## Introduction to Wave Transmission

The vascular system transmits pressure and flow waves that originate at a pulsatile generator - the heart.  You're most likely comfortable with the notion that the heart contracts, generates a pressure pulse, and that there's a brief delay for the pulse to be transmitted to the periphery.  What would you think if you actually were able to feel the pulse before the heart contracted?!  Does that seem like an alien concept?  It's not!  How about the fact that the pulse travels both downstream and upstream in the circulation!?  It's only due to the particular configuration of the circulation that these oddities are not more apparent in everyday clinical experience.  And there are some fascinating nuances about the circulation to be appreciated by studying wave phenomena.

## Why are there Waves?

You're already familiar with the behavior of waves in many contexts.  We use ultrasound waves on a daily basis as a diagnostic tool and many of the concepts of ultrasound wave transmission and reflection will transfer directly to a discussion of pressure and flow waves in the circulation.  But why does this happen?  I'm going to start by explaining how waves arise in the context of hemodynamics.  This is sort of a "proof", but it also demonstrates what I consider to be magic! We will see how arterial pressure waves occur simply be considering 2 basic physical principles - conservation of momentum ($$F=ma$$) and conservation of mass.

The diagram below illustrates a segment of a compliant cylindrical tube of untold length.  We'll consider a parcel of fluid within the tube suggested by the blue color.  The fluid parcel starts at axial coordinate $$z$$ and continues downstream a distance $$\Delta z$$, i.e. to coordinate $$z+\Delta z$$.

The larger arrows within the tube (pointing to the right) represent the velocity of fluid flow in the tube.  For starters we'll assume that the velocity doesn't change across the diameter of the tube (a so-called uniform velocity profile); we'll see later that this approximation isn't good enough for some purposes, but we're going for the big picture right now.  The arrows DO suggest that the velocity could change with the axial coordinate $$z$$ and we'll assume also that it changes with time.  Consequently velocity is a function of $$z$$ and $$t$$ (but not radial coordinate $$r$$ for now); we represent this with the symbol $$v(z,t)$$ for the fluid velocity.

We now write the equation of motion for the blue parcel of fluid assuming that pressure is the only force acting on it.  There is pressure acting on the parcel where it contacts the wall, but the pressure at each spot on the wall is canceled exactly by an identical spot on the opposite side of the tube.  Hence we're left with only the pressure at the upstream side of the parcel and the downstream side as exerting any net force.  Pressure multiplies the cross sectional area of the tube to obtain force.

$$\Large F=p|_{z}A-p|_{z+\Delta z}A$$

$$p|_{z}$$ is the pressure at the $$z$$ location; $$p|_{z+\Delta z}$$ is the pressure at the  $$z+\Delta z$$ location and $$A$$ is the cross-sectional area. This force is equal to the mass of the parcel ($$\rho A \Delta z$$) multiplying it's acceleration, $$\partial v/\partial t$$. $$A \Delta z$$ is the volume of the parcel and $$\rho$$ is the density (mass/volume) of the fluid.

$$\Large F=(p|_{z}-p|_{z+\Delta z})A=\rho A \Delta z \Large \frac{\partial v}{\partial t}$$

Since $$v$$ is some kind of average velocity, I'll replace it with the flow rate normalized by the cross-sectional area $$v=q/A$$ where $$q(z,t)$$ is the flow rate (e.g. mL/sec).

$$\Large F=(p|_{z}-p|_{z+\Delta z})A=\rho \Delta z \Large \frac{\partial q}{\partial t}$$

We now divide the equation by $$\Delta z$$ and allow $$\Delta z$$ to approach 0, recognizing that

$$\Large \frac{p|_{z}-p|_{z+\Delta z}}{\Delta z}$$

is the (negative) derivative of $$p$$ with respect to the axial coordinate.  We've now derived the first of 2 equations expressing conservation of axial momentum:

$$-\Large \frac{\partial p}{\partial z}=\frac{\rho}{A} \Large \frac{\partial q}{\partial t}$$

or

$$\Large \frac{A}{\rho} \frac{\partial p}{\partial z}=-\Large \frac{\partial q}{\partial t}$$

As an aside, $$\rho/A$$ will be call the inertance in this equation and is usually represented by the symbol $$L$$ . It multiplies the rate of change of flow and "resists" flow acceleration just as resistance resists flow (sort of).  Clinicians have adopted the term inertance when describing a heart valve that doesn't open very well.  I won't propose an alternate term here but I consider inertance to be a misnomer as applied in this setting since it isn't the mass of valve that's the problem.  (Words matter! They cause people to think in particular ways, right or wrong.  Ask any politician!)

