Note on scale invariance and self-similar evolution in (3+1)-dimensional signum-Gordon model

0.15 MB·English

by  H. Arodz

Note on scale invariance and self-similar
evolution in (3+1)-dimensional
2
1
0 signum-Gordon model
2

n
a
H. Arodz´, J. Karkowski
J

1 Institute of Physics, Jagiellonian University∗
1
´
Z. Swierczyn´ski
]
h
Institute of Computer Science and Computer Methods,
t
-
p Pedagogical University, Cracow†
e
h
[

1
v
9
7
Abstract
2
2
Several classes of self-similar, spherically symmetric solutions of rela-
.
1
tivistic wave equation with a nonlinear term of the form sign(ϕ) are pre-
0
2 sented. They are constructed from cubic polynomials in the scale invariant
1 variable t/r. One class of solutions describes a process of wiping out the
:
v initial field, another anaccumulation of fieldenergy in afiniteand growing
Xi regionofspace.
r
a
PACS: 11.27.+d,98.80.Cq,11.10.Lm
∗Reymonta4,30-059Cracow,Poland
†Podchora¸z˙ych2,30-084Cracow,Poland
1
1 Introduction
Symmetry transformations for a given equation act within the space of its solu-
tions. Especiallyinterestingarefixedpointsofsuchtransformations,i.e.,thesolu-
tions that remain invariant. In the case of continuous symmetries they depend on
a reduced number of independent variables. This significantly simplifies the task
of finding the solutions, the task that so often is formidably difficult in the case
of nonlinear field equations. Apart from the rudimentary examples of rotational
andtranslationalsymmetries,incertainmodelsthereisalsoaone-parameterscal-
ingsymmetry. Thecorrespondinginvariantsolutions,calledtheself-similarones,
play an importantrolein mathematicalanalysis ofnonlinearevolutionequations,
as well as in physics oriented investigations, see, e.g., the books [1], [2] and a
sampleofrecent research papers [3].
Thepresent paper isa sequelto papers [4], whereself-similarsolutionsin the
signum-Gordon model with a real scalar field in 1+1 dimensions were analyzed.
Rich variety of such solutions was found. Certain related models were investi-
gated in [5]. Derivation and a general discussion of the signum-Gordon model
can be found in [4]. The model is very interesting, there is no doubt that its in-
vestigations should be continued in several directions, such as finding classical
solutionsofvariouskinds,applications,orquantization.
As is well-known, field theories in 1+1 dimensions have rather special prop-
erties. Therefore, it is not clear whether the amazing picture of the space of self-
similarsolutionsfound in [4]is valid also in higherdimensions. This is ourmain
motivationfor the present work. Another attraction is thepossibilityof obtaining
several exact analytic solutions to the signum-Gordon equation in 3+1 dimen-
sions, and to have new insights into rather intriguing dynamics of the scalar field
inthat model.
Wehavefoundseveralfamiliesofself-similar,sphericallysymmetricanalytic
solutions. They are composed from cubic polynomialsin the scale invariantvari-
able u = t/r. Our results show that there is a qualitative similarity between the
3+1 and 1+1 dimensional cases, but significant differences appear. First, the lack
oftranslationalinvarianceintheradialvariablerresultsintheabsenceofcounter-
parts of certain classes of 1+1 dimensional solutions. Second difference, perhaps
not so unexpected, is that the pertinent calculations are a bit more complicated
because we have to look for zeros of cubic polynomials,whilein the1+1 dimen-
sional case only the second order polynomials were present. The exact solutions
wehavefoundprovideexamplesofnon-trivialevolutionofthescalarfieldϕ. Es-
pecially interesting seem to be solutions of the types II and III, presented in
2
Section 3,which showhowthefield settlesexactlyat thevacuumvalueϕ = 0.
The plan of our paper is as follows. In Section 2 we describe the method
of constructing the self-similar solutions in the signum-Gordon model. Explicit
solutions in the 3+1 dimensional case are presented in Section 3. Section 4 is
devotedtodiscussionofourresults.
2 Scale invariant Ansatz and the method of construct-
ing the solutions
The signum-Gordon equation for the real scalar field ϕ in D + 1 dimensional
space-timereads
∂ ∂µϕ(x)+sign(ϕ(x)) = 0. (1)
µ
The sign function takes the values ±1,0, sign(0) = 0. It is clear that Eq. (1) im-
pliesthatthesecondderivativesofϕmaynotexistifϕvanishesatisolatedpoints,
but even at such points one-sided second order derivatives do exist. The proper
mathematical framework for discussing equations of such kind is well-known:
one should consider the so called weak solutions, see, e.g., [6]. All solutions
constructedbelowhavebeen checked inthisrespect.
