Here is a proof that the ratio of the probability density in the middle of the beam must be equal to 2x the probability density at the end of the beam assuming a trapezoidal from of the pdf and the particular “soft reflection” jump pdf near the ends of the beam.
We first note that the steady state condition implies zero net probability flux into and out of any region on the beam. We examine the flux at an infinitesimal region dx’ from the left end of the beam. The jump pdf half width is w. The grasshopper’s steady state probability density at the end of the beam is p(e) and in the middle region it is p(m).
![\text{Probability flux in} = \int_{0}^{w} [p(e) +(p(m)-p(e))\cdot \dfrac{x}{w} ]\cdot \dfrac{dx'}{w+x} \,dx \](https://s0.wp.com/latex.php?latex=%5Ctext%7BProbability+flux+in%7D+%3D+%5Cint_%7B0%7D%5E%7Bw%7D+%5Bp%28e%29+%2B%28p%28m%29-p%28e%29%29%5Ccdot+%5Cdfrac%7Bx%7D%7Bw%7D+%5D%5Ccdot+%5Cdfrac%7Bdx%27%7D%7Bw%2Bx%7D+%5C%2Cdx+%5C++&bg=ffffff&fg=0000ff&s=0&c=20201002)
Evaluating the second integral and setting in and out flux equal:
![\big|_{0}^{w} [ (2\cdot p(e)-p(m)) \cdot log(x+w) + ((p(m)-p(e))\cdot \dfrac{x}{w}] \cdot dx' = p(e) \cdot dx'](https://s0.wp.com/latex.php?latex=%5Cbig%7C_%7B0%7D%5E%7Bw%7D+%5B+%282%5Ccdot+p%28e%29-p%28m%29%29+%5Ccdot+log%28x%2Bw%29+%2B+%28%28p%28m%29-p%28e%29%29%5Ccdot+%5Cdfrac%7Bx%7D%7Bw%7D%5D+%5Ccdot+dx%27+%3D+p%28e%29+%5Ccdot+dx%27+&bg=ffffff&fg=0000ff&s=0&c=20201002)
Simplifying and eliminating dx’^2 terms:
![[(2\cdot p(e)-p(m)) \cdot log(2) + p(m)-p(e)] \cdot dx' = p(e) \cdot dx'](https://s0.wp.com/latex.php?latex=%5B%282%5Ccdot+p%28e%29-p%28m%29%29+%5Ccdot+log%282%29+%2B+p%28m%29-p%28e%29%5D+%5Ccdot+dx%27+%3D+p%28e%29+%5Ccdot+dx%27++&bg=ffffff&fg=0000ff&s=0&c=20201002)
And we see that the probability flux conservation condition is met when:
Riddler Solutions 