The basic geometric model of the PAS
In order to estimate the hydraulic resistance of PASs, we need to know the various sizes and shapes of these spaces in vivo. Recent measurements of periarterial flows in the mouse brain by Mestre et al. [8] show that the PAS around the pial arteries is much larger than previously estimated—comparable to the diameter of the artery itself. In vivo experiments using fluorescent dyes show similar results [36]. The size of the PAS is substantially larger than that shown in previous electron microscope measurements of fixed tissue. Mestre et al. demonstrate that the PAS collapses during fixation: they find that the ratio of the cross-sectional area of the PAS to that of the artery itself is on average about 1.4 in vivo, whereas after fixation this ratio is only about 0.14.
The in vivo observation of the large size of the PAS around pial arteries is important for hydraulic models because the hydraulic resistance depends strongly on the size of the channel cross-section. For a concentric circular annulus of inner and outer radii \(r_1\) and \(r_2\), respectively, for fixed \(r_1\) the hydraulic resistance scales roughly as \((r_2/r_1)^{-4}\), and hence is greatly reduced in a wider annulus. As we demonstrate below, accounting for the actual shapes and eccentricities of the PASs will further reduce the resistance of hydraulic models.
Figure 1 shows images of several different cross-sections of arteries and surrounding PASs in the brain, measured in vivo using fluorescent dyes [6, 8, 36, 37] or optical coherence tomography [7]. The PAS around a pial artery generally forms an annular region, elongated in the direction along the skull. For an artery that penetrates into the parenchyma, the PAS is less elongated, assuming a more circular shape, but not necessarily concentric with the artery. Note that similar geometric models have been used to model CSF flow in the cavity (ellipse) around the spinal cord (circle) [21, 22].
We need a simple working model of the configuration of a PAS that is adjustable so that it can be fit to the various shapes that are actually observed, or at least assumed. Here we propose the model shown in Fig. 2. This model consists of an annular channel whose cross-section is bounded by an inner circle, representing the outer wall of the artery, and an outer ellipse, representing the outer wall of the PAS. The radius \(r_1\) of the circular artery and the semi-major axis \(r_2\) (x-direction) and semi-minor axis \(r_3\) (y-direction) of the ellipse can be varied to produce different cross-sectional shapes of the PAS. With \(r_2 = r_3 > r_1\), we have a circular annulus. Generally, for a pial artery, we have \(r_2 > r_3 \approx r_1\): the PAS is annular but elongated in the direction along the skull. For \(r_3 = r_1 < r_2\), the ellipse is tangent to the circle at the top and bottom, and for \(r_3 \le r_1 < r_2\) the PAS is split into two disconnected regions, one on either side of the artery, a configuration that we often observe for a pial artery in our experiments. We also allow for eccentricity in this model, allowing the circle and ellipse to be non-concentric, as shown in Fig. 2b. The center of the ellipse is displaced from the center of the circle by distances c and d in the x and y directions, respectively. Using these parameters, we have fit circles and ellipses to the images shown in Fig. 1b–f. Specifically, the fitted circles and ellipses have the same centroids and the same normalized second central moments as the dyed regions in the images. The parameters for the fits are provided in Table 1, and the goodness of these fits can be quantified via the residuals. We define \(A_{out}\) as the image area excluded from the fitted PAS shape even though its color suggests it should be included, and \(A_{in}\) as the image area included in the fitted PAS shape even though its color suggests it should be excluded. Those residuals, normalized by the PAS area, are also listed in Table 1. The model is thus able to match quite well the various observed shapes of PASs. To illustrate the fits, in Fig. 1 we have drawn the inner and outer boundaries (thin and thick white curves, respectively) of the geometric model. We have drawn the full ellipse indicating the outer boundary of the PAS to clearly indicate the fit, but the portion which passes through the artery is plotted with a dotted line to indicate that this does not represent an anatomical structure.
Table 1 Dimensional parameters, residuals, nondimensional parameters, and hydraulic resistance of our model fit to periarterial spaces visualized in vivo
Steady laminar flow in the annular tube
We wish to find the velocity distribution for steady, fully developed, laminar viscous flow in our model tube, driven by a uniform pressure gradient in the axial (z) direction. The velocity u(x, y) is purely in the z-direction and the nonlinear term in the Navier–Stokes equation is identically zero. The basic partial differential equation to be solved is the z-component of the Navier–Stokes equation, which reduces to
$$\begin{aligned} \frac{\partial ^2 u}{\partial x^2} + \frac{\partial ^2 u}{\partial y^2} = \frac{1}{\mu } \frac{dp}{dz} \equiv - C = \mathrm{constant}, \end{aligned}$$
(1)
where \(\mu \) is the dynamic viscosity of the CSF. (Note that the pressure gradient dp/dz is constant and negative, so the constant C we have defined here is positive.) If we introduce the nondimensional variables
$$\begin{aligned} \xi = \frac{x}{r_1}, \quad \eta = \frac{y}{r_1}, \quad U = \frac{u}{Cr_1^2} , \end{aligned}$$
(2)
then Eq. (1) becomes the nondimensional Poisson’s equation
$$\begin{aligned} \frac{\partial ^2 U}{\partial \xi ^2} + \frac{\partial ^2 U}{\partial \eta ^2} = - 1. \end{aligned}$$
(3)
We want to solve this equation subject to the Dirichlet (no-slip) condition \(U=0\) on the inner (circle) and outer (ellipse) boundaries. Analytic solutions are known for simple geometries, and we can calculate numerical solutions for a wide variety of geometries, as described below.
