# Thread: DE Power Series

1. ## 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?

2. Try $c_1=3, c_2 = -1, c_3 = 1$

3. 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?

4. Same as your previous question:

$
\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
$

5. Originally Posted by snowtea
Same as your previous question:

$
\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.

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

6. 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.
It becomes the form of the summation you are asking about.

7. Now, my trouble is how do I use the initial condition?

8. You do realize that the summation you wrote gives you the DE:

$3P_s - P_t + P = 0$

Solve it like any other DE.

9. Yes.

10. I can just solve that like any other PDE than?

11. 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?

12. 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.

13. 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.

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

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

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

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

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

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