Question

Find the generating function \(\varphi(t ; s)\) of the continuous time branching process with infinitesimal generating function $$ u(s)=s^{k}-s \quad(k \geq 2, \text { integer }) $$

Short answer

The generating function \(\varphi(t; s)\) of the continuous time branching process with the given infinitesimal generating function \(u(s) = s^k - s\) is: $$ \varphi(t; s) = \, \, \text{solution of }\,\int_{1}^{\varphi}\frac{d\varphi'}{k\varphi'^{k-1} - 1} = t + s - 1 $$

Step-by-step solution

Rewrite the infinitesimal generating function using its derivatives

We can rewrite the given infinitesimal generating function, $$u(s) = s^k - s$$, by expressing it in terms of its first and second derivatives. Differentiate \(u(s)\) with respect to \(s\) to get, $$ u'(s) = k s^{k - 1} - 1 $$ Further differentiate \(u'(s)\) with respect to \(s\) to get the second derivative: $$ u''(s) = k(k - 1) s^{k - 2} $$

Definition of the generating function

The generating function \(\varphi(t; s)\) of a continuous time branching process with infinitesimal generating function \(u(s)\) is defined as the solution to the partial differential equation (PDE): $$ \frac{\partial \varphi}{\partial t}(t; s) = u'\left(\varphi(t; s)\right)\frac{\partial \varphi}{\partial s}(t; s), \quad \text { with } \varphi(0; s) = s. $$ Given our expressions for \(u'(s)\) from Step 1, we now have: $$ \frac{\partial \varphi}{\partial t}(t; s) = \left[k \varphi(t; s)^{k - 1} - 1\right]\frac{\partial \varphi}{\partial s}(t; s) $$

Solve the partial differential equation

To solve the given PDE, we use the method of characteristics. Define a new parameter \(\tau\) and functions \(P(\tau) := \varphi((\tau, Q(\tau))\), and \(Q(\tau) := s - \tau\). Then, we have: $$ \frac{dP}{d\tau} = \frac{\partial \varphi}{\partial t} + \frac{\partial \varphi}{\partial s}\frac{dQ}{d\tau} $$ Plugging the PDE into this equation yields: $$ \frac{dP}{d\tau} = k P^{k - 1} - 1,\qquad \mbox{with }P(0) = \varphi(0; Q(0)) = 1 $$ This is now an ordinary differential equation (ODE) for the function \(P(\tau)\). Separate the variables: $$ \int_{1}^{P}\frac{dP'}{kP'^{k-1} - 1} = \int_{0}^{\tau}d\tau' $$ By integrating both sides, we obtain: $$ \int_{1}^{P}\frac{dP'}{kP'^{k-1} - 1} = \tau + C $$ where \(C\) is the constant of integration. To find \(C\), we use the initial condition \(P(0)=1\): $$ \int_{1}^{1}\frac{dP'}{kP'^{k-1}-1} = C \Rightarrow C = 0 $$ Thus, the solution to the ODE is: $$ \int_{1}^{P}\frac{dP'}{kP'^{k-1} - 1} = \tau $$

Obtain generating function \(\varphi(t; s)\)

Now, we need to express the generating function \(\varphi(t; s)\) in terms of the solution \(P(\tau)\) we obtained in Step 3. Recall that $$ \varphi(t; s) = P(t + Q(t)) \quad \text{ and } \quad Q(t) = s - t $$ Rewrite the equation for \(P(\tau)\) in terms of \(\varphi(t; s)\): $$ \int_{1}^{\varphi}\frac{d\varphi'}{k\varphi'^{k-1}-1} = t + s - 1 $$ Solve this equation for \(\varphi(t; s)\) to obtain the generating function of the continuous time branching process with the given infinitesimal generating function: $$ \varphi(t; s) = \, \, \text{solution of }\,\int_{1}^{\varphi}\frac{d\varphi'}{k\varphi'^{k-1} - 1} = t + s - 1 $$