Let \(A_{pas}\) and \(A_{art}\) denote the cross-sectional areas of the PAS and the artery, respectively. Now, define the nondimensional parameters
$$\begin{aligned} \alpha = \frac{r_2}{r_1}, \quad \beta = \frac{r_3}{r_1} , \quad K = \frac{A_{pas}}{A_{art}} . \end{aligned}$$
(4)
(Note that K is also equal to the volume ratio \(V_{pas}/V_{art}\) of a fixed length of our tube model.) When \(r_1\), \(r_2\), \(r_3\), c, and d have values such that the ellipse surrounds the circle without intersecting it, the cross-sectional areas of the PAS and the artery are given simply by
$$\begin{aligned} A_{pas} = \pi (r_2 r_3 - r_1^2) = \pi r_1^2 (\alpha \beta - 1), \quad A_{art} = \pi r_1^2 , \end{aligned}$$
(5)
and the area ratio is
$$\begin{aligned} K = \frac{A_{pas}}{A_{art}} = \alpha \beta - 1. \end{aligned}$$
(6)
In cases where the ellipse intersects the circle, the determination of \(A_{pas}\) is more complicated: in this case, Eqs. (5) and (6) are no longer valid, and instead we compute \(A_{pas}\) numerically, as described in more detail below.
For our computations of velocity profiles in cases with no eccentricity (\(c = d = 0\)), we can choose a value of the area ratio K, which fixes the volume of fluid in the PAS, and then vary \(\alpha \) to change the shape of the ellipse. Thus we generate a two-parameter family of solutions: the value of \(\beta \) is fixed by the values of K and \(\alpha \). In cases where the circle does not protrude past the boundary of the ellipse, the third parameter \(\beta \) varies according to \(\beta = (K+1)/\alpha \). For \(\alpha = 1\) the ellipse and circle are tangent at \(x= \pm r_2\), \(y=0\) and for \(\alpha = K+1\) they are tangent at \(x=0\), \(y = \pm r_3\). Hence, for fixed K, the circle does not protrude beyond the ellipse for \(\alpha \) in the range \(1 \le \alpha \le K+1\). For values of \(\alpha \) outside this range, we have a two-lobed PAS, and the relationship among K, \(\alpha \), and \(\beta \) is more complicated.
The dimensional volume flow rate \(\overline{Q}\) is found by integrating the velocity-profile
$$\begin{aligned} \overline{Q} = \int _{A_{pas}} u(x, y) \, dx \, dy = Cr_1^4 \int _{A_{pas}} U(\xi , \eta ) \, d\xi \, d\eta \equiv Cr_1^4 Q , \end{aligned}$$
(7)
where \(Q = \overline{Q}/Cr_1^4\) is the dimensionless volume flow rate. The hydraulic resistance \(\overline{\mathcal {R}}\) is given by the relation \(\overline{Q} = \Delta p / \overline{\mathcal {R}}\), where \(\Delta p = (-dp/dz) L\) is the pressure drop over a length L of the tube. For our purposes, it is better to define a hydraulic resistance per unit length, \(\mathcal {R} = \overline{\mathcal {R}}/L\), such that
$$\begin{aligned} \overline{Q} = \frac{(-dp/dz)}{\mathcal {R}} , \quad \mathcal {R} = \frac{(-dp/dz)}{\overline{Q}} = \frac{\mu C}{\overline{Q}} . \end{aligned}$$
(8)
We can use computed values of Q to obtain values of the hydraulic resistance \(\mathcal {R}\). From Eqs. (7) and (8), we have
$$\begin{aligned} \mathcal {R} = \frac{\mu C}{\overline{Q}} = \frac{\mu C}{C r_1^4 Q} = \frac{\mu }{r_1^4} \frac{1}{Q}. \end{aligned}$$
(9)
We can then plot the scaled, dimensionless resistance \(r_1^4 \mathcal {R}/\mu = 1/Q\) as a function of \((\alpha - \beta )/K\) (shape of the ellipse) for different values of K (area ratio). We choose the quantity \((\alpha - \beta )/K\) because it is symmetric with respect to exchange of \(\alpha \) and \(\beta \), larger values of this quantity correspond to a more elongated ellipse, and \((\alpha - \beta )/K=\pm 1\) corresponds to the case in which the ellipse is tangent with the circle.
For viscous flows in ducts of various cross-sections, the hydraulic resistance is often scaled using the hydraulic radius \(r_{\text{h}} = 2A/P\), where A is the cross-sectional area of the duct and P is the wetted perimeter. In the case of our annular model, however, the hydraulic radius \(r_{\text{h}} = 2A_{pas}/P\) is not a useful quantity: when the inner circle lies entirely within the outer ellipse, both \(A_{pas}\) and P, and hence \(r_{\text{h}}\), are independent of the eccentricity, but (as shown below) the hydraulic resistance varies with eccentricity.
