# Homework Help: Help with simplifying a 2nd order pde

1. Sep 7, 2011

### lohanlotter

I was given the equation
dp/ds = 4 + 1/e*d/de(e*dp/de)

The derivatives in the equation are partial derivatives

the values of p,s,e are dimensionless numbers.

I am to assume that the solution is separable and then use finite difference method to solve for p, the finite difference method is not a problem. This is where i am having problems. What will the equation be after the assumption is made and the equation is simplified.

I have attempted the question:

I equated the right hand side = 0:

4 + 1/e*d/de(e*dp/de) = 0

and made the partial derivatives total derivatives and then applied the chain rule:

4 + 1/e*dp/de + d/de(dp/de). Is this correct????????

2. Sep 7, 2011

### Hootenanny

Staff Emeritus
I'm not sure why you have set the right hand side to zero. If you have a separable solution, this will not be the case for non-trivial solutions. The method of separation of variables should be applied as follows.

Let $p(e,s) = q(e)r(s)$. Then we have

$$\frac{\partial}{\partial s}q(e)r(s) = 4 + \frac{1}{e}\frac{\partial}{\partial e}\left(e\frac{\partial}{\partial e} q(e)r(s)\right)\;.$$

Can you take it from here?

3. Sep 7, 2011

### lohanlotter

I have attempted it again:

using the equation given i ended up with:

q*dr/ds = 4 + r(1/e*dq/de+d/de(dq/de))

what happens to the 4 then??

4. Sep 7, 2011

### Hootenanny

Staff Emeritus
Let's re-write it so that we can see what's happening more clearly:

$$q(e)r^\prime(s) = 4 + r(s)\left[\frac{q^\prime(e)}{e} + q^{\prime\prime}(e)\right]$$

This is still a separable equation, the fact that you have an additional constant term doesn't matter. If you're struggling to see it, let $R(s) = 4+r(s)$. Then $R^\prime(s) = r^\prime(s)$ leading to

$$q(e)R^\prime(s) = R(s)\left[\frac{q^\prime(e)}{e} + q^{\prime\prime}(e)\right]$$

Can you now continue?

5. Sep 7, 2011

### lohanlotter

Yes thank you very much