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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2005

Paper 83

For a straight unit-winding <quantum vortex>, the real radial amplitude satisfies
$$
R''+\frac1sR'-\frac R{s^2}+(1-R^2)R=0,
\qquad R(0)=0,\quad R(\infty)=1.
$$
In the frame translating with the ring, put $Z=z-ut$ and $\Phi=e^{i\theta}(R+\phi)$, where $\phi$ is independent of the angular coordinate. The cylindrical <Laplacian> has the exact identity
$$
\nabla^2[e^{i\theta}(R+\phi)]
=e^{i\theta}\left[\frac1s\partial_s(s\partial_s(R+\phi))+\partial_Z^2\phi-\frac{R+\phi}{s^2}\right].
$$
After inserting this into the translating <Gross–Pitaevskii equation>, subtract the straight-vortex equation. Because $R$ is real,
$$
|R+\phi|^2=R^2+R(\phi+\phi^*)+|\phi|^2,
$$
and direct expansion of the nonlinear difference gives
$$
(1-|R+\phi|^2)(R+\phi)-(1-R^2)R
=(1-2R^2)\phi-R^2\phi^*-R\phi^2-2R|\phi|^2-|\phi|^2\phi.
$$
Thus the exact ring-on-line perturbation equation, including its quadratic and cubic terms, is
$$
\boxed{2iu\phi_Z=\frac1s\partial_s(s\phi_s)+\phi_{ZZ}-\frac\phi{s^2}
+\bigl[1-2R^2-R(\phi+2\phi^*)-|\phi|^2\bigr]\phi-R^2\phi^*.}
$$
The two appearances of $R|\phi|^2$ are responsible for the coefficient $2$ multiplying $\phi^*$ inside the bracket. No small-amplitude approximation is needed for this identity.

Solution

Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 83 / 1 / iv / Solution

Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 83 / 1 / iv / Solution

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Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 83 / 1 / iv / Solution

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