Hi

I needed help understand the partial derivatives of a solution.

I have been given the chain rule.

I have attached my problem.

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- Jan 27th 2011, 01:31 AMcooltownsPartial Derivatives satisfying laplace equation.
Hi

I needed help understand the partial derivatives of a solution.

I have been given the chain rule.

I have attached my problem. - Jan 27th 2011, 02:26 AMFernandoRevilla
What concrete doubts have you had?

Fernando Revilla - Jan 27th 2011, 02:34 AMcooltowns
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 - Jan 27th 2011, 06:09 AMHallsofIvy
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 $\displaystyle u_{xx}$ 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 $\displaystyle cos(\theta)\frac{\partial}{\partial r}$ to the $\displaystyle cos(\theta)$ in $\displaystyle \frac{\partial u}{\partial r}cos(\theta)$: $\displaystyle cos(\theta)\frac{\partial cos(\theta)}{\partial r}u_r$. But u and r are independent variables: $\displaystyle \frac{\partial cos(\theta)}{\partial r}= 0$ so that is $\displaystyle cos(\theta) (0) u_r= 0$. That "$\displaystyle u_r(0)$" is $\displaystyle u_r$**times**0, not "$\displaystyle u_r$**of**0".

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

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

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

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

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

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

in $\displaystyle \frac{\partial u}{\partial r}sin(\theta)$: $\displaystyle \frac{cos(\theta)}{r}\frac{\partial u}{\partial r}\frac{\partial sin(\theta)}{\partial \theta}= \frac{cos(\theta)}{r}u_r (cos(\theta)= \frac{cos^2(\theta)}{r}u_r$ - Jan 27th 2011, 08:08 AMcooltowns
thank you for your help. Very much appreciated :)