Singularity formation and blowup of complex-valued solutions of the modified KdV equation

The dynamics of the poles of the two--soliton solutions of the modified Korteweg--de Vries equation $$ u_t + 6u^2u_x + u_{xxx} = 0 $$ are determined. A consequence of this study is the existence of classes of smooth, complex--valued solutions of this equation, defined for $-\infty<x<\infty$, exponentially decreasing to zero as $|x| \to \infty$, that blow up in finite time.


Introduction
Studied here is the modified Korteweg-de Vries equation u t + 6u 2 u x + u xxx = 0, (1.1) which has been derived as a rudimentary model for wave propagation in a number of different physical contexts. The present paper is a sequel to the recent work [9] wherein the dynamics of the complex singularities of the two-soliton solution of the Korteweg de Vries equation, were examined in detail.
The study of the pole dynamics of solutions of the Korteweg-de Vries equation and its near relatives began with some remarks of Kruskal [14] in the early 1970's. More comprehensive work was carried out later, see [18], [11] and [12]. One goal in our preceding paper [9] was to understand in more detail the propagation of solitons in a neighborhood of the interaction time.
Another motivation was an idea to be explained presently concerning singularity formation in nonlinear, dispersive wave equations. More particularly, we are interested in both the generalized Korteweg-de Vries equation and coupled systems of Korteweg-de Vries type, viz.
where P and Q are, say, homogeneous polynomials. Concerning singularity formation, it is an open question whether or not smooth, rapidly decaying, real-valued solutions of (1.3) develop singularities in finite time in the supercritical case p ≥ 5. In the critical case p = 4, blowup in finite time has been established by Martel and Merle [17], [16] whilst for subcritical values p = 1, 2, 3, there is no singularity formation for data that lies at least in the Sobolev space H 1 (R). (However, solutions corresponding to infinitely smooth initial values lying only in L 2 (R) can develop singularities; see [7].) Numerical simulations reported in [4] of solutions of (1.3) initiated with an amplitude-modified solitary wave indicate blowup in finite time. Such initial data has an analytic extension to a strip symmetric about the real axis in the complex plane. The results of Bona, Grujic and Kalisch [13], [5] and [6] indicate that blowup at time t has to be accompanied by the width of the strip of analyticity shrinking to zero at the same time. This points to the prospect of a pair of complex conjugate singular points coliding at a spatial point on the real axis, thereby producing non-smooth behavior of the real-valued solution. It was shown in [9] that, in certain cases, curves of singularities do merge together. This happens at the moment of interaction of the two solitons when the amplitudes are related in a particular way. This result provides some indication that the blowup seen in the numerical simulations might occur because of the coalescence of curves of complex singularities. Such ruminations seem to justify interest in the pole dynamics in the context of (1.3), even in the case where the initial data is real-valued.
If one considers instead complex-valued solutions, it was shown in [8] (and see also [10]) that, in the case of spatially periodic boundary conditions, equation (1.3) has solutions which blow up in finite time for all integers p ≥ 1. Explicit examples of smooth, complex-valued solutions defined for x, t ∈ R of the Korteweg-de Vries equation (1.2) that blow up in finite time have been given in [2,9,15,19]. One outcome of the present paper is an explicit example of a blowing-up soution of the modified Korteweg-de Vries equation (1.1).
For the system (1.4) where P and Q are homogeneous quadratic polynomials, conditions on the coefficients are known that imply global wellposedness for real-valued initial data (u 0 , v 0 ) (see [3]). And, the pole dynamics investigated in [9] for the Korteweg-de Vries equation (1. The latter result is obtained by a careful study of the pole dynamics of the explicit two-soliton solution U = U(z, t) = U(x + iy, t) of the KdV-equation in the complex z-plane. It transpires that as a function of time, most of the singularities of this exact solution, which are all poles, move vertically in the y-direction in the complex plane as well as propagating horizontally in the x-direction. As a consequence, by choosing y 0 appropriately, one can arrange that the function u(x, t) = U(x + iy 0 , t) is a complex-valued solution of the KdV-equation that, at t = 0, is infinitely smooth and decays to zero exponentially rapidly as x → ±∞, but which blows up for a positive value t > 0. If we write u(x, t) = v(x, t) + iw(x, t), then the pair (v, w) is a solution of (1.4) with the choice (1.5) that starts at (v 0 , w 0 ) = (v(·, 0), w(·, 0)) smooth and rapidly decaying, but which forms a singularity in finite time. It is worth noting that by an appropriate choice of the particular two-soliton solution, the initial data (v 0 , w 0 ) can be taken to be arbitrarily small in any of the usual function spaces used in the analysis of such equations.
The motivation for the current paper was to see if the phenomenon just described, i.e. curves of singularities merging together, can occur for the modified KdV-equation, and if this behavior might provide additional insight into the possible ways a singularity is produced in nonlinear, dispersve wave equaions. Indeed, it turns out that the same phenomenon does occur. In certain cases, curves of singularities do converge together at one point. On the other hand, the behavior of these curves is not so different from what occurs in the context of the KdV-equation.

