# Solving infinite ladder of differential equations using generating functions.

I am interested in solving the following infinite ladder of coupled differential equations. For any integer $$k \geq 0$$, we have a real-valued function of a single real variable, $$p_k (t)$$, which satisfies

$$\dot{p}_k(t) = (k+1)p_{k+1}(t) - kp_k(t)$$

Here, $$t \geq 0$$ ("time"), and the dot denotes a derivative. The choice of notation $$p_k$$ is intentional, as these form a set of probabilities. That is,

$$\forall t\geq 0, \quad p_k(t) \geq 0 \, \,\text{and}\,\, \sum_{k=0}^\infty p_k(t)=1$$

(One can show that the differential equations conserve this sum.) To solve this problem, I attempted to introduce a generating function of the form

$$g(z, t) \equiv \sum_{k=0}^\infty z^k p_k(t).$$

This function has the property that $$g(0, t) = 0$$ and $$g(1,t) = 1$$. Moreover, by differentiating the equation with respect to $$t$$, I found that it satisfies the following first-order, linear partial differential equation.

$$\partial_t g(z,t) + (z-1)\partial_z g(z,t) = 0$$

This seems promising to me, as I seem to have a well defined boundary value problem. Namely, letting $$t \in [0,\infty)$$ and $$z \in [0,1]$$, I set the values of $$g$$ at the boundaries $$z = 0, 1$$, and with the corresponding initial condition $$g(z,0)$$. This seems like a well-posed problem. However, I'm having trouble finding the solution. I believe the general solution to the differential equation is

$$g(z,t) = f(e^{-t}(1-z))$$

where $$f$$ is any differentiable function of a single variable. But when I try to satisfy the boundary conditions, I hit a snag. The $$z = 1$$ condition implies $$f(0) = 1$$, but the $$z = 0$$ condition implies $$f(e^{-t}) = 0$$. I'm pretty sure this breaks the camel's back: it seems to be saying $$f = 0$$ for all values!

Am I missing something? Are there modifications to this process that can lead me to a solution? Thanks in advance!

• The boundary condition $g(0,t) = 0$ can't be right. It would say $p_0(t) = 0$. But then your ladder of equations would give you $p_k(t) = 0$ for all $k$, as is easily seen by induction. Oct 18, 2021 at 3:25
• Use the Laplace transform and solve the recurrence $(s+k)P_k(s)=(k+1)P_{k+1}(s)+p_k(0)$. The solution is a polynomial fraction amenable to inversion. Oct 18, 2021 at 7:34
• $p_k(t)$ looks like a non-causal distribution. Oct 18, 2021 at 8:28
• @RobertIsrael You are absolutely right, thank you for the observation. That is certainly the source of my problems. Thanks to yours and Cesareo's comments, I can try to solve and post my own solution. Oct 18, 2021 at 13:38

Thanks to some feedback in the comments to my original post, I was able to come to a solution to the problem. As pointed out by Robert Israel, my error was assuming that $$g(0, t) = 0$$, when, in fact, there is still a contribution from the $$k = 0$$ term in the defining sum. That is, $$g(0, t) = p_0(t)$$. My general solution to the PDE above was correct, so now we have $$f(1-z) = p_0(t)$$.

The form of the function $$f$$ is going to be determined from our initial condition: $$g(z, 0) = f(1-z) = \sum_{k=0}^\infty z^k p_k(0)$$. Let's consider a power series expansion of $$f(x)$$ about $$x =0$$, with coefficients $$f_k$$.

$$f(x) = \sum_{k=0}^\infty f_k x^k$$

Plugging in $$x = 1-z$$ and reexpanding in terms of powers of $$z$$, we can express $$f_k$$ in terms of the various $$p_j(0)$$. I found the result to be

$$f_k = (-1)^k \sum_{n=k}^\infty \binom{n}{k} p_n(0)$$.

Now we may return to our solution for the generating function. I found

\begin{aligned} g(z, t) &= f(e^{-t} (1-z)) = \sum_{k=0}^\infty f_k e^{-kt}(1-z)^k \\ &= \sum_{k=0}^\infty (-1)^k \left(\sum_{n=k}^\infty \binom{n}{k} p_n(0)\right) e^{-kt}(1-z)^k \end{aligned}

Finally, I want to relate this to the functions $$p_k(t)$$, which should now solve the original ladder of differential equations. I can do this by, once more, expanding the $$(1-z)^k$$ in each term and regrouping into powers $$z^j$$. Then I can match the coefficients. Going through this process, I find that

$$\boxed{p_k(t) = (-1)^k \sum_{n=k}^\infty \binom{n}{k}e^{-nt}(-1)^n \sum_{m=n}^\infty \binom{m}{n} p_m(0)}$$.

I should double check that this answer indeed satisfies the original equation, and that it satisfies the basic properties of probabilities for all $$t$$. Perhaps it can be simplified. Nevertheless, I tested this solution for a few simple initial probability distributions $$\{p_m(0)\}$$ and they all satisfied the requirements. So I'm pretty convinced and I'm going to call that good.

I decided to finish the solution using the method I began with, but as Cesareo suggested in the comments, this problem might be easier solved by taking a Laplace transform from the beginning and solving the resulting recurrence relation.