The second equation comes from an expression for the conservation of mass of fluid in the tube which also relates to the pressure.  We already saw that the mass of fluid in blue is the volume multiplied by its density.

$$\Large m=\rho A \Delta z$$

The rate of change of the mass in the blue parcel is the mass flow rate into the segment minus the outflow rate.

$$\Large \rho (q|_{z}-q|_{z+\Delta z})$$

We're using $$q|_{z}$$ to represent the flow rate at axial location $$z$$ and $$q|_{z+\Delta z}$$ for the value at $$z+\Delta z$$.  The first expression becomes a rate of change by differentiating with respect to time and then the two are equal:

$$\Large \rho \Delta z \frac{\partial A}{\partial t}=\rho (q|_{z}-q|_{z+\Delta z})$$

We now introduce the compliance of the tube, $$C=dV/dP$$, as a physical parameter that relates the change in volume to a change in distending pressure.  For the moment we'll assume this is a constant that doesn't change with tube volume, i.e. the compliance is linear.  We also assume that the tube is wholly elastic, i.e. that a change in pressure results in an instantaneous  change in tube volume (no viscosity of the wall material and negligible mass).  As a matter of convenience, it's easier to work with the compliance per unit of tube length $$C'=dA/dP=C/\Delta z$$. Consequently:

$$\Large \frac{\partial A}{\partial t}=C' \frac{\partial p}{\partial t}$$

The equation now appears as:

$$\Large \Delta z C' \frac{\partial p}{\partial t}=(q|_{z}-q|_{z+\Delta z})$$

The fluid density canceled on both sides. We divide through by $$\Delta z$$ which is allowed to approach 0, and replace the right hand side with a derivative expression from the definition (calculus).

$$\Large C' \frac{\partial p}{\partial t} =-\frac{\partial q}{\partial z}$$

This is the second equation of 2 that we need to solve the problem.  Each equation involves both pressure and flow rate ( $$p(z,t)$$ and $$q(z,t)$$ ).  To eliminate the flow rate. we differentiate the first expressiion with respect to $$z$$ and the second wrt $$t$$:

$$\Large \frac{A}{\rho} \frac{\partial^2 p}{\partial z^2}=-\Large \frac{\partial^2 q}{\partial t \partial z}$$

and

$$\Large C' \frac{\partial^2 p}{\partial t^2} =-\frac{\partial^2 q}{\partial z \partial t}$$

The order of differentiation of $$q$$ is immaterial so these equations can be combined, eliminating the flow term:

$$\Large \frac{A}{\rho} \frac{\partial^2 p}{\partial z^2}=\Large C' \frac{\partial^2 p}{\partial t^2}$$

and finally

$$\Large \frac{\partial^2 p}{\partial t^2}-\frac{A}{\rho C'}\frac{\partial^2 p}{\partial z^2}=0$$

I'm assuming that the forgoing looks scary to some and this final result may look like the worst part of it.  However, this is a beautiful thing!  We've distilled a physical concept down to a single equation that explains how pressure varies with location and time in the compliant tube.  This is a partial differential equation and I can appreciate that you might not know what happens next.  But others will see this equation and recognize it; it's a linear one dimensional wave equation and someone has already worked out the solution for us.  The solution consists of TWO independent functions of $$z$$ and $$t$$ :

$$\Large p_1(z+\sqrt{\frac{A}{\rho C'}} t)$$

and

$$\Large p_2(z-\sqrt{\frac{A}{\rho C'}} t)$$

That's an answer!?  YES, it is!!  The interpretation of these equations is that the pressure solution consists of two arbitrary pressure distributions (pressure as a function of $$z$$), one propagating in the positive $$z$$ direction, the other in the negative direction.  The term $$t \sqrt{(A/\rho C')}$$ must have physical units of length (since it's being added directly to $$z$$).  Consequently $$\sqrt{(A/\rho C')}$$ is a velocity (length/time) -- the speed that the pressure wave propagates (same velocity in each direction).  The forward propagating wave (positive $$z$$ direction) is $$p_2$$.

When I say "arbitrary" functions $$p_1(z,t)$$ and $$p_2(z,t)$$, I mean the shape of the wave could be anything; the equation only describes how the wave is transmitted.  You would determine the boundary conditions depending on the specific circumstances of your particular problem.  You would have to know the pressure distribution as a function of $$z$$ at a specific moment in time (how about $$t=0$$?) and pressures at the upstream and downstream ends of the tube as a function of time going forward.  Given this information the equations describe how the pressure propagates through the tube.  While we're at it I'll make one other point: the velocity of propagation of the pressure wave is largely unrelated to the velocity of fluid flow in the tube.  As you see, the expression for the wave velocity $$\sqrt{(A/\rho C')}$$ has nothing in it about the flow velocity.  In the proximal mammalian arterial system, the wave speed is typically an order of magnitude greater than the velocity of flow!

