/sci/ - Science & Math

ps. that alpha should be squared

if you sub in alpha = F(r,ø,θ) T(t) where T(t) is a function of time and F(r,ø,θ) is a a spatial function in spherical coordinates, you can simplify it to:
(1/F) Grad^2(F) = (1/alpha^2) (1/T) ∂T/∂t
then set each of those equal to k^2

This gives you your T(t) = Ae^(k^2[(alpha)^2]t)

Then set F(r,ø,θ) = R(r) Ø(ø) Θ(θ) and expand out the Grad^2(F) in spherical coordinates and use separation of variables.

This will let you get to (1/Ø) ∂^2Ø/(∂ø)^2 = J^2
From there:
Ø(ø) = Asin(Jø)+Bcos(Jø) but Ø(ø) = Ø(ø+2π) so J=m where m is an integer
Ø(ø) = Asin(mø)+Bcos(mø)

>>1141250
yes but if ø is around the z-axis, then it will not have any ø dependence so m=0 so sin(0) = 0 and cos(0) = 1 so Ø(ø)=B

yeah, then continue with the separation of variables so you get
(1/Θ)(1/sin(θ))∂/∂θ(sin(θ)∂Θ/∂θ) = k^2r^2 - (1/R)∂/∂r(r^2 ∂R/∂r)

Then use the Legendre Equation to get Θ(θ) = PL(cos(θ)) where PL is the Lth Legendre Polynomial

Holy...Shit.

set the right side of that equation with bothe the thetas and the rs equal to L(L+1) [which you got from the thetas part equaling a constant] and expand it out.

i know you have to use the method of frobenius to get the R(r) terms, and that R=∑[n=0->∞] (an)r^(n+s) where an is the nth coefficient and does not equal zero. Take two derivatives of this with respect to r and plug that into the equation.

now the part that throws me off is when you expand it all out and make them all have an r^(n+s) term in them you have ∑[n=2 -> ∞] for the k^2 term and from 0 to ∞ for all the others, the problem is, setting n=0 and solving for s, i get s=L or s=-L-1 and when i set n=1 i get s=-L-2 or s=L-1 but that means that s is a different number in each case so there is no solution, where am i going wrong?

My idea is that afterwards, find a recursive solution for a(n-2) in terms of an to solve for k^2 to put that back into the time portion of T(t) = e^(k^2[(alpha)^2]t) and also to solve for R(r)

>>1141346
Wow are you fucking serious? That shit is the easy part. Actually, this whole question is fucking easy mode. Fuck, I could probably do this in high school.

>>1141367
so whats the answer for R(r) and how do you get it?

>>1141372
Fuck you, NO HOME WORK THREADS. Idiot.

>>1141367
Agreed, heat equation is easy shit!
Also, yeah that alpha is fine the way it is, dont worry about it (it SHOULDN'T be squared for the standard heat equation)

>>1141377
lol, its not a homework problem, something i decided to work on on my own, sorry if you aren't good enough at this to help me :(

>>1141379
Trying to use reverse phschology are, we?

>>1141385
trying to spell are we? no but seriously, do you have a link to an answer for this then if its so easy?

>>1141394
Again, I'm not going to do your home work.

>>1141419
again, its not homework

anyone know this stuff and feel like helping?

alright after trying it around for abit, I suggest trying to take the dervivative and then multiply that of the function for r.

take the derivative of what?

So far i have it in the form:

-k^2 ∑[n=2 ->∞] a(n-2) r^(n+s) -L(L+1)∑[n=0 ->∞] an r^(n+s) +2∑[n=0 ->∞] an(n+s) r^(n+s) +∑[n=0 ->∞] an (n+s) (n+s-1) r^(n+s) =0

and then setting n=0 i get s=L or s= -L-1
and setting n=1 i get s= -L-2 or s= L-1
and there should be at least one common one for s in both cases

wolframalphadotcom, enter it in, or use matematica