Shock Waves in the Aorta

Although the linear wave equation can be used to gain an understanding of many features of the arterial pulse, such as reflected waves and waves in an arterial network (Lighthill, 1975), there are experimental indications that nonlinear effects are also important (Anliker et al., 1971a,b). One particular nonlinear effect that we investigate here is the steepening of the wave front as it moves away from the heart. If the wave front becomes too steep, the top of the front overtakes the bottom, and a shock, or discontinuity, forms, a solution typical of hyperbolic equations. Of course, physiologically, a true shock is not possible, as blood viscosity and the elastic properties of the arterial wall preclude the formation of a discontinuous solution. Nevertheless, it might be possible to generate very steep pressure gradients within the aorta.

Under normal conditions, no such shocks develop. However, in conditions where the aorta does not function properly, allowing considerable backflow into the heart, the heart compensates by an increase in the ejection volume, thus generating pressure waves that are steeper and stronger than those observed normally. Furthermore, the pistol-shot phenomenon, a loud cracking sound heard through a stethoscope placed at the radial or femoral artery, often occurs in patients with aortic insufficiency. It has been postulated that the pistol-shot is the result of the formation of a shock wave within the artery, a shock wave that is possible because of the increased amplitude of the pressure pulse.

To model this phenomenon, recall that the governing equations are c(Pt + uPx ) + A(P)ux = 0, (15.147)

which can be written in the form wt + Bwx = 0, (15.149)

1

u

P

A(P)

u

c

Using the method of characteristics (Whitham 1974; Pedley, 1980; Peskin, 1976), we can determine some qualitative features of the solution. Roughly speaking, a characteristic is a curve C in the (x, t) plane along which information about the solution propagates. For example, the equation ut + cux = 0 has solutions of the form u(x, t) = U(x — ct), so that information about the solution propagates along curves x - ct = constant in the (x, t) plane. Similarly, characteristics for the wave equation (15.144) are curves of the form t ± x/s = constant, because it is along these curves that information about the solution travels.

To find characteristic curves, we look for curves in x, t along which the original partial differential equation behaves like an ordinary differential equation. Suppose a characteristic curve C is defined by x = x(X), t = X.

Derivatives of functions w(x, t) along this curve are given by dw dx

Notice that with dx/dX = c, the partial differential equation ut + cux = 0 reduces to the simple ordinary differential equation uX = 0. Thus, curves with dx/dt = c are characteristic curves for this simple equation.

To reduce the system (15.149) to characteristic form, we try to find appropriate linear combinations of the equations that transform the system to an ordinary differential equation. Thus, suppose the matrix B has a left eigenvector £T with corresponding eigenvalue s, so that £TB = sfT. If we multiply (15.149) by fT, we find that with the identification dx/dX = s,

In other words, along the curve dx/dt = s, the original system of equations reduces to the simple ordinary differential equation fTwX = 0.

It is an easy matter to determine that the eigenvalues of B are s = u ± K(P), (15.155)

where

V PC

with corresponding left eigenvector fT = (£,2) = (PK(P), ±1). (15.157)

It follows from fTwX = 0 that uX ±—^PX = 0 (15.158)

along the characteristic curve dx/dX = u ± K(P), which we denote by C±. Now, notice that d 11

so that d

In other words, u + 2K(P) is conserved (remains constant) along C+, the characteristic curve with slope dx/dt = u+K(P), and u-2K(P) is conserved along C-, the characteristic curve with slope dx/dt = u — K(P).

Now, to see how this reduction allows us to solve a specific problem, consider the region x > 0, t > 0 with u(0, t) > 0 specified. For example, u(0, t) could be the velocity

Steps Action Potential
velocity generated by the heartbeat Figure 15.15 Diagram of the characteristics of the arterial pulse equations in the (t, x) plane.

pulse generated by a single heartbeat. We suppose that initially, u(x, 0) = 0, P(x, 0) = P0 for all x > 0, where P0 is the diastolic pressure.

Pick any point A in the region x > 0, t > 0 (Fig. 15.15). There are two characteristics passing through A, one, C+, with positive slope u + K(P) and one, C_, with negative slope u _ K(P). (Here and in the following we assume that u is small enough so that C_ always has negative slope.) Following C_ up and to the left, we see that it intersects the vertical axis, where u = 0 and P = P0 (because of the specified initial data). Since the quantity u _ 2K(P) is conserved on C_, it must be that u _ 2K(P) = _2K(P0) at the point A. Since A is arbitrary, it follows that u = 2K(P) _ 2K(P0) everywhere in the first quadrant. Thus, we know that u + 2K(P) = 4K(P) _ 2K(P0) is constant along C+. Hence, K(P) is conserved along C+, as are both P and u, so that C+ is a straight line. The slope of C+ is the value of u + K(P) at the intersection of C+ with the horizontal axis.

To be specific, suppose u(0, t) first increases and then decreases as a function of t, as shown in Fig. 15.16A. (In this figure u(0, t) is shown as piecewise linear, but this is simply for ease of illustration.) To be consistent, since u = 2K(P)_2K(P0) everywhere in the first quadrant, K(P(0, t)) = K(P0)+1 u(0, t), so that the slope of the C+ characteristics is s(t) = 2u(0, t) + K(P0), which also increases and then decreases as a function of t. With increasing slopes, the characteristics converge, resulting in a steepening of the wave front. If characteristics meet, the solution is not uniquely defined by this method, and shocks develop.

The place a shock first develops can be found by determining the points of intersection of the characteristics. Suppose we have two C+ characteristics, one emanating from the t-axis at t = t\, described by x = s(tO(t _ tO, and the other emanating at t = t2,

Steps Action Potential
Figure 15.16 A: Sketch of u(0,t). B: Characteristics generated by u(0,t) in the previous figure. C: Plots of u(x,t) for x = 0 and x = x0, obtained by taking cross-sections for a fixed x (as indicated by the dotted line in B).

described by x = s(t2)(t - t2). They intersect at any point (ti,xi), where s(t2)t2 - s(t1)t1

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