The two-soliton solutions
Preliminary analysis of the two-soliton solutions of the mKdV-equation are set forth here in preparation for the investigation of their pole dynamics. We begin with a standard transformation enabling one to express solutions of (2.1) in terms of solutions of another equation. Let u = v x where u is a solution of (2.1). Then v satisfies the equation Assume v and its derivatives vanish at infinity and search for solutions of the latter equation of the form v = 2 arctan(g). A calculation shows that v This yields a solution to (2.1) having the form It is important to note that equation ( (2.5) The well-known soliton solution of (2.1) has a hyperbolic secant profile. In detail, for any amplitude value k > 0, it is straightforward to check that is a solution of (2.3). The corresponding solution u of (2.1) is the soliton solution with speed k 2 and is given explicitly as Note that the choice of sign in the exponential function g in (2.6) produces the negative soliton. Replacing g by either −g or 1/g will produce the positive soliton.
The two-soliton solutions are a little more complicated. The formulation presented here is based on that appearing in [1]. As just noted, there are both positive and negative soliton solutions of (2.1). Consequently, there are two types of two-soliton solutions, namely interacting solitons of the same or opposite signs.
Suppose first that 0 < k 1 < k 2 . Define the functions f j by Of course, this definition omits two arbitary spatial translations; these will be added later. Define two auxilliary functions, g + and g − by Proposition 2.1. The functions g + and g − defined in (2.10)-(2.11) are solutions to (2.3).
Proof. It suffices to provide the proof for g − . Indeed, if one replaces f 1 by −f 1 , then g − is transformed into g + , and all the calculations below remain valid in this case. Temporarily, set g = g − . Notice that f jx = −k j f j and f jt = k 3 j f j for j = 1, 2. Thus, the quantities g t , g x , g xx and g xxx may be expressed in terms of f j and k j . First, differentiate with respect to time and come to the expression The above proposition implies that are solutions to (2.1). A further calculations reveals that and where the new combinations have been introduced. Note that the functions with a superscript "+" are obtained from the functions with a superscript "−" simpy by replacing f 1 by −f 1 . If every occurence of k 1 is replaced by −k 1 in formula (2.14) for u − , then f 1 is replaced by 1/f 1 and γ is replaced by 1/γ. Simplifying the resulting expression gives exactly the formula (2.13) for u + . In other words, the formulas for u + and u − can be obtained from each other by replacing every occurence of k 1 by −k 1 . The function u + is a two-soliton solution of (2.1) having two positive interacting solitons whist u − is a two-soliton solution with two interacting solitons of opposite sign, the faster one being the positive one. One can see this graphically using MAPLE or Mathematica, for example. Analytically, there is an explicit relationship between the fomulas for u ± given by (2.13) and (2.14) and the single soliton solutions of speeds k 2 1 and k 2 2 . As in (2.7), let u j be the centered, positive, soliton solution of speed k 2 j , j = 1, 2. If the formulas for (2.13) and (2.14) are both divided by (1 + f 2 1 )(1 + f 2 2 ), there obtains , .
Formulas (2.20) and (2.21) show in particular that u + (x, t) > 0 for all x ∈ R and that u − (x, t) > 0 precisely for those x ∈ R and t ∈ R for which u 2 (x, t) > u 1 (x, t).
In Section 4, the asymptotic behavior of the singularities of u ± are examined for large positive and negative time. We will see that they separate into two groups, corresponding to the two single solitons. More remarks on the shape of the two-soliton solutions during their interaction are to be found in Section 8.
Before ending this section, we return to the issue of spatial shifts. For 0 < k 1 < k 2 and x 1 , x 2 ∈ R, let Define the interaction time t 0 and the interaction center x 0 of g ± (x, t) to be respectively.
Proposition 2.2. Let t 0 and x 0 be the interaction time and interaction center for g + (x, t) and g − (x, t). Then, for any t ∈ R and x ∈ C, we have where g ± are defined in (2.10)-(2.11). Moreover, the functions u ± (·, t 0 ) = 2(arctan g ± (·, t 0 )) x are symmetric about the point x 0 on both R and C.
Proof. It suffices to find ( Equivalently, (2.24) will be satisfied if Since (x 0 , t 0 ) given by (2.22)-(2.23) is the solution for this system, the first assertion is proved. To see the symmetry of the functions u ± (·, t 0 ) about x 0 , it is only necessary to deduce from (2.24) that u ± (·, 0) = 2(arctan(g ± (·, 0)) x is an even function. This is easily verified since g ± (−x, 0) = −g ± (x, 0) for all x ∈ C.
Since (mKdV) is invariant under time-and space-translation, it is concluded from Proposition 2.2 that the functions u ± (x, t) = 2(arctan( g ± (x, t)) x are also solutions to (2.1). The point is that, for a general two-soliton solution, the time and place of the interaction are given by t 0 and x 0 in (2.22) and (2.23), respectively. In particular, the solutions u ± are already normalized so that the interaction time is t = 0, and the center of the interaction is at x = 0. It is interesting to note that the values of t 0 and x 0 in the above proposition are precisely the same as for the two-soliton solution of the Korteweg deVries equation, as given in Theorem 1 of [9].
