# Question in Partial differential equation.

1. Oct 27, 2013

### yungman

For polar coordinates, $u(x,y)=u(r,\theta)$. Using Chain Rule:

$$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x}$$
The book gave

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

But should it be like this according to the first equation using Chain Rule? That it has to include the $\theta$ part?:
$$\frac{\partial ^2 u}{\partial x^2}=\frac{\partial }{\partial r}\left(\frac{\partial u}{\partial x}\right)\frac{\partial r}{\partial x}+\frac{\partial }{\partial \theta}\left(\frac{\partial u}{\partial x}\right)\frac{\partial \theta}{\partial x}$$
I just follow the form like the first equation. If I am wrong, can you explain why?

Thanks

Last edited: Oct 27, 2013
2. Oct 27, 2013

### arildno

If your book actually wrote that nonsense, throw it in the fire.
You are most certainly correct that it is the latter expression one should use.

3. Oct 27, 2013

### yungman

Thanks for the reply. This is the scanned copy of the book. this is "Partial Differential Equation" by Asmar. A book used by San Jose State.

This is part of the derivation of
$$\nabla^2u=\frac{\partial^2 u}{\partial r^2}+\frac {1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial ^2 u}{\partial r^2}$$

#### Attached Files:

• ###### PDE1.jpg
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4. Oct 27, 2013

### arildno

The book does NOT say what you say it said in your first post.

5. Oct 27, 2013

### arildno

In your FIRST post, you say that your book used merely d/dr, but it doesn't do that at all.

It performs x-derivations on the WHOLE expression, and that's completely different.

6. Oct 27, 2013

### arildno

Now, YOUR suggestion is STILL correct:

What you say is that we basically CAN replace operators:
$$\frac{\partial}{\partial{x}}=\frac{\partial{r}}{\partial{x}} \frac{\partial}{\partial{r}} +\frac{\partial\theta} {\partial{x}}\frac{\partial}{\partial\theta}$$

Your book, however, choose to use the second x-differentiation in full, rather than switching operators in the middle of the process.

BOTH strategies are correct

7. Oct 27, 2013

### yungman

If you see my attachment, I wrote it in green, it's exactly what I said. the book just use product rule and only differentiated by x, not with $\theta$.

8. Oct 27, 2013

### arildno

Yes, it differentiates (du/dx) with respect to x, but NOT what you wrote in your first post, with respect to "r".
So, NO, it is NOT exactly what you said.

9. Oct 27, 2013

### yungman

I think I understand now. You only use Chain Rule ONCE in the operation. That is you only perform this once per operation:
$$\frac{\partial}{\partial{x}}=\frac{\partial{r}}{\partial{x}} \frac{\partial}{\partial{r}} +\frac{\partial\theta} {\partial{x}}\frac{\partial}{\partial\theta}$$

So if I do it my way:
$$\frac{\partial ^2 u}{\partial x^2}=\frac{\partial }{\partial r}\left(\frac{\partial u}{\partial x}\right)\frac{\partial r}{\partial x}+\frac{\partial }{\partial \theta}\left(\frac{\partial u}{\partial x}\right)\frac{\partial \theta}{\partial x}$$
Then the next step is just a simple differentiation respect to $r$
$$\frac{\partial }{\partial r}\left(\frac{\partial u}{\partial x}\right)$$
and

$$\frac{\partial }{\partial \theta}\left(\frac{\partial u}{\partial x}\right)$$
where it is a simple differentiation respect to $\theta$. No more Chain Rule as I use it right at the beginning.

What the book did, it did the simple differentiation respect to x first,
$$\frac{\partial ^2 u}{\partial x^2}=\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial x}\right)$$
then apply the chain rule to each part.

10. Oct 27, 2013

### arildno

Remember that you have in your method, for example,
$$\frac{\partial{r}}{\partial{x}}=\cos\theta$$
So that:
$$\frac{\partial}{\partial\theta} (\frac{\partial{r}}{\partial{x}})=-\sin\theta$$
and so on.