DE Power Series

• Jan 11th 2011, 01:50 PM
dwsmith
DE Power Series
The coefficients of a certain power series

$\displaystyle\mbox{P(s,t)}=\sum_{m=0}^{\infty}\sum _{n=0}^{\infty}a_{m,n}s^mt^n$

satisfy

$3(m+1)a_{m+1,n}-(n+1)a_{m,n+1}+a_{m,n}=0$

and it is known that $\mbox{P(t,t)}=\exp{(2t)}$. Find $\mbox{P(s,t)}$.

$\displaystyle\mbox{P}_s=\sum_{m=1}^{\infty}\sum_{n =0}^{\infty}a_{m,n}ms^{m-1}t^n$

$\displaystyle\mbox{P}_t=\sum_{m=0}^{\infty}\sum_{n =1}^{\infty}a_{m,n}ns^{m}t^{n-1}$

$\displaystyle c_1\sum_{m=1}^{\infty}\sum_{n=0}^{\infty}a_{m,n}ms ^{m-1}t^n+c_2\sum_{m=0}^{\infty}\sum_{n=1}^{\infty}a_{ m,n}ns^{m}t^{n-1}+c_3\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{m,n }s^mt^n=0$

Is this how it should be approached?
• Jan 11th 2011, 02:27 PM
snowtea
Try $c_1=3, c_2 = -1, c_3 = 1$
• Jan 11th 2011, 02:28 PM
dwsmith
Quote:

Originally Posted by snowtea
Try $c_1=3, c_2 = -1, c_3 = 1$

Ok, so that is how it is done with those types of questions?
• Jan 11th 2011, 02:30 PM
snowtea

$
\displaystyle \sum_{m=1}^{\infty}\sum_{n=0}^{\infty}a_{m,n}ms^{m-1}t^n
= \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{m+1,n}(m+ 1)s^mt^n
$
• Jan 11th 2011, 02:31 PM
dwsmith
Quote:

Originally Posted by snowtea

$
\displaystyle \sum_{m=1}^{\infty}\sum_{n=0}^{\infty}a_{m,n}ms^{m-1}t^n
= \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{m+1,n}(m+ 1)s^mt^n
$

I understand that. I wasn't sure if doing this (see below) was the correct procedure.

$\displaystyle c_1\sum_{m=1}^{\infty}\sum_{n=0}^{\infty}a_{m,n}ms ^{m-1}t^n+c_2\sum_{m=0}^{\infty}\sum_{n=1}^{\infty}a_{ m,n}ns^{m}t^{n-1}+c_3\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{m,n }s^mt^n=0$
• Jan 11th 2011, 02:34 PM
snowtea
Quote:

Originally Posted by dwsmith
I understand that. I wasn't sure if doing this (see below) was the correct procedure.

$\displaystyle c_1\sum_{m=1}^{\infty}\sum_{n=0}^{\infty}a_{m,n}ms ^{m-1}t^n+c_2\sum_{m=0}^{\infty}\sum_{n=1}^{\infty}a_{ m,n}ns^{m}t^{n-1}+c_3\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{m,n }s^mt^n=0$

Well, it is the correct approach for this problem.

If you do not know the form it is better to take your recurrence:
$3(m+1)a_{m+1,n}-(n+1)a_{m,n+1}+a_{m,n}=0$

Multiply every term by $s^mt^n$, and do the double summation of the entire expression, then simplify.
• Jan 11th 2011, 02:36 PM
dwsmith
Now, my trouble is how do I use the initial condition?
• Jan 11th 2011, 02:38 PM
snowtea
You do realize that the summation you wrote gives you the DE:

$3P_s - P_t + P = 0$

Solve it like any other DE.
• Jan 11th 2011, 02:38 PM
dwsmith
Yes.
• Jan 11th 2011, 02:40 PM
dwsmith
I can just solve that like any other PDE than?
• Jan 11th 2011, 02:41 PM
snowtea
So do you already have the general form of the solutions for the DE?

If so, just plug-in P(t,t) to solve with initial conditions right? Perhaps I'm missing something?
• Jan 11th 2011, 02:42 PM
dwsmith
Quote:

Originally Posted by snowtea
So do you already have the general form of the solutions for the DE?

If so, just plug-in P(t,t) to solve with initial conditions right? Perhaps I'm missing something?

You aren't. I was just confused for no reason.
• Jan 11th 2011, 06:07 PM
dwsmith
I am struggling to obtain the book solution.

$\displaystyle\exp{\left(\frac{s+7t}{4}\right)}$

$P(s,t)=P(\xi\cos{\alpha}-\eta\sin{\alpha},\xi\sin{\alpha}+\eta\cos{\alpha}) =u(\xi,\eta)$

$u_{\xi}(3\cos{\alpha}-\sin{\alpha})-u_{\eta}(3\sin{\alpha}+\cos{\alpha})+u=0$

$\displaystyle\sqrt{10}u_{\xi}+u=0\Rightarrow u_{\xi}+\frac{u}{\sqrt{10}}=0$

$s=3\xi+B\eta \ \ \ t=-\xi+D\eta$

$\exp{(\xi)u=g(\eta)\Rightarrow u(\xi,\eta)=\exp{(-\xi)}g(\eta)$

I tried B = 0 and D = 1/3; and D = 0 and B = 3 but nothing is happening.

Never mind I forgot to divide out a 2.

$s=3\xi+3\eta \ \ \ t=-\xi$

$P(s,t)=\exp{(t)}g\left(\frac{s}{3}+t\right)$

$P(t,t)=\exp{(t)}g\left(\frac{t}{3}+t\right)=\exp{( 2t)}$

$P(t,t)=g\left(\frac{t}{3}+t\right)=\exp{(t)}$

$\displaystyle g(t)=\exp{\left(\frac{3}{4}t\right)}$

$\displaystyle P(s,t)=\exp{(t)}\exp{\left[\frac{3}{4}*\left(\frac{s}{3}+t\right)\right]}=\exp{\left(\frac{s+7t}{4}\right)}$