3 Singularities of the two-soliton solutions in the complex plane As mentioned in the introduction, the two-soliton solutions u(x, t) of (2.1) are viewed as meromorphic functions in the complex variable x. We aim to determine how the dynamics of the singularities in C reflect the behavior of u(x, t) when x is restricted to the real axis.
Consider first the (one)-soliton solution, given by (2.7). It is immediate that the singularities of these solutions are precisely where m runs through the odd integers. Moreover, these singularities of u are all simple poles. Thus, the speed of the soliton is exactly the speed of its poles in the complex plane, while the position of the maximum point of the soliton at time t is the real part of the position of these poles. The imaginary part of the singularity remains constant in time of course. The poles of the two-solitons solutions (2.13) and (2.14), correspond to the zeros of the functions F ± and G ± defined in (2.15), (2.16), (2.17) and (2.18). It will turn out that the singularities of u ± are all simple poles, just as for the one-solitons. In all but a specfic class of exceptional cases, these poles correspond to simple zeros of F ± .
Recall that k 1 and k 2 are called commensurable if there exist positive integers p 1 and p 2 such that Without loss of generality, we may take it that gcd(p 1 , p 2 ) = 1. In what follows, we will always assume that the integers p 1 and p 2 appearing in (3.2) are without common prime factors. In this case, F ± (x, t) and G ± (x, t) (and thus u ± (x, t)) are periodic in x with minimal imaginary period 2πλi where Lemma 3.1. For any t ∈ R, the zeros of F ± (·, t) are simple, except for the following special case. If k 1 and k 2 are commensurable, and if p 1 , p 2 ∈ N and λ > 0 are given by (3.2) and (3. . Noticing that f jx = −k j f j , (j = 1, 2) and setting X = f 1 and Y = f 2 , it follows that if x is a zero of F − (·, t) of order greater than or equal to 2, then Let (X, Y ) be a solution of this system. Then, from the first equation we deduce that where ǫ is either 1 or -1. Inserting this into the second equation, it is found that Then formla (3.5) implies that X must satisfy t are both purely imaginary. Therefore, t = 0 and x is purely imaginary. Additionally, we have Thus k 1 and k 2 are commensurable and in fact, k 2 /k 1 = r 2 /r 1 with r j = 1 + 2q j . Noticing that q 2 − q 1 ∈ 2Z (because k 1 x − k 2 x ∈ 2iπZ), it transpires that r 2 − r 1 = 4k for some k ∈ Z. It follows that d = gcd(r 1 , r 2 ) divides 4k and therefore d divides k since d is necessarily an odd integer. Define p 1 and p 2 by k 2 /k 1 = p 2 /p 1 with gcd(p 1 , p 2 ) = 1. With this definition, There are precisely two such complex numbers in the fundamental strip S = {x ∈ C, −λπ < Im x < λπ}, and they are x = ± λ 2 πi. It is straightforward to ascertain that for such values of The same arguments hold for F + . Lemma 3.2. For any t ∈ R, the zeros of F − (·, t) and G − (·, t) are distinct, except the following special case. If k 1 and k 2 are commensurable, and if p 1 , p 2 ∈ N and λ > 0 are given by (3.2) and (3.3), with p 1 ± p 2 ∈ 4N, then there is a third order zero of F − (·, 0) at x = ( 1 Proof. In the notation of the proof of Lemma 3.1, the result follows if the system admits as its only solution X = Y = ±i. As before, if (X, Y ) is a solution, there exists ε ∈ {−1, 1} for which Extracting from this the variable Y and computing 1 + Y 2 leads to Now insert this into the second equation of the system (3.6) and, assuming by contradiction that 1 + X 2 is not zero, simplify the outcome. It follows that After further simplifications, this becomes and the claim follows.
Definition 3.1. The values of k 1 and k 2 for which F ± (·, 0) has multiple zeros are collectively referred to as the exceptional case. This occurs when k 1 and k 2 are commensurable, p 1 and p 2 are odd, and when considering F − , and when considering F + .
Lemmas 3.1 and 3.2 show that all singularities of u ± (·, t) are simple poles. In all but the exceptional case, they correspond to simple zeros of F ± . In the exceptional case, they correspond to a third order zero of g ± coinciding with a fourth order zero of F ± . It follows from Rouché's Theorem that in the exceptional case, four simple poles converge, as t → 0, to the simple pole at ( 1 2 + q)λπi for all q ∈ Z. Moreover, by the residue theorem, the sum of the residues of u ± (·, t) at these four poles converge, as t → 0, to the residue of the simple pole at ( 1 2 + q)λπi for all q ∈ Z. Proposition 3.1. In all but the exceptional case, the poles of u ± (·, t), i.e. the zeros of F ± (·, t), are described by analytic curves x : R → C. In the exceptional case, these curves are defined and analytic separately for t < 0 and t > 0.
