# Vector Equations to sys of diff eq

1. Aug 31, 2011

### nkk2008

1. The problem statement, all variables and given/known data
The problem has four very similar parts:

A)Rewrite the following vector equations as systems of differential equations:
$\frac{q}{A}=-k \nabla T$ (q is a vector) (spherical coordinates; k and A are constants)

B)Rewrite the following vector equations as systems of differential equations:
$\nabla ^{2} T + \frac{a}{k}=0$ (Cartesian coordiatnes; a and k are constants)

C)Solve the following diff eqs:

C1) $q + \frac{k}{r} \frac{d}{dr}(r \frac{dT}{dr}) =0$

q and k are constant
Hint: integrate and use the constants of integration A and B

C2) $\frac{d^{2}\varphi}{dx^{}2} + s \varphi =0$

Boundary conditions: $\frac{d\phi}{dx}+0 @ x=0 ; \phi=c @ x= \pm L$
c,L are constant s is a positive constant.
Hint: use sin and cos functions

2. Relevant equations

None that I know of

3. The attempt at a solution

I do not have one. I am thouroughly confused. I am asking a TA tomorrow, but if someone could just nudge me in the right direction before that I would be appreciative. I know this is not terribly hard, but for some reason it is stopping me.

Thanks,
Nkk

Last edited: Aug 31, 2011
2. Aug 31, 2011

### lineintegral1

Here are a few hints on some of them:

A) Think about how you can equate components together (use/derive the gradient in spherical coordinates).

B) Using linear algebra techniques, can you rewrite this system? Have you seen this done before (fairly elementary).

C) It's difficult to help you with this one if you haven't mentioned what you have tried yet. They give you an explicit hint; have you applied it?

D) This is an elementary second order differential equation of homogeneous form. If you don't know a shortcut as to how to solve this, perhaps you can guess a solution of the form $Ae^{ix}$ and solve for A. Or, choose to use only a real guess, like $A\sin x+B\cos x$ and solve for A and B by substitution.

If you have any questions, let us know.

3. Sep 1, 2011

### nkk2008

So for 1A I would just break it into three eqs like:

$\frac{q_{r}}{A}=-k \frac{d}{dr}T_{r}$

$\frac{q_{\phi}}{A}=-k \frac{1}{r} \frac{d}{d \phi}T_{\phi}$

$\frac{q_{\theta}}{A}=-k \frac{1}{r sin(\phi)} \frac{d}{d \theta}T_{\theta}$

Is that right? That was easy....

Maybe I am both overthinking them AND just plain confused.

Thanks (and I will probably be back to ask about the other ones),
Nkk