Fromagivensolutionϕ(x) ofEq.(1)onecan obtainone-parameterfamilyof
solutionsoftheform
x
ϕ (x) = λ2ϕ( ), (2)
λ
λ
where λ > 0 is a constant. The solution is self-similar if ϕ (x) = ϕ(x) for all
λ
λ > 0. Takingλ = r,wherer istheradius,andassumingthesphericalsymmetry,
wemaythen write
t
ϕ(x) = r2 G( ), (3)
r
wheret = x0 is thetime. WiththisAnsatz, Eq.(1)gives
(u2 −1)G′′ −(D +1)uG′+2DG−sign(G) = 0, (4)
whereu = t/r and G′ = dG/du. Thisequationhas theparticularsolutions
1
G = 0, G = ± ,
2D
whichare trivial,butneverthelessplayimportantrolebelow.
3
Let g(u)beasolutionoftheauxiliarylinear(!) equation
(u2 −1)g′′−(D +1)ug′+2Dg = 0. (5)
Then G (u) = g(u)+1/(2D) is a solutionof (4) on the interval ofu defined by
+
the condition G+(u) > 0, and G−(u) = g(u)−1/(2D) on the interval in which
G−(u) < 0. It turns out that patching together a number of such partial solutions
and, in some cases, including also the trivial solution G = 0, one can cover the
whole interval 0 ≤ u < ∞. In this way we obtain a self-similar solution of Eq.
(1) valid for all r ∈ [0,∞) and t ∈ [0,∞). The values G(0),G′(0) determine the
initialdataforϕ(t,r):
ϕ(0,r) = r2G(0), (∂ ϕ)(0,r) = rG′(0). (6)
t
WhenpatchingthepartialsolutionswedemandcontinuityofG(u),andalsocon-
tinuity of G′(u), unless u = 1. G′(u) at u = 1 does not have to be continuous
because G′′ in Eq. (1) is multiplied by the factor u2 − 1 which vanishes at that
point. This is not surprising because u = 1 corresponds to the light-cone r = t,
thecharacteristichypersurfaceforequation(1).
Equation(5)has twolinearlyindependentsolutions:
∞
g (u) = Du2 +1, g (u) = c u2l+1, (7)
1 2 l
l=0
X
wherethecoefficientsc are determinedfrom therecurrence relation
l
(2l −1)(2l+1−D)
c = c . (8)
l+1 l
2(l+1)(2l+3)
Note that in the case of odd space dimensionD the series for g is in fact a poly-
2
nomial of order D, because c = 0 for all l > (D − 1)/2. In particular, g = u
l 2
forD = 1, g (u) = u3 +3u when D = 3, and g = u5 −10u3 −15ufor D = 5,
2 2
apart fromoverallmultiplicativeconstants.
In thenextSectionweconsiderindetail theD = 3 case.
= 3
3 Explicit self-similar solutions in the case D
In the case of three space dimensions the general form of the partial solutions
reads
1
G = ± +α(3u2+1)+β(u3+3u), (9)
6
4
orequivalently,
1
G = ± +γ(u−1)3 +δ(u+1)3, (10)
6
whereα,β,γ,δ areconstants. Thesigninfrontof1/6hasto beequal tosign(G).
Letusbeginourexplorationofthespaceoftheself-similarsolutionsbycheck-
ingtheirasymptoticbehavioratu → ∞. ItturnsoutthatGhastohaveaconstant
sign if values of u are large enough – we prove in the Appendix the following
lemma.
Lemma. Suppose that u > 1 is a real zero of the polynomial (9). Then, that
1
polynomialdoesnothaveanyreal zeros larger thanu .
1
Itfollowsthatintheregionu ≥ u thesolutionhasthefixedform(9)withcertain
1
fixed coefficients α,β. The just excluded option was that of having an infinite
sequence of polynomialswith alternating signs. Now, let us recall that the region
u → ∞ corresponds to r → 0 (t > 0). Therefore, thepolynomialu3 +3u would
givea singularfield ϕ = r2G ∼ 1/r witha divergent,non integrableat r = 0en-
ergy density. Forthisreason, weputβ = 0. Notealsothat by takingintoaccount
the symmetry ϕ → −ϕ, we may consider just two cases of the asymptotic form
of G at large values of u: G(u) > 0, or G(u) = 0. The latter case is considered
in the second and third subsections (solutionsof the types II and III). The former
caseis namedtypeI.