Proof. Consider the case of u + and F + . Since all the zeros of F + (·, t) are simple, the implicit-function theorem shows that, for a fixed time t 0 , a zero of F + (·, t 0 ) can be locally and uniquely continued as an analytic curve x(t) such that F + (x(t), t) = 0. Such a curve x(t) can be continued as long as it remains in a bounded region of C. Thus, we need to show that |x(t)| must stay bounded as long as t remains in a bounded interval of R. First, it follows from Proposition 6.3 in Section 6 below that the imaginary part of x(t) must remain bounded. Furthermore, if Re x(t) → ∞ in finite time, then the equation F + (x(t), t) = 0 implies 1 = 0. If Re x(t) → −∞ in finite time, then the equation This section is closed with a specific example of an element in the exceptional case, namely k 1 = 1, k 2 = 5 (calculations done with MAPLE). In this case, λ = 1 and F − and G − may be rewritten as where y = e −x . Note that at time t = 0, F − and G − are symmetric polynomials in y, and that i and −i are third order zeros of G − , and fourth order zeros of F − . All the other zeros are simple. We also have the decomposition

Large-time asymptotic behavior of the singularities
In this section we show that the poles of u ± , that is the zeros of F ± , separate out into two groups, as t → ±∞, behaving asymptotically as poles of single solitons, of speeds k 2 1 and k 2 2 , respectively. This, of course, reflects the fact that u ± are in fact "two-soliton" solutions of (2.1).
The situation is nearly identical to that obtaining for the two-soliton solution for the Korteweg-de Vries equation (compare the following theorem with Theorem 2 in [9]). In particular, the backward shift of the slower wave, well-known in the case of the KdV-solitons, is also present for the modified KdV-equation.
Theorem 4.1. The asymptotic behaviors as t → ±∞ of the curves x(t) of zeroes of F ± (x, t) whose existence was determined in Proposition 3.1 are completely described as follows.

For every odd integer
2. For every odd integer m ∈ Z, there exists a unique curve x m,s + (t) of zeros of F ± (·, t) such that 3. For every odd integer n ∈ Z, there exists a unique curve 4. For every odd integer n ∈ Z, there exists a unique curve x n,f + (t) of zeros of F ± (·, t) such that To prove Theorem 4.1, it is necessary to study the zeros of F ± (·, t) with respect to frames of reference which move at the speed of each constituant soliton. As in Section 4 of [9], we set and then F ± (x, t) = H ± (z, r) = s −2 I ± (w, s).
Solutions z(r) of H ± (·, r) which remain bounded in C as r → 0 and as r → ∞ correspond to zeros x(t) of F ± (·, t) asymptotically traveling at speed k 2 1 as t → ±∞. Likewise, solutions w(s) of I ± (·, s) which remain bounded in C as s → 0 and as s → ∞ correspond to zeros x(t) of F ± (·, t) asymptotically traveling at speed k 2 2 as t → ±∞. The following is true of the curves z(r). Proposition 4.1. For every odd integer m ∈ Z, there exists a smooth curve z ± m (r) defined in some interval of r ≥ 0, such that H ± (z ± m (r), r) = 0 and as r → 0 + . For every odd integer n ∈ Z, there exists a smooth curve w ± n (s) defined in some interval of s ≥ 0, such that I ± (w ± n (s), s) = 0 and and similarly, I ± (w, 0) = 0 is equivalent to for some odd integer n ∈ Z. Applying the implicit function theorem, there exist smooth curves z ± m (r) and w ± n (s) defined in a neighborhood of z 0 = 1/k 1 log γ + mπi/2k 1 and w 0 = −1/k 2 log γ + nπi/2k 2 , respectively, such that H ± (z ± m (r), r) = 0, z ± m (0) = z 0 and I(w ± n (s), s) = 0, w ± n (0) = w 0 . It remains to calculate (z ± m ) ′ (r) and (w ± n ) ′ (r). Differentiating the equation H ± (z ± m (r), r) = 0 with respect to r yields where z = z ± m (r). Taking r = 0 gives and using that The derivative (w ± n ) ′ (0) is similarly calculated. Differentiating the equation I ± (w ± n (s), s) = 0 with respect to s leads to At s = 0, there obtains and since A proof of Theorem 4.1 is now readily available. The last proposition immediately gives all the curves of zeros of F ± (·, t) claimed in Theorem 4.1. The only remaining issue is to prove that there are no additional zeros of F ± (·, t). If k 1 and k 2 are commensurable, this is straightforward. For each t ∈ R, F ± (·, t) is a polynomial in e −xλ of degree 2(p 1 + p 2 ), and thus must have precisely 2(p 1 + p 2 ) zeros in the complex plane. The zeros of F ± (·, t) described in Theorem 4.1 account for all of them, for large |t|, and so no other zeros can exist.

The nature of the singularity in the exceptional case
In this section a more detailed analysis is undertaken of the singularity of u ± in the exceptional case (see Definition 3.1). As described just after this definition, the singularity at ( 1 2 + q)λπi, for any q ∈ Z, corresponds to a fourth order zero of F ± (·, 0) and is approached by four simple zeros of F ± as t → 0. The goal is to understand the behavior of a smooth curve of zeros of F (·, t) in a neighborhood of such a fourth order zero.
