamiltonian of the system (2.1).

Denote the minimum and maximum values of a closed orbit by *g*_{1} and *g*_{2}, respectively. Then, from (3.1), we have(3.2)It is easy to see from figure 2*c* that(3.3)where *g*_{1min} is the point at which the homoclinic orbit crosses the *U*-axis. The value of *g*_{1min} can be deduced from (3.1) and is given by(3.4)Substituting (3.2) into (3.1), we obtain(3.5)(3.6)where(3.7)It should be noted that for a closed orbit to represent a physical solution (satisfying (1.2)), we must have(3.8)It can be seen that *α*_{3} is positive, and further it can be shown that *α*_{1}<*g*_{1} and *α*_{2}<*g*_{2}. As a result, from (3.5) and (3.6), we obtain(3.9)(3.10)where *F*(*ϕ*, *κ*) is the incomplete elliptic integral of the first kind and(3.11)By using the Jacobian elliptic function, from (3.9), we can further obtain(3.12)where(3.13)We have used the boundary condition at *Z*=0, and by further using the boundary condition at *Z*=1 (cf. (1.2)) we can determine *g*_{1} (thus, *g*_{2} and *H*). Similar to (3.8), we need that(3.14)Combining (3.8) and (3.14), we see that there are four possibilities(3.15)(3.16)(3.17)(3.18)In cases (i) and (ii) (we call them ‘symmetrical cases’), the closed orbit can go around the critical point 1, 2, 3,… times; that is, integer times, say *k*=2*k*/2 times (*k*=1, 2,…).

In cases (iii) and (iv) (we call them ‘non-symmetrical cases’), the closed orbit can go around the critical point times; that is, (2*k*−1)/2 times.

Denote *n*=2*k*−1 or 2*k* (i.e. *n* is a natural number). Then, for all four cases, using (3.14) in (3.9), we have(3.19)where *K*(.) is the complete elliptic integral of the first kind. Alternatively, we may rewrite (3.19) as(3.20)Substituting (3.7) and (3.11) into (3.19), one obtains an equation in terms of *n*, *g*_{1} and *g*_{2}. Then, for a chosen natural number *n*, combing this equation with (3.2), one can determine *g*_{1} and *g*_{2} numerically.

It should be noted that for each *n*, there are two subcases: *U*=*g*_{1} at *Z*=0 or *U*=*g*_{2} at *Z*=0. In the following, we only concentrate on the subcase of *U*=*g*_{1} at *Z*=0, and other subcases can be considered similarly.

For a chosen *n*, the above discussion implies that within the length *Z*∈[0,1], the solution is of *n*/2-period long with one period equal to 4*gK*(*κ*). However, one cannot choose *n* to be arbitrarily large, because an upper bound for *n* can be established. As the complete first integral of the first kind *K*(*κ*) is an increasing function of *κ* for 0≤*κ*<1 and has the minimum value *K*(0)=*π*/2, from (3.19), we have(3.21)In deriving (3.21), use has been made of (3.11) and *g*_{1}<*U*_{b} and *g*_{2}>*U*_{b}.

This may be a very important result, because it implies that if the radius–length ratio (here, *a* stands for this ratio as we have taken the length to be 1) is relatively large (not too large as our model equation is derived for a slender cylinder), then the only possible choice for *n* is 1. In this case, there is one non-trivial solution with a one-half period with one thicker part and one thinner part (cf. figure 3 for the case of *n*=1). This may have an important practical application in the cold drawing industry. To draw a thicker cylinder into a thinner cylinder, of course, one wants the lateral surface of the thinner cylinder to be as smooth as possible, and thus one should avoid a periodic profile for the lateral surface of a thinner cylinder. This is equivalent to avoid the possibility that *n* can be 2 or 3 or 4… in the present problem. Thus, if one draws the thicker cylinder at a position with a proper radius–length ratio such that(3.22)then *n*=1, which then eliminates the possibility of a periodic profile of the thinner cylinder.

