Confused on partial differentiation.

[yungman]

In the process of deriving .$$\nabla^2u(x,y)$$. in polar coordinates I am confuse how to work through the steps. The first step from the book is to find $$\frac{\partial u}{\partial x}$$

Standard partial differentiation is at follow:
$$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial r} \frac{\partial r}{\partial x}\;+\; \frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x}$$

If we use the same method of partial differentiation to continue to find .$$\frac{\partial^2 u}{\partial x^2}$$. we should perform:

$$\frac{\partial^2 u}{\partial x^2}= \frac{\partial}{\partial r}(\frac{\partial u}{\partial x}) \;\; \frac{\partial r}{\partial x} \;+\; \frac{\partial}{\partial \theta}(\frac{\partial u}{\partial x}) \;\; \frac{\partial \theta}{\partial x}$$

$$\Rightarrow \frac{\partial^2 u}{\partial x^2}= \frac{\partial}{\partial r}[\frac{\partial u}{\partial r} \frac{\partial r}{\partial x}\;+\; \frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x}] \frac{\partial r}{\partial x} \;+\; \frac{\partial}{\partial \theta}[\frac{\partial u}{\partial r} \frac{\partial r}{\partial x}\;+\; \frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x}] \frac{\partial \theta}{\partial x}$$




But the book did this differently:

$$\frac{\partial^2 u}{\partial x^2}= \frac{\partial}{\partial x}[\frac{\partial u}{\partial r} \frac{\partial r}{\partial x}\;+\; \frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x}]$$

Then apply product rule:

$$\frac{\partial^2 u}{\partial x^2}= [\frac{\partial r}{\partial x}\frac{\partial}{\partial x}(\frac{\partial u}{\partial r}) + \frac{\partial u}{\partial r}\frac{\partial^2r}{\partial x^2}]\;+\; [\frac{\partial \theta}{\partial x}\frac{\partial}{\partial x}(\frac{\partial u}{\partial \theta}) + \frac{\partial u}{\partial \theta}\frac{\partial^2 \theta}{\partial x^2}]$$

Obvious the two don't give the same answer. Why don't the book need to perform the second derivative like the first one like what I did?

 

[pbandjay]

I am confused on what you did. When I calculated the second partial, I got the same answer as your book. The way your book did it is the easiest way I can see. By definition:

$$\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x}\left( \frac{\partial u}{\partial x} \right)$$

Can you explain this statement more:

$$\frac{\partial^2 u}{\partial x^2}= \frac{\partial}{\partial r}\left(\frac{\partial u}{\partial x}\right) \frac{\partial r}{\partial x} + \frac{\partial}{\partial \theta}\left(\frac{\partial u}{\partial x}\right) \frac{\partial \theta}{\partial x}$$

I am just not seeing the transition...

 

[yungman]

Thanks so much for the reply. I am sure I am wrong and the book is right...so are you. I am just looking at the first partial differentiation:

$$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial r} \frac{\partial r}{\partial x}\;+\; \frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x}$$

You notice even though it is .$$\frac{\partial u}{\partial x}$$. , you still see differentiation of theta in the second part. All I did is treating $$\frac{\partial u}{\partial x}$$ as u of the first differentiation and so there is a term of differentiation by theta.

I am just following the partial differentiation rule by the book that if u is function of .$$r \;and\; \theta$$. and both are function of x, then $$\frac{\partial u}{\partial x}$$. has to have terms differentiation by both .$$r \;and\; \theta$$

Tell me why it does not work the same way as the first differentiation?

Thanks

Alan

 

[pbandjay]

Oh I see what you did there. That substitution is a bit of a risky move. When you substitute the first partial of u into the equation, you seem to be ignoring the product rule.

The function u(r(x),θ(x)) is easy to differentiate with respect to x and find the formula you have, just by applying the chain rule. But now your partial u_x is a completely new function, which includes products of functions (the other partials), therefore you have to apply the product rule.