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The differential equations are :

$$ ( n_{j,k,0} )'(x) = - \frac {jn_{j,k,0}(x)} {a-x}, $$

$$ ( n_{j,k,b} )'(x) = \frac { (j-b+1)n_{j,k,b-1}(x) - (j-b)n_{j,k,b}(x) } {a-x}, $$

for $ 0\lt b\lt c $.

$$ (f_{j,k})'(x) = \frac{ (j-c+1)n_{j,k,c-1}(x) }{a-x}.$$ Here, the second equation holds with $0\lt b\lt c$.

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just want to comment that the second equation holds with all 0 less than b and b less than c. Sorry that I donot know how to type it here. –  nana Jun 23 '11 at 1:47
as stated, this is unclear. For what $x,b,k,j$ do you seek the solution? Is $x$ allowed to equal $a$? are the $b$ real-valued, or assumed integers? I'm assuming the latter since you are using it as an index. If not, this looks like a functional differential equation. Also, you need additional conditions on $n$ and $f$ to specify them. –  Nilima Nigam Jun 23 '11 at 2:28
Thanks very much for your comment. Appreciate that ! I seek the solution for $x$. $ n_{j,k,0} $, $n_{j,k,b}$ and $f_{j,k}$ are all functions with respect of $x$. $x$ is a variable. Except $x$, all others are supposed to know. $b$ is an integer. Thanks again ! –  nana Jun 23 '11 at 18:58
I will state the question again : $$(n_{j,k,0}'(\tau))= \frac{jn_{j,k,0}(\tau)}{\lambda -\tau}.$$ $$(n_{j,k,\theta})'(\tau)=\frac{(j-\theta +1)n_{j,k,\theta -1}(\tau)-(j-\theta)n_{j,k,\theta}(\tau)}{\lambda - \tau}$$ for $(0<\theta < \Omega).$ –  nana Jun 23 '11 at 19:40
and $$f_{j,k}'(\tau) = \frac{ (j-\Omega +1)n_{j,k,\Omega -1}(\tau) }{\lambda - \tau}.$$ with $\tau \in [0,\lambda)$, and the initial conditions: $n_{j,k,0}=(1-\alpha)P(j,k),$ $n_{j,k,\theta}(0)=0$ for $0<\theta <\Omega$, and $f_{j,k}(0)=\alpha P(j,k).$ –  nana Jun 23 '11 at 19:44

1 Answer 1

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If b is an integer, you can simply solve the equations successively, starting with b=0. If b is supposed to be real, you need to explain what $n_{j,k,b-1}$ is supposed to mean for b between 0 and 1. Also, what is the point of the index k? Nothing in the equations depends on k!

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Thanks very much for your comment. $b$ is an integer, I do not quite understand how to solve it successively. index $k$ could be anything, although it really does not matter too much here. Can just assume it is fixed. –  nana Jun 23 '11 at 19:00
By solving successively, I mean that you first solve the equation for b=0, then you use that solution to solve for b=1, etc. At each step, you only need to solve a linear equation. –  Michael Renardy Jun 23 '11 at 20:34
Thanks very much ! Michael –  nana Jun 25 '11 at 18:22

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