If we'd decided to eliminate the pressure from the equations instead of the flow, we'd have gotten -- well -- the exact same equation with flow in place of pressure:

$$\Large \frac{\partial^2 q}{\partial t^2}-\frac{A}{\rho C'}\frac{\partial^2 q}{\partial z^2}=0$$

and the solution(s):

$$\Large q_1(z+\sqrt{\frac{A}{\rho C'}} t)$$

and

$$\Large q_2(z-\sqrt{\frac{A}{\rho C'}} t)$$

There are flow waves that coincide with the pressure waves, each traveling at the exact same velocity as their counterpart.  However the shape of the flow wave is not the same as the pressure wave in general.

In case you're underwhelmed by what just happened, let me point out a couple of aspects of this process which I regard as one example of the most amazing magic trick that mankind has ever devised! (The mathematical derivation of scientific knowledge.)

• Using a conceptual device only (no experiments), we extracted a physical situation onto "paper" into the form of equations.  Once this is done, we can devise new concepts merely by manipulating the equations.  IF we start with facts on paper, they remain facts through this manipulation as long as we follow the mathematical rules.  (Did we start with facts?  That's the big IF!)
• The solution of the equations told us that there are 2 pressure waves, one traveling in each direction.  WHO KNEW?  (There are 2 flow waves also.)  We derived a "new" fact about the universe using 2 known ones (conservation of mass and momentum).
• The velocity of the wave propagation drops out of the sky - a gift that not only shows us how the pressure at each location changes with time but also what physical factors the velocity depends on.

What would happen if we go to the laboratory and test this new found knowledge?  We'd be wrong.  We would see that basic concepts are correct but that there are quite a few discrepancies.  One of the big ones is that the solution above indicates that the 2 waves just keep going forever; they never change their shape or run out of energy.  The simplified model left out friction!!  We're going to take some of that into account in a second stab at this problem elsewhere.  Models are NEVER perfect.  They always invite further investigation.

### Show the Math

This is to show that the partial differential equation

$$\Large \frac{\partial^2 p}{\partial t^2}-\frac{A}{\rho C'}\frac{\partial^2 p}{\partial z^2}=0$$

has the following 2 general solutions:

$$\Large p_1(z+\sqrt{\frac{A}{\rho C'}} t)$$

and

$$\Large p_2(z-\sqrt{\frac{A}{\rho C'}} t)$$

Knowing that interest in this will be small, I admittedly sometimes put things on the website just so I'll be able to get to them myself.

We need 2 bits of calculus background information to show the above.  First is the chain rule of differentiation:

$$\Large \frac{dF(u)}{dx}=\frac{dF}{du}\frac{du}{dx}$$

where $$u$$ is a function of $$x$$.  Second is the product rule:

$$\Large \frac{d}{dx}U(x)V(x)=\frac{dU}{dx}V+U\frac{dV}{dx}$$

Let:

$$\Large u(z,t) = z+\sqrt{\frac{A}{\rho C'}}t$$

and

$$\Large v(z,t) = z-\sqrt{\frac{A}{\rho C'}}t$$

Then:

$$\Large \frac{\partial u}{\partial z}=1$$

$$\Large \frac{\partial v}{\partial z}=1$$

$$\Large \frac{\partial u}{\partial t}=\sqrt{\frac{A}{\rho C'}}$$

$$\Large \frac{\partial v}{\partial t}=-\sqrt{\frac{A}{\rho C'}}$$

Second derivatives of $$u$$ and $$v$$ are identically 0 and terms involving them aren't shown below.

$$\Large p=p_1(u)+p_2(v)$$

$$\Large\frac{\partial}{\partial z}p(z,t) = p'_1(u)\frac{du}{dz}+p'_2(v)\frac{dv}{dz}$$

$$\Large\frac{\partial^2}{\partial z^2}p(z,t) = (p''_1(u)\frac{du}{dz})\frac{du}{dz}+(p''_2(v)\frac{dv}{dz})\frac{dv}{dz}$$

$$\Large\frac{\partial}{\partial t}p(z,t) = p'_1(u)\frac{du}{dt}+p'_2(v)\frac{dv}{dt}$$

$$\Large\frac{\partial^2}{\partial t^2}p(z,t) = (p''_1(u)\frac{du}{dt})\frac{du}{dt}+(p''_2(v)\frac{dv}{dt})\frac{dv}{dt}$$

In the forgoing, primes indicate that a function has been differentiated once (') or twice (") with respect to the (single) independent variable. Substituting derivatives of $$u$$ and $$v$$ into the expressions above:

$$\Large\frac{\partial^2}{\partial z^2}p(z,t) = p''_1(u)+p''_2(v)$$

and

$$\Large\frac{\partial^2}{\partial t^2}p(z,t) = \frac{A}{\rho C'}(p''_1(u)+p''_2(v))$$

Hence the original assertion is true.

$$\Large \frac{\partial^2 p}{\partial t^2}-\frac{A}{\rho C'}\frac{\partial^2 p}{\partial z^2}=0$$