# Second order linear 'ODE'

1. ### mathy_girl

22
Can anyone help me to solve the following second order linear 'ODE' for V(x,s,t):

$$\frac{\partial^2 V(x,s,t)}{\partial s^2} = g(s) V(x,s,t)$$
where
$$g(s)=\frac{a^2}{B^4}+\frac{s^2 w^2}{B^2}$$.

Here, a, b and w are (real) constants.

2. ### HallsofIvy

40,203
Staff Emeritus
That is, basically,
$$\frac{d^2V}{ds^2}= (A+ Bs^2)V$$
a linear equation with variable coefficients. The standard way of solving such an equation is to use infinite series.

If
$$V= \sum_{n=0}^\infty a_ns^n$$
then
$$V'= \sum_{n=1}^\infty na_ns^{n-1}$$
$$V"= \sum_{n= 2}^\infty n(n-1)a_ns^{n-2}$$

Putting that into the equation
$$\sum_{n=2}^\infty n(n-1)a_ns^{n-2}= \sum_{n=0}^\infty Aa_n x^n+ \sum_{n=0}^\infty Ba_nx^{n+2}[/itex] In order to be able to compare "like powers", let i= n-2 in the first sum and third sums and i= n in the other second sum: If i= n-2, then n= i+ 2 and when n= 2, i= 4 so the first sum becomes [tex]\sum_{i= 4}^\infty (i+2)(i+1)a_{i+2}x^i$$
and, when n= 0, i= 2 so the third sum becomes
$$\sum_{i= 2}^\infty Aa_{i+2}x^i$$
and the equation becomes
$$\sum_{i= 4}^\infty (i+2)(i+1)a_{i+2}x^i= \sum_{i=0}^\infty Aa_i x^i+ [tex]\sum_{i= 2}^\infty Ba_{i+2}x^i$$[/itex]
For i= 0, that gives $Aa_0= 0$. For i= 1, $Aa_1= 0$. For i= 2, $Aa_2+ a_{3}= 0$ so $a_3= 0$. For i= 3, $Aa_3+ a_{4}= 0$ so $a_4= 0$. For i> 3, $(i+2)(i+1)a_{i+2}= Aa_i+ Ba_{i+ 2}$ or $[(i+2)(i+1)- B]a_{i+2}= Aa_i$.

(Better check my arithmetic!)

Last edited by a moderator: Jul 25, 2009
3. ### mathy_girl

22
I forgot to say, the problem is on an infinite domain, $$-\infty < s < \infty$$. Can we still use this expansion in a series like you do?

4. ### vaibhav1803

69
@mathy_girl
the infinte series is domain independant,flexible, mathod to solve a DE but it is a very painful task
did you try substitutions for V(try thinking a kind of function that coul give this PDE.?my claim is it should be the product of three fuctions..in x,s,t)
wat do u say HallsOfIvy.?

5. ### Thaakisfox

263
This equation in the variable s is basically one of the Weber equations, whose solutions are the parabolic cylinder functions.

So the solution to your equation are the parabolic cylinder functions. These have connections with Bessel functions, and hermite polynomials, and can be expressed using the confluent hypergeometric function.

Let me guess, that you got this equation from writing Laplace's equation in parabolic cylindrical coordinates

a good reference for the harmonics in the 11 separable coordinate systems is Morse-Feshbach: Methods of theoretical physics, another book concerning these kind of functions is: Erdelyi: Higher transcendental functions (this is a three volume tome..)

6. ### HallsofIvy

40,203
Staff Emeritus
Certainly that would be the standard method of solving a partial differential equation- though the result would then typically be an infinite sum of such solutions.

But "x" and "t" don't really figure in this. What was given was really an ordinary differential equation in s. There are no derivatives with respect to x and t so they just "go along for the ride"- they can be treated as constants- so separation of variables is irrelevant.

7. ### vaibhav1803

69
OK apologies for my vague and torturous answer...xDD
(hey can u help me solve the Schrödinger the ie colatitude and radial equations im stuck badly..??)

8. ### mathy_girl

22
Nope, I didn't..

9. ### vaibhav1803

69
did you get this while doing cylider harmonics..?

10. ### vaibhav1803

69
ohyeah..!!..i got the solution its really simple...
you dont even need 2 do a self torture by the infinite series..
didn't you try inspection..??
as in a function giving back on two diferentials: (depressed quadratic)*(dependant function)..with no -ve coefficients...??..doesn't it seem too peculiar a format..?
thus "V" is of the undying kind..hence obviously not a polynomial or anything close to it..this either exponential or fractional power..thats all of the solution i want to tell the remaining is just 4 you to learn...(HAHAHAHA)
$$V=(1+s)^{-n}$$
and also
$$V=\alpha e^{f(s)}$$
..the function f(s) is an obvious guess in this case but do try..xD
ciao if u need any oder help relating 2 this then contact me
and i bet ul be laughin at urself after solving this DE...the solution is soo obvious

11. ### mathy_girl

22
Then why does Maple give me such complicated solutions? Some Whittaker M function..? I don't really get it...

12. ### vaibhav1803

69
but did it work...?the solution is the exponential one..$$f(s)=As^{2}$$
im just a highschool kid i may be wrong after all..?
do i mail u my solution to this one..?
$$f(s)=As^{2}$$
$$D_{2}(e^{As^{2}})= (A+2(As)^{2})e^{As^{2}$$
"D" is d/dx and D2 is 2nd derivative
so we get our condition that solutions are possible IF $$B=2A^{2}$$

Last edited: Jul 27, 2009
13. ### Mute

1,391
Perhaps Maple gives you such complicated solutions because those are the solutions. Just because someone claims to have found the solution and that "it's obvious" doesn't mean it is! You should check their claims yourself!

In this case, you should actually have listened to Thaakisfox, and looked up Parabolic Cylinder Functions. If you go to the wikipedia article,

http://en.wikipedia.org/wiki/Parabolic_cylinder_functions

you'll notice that the DE there looks remarkably like yours.

vaibhav1803's exponential solution only works if the coefficients A and B in (A + Bs^2) are related in a specific manner. If $f(s) = Cs^2$, you find $4C^2 = B$ and $2C = A$, which implies $B = A^2$. Unless the constants in your equation just happen to satisfy this condition, then this solution fails. (The more complicated solution should, of course, reduce to such a solution in when this condition is satisfied).

14. ### mathy_girl

22
Thanks for your explanation, I already didn't believe it would be that simple. In my case A and B are undetermined constants, so the solution in case the relation $B = A^2$ holds, cannot be used.

I'll take a look at those Parabolic Cylinder Functions :-).