Remark that From this, it is deduced that F ± (x(t), t) = 0 if and only if F ± (−x(t), −t) = 0. In other words, if x(t) is a curve of zeros approaching λ 2 πi as t ր 0, then −x(t) is a curve of zeros approaching λ 2 πi as t ց 0. Theorem 5.1. Suppose we are in the exceptional case wherein k 1 and k 2 are commensurable and the odd integers p 1 , p 2 ∈ N and λ > 0 are as in (3.2) and (3.3). In the case of F + , it is assumed that p 1 + p 2 ∈ 4N and in the case of F − , it is presumed that p 1 − p 2 ∈ 4N. Let x(t) be a smooth curve, defined for t close to 0, such that F ± (x(t), t) = 0 and x(t) = λ 2 πi but x(t) → λ 2 πi as t → 0. Then, either Remark 5.1. The similarity of the result in Theorem 5.1 with the "exceptional case" for the two-soliton solution of sthe KdV-equation is quite striking. The behavior of the curves described in 5.1 is exactly the same as given by Proposition 4.9 in [9]. However, the exceptional case for the KdV-equation corresponds to p 1 being an odd integer, and p 2 being an even integer. For the mKdV-equation, the "exceptional pole" is located at λ 2 πi, rather than at λπi as it is for the KdV-equation.
Also, how does one explain the behavior described by (5.2)? In the case of the two-soliton solution of the KdV-equation, if p 1 is even and p 2 is odd, there is a horizontally moving pole approaching the singularity at πλi exactly as described by (5.2). This calculation was not carried out in [9], but can be obtained by a simple modification of the proof of Proposition 4.9 in [9]. Thus, it appears that the exceptional case for the two-soliton solution of the mKdV-equation (2.1) includes the behavior of curves of singularities from two different cases of the two-soliton solution of the KdV-equation, namely the "exceptional case", where p 1 is odd and p 2 is even, as well as the case where p 1 is even and p 2 is odd. It is precisely in these two cases that there is a pole located at the same place πλi (at time 0).
Proof. The proof is provided for F − . Similar arguments apply for F + . Let z(r) = x(t) − k 2 1 t where r is defined in (4.3). It follows that z(r) is a smooth curve such that H − (z(r), r) = 0, where H is as in (4.5). Notice that z(r) → If the minus sign is chosen on the right-hand side of (5.3), we come to Differentiating this relation with respect to r, it follows that This may be rewritten as or what is the same, Since p 1 and p 2 are both odd with p 2 − p 1 ∈ 4N and z(r) → λ 2 πi as r → 1, it must be that e −k 1 z → e −p 1 πi/2 and e −k 2 z → e −p 2 πi/2 as r → 1. If p 1 and p 2 are both in 4N + 1, then e −k 1 z and e −k 1 z both converge to −i as r → 1, while if p 1 and p 2 are both in 4N + 3, then e −k 1 z and e −k 1 z both converge to i as r → 1. We suppose first that p 1 and p 2 are both in 4N + 1. In this case, it follows from (5.4) that as r → 1, and thus Turning back to x(t), and using the fact that (r − 1)/t → k 2 (k 2 2 − k 2 1 ) as t → 0, it follows that If instead the plus sign is chosen on the right-hand side of (5.3), then it is immediately inferred that Differentiating this equation with respect to r, we get which may be simplified to From this, it is inferred that Since p 1 and p 2 are both in 4N + 1, it is concluded that as r → 1. A consequence of this is that .
L'Hopital's rule comes to the rescue and it is found that .
Reverting to the original variable x(t) and using again that It remains to treat the situation where p 1 , p 2 ∈ 4N + 3, in which case e −k j z(r) → i as r → 1. In fact, this case is "dual" to the one just treated and the same calculations lead to the result. To see this, assume first that the minus sign obtains on the right-hand side of (5.3). Then (5.4) implies that Following the same line of development as pursued for the positive sign in the previous situation where both p 1 and p 2 lie in 4N + 1 leads to Now assume we have the positive sign on the right-hand side of (5.3). From (5.6) we deduce

Vertical movement of poles
In this section we study the vertical motion of the singularities of the twosoliton solutions u ± of (2.1). As seen in Section 2, this comes down to studying the zeroes of F ± given by (2.17)- (2.18). Recall the solution u + represents two interacting solitons of the same sign, while u − represents two interacting solitons of opposite sign. Observe that and f 1 and f 2 are as in (2.9). Note the similarity in form between the above functions and the function F defined by formula (2.13) in [9]. Since F + 1 (x, t) = 0 if and only if F + 2 (x, t) = 0, to study the zeros of F + , it suffices to study the zeros of F + 1 . Similarly, since F − 1 (x, t) = 0 if and only if F − 2 (x, t) = 0, to study the zeros of F − , it suffices to study the zeros of F − 1 . As before, it is sometimes necessary to distinguish between the cases where k 1 and k 2 are commensurable and the cases where they are not. In the former case, denote by p 1 and p 2 the relatively prime positive integers such that (3.2) holds, and let λ be defined as in (3.3). Hence, if k 1 and k 2 are commensurable, the functions f 1 and f 2 are 2πλi periodic in x. Thus, in the commensurable case, all four of the functions F ± 1 , F ± 2 , must also be 2πλi periodic in x.
Proposition 6.1. The zeroes of F + and F − in the complex plane lie off the real axis. Moreover, in the commensurable case, if either F + (x, t) = 0 or F − (x, t) = 0, then Im x = 2mπλ for all m ∈ Z.