3.1 Solutions of type I
In thiscase, forarbitrary largevaluesofu
1
2
G(u) = +α(3u +1) > 0,
6
aspointedoutinthepreceding paragraph. Thisimpliesthat α ≥ 0,and thensuch
G(u)doesnothaveanyzeroesdowntou = 1,whereitcanbegluedwithanother
polynomial of the form (9) or (10). We conclude that all type I solutions on the
intervalu ∈ [1,∞)havetheform
1
G∞(u) = +α∞(3u2 +1) (11)
6
withan arbitrary constantα∞ ≥ 0.
At the pointu = 1 thesolutionG∞ can be glued with anotherpolynomial(9)
or (10), taken with the +1/6 because G∞(1) = 1/6 + 4α∞ > 0. Let us denote
5
suchpolynomialbyG anditsconstantsbyγ ,δ (hereweprefertheform(10)).
+ + +
Thegluingconditionat u = 1, G∞(1) = G+(1) givesδ+ = α∞/2. Hence,
1 1
G+(u) = +γ+(u−1)3 + α∞(u+1)3. (12)
6 2
Nowwe haveto determinethelowerend ofthe intervalon which G is thesolu-
+
tion (the upper end is u = 1). The first possibility,denoted as Ia, is that G > 0
+
forallu ∈ [0,1). Thisoccursifγ+ < 1/6+α∞/2. InthiscaseG+ andG∞ cover
thewholeinterval[0,∞),andthesefunctionstogethergivethecompletesolution.
It hastheshapesketchedin Fig.1 withthedashedline.
Figure1: SolutionsofthetypeIa(thedashedline)and Ib(thesolidline)
The other possibility, denoted as Ib, is that there exists u ∈ (0,1) at which
0
G (u ) = 0. Simplecalculationgivesinthiscase
+ 0
1+3α∞(1+u0)3
γ = . (13)
+ 6(1−u )3
0
6
Moreover, it turns out that u has to be the first order zero of G . Therefore, at
0 +
thepointu wemay glueG withanegativepolynomial
0 +
1
G− = − +γ−(u−1)3 +δ−(u+1)3 ≤ 0, (14)
6
and notwiththetrivialsolutionG = 0. Thematchingconditions
′ ′
G−(u0) = 0, G−(u0) = G+(u0)
give
u0 +3α∞(1+u0)3 α∞ 1
γ− = , δ− = + . (15)
6(1−u )3 2 6(1+u )2
0 0
It turns out that G−(u) found above does not have any zeros in the interval
[0,u0). Therefore, the three functions: G∞ for u ≥ 1, G+ for u0 ≤ u ≤ 1, and
G− for0 ≤ u ≤ u0,form thecompletesolutionofEq.(4). ItissketchedinFig. 1
withthesolidline.
The values of G(0),G′(0), which specify the initial values of the scalar field
throughformula(6), read as follows.
TypeIa:
1 α∞ ′ 3α∞
G (0) = −γ + , G (0) = 3γ + .
+ + − +
6 2 2
TypeIb:
1
′
G−(0) = − −γ− +δ−, G−(0) = 3(γ− +δ−).
6
Varying α∞ in the interval [0,∞) and u0 in the interval [0,1) we obtain certain
sets in the(G(0),G′(0)) plane. They are shown in Fig. 2. We have denoted them
Ia,Ib, identicallyas thecorrespondingsolutions.
Theregions−Ia,−Ibareobtainedbythereflectionintheorigin,(G(0),G′(0)) →
(−G(0),−G′(0)), related to the symmetry ϕ → −ϕ. The half-infinite curve that
separates the region Ib from IIb is obtained for α∞ = 0. It has the following
parametricformwithu ∈ [0,1)as theparameter:
0
1 u 1 1 u
0 ′ 0
G(0) = − − , G(0) = + .
6(1+u )2 6(1−u )3 6 2(1+u )2 2(1−u )3
0 0 0 0
Thecurvestartsatthepoint(0,1/2)thatcorrespondstou = 0,anditapproaches
0
itsasymptote,that isthestraighthalf-line
1 1
′
G(0)(σ) = − −σ, G(0)(σ) = +3σ, σ ∈ [0,∞),
8 8
whenu → 1.