Indeed, for the case of *D*_{1}=−18, *D*_{2}=100, *γ*=0.0165 and *a*=0.07, we find that(3.23)Thus, in this case, there are two non-trivial solutions with one-half and one periods in the interval *Z*∈[0,1]. On the other hand, for a different radius *a*=0.1, we find that(3.24)Thus, in this case, there is only one non-trivial solution with one-half period in the interval *Z*∈[0,1].

The solution profiles for the case of *γ*=0.0165 and *a*=0.07 are shown in figure 3. The corresponding values of *g*_{1}, *g*_{2} and *H* are given belowIt should be noted that as the external stress varies, the number of non-trivial solutions can also vary. Indeed, for the above-chosen constants, there is no non-trivial solution if *γ*<0.015 936 95, there is one non-trivial solution if 0.015 936 95≤*γ*<0.016 086 75 and there are two non-trivial solutions if 0.016 086 75≤*γ*<*γ*_{m}.

To conclude this case of *γ*_{1}<*γ*<*γ*_{m}, we consider the total potential energy values of all the possible solutions. The potential energy is given by (cf. (5.4) of part I)(3.25)For the constant solutions *U*=*U*_{a}, *U*=*U*_{b} and *U*=*U*_{c}, we find that the corresponding potential energy values are(3.26)where the values of *D*_{1}, *D*_{2} and *γ* are the same as those above (3.23).

For the case of *a*=0.07, there are also two non-trivial solutions (*n*=1 and 2). The corresponding potential energy values are(3.27)Thus, among these five solutions, *U*=*U*_{a} represents a preferred configuration, *U*=*U*_{b} is an unstable solution and the other three solutions may be metastable solutions.

*Case* (*d*) *γ*=*γ*_{m}

From figure 2*d*, it can be seen that there are three constant solutions: *U*=*U*_{2}, *U*=*U*_{4} and *U*=*U*_{6}.

There are also non-trivial solutions, which are represented by certain closed orbits in figure 2*d*. Let us still use *g*_{1} and *g*_{2} to denote the minimum and maximum values of *U* in one period. Then, it is easy to see from figure 2*d* that(3.28)In this case, the values of *g*_{1} and *g*_{2} are still determined from (3.2) and (3.19). Now, the number *n* of non-trivial solutions satisfies(3.29)The solution expression is still given by (3.12). For *D*_{1}=−18, *D*_{2}=100, *a*=0.07, *γ*=*γ*_{m}=0.0168, we find that(3.30)Thus, there are two non-trivial solutions with one-half and one periods in the interval *Z*∈[0,1]. The corresponding values of *g*_{1}, *g*_{2} and *H* are given below:The cylinder profiles corresponding to these two non-trivial solutions are shown in figure 4.

Since we have obtained the mathematical solutions, we know all the information about the ‘phase boundary’ (which is a narrow smooth region); its structure, width and position and so on. In particular, from (3.12), we can see that the width of a phase boundary is inversely proportional to *a*, which is in agreement with experimental observations (Q. P. Sun 2004, private communication).

The total potential energy values for three constant solutions *U*=*U*_{2}, *U*=*U*_{4} and *U*=*U*_{6} are(3.31)The total corresponding potential energy values for two non-trivial solutions can be calculated from (3.25) and are(3.32)Thus, among these five solutions, *U*=*U*_{2} and *U*=*U*_{6} represent preferred configurations, *U*=*U*_{4} is an unstable solution and the two non-trivial solutions may be metastable solutions. We also note that the solution with one period (*n*=2) has a larger energy value comparing with the solution with one-half period (*n*=1).

We point out that, in this case, for a given material (i.e. *D*_{1} and *D*_{2} are given) the upper bound for the number of non-trivial solutions is uniquely determined by *a* (see (3.29)). The smaller *a* is, the larger *n* is. For *D*_{1}=−18 and *D*_{2}=100, we findFor example, if *a*=0.03, *n*<6.003 and, correspondingly, there are six non-trivial solutions.

If *a*>0.1801, this inequality implies no non-trivial solution. However, for a relatively large *a*, the condition under which the model equation is derived may not be valid (cf. the paragraph above (4.16) of part I).

