Consider the \(n\)-dimensional diffusion equation or Fick’s law.

$$\frac_}=D^_=D\sum_^\frac^_}_^}$$

(A1)

In the diffusion equation, \(_\) is \(n\)-dimensional concentration (in units of mass per unit of \(n\)-dimensional volume), \(t\) is time, and \(x=_,\dots _)}^\) is a vector of Cartesian coordinates in \(n\)-dimensional space, \(D\) is the diffusion coefficient, and \(^\) is the Laplacian operator.

Consider as an initial condition for Eq. (2), a bolus dose of \(m\) injected at the origin \(x=,0\right)}^\) at time \(t=0\):

$$_(x,0+)=m\delta (x)$$

(A2)

where \(\delta (x)\) is the \(n\)-variable Dirac delta function centered at \(x=,0\right)}^\).

We show that a solution to Eqs. (A1) and (A2) is given by:

$$_\left(x,t\right)=\frac\right)}^}^^/(4Dt)}$$

(A3)

where \(r\) is the distance from the origin \(r=\sqrt^x}\).

Further we consider the solution of Eq. (A3) at \(r=\mathcal\), where \(\mathcal\) represents the effective radial distance a drug molecule must diffuse to reach the central compartment

$$_\left(t\right)=_\left(\mathcal,t\right)=\frac\frac\right)}^}^$$

(A4)

where the parameter \(\tau =}^/2D\) represents the effective diffusion time, and \(\upsilon\) is a volume of distribution \(}^\).

To show that \(_(x,t)\) given by Eq. (A3) is a solution to Eqs. (A1)–(A2), we will utilize the properties of the probability density function of multivariate normal distribution with the mean \(\mu =^\) and the variance matrix \(\Sigma =\text\left(^,\dots ,^\right)=^_\):

$$\begin pdf\left( x \right) & = \left( }}} \right)^ \frac }}e^x^ \Sigma^ x}} \\ & = \left( }}} \right)^ \frac } }}e^ }}x^ x}} \\ & = \left( } }}} \right)^ e^ }} }}}} \\ \end$$

(A5)

since \(^x=^.\) Let \(s=^\), then

$$pdf\left(x\right)=p(x,s)=}}\right)}^^^}}}$$

(A6)

If \(\upsigma \to 0+\), then \(pdf\left(x\right)\) approaches the point distribution centered at \(\mu =^\), which is the Dirac delta function. Hence:

$$p\left(x,0+\right)=\delta (x)$$

(A7)

The differentiation of \(p(x,s)\) with respect to \(s\) yields:

$$\frac=\frac\left(-\frac+\frac^}^}\right)p$$

(A8)

The differentiation of \(p(x,s)\) with respect to \(_\) yields:

$$\frac_}=-\frac_}p$$

(A9)

Hence:

$$\frac^p}_}^}=\left(-\frac+\frac_}^}^}\right)p$$

(A10)

and:

$$\sum_^\frac^p}_}^}=\left(-\frac+\frac^}^}\right)p$$

(A11)

Comparing Eqs. (A9) and (A12) we get:

$$\frac=\frac\sum_^\frac^p}_}^}$$

(A12)

Define \(C\left(x,t\right)=mp\left(x,2Dt\right).\) Equations (A7) and (A12) imply that \(p(r,s)\) is a solution to Eqs. (A1) and (A2). From Eq. (A6), we get:

$$_\left(x,t\right)=mp\left(x,2Dt\right)=\frac\right)}^}^^/(4Dt)}$$

(A13)

which proves Eq. (A3).

We make the substitutions using parameter \(\tau =}^/2D\) to represents an effective diffusion time, the parameter \(\upsilon \propto }^\) to represent a diffusion volume, and \(T=t/\tau\) as a non-dimensional time variable to obtain Eq. (A4).