0
7
Figure2: Themapoftheself-similarsolutions. TheregionsIa,IIaareseparated
from Ib, IIb, respectively, by the G′(0) axis. The infinite region III extends
over all area encompassed by the stripes IIa,IIb,−IIa,−IIb, except for the
origin (0,0) which corresponds to the trivial solution G = 0. The stripe IIb
asymptoticallyapproaches the stripe −IIa, similarly IIa approaches −IIb. The
meaningofthedashed straightlineisexplainedin Section4
8
3.2 Solutions of type II
Solutions of the type II are obtained by taking the trivial solution G(u) = 0 for
u ≥ u > 1, and gluing it at thepoint u with the polynomial(10) taken with the
1 1
+sign(theothersigncanbeobtainedbythesymmetrytransformationϕ → −ϕ).
Because u 6= 1,thematchingconditionincludesthefirst derivative,
1
′
G(u ) = 0, G(u ) = 0.
1 1
Solvingtheseconditionsweobtainthefollowingpolynomial
(u −u)2(2u u+u2 −3)
G (u) = 1 1 1 . (16)
d 6(u2 −1)2
1
It has strictly positive values in the whole interval u ∈ [1,u ). At u = 1 that
1
polynomialisgluedwithanotherpositivepolynomialG (u)oftheform(9). The
+
matchingconditionG (1) = G (1) gives
d +
1 u(u2+3)
G (u) = −α (u−1)3 − , (17)
+ 6 + 6(1+u )2
1
where α is an arbitrary constant. Similarly as for the type I solutions, there are
+
two possibilities. The first, denoted as IIa, with G (u) strictly positive on the
+
whole interval (0,1], occurs when α ≥ −1/6. In this case we already have the
+
complete solution: G = 0 for u ≥ u , G for u ∈ [u ,1], and G given by
1 d 1 +
formula (17) for u ∈ [0,1]. It is sketched in Fig. 3 with the dashed line. The
valuesofG ,G′ at u = 0,i.e.,
+ +
1 1
′
G (0) = +α , G (0) = −3α − ,
+ 6 + + + 2(1+u )2
1
where α ∈ [−1/6,∞), u ∈ (1,∞), fill a semi-infinite straight-linear stripe in
+ 1
the(G(0),G′(0))plane. It isdenotedas IIain Fig.2.
The second possibilityis that G (u) has a zero at some u ∈ (0,1). If this is
+ 0
thecase, then
u (u2 +3) 1
0 0
α = − .
+ 6(1−u )3(1+u )2 6(1−u )3
0 1 0
Formula (17) implies that the zero can only be of the first order. The matching
conditionsat u determinethetwo coefficients inthenegativepolynomial
0
1
G−(u) = − +γ−(u−1)3 +δ−(1+u)3 (18)
6
9
whichis gluedat thatpointwithG (u). Simplecalculationsgive
+
u (1+u )3 1 1
0 0
γ− = − , δ− = − .
6(1−u )3 12(1+u )2(1−u )3 6(1+u )2 12(1+u )2
0 1 0 0 1
It turns out that G−(u) is strictly negative in the whole interval u ∈ [0,u0)
for all choices of u ∈ (0,1), u ∈ (1,∞). Thus, we have obtained again the
0 1
complete solution: G = 0 for u ≥ u , G for u ∈ [1,u ], G for u ∈ [u ,1], and
1 d 1 + 0
G− foru ∈ [0,u1]. WedenoteitasIIb. ItissketchedinFig.3withthesolidline.
Figure3: SolutionsofthetypeIIa(thedashed line)andIIb(thesolidline)
The points (G−(0),G′−(0)) with u0 ∈ (0,1),u1 > 1 fill in the (G(0),G′(0))
plane the region denoted as IIb, see Fig. 2. It borders the region IIa along the
segment (3/8,1/2) of the G′(0) axis. The region IIb is bounded from below
by the curve obtained by putting u1 = 1 in formulas for G−(0),G′−(0) for the
solution(18). Thecurveisobtainedin theparametricform, namely
u (2+u ) u (1+u ) 1 3u −1
0 0 0 0 ′ 0
G−(0) = −6(1+u )2 − 24(1−u )2, G−(0) = 2(1+u )2 + 8(1−u )2,
0 0 0 0
10