Solution of the Delay Differential Equation

A more satisfying approach is Rényi's solution to the delay differential equation derived from the master equation (see ¹). We start with the familiar form:

$$ \begin{equation} M(x + 1) = 1 + \frac{2}{x} \int_{0}^{x} M(t) dt \end{equation} $$

then multiply both sides by $x$:

$$ x M(x + 1) = x + 2 \int_{0}^{x} M(t) dt $$

and differentiate the result with respect to x:

$$ \begin{equation} x M^{\prime}(x + 1) + M(x + 1) = 1 + 2 M(x) \quad \text{for} \quad x > 0 \end{equation} $$

We now consider the following:

$$ \begin{equation} \varphi(s) = \int_{0}^{\infty} M(x) e^{-sx} dx \end{equation} $$

which is the Laplace transform of $M(x)$. We will show that $\varphi(s)$ takes the form:

$$ \begin{equation} \varphi(s) = \frac{e^{-s}}{s^2} \int_{s}^{\infty} \exp \left( -2 \int_{s}^{t} \frac{1 - e^{-u}}{u} du \right) dt \end{equation} $$

Returning to equation $2$ with the initial condition:

$$ M(x) = 0 \quad \text{for} \quad 0 \leq x < 1 $$

we multiply across by $e^{-sx}$ and integrating:

$$ \begin{align*} \int_{0}^{\infty} x M^{\prime}(x + 1) e^{-sx} dx + \int_{0}^{\infty} M(x + 1) e^{-sx} dx & = \int_{0}^{\infty} e^{-sx} dx + \int_{0}^{\infty} 2 M(x) e^{-sx} dx \\ & = -\frac{ e^{-sx} }{s} \bigg\vert_{x = 0}^{\infty} + 2 \int_{0}^{\infty} M(x) e^{-sx} dx \\ & = \frac{1}{s} + 2 \varphi(s) \end{align*} $$

We take each part of the left hand side separately:

$$ \begin{align*} \int_{0}^{\infty} M(x + 1) e^{-sx} dx & = \int_{1}^{\infty} M(t) e^{-s(t - 1)} dt \\ & = e^{s} \int_{1}^{\infty} M(t) e^{-st} dt \\ & = e^{s} \varphi(s) \end{align*} $$

making use of our initial condition. Looking at the other integral, and integrating by parts:

$$ \begin{align*} \int_{0}^{\infty} x M^{\prime}(x + 1) e^{-sx} dx & = - \frac{d}{ds} \left( \int_{0}^{\infty} M^{\prime}(x + 1) e^{-sx} dx \right) \\ & = - \frac{d}{ds} \left( M(x + 1) e^{-sx} \bigg\vert_{x = 0}^{\infty} + s \int_{0}^{\infty} M(x + 1) e^{-sx} dx \right) \\ & = - \frac{d}{ds} \left( - M(1) + s e^{s} \varphi(s) \right) \\ & = - \frac{d}{ds} \left( s e^{s} \varphi(s) \right) \end{align*} $$

so our delay differential equation becomes:

$$ \begin{align*} - \frac{d}{ds} \left( s e^{s} \varphi(s) \right) + e^{s} \varphi(s) & = \frac{1}{s} + 2 \varphi(s) \\ - \frac{d}{ds} \left( s e^{s} \varphi(s) \right) & = \frac{1}{s} + 2 \varphi(s) - e^{s} \varphi(s) \\ \frac{d}{ds} \left( s e^{s} \varphi(s) \right) & = \varphi(s) ( e^{s} - 2 ) - \frac{1}{s} \end{align*} $$

we now make the substitution $w(s) = e^{s} \varphi(s)$. Our equation becomes:

$$ \begin{align*} \frac{d}{ds} \left( s w(s) \right) & = w(s) ( 1 - 2 e^{-s}) - \frac{1}{s} \\ w(s) + s w^{\prime}(s) & = w(s) - 2 w(s) e^{-s} - \frac{1}{s} \\ s w^{\prime}(s) & = - 2 w(s) e^{-s} - \frac{1}{s} \end{align*} $$

solving this inhomogeneous equation gives us:

$$ w(s) = \frac{1}{s^2} \int_{s}^{\infty} \exp \left( -2 \int_{s}^{t} \frac{1 - e^{-u}}{u} du \right) dt $$

which is another form of equation $4$.

The Computation of the Limit

R'enyi provides the following theorem:

Theorem. One has the limit $$ \lim_{x \to \infty} \frac{M(x)}{x} = C_R $$ with $$ C_R = \int_{0}^{\infty} \exp \left( -2 \int_{0}^{t} \frac{1 - e^{-u}}{u} du \right) dt $$

In order to compute the limit, R'enyi makes use of the following Tauberian theorem (stated without proof):

Proof. If we re-arrange equation $4$ we have:

$$ s^2 \varphi(s) = e^{-s} \int_{s}^{\infty} \exp \left( -2 \int_{s}^{t} \frac{1 - e^{-u}}{u} du \right) dt $$

taking the limit as $s \to 0$:

$$ \begin{align*} \lim_{s \to 0} s^2 \varphi(s) & = \int_{0}^{\infty} \exp \left( -2 \int_{0}^{t} \frac{1 - e^{-u}}{u} du \right) dt \\ & = C_R \end{align*} $$

from the definition of $\varphi(s)$ ( equation $3$ ):

$$ \int_{0}^{\infty} M(x) e^{-sx} dx = \frac{C_R}{s^2} $$

as $s \to 0$. Applying the above theorem, we get:

$$ \begin{align*} \int_{0}^{x} M(t) dt & = \frac{C_R x^2}{\Gamma(3)} \\ & = \frac{C_R x^2}{2} \\ \frac{2}{x^2} \int_{0}^{x} M(t) dt & = C_R \end{align*} $$

as $x \to \infty$. Returning to the familiar equation $1$ and dividing across by $x$, and taking the limit as $x \to \infty$, we have:

$$ \begin{align*} \lim_{x \to \infty} \frac{M(x)}{x} & = \lim_{x \to \infty} \left( \frac{1}{x} + \frac{2}{x (x - 1)} \int_{0}^{x - 1} M(t) dt \right) \\ & = C_R \end{align*} $$

which is the required limit. $\square$

Remarks

While certainly more interesting than Weiner's approach - indeed an expression for the jamming limit is derived - this author is still left feeling dissatisfied by the approach taken. The Laplace transform of the master equation is introduced without any of it's specific features being used - for example, the reduction of an analytic problem to an algebraic problem, and no inverse step is performed - it is merely used as a convenient form in order to apply a Tauberian theorem, and hence the reference to the Laplace Transform could have been omitted.