Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 353 / 2 / b / ii

Table of contents
- Solution ii

Solution

Under the potential condition, integrate

\partial_{i} Ψ = a_{i} / b_{i}^{2}

to obtain

Ψ = \frac{r _{1} x _{1}^{2}}{2 b _{1}^{2}} - \frac{d _{1} x _{1}^{3}}{3 b _{1}^{2}} + \frac{r _{2} x _{2}^{2}}{2 b _{2}^{2}} - \frac{d _{2} x _{2}^{3}}{3 b _{2}^{2}} - \frac{r _{1} c _{1} x _{1}^{2} x _{2}^{2}}{2 b _{1}^{2}} .

Since

V = - 2Ψ

, the zero-current steady density on the positive quadrant is

P_{ss} (x_{1}, x_{2}) = \frac{1}{Z} exp [\frac{r _{1} x _{1}^{2}}{b _{1}^{2}} - \frac{2 d _{1} x _{1}^{3}}{3 b _{1}^{2}} + \frac{r _{2} x _{2}^{2}}{b _{2}^{2}} - \frac{2 d _{2} x _{2}^{3}}{3 b _{2}^{2}} - \frac{r _{1} c _{1} x _{1}^{2} x _{2}^{2}}{b _{1}^{2}}] .

The equivalent coefficient

r_{2} c_{2} / b_{2}^{2}

may be used for the cross term.

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