*Case* (*e*) *γ*_{m}<*γ*<*γ*_{2}

From figure 2*e*, it is easy to see that there are three constant solutions: *U*=*U*_{a}, *U*=*U*_{b} and *U*=*U*_{c}.

There are also non-trivial solutions, which are represented by certain closed orbits in figure 2*e*. We still use *g*_{1} and *g*_{2} to represent the minimum and maximum values of a solution. Then, we have(3.33)where *g*_{2max} is the point at which the homoclinic orbit crosses the *U*-axis, and it is given by(3.34)In this case, once again, *g*_{1} and *g*_{2} can be determined from (3.2) and (3.19) and *n* still has the upper bound shown in (3.21). In addition, the solution expression for these *n* non-trivial solutions is given by (3.12).

For *D*_{1}=−18, *D*_{2}=100, *a*=0.07, it can be shown that there is no non-trivial solution for *γ*>0.017 663 05, there is one non-trivial solution for 0.017 513 25<*γ*≤0.017 663 05 and there are two non-trivial solutions for *γ*≤0.017 513 25. For *γ*=0.0175, we find that(3.35)Thus, in this case, there are two non-trivial solutions corresponding to *n*=1 and *n*=2.

The solution profiles are shown in figure 5. The corresponding values of *g*_{1}, *g*_{2} and *H* are given below:The total potential energy values for three constant solutions *U*=*U*_{a}, *U*=*U*_{b} and *U*=*U*_{c} are(3.36)For the two non-trivial solutions, the corresponding total potential energy values are(3.37)Thus, *U*=*U*_{c} represents a preferred configuration, *U*=*U*_{b} is an unstable solution and the other three may be metastable solutions.

*Case* (*f*) *γ*=*γ*_{2}

There are only two constant solutions: *U*=*U*_{3} and *U*=*U*_{7}. The corresponding total potential energy values are *P*_{1}=−6.8077*μ*×10^{−4} and *P*_{2}=−8.2477*μ*×10^{−4}, respectively. Thus, *U*=*U*_{7} represents a preferred configuration and *U*=*U*_{3} is an unstable solution.

*Case* (*g*) *γ*>*γ*_{2}

There is only one constant solution: *U*=*U*_{c}.

From the above results, it can be seen that there are multiple solutions for *γ* in certain intervals. To determine which solution is the most preferred one, we have plotted the total potential energies for all possible solutions in figure 6.

The energy curves corresponding to the constant solutions *U*=*U*_{a}, *U*=*U*_{b} and *U*=*U*_{c} and the two non-trivial solutions with one-half and one periods are labelled by *a*, *b*, *c*, *n*=1 and *n*=2, respectively. The enlarged part around R is shown at the top of figure 6.

From figure 6, it can be seen that the preferred configuration is given by *U*=*U*_{a} for 0≤*γ*≤*γ*_{m} and is given by *U*=*U*_{c} for *γ*>*γ*_{m}. However, in a loading process, after *γ*>*γ*_{m}, the change from *U*=*U*_{a} to *U*=*U*_{c} can only be realized for a relatively large perturbation because the difference between the energies of *U*=*U*_{a} and *U*=*U*_{c} is relatively large. On the other hand, the change from *U*=*U*_{a} to a one-half-period solution or to the one-period solution can take place with a very small perturbation, which may well be what happens in reality.

In the case of a symmetrical condition at *Z*=0 and 1 (cf. the discussion below (3.18), in this case, *n* can only be an even number), during the loading process, after *γ*>*γ*_{m}, the constant solution *U*=*U*_{a} may change to the one-period solution due to a small perturbation. Then along this solution, the stress can at most reach *γ*_{B}, the stress value at point B. A further increase in the stress results in a jump from one-period solution to the solution *U*=*U*_{c}. Thus, *γ*_{B} may be regarded as the nucleation stress. We point out that the value of *γ*_{B} depends on the radius–length ratio *a*. Actually, *γ*_{B} can be determined from (3.19). For a given *a* (with *n*=2), the largest possible value of *γ* which (3.19) can be satisfied is just *γ*_{B}. The *γ*_{B} curve as *a* varies is shown in figure 7.

