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For a compactly supported variation $\varphi \mapsto \varphi +\epsilon \eta$, the first variation of the energy functional isIntegration by parts has no endpoint term. Since $\eta$ is arbitrary, the static Euler-Lagrange field equation isIntegrating the specified polynomial derivative gives $U(\varphi )={\varphi }^2/2-{\varphi }^4+{\varphi }^6/2+C$. The polynomial factors as a nonnegative square times ${\varphi }^2$; its global minimum is zero when $C=0$. ThereforeThese are the three constant scalar-field vacua. For a localized finite-energy kink, the two ends approach vacua and the derivative tends to zero. Multiplying the field equation by ${\varphi }^{\prime }$ gives the first integralA nonconstant kink must stay within one interval between neighboring vacua. A proposed direct connection from $-1$ to $1$ would reach $\varphi =0$ at some finite point, where the first integral forces ${\varphi }^{\prime }=0$. Uniqueness for the smooth second-order differential equation then forces the identically zero solution, a contradiction. This is the intermediate-vacuum obstruction to a kink; two elementary kinks can span the outer vacua only in an infinite-separation limit, not as one finite-width static kink.

There are consequently four oriented elementary sectors: $-1\to 0$, $0\to -1$, $0\to 1$ and $1\to 0$. If opposite orientations are grouped together, there are two kink families and their antikinks. Spatial reflection reverses orientation, and $\varphi \mapsto -\varphi$ exchanges the two adjacent intervals; both preserve the energy. All four therefore have the same rest mass, even though the central and outer vacua have different fluctuation curvatures.

For the sector $0\to 1$, define $W(\varphi )={\varphi }^2/2-{\varphi }^4/4$, so $U=(W^{\prime }{)}^2/2$ and $W^{\prime }=\varphi (1-{\varphi }^2)$. The Bogomolny bound follows by completing the square:It is saturated by the Bogomolny equationLet $y={\varphi }^2$ on this positive branch. Then $y^{\prime }=2y(1-y)$, so $\log [y/(1-y)]=2(x-x_0)$. Thus the explicit kink in a phi-six model isIt has the required limits and satisfies the first integral. The translated center $x_0$ is arbitrary. All oriented elementary solutions areFor $\eta =1$ the endpoints are $0\to \sigma$, and for $\eta =-1$ they are $\sigma \to 0$. Direct evaluation ${\int }_0^1\sqrt{2U(\varphi )}d\varphi ={\int }_0^1\varphi (1-{\varphi }^2)d\varphi =1/4$ independently gives their common mass.

In this static scalar problem, a classical field-theory soliton is a nontrivial smooth localized finite-energy field configuration which is static in its rest frame and stable against admissible small perturbations. Moving examples are obtained by a Lorentz boost. The relevant static energy isUse the usual vacuum normalization $U\ge 0$, with zero energy at a vacuum, and impose finite-energy vacuum behavior at spatial infinity. These assumptions, together with the unconstrained real target and the canonical two-derivative kinetic term, are the hypotheses of the following Derrick theorem argument.

For Derrick scaling ${\varphi }_{\lambda }(x)=\varphi (\lambda x)$, change variables to obtainA static solution must be stationary under this admissible scale variation. The Derrick virial identity isFor $D>2$, both terms are nonpositive and the gradient term is strictly negative for a nonconstant field, so no nontrivial static solution is possible. At $D=2$, the identity forces $V=0$, but the gradient term is scale invariant and a further argument is needed.

Since $U\ge 0$, $V=0$ implies $U(\varphi (x))=0$ everywhere. Every zero of a differentiable nonnegative potential has $U^{\prime }=0$. The static Euler-Lagrange field equation $\Delta \varphi =U^{\prime }(\varphi )$ therefore makes $\varphi$ a harmonic function. Each derivative ${\partial }_i\varphi$ is harmonic and belongs to L2 space by finite energy. Its mean-value property and the Cauchy-Schwarz inequality on a ball of radius $R$ implyLetting $R\to \infty$ gives ${\partial }_i\varphi =0$. This proves the two-dimensional flat-target Derrick obstruction. Thus a nontrivial static real-scalar soliton in this class can occur only for $D=1$. The argument does not apply unchanged to curved-target sigma models or gauge-field energies.

