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.

 

[blue_raver22]

It is important to understand that there are multiple ways to approach and solve a problem, especially in mathematics. In this case, both methods are valid and will ultimately give the same answer. The difference lies in the order in which the partial derivatives are taken.

In the first method, the partial derivatives with respect to r and theta are taken first and then the second derivative with respect to x is taken. This is known as the chain rule and is often used in situations where the variables are not explicitly defined in terms of each other, such as in polar coordinates where r and theta are functions of x and y.

In the second method, the partial derivative with respect to x is taken first, using the product rule, and then the second derivative with respect to x is taken. This method is often used when the variables are explicitly defined in terms of each other, such as in Cartesian coordinates where x and y are independent variables.

It is important to familiarize yourself with both methods and understand when to use each one. Both methods are equally valid and will give the same answer, so it is ultimately a matter of personal preference and which method is easier to understand and apply in a given situation.