Similarly, in the case of a non-symmetrical condition at *Z*=0 and 1 (in this case, *n* can only be an odd number), *γ*_{A}, the stress value at point A, may be regarded as the nucleation stress. Its curve as *a* varies is also shown in figure 7.

Actually, it is possible to deduce an approximate formula for the nucleation stress. At the onset of the nucleation, the non-trivial solution should be very close to the constant solution, and this implies that *g*_{1}≈*U*_{b}≈*g*_{2}. Then, from (3.11) and (3.19), we have(3.38)Once *U*_{b} is known, *γ*_{B} (*n*=2) and *γ*_{A} (*n*=1) can be determined from (2.2) by letting *U*=*U*_{b}.

Sun *et al*. (2000) conducted an experiment on NiTi polycrystal SMA wires under tension. They used wires with seven different diameters and found that the nucleation stress is strongly size dependent: the smaller the diameter is, the larger the nucleation stress. From figure 7, it can be seen that the nucleation stress calculated from our asymptotic model equation is indeed a decreasing function of *a*. It seems that our results provide a mathematical description of this size effect.

In their view (Sun *et al*. 2000), whether the non-local gradient theory can describe this ‘size effect’ is an open issue and is a research topic of great interest for both mechanics of materials and intelligent and microsystems such as MEMS communities. What we have developed is based on a local theory but takes into account the radial deformation and traction-free boundary conditions. It is interesting that such a local theory can capture this size effect.

## 4. Solutions in a displacement-controlled problem

We now consider the case that the total elongation *Δ* is given (cf. (1.4)). The governing equation is still (1.1) but now *γ* is an unknown parameter.

Because, in §3, we have solved the problem of a given *γ*, if for a given *Δ*, we can find the corresponding *γ*, then we can obtain the solutions for the present problem. The difficulty is that for a given *Δ*, there may be multiple values for *γ*. In the following, we take the value *a*=0.07.

We first plot the *γ*–*Δ* curves for all the solutions obtained in §3 (see figure 8). In this figure, the dotted line segment in the interval [0,*U*_{3}] comes from the first constant solution *U*=*U*_{a} in cases (*a*)–(*g*) of §3, dotted line segment in the interval [*U*_{3},*U*_{5}] comes from the second constant solution *U*=*U*_{b} in cases (*e*), (*d*) and (*c*) of §3 (corresponding to (*U*_{3}, *U*_{4}), *U*_{4}, (*U*_{4}, *U*_{5}), respectively). The dotted line segment in the interval *U*≥*U*_{5} comes from the last constant solution in cases (*b*)–(*g*) of §3 (corresponding to *U*_{5}, (*U*_{5}, *U*_{6}), *U*_{6}, (*U*_{6}, *U*_{7}), *U*_{7}, (*U*_{7}, ∞), respectively). Thick lines marked by *n*=1 and *n*=2 come from the one-half-period and one-period solutions, respectively.

It should be noted that, for the one-half-period solution, there are more complicated structures at two ends of the curve. We mark these two ends by L and R. The local structures near the two ends are shown at the top of figure 8.

From figure 8, it can be seen that there are eight critical points for the solution of a given displacement *Δ*. We denote these points by: *Δ*_{1}, the minimum value of *Δ* for the one-half-period solution; *Δ*_{2}, the displacement value for the one-half-period solution at point *a*; *Δ*_{3}, the displacement value for the one-half-period solution at point *b*; *Δ*_{4}, the minimum value of *Δ* for the one-period solution; *Δ*_{5}, the maximum value of *Δ* for the one-period solution; *Δ*_{6}, the displacement value for the one-half-period solution at point *c*; *Δ*_{7}, the displacement value for the one-half-period solution at point *d*; *Δ*_{8}, the maximum value of *Δ* for the one-half-period solution. It is interesting to note that for both solutions of one-half and one periods, the curves start and end at points which are very close to (numerically not distinguishable) the curve of the constant solution *U*=*U*_{b}.

