Under the potential condition, integrate $\partial_i\Psi=a_i/b_i^2$ to obtain
$$
\Psi=\frac{r_1x_1^2}{2b_1^2}
-\frac{d_1x_1^3}{3b_1^2}
+\frac{r_2x_2^2}{2b_2^2}
-\frac{d_2x_2^3}{3b_2^2}
-\frac{r_1c_1x_1^2x_2^2}{2b_1^2}.
$$
Since $V=-2\Psi$, the zero-current steady density on the positive quadrant is
$$
\boxed{P_{\rm ss}(x_1,x_2)=\frac1Z\exp\!\left[
\frac{r_1x_1^2}{b_1^2}-\frac{2d_1x_1^3}{3b_1^2}
+\frac{r_2x_2^2}{b_2^2}-\frac{2d_2x_2^3}{3b_2^2}
-\frac{r_1c_1x_1^2x_2^2}{b_1^2}
\right]}.
$$
The equivalent coefficient $r_2c_2/b_2^2$ may be used for the cross term.

Solved by gpt-5.6-sol high.


Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-353/2/b/ii/solution