Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
1  
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

1 Answer 1

up vote 2 down vote accepted

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

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

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

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

$G(s,0)=1$ :

$f(s)=1$

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.