Proof. Suppose x ∈ R and F + (x, t) = 0. It follows that F + 1 (x, t) = F + 2 (x, t) = 0. Taking real and imaginary parts produces the coupled system from which it follows that f 1 (x, t) 2 = −1, which is impossible. If x ∈ R and F − (x, t) = 0, it similarly follows that from which one observes that f 1 (x, t) 2 = −1, which is again impossible.
The last statement follows from the 2πλi periodicity which holds in the commensurable case.
If k 1 and k 2 are commensurable, further information about the location of the zeros of F ± can be obtained. Proposition 6.2. Suppose that k 1 and k 2 are commensurable. If either F + (x, t) = 0 or F − (x, t) = 0, then Im x = (2m + 1)πλ for any m ∈ Z.
Proof. Because of 2πλi-periodicity, it suffices to consider m = 0. Suppose Im x = πλ and F + 1 (x, t) = 0. Since x − x = 2πλi, it follows from 2πλiperiodicity that F + 1 (x, t) = 0. But this says precisely that F + 2 (x, t) = 0. Adding and subtracting the two equations, F + 1 (x, t) = 0 and F + 2 (x, t) = 0 gives 1 (x, t) = ±i and f 2 (x, t) = ∓i. The latter system together with the definition (2.9) of the f j 's implies and so t = Re x = 0. Thus x = πλi. However, since p 1 is an integer. This contradiction shows that F + 1 (x, t) can not be zero if Im x = πλ. A similar argument applies to the other functions F + 2 , F − 1 and F − 2 .
To show that most of the poles of the two-soliton solution of (2.1) move vertically, we need to reproduce calculations which are similar to those found in Section 3 of [9]. Unfortunately, it seems that the calculations in [9] do not directly apply to the present situation in all cases. In an effort at economy, we only prove here that, with certain very specific exceptions, the poles in the two-soliton solution always feature vertical movement. Consequently, we do not need to reproduce the entirety of Section 3 of [9]. Nonetheless, we cannot avoid certain, somewhat tedious calculations, closely modeled on those in [9].
The following notation α = − Im x, is taken from [9]. The function F + 1 may be rewritten in this notation, viz.
Attention is first focussed on the solution u + . To investigate possible vertical movement of poles of u + , we examine more closely the zeros of F + 1 (x, t).
Proof. First, multiply the equation F + 1 (x, t) = 0 by 1 − iγA 1 e −ik 1 α , express this using the representation (6.7) and take the imaginary part of the result. This leads to the formula In the same way, multiplying by 1 − iγA 2 e −ik 2 α and subsequently extracting the imaginary part yields If cos k 2 α = 0, so that in particular sin k 2 α = ±1, it follows from (6.9), using the formulas for the sine of the sum and difference, that cos k 1 α = 0. In the same way, using (6.10), if cos k 1 α = 0, then cos k 2 α = 0. This proves the first assertion in the proposition. For future reference, notice that it follows by subtracting (6.9) from (6.10) that Taking the real and imaginary parts of (6.7) and setting them equal to zero yields 1 − γA 1 sin k 1 α − γA 2 sin k 2 α − A 1 A 2 cos(k 1 + k 2 )α = 0, (6.12) γA 1 cos k 1 α + γA 2 cos k 2 α − A 1 A 2 sin(k 1 + k 2 )α = 0. (6.13) Multiply the first equation above by sin(k 1 + k 2 )α, the second equation by cos(k 1 + k 2 )α and form the difference of the outcomes. The formula sin(k 1 + k 2 )α = γA 2 cos k 1 α + γA 1 cos k 2 α (6.14) emerges from these machinations. But, (6.13) implies The last two equations taken together imply (6.8) This proves the second assertion of the proposition. Finally, it is clear that A 1 = A 2 = 1 if and only if t = 0 and Re x = 0. Proposition 6.4. Suppose that cos k 1 α = cos k 2 α = 0. It follows that k 1 and k 2 are commensurable, that p 1 and p 2 are both odd and that α is an odd multiple of πλ/2.
Proposition 6.5. Let x(t) be a smooth curve such that F + 1 (x(t), t) = 0 and ∂ x F + 1 (x(t), t) = 0. It follows that Im x ′ (t) has the same sign as Proof. Let x(t) be a smooth curve such that F + 1 (x(t), t) = 0 and ∂ x F + 1 (x(t), t) = 0. Implicit differentiation provides the relation The imaginary part of the numerator in this last expression is equal to which is a positive mulitple of, and so has the same sign as, Finally, formulas (6.11) and (6.8) reveal that the expression in (6.17) is equal to Corollary 6.1. Suppose either that k 1 and k 2 are not commensurable, or that they are commensurable with p 1 and p 2 having opposite parity. Let x(t) be a smooth curve such that F + 1 (x(t), t) = 0. If either t = 0 or if Re x = 0, then Im x ′ (t) = 0.
Proof. We know from Lemma 3.1 that the roots of F + (·, t) are all simple unless k 1 and k 2 are commensurable and p 1 and p 2 are both odd. It follows that the same is true for F + 1 (·, t). Thus, it must be the case that ∂ x F + 1 (x(t), t) = 0. The last three propositions can now be brought to bear to establish the claim.
