Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 81 / 1

Take positive volumetric flow rate in the direction of increasing $x$. For an arterial pressure wave, the characteristic admittance of an arterial wave givesThe backward wave contributes the opposite flow sign. Write $p_j$ for each junction's complex pressure amplitude, suppressing the common $e^{i\omega t}$ factor, and put $r=RY$, $C_{\beta }=\cos \beta$, $S_{\beta }=\sin \beta$. The topology is the two paths $K\to L\to N$ and $K\to M\to N$; their intermediate and common distal junctions each have a hydraulic resistance. The coordinate convention in the source puts the upstream end at $-\ell$ and the downstream end at zero in each segment. Consequently the wave transfer matrix of an arterial segment isEquivalently $q_u=Y(C_{\beta }{p}_u-p_d)/(i{S}_{\beta })$ and $q_d=Y(p_u-C_{\beta }{p}_d)/(i{S}_{\beta })$, with their continuous limits understood when $S_{\beta }=0$.

Symmetry gives $p_L=p_M=p_s$. Distal mass conservation and the resistive relation $Q_N=p_N/R$ giveAt $L$, the incoming flow supplies both its bed and segment $D$, soChoose the incident amplitude $P_A$ at $K$. If $P_{AR}$ is the reflected amplitude in vessel $A$, pressure continuity equation gives $p_K=P_A+P_{AR}$, while inlet flow is $2Y(P_A-P_{AR})=2Y(2P_A-p_K)$. Equating this to the two outgoing segment flows yieldsEliminating $p_s,p_K$ from these three relations gives the wave transmission through a resistively loaded arterial loop resultFor large $r$ at fixed phase, $p_N=2P_A{e}^{-2i\beta }+O(r^{-1})$ and $p_M=C_{\beta }{p}_N+O(r^{-1})$. HenceThese leading terms must not be used as nonzero denominators near a phase at which the normal intermediate flow vanishes.

Once $C$ is removed, the remaining connected segments form the chain $K\to L\to N\to M$, of total length $3\ell$. For $RY\gg 1$, each hydraulic resistance draws a small flow, so to leading order the far end at $M$ is reflecting, with zero segment flow there. Successive wave transfer matrices of an arterial segment giveThe incident-plus-reflected inlet relation remains $q_K=2Y(2P_A-p_K)$. Therefore, defining $D=3e^{3i\beta }+e^{-3i\beta }$,The modulus of $D$ equals $\gamma =|3e^{4i\beta }+e^{-2i\beta }|$, because the expressions differ by a unit-modulus factor. Comparison with the normal flows gives the amplitude ratiosFor small phases the cosines are positive and the absolute values can be removed. The original PDF prints an additional $\cos (\beta /2)$ in the middle numerator. That factor does not follow from the displayed equal-length network: the transfer relation from $M$ to $K$ above fixes $p_M=8P_A/D$, with no such factor. The corrected middle ratio is shown here.

Since ${\gamma }^2=10+6\cos 6\beta$, one has $\gamma =4[1-27{\beta }^2/8+O({\beta }^4)]$. Thus the three corrected ratios, in the order $N,M,L$, areAll three exceed one for sufficiently small nonzero $\beta$; they equal one at zero frequency. This is an oscillatory arterial wave reflection effect, not an increase in steady conductance. Removing a branch changes the standing wave pattern and the total inlet pressure associated with a fixed incident wave. Reflected waves can enhance junction pressures and hence peripheral flow amplitudes. Neither total inlet pressure nor total inlet flow was held fixed, and no extra energy source is implied.