Moving Infinite Plate at a Constant Velocity

[jhuleea]

Hi all!

I am required to find the velocity distribution of the flow around an infinite plate that suddenly starts moving with a constant speed U_o. The solution has already been worked out, but I still do not understand all of it. The part that is perplexing to me is where they use the transformation variable to rewrite the terms (del u/del t) and (del^2 u/del y^2) with respect to eta.

Simple rearrangement of the terms in eta certainly does not yield the equation in (del u/del t), although this is true for (del^2 u/del y^2). I would think that there would be more of a consistency?

Please help! Thanks!

 

[AlephZero]

The part that is perplexing to me is where they use the transformation variable to rewrite the terms (del u/del t) and (del^2 u/del y^2) with respect to eta.

It's just the chain rule for differentiation

del u/del t = (del u/del eta)(del eta/del t)

and a similar equation for del^2 u/del y^2

 

[jhuleea]

It's just the chain rule for differentiation

del u/del t = (del u/del eta)(del eta/del t)

and a similar equation for del^2 u/del y^2

Why of course! Chain Rule, so elementary yet I didn't even consider it. Thanks!

This brings up one more question:

For (del^2 eta/del y^2), why is it that the answer is the square of the first derivative of eta instead of the second derivative of eta? That is,

(del eta/del y) = 1/(2*sqrt(nu*t))

=> (del^2 eta/del y^2) = 0


But according to the solution,

(del^2 eta/del y^2) = [(del eta/del y)]^2 = 1/(4*nu*t)

 

[AlephZero]

$$\frac{\partial}{\partial y} = \frac{\partial \eta}{\partial y}\frac{\partial}{\partial \eta}$$

$$\frac{\partial^2}{\partial y^2} = <br /> \frac{\partial \eta}{\partial y} \frac{\partial}{\partial \eta} <br /> \left( \frac{\partial \eta}{\partial y} \frac {\partial}{\partial \eta} \right)$$

 

[jhuleea]

$$\frac{\partial}{\partial y} = \frac{\partial \eta}{\partial y}\frac{\partial}{\partial \eta}$$

$$\frac{\partial^2}{\partial y^2} = <br /> \frac{\partial \eta}{\partial y} \frac{\partial}{\partial \eta} <br /> \left( \frac{\partial \eta}{\partial y} \frac {\partial}{\partial \eta} \right)$$

I'm sorry, but I'm not familiar with your notation. For the expression for
$$\frac{\partial}{\partial y}$$, you used $$\frac{\partial}{\partial \eta}$$. What function are you taking the partial (with respect to eta)??

Also, for your expression of $$\frac{\partial^2}{\partial y^2}$$, are you just restating what I had posted in my second question? I'm looking for why the partial of second order (that is, $$\frac{\partial^2 \eta}{\partial y^2}$$) is not zero, but rather just the square of the the partial of the first order.

Thank you for your patience!

 

[AlephZero]

I was using $$\frac{\partial}{\partial y}$$ as an operator, that is something that can operate on any function.

If you don't like that standard calculus notation, then

Differentiate u once:

$$\frac{\partial u}{\partial y} = \frac{\partial \eta}{\partial y}\frac{\partial u}{\partial \eta}$$

Then differentiate again:

$$\frac{\partial^2 u}{\partial y^2} = \frac{\partial \eta}{\partial y} \partial \left( \frac{\partial \eta}{\partial y} \frac {\partial u}{\partial \eta} \right) / \partial \eta}$$

So, because $$\frac{\partial \eta}{\partial y}$$ is only a function of $$\nu$$ and t so it's just a constant when differentiating with respect to $$\eta$$

$$\frac{\partial^2 u}{\partial y^2} =<br /> \frac{\partial \eta}{\partial y}<br /> \frac{\partial \eta}{\partial y}<br /> \frac {\partial^2 u}{\partial \eta^2}$$

 

[AlephZero]

A simple example to cut out all the partial differentials, if they are confusing you:

Let y = ax, where a is a constant. Then dy/dx = a

For any function u(x), $$\frac{du}{dx} = \frac{dy}{dx} \frac{du}{dy} = a \frac{du}{dy}$$

$$\frac{d^2u}{dx^2} = \frac{d}{dx}\left(\frac{du}{dx}\right) = \frac{dy}{dx} \frac{d}{dy}\left(a \frac{du}{dy}\right)$$
$$= a^2 \frac{d^2u}{dy^2}$$

 

[jhuleea]

AlephZero, THANK YOU!

Your explanantion with the example was amazing! I truly appreciate your patience with me, as I realize that my math is a little rusty (since I just started school again recently).

Once again, many thanks!