For *D*_{1}=−18, *D*_{2}=100 and *a*=0.07, the values of the displacement and stress at these critical points areIt is easy to see that there are multiple solutions in the case of *Δ*_{1}≤*Δ*≤*Δ*_{8}. Now, we consider all the possible solutions for a given *Δ*.

*Case* (*a*) *Δ*<*Δ*_{1}

In this case, from figure 8, it can be seen that there is only one constant solution *U*=*Δ* and therefore only one possible stress(4.1)

*Case* (*b*) *Δ*=*Δ*_{1}

In this case, there is a constant solution *U*=*Δ*_{1} and the stress is still determined by (4.1).

Further, there is a one-half-period solution with the stress value . This belongs to case (*e*) (*γ*_{m}<*γ*<*γ*_{2}) of §3 and the corresponding solution expression is the same as that in that case.

*Case* (*c*) *Δ*_{1}<*Δ*<*Δ*_{2}

In this case, there is a constant solution *U*=*Δ*. Furthermore, it can be seen from figure 8 that there are two one-half-period solutions with two different stress values. There are two ways to get the solution expressions. First, for a given displacement *Δ*_{0} (*Δ*_{1}<*Δ*_{0}<*Δ*_{2}), draw the vertical line *Δ*=*Δ*_{0} in figure 8. Then, find the stress values at the intersection points with *n*=1 curve. Once the two stress values are known, the solution expression can be obtained as in case (*e*) of §3.

Alternatively, by using (1.4) and (3.12), we have(4.2)where *Π*(.) is the complete elliptic integral of the third kind. From (3.2) we can obtain(4.3)As a result, *H*, *α*_{1}, *α*_{2}, *g*, *A*_{1}, *A*_{2}, *κ* in (3.1), (3.7), (3.11) and (3.13) can all be expressed in terms of *g*_{1} and *g*_{2}. Then, for a given *Δ*_{0}, (3.19) and (4.2) (with *n*=1) provide two equations for (*g*_{1}, *g*_{2}). Numerically solving these two equations should yield two groups of values for (*g*_{1}, *g*_{2}). Equation (3.12) then gives the solution expressions for two different values of (*g*_{1}, *g*_{2}).

For *Δ*_{0}=0.0449, we find that(4.4)The corresponding solution profiles are shown in figure 9.

*Case* (*d*) *Δ*_{2}≤*Δ*<*Δ*_{3}

In this case, there is a constant solution *U*=*Δ*. In addition, there are three one-half-period solutions. The solution expressions can be determined in the same way given in case (*c*). The solution profiles for *Δ*=0.045 are shown in figure 10.

*Case* (*e*) *Δ*=*Δ*_{3}

In this case, there is a constant solution *U*=*Δ*_{3}. There is a one-half-period solution with the stress value . The solution expression is the same as that given in case (*e*) of §3. Further, there is another one-half-period solution whose expression can be determined in the same way given in case (*c*).

*Case* (*f*) *Δ*_{3}<*Δ*<*Δ*_{4}

In this case, there are two solutions: one constant solution and one one-half-period solution.

*Case* (*g*) *Δ*_{4}≤*Δ*≤*Δ*_{5}

In this case, there are three solutions: one constant solution, one one-half-period solution and one one-period solution.

For the one-period solution, *n*=2, and we have(4.5)Coupling the above equation with (3.19), (*g*_{1}, *g*_{2}) can be determined. As a result, from (3.12), we have the solution. The solution profiles for *Δ*=0.06 are shown in figure 11.

*Case* (*h*) *Δ*_{5}<*Δ*<*Δ*_{6}

In this case, there are two solutions: one constant solution and one one-half-period solution.

*Case* (*i*) *Δ*=*Δ*_{6}

In this case, there are three solutions: one constant solution and two one-half-period solutions.

*Case* (*j*) *Δ*_{6}<*Δ*≤*Δ*_{7}

In this case, there are four solutions: one constant solution and three one-half-period solutions.

