Write
$$
V(\phi)=\frac12(q^2-\phi^2)^2(p^2-\phi^2)^2,
\qquad q>p>0.
$$
The four vacua are $-q,-p,p,q$. Hence there are three adjacent <scalar-field kink> sectors, $(-q,-p)$, $(-p,p)$ and $(p,q)$, together with their three reversed antikinks. With
$$
W'(\phi)=(q^2-\phi^2)(p^2-\phi^2),
\qquad
W(\phi)=p^2q^2\phi-\frac{p^2+q^2}{3}\phi^3+\frac15\phi^5,
$$
the <Bogomolny bound> gives $E=|W(\phi_+)-W(\phi_-)|$. The central kink obeys $\phi'=W'(\phi)$ and has
$$
E_c=\frac{4p^3}{15}(5q^2-p^2).
$$
The two outer kinks obey $\phi'=-W'(\phi)$ and have equal energy
$$
E_o=\frac{2}{15}(q-p)^3(p^2+3pq+q^2).
$$
For the kink passing through zero, choose $\phi(0)=0$. Its profile is determined implicitly by
$$
\boxed{
\frac1{q^2-p^2}\left[\frac1p\operatorname{artanh}\frac{\phi}{p}-\frac1q\operatorname{artanh}\frac{\phi}{q}\right]=x
}.
$$
If $q=p$, the central equation becomes $\phi'=(p^2-\phi^2)^2$, so
$$
\boxed{
\frac{\phi}{2p^2(p^2-\phi^2)}+\frac1{2p^3}\operatorname{artanh}\frac\phi p=x
}.
$$
For $q=p+\varepsilon$,
$$
E_o=\frac23p^2\varepsilon^3+O(\varepsilon^4),
\qquad
E_c=\frac{16}{15}p^5+O(\varepsilon),
$$
so the leading ratio is
$$
\boxed{E_o:E_c:E_o=1:\frac85(p/\varepsilon)^3:1}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 308 / 1 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-308/1/solution