In the process of solving a continuous-time Markov chain problem, I derived the following PDE for the probability distribution $P_0(t,T)$, $t\in[0, T], T\in \mathbb{R}$: \begin{align} \frac{\partial^2 P_0}{\partial T^2} + \frac{\partial^2 P_0}{\partial t \, \partial T} + (k_{\text{on}} + k_{\text{off}}) \frac{\partial P_0}{\partial T} + k_{\text{on}} \frac{\partial P_0}{\partial t} = 0. \end{align} The rates $k_{\text{on}}$ and $k_{\text{off}}$ are positive numbers. This equation is augmented by the condition \begin{align} P_0(0,T) = p_0^{\text{ss}} e^{-k_{\text{on}}T}. \end{align} I don't have much experience solving PDEs, so I looked up some worked out examples and following their approach reduced the equation into its canonical form given by \begin{align} \frac{\partial^2 P_0}{\partial \eta \partial t} + (k_{\text{on}} + k_{\text{off}}) \frac{\partial P_0}{\partial \eta} + k_{\text{on}} \frac{\partial P_0}{\partial t} = 0, \end{align} where $\eta = T-t$. Then, I used the separation of variables approach ($P_0(t, \eta) = Q(t)E(\eta)$), which is what I would normally do to solve, say, the diffusion equation. Unfortunately, the solution I get leads to unphysical results when I impose the 'initial' condition. Skipping some of the steps, I find that \begin{align} Q(t)E(\eta) = C \exp\left( -\frac{k_{\text{on}} + k_{\text{off}}}{1+c_e}t + \frac{k_{\text{on}}}{c_e} (T-t) \right), \end{align} where $C$ and $c_e$ are constants. Imposing the condition leads to $C=p_0^{\text{ss}}$ and $c_e=-1$. This second number is what blows the solution. It suggests that $1+c_e=0$ and the denominator in the exponent makes the whole expression diverge.

Any suggestions on how to approach this problem? Can it be that the separation of variables does not apply in this case? What other methods can I try? Thanks in advance!

[Edit] There is also a second condition that can be helpful in solving this: $P_0(T,T)=0$.

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    $\begingroup$ An interesting problem to be sure. Take the time to study it and understand it and, as always, check your work, since your edit that $P(T, T) = 0$ suggests something's amiss somewhere, and probably worth your time tracking down. Why? --- Because if $P_0(T, T) = 0$ and $P_0(0, T) = p_0^{\text{ss}} e^{-k_{\text{on}}T}$, what does $P_0(0, 0)$ tell you about $p_0^{\text{ss}}$? $\endgroup$ Commented Feb 14, 2021 at 3:29
  • $\begingroup$ Hmm.. I see. Great point! Yeah, looks like there's something deeper going on. These conditions I stated from my intuitive understanding of the problem, but my intuition may be flawed. I'll keep digging. $\endgroup$ Commented Feb 14, 2021 at 3:38
  • $\begingroup$ Dig like the Devil's after you, it'll pay dividends. $\endgroup$ Commented Feb 14, 2021 at 3:40

1 Answer 1


Simplifying your notation to make the typing a bit easier you have $$P_{TT} + P_{tT} + (k_1 + k_2)P_T + k_1P_t = 0$$ subject to $P(0, T) = p_0 e^{-k_1 T}$. You're on the right track separating the variables, $P(t, T) = A(t)B(T)$.

Doing this I get $B' + cB = 0$, where $c$ is some constant, and $ A'' + (k_1 + k_2 - c)A' - ck_1 A = 0$. Using the initial condition, the former tells us $c = k_1$, so the latter becomes $A'' + k_2 A' - k_1^2 = 0$.

Solve the characteristic equation associated with this ordinary differential equation. You'll see it has roots $-\tfrac{1}{2}k_2 \pm \tfrac{1}{2}\sqrt{k_2^2 + 4k_1^2}$. Since your problem seems to deal with probablility distributions I imagine you'll want to pick the negative root so that $A \rightarrow 0$ as $t \rightarrow \infty$, so with $\kappa = \tfrac{1}{2}k_2 + \tfrac{1}{2}\sqrt{k_2^2 + 4k_1^2}$ your solution is $$ P(t, T) = p_0 e^{-k_1T} e^{-\kappa t}. $$

  • $\begingroup$ I followed the strategy you proposed (separating the variables right away) but the equations I got have A(t) and B(T) swapped. I think in your notation it should have been P(t,T)=A(T)B(t). Do you agree? With this change, I get to a similar problematic situation where imposing the initial condition on one of the solutions results in an equation that has no solution for c, namely, 2 k_1 k_2 = c * 0. I'm thinking that maybe I need to be looking for some kind of a series solution. I realized that there is also a second boundary condition that can be useful (P(T,T)=0) but not sure in what way yet. $\endgroup$ Commented Feb 14, 2021 at 2:59
  • $\begingroup$ Best of luck and God bless, and in the meantime the solution above might be helpful. $\endgroup$ Commented Feb 14, 2021 at 3:03
  • $\begingroup$ I appreciate you taking the time to look at this! $\endgroup$ Commented Feb 14, 2021 at 3:06

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