In one dimension the scale identity is $T=V$, and the second scale derivative is $2V>0$ for a nontrivial solution; scale variation does not rule it out. To derive the Bogomolny equations, write $U=\frac{1}{2}{W}^{\prime }(\varphi {)}^2$ on the field interval of interest. Completing the square givesTherefore the Bogomolny bound and its saturation equation areEquivalently, for a monotone field connecting adjacent vacua, ${\varphi }^{\prime }=\pm \sqrt{2U(\varphi )}$, with the sign chosen for its direction. Differentiating the first-order equation gives ${\varphi }^{\prime \prime }=W^{\prime }{W}^{\prime \prime }=U^{\prime }$, verifying the static Euler-Lagrange field equation. Saturation minimizes energy within the fixed-boundary topological sector.

For the given sextic potential, factorization gives $U(\varphi )={\varphi }^2({\varphi }^2-4{)}^2$, so the vacua are $-2,0,2$. ChooseThis choice satisfies $U=W^{\prime 2}/2$. On $0<\varphi <2$, the increasing kink in a phi-six model satisfies ${\varphi }^{\prime }=\sqrt{2}\varphi (4-{\varphi }^2)$. Setting $y={\varphi }^2$ gives $y^{\prime }=2\sqrt{2}y(4-y)$. Separation givesand henceThe endpoints are zero and two, and the mass follows from $W(2)-W(0)=4\sqrt{2}$. The other increasing elementary scalar-field kink isIt uses the opposite sign of $W^{\prime }$ and has the same mass. Reflecting $x-x_0$ yields the two decreasing antikinks; $x_0$ is the free center.

There is no single finite-width static kink directly connecting $-2$ to $2$. Multiplying the static equation by ${\varphi }^{\prime }$ gives the first integral ${\varphi }^{\prime 2}/2-U=0$ for a finite-energy vacuum-to-vacuum solution. At the intermediate vacuum $\varphi =0$, this forces ${\varphi }^{\prime }=0$. Uniqueness for the smooth second-order field equation makes data $(\varphi ,{\varphi }^{\prime })=(0,0)$ stay at that vacuum, so a nontrivial solution cannot cross it at finite $x$. This is the intermediate-vacuum obstruction to a kink; the two elementary transitions can only separate by an infinite interval.

Write the scalar field energy as $E=T+V$, whereUnder the Derrick scaling ${\varphi }_{\lambda }(x)=\varphi (\lambda x)$, a change of variables givesA finite-energy solution of the Euler-Lagrange equation is stationary under this admissible variation. The Derrick virial identity is therefore

The static field equation is ${\varphi }^{\prime \prime }=U^{\prime }(\varphi )$, so integration givesThe polynomial before $C$ is nonnegative and vanishes, so requiring the minimum to be zero fixes $C=0$. Hence the vacuum manifold iswhich has three elements. A finite-energy scalar-field kink can join only adjacent vacua: a solution cannot cross the intermediate vacuum at finite $x$ because its first integral would have $\varphi ={\varphi }^{\prime }=0$ there and the Picard-Lindelöf theorem would make it constant. There are therefore four oriented topological sectors,comprising two increasing kinks and their two antikinks. Symmetry under $\varphi \mapsto -\varphi$ and spatial reflection generates all four from one profile.

For the $0\to 1$ sector, completing the square gives the Bogomolny boundEquality holds for the Bogomolny equationWith $y={\varphi }^2$, this becomes the logistic differential equation $y^{\prime }=2y(1-y)$. Translation invariance supplies an arbitrary center $x_0$, and the explicit kink in a phi-six model isIt tends to $0$ and $1$ at the two spatial ends and saturates the bound. Its mass is consequently

For the kink in a phi-six model, choose the sector from $\varphi =0$ to $\varphi =2$ and the increasing sign of the Bogomolny equation:For $y={\varphi }^2$ this becomes the logistic differential equationAfter translating the centre to $x_0$, its solution isIt tends to $0$ as $x\to -\infty$ and to $2$ as $x\to +\infty$. Spatial reflection and $\varphi \mapsto -\varphi$ generate the other kink and antikink sectors.

Because the first-order equation saturates the Bogomolny bound, its mass is