To recapitulate, it has been shown, except in the commensurable case with p 1 and p 2 both odd, that if either t = 0 or if Re x = 0, then Im x ′ (t) = 0, so the poles of the solution u + are always moving vertically. The same is true in the case that p 1 and p 2 are both odd, so long as the imaginary part of the pole is not an odd multiple of πλ/2. The same conclusions are true about the poles of u − . The calculations leading to this conclusion, while not the same as those for u + , are analogous enough that we pass over the details. Interestingly, in the commensurable case when p 1 and p 2 have opposite parity, it turns out that the poles of u − and those of u + bear a very simple relationship to one another. Proposition 6.6. Suppose that p 1 and p 2 have opposite parity (one odd, one even). It follows that there exists θ ∈ R such that for all x ∈ C and t ∈ R. In other words, the poles of u − are precisely given by a vertical translation of the poles of u + .
Proof. Suppose θ ∈ R is such that for all x ∈ C and t ∈ R. It would then follow that (6.18) holds for all x ∈ C and t ∈ R. The same would be true if we had instead For these latter conditions to hold, it is necessary for there to be two integers m and n such that p 1 θ/λ = 2mπ, (6.25) p 2 θ/λ = (2n + 1)π, (6.26) or, what is the same, θ λπ = 2m If p 1 is even and p 2 is odd, it is clear that one may choose appropriate values of θ, m and n so that the last equation holds In the opposite case, if p 1 is odd and p 2 is even, a similar argument shows that there exists θ, mand n such that (6.21) and (6.22) hold.
It remains to consider the case where k 1 and k 2 are commensurable, with p 1 and p 2 odd. To establish the existence of poles with non-trivial vertical movement, it suffices by Proposition 6.4 to establish the existence of poles whose imaginary parts are not an odd multiple of πλ/2 (with either t = 0 or with non-zero real part).
As shown in [9], for the two-soliton solution of the KdV-equation in the commensurable case, there are always poles with vertical movement, and at least one pole moving precisely horizontally. The proof of this fact requires the full force of the delicate and technical analysis in [9]. We would like to avoid such technical calculations in this paper to the extent possible.
Thus, for the modified KdV-equation (2.1), we should not expect that all the poles will be moving vertically in the remaining cases. The fact that in the case of opposite parity, all the poles move vertically (except if t = 0 and Re x = 0) is already an interesting difference in behavior between the two equations.
It turns out that in the commensurable case, with p 1 and p 2 both odd, the movement of the poles of u + can in fact be reduced to the movement of the poles of the 2-soliton solution of the KdV-equation if Here are the precise statements. Proposition 6.7. Suppose that p 1 and p 2 are both odd. If p 2 − p 1 ∈ 4N then there exist θ 1 , θ 2 ∈ R such that for all x ∈ C and t ∈ R.
Proof. Suppose θ 1 ∈ R is such that for all x ∈ C and t ∈ R. It would follow that (6.28) holds for all x ∈ C and t ∈ R. In addition, if we have (6.33) then (6.29) would be true.
For the system (6.30)-(6.31) to be valid, it is necessary and sufficient that exp(ik 1 θ) = exp(ip 1 θ/λ) = −i, (6.34) exp(ik 2 θ) = exp(ip 2 θ/λ) = −i. (6.35) For this, we need to find two integers m and n such that which is the same as asking for two integers m and n such that For the latter to hold true, it must be the case that Since p 1 and p 2 are relatively prime, there exist integers r and s such that hence, simply take The proof for F + 2 is similar, but with 4m + 1 replacing 4m − 1 and 4n + 1 replacing 4n − 1.
Proposition 6.8. Suppose that p 1 and p 2 are both odd. If p 2 + p 1 ∈ 4N then there exist θ 1 , θ 2 ∈ R such that for all x ∈ C and t ∈ R.
is exactly the function F in formula (2.13) of [9] whose zeros correspond to the poles of the 2-soliton solution of the KdV-equation. Thus, we may use the results of [9] to describe the behavior of the poles of u ± in the cases under consideration.
More precisely, we may now affirm that in the case where p 1 and p 2 are both odd and p 2 −p 1 ∈ 4N, the solution u + has two poles moving horizontally on the line Im x = πλ/2 and also on the line Im x = −πλ/2. There are 2(p 1 + p 2 − 2) other poles with imaginary part between ±πλ, and they will move vertically for all t = 0. The same will be true for u − in the case p 1 and p 2 are both odd if p 2 + p 1 ∈ 4N. These configurations will repeat with 2πλi periodicity, so that horizontally moving poles are found with imaginary part equal to every odd multiple of πλ/2.
The last case is one where four poles meet at t = 0, as described in the previous section, at poles on the imaginary axis whose imaginary part is an odd multiple of πλ/2. At each of these points, the analysis shows that (at least) two of these poles have vertical movement. We refrain from going into the detailed considerations needed to establish that all the poles move vertically for all t = 0, excepting the two poles approaching the singular points horizontally. In any event, we have shown in this case the existence of poles with vertical movement.

