# Solution to $\frac{\partial G}{\partial t}=(1-s)\left(-k_1G+k_2\frac{\partial G}{\partial s}\right)$

I would like to solve the PDE

$$\frac{\partial G}{\partial t}=(1-s)\left(-k_1G+k_2\frac{\partial G}{\partial s}\right)$$

subject to $G(s,0)=1$ .

The only thing I know to try is assuming $G(s,t)=f(s)g(t)$ and separating variables. With this method, I found

$$G(s,t)=\int_{-\infty}^\infty C(x)e^{xt}e^{\frac{k_1}{k_2}s}(1-s)^{-D/k_2}~\mathrm{d}x$$

However, I don't see how appropriate choice of $C(x)$ could satisfy the initial condition. What's the next step?

I'd be happy to have an answer that either solves the equation or just mentions what technique I should learn and apply, or points out a mistake I made.

The differential equation itself is an equation for the generating function of the master equation

$$\frac{\partial P(n,t)}{\partial t}=k_1(P(n-1,t)-P(n,t))+k_2((n+1)P(n+1,t)-nP(n,t))$$

defined as

$$G(s,t)=\sum_{n=0}^\infty s^nP(n,t)$$

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The solution method often used for first-order PDEs is the method of characteristics. I could write something up about it if you like. Just to clarify though -- is $s \in \mathbb{R}$ and $t \geq 0$? – Kyle Dec 9 '13 at 22:53
Yes, $s\in \mathbb{R}$ and $t \ge 0$. I'll look up the method of characteristics. – Mark Eichenlaub Dec 9 '13 at 22:59

$\dfrac{\partial G}{\partial t}=(1-s)\left(-k_1G+k_2\dfrac{\partial G}{\partial s}\right)$

$\dfrac{\partial G}{\partial t}+k_2(s-1)\dfrac{\partial G}{\partial s}=k_1(s-1)G$

$\dfrac{dt}{du}=1$ , letting $t(0)=0$ , we have $t=u$

$\dfrac{ds}{du}=k_2(s-1)$ , letting $s(0)=s_0$ , we have $s=(s_0-1)e^{k_2u}+1=(s_0-1)e^{k_2t}+1$

$\dfrac{dG}{du}=k_1(s-1)G=k_1(s_0-1)e^{k_2u}G$ , letting $G(0)=f(s_0)$ , we have $G(s,t)=f(s_0)e^\frac{k_1(s_0-1)(e^{k_2u}-1)}{k_2}=f((s-1)e^{-k_2t}+1)e^{-\frac{k_1(s-1)(e^{-k_2t}-1)}{k_2}}$

$G(s,0)=1$ :

$f(s)=1$

$\therefore G(s,t)=e^{-\frac{k_1(s-1)(e^{-k_2t}-1)}{k_2}}$

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