Numerical methods
In order to solve Poisson’s Eq. (3) subject to the Dirichlet condition \(U=0\) on the inner and outer boundaries of the PAS, we employ the Partial Differential Equation (PDE) Toolbox in MATLAB. This PDE solver utilizes finite-element methods and can solve Poisson’s equation in only a few steps. First, the geometry is constructed by specifying a circle and an ellipse (the ellipse is approximated using a polygon with a high number of vertices, typically 100). Eccentricity may be included by shifting the centers of the circle and ellipse relative to each other. We specify that the equation is to be solved in the PAS domain corresponding to the part of the ellipse that does not overlap with the circle. We next specify the Dirichlet boundary condition \(U=0\) along the boundary of the PAS domain and the coefficients that define the nondimensional Poisson’s Eq. (3). Finally, we generate a fine mesh throughout the PAS domain, with a maximum element size of 0.02 (nondimensionalized by \(r_1\)), and MATLAB computes the solution to Eq. (3) at each mesh point. The volume flow rate is obtained by numerically integrating the velocity profile over the domain. Choosing the maximum element size of 0.02 ensures that the numerical results are converged. Specifically, we compare the numerically obtained value of the flow rate Q for a circular annulus to the analytical values given by Eq. (11) or Eq. (12) below to ensure that the numerical results are accurate to within 1%.
For the case where the circle protrudes beyond the boundary of the ellipse, Eqs. (5) and (6) do not apply. We check for this case numerically by testing whether any points defining the boundary of the circle extend beyond the boundary of the ellipse. If so, we compute the area ratio K numerically by integrating the area of the finite elements in the PAS domain (\(A_{art}\) is known but \(A_{pas}\) is not). In cases where we want to fix K and vary the shape of the ellipse (e.g. Fig. 5a), it is necessary to change the shape of the ellipse iteratively until K converges to the desired value. We do so by choosing \(\alpha \) and varying \(\beta \) until K converges to its desired value within 0.01%.
Analytical solutions
There are two special cases for which there are explicit analytical solutions, and we can use these solutions as checks on the numerical method.
The concentric circular annulus
For a concentric circular annulus we have \(c=d=0\), \(r_2 = r_3 > r_1\), \(\alpha = \beta >1\), and \(K = \alpha ^2 -1\). Let r be the radial coordinate, and \(\rho = r/r_1\) be the corresponding dimensionless radial coordinate. The dimensionless velocity profile is axisymmetric, and is given by White [38], p. 114:
$$\begin{aligned} U(\rho ) = \frac{1}{4} \left[ (\alpha ^2 - \rho ^2) - (\alpha ^2 - 1) \frac{\ln (\alpha /\rho )}{\ln (\alpha )} \right] , \quad 1<\rho < \alpha , \end{aligned}$$
(10)
and the corresponding dimensionless volume flux rate is given by:
$$\begin{aligned} {Q} = \frac{\pi }{8} \left[ (\alpha ^4 - 1) - \frac{(\alpha ^2 - 1)^2}{\ln (\alpha )} \right] = \frac{\pi }{8} \left[ (K+1)^2 -1 - \frac{2K^2}{\ln (K+1)} \right] . \end{aligned}$$
(11)
The eccentric circular annulus
There is also an analytical solution for the case of an eccentric circular annulus, in which the centers of the two circles do not coincide [38, 39]. Let c denote the radial distance between the two centers. Then, in cases where the two circles do not intersect, the dimensionless volume flow rate is given by White [38], p. 114:
$$\begin{aligned} Q = \frac{\pi }{8} \left[ (\alpha ^4 - 1) - \frac{4 \epsilon ^2 \mathcal {M}^2}{(B-A)} - 8 \epsilon ^2 \mathcal {M}^2 \sum _{n=1}^{\infty } \frac{n \exp (-n[B+A])}{\sinh (n[B-A])} \right] , \end{aligned}$$
(12)
where \(\epsilon = c/r_1\) is the dimensionless eccentricity and
$$\begin{aligned} \mathcal {M} = (\mathcal {F}^2 - \alpha ^2)^{1/2}, \quad \mathcal {F} = \frac{\alpha ^2 - 1 + \epsilon ^2}{2 \epsilon }, \nonumber \\ A = \frac{1}{2} \ln \left( \frac{\mathcal {F} + \mathcal {M}}{\mathcal {F}-\mathcal {M}} \right) , \quad B = \frac{1}{2} \ln \left( \frac{\mathcal {F} -\epsilon + \mathcal {M}}{\mathcal {F} -\epsilon - \mathcal {M}} \right) . \end{aligned}$$
(13)
From this solution, it can be shown that increasing the eccentricity substantially increases the flow rate (see Fig. 3-10 in [38]). This solution can be used as a check on the computations of the effect of eccentricity in our model PAS in the particular case where the outer boundary is a circle.