Partial Derivatives satisfying laplace equation.

**cooltowns** · January 27th 2011, 01:31 AM

Hi

I needed help understand the partial derivatives of a solution.

I have been given the chain rule.

I have attached my problem.

**FernandoRevilla** · January 27th 2011, 02:26 AM

What concrete doubts have you had?

Fernando Revilla

**cooltowns** · January 27th 2011, 02:34 AM

Please find attached the the bits in red which I don't understand how they appeared and how they were calculated.

I know that the rest of the terms came from the expansion of the brackets.

Thanks

**HallsofIvy** · January 27th 2011, 06:09 AM

Remember that each of those derivatives involves a derivative of a product and so, by the product rule, the sum of two terms. That's why, in the top line, you have four terms but in the next eight. Each of the terms you have in red for $u_{xx}$ is the result of applying the derivative to the part of the product that does not already involve a derivative of u.

The first term in red I wouldn't have written at all. It is from applying the derivative in $cos(\theta)\frac{\partial}{\partial r}$ to the $cos(\theta)$ in $\frac{\partial u}{\partial r}cos(\theta)$ : $cos(\theta)\frac{\partial cos(\theta)}{\partial r}u_r$ . But u and r are independent variables: $\frac{\partial cos(\theta)}{\partial r}= 0$ so that is $cos(\theta) (0) u_r= 0$ . That " $u_r(0)$ " is $u_r$ times 0, not " $u_r$ of 0".

The second term is red is from applying the derivative in $cos(\theta)\frac{\partial}{\partial r}$ to the $\frac{1}{r}$ in $-\frac{\partial u}{\partial r}\frac{sin(\theta)}{r}$ : $\left(\cos(\theta)\frac{\partial u}{\partial \theta}sin(\theta)\right)\frac{\partial}{\partial r}\frac{1}{r}= \left(sin(\theta)cos(\theta)u_\theta\right)\left(-\frac{1}{r^2}\right)= -\frac{sin(\theta)cos(\theta)}{r^2}u_\theta$ .

The third is from applying the derivative in $-\frac{sin(\theta)}{r}\frac{\partial}{\partial\thet a}$ to the $cos(\theta)$ in $\frac{\partial u}{\partial\theta}cos(\theta)$ : $-\frac{sin(\theta)}{r}\frac{\partial u}{\partial r}\frac{\partial cos(\theta)}{\partial \theta}= -\frac{sin(\theta)}{r}u_r\left(- sin(\theta)\right)= \frac{sin^2(\theta)}{r}u_r$

The fourth is from applying the derivative in $-\frac{sin(\theta)}{r}\frac{\partial}{\partial\thet a}$ to the $sin(\theta)$ in $\frac{\partial u}{\partial \theta}\frac{sin(\theta)}{r}$ : $-\frac{sin(\theta)}{r}\left(-\frac{\partial u}{\partial r}\right)\frac{1}{r}\frac{\partial sin(\theta)}{\partial \theta}= \frac{sin(\theta)cos(\theta)}{r^2}u_r$

In the last, you have encircled two separate terms from two separate derivatives.

The first, $-2\frac{sin(\theta)cos(\theta)}{r^2}u_\theta$ , is from applying the derivative in $sin(\theta)\frac{\partial }{\partial r}$ to the $\frac{1}{r}$ in $\frac{\partial u}{\partial \theta}\frac{cos(\theta)}{r}$ : $sin(\theta)cos(\theta)u_\theta\frac{\partial \frac{1}{r}}{\partial r}= -\frac{sin(\theta)cos(\theta)}{r^2}u_\theta$ .

The second, $\frac{cos^2(\theta)}{r}u_r$ , is from applying the derivative in $\frac{cos(\theta)}{r}\frac{\partial}{\partial \theta}$ to the $sin(\theta)$
in $\frac{\partial u}{\partial r}sin(\theta)$ : $\frac{cos(\theta)}{r}\frac{\partial u}{\partial r}\frac{\partial sin(\theta)}{\partial \theta}= \frac{cos(\theta)}{r}u_r (cos(\theta)= \frac{cos^2(\theta)}{r}u_r$

**cooltowns** · January 27th 2011, 08:08 AM

thank you for your help. Very much appreciated

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