Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 313 / 1

A classical field-theory soliton is a smooth, spatially localized, finite-energy solution which retains its identity under time evolution and is stable against small perturbations, commonly because of a topological charge or a balance between energy terms with different scaling behavior.

For a static field writewhereUnder the Derrick scaling ${\varphi }_{\lambda }(\mathbf{x})=\varphi (\lambda \mathbf{x})$, a change of variables givesA static solution must be stationary under this variation, so the Derrick virial identity isAll three energies are nonnegative. For $D\ge 4$, every coefficient is nonpositive and the coefficient of the strictly positive $E_2$ of any nonconstant field is negative. The identity is impossible. Thus, when the quartic-gradient term is available,For $D=2$ or $3$, the $E_4$ term has the opposite sign to at least one other term and the Derrick theorem does not rule out a soliton. If $\kappa =0$ from the outset, the identity reduces to $(2-D)E_2-D{E}_0=0$, recovering the stronger standard obstruction for $D\ge 2$.

The potential factorizes asIt is even and nonnegative, with three degenerate vacua at $\varphi =-2,0,2$. Its two intervening maxima occur at $\varphi =\pm 2/\sqrt{3}$ and have height $256/27$.

A finite-energy static solution in one dimension satisfiesMultiplication by ${\varphi }^{\prime }$ and use of the vacuum boundary conditions gives the first integralAny path from a negative vacuum to a positive vacuum must pass through the intermediate vacuum $\varphi =0$. There both $U$ and ${\varphi }^{\prime }$ vanish. The Picard-Lindelof theorem then forces a solution reaching $(\varphi ,{\varphi }^{\prime })=(0,0)$ at finite $x$ to remain there. Equivalently, the first-order orbit approaches $\varphi =0$ only as $|x|\to \infty$. This intermediate-vacuum obstruction to a kink means that a single kink cannot connect $-2$ to $2$; it splits into two elementary kinks at infinite separation.

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