*Case* (*k*) *Δ*_{7}<*Δ*<*Δ*_{8}

In this case, there are three solutions: one constant solution and two one-half-period solutions.

*Case* (*l*) *Δ*=*Δ*_{8}

In this case, there are two solutions: one constant solution and one one-half-period solution.

*Case* (*m*) *Δ*>*Δ*_{8}

In this case, there is only one constant solution.

For the displacement in several intervals, there are multiple solutions. To find the most preferred solution, we have plotted the differences of the total strain energies between the non-trivial solutions and the constant solution in figure 12. The enlarged parts around R and L are shown at the top of the figure.

We take the solution with the smallest energy value as the preferred one. Then, from figure 12, we can see that for *Δ*<*Δ*_{1} and *Δ*≥*Δ*_{9}(≈0.076 4623), the constant solution is the preferred one and for *Δ*_{1}≤*Δ*<*Δ*_{9}, the one-half-period solution of the branch with the least stress is the preferred one. The structure response curve (*γ*–*Δ* curve) corresponding to the preferred solution is shown in figure 13. We point out that this curve is not necessarily the actual curve measured in experiments, as, in reality, metastable solutions can also appear. Nevertheless, this curve already captures the key features of the engineering strain–stress curves measured in a few experiments (Shaw & Kyriakides 1995, 1997; Sun *et al*. 2000; Favier *et al*. 2001; Li & Sun 2002). The nucleation stress appears at a stress peak; following it, there is a sharp stress drop and afterwards there is stress plateau (with stress value equal to the Maxwell stress). This gives good evidences of the validity of the asymptotic model equation derived by us in a rational manner as well as using a non-convex strain energy function to model a phase-transforming material. It is worthwhile to note that the curve obtained by us contains another stress drop near the end of the stress plateau and in the experiments (Shaw & Kyriakides 1997) such a small stress drop was indeed observed.

## 5. Conclusions

Based on the asymptotic model equation of a slender elastic cylinder derived in part I, by considering the free-end boundary condition and using the method of phase-plane analysis, the bifurcation properties of its solutions are revealed. Then, the analytic solutions for a force-controlled boundary-value problem are obtained. These solutions show that the radius–length ratio *a* plays a significant role in the nucleation stress: the smaller *a* is, the larger the nucleation stress. A simple approximate formula for the nucleation stress is also given. It is found that, for a force-controlled boundary-value problem, the number of non-trivial solutions depends on the radius–length ratio *a* as well. Based on the solutions for the force-controlled boundary-value problem, the analytic solutions for a displacement-controlled boundary-value problem are obtained. The engineering stress–strain curve for the preferred solutions is also plotted. By comparing the analytical solutions and the experimental results, it can be concluded that the asymptotic model equation derived by us for non-convex strain energy functions, which takes into account the radial deformation and traction-free boundary conditions, can capture most of the key features of phase transitions in slender elastic cylinders.

This paper concentrates on the solutions and their behaviour, and we shall leave the studies on the stabilities of these solutions for a future investigation. However, it is worthwhile to make some comments on the energy convexity. The strain energy function used by us is non-convex in the case of a uniaxial tension. Owing to the influences of the radial deformation as well as the traction-free boundary conditions, the strain gradient also makes a contribution to the energy (see (5.4) of part I). The total potential energy in a displacement-controlled problem is a function of the given end displacement (or the engineering strain *Δ*); that is, *P*=*P*(*Δ*). It is interesting to point out that, through numerical computations, we find that this energy, for the non-trivial solution (e.g. *n*=1), is actually a convex function of the engineering strain (up to the order of the numerical error).

## Acknowledgments

The work described in this paper is fully supported by a grant from the Research Grants Council of the HKSAR, China (project no. Cityu 1225/03E). The authors would like to thank professors Y. B. Fu and Q. P. Sun for valuable discussions.

## Footnotes

- Received September 24, 2004.
- Accepted August 1, 2005.

- © 2005 The Royal Society