Finite time blowup of solutions
The results of the previous section immediately imply that there exist complexvalued solutions to (2.1) on R which blow up in finite time. Indeed, fix α ∈ R, and let u be given by where we may use either of the two soliton solutions u + or u − . As long as the set {(x − iα, t 0 ) : x ∈ R} does not contain a pole of u ± , u is a smooth, complex-valued solution of (2.1) for t in a neighborhood of t 0 . Since at any given time, the imaginary parts of the collection of all poles of u + or u − form a discrete set, and since both u + and u − have poles whose imaginary parts move continuously in time, there exist α and t 0 such that (x − iα, t 0 ) is a pole of u + , say, but (x − iα, t) is not a pole of u + if t is sufficiently close to, but not equal to, t 0 . It follows that the solution u defined in (7.1) with this choice of α is a regular complex-valued solution of (2.1) on R, decaying exponentially to zero as x → ±∞ for t close to t 0 , but which is singular at t = t 0 . In other words, the solution blows up in finite time.
It is interesting to note that this result of singularity formation for complexvalued solutions of the mKdV-equation can be interpreted as a blow-up result for real-valued solutions of a system of dispersive equations. Let u be a solution of (2.1) and let r = Re u and s = Im u. It follows that r and s satisfy the real-valued system r t + r xxx + 6(r 2 − s 2 )r x − 2rss x = 0, (7.2) s t + s xxx + 2rsr x + 6(r 2 − s 2 )s x = 0. (7.3) Thus, we have shown that this coupled, dispersive system admits real-valued solutions (exponentially decaying in space) which blow up in finite time.
8 Some formal calculations (All the computations in this section have been carried out using MAPLE.) As noted at the end of Section 2, the interaction time for the two-soliton solutions u ± , given in (2.13) and (2.14), is t = 0, the center of the interaction is x = 0 and at t = 0, the solution is even in x. The explicit formulas for u ± allow us to observe and calculate certain aspects of these solutions at the moment of interaction. In particular, it is interesting to know whether there is a single maxima during the interaction or not, and it is likewise interesting to know the speed of the two solitons at the moment of interaction.
In the case of u + , one observes that that the solution has one centered maximum if the ratio k 2 /k 1 is large enough (bigger than around 2.6), and two symmetrically located maxima for smaller values of k 2 /k 1 . Partial confirmation of this can be obtained by computing u + xx (0, 0). Since, by symmetry, u + x (0, 0) = 0, the sign of the second derivative will tell us if it is a local maximum or a local minimum. A MAPLE-implemented calculation shows that u + xx (0, 0) = −(k 2 − k 1 )(k 2 1 − 3k 1 k 2 + k 2 2 ). Therefore, if 1 < k 2 /k 1 < (3 + √ 5)/2, then u + xx (0, 0) > 0, which means that u + (·, 0) has a local minimum at x = 0. Thus, there are (at least) two maxima at the moment of interaction. If k 2 /k 1 > (3 + √ 5)/2, then u + (·, 0) has a local maximum at x = 0, which is consistent with there being a single maximum at the moment of interaction.
If we are interested in the "speed" of the two-soliton solution at the moment of interaction, one idea is to calculate the speed of the maximum or the local minimum of the solution as (x, t) approaches (0, 0). One might argue that the movement of the maximum is some kind of speed. It must be acknowledged that the interpretation is less clear when one is tracking a local minimum. To calculate this speed, suppose y(t) = y ± (t) is a real-valued curve such that u ± x (y(t), t) = 0, i.e. y ± (t) is always at an extemal point of the solution u = u ± . Suppose also that y ± (0) = 0, which is to say the curve is at the interaction center at the interaction time t = 0. Differentiating with respect to t, we see that u xx (y(t), t)y ′ (t) + u xt (y(t), t) = 0, or y ′ (t) = − u xt (y(t), t) u xx (y(t), t) .
This gives, in turn, y ′ (0) = − u xt (0, 0) u xx (0, 0) , (8.1) which might be thought of as representing the speed of the two-soliton solution at the moment of interaction of the two solitons. The value of y ′ (0) can be calculated explicitly from (2.13) and (2.14). The results are as follows, for both u + and u − , For u − , where there is always a maximum at x = 0 at the moment of interaction, the maximum is moving with a posistive speed. What that speed represents is not entirely clear. In the case of u + , where the midpoint is a local maximum only if k 2 /k 1 > (3 + √ 5)/2, we see that for these values of 0 < k 1 < k 2 , we have indeed y ′ + (0) > 0. On the other hand, y ′ + (0) < 0 for at least some values of 0 < k 1 < k 2 with k 2 /k 1 < (3 + √ 5)/2. (The lower bound on k 2 /k 1 for which this speed is negative is around 2.15.) This negative speed represents the speed of the local minimum, between the two maxima. It is curious that in some cases this minimum is moving backwards.
It is also interesting to do this for the two-soliton solution of the KdVequation, a calculation which was not carried out in [9]. In this case, we obtain As is known, the two-soliton solution of the KdV-equation has one maximum at the interaction time if k 2 /k 1 > √ 3 and two maxima, symmetrically located about the interaction center, if 1 < k 2 /k 1 < √ 3. Here we see that y ′ (0) < 0 for 1 + √ 2 < k 2 /k 1 < √ 3. In these case, of course, it is the minimum which is moving backward.