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\textit{2.6 minimising the Jensen-Shannon distance between $\phi(dP)$ and $\phi(\rho cdU)$}  We observed when calculating the net Jensen-Shannon distance in eq.(18) that the minimum of $\Delta$ frequently coincided with the minimum Jensen-Shannon distance $\Delta_{JS}(dP; \rho cdU)$. Since this depends directly on the measured $dP$ and $dU$ and the assumed value of $\rho c$, it is much easier to calculate. For this reason we investigate the conditions under which the two minama minima  would coincide. From the mechanistic theory, we expect $\rho cdu = dP_+ + dP_-$ and therefore compare the distributions $\phi(\rho cdU$ and $\phi(dP_+ + dP_-)$. Since $\phi(dP_+ - dP_-) = \frac{1}{2}\big(\phi(dP_+) + \phi(-dP_-)\big)$, we see that if $\phi(dP_-)$ is symmetrical then $\phi(dP_+ - dP_-) = \frac{1}{2}\big(\phi(dP_+) + \phi(dP_-)\big) = $\phi(dP_+ \phi(dP_+  + dP_-)$. Thus, $\Delta$ is the sum of the distances from $\phi(dP)$ and $\phi(\rho cdU)$ to the same distribution. By the triangular inequality This distance will be minimum when the distance between $\phi(dP)$ and $\phi(\rho cdU)$ is minimum. We have found no compelling argument that $\phi(dP_-)$ should be symmetrical. If $P$ and $U$ are periodic then we know that $\sum dP_- = 0$ if the sum is taken over an integral number of periods. This is satisfied if $\phi(dP_-)$ is symmetrical but it is a sufficient, not necessary condition. In the other systemic arteries it can be argued that the backward pressure waves arise from reflections and re-reflections from widely dispersed reflection sites which could lead to homogeneity of their distribution, the inhomogeneity due to the forward waves introduced by the contraction of the left ventricle (LV) not being present in the backward waves. Because contraction of the LV also introduces backward waves in the coronary arteries due to compression of the distal, intra-myocardial blood vessels, this argument is not as strong in the coronary arteries.