Talk:Advanced Mathematics for Engineers and Scientists/Parallel Plate Flow: Easy IC

Negative Constant
It isn't /completely/ arbitrary, because one needs to rule out that they might equal a positive constant or zero, leaving only the negative constant case shown in this module. Well, it /can/ be a positive or zero constant; but this leaves you with the trivial solution. Fephisto 11:29, 18 December 2006 (UTC)

In context, how the constant is described is arbitrary. -k2 will work, so will k or k2 or k-300. Leaving out the negative sign would lead to complex numbers which nobody seams to like.

Ben pcc 04:42, 19 December 2006 (UTC)

I'm talking about clarifying the k>0 for your -k^2 used, and the reason behind this without losing generality.

You want to avoid getting the trivial solutions:


 * u_t=u_xx
 * u(0,t) = 0
 * u(1,t) = 0
 * T'/T=(d^2X/dx^2)/X=A^2 (say A>0)
 * (d^2X/dx^2)-AX = 0 -> X=0, thus we don't assume the constant is positive because we aren't looking for the trivial solution
 * (d^2X/dx^2)/X=-A^2 -> X!=0 which is a non-trivial solution, which is what we're looking for. Fephisto 10:51, 19 December 2006 (UTC)

"If we took A = 0, the solution would have just been u(y, t) = 0, which would satisfy the BCs and the PDE but couldn't possibly satisfy the IC. So, we take k = nπ instead, where n is any integer. So far: [...]"

In this first example you don't construct series, you determine that A is not zero and so k = nπ. n is later identified. No trivial solution; the IC picks n.

Ben pcc 17:22, 5 January 2007 (UTC)

You're right, in this case the IC takes care of it. I took my PDE class with a bunch of engineers, so it was rather cookbookish: check that the sln. with k>0, k<0, k=0. Fephisto 03:16, 14 January 2007 (UTC)

Sides are Constant Proof
Fephisto, I'm afraid I removed your proof because it was wrong. Look:


 * $$\frac{\partial}{\partial t} \left(\frac{T'(t)}{\nu T(t)}\right) = \frac{\partial}{\partial t} \left(\frac{Y''(y)}{Y(y)}\right)\,$$


 * $$\frac{\partial}{\partial t} \left(\frac{T'(t)}{\nu T(t)}\right) = 0\,$$


 * $$\frac{T'(t)}{\nu T(t)} = f(y)\,$$

The integration performed with respect to t produces a constant with respect to t, in other words a function of y.

Thanks for the many fixes. I always thought it was spelled "independant". I'll be careful not to use "unkown" also.

Ben pcc 00:51, 7 January 2007 (UTC)

Spelling is pretty minor, don't worry about it.

Huh, I'm kind of skeptical of this whole thing now, because the above proof I gave you is the /exact/ proof my old textbook gave me (I had, in fact, copied it directly from there). And, no offense, I trust the textbook more than you. Let me sit back and think about it for a bit. Fephisto 02:58, 14 January 2007 (UTC)


 * $$\frac{\partial}{\partial t} \left(\frac{T'(t)}{\nu T(t)}\right) = \frac{d}{dt} \left(\frac{T'(t)}{\nu T(t)}\right),$$

Since we have defined T as a function solely of t, correct? From that step the rest follows. Fephisto 03:02, 14 January 2007 (UTC)

Ahhh, ok, I see. You take a partial derivative which later becomes a ordinary derivative (but couldn't take total to begin with). I'll notate it right now. I did always kind of feel unsatisfied saying "the sides are obviously equal to a constant".

Ben pcc 18:13, 22 January 2007 (UTC)

Can't the proof be simplified to just taking $$\frac{d}{dt}$$ of both sides to begin with, instead of partialing and then reducing to a $$\frac{d}{dt}$$? I think this simplifies the proof a bit, so I'll go ahead and change that now. Fephisto 20:39, 1 February 2007 (UTC)

I'm not sure if it's correct. t and y are related via u, so taking a total derivative would require application of the chain rule, wouldn't it?

Or maybe not, since they're both independant variables. Not sure. Oh well.

Ben pcc 19:14, 4 February 2007 (UTC)